https://wiki.swarma.org/api.php?action=feedcontributions&feedformat=atom&user=Smile集智百科 - 复杂系统|人工智能|复杂科学|复杂网络|自组织 - 用户贡献 [zh-cn]2026-11-03T19:25:07Z用户贡献MediaWiki 1.35.0https://wiki.swarma.org/index.php?title=%E9%87%8D%E5%B0%BE%E5%88%86%E5%B8%83&diff=21353重尾分布2021-01-26T13:15:38Z<p>Smile：</p>
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<div>此词条Jie翻译。已由Smile审校。<br />
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{{too technical|date=May 2020}}<br />
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{{mergefrom|Fat-tailed distribution|date=May 2020}}<br />
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In [[probability theory]], '''heavy-tailed distributions''' are [[probability distribution]]s whose tails are not exponentially bounded:<ref name="Asmussen">{{Cite book | doi = 10.1007/0-387-21525-5_10 | first = S. R. | last = Asmussen| chapter = Steady-State Properties of GI/G/1 | title = Applied Probability and Queues | series = Stochastic Modelling and Applied Probability | volume = 51 | pages = 266–301 | year = 2003 | isbn = 978-0-387-00211-8 | pmid = | pmc = }}</ref> that is, they have heavier tails than the [[exponential distribution]]. In many applications it is the right tail of the distribution that is of interest, but a distribution may have a heavy left tail, or both tails may be heavy.<br />
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In probability theory, heavy-tailed distributions are probability distributions whose tails are not exponentially bounded: that is, they have heavier tails than the exponential distribution. In many applications it is the right tail of the distribution that is of interest, but a distribution may have a heavy left tail, or both tails may be heavy.<br />
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在概率论中，<font color="#ff8000">重尾分布 Heavy-tailed distributions</font>是指其尾部呈现出不受指数限制的概率分布<ref name="Asmussen">{{Cite book | doi = 10.1007/0-387-21525-5_10 | first = S. R. | last = Asmussen| chapter = Steady-State Properties of GI/G/1 | title = Applied Probability and Queues | series = Stochastic Modelling and Applied Probability | volume = 51 | pages = 266–301 | year = 2003 | isbn = 978-0-387-00211-8 | pmid = | pmc = }}</ref>：也就是说，它们的尾部比<font color="#ff8000">指数分布 exponential distribution </font> “重”。在许多应用中，关注的是分布的右尾，但是分布的左尾可能也很重，或者两个尾都很重。<br />
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There are three important subclasses of heavy-tailed distributions: the [[fat-tailed distribution]]s, the [[long-tailed distribution]]s and the '''subexponential distributions'''. In practice, all commonly used heavy-tailed distributions belong to the subexponential class.<br />
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There are three important subclasses of heavy-tailed distributions: the fat-tailed distributions, the long-tailed distributions and the subexponential distributions. In practice, all commonly used heavy-tailed distributions belong to the subexponential class.<br />
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重尾分布有三个重要的子类：<font color="#ff8000">胖尾分布 Fat-tailed distribution</font>，<font color="#ff8000">长尾分布 Long-tailed distribution</font>和<font color="#ff8000">次指数分布 Subexponential distributions</font>。实际上，所有常用的重尾分布都属于<font color="#ff8000">次指数分布类 subexponential class </font>。<br />
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There is still some discrepancy over the use of the term '''heavy-tailed'''. There are two other definitions in use. Some authors use the term to refer to those distributions which do not have all their power [[Moment (mathematics)|moments]] finite; and some others to those distributions that do not have a finite [[variance]]. The definition given in this article is the most general in use, and includes all distributions encompassed by the alternative definitions, as well as those distributions such as [[log-normal]] that possess all their power moments, yet which are generally considered to be heavy-tailed. (Occasionally, heavy-tailed is used for any distribution that has heavier tails than the normal distribution.)<br />
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There is still some discrepancy over the use of the term heavy-tailed. There are two other definitions in use. Some authors use the term to refer to those distributions which do not have all their power moments finite; and some others to those distributions that do not have a finite variance. The definition given in this article is the most general in use, and includes all distributions encompassed by the alternative definitions, as well as those distributions such as log-normal that possess all their power moments, yet which are generally considered to be heavy-tailed. (Occasionally, heavy-tailed is used for any distribution that has heavier tails than the normal distribution.)<br />
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在使用<font color="#ff8000">“重尾” Heavy-tailed</font>一词时仍存在一些歧义。于是就出现了另外两种定义。一些作者使用该术语来指代并非所有幂矩都是有限的那些分布，以及其它一些没有有限方差的分布。本文中给出的是最常用的定义，包括替代定义所涵盖的所有分布，以及具有所有幂矩的<font color="#ff8000">对数正态分布 long-normal distributions </font>，但通常被认为是重尾的。（有时“重尾”用于任何具有比正态分布更重的尾巴的分布。）<br />
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== Definitions 定义 ==<br />
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=== Definition of heavy-tailed distribution 重尾分布的定义 ===<br />
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The distribution of a [[random variable]] ''X'' with [[cumulative distribution function|distribution function]] ''F'' is said to have a heavy (right) tail if the [[moment generating function]] of ''X'', ''M<sub>X</sub>''(''t''), is infinite for all ''t''&nbsp;>&nbsp;0.<ref name="ReferenceA">Rolski, Schmidli, Scmidt, Teugels, ''Stochastic Processes for Insurance and Finance'', 1999</ref><br />
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The distribution of a random variable X with distribution function F is said to have a heavy (right) tail if the moment generating function of X, MX(t), is infinite for all t&nbsp;>&nbsp;0.<br />
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如果''X''的矩生成函数, ''M<sub>X</sub>''(''t'')对于所有''t''&nbsp;>&nbsp;0都是无限的，则具有分布函数''F''的随机变量''X''的分布被称为重尾（右）。<ref name="ReferenceA">Rolski, Schmidli, Scmidt, Teugels, ''Stochastic Processes for Insurance and Finance'', 1999</ref><br />
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That means<br />
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也就是说<br />
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:<math><br />
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\int_{-\infty}^\infty e^{t x} \,dF(x) = \infty \quad \mbox{for all } t>0.<br />
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</math><br />
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An implication of this is that<br />
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这意味着<br />
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:<math><br />
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\lim_{x \to \infty} e^{t x}\Pr[X>x] = \infty \quad \mbox{for all } t>0.\,<br />
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</math><br />
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This is also written in terms of the tail distribution function<br />
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也可以写成<font color="#ff8000">尾分布函数 the tail distribution function </font>：<br />
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<math><br />
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\overline{F}(x) ≡ \Pr[X>x]<br />
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</math><br />
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as<br />
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:<math><br />
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\lim_{x \to \infty} e^{t x}\overline{F}(x) = \infty \quad \mbox{for all } t >0.\,<br />
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</math><br />
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=== Definition of long-tailed distribution 长尾分布的定义 ===<br />
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The distribution of a [[random variable]] ''X'' with [[cumulative distribution function|distribution function]] ''F'' is said to have a long right tail if for all ''t'' > 0,<br />
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The distribution of a random variable X with distribution function F is said to have a long right tail[1] if for all t > 0,<br />
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如果对于所有''t''>0，则称具有分布函数''F''的随机变量''X''的分布为有较长的右尾，<br />
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:<math><br />
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\lim_{x \to \infty} \Pr[X>x+t\mid X>x] =1, \,<br />
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</math><br />
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or equivalently<br />
或等同于<br />
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:<math><br />
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\overline{F}(x+t) \sim \overline{F}(x) \quad \mbox{as } x \to \infty. \,<br />
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</math><br />
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This has the intuitive interpretation for a right-tailed long-tailed distributed quantity that if the long-tailed quantity exceeds some high level, the probability approaches 1 that it will exceed any other higher level.<br />
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This has the intuitive interpretation for a right-tailed long-tailed distributed quantity that if the long-tailed quantity exceeds some high level, the probability approaches 1 that it will exceed any other higher level.<br />
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对于右尾长尾分布量具有直观的解释，即如果长尾量超过某个高水平，则概率将接近1，它将超过其他更高的水平。<br />
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All long-tailed distributions are heavy-tailed, but the converse is false, and it is possible to construct heavy-tailed distributions that are not long-tailed.<br />
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All long-tailed distributions are heavy-tailed, but the converse is false, and it is possible to construct heavy-tailed distributions that are not long-tailed.<br />
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所有长尾分布都是重尾分布，但反过来不一定成立，且可以构造出非长尾分布的重尾分布。<br />
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=== Subexponential distributions 次指数分布 ===<br />
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Subexponentiality is defined in terms of [[Convolution of probability distributions|convolutions of probability distributions]]. For two independent, identically distributed [[random variables]] <math> X_1,X_2</math> with common distribution function <math>F</math> the convolution of <math>F</math> with itself, <math>F^{*2}</math> is convolution square, using [[Lebesgue–Stieltjes integration]], by:<br />
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Subexponentiality is defined in terms of [[Convolution of probability distributions|convolutions of probability distributions]]. For two independent, identically distributed [[random variables]] <math> X_1,X_2</math> with common distribution function <math>F</math> the convolution of <math>F</math> with itself, <math>F^{*2}</math> is convolution square, using [[Lebesgue–Stieltjes integration]], by:<br />
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次指数性是根据概率分布的<font color="#ff8000">卷积 Convolution </font>定义的。对于具有共同分布函数<math>F</math>的两个独立且分布均匀的随机变量<math> X_1,X_2</math>，<math>F</math>与自身的卷积，<math>F^{*2}</math>是卷积的平方，使用Lebesgue–Stieltjes积分，方法如下：<br />
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:<math><br />
\Pr[X_1+X_2 \leq x] = F^{*2}(x) = \int_{0}^x F(x-y)\,dF(y),<br />
</math><br />
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and the ''n''-fold convolution <math>F^{*n}</math> is defined inductively by the rule:<br />
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''n''倍卷积<math>F^{*n}</math>定义如下：<br />
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:<math><br />
F^{*n}(x) = \int_{0}^x F(x-y)\,dF^{*n-1}(y).<br />
</math><br />
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The tail distribution function <math>\overline{F}</math> is defined as <math>\overline{F}(x) = 1-F(x)</math>.<br />
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尾分布函数<math>\overline{F}</math>定义为<math>\overline{F}(x) = 1-F(x)</math>。<br />
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A distribution <math>F</math> on the positive half-line is subexponential<ref name="Asmussen"/><ref>{{Cite web|url=https://www.researchgate.net/publication/242637603_A_Theorem_on_Sums_of_Independent_Positive_Random_Variables_and_Its_Applications_to_Branching_Random_Processes|title=A Theorem on Sums of Independent Positive Random Variables and Its Applications to Branching Random Processes|last=Chistyakov|first=V. P.|date=1964|website=ResearchGate|language=en|archive-url=|archive-date=|access-date=April 7, 2019}}</ref><ref>{{Cite web|url=https://projecteuclid.org/download/pdf_1/euclid.aop/1176996225|title=The Class of Subexponential Distributions|last=Teugels|first=Jozef L.|authorlink=|date=1975|website=|publisher=Annals of Probability|publication-place=[[KU Leuven|University of Louvain]]|archive-url=|archive-date=|access-date=April 7, 2019}}</ref> if<br />
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如果满足以下条件，则正半线上的分布<math>F</math>为次指数<ref name="Asmussen"/><ref>{{Cite web|url=https://www.researchgate.net/publication/242637603_A_Theorem_on_Sums_of_Independent_Positive_Random_Variables_and_Its_Applications_to_Branching_Random_Processes|title=A Theorem on Sums of Independent Positive Random Variables and Its Applications to Branching Random Processes|last=Chistyakov|first=V. P.|date=1964|website=ResearchGate|language=en|archive-url=|archive-date=|access-date=April 7, 2019}}</ref><ref>{{Cite web|url=https://projecteuclid.org/download/pdf_1/euclid.aop/1176996225|title=The Class of Subexponential Distributions|last=Teugels|first=Jozef L.|authorlink=|date=1975|website=|publisher=Annals of Probability|publication-place=[[KU Leuven|University of Louvain]]|archive-url=|archive-date=|access-date=April 7, 2019}}</ref><br />
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:<math><br />
\overline{F^{*2}}(x) \sim 2\overline{F}(x) \quad \mbox{as } x \to \infty. <br />
</math><br />
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This implies<ref name="Embrechts">{{cite book |author1=Embrechts P. |author2=Klueppelberg C. |author3=Mikosch T. |title=Modelling extremal events for insurance and finance |publisher=Springer | series = Stochastic Modelling and Applied Probability|location=Berlin |year=1997 | volume=33| doi = 10.1007/978-3-642-33483-2|isbn=978-3-642-08242-9 }}</ref> that, for any <math>n \geq 1</math>,<br />
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这意味着<ref name="Embrechts">{{cite book |author1=Embrechts P. |author2=Klueppelberg C. |author3=Mikosch T. |title=Modelling extremal events for insurance and finance |publisher=Springer | series = Stochastic Modelling and Applied Probability|location=Berlin |year=1997 | volume=33| doi = 10.1007/978-3-642-33483-2|isbn=978-3-642-08242-9 }}</ref>，对于任何<math>n \geq 1</math>，<br />
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:<math><br />
\overline{F^{*n}}(x) \sim n\overline{F}(x) \quad \mbox{as } x \to \infty. <br />
</math><br />
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The probabilistic interpretation<ref name="Embrechts"/> of this is that, for a sum of <math>n</math> [[statistical independence|independent]] [[random variables]] <math>X_1,\ldots,X_n</math> with common distribution <math>F</math>,<br />
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对此的概率解释<ref name="Embrechts"/>是，对于具有共同分布<math>F</math>的<math>n</math>个独立随机变量<math>X_1,\ldots,X_n</math>的总和<br />
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:<math><br />
\Pr[X_1+ \cdots +X_n>x] \sim \Pr[\max(X_1, \ldots,X_n)>x] \quad \text{as } x \to \infty. <br />
</math><br />
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This is often known as the principle of the single big jump<ref>{{Cite journal | last1 = Foss | first1 = S. | last2 = Konstantopoulos | first2 = T. | last3 = Zachary | first3 = S. | doi = 10.1007/s10959-007-0081-2 | title = Discrete and Continuous Time Modulated Random Walks with Heavy-Tailed Increments | journal = Journal of Theoretical Probability| volume = 20 | issue = 3 | pages = 581 | year = 2007 | arxiv = math/0509605| pmid = | url = http://www.math.nsc.ru/LBRT/v1/foss/fkz_revised.pdf| pmc = | citeseerx = 10.1.1.210.1699 }}</ref> or catastrophe principle.<ref>{{cite web| url = http://rigorandrelevance.wordpress.com/2014/01/09/catastrophes-conspiracies-and-subexponential-distributions-part-iii/ | title = Catastrophes, Conspiracies, and Subexponential Distributions (Part III) | first = Adam | last = Wierman | authorlink = Adam Wierman | date = January 9, 2014 | accessdate = January 9, 2014 | website = Rigor + Relevance blog | publisher = RSRG, Caltech}}</ref><br />
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这通常被称为<font color="#ff8000">单跳 single big jump</font><ref>{{Cite journal | last1 = Foss | first1 = S. | last2 = Konstantopoulos | first2 = T. | last3 = Zachary | first3 = S. | doi = 10.1007/s10959-007-0081-2 | title = Discrete and Continuous Time Modulated Random Walks with Heavy-Tailed Increments | journal = Journal of Theoretical Probability| volume = 20 | issue = 3 | pages = 581 | year = 2007 | arxiv = math/0509605| pmid = | url = http://www.math.nsc.ru/LBRT/v1/foss/fkz_revised.pdf| pmc = | citeseerx = 10.1.1.210.1699 }}</ref>或<font color="#ff8000">浩劫原则 catastrophe principle</font> <ref>{{cite web| url = http://rigorandrelevance.wordpress.com/2014/01/09/catastrophes-conspiracies-and-subexponential-distributions-part-iii/ | title = Catastrophes, Conspiracies, and Subexponential Distributions (Part III) | first = Adam | last = Wierman | authorlink = Adam Wierman | date = January 9, 2014 | accessdate = January 9, 2014 | website = Rigor + Relevance blog | publisher = RSRG, Caltech}}</ref>。<br />
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A distribution <math>F</math> on the whole real line is subexponential if the distribution<br />
<math>F I([0,\infty))</math> is.<ref>{{cite journal | last = Willekens | first = E. | title = Subexponentiality on the real line | journal = Technical Report | publisher = K.U. Leuven | year = 1986}}</ref> Here <math>I([0,\infty))</math> is the [[indicator function]] of the positive half-line. Alternatively, a random variable <math>X</math> supported on the real line is subexponential if and only if <math>X^+ = \max(0,X)</math> is subexponential.<br />
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如果分布<math>F I([0,\infty))</m4ath>为实数，则<math>F</math>为整个实数上的次指数分布。<ref>{{cite journal | last = Willekens | first = E. | title = Subexponentiality on the real line | journal = Technical Report | publisher = K.U. Leuven | year = 1986}}</ref>此时<math>I([0,\infty))</math>是正半轴的指标函数。或者，当且仅当<math>X^+ = \max(0,X)</math>是次指数时，实数上支持的随机变量<math>X</math>才是次指数。<br />
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All subexponential distributions are long-tailed, but examples can be constructed of long-tailed distributions that are not subexponential.<br />
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所有次指数分布都是长尾分布，但可以构造出非次指数分布的长尾分布的示例。<br />
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== Common heavy-tailed distributions 常见的重尾分布 ==<br />
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All commonly used heavy-tailed distributions are subexponential.<ref name="Embrechts"/><br />
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Those that are one-tailed include:<br />
*the [[Pareto distribution]];<br />
*the [[Log-normal distribution]];<br />
*the [[Lévy distribution]];<br />
*the [[Weibull distribution]] with shape parameter greater than 0 but less than 1;<br />
*the [[Burr distribution]];<br />
*the [[log-logistic distribution]];<br />
*the [[log-gamma distribution]];<br />
*the [[Fréchet distribution]];<br />
*the [[log-Cauchy distribution]], sometimes described as having a "super-heavy tail" because it exhibits [[logarithmic growth|logarithmic decay]] producing a heavier tail than the Pareto distribution.<ref>{{cite book|title=Laws of Small Numbers: Extremes and Rare Events|author=Falk, M., Hüsler, J. & Reiss, R.|page=80|year=2010|publisher=Springer|isbn=978-3-0348-0008-2}}</ref><ref>{{cite web|title=Statistical inference for heavy and super-heavy tailed distributions|url=http://docentes.deio.fc.ul.pt/fragaalves/SuperHeavy.pdf|author=Alves, M.I.F., de Haan, L. & Neves, C.|date=March 10, 2006|access-date=November 1, 2011|archive-url=https://web.archive.org/web/20070623175435/http://docentes.deio.fc.ul.pt/fragaalves/SuperHeavy.pdf|archive-date=June 23, 2007|url-status=dead}}</ref><br />
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Those that are two-tailed include:<br />
*The [[Cauchy distribution]], itself a special case of both the stable distribution and the t-distribution;<br />
*The family of [[stable distributions]],<ref>{{cite web |author=John P. Nolan | title=Stable Distributions: Models for Heavy Tailed Data| year=2009 | url=http://academic2.american.edu/~jpnolan/stable/chap1.pdf | accessdate=2009-02-21}}</ref> excepting the special case of the normal distribution within that family. Some stable distributions are one-sided (or supported by a half-line), see e.g. [[Lévy distribution]]. See also ''[[financial models with long-tailed distributions and volatility clustering]]''.<br />
*The [[Student's t-distribution|t-distribution]].<br />
*The skew lognormal cascade distribution.<ref>{{cite web | author=Stephen Lihn | title=Skew Lognormal Cascade Distribution | year=2009 | url=http://www.skew-lognormal-cascade-distribution.org/ | access-date=2009-06-12 | archive-url=https://web.archive.org/web/20140407075213/http://www.skew-lognormal-cascade-distribution.org/ | archive-date=2014-04-07 | url-status=dead }}</ref><br />
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All commonly used heavy-tailed distributions are subexponential.[6]<br />
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所有常用的重尾分布都是次指数的。<ref name="Embrechts"/><br />
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Those that are one-tailed include:<br />
单尾的包括：<br />
* <font color="#ff8000">帕累托分布 Pareto distribution</font>;<br />
* <font color="#ff8000">对数正态分布 Log-normal distribution</font>;<br />
* <font color="#ff8000">莱维分布 Lévy distribution</font>;<br />
* 形状参数大于0但小于1的<font color="#ff8000">韦布尔分布 Weibull distribution</font>;<br />
* <font color="#ff8000">伯尔分布 Burr distribution</font>;<br />
* <font color="#ff8000">对数逻辑分布 log-logistic distribution</font>;<br />
* <font color="#ff8000">对数伽玛分布 log-gamma distribution</font>;<br />
* <font color="#ff8000">弗雷歇分布 Fréchet distribution</font>;<br />
* <font color="#ff8000">对数柯西分布 log-Cauchy distribution</font>，有时被描述为“超重尾”分布，因为它表现出对数衰减，从而产生比帕累托分布更重的尾。<ref>{{cite book|title=Laws of Small Numbers: Extremes and Rare Events|author=Falk, M., Hüsler, J. & Reiss, R.|page=80|year=2010|publisher=Springer|isbn=978-3-0348-0008-2}}</ref><ref>{{cite web|title=Statistical inference for heavy and super-heavy tailed distributions|url=http://docentes.deio.fc.ul.pt/fragaalves/SuperHeavy.pdf|author=Alves, M.I.F., de Haan, L. & Neves, C.|date=March 10, 2006|access-date=November 1, 2011|archive-url=https://web.archive.org/web/20070623175435/http://docentes.deio.fc.ul.pt/fragaalves/SuperHeavy.pdf|archive-date=June 23, 2007|url-status=dead}}</ref><br />
* <br />
Those that are two-tailed include:<br />
双尾的包括：<br />
* <font color="#ff8000">柯西分布 Cauchy distribution</font>本身就是稳定分布和t分布的特例；<br />
* <font color="#ff8000">稳定分布族 The family of stable distributions</font><ref>{{cite web |author=John P. Nolan | title=Stable Distributions: Models for Heavy Tailed Data| year=2009 | url=http://academic2.american.edu/~jpnolan/stable/chap1.pdf | accessdate=2009-02-21}}</ref>，但该族中正态分布的特殊情况除外。一些稳定的分布是单面的（或有半线的支持），例如莱维分布。另请参见具有长尾分布和波动性聚类的财务模型。<br />
* t分布<br />
*<font color="#ff8000">偏对数正态级联分布 The skew lognormal cascade distribution</font>。<ref>{{cite web | author=Stephen Lihn | title=Skew Lognormal Cascade Distribution | year=2009 | url=http://www.skew-lognormal-cascade-distribution.org/ | access-date=2009-06-12 | archive-url=https://web.archive.org/web/20140407075213/http://www.skew-lognormal-cascade-distribution.org/ | archive-date=2014-04-07 | url-status=dead }}</ref><br />
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<br />
== Relationship to fat-tailed distributions 与胖尾分布的关系 ==<br />
A [[fat-tailed distribution]] is a distribution for which the probability density function, for large x, goes to zero as a power <math>x^{-a}</math>. Since such a power is always bounded below by the probability density function of an exponential distribution, fat-tailed distributions are always heavy-tailed. Some distributions, however, have a tail which goes to zero slower than an exponential function (meaning they are heavy-tailed), but faster than a power (meaning they are not fat-tailed). An example is the [[log-normal distribution]] <ref>{{Contradict-inline|article=fat-tailed distribution|reason=Fat-tailed page says log-normals are in fact fat-tailed.|date=June 2019}}</ref>. Many other heavy-tailed distributions such as the [[log-logistic distribution|log-logistic]] and [[Pareto distribution|Pareto]] distribution are, however, also fat-tailed.<br />
<br />
胖尾分布是这样的分布，对于较大的x，概率密度函数为<math>x^{-a}</math>趋于零。由于这样的幂总是受到指数分布概率密度函数的限制，因此，胖尾分布始终是重尾分布。但是，某些分布的尾部趋近于零的速率比指数函数慢（表示它们是重尾），而比幂快（表示它们不是胖尾）。例如对数正态分布<ref>{{Contradict-inline|article=fat-tailed distribution|reason=Fat-tailed page says log-normals are in fact fat-tailed.|date=June 2019}}</ref>。当然，许多其他的重尾分布，例如对数逻辑分布和帕累托分布也属于胖尾分布。<br />
<br />
<br />
<br />
== Estimating the tail-index{{definition|date=January 2018}} 尾指数估计 ==<br />
<br />
There are parametric (see Embrechts et al.<ref name="Embrechts"/>) and non-parametric (see, e.g., Novak<ref name="Novak2011">{{cite book<br />
| author=Novak S.Y. <br />
| title=Extreme value methods with applications to finance<br />
| year=2011<br />
| series=London: CRC<br />
| isbn=978-1-43983-574-6 <br />
}}</ref>) approaches to the problem of the tail-index estimation.<br />
<br />
对于尾指数估计的问题，有参数方法（参见Emprechts等人<ref name="Embrechts"/>）和非参数方法（例如，Novak<ref name="Novak2011">{{cite book<br />
| author=Novak S.Y. <br />
| title=Extreme value methods with applications to finance<br />
| year=2011<br />
| series=London: CRC<br />
| isbn=978-1-43983-574-6 <br />
}}</ref>）两种。<br />
<br />
<br />
<br />
To estimate the tail-index using the parametric approach, some authors employ [[GEV distribution]] or [[Pareto distribution]]; they may apply the maximum-likelihood estimator (MLE).<br />
<br />
为了使用参数化方法估计尾指数，有些作者采用了GEV分布或帕累托分布；他们可能会运用极大似然估计方法（MLE）。<br />
<br />
<br />
<br />
=== Pickand's tail-index estimator Pickand的尾指数估算器===<br />
<br />
With <math>(X_n , n \geq 1)</math> a random sequence of independent and same density function <math>F \in D(H(\xi))</math>, the Maximum Attraction Domain<ref name=Pickands>{{cite journal|last=Pickands III|first=James|title=Statistical Inference Using Extreme Order Statistics|journal=The Annals of Statistics|date=Jan 1975|volume=3|issue=1|pages=119–131|jstor=2958083|doi=10.1214/aos/1176343003|doi-access=free}}</ref> of the generalized extreme value density <math> H </math>, where <math>\xi \in \mathbb{R}</math>. If <math>\lim_{n\to\infty} k(n) = \infty </math> and <math>\lim_{n\to\infty} \frac{k(n)}{n}= 0</math>, then the ''Pickands'' tail-index estimation is<ref name="Embrechts"/><ref name="Pickands"/><br />
<br />
对于<math>(X_n , n \geq 1)</math>的独立且相同的密度函数<math>F \in D(H(\xi))</math>的随机序列，是<font color="#ff8000">广义极值密度 the generalized extreme value density </font><math>H</math>的<font color="#ff8000">最大吸引域 the Maximum Attraction Domain </font><ref name=Pickands>{{cite journal|last=Pickands III|first=James|title=Statistical Inference Using Extreme Order Statistics|journal=The Annals of Statistics|date=Jan 1975|volume=3|issue=1|pages=119–131|jstor=2958083|doi=10.1214/aos/1176343003|doi-access=free}}</ref>，其中<math>\xi \in \mathbb{R}</math>。如果<math>\lim_{n\to\infty} k(n) = \infty </math>和<math>\lim_{n\to\infty} \frac{k(n)}{n}= 0</math>，则Pickands尾部指数估计为<ref name="Embrechts"/><ref name="Pickands"/><br />
<br />
<br />
:<math><br />
\xi^\text{Pickands}_{(k(n),n)} =\frac{1}{\ln 2} \ln \left( \frac{X_{(n-k(n)+1,n)} - X_{(n-2k(n)+1,n)}}{X_{(n-2k(n)+1,n)} - X_{(n-4k(n)+1,n)}}\right)<br />
</math><br />
<br />
<br />
where <math>X_{(n-k(n)+1,n)}=\max \left(X_{n-k(n)+1},\ldots ,X_{n}\right)</math>. This estimator converges in probability to <math>\xi</math>.<br />
<br />
其中<math>X_{(n-k(n)+1,n)}=\max \left(X_{n-k(n)+1},\ldots ,X_{n}\right)</math>。 此估计量的概率收敛到<math>\xi</math>。<br />
<br />
<br />
<br />
=== Hill's tail-index estimator 希尔 Hill的尾指数估算器 ===<br />
<br />
Let <math>(X_t , t \geq 1)</math> be a sequence of independent and identically distributed random variables with distribution function <math>F \in D(H(\xi))</math>, the maximum domain of attraction of the [[generalized extreme value distribution]] <math> H </math>, where <math>\xi \in \mathbb{R}</math>. The sample path is <math>{X_t: 1 \leq t \leq n}</math> where <math>n</math> is the sample size. If <br />
<math>\{k(n)\}</math> is an intermediate order sequence, i.e. <math>k(n) \in \{1,\ldots,n-1\}, </math>, <math>k(n) \to \infty</math> and <math>k(n)/n \to 0</math>, then the Hill tail-index estimator is<ref>Hill B.M. (1975) A simple general approach to inference about the tail of a distribution. Ann. Stat., v. 3, 1163–1174.</ref><br />
<br />
令<math>(X_t , t \geq 1)</math>为具有分布函数<math>F \in D(H(\xi))</math>独立且均匀分布的随机变量序列，其分布函数为广义极值分布<math> H </math>的最大吸引域，其中<math>\xi \in \mathbb{R}</math>。样本路径为<math>{X_t: 1 \leq t \leq n}</math>，其中<math>n</math>为样本大小。 如果<math>\{k(n)\}</math>是中间阶数序列，即<math>k(n) \in \{1,\ldots,n-1\}, </math>，<math>k(n) \to \infty</math>和<math>k(n)/n \to 0</math>，则Hill尾指数估计器为<ref>Hill B.M. (1975) A simple general approach to inference about the tail of a distribution. Ann. Stat., v. 3, 1163–1174.</ref>：<br />
<br />
<br />
: <math><br />
\xi^\text{Hill}_{(k(n),n)} = \left(\frac 1 {k(n)} \sum_{i=n-k(n)+1}^n \ln(X_{(i,n)}) - \ln (X_{(n-k(n)+1,n)})\right)^{-1},<br />
</math><br />
<br />
<br />
where <math>X_{(i,n)}</math> is the <math>i</math>-th [[order statistic]] of <math>X_1, \dots, X_n</math>.<br />
This estimator converges in probability to <math>\xi</math>, and is asymptotically normal provided <math>k(n) \to \infty </math> is restricted based on a higher order regular variation property<ref>Hall, P.(1982) On some estimates of an exponent of regular variation. J. R. Stat. Soc. Ser. B., v. 44, 37–42.</ref> <br />
.<ref>Haeusler, E. and J. L. Teugels (1985) On asymptotic normality of Hill's estimator for the exponent of regular variation. Ann. Stat., v. 13, 743–756.</ref> Consistency and asymptotic normality extend to a large class of dependent and heterogeneous sequences,<ref>Hsing, T. (1991) On tail index estimation using dependent data. Ann. Stat., v. 19, 1547–1569.</ref><ref>Hill, J. (2010) On tail index estimation for dependent, heterogeneous data. Econometric Th., v. 26, 1398–1436.</ref> irrespective of whether <math>X_t</math> is observed, or a computed residual or filtered data from a large class of models and estimators, including mis-specified models and models with errors that are dependent.<ref>Resnick, S. and Starica, C. (1997). Asymptotic behavior of Hill’s estimator for autoregressive data. Comm. Statist. Stochastic Models 13, 703–721.</ref><ref>Ling, S. and Peng, L. (2004). Hill’s estimator for the tail index of an ARMA model. J. Statist. Plann. Inference 123, 279–293.</ref><ref>Hill, J. B. (2015). Tail index estimation for a filtered dependent time series. Stat. Sin. 25, 609–630.</ref><br />
<br />
其中<math>X_{(i,n)}</math>是<math>X_1, \dots, X_n</math>的第<math>i</math>次序统计量。该估计量依概率收敛于<math>\xi</math>，并且在基于高阶的正则变化性质的情况下，是限制<math>k(n) \to \infty </math>的渐近正态<ref>Hall, P.(1982) On some estimates of an exponent of regular variation. J. R. Stat. Soc. Ser. B., v. 44, 37–42.</ref>.<ref>Haeusler, E. and J. L. Teugels (1985) On asymptotic normality of Hill's estimator for the exponent of regular variation. Ann. Stat., v. 13, 743–756.</ref>。一致性和渐近正态性适用于一大类相关序列和异类序列<ref>Hsing, T. (1991) On tail index estimation using dependent data. Ann. Stat., v. 19, 1547–1569.</ref><ref>Hill, J. (2010) On tail index estimation for dependent, heterogeneous data. Econometric Th., v. 26, 1398–1436.</ref>，而不管是否观测到<math>X_t</math>，或者来自大量模型和估计量（包括错误指定的模型和具有相关误差的模型）计算出的残差或筛选数据。<ref>Resnick, S. and Starica, C. (1997). Asymptotic behavior of Hill’s estimator for autoregressive data. Comm. Statist. Stochastic Models 13, 703–721.</ref><ref>Ling, S. and Peng, L. (2004). Hill’s estimator for the tail index of an ARMA model. J. Statist. Plann. Inference 123, 279–293.</ref><ref>Hill, J. B. (2015). Tail index estimation for a filtered dependent time series. Stat. Sin. 25, 609–630.</ref><br />
<br />
<br />
<br />
=== Ratio estimator of the tail-index 尾部指数的比率估计器 ===<br />
<br />
The ratio estimator (RE-estimator) of the tail-index was introduced by Goldie <br />
and Smith.<ref>Goldie C.M., Smith R.L. (1987) Slow variation with remainder:<br />
theory and applications. Quart. J. Math. Oxford, v. 38, 45–71.</ref> <br />
It is constructed similarly to Hill's estimator but uses a non-random "tuning parameter".<br />
<br />
尾指数的比率估计器（RE估计器）由Goldie和Smith提出<ref>Goldie C.M., Smith R.L. (1987) Slow variation with remainder:<br />
theory and applications. Quart. J. Math. Oxford, v. 38, 45–71.</ref>。它的构造类似于Hill估计器，但使用了非随机的“调整参数”<br />
<br />
<br />
<br />
A comparison of Hill-type and RE-type estimators can be found in Novak.<ref name="Novak2011"/><br />
<br />
在Novak中可以找到Hill型和RE型估计量的比较。<ref name="Novak2011"/><br />
<br />
=== Software 应用软件===<br />
* [http://www.cs.bu.edu/~crovella/aest.html aest], [[C (programming language)|C]] tool for estimating the heavy-tail index.<ref>{{Cite journal | last1 = Crovella | first1 = M. E. | last2 = Taqqu | first2 = M. S. | title = Estimating the Heavy Tail Index from Scaling Properties| journal = Methodology and Computing in Applied Probability | volume = 1 | pages = 55–79 | year = 1999 | doi = 10.1023/A:1010012224103 | url = http://www.cs.bu.edu/~crovella/paper-archive/aest.ps| pmid = | pmc = }}</ref><br />
<br />
* 用于估计重尾指数的软件[http://www.cs.bu.edu/~crovella/aest.html aest]和C。<ref>{{Cite journal | last1 = Crovella | first1 = M. E. | last2 = Taqqu | first2 = M. S. | title = Estimating the Heavy Tail Index from Scaling Properties| journal = Methodology and Computing in Applied Probability | volume = 1 | pages = 55–79 | year = 1999 | doi = 10.1023/A:1010012224103 | url = http://www.cs.bu.edu/~crovella/paper-archive/aest.ps| pmid = | pmc = }}</ref><br />
<br />
== Estimation of heavy-tailed density 重尾密度的估计 ==<br />
<br />
Nonparametric approaches to estimate heavy- and superheavy-tailed probability density functions were given in <br />
Markovich.<ref name="Markovich2007">{{cite book<br />
| author=Markovich N.M. <br />
| title=Nonparametric Analysis of Univariate Heavy-Tailed data: Research and Practice<br />
| year=2007<br />
| series=Chitester: Wiley<br />
| isbn=978-0-470-72359-3<br />
}}</ref> These are approaches based on variable bandwidth and long-tailed kernel estimators; on the preliminary data transform to a new random variable at finite or infinite intervals which is more convenient for the estimation and then inverse transform of the obtained density estimate; and "piecing-together approach" which provides a certain parametric model for the tail of the density and a non-parametric model to approximate the mode of the density. Nonparametric estimators require an appropriate selection of tuning (smoothing) parameters like a bandwidth of kernel estimators and the bin width of the histogram. The well known data-driven methods of such selection are a cross-validation and its modifications, methods based on the minimization of the mean squared error (MSE) and its asymptotic and their upper bounds.<ref name="WandJon1995">{{cite book<br />
| author=Wand M.P., Jones M.C. <br />
| title=Kernel smoothing<br />
| year=1995<br />
| series=New York: Chapman and Hall<br />
| isbn=978-0412552700<br />
}}</ref> A discrepancy method which uses well-known nonparametric statistics like Kolmogorov-Smirnov's, von Mises and Anderson-Darling's ones as a metric in the space of distribution functions (dfs) and quantiles of the later statistics as a known uncertainty or a discrepancy value can be found in.<ref name="Markovich2007"/> Bootstrap is another tool to find smoothing parameters using approximations of unknown MSE by different schemes of re-samples selection, see e.g.<ref name="Hall1992">{{cite book<br />
| author=Hall P. <br />
| title=The Bootstrap and Edgeworth Expansion<br />
| year=1992<br />
| series=Springer<br />
| isbn=9780387945088<br />
}}</ref><br />
<br />
Markovich中给出了估计重尾和超重尾概率密度函数的非参数方法。<ref name="Markovich2007">{{cite book<br />
| author=Markovich N.M. <br />
| title=Nonparametric Analysis of Univariate Heavy-Tailed data: Research and Practice<br />
| year=2007<br />
| series=Chitester: Wiley<br />
| isbn=978-0-470-72359-3<br />
}}</ref>这些是基于<font color="#ff8000">可变带宽 variable bandwidth</font>和<font color="#ff8000">长尾核估计器 long-tailed kernel estimators</font>的方法。将初步数据以有限或无限间隔变换为新的随机变量，这样更便于估计，然后对获得的密度估计进行逆变换；以及“拼合方法”，它为密度的尾部提供了确定的参数模型，并为近似密度模型提供了非参数模型。非参数估计器需要适当选择调整（平滑）参数，例如内核估计器的带宽和直方图的组距。这种选择大众化数据驱动方法是基于均方误差（MSE）及其渐近或上限的最小化的交叉验证及修改方法。<ref name="WandJon1995">{{cite book<br />
| author=Wand M.P., Jones M.C. <br />
| title=Kernel smoothing<br />
| year=1995<br />
| series=New York: Chapman and Hall<br />
| isbn=978-0412552700<br />
}}</ref>可以找到一种差异方法，通过使用著名的非参数统计数据（例如Kolmogorov-Smirnov's，von Mises和Anderson-Darling的统计量）作为分布函数（dfs）空间中的度量，并将后来的统计量的分位数作为已知的不确定性或差异值。<ref name="Markovich2007"/><font color="#ff8000">自助法 Bootstrap</font>是另一种工具，可以通过不同的重抽样方案使用未知MSE的近似值来查找平滑参数。<ref name="Hall1992">{{cite book<br />
| author=Hall P. <br />
| title=The Bootstrap and Edgeworth Expansion<br />
| year=1992<br />
| series=Springer<br />
| isbn=9780387945088<br />
}}</ref><br />
<br />
<br />
<br />
== See also 其他参考资料 ==<br />
*[[Leptokurtic distribution]]<br />
*[[Generalized extreme value distribution]]<br />
*[[Outlier]]<br />
*[[Long tail]]<br />
*[[Power law]]<br />
*[[Seven states of randomness]]<br />
*[[Fat-tailed distribution]]<br />
**[[Taleb distribution]] and [[Holy grail distribution]]<br />
<br />
<br />
* <font color="#ff8000">尖峭态分布 Leptokurtic distribution</font><br />
* <font color="#ff8000">广义极值分布 Generalized extreme value distribution</font><br />
* <font color="#ff8000">离群值 Outlier</font><br />
* <font color="#ff8000">长尾 Long tail</font><br />
* <font color="#ff8000">幂律 Power law</font><br />
* <font color="#ff8000">随机的七个状态 Seven states of randomness</font><br />
* <font color="#ff8000">胖尾分布 Fat-tailed distribution</font><br />
**<font color="#ff8000">塔勒布分布 Taleb distribution</font>和<font color="#ff8000">圣杯分布 Holy grail distribution</font><br />
<br />
<br />
<br />
== References 参考文献 ==<br />
<br />
<references/><br />
<br />
[[Category:Tails of probability distributions]]<br />
[[Category:Types of probability distributions]]<br />
[[Category:Actuarial science]]<br />
[[Category:Risk]]<br />
<br />
<br />
<br />
Category:Tails of probability distributions<br />
<br />
类别: 概率分布的尾部<br />
<br />
*the [[Burr distribution]];<br />
<br />
Category:Types of probability distributions<br />
<br />
类别: 概率分布的类型<br />
<br />
*the [[log-logistic distribution]];<br />
<br />
Category:Actuarial science<br />
<br />
类别: 精算<br />
<br />
*the [[log-gamma distribution]];<br />
<br />
Category:Risk<br />
<br />
类别: 风险<br />
<br />
<noinclude><br />
<br />
<small>This page was moved from [[wikipedia:en:Heavy-tailed distribution]]. Its edit history can be viewed at [[重尾分布/edithistory]]</small></noinclude><br />
<br />
[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E5%9B%9A%E5%BE%92%E5%9B%B0%E5%A2%83&diff=21000囚徒困境2021-01-19T16:17:41Z<p>Smile：</p>
<hr />
<div>此词条由Henry初步翻译。已由Smile审校。<br />
<br />
{{other uses}}<br />
<br />
{{short description|Canonical example of a game analyzed in game theory}}<br />
<br />
{| class="wikitable floatright"<br />
<br />
|+ Prisoner's dilemma payoff matrix<br />
<br />
! {{diagonal split header|A|B}}<br />
<br />
! B stays<br />silent<br />
<br />
! B<br />betrays<br />
<br />
|-<br />
<br />
! A stays<br />silent <br />
<br />
| {{diagonal split header|-1|-1|transparent}}<br />
<br />
| {{diagonal split header|-3|0|transparent}}<br />
<br />
|-<br />
<br />
! A<br />betrays<br />
<br />
| {{diagonal split header|0|-3|transparent}}<br />
<br />
| {{diagonal split header|-2|-2|transparent}}<br />
<br />
|}<br />
<br />
<br />
<br />
{| class="wikitable floatright"<br />
<br />
|+ <font color="#ff8000"> 囚徒困境支付矩阵 Prisoner's dilemma payoff matrix</font><br />
<br />
! {{diagonal split header|A|B}}<br />
<br />
! B 保持<br />缄默<br />
<br />
! B<br />背叛<br />
<br />
|-<br />
<br />
! A 保持<br />缄默 <br />
<br />
| {{diagonal split header|-1|-1|transparent}}<br />
<br />
| {{diagonal split header|-3|0|transparent}}<br />
<br />
|-<br />
<br />
! A<br />背叛<br />
<br />
| {{diagonal split header|0|-3|transparent}}<br />
<br />
| {{diagonal split header|-2|-2|transparent}}<br />
<br />
|}<br />
<br />
<br />
<br />
The '''prisoner's dilemma''' is a standard example of a game analyzed in [[game theory]] that shows why two completely [[Rationality#Economics|rational]] individuals might not cooperate, even if it appears that it is in their best interests to do so. It was originally framed by [[Merrill Flood]] and [[Melvin Dresher]] while working at [[RAND Corporation|RAND]] in 1950. [[Albert W. Tucker]] formalized the game with prison sentence rewards and named it "prisoner's dilemma",<ref>Poundstone, 1992</ref> presenting it as follows:<br />
<br />
The prisoner's dilemma is a standard example of a game analyzed in game theory that shows why two completely rational individuals might not cooperate, even if it appears that it is in their best interests to do so. It was originally framed by Merrill Flood and Melvin Dresher while working at RAND in 1950. Albert W. Tucker formalized the game with prison sentence rewards and named it "prisoner's dilemma", prensenting it as follows:<br />
<br />
<font color="#ff8000"> 囚徒困境prisoner's dilemma</font>是<font color="#ff8000"> 博弈论game theory</font>分析博弈的一个代表性例子，它揭示了为什么两个完全理性的个体可能不会合作，即使这样做符合他们的最大利益。它最初是由梅里尔·弗勒德 Merrill Flood和 梅文·加舍尔 Melvin Dresher于1950年在兰德公司工作时构建的。阿尔伯特.W.塔克 Albert W. Tucker将这种博弈以监禁刑罚奖励的方式正式化，并将其命名为囚徒困境，具体阐述如下：<br />
<br />
<br />
<br />
{{quote|Two members of a criminal gang are arrested and imprisoned. Each prisoner is in solitary confinement with no means of communicating with the other. The prosecutors lack sufficient evidence to convict the pair on the principal charge, but they have enough to convict both on a lesser charge. Simultaneously, the prosecutors offer each prisoner a bargain. Each prisoner is given the opportunity either to betray the other by testifying that the other committed the crime, or to cooperate with the other by remaining silent. The possible outcomes are:<br />
<br />
{{quote|Two members of a criminal gang are arrested and imprisoned. Each prisoner is in solitary confinement with no means of communicating with the other. The prosecutors lack sufficient evidence to convict the pair on the principal charge, but they have enough to convict both on a lesser charge. Simultaneously, the prosecutors offer each prisoner a bargain. Each prisoner is given the opportunity either to betray the other by testifying that the other committed the crime, or to cooperate with the other by remaining silent. The possible outcomes are:<br />
<br />
* If A and B each betray the other, each of them serves two years in prison<br />
* If A betrays B but B remains silent, A will be set free and B will serve three years in prison<br />
* If A remains silent but B betrays A, A will serve three years in prison and B will be set free<br />
* If A and B both remain silent, both of them will serve only one year in prison (on the lesser charge).}}<br />
<br />
{{一个犯罪团伙的两名成员被捕入狱。每个囚犯都被单独监禁，与他人无法沟通。检察官缺乏足够的证据来对这两个人定罪，但有足够的证据以较低的罪名定罪。同时，检察官向每个犯人提供了一个交易。每个囚犯都有机会出卖对方，证明对方犯下的罪行，或者他们可以合作，保持沉默。可能的结果有:<br />
<br />
*如果A和B都背叛了对方，他们都会在监狱服刑两年。<br />
*如果A背叛了B但B保持沉默，A会被无罪释放而B会服刑三年。<br />
*如果A保持沉默但B背叛了A，A会服刑三年而B会无罪释放。<br />
*如果A和B都保持沉默，他们就只用服刑一年（以较低的罪名）。}}<br />
<br />
It is implied that the prisoners will have no opportunity to reward or punish their partner other than the prison sentences they get and that their decision will not affect their reputation in the future. Because betraying a partner offers a greater reward than cooperating with them, all purely rational self-interested prisoners will betray the other, meaning the only possible outcome for two purely rational prisoners is for them to betray each other.<ref>{{cite web|last=Milovsky|first=Nicholas|title=The Basics of Game Theory and Associated Games|url=https://issuu.com/johnsonnick895/docs/game_theory_paper|accessdate=11 February 2014}}</ref> In reality, humans display a [[systemic bias]] towards cooperative behavior in this and similar games despite what is predicted by simple models of "rational" self-interested action.<ref name = Fehr>{{cite journal | last1=Fehr | first1= Ernst | last2=Fischbacher | first2=Urs | date= Oct 23, 2003 | title=The Nature of human altruism |journal=Nature | volume=425 | pages=785–91 | doi=10.1038/nature02043 | url=http://www.iwp.jku.at/born/mpwfst/04/nature02043_f_born.pdf | accessdate=February 27, 2013 | pmid=14574401 | issue=6960|bibcode = 2003Natur.425..785F }}</ref><ref name = Amos>{{cite book | title=Preference, belief, and similarity: selected writings. | publisher=Massachusetts Institute of Technology Press | first1= Amos | last1=Tversky | first2=Eldar | last2=Shafir | url=http://cseweb.ucsd.edu/~gary/PAPER-SUGGESTIONS/Preference,%20Belief,%20and%20Similarity%20Selected%20Writings%20(Bradford%20Books).pdf | year=2004 | isbn=9780262700931 | accessdate=February 27, 2013}}</ref><ref name="Ahn">{{cite journal |last1 = Toh-Kyeong|first1 = Ahn|last2 = Ostrom|first2 = Elinor|last3 = Walker|first3 = James|date = Sep 5, 2002|title = Incorporating Motivational Heterogeneity into Game-Theoretic Models of Collective Action|journal = Public Choice|volume = 117|issue = 3–4|pages = 295–314|doi =10.1023/b:puch.0000003739.54365.fd |url = http://www.indiana.edu/~workshop/seminars/papers/ahnostromwalker_092402.pdf|accessdate = June 27, 2015|hdl = 10535/4697}}</ref><ref name="Hessel">{{cite journal|last1 = Oosterbeek|first1 = Hessel|last2 = Sloof|first2 = Randolph|last3 = Van de Kuilen|first3 = Gus|date = Dec 3, 2003|title = Cultural Differences in Ultimatum Game Experiments: Evidence from a Meta-Analysis|journal = Experimental Economics|volume = 7|issue = 2|pages = 171–88|doi = 10.1023/B:EXEC.0000026978.14316.74|url = http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|accessdate = February 27, 2013|url-status = dead|archiveurl = https://web.archive.org/web/20130512175243/http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|archivedate = May 12, 2013}}</ref> This bias towards cooperation has been known since the test was first conducted at RAND; the secretaries involved trusted each other and worked together for the best common outcome.<ref>{{Cite book | url=https://books.google.com/?id=WIhZlB86nJwC&pg=PT96&lpg=PT96&dq=rand+secretaries+prisoner%27s+dilemma#v=onepage |title = Why Most Things Fail|isbn = 9780571266142|last1 = Ormerod|first1 = Paul|date = 2010-12-22}}</ref> The prisoner's dilemma became the focus of extensive experimental research.<ref>Deutsch, M. (1958). Trust and suspicion. Journal of Conflict Resolution, 2(4), 265–279. https://doi.org/10.1177/002200275800200401</ref> <ref>Rapoport, A., & Chammah, A. M. (1965). Prisoner’s Dilemma: A study of conflict and cooperation. Ann Arbor, MI: University of Michigan Press.</ref><br />
<br />
It is implied that the prisoners will have no opportunity to reward or punish their partner other than the prison sentences they get and that their decision will not affect their reputation in the future. Because betraying a partner offers a greater reward than cooperating with them, all purely rational self-interested prisoners will betray the other, meaning the only possible outcome for two purely rational prisoners is for them to betray each other. In reality, humans display a systemic bias towards cooperative behavior in this and similar games despite what is predicted by simple models of "rational" self-interested action. This bias towards cooperation has been known since the test was first conducted at RAND; the secretaries involved trusted each other and worked together for the best common outcome. The prisoner's dilemma became the focus of extensive experimental research. <br />
<br />
这意味着，囚犯除了监禁刑罚之外，没有机会奖励或惩罚他们的同伴，他们的决定也不会影响他们未来的声誉。因为背叛一个同伴比与他们合作能得到更大的回报，所以所有纯粹理性的、自私自利的囚犯都会背叛对方，这意味着，对于两个纯粹理性的囚犯来说，唯一可能的结果就是他们相互背叛。<ref>{{cite web|last=Milovsky|first=Nicholas|title=The Basics of Game Theory and Associated Games|url=https://issuu.com/johnsonnick895/docs/game_theory_paper|accessdate=11 February 2014}}</ref>实际上，尽管“理性的”自利行为的简单模型已经预测到了这一点，但人类在这种和类似的博弈中仍然表现出对合作行为的<font color="#ff8000">系统性偏差 systemic bias </font>。<ref name = Fehr>{{cite journal | last1=Fehr | first1= Ernst | last2=Fischbacher | first2=Urs | date= Oct 23, 2003 | title=The Nature of human altruism |journal=Nature | volume=425 | pages=785–91 | doi=10.1038/nature02043 | url=http://www.iwp.jku.at/born/mpwfst/04/nature02043_f_born.pdf | accessdate=February 27, 2013 | pmid=14574401 | issue=6960|bibcode = 2003Natur.425..785F }}</ref><ref name = Amos>{{cite book | title=Preference, belief, and similarity: selected writings. | publisher=Massachusetts Institute of Technology Press | first1= Amos | last1=Tversky | first2=Eldar | last2=Shafir | url=http://cseweb.ucsd.edu/~gary/PAPER-SUGGESTIONS/Preference,%20Belief,%20and%20Similarity%20Selected%20Writings%20(Bradford%20Books).pdf | year=2004 | isbn=9780262700931 | accessdate=February 27, 2013}}</ref><ref name="Ahn">{{cite journal |last1 = Toh-Kyeong|first1 = Ahn|last2 = Ostrom|first2 = Elinor|last3 = Walker|first3 = James|date = Sep 5, 2002|title = Incorporating Motivational Heterogeneity into Game-Theoretic Models of Collective Action|journal = Public Choice|volume = 117|issue = 3–4|pages = 295–314|doi =10.1023/b:puch.0000003739.54365.fd |url = http://www.indiana.edu/~workshop/seminars/papers/ahnostromwalker_092402.pdf|accessdate = June 27, 2015|hdl = 10535/4697}}</ref><ref name="Hessel">{{cite journal|last1 = Oosterbeek|first1 = Hessel|last2 = Sloof|first2 = Randolph|last3 = Van de Kuilen|first3 = Gus|date = Dec 3, 2003|title = Cultural Differences in Ultimatum Game Experiments: Evidence from a Meta-Analysis|journal = Experimental Economics|volume = 7|issue = 2|pages = 171–88|doi = 10.1023/B:EXEC.0000026978.14316.74|url = http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|accessdate = February 27, 2013|url-status = dead|archiveurl = https://web.archive.org/web/20130512175243/http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|archivedate = May 12, 2013}}</ref>自从在兰德公司首次进行这项测试以来，人们就已经知道了这种对合作的偏见; 参与测试的秘书们相互信任，为了最好的共同的目标而努力。<ref>{{Cite book | url=https://books.google.com/?id=WIhZlB86nJwC&pg=PT96&lpg=PT96&dq=rand+secretaries+prisoner%27s+dilemma#v=onepage |title = Why Most Things Fail|isbn = 9780571266142|last1 = Ormerod|first1 = Paul|date = 2010-12-22}}</ref>囚徒困境成为大量实验研究的焦点。<ref>Deutsch, M. (1958). Trust and suspicion. Journal of Conflict Resolution, 2(4), 265–279. https://doi.org/10.1177/002200275800200401</ref> <ref>Rapoport, A., & Chammah, A. M. (1965). Prisoner’s Dilemma: A study of conflict and cooperation. Ann Arbor, MI: University of Michigan Press.</ref><br />
<br />
<br />
<br />
An extended "iterated" version of the game also exists. In this version, the classic game is played repeatedly between the same prisoners, who continuously have the opportunity to penalize the other for previous decisions. If the number of times the game will be played is known to the players, then (by [[backward induction]]) two classically rational players will betray each other repeatedly, for the same reasons as the single-shot variant. In an infinite or unknown length game there is no fixed optimum strategy, and prisoner's dilemma tournaments have been held to compete and test algorithms for such cases.<ref>{{cite journal|url = https://egtheory.wordpress.com/2015/03/02/ipd/|title = Short history of iterated prisoner's dilemma tournaments|date = March 2, 2015|access-date = February 8, 2016|journal = Journal of Conflict Resolution|volume = 24|issue = 3|pages = 379–403|last = Kaznatcheev|first = Artem|doi = 10.1177/002200278002400301}}</ref><br />
<br />
An extended "iterated" version of the game also exists. In this version, the classic game is played repeatedly between the same prisoners, who continuously have the opportunity to penalize the other for previous decisions. If the number of times the game will be played is known to the players, then (by backward induction) two classically rational players will betray each other repeatedly, for the same reasons as the single-shot variant. In an infinite or unknown length game there is no fixed optimum strategy, and prisoner's dilemma tournaments have been held to compete and test algorithms for such cases.<br />
<br />
一个扩展的<font color="#ff8000">迭代iterated</font>版本的博弈由此衍生出来。在这个版本中，经典博弈会在在同一组囚犯之间重复进行，他们不断有机会为了之前的决定对其他囚犯进行惩罚。如果参与者知道博弈的次数，那么(通过<font color="#ff8000">逆向归纳法 backward induction </font>)两个经典的理性的玩家就会因为和在单次博弈中相同的原因反复背叛对方。在无限次或未知次数的博弈中，没有固定的最优策略，因而，举办囚徒困境竞赛来竞争和检验这种情况下的算法。<ref>{{cite journal|url = https://egtheory.wordpress.com/2015/03/02/ipd/|title = Short history of iterated prisoner's dilemma tournaments|date = March 2, 2015|access-date = February 8, 2016|journal = Journal of Conflict Resolution|volume = 24|issue = 3|pages = 379–403|last = Kaznatcheev|first = Artem|doi = 10.1177/002200278002400301}}</ref><br />
<br />
<br />
<br />
The prisoner's dilemma game can be used as a model for many [[#Real-life examples|real world situations]] involving cooperative behavior. In casual usage, the label "prisoner's dilemma" may be applied to situations not strictly matching the formal criteria of the classic or iterative games: for instance, those in which two entities could gain important benefits from cooperating or suffer from the failure to do so, but find it difficult or expensive—not necessarily impossible—to coordinate their activities.<br />
<br />
The prisoner's dilemma game can be used as a model for many real world situations involving cooperative behavior. In casual usage, the label "prisoner's dilemma" may be applied to situations not strictly matching the formal criteria of the classic or iterative games: for instance, those in which two entities could gain important benefits from cooperating or suffer from the failure to do so, but find it difficult or expensive—not necessarily impossible—to coordinate their activitie <br />
<br />
囚徒困境博弈可以作为许多现实中涉及合作行为的模型。在非正式用法中，“囚徒困境”一词可适用于不严格符合经典或迭代博弈的形式标准的情况: 例如，两个实体可以从合作中获得巨大利益或者会因为合作失败而遭受损失，但发现协调他们的活动很困难或者代价昂贵（并非是不可能的）。<br />
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<br />
<br />
==Strategy for the prisoner's dilemma==<br />
囚徒困境的策略<br />
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Two prisoners are separated into individual rooms and cannot communicate with each other.<br />
<br />
Two prisoners are separated into individual rooms and cannot communicate with each other.<br />
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两名囚犯被分开关押在各自的房间里，不能相互交流。<br />
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The normal game is shown below:<br />
<br />
The normal game is shown below:<br />
<br />
正常的博弈如下:<br />
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<br />
<br />
{| class="wikitable"<br />
<br />
{| class="wikitable"<br />
<br />
{ | class“ wikitable”<br />
<br />
|-<br />
<br />
|-<br />
<br />
|-<br />
<br />
! {{diagonal split header|<br />Prisoner A|Prisoner B}} !! Prisoner B stays silent<br>(''cooperates'') !! Prisoner B betrays<br>(''defects'')<br />
<br />
! !! Prisoner B stays silent<br>(cooperates) !! Prisoner B betrays<br>(defects)<br />
<br />
! {{diagonal split header|<br />犯人A|犯人B}}！！犯人B保持沉默<br>(''合作'') !! 犯人B背叛<br>(''背叛'')<br />
<br />
|-<br />
<br />
|-<br />
<br />
|-<br />
<br />
! Prisoner A stays silent<br>(''cooperates'')<br />
<br />
! Prisoner A stays silent<br>(cooperates)<br />
<br />
!犯人 a 保持沉默<br>(''合作'')<br />
<br />
| Each serves 1 year|| Prisoner A: 3 years<br />Prisoner B: goes free<br />
<br />
| Each serves 1 year|| Prisoner A: 3 years<br />Prisoner B: goes free<br />
<br />
|每人服刑1年||囚犯 A: 3年<br />囚犯 B: 无罪释放<br />
<br />
|-<br />
<br />
|-<br />
<br />
|-<br />
<br />
! Prisoner A betrays<br>(''defects'')<br />
<br />
! Prisoner A betrays<br>(defects)<br />
<br />
!犯人A背叛<br>(''背叛'')<br />
<br />
| Prisoner A: goes free<br />Prisoner B: 3 years || Each serves 2 years<br />
<br />
| Prisoner A: goes free<br />Prisoner B: 3 years || Each serves 2 years<br />
<br />
|囚犯 A: 获释<br />囚犯 B: 3年||每人服刑2年<br />
<br />
|}<br />
<br />
|}<br />
<br />
|}<br />
<br />
<br />
<br />
It is assumed that both prisoners understand the nature of the game, have no loyalty to each other, and will have no opportunity for retribution or reward outside the game. Regardless of what the other decides, each prisoner gets a higher reward by betraying the other ("defecting"). The reasoning involves an argument by [[Dilemma#Use in logic|dilemma]]: B will either cooperate or defect. If B cooperates, A should defect, because going free is better than serving 1 year. If B defects, A should also defect, because serving 2 years is better than serving 3. So either way, A should defect. Parallel reasoning will show that B should defect.<br />
<br />
It is assumed that both prisoners understand the nature of the game, have no loyalty to each other, and will have no opportunity for retribution or reward outside the game. Regardless of what the other decides, each prisoner gets a higher reward by betraying the other ("defecting"). The reasoning involves an argument by dilemma: B will either cooperate or defect. If B cooperates, A should defect, because going free is better than serving 1 year. If B defects, A should also defect, because serving 2 years is better than serving 3. So either way, A should defect. Parallel reasoning will show that B should defect.<br />
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假设两个囚犯都了解博弈的本质，对彼此不忠诚，且在博弈之外没有机会得到报复或奖励。那么不管对方怎么决定，每个犯人背叛对方都会得到更高的奖励(“叛变”)。推理涉及一个进退两难的论点:B 要么合作，要么叛变。如果B合作，A 应该叛变，因为得到释放总比服刑1年好。如果 B叛变，A也应该叛变，因为服刑2年总比服刑3年好。所以不管怎样，A都应该叛变。并行推理表明B应该选择叛变。<br />
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Because defection always results in a better payoff than cooperation regardless of the other player's choice, it is a [[dominant strategy]]. Mutual defection is the only strong [[Nash equilibrium]] in the game (i.e. the only outcome from which each player could only do worse by unilaterally changing strategy). The dilemma, then, is that mutual cooperation yields a better outcome than mutual defection but is not the rational outcome because the choice to cooperate, from a self-interested perspective, is irrational.<br />
<br />
Because defection always results in a better payoff than cooperation regardless of the other player's choice, it is a dominant strategy. Mutual defection is the only strong Nash equilibrium in the game (i.e. the only outcome from which each player could only do worse by unilaterally changing strategy). The dilemma, then, is that mutual cooperation yields a better outcome than mutual defection but is not the rational outcome because the choice to cooperate, from a self-interested perspective, is irrational.<br />
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因为不管对方的选择如何，背叛总是比合作带来更好的回报，所以这是一个<font color="#ff8000">占优策略 dominant strategy</font>。相互背叛是博弈中唯一的强<font color="#ff8000">纳什均衡点 dominant strategy </font>（即每个参与者单方面的改变策略只能使自己的情况变糟)。因此，困境在于，虽然相互合作比相互背叛产生更好的结果，但它却不是理性的结果，因为从自我利益的角度来看，合作的选择是非理性的。<br />
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==Generalized form==<br />
<font color="#ff8000">广泛形态 Generalized form </font><br />
The structure of the traditional prisoner's dilemma can be generalized from its original prisoner setting. Suppose that the two players are represented by the colors red and blue, and that each player chooses to either "cooperate" or "defect".<br />
<br />
The structure of the traditional prisoner's dilemma can be generalized from its original prisoner setting. Suppose that the two players are represented by the colors red and blue, and that each player chooses to either "cooperate" or "defect".<br />
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传统囚徒困境的结构可以从其最初的囚徒环境中概括出来。假设两个玩家用红色和蓝色表示，并且每个玩家选择“合作”或“背叛”。<br />
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If both players cooperate, they both receive the reward ''R'' for cooperating. If both players defect, they both receive the punishment payoff ''P''. If Blue defects while Red cooperates, then Blue receives the temptation payoff ''T'', while Red receives the "sucker's" payoff, ''S''. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff ''S'', while Red receives the temptation payoff ''T''.<br />
<br />
If both players cooperate, they both receive the reward R for cooperating. If both players defect, they both receive the punishment payoff P. If Blue defects while Red cooperates, then Blue receives the temptation payoff T, while Red receives the "sucker's" payoff, S. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff S, while Red receives the tem punishment payoffptation payoff T.<br />
<br />
如果两个玩家合作，他们都会因为合作而获得奖励''R''。如果两个参与人都叛变，他们都会受到惩罚''P''。 如果蓝方叛变而红方合作，那么蓝方得到诱惑回报''T''，而红方受到“上当受骗者”损失''S''。同样地，如果蓝方合作而红方叛变，那么蓝方得到上当受骗者的损失''S''，而红方得到诱惑支付''T''。<br />
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<br />
<br />
This can be expressed in [[Normal-form game|normal form]]:<br />
<br />
This can be expressed in normal form:<br />
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这可以用标准形式的博弈来表示:<br />
<br />
<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
|+ Canonical PD payoff matrix<br />
<br />
|+ Canonical PD payoff matrix<br />
<br />
| + <font color="#ff8000">正则 PD 支付矩阵 Canonical PD payoff matrix </font> <br />
<br />
! {{diagonal split header|{{color|#009|Blue}}|{{color|#900|Red}}}}<br />
<br />
! |}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|Cooperate}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|Defect}}<br />
<br />
|-<br />
<br />
! scope="row" style="width:60px;" | {{color|#009|Cooperate}}<br />
<br />
| {{diagonal split header|{{color|#009|''R''}}|{{color|#900|''R''}}|transparent}}<br />
<br />
| ||transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''S''}}|{{color|#900|''T''}}|transparent}}<br />
<br />
|-<br />
<br />
! scope="row" | {{color|#009|Defect}}<br />
<br />
| {{diagonal split header|{{color|#009|''T''}}|{{color|#900|''S''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''P''}}|{{color|#900|''P''}}|transparent}}<br />
<br />
|}<br />
<br />
<br />
! {{diagonal split header|{{color|#009|蓝方}}|{{color|#900|红方}}}}<br />
<br />
! |}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|合作}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|背叛}}<br />
<br />
|-<br />
<br />
! scope="row" style="width:60px;" | {{color|#009|合作}}<br />
<br />
| {{diagonal split header|{{color|#009|''R''}}|{{color|#900|''R''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''S''}}|{{color|#900|''T''}}|transparent}}<br />
<br />
|-<br />
<br />
! scope="row" | {{color|#009|背叛}}<br />
<br />
| {{diagonal split header|{{color|#009|''T''}}|{{color|#900|''S''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''P''}}|{{color|#900|''P''}}|transparent}}<br />
<br />
|}<br />
<br />
<br />
<br />
and to be a prisoner's dilemma game in the strong sense, the following condition must hold for the payoffs:<br />
<br />
and to be a prisoner's dilemma game in the strong sense, the following condition must hold for the payoffs:<br />
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要成为强意义下的囚徒困境博弈，收益必须满足以下条件:<br />
<br />
<br />
<br />
:{{tmath|T > R > P > S}}<br />
<br />
<br />
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The payoff relationship {{tmath|R > P}} implies that mutual cooperation is superior to mutual defection, while the payoff relationships {{tmath|T > R}} and {{tmath|P > S}} imply that defection is the [[dominant strategy]] for both agents.<br />
<br />
The payoff relationship implies that mutual cooperation is superior to mutual defection, while the payoff relationships and imply that defection is the dominant strategy for both agents.<br />
<br />
回报关系{{tmath|R > P}}意味着相互合作优于相互背叛，然而回报关系{{tmath|T > R}}和{{tmath|P > S}}也意味着相互背叛是双方的占优策略。<br />
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<br />
===Special case: donation game===<br />
特例：<font color="#ff8000">捐赠博弈 donation game </font><br />
The "donation game"<ref name=Hilbe2013>{{cite journal|last=Hilbe|first=Christian |author2=Martin A. Nowak |author3=Karl Sigmund|title=Evolution of extortion in Iterated Prisoner's Dilemma games|journal=PNAS|date=April 2013|volume=110|issue=17|pages=6913–18|doi=10.1073/pnas.1214834110|pmid=23572576 |pmc=3637695 |bibcode=2013PNAS..110.6913H |arxiv=1212.1067}}</ref> is a form of prisoner's dilemma in which cooperation corresponds to offering the other player a benefit ''b'' at a personal cost ''c'' with ''b'' > ''c''. Defection means offering nothing. The payoff matrix is thus<br />
<br />
The "donation game" is a form of prisoner's dilemma in which cooperation corresponds to offering the other player a benefit b at a personal cost c with b > c. Defection means offering nothing. The payoff matrix is thus<br />
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捐赠博弈<ref name=Hilbe2013>{{cite journal|last=Hilbe|first=Christian |author2=Martin A. Nowak |author3=Karl Sigmund|title=Evolution of extortion in Iterated Prisoner's Dilemma games|journal=PNAS|date=April 2013|volume=110|issue=17|pages=6913–18|doi=10.1073/pnas.1214834110|pmid=23572576 |pmc=3637695 |bibcode=2013PNAS..110.6913H |arxiv=1212.1067}}</ref>是囚徒困境的一种形式，在这种博弈中，合作相当于以''b'' > ''c''条件下的个人成本''c''为另一方提供一个收益''b''，而叛变意味着什么也不提供。收益矩阵是这样的<br />
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{| class="wikitable" style="text-align:center"<br />
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! {{diagonal split header|{{navy (color)|Blue}}|{{color|#900|Red}}}}<br />
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! scope="col" style="width:60px;" | {{color|#900|Cooperate}}<br />
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! scope="col" style="width:60px;" | {{color|#900|Defect}}<br />
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|-<br />
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! scope="row" style="width:60px;" | {{color|#009|Cooperate}}<br />
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| {{diagonal split header|{{color|#009|''b''-''c''}}|{{color|#900|''b''-''c''}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|-''c''}}|{{color|#900|''b''}}|transparent}}<br />
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|-<br />
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! scope="row" | {{color|#009|Defect}}<br />
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| {{diagonal split header|{{color|#009|''b''}}|{{color|#900|-''c''}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
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|}<br />
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{| class="wikitable" style="text-align:center"<br />
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! {{diagonal split header|{{navy (color)|蓝方}}|{{color|#900|红方}}}}<br />
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! scope="col" style="width:60px;" | {{color|#900|合作}}<br />
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! scope="col" style="width:60px;" | {{color|#900|背叛}}<br />
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|-<br />
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! scope="row" style="width:60px;" | {{color|#009|合作}}<br />
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| {{diagonal split header|{{color|#009|''b''-''c''}}|{{color|#900|''b''-''c''}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|-''c''}}|{{color|#900|''b''}}|transparent}}<br />
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|-<br />
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! scope="row" | {{color|#009|背叛}}<br />
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| {{diagonal split header|{{color|#009|''b''}}|{{color|#900|-''c''}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
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|}<br />
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Note that {{tmath|2R>T+S}} (i.e. {{tmath|2(b-c)>b-c}}) which qualifies the donation game to be an iterated game (see next section).<br />
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Note that (i.e. ) which qualifies the donation game to be an iterated game (see next section).<br />
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请注意{{tmath|2R>T+S}}（即{{tmath|2(b-c)>b-c}}）这使得捐赠博弈成为一个迭代博弈(见下一节)。<br />
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The donation game may be applied to markets. Suppose X grows oranges, Y grows apples. The [[marginal utility]] of an apple to the orange-grower X is ''b'', which is higher than the marginal utility (''c'') of an orange, since X has a surplus of oranges and no apples. Similarly, for apple-grower Y, the marginal utility of an orange is ''b'' while the marginal utility of an apple is ''c''. If X and Y contract to exchange an apple and an orange, and each fulfills their end of the deal, then each receive a payoff of ''b''-''c''. If one "defects" and does not deliver as promised, the defector will receive a payoff of ''b'', while the cooperator will lose ''c''. If both defect, then neither one gains or loses anything.<br />
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The donation game may be applied to markets. Suppose X grows oranges, Y grows apples. The marginal utility of an apple to the orange-grower X is b, which is higher than the marginal utility (c) of an orange, since X has a surplus of oranges and no apples. Similarly, for apple-grower Y, the marginal utility of an orange is b while the marginal utility of an apple is c. If X and Y contract to exchange an apple and an orange, and each fulfills their end of the deal, then each receive a payoff of b-c. If one "defects" and does not deliver as promised, the defector will receive a payoff of b, while the cooperator will lose c. If both defect, then neither one gains or loses anything.<br />
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捐赠博弈可能适用于市场。假设 X种橘子，Y 种苹果。苹果对橙子种植者 X 的<font color="#ff8000">边际效用 marginal utility</font>是''b''，这比橙子的边际效用''c''高，因为X有橙子剩余而没有苹果。同样地，对于苹果种植者Y来说，橙子的边际效用是''b''，而苹果的边际效用是''c''。 如果X和Y签约交换一个苹果和一个橙子，并且每个人都完成了交易，那么每个人都会得到''b-c''的收益。如果一方违约没有按照承诺交货，那么这个违约者将得到''b''的收益，而合作者将失去''c''的收益。 如果两者都违约，那么谁也不会得到或失去任何东西。<br />
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==The iterated prisoner's dilemma==<br />
<font color="#ff8000">迭代囚徒困境 iterated prisoner's dilemma </font> {{more citations needed section|date=November 2012}}<br />
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If two players play prisoner's dilemma more than once in succession and they remember previous actions of their opponent and change their strategy accordingly, the game is called iterated prisoner's dilemma.<br />
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If two players play prisoner's dilemma more than once in succession and they remember previous actions of their opponent and change their strategy accordingly, the game is called iterated prisoner's dilemma.<br />
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如果两个参与者连续进行多次囚徒困境博弈，他们记住对手先前的行动并相应地改变策略，这种博弈被称为迭代囚徒困境。<br />
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In addition to the general form above, the iterative version also requires that {{tmath|2R > T + S}}, to prevent alternating cooperation and defection giving a greater reward than mutual cooperation.<br />
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In addition to the general form above, the iterative version also requires that , to prevent alternating cooperation and defection giving a greater reward than mutual cooperation.<br />
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除了上面的一般形式之外，迭代版本还要求{{tmath|2R > T + S}}，防止交替合作和背叛比相互合作有更大的回报。<br />
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The iterated prisoner's dilemma game is fundamental to some theories of human cooperation and trust. On the assumption that the game can model transactions between two people requiring trust, cooperative behaviour in populations may be modeled by a multi-player, iterated, version of the game. It has, consequently, fascinated many scholars over the years. In 1975, Grofman and Pool estimated the count of scholarly articles devoted to it at over 2,000. The iterated prisoner's dilemma has also been referred to as the "[[Peace war game|peace-war game]]".<ref name = Shy>{{cite book | title= Industrial Organization: Theory and Applications | publisher=Massachusetts Institute of Technology Press | first1= Oz | last1=Shy |url=https://books.google.com/?id=tr4CjJ5LlRcC&pg=PR13&dq=industrial+organization+theory+and+applications | year=1995 | isbn=978-0262193665 | accessdate=February 27, 2013}}</ref><br />
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The iterated prisoner's dilemma game is fundamental to some theories of human cooperation and trust. On the assumption that the game can model transactions between two people requiring trust, cooperative behaviour in populations may be modeled by a multi-player, iterated, version of the game. It has, consequently, fascinated many scholars over the years. In 1975, Grofman and Pool estimated the count of scholarly articles devoted to it at over 2,000. The iterated prisoner's dilemma has also been referred to as the "peace-war game".<br />
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迭代囚徒困境博弈是人类合作与信任的理论基础。假设博弈可以为两个需要信任的人之间的交易建模，那么群体中的合作行为也可以由多个参与者迭代的博弈模型来建模。因此，这些年来，它吸引了许多学者。1975年，葛夫曼 Grofman和普尔 Pool估计专门撰写有关该领域的学术文章超过2000篇。迭代囚徒困境也被称为“和平-战争博弈”。<ref name = Shy>{{cite book | title= Industrial Organization: Theory and Applications | publisher=Massachusetts Institute of Technology Press | first1= Oz | last1=Shy |url=https://books.google.com/?id=tr4CjJ5LlRcC&pg=PR13&dq=industrial+organization+theory+and+applications | year=1995 | isbn=978-0262193665 | accessdate=February 27, 2013}}</ref><br />
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If the game is played exactly ''N'' times and both players know this, then it is optimal to defect in all rounds. The only possible [[Nash equilibrium]] is to always defect. The proof is [[Mathematical induction|inductive]]: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on. The same applies if the game length is unknown but has a known upper limit.<br />
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If the game is played exactly N times and both players know this, then it is optimal to defect in all rounds. The only possible Nash equilibrium is to always defect. The proof is inductive: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on. The same applies if the game length is unknown but has a known upper limit.<br />
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如果这个游戏正好玩了N次，并且两个玩家都知道这一点，那么在所有回合中最佳的策略就是叛变。唯一可能的纳什均衡点就是永远叛变。证明是通过归纳法证出来的: 不妨假设一个人在最后一回合叛变，因为对手之后没有机会反击。因此，双方都会在最后一个回合叛变。所以玩家同样也会在倒数第二回合时叛变，因为无论采取什么策略，对手都会在倒数第一回合叛变，依此类推。如果博弈次数未知但次数有限的情况也同样如此。<br />
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Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer. The [[superrational]] strategy in the iterated prisoner's dilemma with fixed ''N'' is to cooperate against a superrational opponent, and in the limit of large ''N'', experimental results on strategies agree with the superrational version, not the game-theoretic rational one.<br />
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Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer. The superrational strategy in the iterated prisoner's dilemma with fixed N is to cooperate against a superrational opponent, and in the limit of large N, experimental results on strategies agree with the superrational version, not the game-theoretic rational one.<br />
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与标准的囚徒困境不同，在迭代囚徒困境中，叛变策略是严重违反直觉的，以至于不能很好地预测人类玩家的行为。然而，在标准的经济理论中，这是唯一正确的答案。具有固定次数 N的迭代囚徒困境中的<font color="#ff8000">超理性 superrational</font>策略是与超理性对手进行合作，在N很大的限制下，实验结果的策略与超理性结果的策略一致，而不是博弈论的理性结果。<br />
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For [[cooperation]] to emerge between game theoretic rational players, the total number of rounds ''N'' must be unknown to the players. In this case "always defect" may no longer be a strictly dominant strategy, only a Nash equilibrium. Amongst results shown by [[Robert Aumann]] in a 1959 paper, rational players repeatedly interacting for indefinitely long games can sustain the cooperative outcome.<br />
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For cooperation to emerge between game theoretic rational players, the total number of rounds N must be unknown to the players. In this case "always defect" may no longer be a strictly dominant strategy, only a Nash equilibrium. Amongst results shown by Robert Aumann in a 1959 paper, rational players repeatedly interacting for indefinitely long games can sustain the cooperative outcome.<br />
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为了使合作在博弈论的理性参与者之间出现，参与者必须不知道回合总数N。在这种情况下，“总是叛变”可能不再是一个严格占优策略，而只是一个纳什均衡。罗伯特·奥曼 Robert Aumann在1959年的一篇论文中表明，理性参与者在无限多次的博弈中通过反复互动可以维持合作的结果。<br />
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According to a 2019 experimental study in the ''American Economic Review'' which tested what strategies real-life subjects used in iterated prisoners' dilemma situations with perfect monitoring, the majority of chosen strategies were always defect, [[Tit for tat|tit-for-tat]], and [[Grim trigger]]. Which strategy the subjects chose depended on the parameters of the game.<ref>{{Cite journal|last=Dal Bó|first=Pedro|last2=Fréchette|first2=Guillaume R.|date=2019|title=Strategy Choice in the Infinitely Repeated Prisoner's Dilemma|journal=American Economic Review|language=en|volume=109|issue=11|pages=3929–3952|doi=10.1257/aer.20181480|issn=0002-8282}}</ref><br />
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According to a 2019 experimental study in the American Economic Review which tested what strategies real-life subjects used in iterated prisoners' dilemma situations with perfect monitoring, the majority of chosen strategies were always defect, tit-for-tat, and Grim trigger. Which strategy the subjects chose depended on the parameters of the game.<br />
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根据《美国经济评论》于2019年进行的一项实验研究，该实验中通过完美的监控测试了现实中被用在迭代囚徒困境情况下的策略，监测选择的策略总是背叛，针锋相对的和 <font color="#ff8000"> 冷酷触发策略 Grim trigger</font>。受试者选择的策略取决于博弈的参数。<ref>{{Cite journal|last=Dal Bó|first=Pedro|last2=Fréchette|first2=Guillaume R.|date=2019|title=Strategy Choice in the Infinitely Repeated Prisoner's Dilemma|journal=American Economic Review|language=en|volume=109|issue=11|pages=3929–3952|doi=10.1257/aer.20181480|issn=0002-8282}}</ref><br />
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===Strategy for the iterated prisoner's dilemma===<br />
迭代囚徒困境下的策略<br />
Interest in the iterated prisoner's dilemma (IPD) was kindled by [[Robert Axelrod]] in his book ''[[The Evolution of Cooperation]]'' (1984). In it he reports on a tournament he organized of the ''N'' step prisoner's dilemma (with ''N'' fixed) in which participants have to choose their mutual strategy again and again, and have memory of their previous encounters. Axelrod invited academic colleagues all over the world to devise computer strategies to compete in an IPD tournament. The programs that were entered varied widely in algorithmic complexity, initial hostility, capacity for forgiveness, and so forth.<br />
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Interest in the iterated prisoner's dilemma (IPD) was kindled by Robert Axelrod in his book The Evolution of Cooperation (1984). In it he reports on a tournament he organized of the N step prisoner's dilemma (with N fixed) in which participants have to choose their mutual strategy again and again, and have memory of their previous encounters. Axelrod invited academic colleagues all over the world to devise computer strategies to compete in an IPD tournament. The programs that were entered varied widely in algorithmic complexity, initial hostility, capacity for forgiveness, and so forth.<br />
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罗伯特·阿克塞尔罗德 Robert Axelrod在他的著作《合作的进化》(1984)中激起了人们对迭代囚徒困境(IPD)的兴趣。在这篇文章中，他报道了自己组织的固定N次囚徒困境的比赛，参与者必须一次又一次地选择他们的共同策略，并且要记住他们之前的遭遇。阿克塞尔罗德邀请世界各地的学术界同仁设计计算机策略来参加IPD锦标赛。输入的程序在算法复杂性、最初敌意、宽恕能力等方面有很大差异。<br />
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Axelrod discovered that when these encounters were repeated over a long period of time with many players, each with different strategies, greedy strategies tended to do very poorly in the long run while more [[altruism|altruistic]] strategies did better, as judged purely by self-interest. He used this to show a possible mechanism for the evolution of altruistic behaviour from mechanisms that are initially purely selfish, by [[natural selection]].<br />
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Axelrod discovered that when these encounters were repeated over a long period of time with many players, each with different strategies, greedy strategies tended to do very poorly in the long run while more altruistic strategies did better, as judged purely by self-interest. He used this to show a possible mechanism for the evolution of altruistic behaviour from mechanisms that are initially purely selfish, by natural selection.<br />
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阿克塞尔罗德发现，当这些遭遇长时间在许多玩家身上重复发生时，每个玩家都有不同的策略，从长远来看，贪婪策略往往表现得非常糟糕，而更加利他的策略表现得更好，这完全是根据自身利益来判断的。他利用这一结果揭示了通过自然选择，从最初纯粹自私行为向利他行为进化的可能机制。<br />
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The winning [[deterministic algorithm|deterministic]] strategy was tit for tat, which [[Anatol Rapoport]] developed and entered into the tournament. It was the simplest of any program entered, containing only four lines of [[BASIC]], and won the contest. The strategy is simply to cooperate on the first iteration of the game; after that, the player does what his or her opponent did on the previous move. Depending on the situation, a slightly better strategy can be "tit for tat with forgiveness". When the opponent defects, on the next move, the player sometimes cooperates anyway, with a small probability (around 1–5%). This allows for occasional recovery from getting trapped in a cycle of defections. The exact probability depends on the line-up of opponents.<br />
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The winning deterministic strategy was tit for tat, which Anatol Rapoport developed and entered into the tournament. It was the simplest of any program entered, containing only four lines of BASIC, and won the contest. The strategy is simply to cooperate on the first iteration of the game; after that, the player does what his or her opponent did on the previous move. Depending on the situation, a slightly better strategy can be "tit for tat with forgiveness". When the opponent defects, on the next move, the player sometimes cooperates anyway, with a small probability (around 1–5%). This allows for occasional recovery from getting trapped in a cycle of defections. The exact probability depends on the line-up of opponents.<br />
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最终获胜的决定性策略是针锋相对策略，这是阿纳托尔·拉波波特 Anatol Rapoport开发并参加比赛的策略。这是所有参赛程序中最简单的一个，只有四行 BASIC 语言，并且赢得了比赛。策略很简单，就是在游戏的第一次迭代中进行合作; 在此之后，玩家将执行做他的对手在前一步中所做的事情。根据具体情况，一个稍微好一点的策略可以是“带着宽恕之心针锋相对”。当对手叛变时，在下一次博弈中，玩家有时还是会合作，但概率很小(大约1-5%)。这允许博弈偶尔能从陷入叛变循环中恢复过来。确切的概率取决于对手的安排。<br />
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By analysing the top-scoring strategies, Axelrod stated several conditions necessary for a strategy to be successful.<br />
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By analysing the top-scoring strategies, Axelrod stated several conditions necessary for a strategy to be successful.<br />
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通过分析得分最高的战略，阿克塞尔罗德阐述了战略成功的几个必要条件。<br />
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; Nice: The most important condition is that the strategy must be "nice", that is, it will not defect before its opponent does (this is sometimes referred to as an "optimistic" algorithm). Almost all of the top-scoring strategies were nice; therefore, a purely selfish strategy will not "cheat" on its opponent, for purely self-interested reasons first.<br />
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Nice: The most important condition is that the strategy must be "nice", that is, it will not defect before its opponent does (this is sometimes referred to as an "optimistic" algorithm). Almost all of the top-scoring strategies were nice; therefore, a purely selfish strategy will not "cheat" on its opponent, for purely self-interested reasons first.<br />
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;友好:最重要的条件是策略必须是好的，也就是说，它不会在对手之前叛变(这有时被称为“乐观”算法)。几乎所有得分最高的策略都是友好的; 因此，一个纯粹的自私策略不会为了纯粹自身的利益而“欺骗”对手。<br />
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; Retaliating: However, Axelrod contended, the successful strategy must not be a blind optimist. It must sometimes retaliate. An example of a non-retaliating strategy is Always Cooperate. This is a very bad choice, as "nasty" strategies will ruthlessly exploit such players.<br />
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Retaliating: However, Axelrod contended, the successful strategy must not be a blind optimist. It must sometimes retaliate. An example of a non-retaliating strategy is Always Cooperate. This is a very bad choice, as "nasty" strategies will ruthlessly exploit such players.<br />
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;报复:然而，阿克塞尔罗德认为，成功的战略决不能是盲目的乐观主义。它有时必须进行报复。非报复策略的一个例子就是永远合作。这是一个非常糟糕的选择，因为“肮脏”的策略会无情地利用这些玩家。<br />
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; Forgiving: Successful strategies must also be forgiving. Though players will retaliate, they will once again fall back to cooperating if the opponent does not continue to defect. This stops long runs of revenge and counter-revenge, maximizing points.<br />
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Forgiving: Successful strategies must also be forgiving. Though players will retaliate, they will once again fall back to cooperating if the opponent does not continue to defect. This stops long runs of revenge and counter-revenge, maximizing points.<br />
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;宽容: 成功的策略也必须是宽容的。虽然玩家会报复，但如果对手不继续叛变，他们将再次回到合作的状态。这阻止了长时间的报复和反报复，最大限度地提高积分。<br />
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; Non-envious: The last quality is being non-envious, that is not striving to score more than the opponent.<br />
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Non-envious: The last quality is being non-envious, that is not striving to score more than the opponent.<br />
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;不嫉妒: 最后一个品质是不嫉妒，不强求比对手得分更多。<br />
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The optimal (points-maximizing) strategy for the one-time PD game is simply defection; as explained above, this is true whatever the composition of opponents may be. However, in the iterated-PD game the optimal strategy depends upon the strategies of likely opponents, and how they will react to defections and cooperations. For example, consider a population where everyone defects every time, except for a single individual following the tit for tat strategy. That individual is at a slight disadvantage because of the loss on the first turn. In such a population, the optimal strategy for that individual is to defect every time. In a population with a certain percentage of always-defectors and the rest being tit for tat players, the optimal strategy for an individual depends on the percentage, and on the length of the game.<br />
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The optimal (points-maximizing) strategy for the one-time PD game is simply defection; as explained above, this is true whatever the composition of opponents may be. However, in the iterated-PD game the optimal strategy depends upon the strategies of likely opponents, and how they will react to defections and cooperations. For example, consider a population where everyone defects every time, except for a single individual following the tit for tat strategy. That individual is at a slight disadvantage because of the loss on the first turn. In such a population, the optimal strategy for that individual is to defect every time. In a population with a certain percentage of always-defectors and the rest being tit for tat players, the optimal strategy for an individual depends on the percentage, and on the length of the game.<br />
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对于一次性的囚徒困境博弈，最优(点数最大化)策略就是简单的叛变; 正如上面所说，无论对手的构成如何，这都是正确的。然而，在迭代囚徒困境博弈中，最优策略取决于可能的对手的策略，以及他们对叛变和合作的反应。例如，考虑一个群体，其中每个人每次都会叛变，只有一个人遵循针锋相对的策略。那个人就会由于第一回合的失利而处于轻微的不利地位。在这样一个群体中，个体的最佳策略是每次都叛变。在一定比例的总是选择背叛的玩家和其余组成选择针锋相对策略的玩家的人群中，个人的最佳策略取决于这一比例和博弈的次数。<br />
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In the strategy called Pavlov, [[win-stay, lose-switch]], faced with a failure to cooperate, the player switches strategy the next turn.<ref>http://www.pnas.org/content/pnas/93/7/2686.full.pdf</ref> In certain circumstances,{{specify|date=November 2012}} Pavlov beats all other strategies by giving preferential treatment to co-players using a similar strategy.<br />
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In the strategy called Pavlov, win-stay, lose-switch, faced with a failure to cooperate, the player switches strategy the next turn. In certain circumstances, Pavlov beats all other strategies by giving preferential treatment to co-players using a similar strategy.<br />
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在所谓的<font color="#ff8000">巴甫洛夫策略 Pavlov strategy</font>中，赢-保持，输-变换，面对一次合作失败，玩家将在下一次变换策略。<ref>http://www.pnas.org/content/pnas/93/7/2686.full.pdf</ref>在某些情况下，{{specify|date=November 2012}}巴甫洛夫通过使用类似策略给与合作者优惠待遇打败了其他所有策略。<br />
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Deriving the optimal strategy is generally done in two ways:<br />
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Deriving the optimal strategy is generally done in two ways:<br />
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得出最佳策略通常有两种方法:<br />
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* [[Bayesian Nash equilibrium]]: If the statistical distribution of opposing strategies can be determined (e.g. 50% tit for tat, 50% always cooperate) an optimal counter-strategy can be derived analytically.{{efn|1=For example see the 2003 study<ref>{{cite web|url= http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|title=Bayesian Nash equilibrium; a statistical test of the hypothesis|url-status=dead|archive-url= https://web.archive.org/web/20051002195142/http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|archive-date=2005-10-02|publisher=[[Tel Aviv University]]}}</ref> for discussion of the concept and whether it can apply in real [[economic]] or strategic situations.}}<br />
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<font color="#ff8000">贝叶斯纳什均衡 Bayesian Nash equilibrium</font>：如果可以确定对立策略的统计分布（例如，50%针锋相对，50%总是合作），那么，可以通过分析得出最佳的反策略{{efn|1=例如2003年的研究<ref>{{cite web|url= http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|title=Bayesian Nash equilibrium; a statistical test of the hypothesis|url-status=dead|archive-url= https://web.archive.org/web/20051002195142/http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|archive-date=2005-10-02|publisher=[[Tel Aviv University]]}}</ref>讨论这一概念以及它是否可以应用于实际经济或战略情况。}}<br />
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* [[Monte Carlo method|Monte Carlo]] simulations of populations have been made, where individuals with low scores die off, and those with high scores reproduce (a [[genetic algorithm]] for finding an optimal strategy). The mix of algorithms in the final population generally depends on the mix in the initial population. The introduction of mutation (random variation during reproduction) lessens the dependency on the initial population; empirical experiments with such systems tend to produce tit for tat players (see for instance Chess 1988),{{Clarify|date=August 2016}} but no analytic proof exists that this will always occur.<ref>{{Citation|last=Wu|first=Jiadong|title=Cooperation on the Monte Carlo Rule: Prisoner's Dilemma Game on the Grid|date=2019|work=Theoretical Computer Science|volume=1069|pages=3–15|editor-last=Sun|editor-first=Xiaoming|publisher=Springer Singapore|language=en|doi=10.1007/978-981-15-0105-0_1|isbn=978-981-15-0104-3|last2=Zhao|first2=Chengye|editor2-last=He|editor2-first=Kun|editor3-last=Chen|editor3-first=Xiaoyun}}</ref><br />
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<font color="#ff8000">蒙特卡洛方法 Monte Carlo method </font>已经对种群进行了模拟，分数低的个体死亡，分数高的个体繁殖（<font color="#ff8000">遗传算法 genetic algorithm </font>用于寻找一个最佳策略）。最终群体中的算法组合通常取决于初始总体的组合。引入突变（繁殖过程中的随机变异）可以减少对初始种群的依赖性。使用这种系统进行经验性实验往往会为针锋相对的玩家带来麻烦（见Chess 1988），{{Clarify|date=August 2016}}，但是没有分析证据表明这种情况会一直发生。<ref>{{Citation|last=Wu|first=Jiadong|title=Cooperation on the Monte Carlo Rule: Prisoner's Dilemma Game on the Grid|date=2019|work=Theoretical Computer Science|volume=1069|pages=3–15|editor-last=Sun|editor-first=Xiaoming|publisher=Springer Singapore|language=en|doi=10.1007/978-981-15-0105-0_1|isbn=978-981-15-0104-3|last2=Zhao|first2=Chengye|editor2-last=He|editor2-first=Kun|editor3-last=Chen|editor3-first=Xiaoyun}}</ref><br />
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Although tit for tat is considered to be the most [[robust]] basic strategy, a team from [[Southampton University]] in England introduced a new strategy at the 20th-anniversary iterated prisoner's dilemma competition, which proved to be more successful than tit for tat. This strategy relied on collusion between programs to achieve the highest number of points for a single program. The university submitted 60 programs to the competition, which were designed to recognize each other through a series of five to ten moves at the start.<ref>{{cite press release|url= http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|publisher=University of Southampton|title=University of Southampton team wins Prisoner's Dilemma competition|date=7 October 2004|url-status=dead|archive-url= https://web.archive.org/web/20140421055745/http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|archive-date=2014-04-21}}</ref> Once this recognition was made, one program would always cooperate and the other would always defect, assuring the maximum number of points for the defector. If the program realized that it was playing a non-Southampton player, it would continuously defect in an attempt to minimize the score of the competing program. As a result, the 2004 Prisoners' Dilemma Tournament results show [[University of Southampton]]'s strategies in the first three places, despite having fewer wins and many more losses than the GRIM strategy. (In a PD tournament, the aim of the game is not to "win" matches&nbsp;– that can easily be achieved by frequent defection). Also, even without implicit collusion between [[computer program|software strategies]] (exploited by the Southampton team) tit for tat is not always the absolute winner of any given tournament; it would be more precise to say that its long run results over a series of tournaments outperform its rivals. (In any one event a given strategy can be slightly better adjusted to the competition than tit for tat, but tit for tat is more robust). The same applies for the tit for tat with forgiveness variant, and other optimal strategies: on any given day they might not "win" against a specific mix of counter-strategies. An alternative way of putting it is using the Darwinian [[Evolutionarily stable strategy|ESS]] simulation. In such a simulation, tit for tat will almost always come to dominate, though nasty strategies will drift in and out of the population because a tit for tat population is penetrable by non-retaliating nice strategies, which in turn are easy prey for the nasty strategies. [[Richard Dawkins]] showed that here, no static mix of strategies form a stable equilibrium and the system will always oscillate between bounds.}} this strategy ended up taking the top three positions in the competition, as well as a number of positions towards the bottom.<br />
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Although tit for tat is considered to be the most robust basic strategy, a team from Southampton University in England introduced a new strategy at the 20th-anniversary iterated prisoner's dilemma competition, which proved to be more successful than tit for tat. This strategy relied on collusion between programs to achieve the highest number of points for a single program. The university submitted 60 programs to the competition, which were designed to recognize each other through a series of five to ten moves at the start. Once this recognition was made, one program would always cooperate and the other would always defect, assuring the maximum number of points for the defector. If the program realized that it was playing a non-Southampton player, it would continuously defect in an attempt to minimize the score of the competing program. As a result, the 2004 Prisoners' Dilemma Tournament results show University of Southampton's strategies in the first three places, despite having fewer wins and many more losses than the GRIM strategy. (In a PD tournament, the aim of the game is not to "win" matches&nbsp;– that can easily be achieved by frequent defection). Also, even without implicit collusion between software strategies (exploited by the Southampton team) tit for tat is not always the absolute winner of any given tournament; it would be more precise to say that its long run results over a series of tournaments outperform its rivals. (In any one event a given strategy can be slightly better adjusted to the competition than tit for tat, but tit for tat is more robust). The same applies for the tit for tat with forgiveness variant, and other optimal strategies: on any given day they might not "win" against a specific mix of counter-strategies. An alternative way of putting it is using the Darwinian ESS simulation. In such a simulation, tit for tat will almost always come to dominate, though nasty strategies will drift in and out of the population because a tit for tat population is penetrable by non-retaliating nice strategies, which in turn are easy prey for the nasty strategies. Richard Dawkins showed that here, no static mix of strategies form a stable equilibrium and the system will always oscillate between bounds.}} this strategy ended up taking the top three positions in the competition, as well as a number of positions towards the bottom.<br />
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尽管针锋相对被认为是最有力的基本策略，来自英格兰南安普敦大学的一个团队在20周年的迭代囚徒困境竞赛中提出了一个新策略，这个策略被证明比针锋相对更为成功。这种策略依赖于程序之间的串通，以获得单个程序的最高分数。这所大学提交了60个程序，这些程序的设计目的是在比赛开始时通过一系列的5到10个动作来互相认识。<ref>{{cite press release|url= http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|publisher=University of Southampton|title=University of Southampton team wins Prisoner's Dilemma competition|date=7 October 2004|url-status=dead|archive-url= https://web.archive.org/web/20140421055745/http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|archive-date=2014-04-21}}</ref>一旦认识建立，一个程序总是合作，另一个程序总是叛变，保证叛变者得到最多的分数。如果这个程序意识到它正在和一个非南安普顿的球员比赛，它会不断地叛变，试图最小化与之竞争的程序的得分。因此，2004年囚徒困境锦标赛的结果显示了南安普敦大学战略位居前三名，尽管它比冷酷战略赢得更少，输的更多。(在囚徒困境锦标赛中，比赛的目的不是“赢”比赛——这一点频繁叛变很容易实现)。此外，即使没有软件策略之间的暗中串通(南安普顿队利用了这一点) ，针锋相对并不总是任何特定锦标赛的绝对赢家; 更准确地说，它是在一系列锦标赛中的长期结果超过了它的竞争对手。(在任何一个事件中，一个给定的策略可以比针锋相对稍微更好地适应竞争，但是针锋相对更稳健)。这同样适用于带有宽恕变量的针锋相对，和其他最佳策略: 在任何特定的一天，他们可能不会“赢得”一个特定的混合反战略。另一种方法是使用达尔文 Darwinian的<font color="#ff8000"> ESS模拟 ESS simulation</font>。在这样的模拟中，针锋相对几乎总是占主导地位，尽管讨厌的策略会在人群中漂移，因为使用针锋相对策略的人群可以通过非报复性的好策略进行渗透，这反过来使他们容易成为讨厌策略的猎物。理查德·道金斯 Richard Dawkins指出，在这里，没有静态的混合策略会形成一个稳定的平衡，系统将始终在边界之间振荡。这种策略最终在比赛中获得了前三名的成绩，或者是接近垫底的成绩。<br />
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This strategy takes advantage of the fact that multiple entries were allowed in this particular competition and that the performance of a team was measured by that of the highest-scoring player (meaning that the use of self-sacrificing players was a form of [[minmaxing]]). In a competition where one has control of only a single player, tit for tat is certainly a better strategy. Because of this new rule, this competition also has little theoretical significance when analyzing single agent strategies as compared to Axelrod's seminal tournament. However, it provided a basis for analysing how to achieve cooperative strategies in multi-agent frameworks, especially in the presence of noise. In fact, long before this new-rules tournament was played, Dawkins, in his book ''[[The Selfish Gene]]'', pointed out the possibility of such strategies winning if multiple entries were allowed, but he remarked that most probably Axelrod would not have allowed them if they had been submitted. It also relies on circumventing rules about the prisoner's dilemma in that there is no communication allowed between the two players, which the Southampton programs arguably did with their opening "ten move dance" to recognize one another; this only reinforces just how valuable communication can be in shifting the balance of the game.<br />
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This strategy takes advantage of the fact that multiple entries were allowed in this particular competition and that the performance of a team was measured by that of the highest-scoring player (meaning that the use of self-sacrificing players was a form of minmaxing). In a competition where one has control of only a single player, tit for tat is certainly a better strategy. Because of this new rule, this competition also has little theoretical significance when analyzing single agent strategies as compared to Axelrod's seminal tournament. However, it provided a basis for analysing how to achieve cooperative strategies in multi-agent frameworks, especially in the presence of noise. In fact, long before this new-rules tournament was played, Dawkins, in his book The Selfish Gene, pointed out the possibility of such strategies winning if multiple entries were allowed, but he remarked that most probably Axelrod would not have allowed them if they had been submitted. It also relies on circumventing rules about the prisoner's dilemma in that there is no communication allowed between the two players, which the Southampton programs arguably did with their opening "ten move dance" to recognize one another; this only reinforces just how valuable communication can be in shifting the balance of the game.<br />
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这种策略利用了这样一个事实，即在这场特殊的比赛中允许多个参赛项目，并且团队的表现由得分最高的项目来衡量(这意味着使用自我牺牲的项目是一种分数最大化的形式)。在一个只能控制一个玩家的比赛中，针锋相对当然是一个更好的策略。由于这一新规则的存在，与阿克塞尔罗德的具有深远影响的竞赛相比，这种竞赛在分析单个主体策略时也就没有什么理论意义。然而，它为在分析多主体框架下，特别是在存在干扰的情况下，如何实现协作策略提供了基础。事实上，早在这场新规则锦标赛开始之前，道金斯就在他的《自私的基因》一书中指出，如果允许多次参赛，这种策略就有可能获胜，但他说，如果提交这种策略的话，阿克塞尔罗德很可能不会允许。因为它依赖于规避囚徒困境的规则，即两个参与者之间不允许交流，南安普顿的项目可以说在开场的“十步舞”中就是这样做以认识对方的; 这只是强调了交流在改变游戏平衡方面的价值。<br />
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===Stochastic iterated prisoner's dilemma===<br />
随机的迭代囚徒困境<br />
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In a stochastic iterated prisoner's dilemma game, strategies are specified by in terms of "cooperation probabilities".<ref name=Press2012>{{cite journal|last1=Press|first1=WH|last2=Dyson|first2=FJ|title=Iterated Prisoner's Dilemma contains strategies that dominate any evolutionary opponent|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|date=26 June 2012|volume=109|issue=26|pages=10409–13|doi=10.1073/pnas.1206569109|pmid=22615375|pmc=3387070|bibcode=2012PNAS..10910409P}}</ref> In an encounter between player ''X'' and player ''Y'', ''X'' 's strategy is specified by a set of probabilities ''P'' of cooperating with ''Y''. ''P'' is a function of the outcomes of their previous encounters or some subset thereof. If ''P'' is a function of only their most recent ''n'' encounters, it is called a "memory-n" strategy. A memory-1 strategy is then specified by four cooperation probabilities: <math>P=\{P_{cc},P_{cd},P_{dc},P_{dd}\}</math>, where <math>P_{ab}</math> is the probability that ''X'' will cooperate in the present encounter given that the previous encounter was characterized by (ab). For example, if the previous encounter was one in which ''X'' cooperated and ''Y'' defected, then <math>P_{cd}</math> is the probability that ''X'' will cooperate in the present encounter. If each of the probabilities are either 1 or 0, the strategy is called deterministic. An example of a deterministic strategy is the tit for tat strategy written as ''P''={1,0,1,0}, in which ''X'' responds as ''Y'' did in the previous encounter. Another is the [[win–stay, lose–switch]] strategy written as ''P''={1,0,0,1}, in which ''X'' responds as in the previous encounter, if it was a "win" (i.e. cc or dc) but changes strategy if it was a loss (i.e. cd or dd). It has been shown that for any memory-n strategy there is a corresponding memory-1 strategy which gives the same statistical results, so that only memory-1 strategies need be considered.<ref name="Press2012"/><br />
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In a stochastic iterated prisoner's dilemma game, strategies are specified by in terms of "cooperation probabilities". In an encounter between player X and player Y, X 's strategy is specified by a set of probabilities P of cooperating with Y. P is a function of the outcomes of their previous encounters or some subset thereof. If P is a function of only their most recent n encounters, it is called a "memory-n" strategy. A memory-1 strategy is then specified by four cooperation probabilities: <math>P=\{P_{cc},P_{cd},P_{dc},P_{dd}\}</math>, where <math>P_{ab}</math> is the probability that X will cooperate in the present encounter given that the previous encounter was characterized by (ab). For example, if the previous encounter was one in which X cooperated and Y defected, then <math>P_{cd}</math> is the probability that X will cooperate in the present encounter. If each of the probabilities are either 1 or 0, the strategy is called deterministic. An example of a deterministic strategy is the tit for tat strategy written as P={1,0,1,0}, in which X responds as Y did in the previous encounter. Another is the win–stay, lose–switch strategy written as P={1,0,0,1}, in which X responds as in the previous encounter, if it was a "win" (i.e. cc or dc) but changes strategy if it was a loss (i.e. cd or dd). It has been shown that for any memory-n strategy there is a corresponding memory-1 strategy which gives the same statistical results, so that only memory-1 strategies need be considered.<br />
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在随机迭代<font color="#ff8000"> 囚徒困境prisoner's dilemma</font>博弈中，策略由“合作概率”来确定。<ref name=Press2012>{{cite journal|last1=Press|first1=WH|last2=Dyson|first2=FJ|title=Iterated Prisoner's Dilemma contains strategies that dominate any evolutionary opponent|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|date=26 June 2012|volume=109|issue=26|pages=10409–13|doi=10.1073/pnas.1206569109|pmid=22615375|pmc=3387070|bibcode=2012PNAS..10910409P}}</ref>在玩家''X''和玩家''Y''之间的遭遇中，''X''‘s的策略由一组与''Y''合作的概率''P''确定，''P''是他们之前遭遇的结果的函数，或者是其中的一些子集。如果''P''只是它们最近遇到次数 ''n''的函数，那么它被称为“记忆-n”策略。我们可以由四个联合概率指定一个记忆-1策略: <math>P=\{P_{cc},P_{cd},P_{dc},P_{dd}\}</math>，其中<math>P_{ab}</math>是在当前遭遇中基于先前联合的概率。如果每个概率都是1或0，这种策略称为确定性策略。确定性策略的一个例子是针锋相对策略，写成 p {1,0,1,0} ，其中 x 的反应和 y 在前一次遭遇中的反应一样。另一种是胜-保持-败-转换策略，它被写成 p {1,0,0,1} ，在这种策略中，如果 x 获得胜利(即:cc 或 dc)，x会做出与上一次遭遇一样的反应 ，但如果失败，x会改变策略(即cd 或 dd)。研究表明，对于任何一种记忆-n 策略，存在一个相应的记忆-1策略，这个策略给出相同的统计结果，因此只需要考虑记忆-1策略。<ref name="Press2012"/><br />
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If we define ''P'' as the above 4-element strategy vector of ''X'' and <math>Q=\{Q_{cc},Q_{cd},Q_{dc},Q_{dd}\}</math> as the 4-element strategy vector of ''Y'', a transition matrix ''M'' may be defined for ''X'' whose ''ij'' th entry is the probability that the outcome of a particular encounter between ''X'' and ''Y'' will be ''j'' given that the previous encounter was ''i'', where ''i'' and ''j'' are one of the four outcome indices: ''cc'', ''cd'', ''dc'', or ''dd''. For example, from ''X'' 's point of view, the probability that the outcome of the present encounter is ''cd'' given that the previous encounter was ''cd'' is equal to <math>M_{cd,cd}=P_{cd}(1-Q_{dc})</math>. (The indices for ''Q'' are from ''Y'' 's point of view: a ''cd'' outcome for ''X'' is a ''dc'' outcome for ''Y''.) Under these definitions, the iterated prisoner's dilemma qualifies as a [[stochastic process]] and ''M'' is a [[stochastic matrix]], allowing all of the theory of stochastic processes to be applied.<ref name="Press2012"/><br />
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If we define P as the above 4-element strategy vector of X and <math>Q=\{Q_{cc},Q_{cd},Q_{dc},Q_{dd}\}</math> as the 4-element strategy vector of Y, a transition matrix M may be defined for X whose ij th entry is the probability that the outcome of a particular encounter between X and Y will be j given that the previous encounter was i, where i and j are one of the four outcome indices: cc, cd, dc, or dd. For example, from X 's point of view, the probability that the outcome of the present encounter is cd given that the previous encounter was cd is equal to <math>M_{cd,cd}=P_{cd}(1-Q_{dc})</math>. (The indices for Q are from Y 's point of view: a cd outcome for X is a dc outcome for Y.) Under these definitions, the iterated prisoner's dilemma qualifies as a stochastic process and M is a stochastic matrix, allowing all of the theory of stochastic processes to be applied.<br />
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如果我们将''P''定义为''X''的上述4元策略向量，并将<math>Q=\{Q_{cc},Q_{cd},Q_{dc},Q_{dd}\}</math>定义为''Y''的4元策略向量，则对于''X''可以定义一个转移矩阵''M''，其第ij项是''X''和''Y''之间特定相遇的结果为j的概率，给定i，其中i和j是cc、cd、dc或dd 四个结果索引中的一个。例如，从''X''的角度来看，如果给定''cd''，那么这次的结果是''cd''的概率等于<math>M_{cd,cd}=P_{cd}(1-Q_{dc})</math>。(''Q''的指标是 从''Y''的角度: ''X''的''cd''结果是''Y''的''dc''结果)在这些定义下，重复的囚徒困境被定义为一个随机过程，''M''是一个随机矩阵，允许应用所有的随机过程理论。<ref name="Press2012"/><br />
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One result of stochastic theory is that there exists a stationary vector ''v'' for the matrix ''M'' such that <math>v\cdot M=v</math>. Without loss of generality, it may be specified that ''v'' is normalized so that the sum of its four components is unity. The ''ij'' th entry in <math>M^n</math> will give the probability that the outcome of an encounter between ''X'' and ''Y'' will be ''j'' given that the encounter ''n'' steps previous is ''i''. In the limit as ''n'' approaches infinity, ''M'' will converge to a matrix with fixed values, giving the long-term probabilities of an encounter producing ''j'' which will be independent of ''i''. In other words, the rows of <math>M^\infty</math> will be identical, giving the long-term equilibrium result probabilities of the iterated prisoners dilemma without the need to explicitly evaluate a large number of interactions. It can be seen that ''v'' is a stationary vector for <math>M^n</math> and particularly <math>M^\infty</math>, so that each row of <math>M^\infty</math> will be equal to ''v''. Thus the stationary vector specifies the equilibrium outcome probabilities for ''X''. Defining <math>S_x=\{R,S,T,P\}</math> and <math>S_y=\{R,T,S,P\}</math> as the short-term payoff vectors for the {cc,cd,dc,dd} outcomes (From ''X'' 's point of view), the equilibrium payoffs for ''X'' and ''Y'' can now be specified as <math>s_x=v\cdot S_x</math> and <math>s_y=v\cdot S_y</math>, allowing the two strategies ''P'' and ''Q'' to be compared for their long term payoffs.<br />
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One result of stochastic theory is that there exists a stationary vector v for the matrix M such that <math>v\cdot M=v</math>. Without loss of generality, it may be specified that v is normalized so that the sum of its four components is unity. The ij th entry in <math>M^n</math> will give the probability that the outcome of an encounter between X and Y will be j given that the encounter n steps previous is i. In the limit as n approaches infinity, M will converge to a matrix with fixed values, giving the long-term probabilities of an encounter producing j which will be independent of i. In other words, the rows of <math>M^\infty</math> will be identical, giving the long-term equilibrium result probabilities of the iterated prisoners dilemma without the need to explicitly evaluate a large number of interactions. It can be seen that v is a stationary vector for <math>M^n</math> and particularly <math>M^\infty</math>, so that each row of <math>M^\infty</math> will be equal to v. Thus the stationary vector specifies the equilibrium outcome probabilities for X. Defining <math>S_x=\{R,S,T,P\}</math> and <math>S_y=\{R,T,S,P\}</math> as the short-term payoff vectors for the {cc,cd,dc,dd} outcomes (From X 's point of view), the equilibrium payoffs for X and Y can now be specified as <math>s_x=v\cdot S_x</math> and <math>s_y=v\cdot S_y</math>, allowing the two strategies P and Q to be compared for their long term payoffs.<br />
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随机理论的一个结果是，矩阵''M''存在一个平稳向量''v''使得<math>v\cdot M=v</math>成立。一般地，我们可以指定''v''是标准化的，因此它的4个组成部分之和为1。the equilibrium payoffs for and can now be specified as and, allowing the two strategies ''P'' and ''Q'' to be compared for their long term payoffs.第''ij''项<math>M^n</math>给出了''X''和''Y''相遇的结果的概率为''j''，给定前面相遇''n''步的概率是''i''。当''n''趋于无穷时，''M''收敛于一个具有固定值的矩阵，并且''j''趋向一个长期概率，与''i''独立。换句话说， <math>M^\infty</math>的行将是相同的，从而给出了重复囚徒困境的长期均衡结果概率，而不需要明确地计算大量的相互作用。可以看出，''v''是<math>M^n</math>特别是<math>M^\infty</math>, 的平稳向量，因此<math>M^\infty</math>的每一行都等于''v''。因此平稳向量指定了''X''的均衡结果概率。定义<math>S_x=\{R,S,T,P\}</math>和<math>S_y=\{R,T,S,P\}</math>作为{cc,cd,dc,dd}结果的短期收益向量(从''X''的角度来看) ，现在可以将''X''和''Y''的均衡收益指定为<math>s_x=v\cdot S_x</math>和<math>s_y=v\cdot S_y</math>，使得''P''、''Q''两种策略的长期收益可以比较。<br />
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====Zero-determinant strategies====<br />
<font color="#ff8000">零决定策略 Zero-determinant strategies</font><br />
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[[File:IPD Venn.svg|right|thumb|upright=2.5|The relationship between zero-determinant (ZD), cooperating and defecting strategies in the iterated prisoner's dilemma (IPD) illustrated in a [[Venn diagram]]. Cooperating strategies always cooperate with other cooperating strategies, and defecting strategies always defect against other defecting strategies. Both contain subsets of strategies that are robust under strong selection, meaning no other memory-1 strategy is selected to invade such strategies when they are resident in a population. Only cooperating strategies contain a subset that are always robust, meaning that no other memory-1 strategy is selected to invade and replace such strategies, under both strong and [[weak selection]]. The intersection between ZD and good cooperating strategies is the set of generous ZD strategies. Extortion strategies are the intersection between ZD and non-robust defecting strategies. Tit-for-tat lies at the intersection of cooperating, defecting and ZD strategies.]]<br />
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The relationship between zero-determinant (ZD), cooperating and defecting strategies in the iterated prisoner's dilemma (IPD) illustrated in a [[Venn diagram. Cooperating strategies always cooperate with other cooperating strategies, and defecting strategies always defect against other defecting strategies. Both contain subsets of strategies that are robust under strong selection, meaning no other memory-1 strategy is selected to invade such strategies when they are resident in a population. Only cooperating strategies contain a subset that are always robust, meaning that no other memory-1 strategy is selected to invade and replace such strategies, under both strong and weak selection. The intersection between ZD and good cooperating strategies is the set of generous ZD strategies. Extortion strategies are the intersection between ZD and non-robust defecting strategies. Tit-for-tat lies at the intersection of cooperating, defecting and ZD strategies.]]<br />
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<font color="#ff8000">维恩图 Venn diagram</font>中讨论了<font color="#ff8000">迭代囚徒困境 iterated prisoner's dilemma</font>(IPD)中零决定策略(ZD)、合作策略和背叛策略之间的关系。合作策略总是与其他合作策略相互配合，而背叛策略总是与其他背叛策略相抵触。这两种策略都包都含在强选择下稳健的策略子集，这意味着当它们驻留在一个种群中时，没有选择其他的记忆-1策略来入侵此策略。只有合作策略包含在始终稳健的策略子集，意味着无论选择强项还是弱项，都不会选择其他任何记忆-1策略来入侵和替换此策略。零决定策略和良好的合作策略之间的交集是一组宽松的零决定策略。勒索策略是零决定策略和非稳健背叛策略的交集。针锋相对是合作、背叛和零决定策略的交集。<br />
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In 2012, [[William H. Press]] and [[Freeman Dyson]] published a new class of strategies for the stochastic iterated prisoner's dilemma called "zero-determinant" (ZD) strategies.<ref name="Press2012"/> The long term payoffs for encounters between ''X'' and ''Y'' can be expressed as the determinant of a matrix which is a function of the two strategies and the short term payoff vectors: <math>s_x=D(P,Q,S_x)</math> and <math>s_y=D(P,Q,S_y)</math>, which do not involve the stationary vector ''v''. Since the determinant function <math>s_y=D(P,Q,f)</math> is linear in ''f'', it follows that <math>\alpha s_x+\beta s_y+\gamma=D(P,Q,\alpha S_x+\beta S_y+\gamma U)</math> (where ''U''={1,1,1,1}). Any strategies for which <math>D(P,Q,\alpha S_x+\beta S_y+\gamma U)=0</math> is by definition a ZD strategy, and the long term payoffs obey the relation <math>\alpha s_x+\beta s_y+\gamma=0</math>.<br />
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In 2012, William H. Press and Freeman Dyson published a new class of strategies for the stochastic iterated prisoner's dilemma called "zero-determinant" (ZD) strategies. The long term payoffs for encounters between X and Y can be expressed as the determinant of a matrix which is a function of the two strategies and the short term payoff vectors: <math>s_x=D(P,Q,S_x)</math> and <math>s_y=D(P,Q,S_y)</math>, which do not involve the stationary vector v. Since the determinant function <math>s_y=D(P,Q,f)</math> is linear in f, it follows that <math>\alpha s_x+\beta s_y+\gamma=D(P,Q,\alpha S_x+\beta S_y+\gamma U)</math> (where U={1,1,1,1}). Any strategies for which <math>D(P,Q,\alpha S_x+\beta S_y+\gamma U)=0</math> is by definition a ZD strategy, and the long term payoffs obey the relation <math>\alpha s_x+\beta s_y+\gamma=0</math>.<br />
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2012年，威廉·H·普莱斯 William H. Press和弗里曼·戴森 Freeman Dyson针对随机迭代囚徒困境提出了一类新的策略，称为“零决定”策略。<ref name="Press2012"/>''X''和''Y''之间的长期收益可以表示为一个矩阵的决定因素，它是两个策略和短期收益向量的函数: <math>s_x=D(P,Q,S_x)</math>和<math>s_y=D(P,Q,S_y)</math>，不涉及平稳向量''v''。 由于行列式函数<math>s_y=D(P,Q,f)</math>在''f''中是线性的，因此可以推出<math>\alpha s_x+\beta s_y+\gamma=D(P,Q,\alpha S_x+\beta S_y+\gamma U)</math>(其中''U''={1,1,1,1})。任何策略的<math>D(P,Q,\alpha S_x+\beta S_y+\gamma U)=0</math>被定义为零决定策略，长期收益服从关系式<math>\alpha s_x+\beta s_y+\gamma=0</math>。<br />
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Tit-for-tat is a ZD strategy which is "fair" in the sense of not gaining advantage over the other player. However, the ZD space also contains strategies that, in the case of two players, can allow one player to unilaterally set the other player's score or alternatively, force an evolutionary player to achieve a payoff some percentage lower than his own. The extorted player could defect but would thereby hurt himself by getting a lower payoff. Thus, extortion solutions turn the iterated prisoner's dilemma into a sort of [[ultimatum game]]. Specifically, ''X'' is able to choose a strategy for which <math>D(P,Q,\beta S_y+\gamma U)=0</math>, unilaterally setting <math>s_y</math> to a specific value within a particular range of values, independent of ''Y'' 's strategy, offering an opportunity for ''X'' to "extort" player ''Y'' (and vice versa). (It turns out that if ''X'' tries to set <math>s_x</math> to a particular value, the range of possibilities is much smaller, only consisting of complete cooperation or complete defection.<ref name="Press2012"/>)<br />
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Tit-for-tat is a ZD strategy which is "fair" in the sense of not gaining advantage over the other player. However, the ZD space also contains strategies that, in the case of two players, can allow one player to unilaterally set the other player's score or alternatively, force an evolutionary player to achieve a payoff some percentage lower than his own. The extorted player could defect but would thereby hurt himself by getting a lower payoff. Thus, extortion solutions turn the iterated prisoner's dilemma into a sort of ultimatum game. Specifically, X is able to choose a strategy for which <math>D(P,Q,\beta S_y+\gamma U)=0</math>, unilaterally setting <math>s_y</math> to a specific value within a particular range of values, independent of Y 's strategy, offering an opportunity for X to "extort" player Y (and vice versa). (It turns out that if X tries to set <math>s_x</math> to a particular value, the range of possibilities is much smaller, only consisting of complete cooperation or complete defection.)<br />
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针锋相对是一种零决定策略，在不获得超越其他玩家优势的意义下是“公平”的。然而，零决定策略空间还包含这样的策略：在两个玩家的情况下，可以允许一个玩家单方面设置另一个玩家的分数，或者强迫进化的玩家获得比他自己的分数低一些的收益。被勒索的玩家可能会背叛，但会因此获得较低的回报并且受到伤害。因此，勒索的解决方案将迭代囚徒困境转化为一种<font color="#ff8000">最后通牒博弈 ultimatum game </font>。具体来说，''X''能够选择一种策略，对于这种策略，<math>D(P,Q,\beta S_y+\gamma U)=0</math>单方面地将<math>s_y</math>设置为一个特定值范围内的特定值，与''Y''的策略无关，为''X''提供了“勒索”玩家''Y''的机会(反之亦然)。(事实证明，如果''X''试图将<math>s_x</math>设置为一个特定的值，那么可能的范围要小得多，只包括完全合作或完全叛变。<ref name="Press2012"/>)<br />
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An extension of the IPD is an evolutionary stochastic IPD, in which the relative abundance of particular strategies is allowed to change, with more successful strategies relatively increasing. This process may be accomplished by having less successful players imitate the more successful strategies, or by eliminating less successful players from the game, while multiplying the more successful ones. It has been shown that unfair ZD strategies are not [[evolutionarily stable strategy|evolutionarily stable]]. The key intuition is that an evolutionarily stable strategy must not only be able to invade another population (which extortionary ZD strategies can do) but must also perform well against other players of the same type (which extortionary ZD players do poorly, because they reduce each other's surplus).<ref>{{cite journal|last=Adami|first=Christoph|author2=Arend Hintze|title=Evolutionary instability of Zero Determinant strategies demonstrates that winning isn't everything|journal=Nature Communications|volume=4|year=2013|page=3|arxiv=1208.2666|doi=10.1038/ncomms3193|pmid=23903782|pmc=3741637|bibcode=2013NatCo...4.2193A}}</ref><br />
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An extension of the IPD is an evolutionary stochastic IPD, in which the relative abundance of particular strategies is allowed to change, with more successful strategies relatively increasing. This process may be accomplished by having less successful players imitate the more successful strategies, or by eliminating less successful players from the game, while multiplying the more successful ones. It has been shown that unfair ZD strategies are not evolutionarily stable. The key intuition is that an evolutionarily stable strategy must not only be able to invade another population (which extortionary ZD strategies can do) but must also perform well against other players of the same type (which extortionary ZD players do poorly, because they reduce each other's surplus).<br />
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迭代囚徒困境的一个扩展是进化的随机迭代囚徒困境，其中允许特定策略的相对丰度改变，更成功的策略相对增加。这个过程可以通过让不太成功的玩家模仿更成功的策略，或者通过从游戏中淘汰不太成功的玩家，同时让更成功的玩家成倍增加。研究表明，不公平的零决定策略不是进化稳定策略。关键的直觉告诉我们，进化稳定策略不仅要能够入侵另一个群体(这是勒索零决定策略可以做到的) ，而且还要在同类型的其他玩家面前表现良好(勒索零决定策略玩家表现不佳，因为他们减少了彼此的盈余)。<ref>{{cite journal|last=Adami|first=Christoph|author2=Arend Hintze|title=Evolutionary instability of Zero Determinant strategies demonstrates that winning isn't everything|journal=Nature Communications|volume=4|year=2013|page=3|arxiv=1208.2666|doi=10.1038/ncomms3193|pmid=23903782|pmc=3741637|bibcode=2013NatCo...4.2193A}}</ref><br />
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Theory and simulations confirm that beyond a critical population size, ZD extortion loses out in evolutionary competition against more cooperative strategies, and as a result, the average payoff in the population increases when the population is larger. In addition, there are some cases in which extortioners may even catalyze cooperation by helping to break out of a face-off between uniform defectors and [[win–stay, lose–switch]] agents.<ref name=Hilbe2013 /><br />
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Theory and simulations confirm that beyond a critical population size, ZD extortion loses out in evolutionary competition against more cooperative strategies, and as a result, the average payoff in the population increases when the population is larger. In addition, there are some cases in which extortioners may even catalyze cooperation by helping to break out of a face-off between uniform defectors and win–stay, lose–switch agents.<br />
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理论和模拟证实，超过一个临界种群规模，零决定勒索在与更多合作策略的进化竞争中会失败，因此，种群越大，种群的平均收益就越大。此外，在某些情况下，勒索者甚至可能通过帮助打破统一的背叛者与使用“赢-保持-输”策略的转换玩家之间的对峙而促进合作。<ref name=Hilbe2013 /><br />
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While extortionary ZD strategies are not stable in large populations, another ZD class called "generous" strategies ''is'' both stable and robust. In fact, when the population is not too small, these strategies can supplant any other ZD strategy and even perform well against a broad array of generic strategies for iterated prisoner's dilemma, including win–stay, lose–switch. This was proven specifically for the [[Prisoner's dilemma#Special case: Donation game|donation game]] by Alexander Stewart and Joshua Plotkin in 2013.<ref name=Stewart2013>{{cite journal|last=Stewart|first=Alexander J.|author2=Joshua B. Plotkin|title=From extortion to generosity, evolution in the Iterated Prisoner's Dilemma|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|year=2013|doi=10.1073/pnas.1306246110|pmid=24003115|volume=110|issue=38|pages=15348–53|bibcode=2013PNAS..11015348S|pmc=3780848}}</ref> Generous strategies will cooperate with other cooperative players, and in the face of defection, the generous player loses more utility than its rival. Generous strategies are the intersection of ZD strategies and so-called "good" strategies, which were defined by Akin (2013)<ref name=Akin2013>{{cite arxiv|last=Akin|first=Ethan|title=Stable Cooperative Solutions for the Iterated Prisoner's Dilemma|year=2013|page=9|class=math.DS|eprint=1211.0969}} {{bibcode|2012arXiv1211.0969A}}</ref> to be those for which the player responds to past mutual cooperation with future cooperation and splits expected payoffs equally if he receives at least the cooperative expected payoff. Among good strategies, the generous (ZD) subset performs well when the population is not too small. If the population is very small, defection strategies tend to dominate.<ref name=Stewart2013 /><br />
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While extortionary ZD strategies are not stable in large populations, another ZD class called "generous" strategies is both stable and robust. In fact, when the population is not too small, these strategies can supplant any other ZD strategy and even perform well against a broad array of generic strategies for iterated prisoner's dilemma, including win–stay, lose–switch. This was proven specifically for the donation game by Alexander Stewart and Joshua Plotkin in 2013. Generous strategies will cooperate with other cooperative players, and in the face of defection, the generous player loses more utility than its rival. Generous strategies are the intersection of ZD strategies and so-called "good" strategies, which were defined by Akin (2013) to be those for which the player responds to past mutual cooperation with future cooperation and splits expected payoffs equally if he receives at least the cooperative expected payoff. Among good strategies, the generous (ZD) subset performs well when the population is not too small. If the population is very small, defection strategies tend to dominate.<br />
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虽然勒索零决定策略在人口众多的情况下并不稳定，但另一种宽松的零决定策略既稳定又稳健。事实上，当人口不算太少的时候，这些策略可以取代任何其他零决定策略，甚至在一系列针对迭代囚徒困境的广泛通用策略（包括“获胜-保持-输”的转换策略）中表现良好。亚历山大·斯图尔特 Alexander Stewart和约书亚·普洛特金 Joshua Plotkin在2013年的捐赠博弈中证明了这一点。<ref name=Stewart2013>{{cite journal|last=Stewart|first=Alexander J.|author2=Joshua B. Plotkin|title=From extortion to generosity, evolution in the Iterated Prisoner's Dilemma|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|year=2013|doi=10.1073/pnas.1306246110|pmid=24003115|volume=110|issue=38|pages=15348–53|bibcode=2013PNAS..11015348S|pmc=3780848}}</ref>宽松的策略会与其他合作的玩家合作，面对背叛，慷慨的玩家比他的对手失去更多的效用。宽松策略是零决定策略和所谓的“好”策略的交集，阿金(2013) <ref name=Akin2013>{{cite arxiv|last=Akin|first=Ethan|title=Stable Cooperative Solutions for the Iterated Prisoner's Dilemma|year=2013|page=9|class=math.DS|eprint=1211.0969}} {{bibcode|2012arXiv1211.0969A}}</ref> Among good strategies, the generous (ZD) subset performs well when the population is not too small. If the population is very small, defection strategies tend to dominate.将这两种策略定义为玩家对过去的相互合作作出回应，并在至少获得合作预期收益的情况下平均分配预期收益的策略。在好的策略中，当总体不太小时，宽松(零决定)子集表现良好。如果总体很少，背叛策略往往占主导地位。<ref name=Stewart2013 /><br />
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===Continuous iterated prisoner's dilemma===<br />
<font color="#ff8000">连续迭代囚徒困境 Continuous iterated prisoner's dilemma </font> <br />
Most work on the iterated prisoner's dilemma has focused on the discrete case, in which players either cooperate or defect, because this model is relatively simple to analyze. However, some researchers have looked at models of the continuous iterated prisoner's dilemma, in which players are able to make a variable contribution to the other player. Le and Boyd<ref>{{cite journal | last1 = Le | first1 = S. | last2 = Boyd | first2 = R. |name-list-format=vanc| year = 2007 | title = Evolutionary Dynamics of the Continuous Iterated Prisoner's Dilemma | url = | journal = Journal of Theoretical Biology | volume = 245 | issue = 2| pages = 258–67 | doi = 10.1016/j.jtbi.2006.09.016 | pmid = 17125798 }}</ref> found that in such situations, cooperation is much harder to evolve than in the discrete iterated prisoner's dilemma. The basic intuition for this result is straightforward: in a continuous prisoner's dilemma, if a population starts off in a non-cooperative equilibrium, players who are only marginally more cooperative than non-cooperators get little benefit from [[Assortative mating|assorting]] with one another. By contrast, in a discrete prisoner's dilemma, tit for tat cooperators get a big payoff boost from assorting with one another in a non-cooperative equilibrium, relative to non-cooperators. Since nature arguably offers more opportunities for variable cooperation rather than a strict dichotomy of cooperation or defection, the continuous prisoner's dilemma may help explain why real-life examples of tit for tat-like cooperation are extremely rare in nature (ex. Hammerstein<ref>Hammerstein, P. (2003). Why is reciprocity so rare in social animals? A protestant appeal. In: P. Hammerstein, Editor, Genetic and Cultural Evolution of Cooperation, MIT Press. pp. 83–94. </ref>)<br />
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Most work on the iterated prisoner's dilemma has focused on the discrete case, in which players either cooperate or defect, because this model is relatively simple to analyze. However, some researchers have looked at models of the continuous iterated prisoner's dilemma, in which players are able to make a variable contribution to the other player. Le and Boyd found that in such situations, cooperation is much harder to evolve than in the discrete iterated prisoner's dilemma. The basic intuition for this result is straightforward: in a continuous prisoner's dilemma, if a population starts off in a non-cooperative equilibrium, players who are only marginally more cooperative than non-cooperators get little benefit from assorting with one another. By contrast, in a discrete prisoner's dilemma, tit for tat cooperators get a big payoff boost from assorting with one another in a non-cooperative equilibrium, relative to non-cooperators. Since nature arguably offers more opportunities for variable cooperation rather than a strict dichotomy of cooperation or defection, the continuous prisoner's dilemma may help explain why real-life examples of tit for tat-like cooperation are extremely rare in nature (ex. Hammerstein<ref>Hammerstein, P. (2003). Why is reciprocity so rare in social animals? A protestant appeal. In: P. Hammerstein, Editor, Genetic and Cultural Evolution of Cooperation, MIT Press. pp. 83–94. </ref>)<br />
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关于迭代囚徒困境的研究大多集中在离散情况下，在这种情况下，参与者要么合作，要么背叛，因为这个模型分析起来比较简单。然而，一些研究人员已经研究了连续迭代囚徒困境模型，在这个模型中，玩家能够对另一个玩家做出可变的贡献。乐 Le和博伊德 Boyd<ref>{{cite journal | last1 = Le | first1 = S. | last2 = Boyd | first2 = R. |name-list-format=vanc| year = 2007 | title = Evolutionary Dynamics of the Continuous Iterated Prisoner's Dilemma | url = | journal = Journal of Theoretical Biology | volume = 245 | issue = 2| pages = 258–67 | doi = 10.1016/j.jtbi.2006.09.016 | pmid = 17125798 }}</ref>发现，在这种情况下，合作比离散迭代的囚徒困境更难发展。这个结果的基本直觉很简单: 在一个持续的囚徒困境中，如果一个人群开始处于非合作均衡状态，那么与非合作者相比，合作程度稍高的玩家不会从相互配合中获益。相比之下，在离散的囚徒困境中，相对于非合作者，针锋相对的合作者在非合作均衡中相互配合会获得巨大的回报。由于自然界可以提供更多的机会来进行各种各样的合作，而不是严格地将合作或背叛分为两类，因此连续的囚徒困境可以帮助解释为什么现实生活中针锋相对的合作的例子在自然界中极其罕见。（例如，哈默斯坦 Hammerstein <ref>Hammerstein, P. (2003). Why is reciprocity so rare in social animals? A protestant appeal. In: P. Hammerstein, Editor, Genetic and Cultural Evolution of Cooperation, MIT Press. pp. 83–94. </ref>）。<br />
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even though tit for tat seems robust in theoretical models.<br />
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even though tit for tat seems robust in theoretical models.<br />
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尽管在理论模型中，针锋相对策略似乎是稳健的。<br />
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===Emergence of stable strategies===<br />
<font color="#ff8000">稳定策略的出现 Emergence of stable strategies </font> <br />
Players cannot seem to coordinate mutual cooperation, thus often get locked into the inferior yet stable strategy of defection. In this way, iterated rounds facilitate the evolution of stable strategies.<ref>{{cite book|last=Spaniel|first=William|title=Game Theory 101: The Complete Textbook|year=2011}}</ref> Iterated rounds often produce novel strategies, which have implications to complex social interaction. One such strategy is win-stay lose-shift. This strategy outperforms a simple Tit-For-Tat strategy&nbsp;– that is, if you can get away with cheating, repeat that behavior, however if you get caught, switch.<ref>{{cite journal|last=Nowak|first=Martin|author2=Karl Sigmund|title=A strategy of win-stay, lose-shift that outperforms tit-for-tat in the Prisoner's Dilemma game|journal=Nature|year=1993|volume=364|issue=6432|doi=10.1038/364056a0|pages=56–58|pmid=8316296|bibcode=1993Natur.364...56N}}</ref><br />
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Players cannot seem to coordinate mutual cooperation, thus often get locked into the inferior yet stable strategy of defection. In this way, iterated rounds facilitate the evolution of stable strategies. Iterated rounds often produce novel strategies, which have implications to complex social interaction. One such strategy is win-stay lose-shift. This strategy outperforms a simple Tit-For-Tat strategy&nbsp;– that is, if you can get away with cheating, repeat that behavior, however if you get caught, switch.<br />
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玩家似乎不能协调相互合作，因此常常陷入劣等而稳定的背叛策略。这样，迭代回合可以促进稳定策略的发展。<ref>{{cite book|last=Spaniel|first=William|title=Game Theory 101: The Complete Textbook|year=2011}}</ref>迭代回合往往产生新颖的策略，这对复杂的社会互动有影响。其中一个策略就是“赢-保持-输”的转变。这个策略比一个简单的针锋相对策略要好&nbsp;–也就是说，如果你能逃脱作弊的惩罚，就重复这个行为，如果你被抓住了，就改变策略。<ref>{{cite journal|last=Nowak|first=Martin|author2=Karl Sigmund|title=A strategy of win-stay, lose-shift that outperforms tit-for-tat in the Prisoner's Dilemma game|journal=Nature|year=1993|volume=364|issue=6432|doi=10.1038/364056a0|pages=56–58|pmid=8316296|bibcode=1993Natur.364...56N}}</ref><br />
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The only problem of this tit-for-tat strategy is that they are vulnerable to signal error. The problem arises when one individual cheats in retaliation but the other interprets it as cheating. As a result of this, the second individual now cheats and then it starts a see-saw pattern of cheating in a chain reaction.<br />
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The only problem of this tit-for-tat strategy is that they are vulnerable to signal error. The problem arises when one individual cheats in retaliation but the other interprets it as cheating. As a result of this, the second individual now cheats and then it starts a see-saw pattern of cheating in a chain reaction.<br />
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这种针锋相对策略的唯一问题是它们很容易出现信号错误。当一个人因报复而作弊，而另一个人将其单纯解释为欺骗时，就会出现问题。结果，第二个人现在作弊，然后在接下来的连锁反应中开始了反复交替的作弊模式。<br />
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==Real-life examples==<br />
现实生活的例子<br />
The prisoner setting may seem contrived, but there are in fact many examples in human interaction as well as interactions in nature that have the same payoff matrix. The prisoner's dilemma is therefore of interest to the [[social science]]s such as [[economics]], [[politics]], and [[sociology]], as well as to the biological sciences such as [[ethology]] and [[evolutionary biology]]. Many natural processes have been abstracted into models in which living beings are engaged in endless games of prisoner's dilemma. This wide applicability of the PD gives the game its substantial importance.<br />
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The prisoner setting may seem contrived, but there are in fact many examples in human interaction as well as interactions in nature that have the same payoff matrix. The prisoner's dilemma is therefore of interest to the social sciences such as economics, politics, and sociology, as well as to the biological sciences such as ethology and evolutionary biology. Many natural processes have been abstracted into models in which living beings are engaged in endless games of prisoner's dilemma. This wide applicability of the PD gives the game its substantial importance.<br />
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囚犯的环境似乎是人为的，但实际上，在人类交往以及自然界的交互中有许多具有相同收益矩阵的例子。因此，囚徒困境是经济学、政治学、社会学等社会科学以及动物行为学、进化生物学等生物学研究的热点问题。许多自然过程都被抽象为生物进行无休止的囚徒困境博弈的模型。囚徒困境这种广泛的适用性让博弈变得非常重要。<br />
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===Environmental studies===<br />
环境研究<br />
In [[environmental studies]], the PD is evident in crises such as global [[climate change|climate-change]]. It is argued all countries will benefit from a stable climate, but any single country is often hesitant to curb [[Carbon dioxide|{{Co2}}]] emissions. The immediate benefit to any one country from maintaining current behavior is wrongly perceived to be greater than the purported eventual benefit to that country if all countries' behavior was changed, therefore explaining the impasse concerning climate-change in 2007.<ref>{{cite news|newspaper=[[The Economist]]|url=http://www.economist.com/finance/displaystory.cfm?story_id=9867020|title=Markets & Data|date=2007-09-27}}</ref><br />
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In environmental studies, the PD is evident in crises such as global climate-change. It is argued all countries will benefit from a stable climate, but any single country is often hesitant to curb Carbon dioxide| emissions. The immediate benefit to any one country from maintaining current behavior is wrongly perceived to be greater than the purported eventual benefit to that country if all countries' behavior was changed, therefore explaining the impasse concerning climate-change in 2007.<br />
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在环境研究中，囚徒困境在诸如全球气候变化等危机中显而易见。有人认为，所有国家都将从稳定的气候中受益，但是每一个国家通常都在限制二氧化碳排放方面犹豫不决。人们错误地认为，如果所有国家的行为都改变，任何一个国家保持目前的行为所带来的直接好处都会大于所谓的最终好处，这就解释了2007年气候变化方面的僵局。<ref>{{cite news|newspaper=[[The Economist]]|url=http://www.economist.com/finance/displaystory.cfm?story_id=9867020|title=Markets & Data|date=2007-09-27}}</ref><br />
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An important difference between climate-change politics and the prisoner's dilemma is uncertainty; the extent and pace at which pollution can change climate is not known. The dilemma faced by government is therefore different from the prisoner's dilemma in that the payoffs of cooperation are unknown. This difference suggests that states will cooperate much less than in a real iterated prisoner's dilemma, so that the probability of avoiding a possible climate catastrophe is much smaller than that suggested by a game-theoretical analysis of the situation using a real iterated prisoner's dilemma.<ref>{{cite web|last=Rehmeyer|first=Julie|title=Game theory suggests current climate negotiations won't avert catastrophe|url=https://www.sciencenews.org/article/game-theory-suggests-current-climate-negotiations-won%E2%80%99t-avert-catastrophe|work=Science News|publisher=Society for Science & the Public|date=2012-10-29}}</ref><br />
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An important difference between climate-change politics and the prisoner's dilemma is uncertainty; the extent and pace at which pollution can change climate is not known. The dilemma faced by government is therefore different from the prisoner's dilemma in that the payoffs of cooperation are unknown. This difference suggests that states will cooperate much less than in a real iterated prisoner's dilemma, so that the probability of avoiding a possible climate catastrophe is much smaller than that suggested by a game-theoretical analysis of the situation using a real iterated prisoner's dilemma.<br />
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气候变化政治与囚徒困境之间的一个重要区别是不确定性; 污染对气候变化的影响程度和速度尚不清楚。因此，政府面临的困境不同于囚徒困境，因为合作的回报是未知的。这种差异表明，各国之间的合作远远少于真正的迭代囚徒困境中的合作，因此避免可能发生的气候灾难的可能性远远小于使用真正的迭代囚徒困境博弈论情景分析<ref>{{cite web|last=Rehmeyer|first=Julie|title=Game theory suggests current climate negotiations won't avert catastrophe|url=https://www.sciencenews.org/article/game-theory-suggests-current-climate-negotiations-won%E2%80%99t-avert-catastrophe|work=Science News|publisher=Society for Science & the Public|date=2012-10-29}}</ref><br />
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Osang and Nandy (2003) provide a theoretical explanation with proofs for a regulation-driven win-win situation along the lines of [[Michael Porter]]'s hypothesis, in which government regulation of competing firms is substantial.<ref>{{cite thesis|type=paper|url= http://faculty.smu.edu/tosang/pdf/regln0803.pdf|first=Thomas|last=Osang|first2=Arundhati|last2=Nandyyz|date=August 2003|title=Environmental Regulation of Polluting Firms: Porter's Hypothesis Revisited}}</ref><br />
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Osang and Nandy (2003) provide a theoretical explanation with proofs for a regulation-driven win-win situation along the lines of Michael Porter's hypothesis, in which government regulation of competing firms is substantial.<br />
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欧桑 Osang和南迪 Nandy (2003)提供了一个理论解释，并根据迈克尔·波特 Michael Porter的假设，即政府对竞争企业的监管是实质性的，证明了监管驱动的双赢局面。<ref>{{cite thesis|type=paper|url= http://faculty.smu.edu/tosang/pdf/regln0803.pdf|first=Thomas|last=Osang|first2=Arundhati|last2=Nandyyz|date=August 2003|title=Environmental Regulation of Polluting Firms: Porter's Hypothesis Revisited}}</ref><br />
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===Animals===<br />
动物<br />
Cooperative behavior of many animals can be understood as an example of the prisoner's dilemma. Often animals engage in long term partnerships, which can be more specifically modeled as iterated prisoner's dilemma. For example, [[guppy|guppies]] inspect predators cooperatively in groups, and they are thought to punish non-cooperative inspectors.<br />
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Cooperative behavior of many animals can be understood as an example of the prisoner's dilemma. Often animals engage in long term partnerships, which can be more specifically modeled as iterated prisoner's dilemma. For example, guppies inspect predators cooperatively in groups, and they are thought to punish non-cooperative inspectors.<br />
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许多动物的合作行为可以理解为囚徒困境的一个例子。通常动物会建立长期的伙伴关系，这种关系可以更具体地模拟为迭代囚徒困境。例如，孔雀鱼成群结队地合作监察捕食者，它们被认为是在惩罚不合作的监察者。<br />
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[[Vampire bats]] are social animals that engage in reciprocal food exchange. Applying the payoffs from the prisoner's dilemma can help explain this behavior:<ref>{{cite book|last=Dawkins|first=Richard|title=The Selfish Gene|year=1976|publisher=Oxford University Press}}</ref><br />
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Vampire bats are social animals that engage in reciprocal food exchange. Applying the payoffs from the prisoner's dilemma can help explain this behavior:<br />
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吸血蝙蝠是从事相互的食物交换的群居动物。应用囚徒困境收益可以帮助解释这种行为: <ref>{{cite book|last=Dawkins|first=Richard|title=The Selfish Gene|year=1976|publisher=Oxford University Press}}</ref><br />
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* C/C: "Reward: I get blood on my unlucky nights, which saves me from starving. I have to give blood on my lucky nights, which doesn't cost me too much."<br />
* 合作/合作："回报：我在不幸运的晚上得到了能让我果腹的血，那在幸运的晚上我也应该分出点血，那不会花费多少。"<br />
* D/C: "Temptation: You save my life on my poor night. But then I get the added benefit of not having to pay the slight cost of feeding you on my good night."<br />
* 背叛/合作："诱惑：你在我的不幸的夜里救了我，但在我的幸运夜我不会给你血，那样我会活的更好。"<br />
* C/D: "Sucker's Payoff: I pay the cost of saving your life on my good night. But on my bad night you don't feed me and I run a real risk of starving to death."<br />
* 合作/叛变："可怜者的回报：在我的幸运夜我救了你的命，但在我的不幸夜里你没有救我，我有饿死的风险。"<br />
* D/D: "Punishment: I don't have to pay the slight costs of feeding you on my good nights. But I run a real risk of starving on my poor nights."<br />
* 叛变/叛变："惩罚：我在我的幸运夜里不必付出代价来救你，但我在我的不幸夜里有挨饿的风险。"<br />
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===Psychology===<br />
心理学<br />
In [[addiction]] research / [[behavioral economics]], [[George Ainslie (psychologist)|George Ainslie]] points out<ref>{{cite book |first=George|last=Ainslie |title=Breakdown of Will |year=2001 |isbn=978-0-521-59694-7}}</ref> that addiction can be cast as an intertemporal PD problem between the present and future selves of the addict. In this case, ''defecting'' means ''relapsing'', and it is easy to see that not defecting both today and in the future is by far the best outcome. The case where one abstains today but relapses in the future is the worst outcome&nbsp;– in some sense the discipline and self-sacrifice involved in abstaining today have been "wasted" because the future relapse means that the addict is right back where he started and will have to start over (which is quite demoralizing, and makes starting over more difficult). Relapsing today and tomorrow is a slightly "better" outcome, because while the addict is still addicted, they haven't put the effort in to trying to stop. The final case, where one engages in the addictive behavior today while abstaining "tomorrow" will be familiar to anyone who has struggled with an addiction. The problem here is that (as in other PDs) there is an obvious benefit to defecting "today", but tomorrow one will face the same PD, and the same obvious benefit will be present then, ultimately leading to an endless string of defections.<br />
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In addiction research / behavioral economics, George Ainslie points out that addiction can be cast as an intertemporal PD problem between the present and future selves of the addict. In this case, defecting means relapsing, and it is easy to see that not defecting both today and in the future is by far the best outcome. The case where one abstains today but relapses in the future is the worst outcome&nbsp;– in some sense the discipline and self-sacrifice involved in abstaining today have been "wasted" because the future relapse means that the addict is right back where he started and will have to start over (which is quite demoralizing, and makes starting over more difficult). Relapsing today and tomorrow is a slightly "better" outcome, because while the addict is still addicted, they haven't put the effort in to trying to stop. The final case, where one engages in the addictive behavior today while abstaining "tomorrow" will be familiar to anyone who has struggled with an addiction. The problem here is that (as in other PDs) there is an obvious benefit to defecting "today", but tomorrow one will face the same PD, and the same obvious benefit will be present then, ultimately leading to an endless string of defections.<br />
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在成瘾研究/行为经济学中，乔治·安斯利 George Ainslie指出<ref>{{cite book |first=George|last=Ainslie |title=Breakdown of Will |year=2001 |isbn=978-0-521-59694-7}}</ref>，可以将成瘾视为成瘾者现在和未来自我之间的跨期囚徒困境问题。在这种情况下，背叛意味着复发，很容易看出，目前和未来都没有背叛是迄今为止最好的结果。如果一个人今天戒了，但在将来又复吸，这是最糟糕的结果&nbsp;–从某种意义上来说，今天戒瘾所包含的纪律和自我牺牲已经被“浪费”了，因为未来的复吸意味着瘾君子又回到了他开始的地方，他将被迫重新开始(这相当令人沮丧，也使得重新开始更加困难)。今天和明天复发是一个稍微“好一点”的结果，因为尽管瘾君子仍然上瘾，但他们没有努力去尝试停止。最后一种情况是，现在与成瘾斗争的任何人都会熟悉现在的成瘾行为，而在明天放弃。这里的问题是(和其他囚徒困境问题一样)，背叛“今天”有一个明显的好处，但明天这个人将面临同样的囚徒困境问题，同样明显的好处是背叛，最终导致一连串无休止的背叛。<br />
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[[John Gottman]] in his research described in "the science of trust" defines good relationships as those where partners know not to enter the (D,D) cell or at least not to get dynamically stuck there in a loop.<br />
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John Gottman in his research described in "the science of trust" defines good relationships as those where partners know not to enter the (D,D) cell or at least not to get dynamically stuck there in a loop.<br />
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约翰·高特曼 John Gottman在他的研究《信任的科学》中将良好的关系定义为伙伴知道不进入(背叛，背叛)牢房中或者至少不要陷入这样的动态循环关系中。<br />
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===Economics===<br />
经济学<br />
The prisoner's dilemma has been called the ''[[Escherichia coli|E. coli]]'' of social psychology, and it has been used widely to research various topics such as [[Oligopoly|oligopolistic]] competition and collective action to produce a collective good.<ref>{{Cite journal|last=Axelrod|first=Robert|date=1980|title=Effective Choice in the Prisoner's Dilemma|journal=The Journal of Conflict Resolution|volume=24|issue=1|pages=3–25|issn=0022-0027|jstor=173932|doi=10.1177/002200278002400101|url=https://semanticscholar.org/paper/fd1ab82470446bfb12c39f0c577644291027cf76}}</ref> <br />
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The prisoner's dilemma has been called the E. coli of social psychology, and it has been used widely to research various topics such as oligopolistic competition and collective action to produce a collective good. <br />
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囚徒困境被称为社会心理学中的“大肠杆菌”，它被广泛用于研究寡头垄断竞争和集体行动来产生集体利益等问题。<ref>{{Cite journal|last=Axelrod|first=Robert|date=1980|title=Effective Choice in the Prisoner's Dilemma|journal=The Journal of Conflict Resolution|volume=24|issue=1|pages=3–25|issn=0022-0027|jstor=173932|doi=10.1177/002200278002400101|url=https://semanticscholar.org/paper/fd1ab82470446bfb12c39f0c577644291027cf76}}</ref><br />
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Advertising is sometimes cited as a real-example of the prisoner's dilemma. When [[cigarette advertising]] was legal in the United States, competing cigarette manufacturers had to decide how much money to spend on advertising. The effectiveness of Firm A's advertising was partially determined by the advertising conducted by Firm B. Likewise, the profit derived from advertising for Firm B is affected by the advertising conducted by Firm A. If both Firm A and Firm B chose to advertise during a given period, then the advertisement from each firm negates the other's, receipts remain constant, and expenses increase due to the cost of advertising. Both firms would benefit from a reduction in advertising. However, should Firm B choose not to advertise, Firm A could benefit greatly by advertising. Nevertheless, the optimal amount of advertising by one firm depends on how much advertising the other undertakes. As the best strategy is dependent on what the other firm chooses there is no dominant strategy, which makes it slightly different from a prisoner's dilemma. The outcome is similar, though, in that both firms would be better off were they to advertise less than in the equilibrium. Sometimes cooperative behaviors do emerge in business situations. For instance, cigarette manufacturers endorsed the making of laws banning cigarette advertising, understanding that this would reduce costs and increase profits across the industry. <ref>{{Citation needed|reason=This reference doesn't mention or support the claimed historical account.|date=December 2012}}</ref><ref>{{efn|1=This argument for the development of cooperation through trust is given in ''[[The Wisdom of Crowds]]'', where it is argued that long-distance [[capitalism]] was able to form around a nucleus of [[Religious Society of Friends|Quakers]], who always dealt honourably with their business partners. (Rather than defecting and reneging on promises&nbsp;– a phenomenon that had discouraged earlier long-term unenforceable overseas contracts). It is argued that dealings with reliable merchants allowed the [[meme]] for cooperation to spread to other traders, who spread it further until a high degree of cooperation became a profitable strategy in general [[commerce]]}} </ref> This analysis is likely to be pertinent in many other business situations involving advertising. <ref>{{Citation needed|reason=This doesn't sound like cooperation|date=November 2012}}</ref><br />
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Advertising is sometimes cited as a real-example of the prisoner's dilemma. When cigarette advertising was legal in the United States, competing cigarette manufacturers had to decide how much money to spend on advertising. The effectiveness of Firm A's advertising was partially determined by the advertising conducted by Firm B. Likewise, the profit derived from advertising for Firm B is affected by the advertising conducted by Firm A. If both Firm A and Firm B chose to advertise during a given period, then the advertisement from each firm negates the other's, receipts remain constant, and expenses increase due to the cost of advertising. Both firms would benefit from a reduction in advertising. However, should Firm B choose not to advertise, Firm A could benefit greatly by advertising. Nevertheless, the optimal amount of advertising by one firm depends on how much advertising the other undertakes. As the best strategy is dependent on what the other firm chooses there is no dominant strategy, which makes it slightly different from a prisoner's dilemma. The outcome is similar, though, in that both firms would be better off were they to advertise less than in the equilibrium. Sometimes cooperative behaviors do emerge in business situations. For instance, cigarette manufacturers endorsed the making of laws banning cigarette advertising, understanding that this would reduce costs and increase profits across the industry. This analysis is likely to be pertinent in many other business situations involving advertising.<br />
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广告有时被认为是囚徒困境的一个真实例子。当香烟广告在美国是合法的时候，相互竞争的香烟制造商必须决定在广告上花多少钱。公司A的广告效果部分取决于公司B的广告效果。同样，公司B的广告带来的利润也受到公司A的广告影响。如果公司A和公司B都选择在给定的时间段内做广告，那么一家公司的广告就会抵消另一方的广告，倘若收入保持不变，费用就会因广告成本而增加。两家公司都将从广告减少中获益。然而，如果B公司选择不做广告，A公司就可以通过广告获得巨大的利益。尽管如此，一家公司的最佳广告数量仍取决于另一家公司的广告投放量。由于最佳策略取决于其他公司的选择，因此这里没有占主导地位的策略，这使得它与囚徒困境略有不同。但结果是相似的，如果两家公司的广告都少于均衡状态，他们的处境会更好。有时合作行为确实会在商业环境中出现。例如，香烟制造商支持立法禁止香烟广告，因为这将降低成本并增加整个行业的利润。<ref>{{Citation needed|reason=This reference doesn't mention or support the claimed historical account.|date=December 2012}}</ref><ref>{{efn|1=This argument for the development of cooperation through trust is given in ''[[The Wisdom of Crowds]]'', where it is argued that long-distance [[capitalism]] was able to form around a nucleus of [[Religious Society of Friends|Quakers]], who always dealt honourably with their business partners. (Rather than defecting and reneging on promises&nbsp;– a phenomenon that had discouraged earlier long-term unenforceable overseas contracts). It is argued that dealings with reliable merchants allowed the [[meme]] for cooperation to spread to other traders, who spread it further until a high degree of cooperation became a profitable strategy in general [[commerce]]}} </ref>这种分析可能适用于许多其他涉及广告的商业情况。<ref>{{Citation needed|reason=This doesn't sound like cooperation|date=November 2012}}</ref> <br />
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Without enforceable agreements, members of a [[cartel]] are also involved in a (multi-player) prisoner's dilemma.<ref>{{Cite book|last1=Nicholson|first=Walter|year=2000|title=Intermediate microeconomics and its application|edition=8th|location=Fort Worth, TX|publisher=Dryden Press : Harcourt College Publishers|isbn=978-0-030-25916-6}}</ref> 'Cooperating' typically means keeping prices at a pre-agreed minimum level. 'Defecting' means selling under this minimum level, instantly taking business (and profits) from other cartel members. [[Anti-trust]] authorities want potential cartel members to mutually defect, ensuring the lowest possible prices for [[consumer]]s.<br />
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Without enforceable agreements, members of a cartel are also involved in a (multi-player) prisoner's dilemma. 'Cooperating' typically means keeping prices at a pre-agreed minimum level. 'Defecting' means selling under this minimum level, instantly taking business (and profits) from other cartel members. Anti-trust authorities want potential cartel members to mutually defect, ensuring the lowest possible prices for consumers.<br />
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没有可强制执行的协议，卡特尔 cartel的成员国也会陷入(多玩家)囚徒困境。<ref>{{Cite book|last1=Nicholson|first=Walter|year=2000|title=Intermediate microeconomics and its application|edition=8th|location=Fort Worth, TX|publisher=Dryden Press : Harcourt College Publishers|isbn=978-0-030-25916-6}}</ref> “合作”通常意味着将价格保持在预先商定的最低水平。“背叛”意味着低于最低价格水平销售，并立即从其他卡特尔成员那里获得业务(和利润)。反垄断机构希望潜在的卡特尔成员相互背叛，确保消费者获得尽可能低的价格。<br />
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===Sport===<br />
运动<br />
[[Doping in sport]] has been cited as an example of a prisoner's dilemma.<ref name="wired">{{cite journal|last=Schneier |first=Bruce |url=https://www.wired.com/opinion/2012/10/lance-armstrong-and-the-prisoners-dilemma-of-doping-in-professional-sports/ |title=Lance Armstrong and the Prisoners' Dilemma of Doping in Professional Sports &#124; Wired Opinion |journal=Wired |publisher=Wired.com |date=2012-10-26 |accessdate=2012-10-29}}</ref><br />
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Doping in sport has been cited as an example of a prisoner's dilemma.<br />
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体育运动中的兴奋剂被认为是囚徒困境的一个例子。<ref name="wired">{{cite journal|last=Schneier |first=Bruce |url=https://www.wired.com/opinion/2012/10/lance-armstrong-and-the-prisoners-dilemma-of-doping-in-professional-sports/ |title=Lance Armstrong and the Prisoners' Dilemma of Doping in Professional Sports &#124; Wired Opinion |journal=Wired |publisher=Wired.com |date=2012-10-26 |accessdate=2012-10-29}}</ref><br />
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Two competing athletes have the option to use an illegal and/or dangerous drug to boost their performance. If neither athlete takes the drug, then neither gains an advantage. If only one does, then that athlete gains a significant advantage over their competitor, reduced by the legal and/or medical dangers of having taken the drug. If both athletes take the drug, however, the benefits cancel out and only the dangers remain, putting them both in a worse position than if neither had used doping.<ref name="wired" /><br />
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Two competing athletes have the option to use an illegal and/or dangerous drug to boost their performance. If neither athlete takes the drug, then neither gains an advantage. If only one does, then that athlete gains a significant advantage over their competitor, reduced by the legal and/or medical dangers of having taken the drug. If both athletes take the drug, however, the benefits cancel out and only the dangers remain, putting them both in a worse position than if neither had used doping.<br />
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两名参赛运动员可以选择使用非法或危险药物来提高成绩。如果两个运动员都没有服用这种药物，那么他们都不会获得优势。如果只有一个人这样做，那么这个运动员就比他们的竞争对手获得了明显的优势，但由于法律或服用药物的医疗风险，这种优势会减少。然而，如果两名运动员都服用了这种药物，那么好处就被抵消了，只剩下风险，这使得他们的处境比没有服用兴奋剂的情况更加糟糕。<ref name="wired" /><br />
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===International politics===<br />
国际政治<br />
In [[international politics|international political theory]], the Prisoner's Dilemma is often used to demonstrate the coherence of [[strategic realism]], which holds that in international relations, all states (regardless of their internal policies or professed ideology), will act in their rational self-interest given [[anarchy (international relations)|international anarchy]]. A classic example is an arms race like the [[Cold War]] and similar conflicts.<ref>{{cite journal| title = Arms races as iterated prisoner's dilemma games | author = Stephen J. Majeski | journal = Mathematical and Social Sciences | volume = 7 | issue = 3 | pages = 253–66 | year = 1984 | doi=10.1016/0165-4896(84)90022-2}}</ref> During the Cold War the opposing alliances of [[NATO]] and the [[Warsaw Pact]] both had the choice to arm or disarm. From each side's point of view, disarming whilst their opponent continued to arm would have led to military inferiority and possible annihilation. Conversely, arming whilst their opponent disarmed would have led to superiority. If both sides chose to arm, neither could afford to attack the other, but both incurred the high cost of developing and maintaining a nuclear arsenal. If both sides chose to disarm, war would be avoided and there would be no costs.<br />
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In international political theory, the Prisoner's Dilemma is often used to demonstrate the coherence of strategic realism, which holds that in international relations, all states (regardless of their internal policies or professed ideology), will act in their rational self-interest given international anarchy. A classic example is an arms race like the Cold War and similar conflicts. During the Cold War the opposing alliances of NATO and the Warsaw Pact both had the choice to arm or disarm. From each side's point of view, disarming whilst their opponent continued to arm would have led to military inferiority and possible annihilation. Conversely, arming whilst their opponent disarmed would have led to superiority. If both sides chose to arm, neither could afford to attack the other, but both incurred the high cost of developing and maintaining a nuclear arsenal. If both sides chose to disarm, war would be avoided and there would be no costs.<br />
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在国际政治理论中，囚徒困境经常被用来证明战略现实主义的一致性，这种战略现实主义认为，在国际关系中，由于国际无政府状态，所有国家(无论其国内政策或公开宣称的意识形态如何)都会为了自身的理性利益来行动。一个典型的例子是类似冷战和类似冲突的军备竞赛。<ref>{{cite journal| title = Arms races as iterated prisoner's dilemma games | author = Stephen J. Majeski | journal = Mathematical and Social Sciences | volume = 7 | issue = 3 | pages = 253–66 | year = 1984 | doi=10.1016/0165-4896(84)90022-2}}</ref>在冷战期间，北约和华约组织的对立联盟都可以选择武装或解除武装。从双方的观点来看，解除武装而对手继续武装可能会导致军事劣势和被歼灭。相反，如果选择武装而对手已经解除了武装，那么就会获得优势。如果双方都选择武装自己，那么任何一方都承担不起攻击对方的代价，但是双方都为发展和维持核武库付出了高昂的代价。如果双方都选择裁军，战争就可以避免，也不会有任何代价。<br />
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Although the 'best' overall outcome is for both sides to disarm, the rational course for both sides is to arm, and this is indeed what happened. Both sides poured enormous resources into military research and armament in a [[War of attrition (game)|war of attrition]] for the next thirty years until the Soviet Union could not withstand the economic cost.<ref>{{Citation|last=Kuhn|first=Steven|title=Prisoner's Dilemma|date=2019|url=https://plato.stanford.edu/archives/win2019/entries/prisoner-dilemma/|encyclopedia=The Stanford Encyclopedia of Philosophy|editor-last=Zalta|editor-first=Edward N.|edition=Winter 2019|publisher=Metaphysics Research Lab, Stanford University|access-date=2020-04-12}}</ref> The same logic could be applied in any similar scenario, be it economic or technological competition between sovereign states.<br />
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Although the 'best' overall outcome is for both sides to disarm, the rational course for both sides is to arm, and this is indeed what happened. Both sides poured enormous resources into military research and armament in a war of attrition for the next thirty years until the Soviet Union could not withstand the economic cost. The same logic could be applied in any similar scenario, be it economic or technological competition between sovereign states.<br />
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虽然最好的结果是双方解除武装，但是双方的理性选择是武装起来，事实也的确如此。在接下来的三十年里，双方都在军事研究和武器装备的消耗战上投入了大量的资源，直到苏联无法承受经济损失。<ref>{{Citation|last=Kuhn|first=Steven|title=Prisoner's Dilemma|date=2019|url=https://plato.stanford.edu/archives/win2019/entries/prisoner-dilemma/|encyclopedia=The Stanford Encyclopedia of Philosophy|editor-last=Zalta|editor-first=Edward N.|edition=Winter 2019|publisher=Metaphysics Research Lab, Stanford University|access-date=2020-04-12}}</ref>同样的逻辑也适用于任何类似的情况，无论是主权国家之间的经济竞争还是技术竞争。<br />
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===Multiplayer dilemmas===<br />
<font color="#ff8000">多玩家困境 Multiplayer dilemmas</font><br />
Many real-life dilemmas involve multiple players.<ref>Gokhale CS, Traulsen A. Evolutionary games in the multiverse. Proceedings of the National Academy of Sciences. 2010 Mar 23. 107(12):5500–04.</ref> Although metaphorical, [[Garrett Hardin|Hardin's]] [[tragedy of the commons]] may be viewed as an example of a multi-player generalization of the PD: Each villager makes a choice for personal gain or restraint. The collective reward for unanimous (or even frequent) defection is very low payoffs (representing the destruction of the "commons"). A commons dilemma most people can relate to is washing the dishes in a shared house. By not washing dishes an individual can gain by saving his time, but if that behavior is adopted by every resident the collective cost is no clean plates for anyone.<br />
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Many real-life dilemmas involve multiple players. Although metaphorical, Hardin's tragedy of the commons may be viewed as an example of a multi-player generalization of the PD: Each villager makes a choice for personal gain or restraint. The collective reward for unanimous (or even frequent) defection is very low payoffs (representing the destruction of the "commons"). A commons dilemma most people can relate to is washing the dishes in a shared house. By not washing dishes an individual can gain by saving his time, but if that behavior is adopted by every resident the collective cost is no clean plates for anyone.<br />
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许多现实生活中的困境牵涉到多个参与者。<ref>Gokhale CS, Traulsen A. Evolutionary games in the multiverse. Proceedings of the National Academy of Sciences. 2010 Mar 23. 107(12):5500–04.</ref>尽管具有隐喻性，但哈丁的<font color="#ff8000">公地悲剧 tragedy of the commons</font>可以看作是囚徒困境多个参与者的一个例子: 每个村民做出选择是为了个人利益还是克制。对于一致(甚至频繁)叛变的集体回报是非常低的(代表了对“公共资源”的破坏)。大多数人可能会遇到的公地困境是在一个共用的房子里洗碗。通过不洗碗，个人可以节省时间，但如果每个居民都选择这种行为，那么集体的代价是任何人都没有干净的盘子。<br />
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The commons are not always exploited: [[William Poundstone]], in a book about the prisoner's dilemma (see References below), describes a situation in New Zealand where newspaper boxes are left unlocked. It is possible for people to [[Excludability|take a paper without paying]] (''defecting'') but very few do, feeling that if they do not pay then neither will others, destroying the system. Subsequent research by [[Elinor Ostrom]], winner of the 2009 [[Nobel Memorial Prize in Economic Sciences]], hypothesized that the tragedy of the commons is oversimplified, with the negative outcome influenced by outside influences. Without complicating pressures, groups communicate and manage the commons among themselves for their mutual benefit, enforcing social norms to preserve the resource and achieve the maximum good for the group, an example of effecting the best case outcome for PD.<ref>{{cite web|url=http://volokh.com/2009/10/12/elinor-ostrom-and-the-tragedy-of-the-commons/ |title=The Volokh Conspiracy " Elinor Ostrom and the Tragedy of the Commons |publisher=Volokh.com |date=2009-10-12 |accessdate=2011-12-17}}</ref><br />
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The commons are not always exploited: William Poundstone, in a book about the prisoner's dilemma (see References below), describes a situation in New Zealand where newspaper boxes are left unlocked. It is possible for people to take a paper without paying (defecting) but very few do, feeling that if they do not pay then neither will others, destroying the system. Subsequent research by Elinor Ostrom, winner of the 2009 Nobel Memorial Prize in Economic Sciences, hypothesized that the tragedy of the commons is oversimplified, with the negative outcome influenced by outside influences. Without complicating pressures, groups communicate and manage the commons among themselves for their mutual benefit, enforcing social norms to preserve the resource and achieve the maximum good for the group, an example of effecting the best case outcome for PD.<br />
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公共资源并不总是被利用: 威廉·庞德斯通 William Poundstone在一本关于囚徒困境的书(见下文参考文献)中描述了新西兰的一种情况，信箱没有上锁。人们可以不付钱就拿报纸(背叛) ，但很少有人这样做，他们觉得如果他们不付钱，那么其他人也不会付钱，这会摧毁整个系统。2009年诺贝尔经济学奖获得者埃莉诺·奥斯特罗姆 Elinor Ostrom随后的研究认为公地悲剧过于简单化，其负面结果会受到外部影响。在没有复杂压力的情况下，团体之间为了共同利益进行沟通和管理，执行社会规范以保护资源并为团体实现最大利益，这是实现囚徒困境最佳结果的一个例子。<ref>{{cite web|url=http://volokh.com/2009/10/12/elinor-ostrom-and-the-tragedy-of-the-commons/ |title=The Volokh Conspiracy " Elinor Ostrom and the Tragedy of the Commons |publisher=Volokh.com |date=2009-10-12 |accessdate=2011-12-17}}</ref><br />
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==Related games==<br />
相关博弈<br />
===Closed-bag exchange===<br />
<font color="#ff8000">封闭袋子交换 Closed-bag exchange </font><br />
[[File:Prisoner's Dilemma briefcase exchange (colorized).svg|thumb|The prisoner's dilemma as a briefcase exchange]]<br />
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The prisoner's dilemma as a briefcase exchange<br />
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囚徒困境是一个公文包式的交换<br />
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[[Douglas Hofstadter]]<ref name="dh">{{cite book | first=Douglas R. | last=Hofstadter| authorlink=Douglas Hofstadter | title= Metamagical Themas: questing for the essence of mind and pattern | publisher= Bantam Dell Pub Group| year=1985 | isbn=978-0-465-04566-2|chapter= Ch.29 ''The Prisoner's Dilemma Computer Tournaments and the Evolution of Cooperation''.| title-link=Metamagical Themas}}</ref> once suggested that people often find problems such as the PD problem easier to understand when it is illustrated in the form of a simple game, or trade-off. One of several examples he used was "closed bag exchange":<br />
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Douglas Hofstadter once suggested that people often find problems such as the PD problem easier to understand when it is illustrated in the form of a simple game, or trade-off. One of several examples he used was "closed bag exchange":<br />
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侯世达 Douglas Hofstadter <ref name="dh">{{cite book | first=Douglas R. | last=Hofstadter| authorlink=Douglas Hofstadter | title= Metamagical Themas: questing for the essence of mind and pattern | publisher= Bantam Dell Pub Group| year=1985 | isbn=978-0-465-04566-2|chapter= Ch.29 ''The Prisoner's Dilemma Computer Tournaments and the Evolution of Cooperation''.| title-link=Metamagical Themas}}</ref>曾经指出，人们通常会发现诸如囚徒困境的问题，比如，当它以一个简单囚徒困境博弈的形式，或者以权衡的方式表现出来时，会更容易理解。他使用的几个例子之一是“封闭袋子交换” :<br />
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{{quote|Two people meet and exchange closed bags, with the understanding that one of them contains money, and the other contains a purchase. Either player can choose to honor the deal by putting into his or her bag what he or she agreed, or he or she can defect by handing over an empty bag.}}<br />
两人相遇并交换包裹，事先知道一个包里装着钱，一个装着订单。任一玩家都可选择尊重交易，放入事先约定的东西；也可以选择背叛，交换空的公文包。<br />
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Defection always gives a game-theoretically preferable outcome.<ref>{{Cite web|url=https://users.auth.gr/kehagiat/Research/GameTheory/06GamesToPlay/Prisoner%27s_dilemma.htm#Closed_Bag_Exchange|title=Prisoner's dilemma - Wikipedia, the free encyclopedia|website=users.auth.gr|access-date=2020-04-12}}</ref><br />
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Defection always gives a game-theoretically preferable outcome.<br />
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背叛总是会带来一个理论上更可取的结果。<ref>{{Cite web|url=https://users.auth.gr/kehagiat/Research/GameTheory/06GamesToPlay/Prisoner%27s_dilemma.htm#Closed_Bag_Exchange|title=Prisoner's dilemma - Wikipedia, the free encyclopedia|website=users.auth.gr|access-date=2020-04-12}}</ref><br />
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===''Friend or Foe?''===<br />
朋友还是敌人？<br />
''[[Friend or Foe? (TV series)|Friend or Foe?]]'' is a game show that aired from 2002 to 2005 on the [[Game Show Network]] in the US. It is an example of the prisoner's dilemma game tested on real people, but in an artificial setting. On the game show, three pairs of people compete. When a pair is eliminated, they play a game similar to the prisoner's dilemma to determine how the winnings are split. If they both cooperate (Friend), they share the winnings 50–50. If one cooperates and the other defects (Foe), the defector gets all the winnings and the cooperator gets nothing. If both defect, both leave with nothing. Notice that the reward matrix is slightly different from the standard one given above, as the rewards for the "both defect" and the "cooperate while the opponent defects" cases are identical. This makes the "both defect" case a weak equilibrium, compared with being a strict equilibrium in the standard prisoner's dilemma. If a contestant knows that their opponent is going to vote "Foe", then their own choice does not affect their own winnings. In a specific sense, ''Friend or Foe'' has a rewards model between prisoner's dilemma and the [[Chicken (game)|game of Chicken]].<br />
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Friend or Foe? is a game show that aired from 2002 to 2005 on the Game Show Network in the US. It is an example of the prisoner's dilemma game tested on real people, but in an artificial setting. On the game show, three pairs of people compete. When a pair is eliminated, they play a game similar to the prisoner's dilemma to determine how the winnings are split. If they both cooperate (Friend), they share the winnings 50–50. If one cooperates and the other defects (Foe), the defector gets all the winnings and the cooperator gets nothing. If both defect, both leave with nothing. Notice that the reward matrix is slightly different from the standard one given above, as the rewards for the "both defect" and the "cooperate while the opponent defects" cases are identical. This makes the "both defect" case a weak equilibrium, compared with being a strict equilibrium in the standard prisoner's dilemma. If a contestant knows that their opponent is going to vote "Foe", then their own choice does not affect their own winnings. In a specific sense, Friend or Foe has a rewards model between prisoner's dilemma and the game of Chicken.<br />
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朋友还是敌人？是一个竞赛节目，从2002年至2005年在美国的Game show Network播出。这是囚徒困境博弈在真人身上测试的一个例子，只是在人为的环境中。在游戏节目中，有三对选手参加比赛。当一对被淘汰时，他们会玩一个类似囚徒困境的游戏来决定奖金如何分配。如果他们都合作(朋友) ，他们分享奖金50-50。如果一方合作而另一方背叛(敌人) ，那么叛变者将得到所有的奖金，而合作者将一无所获。如果双方都背叛，那么双方都将一无所有。请注意，奖励矩阵与上面给出的标准矩阵略有不同，因为“双方都背叛”和“合作而对方背叛”情况下的奖励是相同的。与标准囚徒困境中的严格均衡相比，这使得“两个都背叛”情况成为一个弱均衡。如果一个参赛者知道他们的对手将投票给“敌人” ，那么他们自己的选择不会影响他们自己的奖金。从特定意义上讲，“朋友还是敌人”节目在囚徒困境和“胆小鬼”博弈之间有一个奖励模型。<br />
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The rewards matrix is<br />
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The rewards matrix is<br />
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奖励矩阵是<br />
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{| class="wikitable"<br />
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! {{diagonal split header|{{color|#009|Pair 1}}|{{color|#900|Pair 2}}}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"Friend"<br />(cooperate)}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"Foe"<br />(defect)}}<br />
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|-<br />
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! scope="row" style="width:6em;" | {{color|#009|"Friend"<br />(cooperate)}}<br />
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| {{diagonal split header|{{color|#009|1}}|{{color|#900|1}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|2}}|transparent}}<br />
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|-<br />
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! scope="row" | {{color|#009|"Foe"<br />(defect)}}<br />
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| {{diagonal split header|{{color|#009|2}}|{{color|#900|0}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
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|}<br />
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{| class="wikitable"<br />
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! {{diagonal split header|{{color|#009|Pair 1}}|{{color|#900|Pair 2}}}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"朋友"<br />(合作)}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"敌人"<br />(背叛)}}<br />
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|-<br />
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! scope="row" style="width:6em;" | {{color|#009|"朋友"<br />(合作)}}<br />
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| {{diagonal split header|{{color|#009|1}}|{{color|#900|1}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|2}}|transparent}}<br />
<br />
|-<br />
<br />
! scope="row" | {{color|#009|"敌人"<br />(背叛)}}<br />
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| {{diagonal split header|{{color|#009|2}}|{{color|#900|0}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
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|}<br />
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This payoff matrix has also been used on the [[United Kingdom|British]] [[television]] programmes ''Trust Me'', ''[[Shafted]]'', ''[[The Bank Job (TV series)|The Bank Job]]'' and ''[[Golden Balls]]'', and on the [[United States|American]] shows ''[[Bachelor Pad]]'' and ''[[Take It All (game show)|Take It All]]''. Game data from the ''[[Golden Balls]]'' series has been analyzed by a team of economists, who found that cooperation was "surprisingly high" for amounts of money that would seem consequential in the real world, but were comparatively low in the context of the game.<ref>{{cite journal | ssrn=1592456 | title=Split or Steal? Cooperative Behavior When the Stakes Are Large | author=Van den Assem, Martijn J. | journal=Management Science |date=January 2012 | volume=58 | issue=1 | pages=2–20 | doi=10.1287/mnsc.1110.1413| url=http://faculty.chicagobooth.edu/richard.thaler/research/pdf/Split%20or%20Steal%20Cooperative%20Behavior%20When%20the%20Stakes%20Are%20Large.pdf }}</ref><br />
<br />
This payoff matrix has also been used on the British television programmes Trust Me, Shafted, The Bank Job and Golden Balls, and on the American shows Bachelor Pad and Take It All. Game data from the Golden Balls series has been analyzed by a team of economists, who found that cooperation was "surprisingly high" for amounts of money that would seem consequential in the real world, but were comparatively low in the context of the game.<br />
<br />
英国电视节目《相信我》、《阴影》、《银行工作》和《黄金球》以及美国电视节目《单身公寓》和《全部拿走》也采用了这种奖励矩阵。一个经济学家团队分析了“金球奖”系列的游戏数据，他们发现，现实生活中，合作对于金额而言“惊人地高” ，但在游戏的背景下，相对较低。<ref>{{cite journal | ssrn=1592456 | title=Split or Steal? Cooperative Behavior When the Stakes Are Large | author=Van den Assem, Martijn J. | journal=Management Science |date=January 2012 | volume=58 | issue=1 | pages=2–20 | doi=10.1287/mnsc.1110.1413| url=http://faculty.chicagobooth.edu/richard.thaler/research/pdf/Split%20or%20Steal%20Cooperative%20Behavior%20When%20the%20Stakes%20Are%20Large.pdf }}</ref><br />
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<br />
===Iterated snowdrift===<br />
<font color="#ff8000">迭代雪堆 Iterated snowdrift </font><br />
{{main|snowdrift game}}<br />
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Researchers from the [[University of Lausanne]] and the [[University of Edinburgh]] have suggested that the "Iterated Snowdrift Game" may more closely reflect real-world social situations. Although this model is actually a [[chicken game]], it will be described here. In this model, the risk of being exploited through defection is lower, and individuals always gain from taking the cooperative choice. The snowdrift game imagines two drivers who are stuck on opposite sides of a [[snowdrift]], each of whom is given the option of shoveling snow to clear a path, or remaining in their car. A player's highest payoff comes from leaving the opponent to clear all the snow by themselves, but the opponent is still nominally rewarded for their work.<br />
<br />
Researchers from the University of Lausanne and the University of Edinburgh have suggested that the "Iterated Snowdrift Game" may more closely reflect real-world social situations. Although this model is actually a chicken game, it will be described here. In this model, the risk of being exploited through defection is lower, and individuals always gain from taking the cooperative choice. The snowdrift game imagines two drivers who are stuck on opposite sides of a snowdrift, each of whom is given the option of shoveling snow to clear a path, or remaining in their car. A player's highest payoff comes from leaving the opponent to clear all the snow by themselves, but the opponent is still nominally rewarded for their work.<br />
<br />
来自洛桑大学和爱丁堡大学的研究人员认为，“迭代雪堆游戏”可能更能反映现实世界的社会状况。虽然这个模型实际上是一个胆小鬼博弈。在这个模型中，由于背叛可以降低被剥削的风险，个体总是从合作选择中获益。这个雪堆游戏可以设想两个司机被困在雪堆的两侧，每个司机都可以选择铲雪清理道路，或者留在自己的车里。一个玩家的最高回报来自于让对手清除所有的积雪，但是仍然可以从对手的工作中得到回报。<br />
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<br />
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This may better reflect real world scenarios, the researchers giving the example of two scientists collaborating on a report, both of whom would benefit if the other worked harder. "But when your collaborator doesn’t do any work, it’s probably better for you to do all the work yourself. You’ll still end up with a completed project."<ref>{{cite web|last=Kümmerli|first=Rolf|title='Snowdrift' game tops 'Prisoner's Dilemma' in explaining cooperation|url=http://phys.org/news111145481.html|accessdate=11 April 2012}}</ref><br />
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This may better reflect real world scenarios, the researchers giving the example of two scientists collaborating on a report, both of whom would benefit if the other worked harder. "But when your collaborator doesn’t do any work, it’s probably better for you to do all the work yourself. You’ll still end up with a completed project."<br />
<br />
这可能更好地反映了现实世界的情景，研究人员举了两位科学家合作完成一份报告的例子，如果另一位科学家更加努力地工作，这两位科学家都会受益。“但当你的合作者不做任何工作时，你自己完成所有的工作可能会更好。你最终还是会完成一个项目。” <ref>{{cite web|last=Kümmerli|first=Rolf|title='Snowdrift' game tops 'Prisoner's Dilemma' in explaining cooperation|url=http://phys.org/news111145481.html|accessdate=11 April 2012}}</ref><br />
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{|<br />
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|-<br />
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{| class="wikitable" style="text-align: center;"<br />
<br />
|+ Example snowdrift payouts (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! Cooperates !! Defects<br />
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|-<br />
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! Cooperates<br />
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| 200, 200 || 100, 300<br />
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|-<br />
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! Defects<br />
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| 300, 100 || 0, 0<br />
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|}<br />
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||<br />
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{| class="wikitable" style="text-align: center;margin-left:2em;"<br />
<br />
|+ Example PD payouts (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! Cooperates !! Defects<br />
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|-<br />
<br />
! Cooperates<br />
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| 200, 200 || -100, 300<br />
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|-<br />
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! Defects<br />
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| 300, -100 || 0, 0<br />
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|}<br />
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{|<br />
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|-<br />
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|<br />
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{| class="wikitable" style="text-align: center;"<br />
<br />
|+ 迭代雪堆的支出示例 (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! 合作 !! 背叛<br />
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|-<br />
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! 合作<br />
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| 200, 200 || 100, 300<br />
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|-<br />
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! 背叛<br />
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| 300, 100 || 0, 0<br />
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|}<br />
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||<br />
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{| class="wikitable" style="text-align: center;margin-left:2em;"<br />
<br />
|+ 囚徒困境支出示例 (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! 合作 !! 背叛<br />
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|-<br />
<br />
! 合作<br />
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| 200, 200 || -100, 300<br />
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|-<br />
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! 背叛<br />
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| 300, -100 || 0, 0<br />
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|}<br />
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===Coordination games===<br />
<font color="#ff8000">协调博弈 Coordination games</font><br />
{{main|coordination games}}<br />
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In coordination games, players must coordinate their strategies for a good outcome. An example is two cars that abruptly meet in a blizzard; each must choose whether to swerve left or right. If both swerve left, or both right, the cars do not collide. The local [[left- and right-hand traffic]] convention helps to co-ordinate their actions.<br />
<br />
In coordination games, players must coordinate their strategies for a good outcome. An example is two cars that abruptly meet in a blizzard; each must choose whether to swerve left or right. If both swerve left, or both right, the cars do not collide. The local left- and right-hand traffic convention helps to co-ordinate their actions.<br />
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在协调博弈中，参与者必须协调自己的策略以获得一个好的结果。一个例子是两辆车在暴风雪中突然相遇，每辆车必须选择是左转还是右转。如果两辆车都向左转弯，或者都向右转弯，那么两辆车就不会相撞。当地的左右向交通惯例有助于协调他们的行动。<br />
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Symmetrical co-ordination games include [[Stag hunt]] and [[Bach or Stravinsky]].<br />
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Symmetrical co-ordination games include Stag hunt and Bach or Stravinsky.<br />
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对称的协调游戏包括猎鹿 Stag hunt和巴赫 Bach或斯特拉文斯基 Stravinsky。<br />
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===Asymmetric prisoner's dilemmas===<br />
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<font color="#ff8000">不对称的囚徒困境 Asymmetric prisoner's dilemmas</font><br />
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A more general set of games are asymmetric. As in the prisoner's dilemma, the best outcome is co-operation, and there are motives for defection. Unlike the symmetric prisoner's dilemma, though, one player has more to lose and/or more to gain than the other. Some such games have been described as a prisoner's dilemma in which one prisoner has an [[alibi]], whence the term "alibi game".<ref>{{cite conference|last1=Robinson |first1=D.R. |last2=Goforth |first2=D.J. |title=Alibi games: the Asymmetric Prisoner' s Dilemmas |date=May 5, 2004 |url=https://economics.ca/2004/papers/0359.pdf |conference=Meetings of the Canadian Economics Association, Toronto, June 4-6, 2004}}</ref><br />
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A more general set of games are asymmetric. As in the prisoner's dilemma, the best outcome is co-operation, and there are motives for defection. Unlike the symmetric prisoner's dilemma, though, one player has more to lose and/or more to gain than the other. Some such games have been described as a prisoner's dilemma in which one prisoner has an alibi, whence the term "alibi game".<br />
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一个更一般的博弈集是不对称的。就像在囚徒困境中一样，最好的结果是合作，而背叛是有动机的。与对称的囚徒困境不同的是一个玩家比另一个玩家有更多的损失或收获。这样的博弈被描述为囚徒困境，其中一个囚徒有不在场证明，这就是术语“不在场证明游戏”的由来。<ref>{{cite conference|last1=Robinson |first1=D.R. |last2=Goforth |first2=D.J. |title=Alibi games: the Asymmetric Prisoner' s Dilemmas |date=May 5, 2004 |url=https://economics.ca/2004/papers/0359.pdf |conference=Meetings of the Canadian Economics Association, Toronto, June 4-6, 2004}}</ref><br />
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In experiments, players getting unequal payoffs in repeated games may seek to maximize profits, but only under the condition that both players receive equal payoffs; this may lead to a stable equilibrium strategy in which the disadvantaged player defects every X games, while the other always co-operates. Such behaviour may depend on the experiment's social norms around fairness.<ref>{{cite chapter|last1=Beckenkamp |first1=Martin |last2=Hennig-Schmidt |first2=Heike |last3=Maier-Rigaud |first3=Frank P. |chapter=Cooperation in Symmetric and Asymmetric Prisoner's Dilemma Games |date=March 4, 2007 |chapter-url=http://homepage.coll.mpg.de/pdf_dat/2006_25online.pdf |title=[[Max Planck Institute for Research on Collective Goods]]}}</ref><br />
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In experiments, players getting unequal payoffs in repeated games may seek to maximize profits, but only under the condition that both players receive equal payoffs; this may lead to a stable equilibrium strategy in which the disadvantaged player defects every X games, while the other always co-operates. Such behaviour may depend on the experiment's social norms around fairness.<br />
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在实验中，在重复博弈中获得不均等收益的参与者可能会寻求利润最大化，但是前提是两个玩家都必须获得均等的收益。这可能会导致一个稳定的均衡策略，即弱势参与者在每隔X场博弈中都会背叛，而另一个参与者总是保持合作。这种行为可能取决于实验围绕公平的社会规范。<ref>{{cite chapter|last1=Beckenkamp |first1=Martin |last2=Hennig-Schmidt |first2=Heike |last3=Maier-Rigaud |first3=Frank P. |chapter=Cooperation in Symmetric and Asymmetric Prisoner's Dilemma Games |date=March 4, 2007 |chapter-url=http://homepage.coll.mpg.de/pdf_dat/2006_25online.pdf |title=[[Max Planck Institute for Research on Collective Goods]]}}</ref><br />
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==Software==<br />
软件<br />
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Several software packages have been created to run prisoner's dilemma simulations and tournaments, some of which have available source code.<br />
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Several software packages have been created to run prisoner's dilemma simulations and tournaments, some of which have available source code.<br />
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已经有一些可以用来运行囚徒困境模拟和比赛的软件包，其中一些有可用的源代码。<br />
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* The source code for the [[The Evolution of Cooperation|second tournament]] run by Robert Axelrod (written by Axelrod and many contributors in [[Fortran]]) is available [http://www-personal.umich.edu/~axe/research/Software/CC/CC2.html online]<br />
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* [https://web.archive.org/web/19991010053242/http://www.lifl.fr/IPD/ipd.frame.html Prison], a library written in [[Java (programming language)|Java]], last updated in 1998<br />
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* [https://github.com/Axelrod-Python/Axelrod Axelrod-Python], written in [[Python (programming language)|Python]]<br />
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* [http://selborne.nl/ipd/ play the Iterative Prisoner's Dilemma in the browser], play against strategies or let strategies play against other strategies<br />
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==In fiction==<br />
在小说中<br />
[[Hannu Rajaniemi]] set the opening scene of his ''[[The Quantum Thief]]'' trilogy in a "dilemma prison". The main theme of the series has been described as the "inadequacy of a binary universe" and the ultimate antagonist is a character called the All-Defector. Rajaniemi is particularly interesting as an artist treating this subject in that he is a Cambridge-trained mathematician and holds a PhD in [[mathematical physics]]&nbsp;– the interchangeability of matter and information is a major feature of the books, which take place in a "post-singularity" future. The first book in the series was published in 2010, with the two sequels, ''[[The Fractal Prince]]'' and ''[[The Causal Angel]]'', published in 2012 and 2014, respectively.<br />
<br />
Hannu Rajaniemi set the opening scene of his The Quantum Thief trilogy in a "dilemma prison". The main theme of the series has been described as the "inadequacy of a binary universe" and the ultimate antagonist is a character called the All-Defector. Rajaniemi is particularly interesting as an artist treating this subject in that he is a Cambridge-trained mathematician and holds a PhD in mathematical physics&nbsp;– the interchangeability of matter and information is a major feature of the books, which take place in a "post-singularity" future. The first book in the series was published in 2010, with the two sequels, The Fractal Prince and The Causal Angel, published in 2012 and 2014, respectively.<br />
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汉努·拉贾尼埃米 Hannu Rajaniemi将他的《量子窃贼》三部曲的开场场景设置在一个“囚徒困境”中。该系列的主题被描述为“双重宇宙的不足” ，最终的对手是一个叫做全面背叛者的角色。作为一个处理这个问题的艺术家，拉贾尼埃米尤其有趣，因为他是剑桥大学培养的数学家，拥有数学物理学博士学位——物质和信息的可互换性是这本书的一个主要特征，它发生在<font color="#ff8000">后奇点post-singularity</font>的未来。该系列的第一本书于2010年出版，其续集《分形王子》和《因果天使》分别于2012年和2014年出版。<br />
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A game modeled after the (iterated) prisoner's dilemma is a central focus of the 2012 video game ''[[Zero Escape: Virtue's Last Reward]]'' and a minor part in its 2016 sequel ''[[Zero Escape: Zero Time Dilemma]]''.<br />
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A game modeled after the (iterated) prisoner's dilemma is a central focus of the 2012 video game Zero Escape: Virtue's Last Reward and a minor part in its 2016 sequel Zero Escape: Zero Time Dilemma.<br />
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一个（迭代）囚徒困境博弈的模型是2012年电子游戏《零度逃脱: 美德的最后奖励》的重点，也是2016年续集《零度逃脱: 极限脱出刻之困境》的一个次要部分。<br />
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In ''The Mysterious Benedict Society and the Prisoner's Dilemma'' by [[Trenton Lee Stewart]], the main characters start by playing a version of the game and escaping from the "prison" altogether. Later they become actual prisoners and escape once again.<br />
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In The Mysterious Benedict Society and the Prisoner's Dilemma by Trenton Lee Stewart, the main characters start by playing a version of the game and escaping from the "prison" altogether. Later they become actual prisoners and escape once again.<br />
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在特伦顿·李·斯图尔特 Trenton Lee Stewart的《神秘的本尼迪克特社会和囚徒困境》中，主要角色从玩一个版本的游戏开始，然后一起逃离“监狱”。后来他们变成了真正的囚犯，再次越狱。<br />
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In ''[[The Adventure Zone]]: Balance'' during ''The Suffering Game'' subarc, the player characters are twice presented with the prisoner's dilemma during their time in two liches' domain, once cooperating and once defecting.<br />
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In The Adventure Zone: Balance during The Suffering Game subarc, the player characters are twice presented with the prisoner's dilemma during their time in two liches' domain, once cooperating and once defecting.<br />
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在冒险区: 苦难游戏的平衡中，玩家角色在他们在两个领域的时间内两次呈现出囚徒困境，一次是合作，一次是背叛。<br />
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In the 8th novel from the author James S. A. Corey [[Tiamat's Wrath]] . Winston Duarte explains the prisoners dilemma in his 14-year-old daughter, Teresa, to train her in strategic thinking. {{cn|date=April 2020}}<br />
<br />
In the 8th novel from the author James S. A. Corey Tiamat's Wrath . Winston Duarte explains the prisoners dilemma in his 14-year-old daughter, Teresa, to train her in strategic thinking. <br />
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在作者詹姆斯·S·A·科里·提亚玛特 James S. A. Corey Tiamat的《愤怒》中的第八部小说中，温斯顿•杜阿尔特 Winston Duarte向他14岁的女儿特蕾莎 Teresa解释了她面临的囚徒困境，来训练她的战略思维能力。<br />
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==See also==<br />
请参阅<br />
{{div col|colwidth=18em}}<br />
<br />
* [[Abilene paradox]]<br />
<font color="#ff8000"> 阿背伦悖论</font><br />
* [[Centipede game]]<br />
<font color="#ff8000"> 蜈蚣博弈</font><br />
* [[Christmas truce]]<br />
<font color="#ff8000"> 圣诞休战</font><br />
* [[Folk theorem (game theory)]]<br />
<font color="#ff8000"> 无名氏定理（博弈论）/font><br />
* [[Free-rider problem]]<br />
<font color="#ff8000"> 搭便车问题</font><br />
* [[Hobbesian trap]]<br />
]<font color="#ff8000"> 霍布斯主义陷阱</font><br />
* [[Innocent prisoner's dilemma]]<br />
<font color="#ff8000"> 无辜囚徒困局</font><br />
* [[Liar Game]]<br />
<font color="#ff8000">说谎者博弈</font><br />
* [[Optional prisoner's dilemma]]<br />
<font color="#ff8000"> 可选择囚徒困境</font><br />
* [[Robert H. Frank#Prisoner's dilemma and cooperation|Prisoner's dilemma and cooperation]]<br />
罗伯特·H·弗兰克囚徒的困境和合作|<font color="#ff8000"> 囚徒困境</font>和合作<br />
* [[Public goods game]]<br />
<font color="#ff8000"> 公共商品博弈</font><br />
* [[Gift-exchange game]]<br />
<font color="#ff8000"> 互利博弈</font><br />
* [[Reciprocal altruism]]<br />
<font color="#ff8000"> 相互利他行为</font><br />
* [[Social preferences]]<br />
<font color="#ff8000"> 社会偏好</font><br />
* [[Swift trust theory]]<br />
<font color="#ff8000"> 快速信任理论</font><br />
* [[Unscrupulous diner's dilemma]]<br />
]<font color="#ff8000"> 无道德食客困境</font><br />
{{div col end}}<br />
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==References==<br />
==参考==<br />
{{notelist}}<br />
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{{reflist|colwidth=30em}}<br />
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<br />
<br />
==Further reading==<br />
==拓展阅读==<br />
{{refbegin|30em}}<br />
<br />
* [[S.M. Amadae|Amadae, S.]] (2016). 'Prisoner's Dilemma,' ''Prisoners of Reason.'' [[Cambridge University Press]], NY, pp.&nbsp;24–61.<br />
<br />
* {{cite book |first1=Robert |last1=Aumann |authorlink=Robert Aumann |chapter=Acceptable points in general cooperative ''n''-person games |editor1-first=R. D. |editor1-last=Luce |editor2-first=A. W. |editor2-last=Tucker |title=Contributions to the Theory 23 of Games IV |series=Annals of Mathematics Study |volume=40 |pages=287–324 |publisher=Princeton University Press |location=Princeton NJ |year=1959 |mr=0104521}}<br />
<br />
* [[Robert Axelrod|Axelrod, R.]] (1984). ''[[The Evolution of Cooperation]]''. {{isbn|0-465-02121-2}}<br />
<br />
* [[Cristina Bicchieri|Bicchieri, Cristina]] (1993). Rationality and Coordination. [[Cambridge University Press]].<br />
<br />
* {{cite journal |first1=David M. |last1=Chess |date=December 1988 |title=Simulating the evolution of behavior: the iterated prisoners' dilemma problem |url=http://www.complex-systems.com/pdf/02-6-4.pdf |journal=Complex Systems |volume=2 |issue=6 |pages=663–70}}<br />
<br />
* [[Melvin Dresher|Dresher, M.]] (1961). ''The Mathematics of Games of Strategy: Theory and Applications'' [[Prentice-Hall]], Englewood Cliffs, NJ.<br />
<br />
* Greif, A. (2006). ''Institutions and the Path to the Modern Economy: Lessons from Medieval Trade.'' Cambridge University Press, [[Cambridge]], UK.<br />
<br />
* [[Anatol Rapoport|Rapoport, Anatol]] and Albert M. Chammah (1965). ''Prisoner's Dilemma''. [[University of Michigan Press]].<br />
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{{refend}}<br />
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<br />
<br />
==External links==<br />
外部链接<br />
*{{Commonscat-inline}}<br />
<br />
* [http://plato.stanford.edu/entries/prisoner-dilemma/ Prisoner's Dilemma (''Stanford Encyclopedia of Philosophy'')]<br />
<br />
* [http://www.msri.org/ext/larryg/pages/15.htm The Bowerbird's Dilemma] The Prisoner's Dilemma in ornithology&nbsp;– mathematical cartoon by Larry Gonick.<br />
<br />
* [https://www.youtube.com/watch?v=_1SEXTVsxjk The Prisoner's Dilemma] The Prisoner's Dilemma with Lego minifigures.<br />
<br />
* {{cite encyclopedia |last1=Dixit |first1=Avinash |authorlink1=Avinash Dixit |last2= Nalebuff |first2=Barry |authorlink2=Barry Nalebuff |editor=[[David R. Henderson]]|encyclopedia=[[Concise Encyclopedia of Economics]] |title=Prisoner's Dilemma |url=http://www.econlib.org/library/Enc/PrisonersDilemma.html |year=2008 |edition= 2nd |publisher=[[Library of Economics and Liberty]] |location=Indianapolis |isbn=978-0865976658 |oclc=237794267}}<br />
<br />
* [http://gametheory101.com/The_Prisoner_s_Dilemma.html Game Theory 101: Prisoner's Dilemma]<br />
<br />
* [https://www.youtube.com/watch?v=I71mjZefg8g Dawkins: Nice Guys Finish First]<br />
<br />
* [https://axelrod.readthedocs.io/en/stable/ Axelrod] Iterated Prisoner's Dilemma [[Python (programming language)|Python]] library<br />
<br />
* [http://gametheorygames.nl/index.html Play the Iterated Prisoner's Dilemma on gametheorygames.nl]<br />
<br />
* [https://web.archive.org/web/20141011014608/http://demo.otree.org/demo/Prisoner%27s+Dilemma/ Play Prisoner's Dilemma on ''oTree''] (N/A 11-5-17)<br />
<br />
* Nicky Case's [https://web.archive.org/web/20181229222135/https://ncase.me/trust/ Evolution of Trust], an example of the donation game<br />
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* [http://iterated-prisoners-dilemma.info Iterated Prisoner's Dilemma online game] by Wayne Davis<br />
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{{Decision theory paradoxes}}<br />
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{{Game theory}}<br />
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{{Authority control}}<br />
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[[Category:Non-cooperative games]]<br />
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Category:Non-cooperative games<br />
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类别: 非合作性游戏<br />
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[[Category:Thought experiments]]<br />
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Category:Thought experiments<br />
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类别: 思维实验<br />
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[[Category:Dilemmas]]<br />
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Category:Dilemmas<br />
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类别: 困境<br />
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[[Category:Environmental studies]]<br />
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Category:Environmental studies<br />
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类别: 环境研究<br />
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[[Category:Social psychology]]<br />
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Category:Social psychology<br />
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类别: 社会心理学<br />
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[[Category:Moral psychology]]<br />
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Category:Moral psychology<br />
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范畴: 道德心理学<br />
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<small>This page was moved from [[wikipedia:en:Prisoner's dilemma]]. Its edit history can be viewed at [[囚徒困境/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E5%9B%9A%E5%BE%92%E5%9B%B0%E5%A2%83&diff=20999囚徒困境2021-01-19T16:10:57Z<p>Smile：</p>
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<div>此词条由Henry初步翻译。已由Smile审校。<br />
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{{short description|Canonical example of a game analyzed in game theory}}<br />
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The '''prisoner's dilemma''' is a standard example of a game analyzed in [[game theory]] that shows why two completely [[Rationality#Economics|rational]] individuals might not cooperate, even if it appears that it is in their best interests to do so. It was originally framed by [[Merrill Flood]] and [[Melvin Dresher]] while working at [[RAND Corporation|RAND]] in 1950. [[Albert W. Tucker]] formalized the game with prison sentence rewards and named it "prisoner's dilemma",<ref>Poundstone, 1992</ref> presenting it as follows:<br />
<br />
The prisoner's dilemma is a standard example of a game analyzed in game theory that shows why two completely rational individuals might not cooperate, even if it appears that it is in their best interests to do so. It was originally framed by Merrill Flood and Melvin Dresher while working at RAND in 1950. Albert W. Tucker formalized the game with prison sentence rewards and named it "prisoner's dilemma", prensenting it as follows:<br />
<br />
<font color="#ff8000"> 囚徒困境prisoner's dilemma</font>是<font color="#ff8000"> 博弈论game theory</font>分析博弈的一个代表性例子，它揭示了为什么两个完全理性的个体可能不会合作，即使这样做符合他们的最大利益。它最初是由梅里尔·弗勒德 Merrill Flood和 梅文·加舍尔 Melvin Dresher于1950年在兰德公司工作时构建的。阿尔伯特.W.塔克 Albert W. Tucker将这种博弈以监禁刑罚奖励的方式正式化，并将其命名为囚徒困境，具体阐述如下：<br />
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{{quote|Two members of a criminal gang are arrested and imprisoned. Each prisoner is in solitary confinement with no means of communicating with the other. The prosecutors lack sufficient evidence to convict the pair on the principal charge, but they have enough to convict both on a lesser charge. Simultaneously, the prosecutors offer each prisoner a bargain. Each prisoner is given the opportunity either to betray the other by testifying that the other committed the crime, or to cooperate with the other by remaining silent. The possible outcomes are:<br />
<br />
{{quote|Two members of a criminal gang are arrested and imprisoned. Each prisoner is in solitary confinement with no means of communicating with the other. The prosecutors lack sufficient evidence to convict the pair on the principal charge, but they have enough to convict both on a lesser charge. Simultaneously, the prosecutors offer each prisoner a bargain. Each prisoner is given the opportunity either to betray the other by testifying that the other committed the crime, or to cooperate with the other by remaining silent. The possible outcomes are:<br />
<br />
一个犯罪团伙的两名成员被捕入狱。每个囚犯都被单独监禁，与他人无法沟通。检察官缺乏足够的证据来对这两个人定罪，但有足够的证据以较低的罪名定罪。同时，检察官向每个犯人提供了一个交易。每个囚犯都有机会出卖对方，证明对方犯下的罪行，或者他们可以合作，保持沉默。可能的结果有:<br />
<br />
* If A and B each betray the other, each of them serves two years in prison<br />
如果A和B都背叛了对方，他们都会在监狱服刑两年。<br />
* If A betrays B but B remains silent, A will be set free and B will serve three years in prison<br />
如果A背叛了B但B保持沉默，A会被无罪释放而B会服刑三年。<br />
* If A remains silent but B betrays A, A will serve three years in prison and B will be set free<br />
如果A保持沉默但B背叛了A，A会服刑三年而B会无罪释放。<br />
* If A and B both remain silent, both of them will serve only one year in prison (on the lesser charge).}}<br />
如果A和B都保持沉默，他们就只用服刑一年（以较低的罪名）。<br />
<br />
It is implied that the prisoners will have no opportunity to reward or punish their partner other than the prison sentences they get and that their decision will not affect their reputation in the future. Because betraying a partner offers a greater reward than cooperating with them, all purely rational self-interested prisoners will betray the other, meaning the only possible outcome for two purely rational prisoners is for them to betray each other.<ref>{{cite web|last=Milovsky|first=Nicholas|title=The Basics of Game Theory and Associated Games|url=https://issuu.com/johnsonnick895/docs/game_theory_paper|accessdate=11 February 2014}}</ref> In reality, humans display a [[systemic bias]] towards cooperative behavior in this and similar games despite what is predicted by simple models of "rational" self-interested action.<ref name = Fehr>{{cite journal | last1=Fehr | first1= Ernst | last2=Fischbacher | first2=Urs | date= Oct 23, 2003 | title=The Nature of human altruism |journal=Nature | volume=425 | pages=785–91 | doi=10.1038/nature02043 | url=http://www.iwp.jku.at/born/mpwfst/04/nature02043_f_born.pdf | accessdate=February 27, 2013 | pmid=14574401 | issue=6960|bibcode = 2003Natur.425..785F }}</ref><ref name = Amos>{{cite book | title=Preference, belief, and similarity: selected writings. | publisher=Massachusetts Institute of Technology Press | first1= Amos | last1=Tversky | first2=Eldar | last2=Shafir | url=http://cseweb.ucsd.edu/~gary/PAPER-SUGGESTIONS/Preference,%20Belief,%20and%20Similarity%20Selected%20Writings%20(Bradford%20Books).pdf | year=2004 | isbn=9780262700931 | accessdate=February 27, 2013}}</ref><ref name="Ahn">{{cite journal |last1 = Toh-Kyeong|first1 = Ahn|last2 = Ostrom|first2 = Elinor|last3 = Walker|first3 = James|date = Sep 5, 2002|title = Incorporating Motivational Heterogeneity into Game-Theoretic Models of Collective Action|journal = Public Choice|volume = 117|issue = 3–4|pages = 295–314|doi =10.1023/b:puch.0000003739.54365.fd |url = http://www.indiana.edu/~workshop/seminars/papers/ahnostromwalker_092402.pdf|accessdate = June 27, 2015|hdl = 10535/4697}}</ref><ref name="Hessel">{{cite journal|last1 = Oosterbeek|first1 = Hessel|last2 = Sloof|first2 = Randolph|last3 = Van de Kuilen|first3 = Gus|date = Dec 3, 2003|title = Cultural Differences in Ultimatum Game Experiments: Evidence from a Meta-Analysis|journal = Experimental Economics|volume = 7|issue = 2|pages = 171–88|doi = 10.1023/B:EXEC.0000026978.14316.74|url = http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|accessdate = February 27, 2013|url-status = dead|archiveurl = https://web.archive.org/web/20130512175243/http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|archivedate = May 12, 2013}}</ref> This bias towards cooperation has been known since the test was first conducted at RAND; the secretaries involved trusted each other and worked together for the best common outcome.<ref>{{Cite book | url=https://books.google.com/?id=WIhZlB86nJwC&pg=PT96&lpg=PT96&dq=rand+secretaries+prisoner%27s+dilemma#v=onepage |title = Why Most Things Fail|isbn = 9780571266142|last1 = Ormerod|first1 = Paul|date = 2010-12-22}}</ref> The prisoner's dilemma became the focus of extensive experimental research.<ref>Deutsch, M. (1958). Trust and suspicion. Journal of Conflict Resolution, 2(4), 265–279. https://doi.org/10.1177/002200275800200401</ref> <ref>Rapoport, A., & Chammah, A. M. (1965). Prisoner’s Dilemma: A study of conflict and cooperation. Ann Arbor, MI: University of Michigan Press.</ref><br />
<br />
It is implied that the prisoners will have no opportunity to reward or punish their partner other than the prison sentences they get and that their decision will not affect their reputation in the future. Because betraying a partner offers a greater reward than cooperating with them, all purely rational self-interested prisoners will betray the other, meaning the only possible outcome for two purely rational prisoners is for them to betray each other. In reality, humans display a systemic bias towards cooperative behavior in this and similar games despite what is predicted by simple models of "rational" self-interested action. This bias towards cooperation has been known since the test was first conducted at RAND; the secretaries involved trusted each other and worked together for the best common outcome. The prisoner's dilemma became the focus of extensive experimental research. <br />
<br />
这意味着，囚犯除了监禁刑罚之外，没有机会奖励或惩罚他们的同伴，他们的决定也不会影响他们未来的声誉。因为背叛一个同伴比与他们合作能得到更大的回报，所以所有纯粹理性的、自私自利的囚犯都会背叛对方，这意味着，对于两个纯粹理性的囚犯来说，唯一可能的结果就是他们相互背叛。<ref>{{cite web|last=Milovsky|first=Nicholas|title=The Basics of Game Theory and Associated Games|url=https://issuu.com/johnsonnick895/docs/game_theory_paper|accessdate=11 February 2014}}</ref>实际上，尽管“理性的”自利行为的简单模型已经预测到了这一点，但人类在这种和类似的博弈中仍然表现出对合作行为的<font color="#ff8000">系统性偏差 systemic bias </font>。<ref name = Fehr>{{cite journal | last1=Fehr | first1= Ernst | last2=Fischbacher | first2=Urs | date= Oct 23, 2003 | title=The Nature of human altruism |journal=Nature | volume=425 | pages=785–91 | doi=10.1038/nature02043 | url=http://www.iwp.jku.at/born/mpwfst/04/nature02043_f_born.pdf | accessdate=February 27, 2013 | pmid=14574401 | issue=6960|bibcode = 2003Natur.425..785F }}</ref><ref name = Amos>{{cite book | title=Preference, belief, and similarity: selected writings. | publisher=Massachusetts Institute of Technology Press | first1= Amos | last1=Tversky | first2=Eldar | last2=Shafir | url=http://cseweb.ucsd.edu/~gary/PAPER-SUGGESTIONS/Preference,%20Belief,%20and%20Similarity%20Selected%20Writings%20(Bradford%20Books).pdf | year=2004 | isbn=9780262700931 | accessdate=February 27, 2013}}</ref><ref name="Ahn">{{cite journal |last1 = Toh-Kyeong|first1 = Ahn|last2 = Ostrom|first2 = Elinor|last3 = Walker|first3 = James|date = Sep 5, 2002|title = Incorporating Motivational Heterogeneity into Game-Theoretic Models of Collective Action|journal = Public Choice|volume = 117|issue = 3–4|pages = 295–314|doi =10.1023/b:puch.0000003739.54365.fd |url = http://www.indiana.edu/~workshop/seminars/papers/ahnostromwalker_092402.pdf|accessdate = June 27, 2015|hdl = 10535/4697}}</ref><ref name="Hessel">{{cite journal|last1 = Oosterbeek|first1 = Hessel|last2 = Sloof|first2 = Randolph|last3 = Van de Kuilen|first3 = Gus|date = Dec 3, 2003|title = Cultural Differences in Ultimatum Game Experiments: Evidence from a Meta-Analysis|journal = Experimental Economics|volume = 7|issue = 2|pages = 171–88|doi = 10.1023/B:EXEC.0000026978.14316.74|url = http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|accessdate = February 27, 2013|url-status = dead|archiveurl = https://web.archive.org/web/20130512175243/http://www.econ.nagoya-cu.ac.jp/~yhamagu/ultimatum.pdf|archivedate = May 12, 2013}}</ref>自从在兰德公司首次进行这项测试以来，人们就已经知道了这种对合作的偏见; 参与测试的秘书们相互信任，为了最好的共同的目标而努力。<ref>{{Cite book | url=https://books.google.com/?id=WIhZlB86nJwC&pg=PT96&lpg=PT96&dq=rand+secretaries+prisoner%27s+dilemma#v=onepage |title = Why Most Things Fail|isbn = 9780571266142|last1 = Ormerod|first1 = Paul|date = 2010-12-22}}</ref>囚徒困境成为大量实验研究的焦点。<ref>Deutsch, M. (1958). Trust and suspicion. Journal of Conflict Resolution, 2(4), 265–279. https://doi.org/10.1177/002200275800200401</ref> <ref>Rapoport, A., & Chammah, A. M. (1965). Prisoner’s Dilemma: A study of conflict and cooperation. Ann Arbor, MI: University of Michigan Press.</ref><br />
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<br />
<br />
An extended "iterated" version of the game also exists. In this version, the classic game is played repeatedly between the same prisoners, who continuously have the opportunity to penalize the other for previous decisions. If the number of times the game will be played is known to the players, then (by [[backward induction]]) two classically rational players will betray each other repeatedly, for the same reasons as the single-shot variant. In an infinite or unknown length game there is no fixed optimum strategy, and prisoner's dilemma tournaments have been held to compete and test algorithms for such cases.<ref>{{cite journal|url = https://egtheory.wordpress.com/2015/03/02/ipd/|title = Short history of iterated prisoner's dilemma tournaments|date = March 2, 2015|access-date = February 8, 2016|journal = Journal of Conflict Resolution|volume = 24|issue = 3|pages = 379–403|last = Kaznatcheev|first = Artem|doi = 10.1177/002200278002400301}}</ref><br />
<br />
An extended "iterated" version of the game also exists. In this version, the classic game is played repeatedly between the same prisoners, who continuously have the opportunity to penalize the other for previous decisions. If the number of times the game will be played is known to the players, then (by backward induction) two classically rational players will betray each other repeatedly, for the same reasons as the single-shot variant. In an infinite or unknown length game there is no fixed optimum strategy, and prisoner's dilemma tournaments have been held to compete and test algorithms for such cases.<br />
<br />
一个扩展的<font color="#ff8000">迭代iterated</font>版本的博弈由此衍生出来。在这个版本中，经典博弈会在在同一组囚犯之间重复进行，他们不断有机会为了之前的决定对其他囚犯进行惩罚。如果参与者知道博弈的次数，那么(通过<font color="#ff8000">逆向归纳法 backward induction </font>)两个经典的理性的玩家就会因为和在单次博弈中相同的原因反复背叛对方。在无限次或未知次数的博弈中，没有固定的最优策略，因而，举办囚徒困境竞赛来竞争和检验这种情况下的算法。<ref>{{cite journal|url = https://egtheory.wordpress.com/2015/03/02/ipd/|title = Short history of iterated prisoner's dilemma tournaments|date = March 2, 2015|access-date = February 8, 2016|journal = Journal of Conflict Resolution|volume = 24|issue = 3|pages = 379–403|last = Kaznatcheev|first = Artem|doi = 10.1177/002200278002400301}}</ref><br />
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The prisoner's dilemma game can be used as a model for many [[#Real-life examples|real world situations]] involving cooperative behavior. In casual usage, the label "prisoner's dilemma" may be applied to situations not strictly matching the formal criteria of the classic or iterative games: for instance, those in which two entities could gain important benefits from cooperating or suffer from the failure to do so, but find it difficult or expensive—not necessarily impossible—to coordinate their activities.<br />
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The prisoner's dilemma game can be used as a model for many real world situations involving cooperative behavior. In casual usage, the label "prisoner's dilemma" may be applied to situations not strictly matching the formal criteria of the classic or iterative games: for instance, those in which two entities could gain important benefits from cooperating or suffer from the failure to do so, but find it difficult or expensive—not necessarily impossible—to coordinate their activitie <br />
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囚徒困境博弈可以作为许多现实中涉及合作行为的模型。在非正式用法中，“囚徒困境”一词可适用于不严格符合经典或迭代博弈的形式标准的情况: 例如，两个实体可以从合作中获得巨大利益或者会因为合作失败而遭受损失，但发现协调他们的活动很困难或者代价昂贵（并非是不可能的）。<br />
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==Strategy for the prisoner's dilemma==<br />
囚徒困境的策略<br />
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Two prisoners are separated into individual rooms and cannot communicate with each other.<br />
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Two prisoners are separated into individual rooms and cannot communicate with each other.<br />
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两名囚犯被分开关押在各自的房间里，不能相互交流。<br />
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The normal game is shown below:<br />
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The normal game is shown below:<br />
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正常的博弈如下:<br />
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! {{diagonal split header|<br />Prisoner A|Prisoner B}} !! Prisoner B stays silent<br>(''cooperates'') !! Prisoner B betrays<br>(''defects'')<br />
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! !! Prisoner B stays silent<br>(cooperates) !! Prisoner B betrays<br>(defects)<br />
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! {{diagonal split header|<br />犯人A|犯人B}}！！犯人B保持沉默<br>(''合作'') !! 犯人B背叛<br>(''背叛'')<br />
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! Prisoner A stays silent<br>(''cooperates'')<br />
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! Prisoner A stays silent<br>(cooperates)<br />
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!犯人 a 保持沉默<br>(''合作'')<br />
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| Each serves 1 year|| Prisoner A: 3 years<br />Prisoner B: goes free<br />
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| Each serves 1 year|| Prisoner A: 3 years<br />Prisoner B: goes free<br />
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|每人服刑1年||囚犯 A: 3年<br />囚犯 B: 无罪释放<br />
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|-<br />
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|-<br />
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! Prisoner A betrays<br>(''defects'')<br />
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! Prisoner A betrays<br>(defects)<br />
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!犯人A背叛<br>(''背叛'')<br />
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| Prisoner A: goes free<br />Prisoner B: 3 years || Each serves 2 years<br />
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| Prisoner A: goes free<br />Prisoner B: 3 years || Each serves 2 years<br />
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|囚犯 A: 获释<br />囚犯 B: 3年||每人服刑2年<br />
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It is assumed that both prisoners understand the nature of the game, have no loyalty to each other, and will have no opportunity for retribution or reward outside the game. Regardless of what the other decides, each prisoner gets a higher reward by betraying the other ("defecting"). The reasoning involves an argument by [[Dilemma#Use in logic|dilemma]]: B will either cooperate or defect. If B cooperates, A should defect, because going free is better than serving 1 year. If B defects, A should also defect, because serving 2 years is better than serving 3. So either way, A should defect. Parallel reasoning will show that B should defect.<br />
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It is assumed that both prisoners understand the nature of the game, have no loyalty to each other, and will have no opportunity for retribution or reward outside the game. Regardless of what the other decides, each prisoner gets a higher reward by betraying the other ("defecting"). The reasoning involves an argument by dilemma: B will either cooperate or defect. If B cooperates, A should defect, because going free is better than serving 1 year. If B defects, A should also defect, because serving 2 years is better than serving 3. So either way, A should defect. Parallel reasoning will show that B should defect.<br />
<br />
假设两个囚犯都了解博弈的本质，对彼此不忠诚，且在博弈之外没有机会得到报复或奖励。那么不管对方怎么决定，每个犯人背叛对方都会得到更高的奖励(“叛变”)。推理涉及一个进退两难的论点:B 要么合作，要么叛变。如果B合作，A 应该叛变，因为得到释放总比服刑1年好。如果 B叛变，A也应该叛变，因为服刑2年总比服刑3年好。所以不管怎样，A都应该叛变。并行推理表明B应该选择叛变。<br />
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<br />
Because defection always results in a better payoff than cooperation regardless of the other player's choice, it is a [[dominant strategy]]. Mutual defection is the only strong [[Nash equilibrium]] in the game (i.e. the only outcome from which each player could only do worse by unilaterally changing strategy). The dilemma, then, is that mutual cooperation yields a better outcome than mutual defection but is not the rational outcome because the choice to cooperate, from a self-interested perspective, is irrational.<br />
<br />
Because defection always results in a better payoff than cooperation regardless of the other player's choice, it is a dominant strategy. Mutual defection is the only strong Nash equilibrium in the game (i.e. the only outcome from which each player could only do worse by unilaterally changing strategy). The dilemma, then, is that mutual cooperation yields a better outcome than mutual defection but is not the rational outcome because the choice to cooperate, from a self-interested perspective, is irrational.<br />
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因为不管对方的选择如何，背叛总是比合作带来更好的回报，所以这是一个<font color="#ff8000">占优策略 dominant strategy</font>。相互背叛是博弈中唯一的强<font color="#ff8000">纳什均衡点 dominant strategy </font>（即每个参与者单方面的改变策略只能使自己的情况变糟)。因此，困境在于，虽然相互合作比相互背叛产生更好的结果，但它却不是理性的结果，因为从自我利益的角度来看，合作的选择是非理性的。<br />
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<br />
<br />
==Generalized form==<br />
<font color="#ff8000">广泛形态 Generalized form </font><br />
The structure of the traditional prisoner's dilemma can be generalized from its original prisoner setting. Suppose that the two players are represented by the colors red and blue, and that each player chooses to either "cooperate" or "defect".<br />
<br />
The structure of the traditional prisoner's dilemma can be generalized from its original prisoner setting. Suppose that the two players are represented by the colors red and blue, and that each player chooses to either "cooperate" or "defect".<br />
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传统囚徒困境的结构可以从其最初的囚徒环境中概括出来。假设两个玩家用红色和蓝色表示，并且每个玩家选择“合作”或“背叛”。<br />
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<br />
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If both players cooperate, they both receive the reward ''R'' for cooperating. If both players defect, they both receive the punishment payoff ''P''. If Blue defects while Red cooperates, then Blue receives the temptation payoff ''T'', while Red receives the "sucker's" payoff, ''S''. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff ''S'', while Red receives the temptation payoff ''T''.<br />
<br />
If both players cooperate, they both receive the reward R for cooperating. If both players defect, they both receive the punishment payoff P. If Blue defects while Red cooperates, then Blue receives the temptation payoff T, while Red receives the "sucker's" payoff, S. Similarly, if Blue cooperates while Red defects, then Blue receives the sucker's payoff S, while Red receives the tem punishment payoffptation payoff T.<br />
<br />
如果两个玩家合作，他们都会因为合作而获得奖励''R''。如果两个参与人都叛变，他们都会受到惩罚''P''。 如果蓝方叛变而红方合作，那么蓝方得到诱惑回报''T''，而红方受到“上当受骗者”损失''S''。同样地，如果蓝方合作而红方叛变，那么蓝方得到上当受骗者的损失''S''，而红方得到诱惑支付''T''。<br />
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<br />
<br />
This can be expressed in [[Normal-form game|normal form]]:<br />
<br />
This can be expressed in normal form:<br />
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这可以用标准形式的博弈来表示:<br />
<br />
<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
|+ Canonical PD payoff matrix<br />
<br />
|+ Canonical PD payoff matrix<br />
<br />
| + <font color="#ff8000">正则 PD 支付矩阵 Canonical PD payoff matrix </font> <br />
<br />
! {{diagonal split header|{{color|#009|Blue}}|{{color|#900|Red}}}}<br />
<br />
! |}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|Cooperate}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|Defect}}<br />
<br />
|-<br />
<br />
! scope="row" style="width:60px;" | {{color|#009|Cooperate}}<br />
<br />
| {{diagonal split header|{{color|#009|''R''}}|{{color|#900|''R''}}|transparent}}<br />
<br />
| ||transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''S''}}|{{color|#900|''T''}}|transparent}}<br />
<br />
|-<br />
<br />
! scope="row" | {{color|#009|Defect}}<br />
<br />
| {{diagonal split header|{{color|#009|''T''}}|{{color|#900|''S''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''P''}}|{{color|#900|''P''}}|transparent}}<br />
<br />
|}<br />
<br />
<br />
! {{diagonal split header|{{color|#009|蓝方}}|{{color|#900|红方}}}}<br />
<br />
! |}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|合作}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|背叛}}<br />
<br />
|-<br />
<br />
! scope="row" style="width:60px;" | {{color|#009|合作}}<br />
<br />
| {{diagonal split header|{{color|#009|''R''}}|{{color|#900|''R''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''S''}}|{{color|#900|''T''}}|transparent}}<br />
<br />
|-<br />
<br />
! scope="row" | {{color|#009|背叛}}<br />
<br />
| {{diagonal split header|{{color|#009|''T''}}|{{color|#900|''S''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|''P''}}|{{color|#900|''P''}}|transparent}}<br />
<br />
|}<br />
<br />
<br />
<br />
and to be a prisoner's dilemma game in the strong sense, the following condition must hold for the payoffs:<br />
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and to be a prisoner's dilemma game in the strong sense, the following condition must hold for the payoffs:<br />
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要成为强意义下的囚徒困境博弈，收益必须满足以下条件:<br />
<br />
<br />
<br />
:{{tmath|T > R > P > S}}<br />
<br />
<br />
<br />
The payoff relationship {{tmath|R > P}} implies that mutual cooperation is superior to mutual defection, while the payoff relationships {{tmath|T > R}} and {{tmath|P > S}} imply that defection is the [[dominant strategy]] for both agents.<br />
<br />
The payoff relationship implies that mutual cooperation is superior to mutual defection, while the payoff relationships and imply that defection is the dominant strategy for both agents.<br />
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回报关系{{tmath|R > P}}意味着相互合作优于相互背叛，然而回报关系{{tmath|T > R}}和{{tmath|P > S}}也意味着相互背叛是双方的占优策略。<br />
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<br />
<br />
===Special case: donation game===<br />
特例：<font color="#ff8000">捐赠博弈 donation game </font><br />
The "donation game"<ref name=Hilbe2013>{{cite journal|last=Hilbe|first=Christian |author2=Martin A. Nowak |author3=Karl Sigmund|title=Evolution of extortion in Iterated Prisoner's Dilemma games|journal=PNAS|date=April 2013|volume=110|issue=17|pages=6913–18|doi=10.1073/pnas.1214834110|pmid=23572576 |pmc=3637695 |bibcode=2013PNAS..110.6913H |arxiv=1212.1067}}</ref> is a form of prisoner's dilemma in which cooperation corresponds to offering the other player a benefit ''b'' at a personal cost ''c'' with ''b'' > ''c''. Defection means offering nothing. The payoff matrix is thus<br />
<br />
The "donation game" is a form of prisoner's dilemma in which cooperation corresponds to offering the other player a benefit b at a personal cost c with b > c. Defection means offering nothing. The payoff matrix is thus<br />
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捐赠博弈<ref name=Hilbe2013>{{cite journal|last=Hilbe|first=Christian |author2=Martin A. Nowak |author3=Karl Sigmund|title=Evolution of extortion in Iterated Prisoner's Dilemma games|journal=PNAS|date=April 2013|volume=110|issue=17|pages=6913–18|doi=10.1073/pnas.1214834110|pmid=23572576 |pmc=3637695 |bibcode=2013PNAS..110.6913H |arxiv=1212.1067}}</ref>是囚徒困境的一种形式，在这种博弈中，合作相当于以''b'' > ''c''条件下的个人成本''c''为另一方提供一个收益''b''，而叛变意味着什么也不提供。收益矩阵是这样的<br />
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<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
! {{diagonal split header|{{navy (color)|Blue}}|{{color|#900|Red}}}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|Cooperate}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|Defect}}<br />
<br />
|-<br />
<br />
! scope="row" style="width:60px;" | {{color|#009|Cooperate}}<br />
<br />
| {{diagonal split header|{{color|#009|''b''-''c''}}|{{color|#900|''b''-''c''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|-''c''}}|{{color|#900|''b''}}|transparent}}<br />
<br />
|-<br />
<br />
! scope="row" | {{color|#009|Defect}}<br />
<br />
| {{diagonal split header|{{color|#009|''b''}}|{{color|#900|-''c''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
<br />
|}<br />
<br />
<br />
<br />
{| class="wikitable" style="text-align:center"<br />
<br />
! {{diagonal split header|{{navy (color)|蓝方}}|{{color|#900|红方}}}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|合作}}<br />
<br />
! scope="col" style="width:60px;" | {{color|#900|背叛}}<br />
<br />
|-<br />
<br />
! scope="row" style="width:60px;" | {{color|#009|合作}}<br />
<br />
| {{diagonal split header|{{color|#009|''b''-''c''}}|{{color|#900|''b''-''c''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|-''c''}}|{{color|#900|''b''}}|transparent}}<br />
<br />
|-<br />
<br />
! scope="row" | {{color|#009|背叛}}<br />
<br />
| {{diagonal split header|{{color|#009|''b''}}|{{color|#900|-''c''}}|transparent}}<br />
<br />
| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
<br />
|}<br />
<br />
<br />
<br />
Note that {{tmath|2R>T+S}} (i.e. {{tmath|2(b-c)>b-c}}) which qualifies the donation game to be an iterated game (see next section).<br />
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Note that (i.e. ) which qualifies the donation game to be an iterated game (see next section).<br />
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请注意{{tmath|2R>T+S}}（即{{tmath|2(b-c)>b-c}}）这使得捐赠博弈成为一个迭代博弈(见下一节)。<br />
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The donation game may be applied to markets. Suppose X grows oranges, Y grows apples. The [[marginal utility]] of an apple to the orange-grower X is ''b'', which is higher than the marginal utility (''c'') of an orange, since X has a surplus of oranges and no apples. Similarly, for apple-grower Y, the marginal utility of an orange is ''b'' while the marginal utility of an apple is ''c''. If X and Y contract to exchange an apple and an orange, and each fulfills their end of the deal, then each receive a payoff of ''b''-''c''. If one "defects" and does not deliver as promised, the defector will receive a payoff of ''b'', while the cooperator will lose ''c''. If both defect, then neither one gains or loses anything.<br />
<br />
The donation game may be applied to markets. Suppose X grows oranges, Y grows apples. The marginal utility of an apple to the orange-grower X is b, which is higher than the marginal utility (c) of an orange, since X has a surplus of oranges and no apples. Similarly, for apple-grower Y, the marginal utility of an orange is b while the marginal utility of an apple is c. If X and Y contract to exchange an apple and an orange, and each fulfills their end of the deal, then each receive a payoff of b-c. If one "defects" and does not deliver as promised, the defector will receive a payoff of b, while the cooperator will lose c. If both defect, then neither one gains or loses anything.<br />
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捐赠博弈可能适用于市场。假设 X种橘子，Y 种苹果。苹果对橙子种植者 X 的<font color="#ff8000">边际效用 marginal utility</font>是''b''，这比橙子的边际效用''c''高，因为X有橙子剩余而没有苹果。同样地，对于苹果种植者Y来说，橙子的边际效用是''b''，而苹果的边际效用是''c''。 如果X和Y签约交换一个苹果和一个橙子，并且每个人都完成了交易，那么每个人都会得到''b-c''的收益。如果一方违约没有按照承诺交货，那么这个违约者将得到''b''的收益，而合作者将失去''c''的收益。 如果两者都违约，那么谁也不会得到或失去任何东西。<br />
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<br />
==The iterated prisoner's dilemma==<br />
<font color="#ff8000">迭代囚徒困境 iterated prisoner's dilemma </font> {{more citations needed section|date=November 2012}}<br />
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If two players play prisoner's dilemma more than once in succession and they remember previous actions of their opponent and change their strategy accordingly, the game is called iterated prisoner's dilemma.<br />
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If two players play prisoner's dilemma more than once in succession and they remember previous actions of their opponent and change their strategy accordingly, the game is called iterated prisoner's dilemma.<br />
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如果两个参与者连续进行多次囚徒困境博弈，他们记住对手先前的行动并相应地改变策略，这种博弈被称为迭代囚徒困境。<br />
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In addition to the general form above, the iterative version also requires that {{tmath|2R > T + S}}, to prevent alternating cooperation and defection giving a greater reward than mutual cooperation.<br />
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In addition to the general form above, the iterative version also requires that , to prevent alternating cooperation and defection giving a greater reward than mutual cooperation.<br />
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除了上面的一般形式之外，迭代版本还要求{{tmath|2R > T + S}}，防止交替合作和背叛比相互合作有更大的回报。<br />
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<br />
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The iterated prisoner's dilemma game is fundamental to some theories of human cooperation and trust. On the assumption that the game can model transactions between two people requiring trust, cooperative behaviour in populations may be modeled by a multi-player, iterated, version of the game. It has, consequently, fascinated many scholars over the years. In 1975, Grofman and Pool estimated the count of scholarly articles devoted to it at over 2,000. The iterated prisoner's dilemma has also been referred to as the "[[Peace war game|peace-war game]]".<ref name = Shy>{{cite book | title= Industrial Organization: Theory and Applications | publisher=Massachusetts Institute of Technology Press | first1= Oz | last1=Shy |url=https://books.google.com/?id=tr4CjJ5LlRcC&pg=PR13&dq=industrial+organization+theory+and+applications | year=1995 | isbn=978-0262193665 | accessdate=February 27, 2013}}</ref><br />
<br />
The iterated prisoner's dilemma game is fundamental to some theories of human cooperation and trust. On the assumption that the game can model transactions between two people requiring trust, cooperative behaviour in populations may be modeled by a multi-player, iterated, version of the game. It has, consequently, fascinated many scholars over the years. In 1975, Grofman and Pool estimated the count of scholarly articles devoted to it at over 2,000. The iterated prisoner's dilemma has also been referred to as the "peace-war game".<br />
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迭代囚徒困境博弈是人类合作与信任的理论基础。假设博弈可以为两个需要信任的人之间的交易建模，那么群体中的合作行为也可以由多个参与者迭代的博弈模型来建模。因此，这些年来，它吸引了许多学者。1975年，葛夫曼 Grofman和普尔 Pool估计专门撰写有关该领域的学术文章超过2000篇。迭代囚徒困境也被称为“和平-战争博弈”。<ref name = Shy>{{cite book | title= Industrial Organization: Theory and Applications | publisher=Massachusetts Institute of Technology Press | first1= Oz | last1=Shy |url=https://books.google.com/?id=tr4CjJ5LlRcC&pg=PR13&dq=industrial+organization+theory+and+applications | year=1995 | isbn=978-0262193665 | accessdate=February 27, 2013}}</ref><br />
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If the game is played exactly ''N'' times and both players know this, then it is optimal to defect in all rounds. The only possible [[Nash equilibrium]] is to always defect. The proof is [[Mathematical induction|inductive]]: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on. The same applies if the game length is unknown but has a known upper limit.<br />
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If the game is played exactly N times and both players know this, then it is optimal to defect in all rounds. The only possible Nash equilibrium is to always defect. The proof is inductive: one might as well defect on the last turn, since the opponent will not have a chance to later retaliate. Therefore, both will defect on the last turn. Thus, the player might as well defect on the second-to-last turn, since the opponent will defect on the last no matter what is done, and so on. The same applies if the game length is unknown but has a known upper limit.<br />
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如果这个游戏正好玩了N次，并且两个玩家都知道这一点，那么在所有回合中最佳的策略就是叛变。唯一可能的纳什均衡点就是永远叛变。证明是通过归纳法证出来的: 不妨假设一个人在最后一回合叛变，因为对手之后没有机会反击。因此，双方都会在最后一个回合叛变。所以玩家同样也会在倒数第二回合时叛变，因为无论采取什么策略，对手都会在倒数第一回合叛变，依此类推。如果博弈次数未知但次数有限的情况也同样如此。<br />
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Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer. The [[superrational]] strategy in the iterated prisoner's dilemma with fixed ''N'' is to cooperate against a superrational opponent, and in the limit of large ''N'', experimental results on strategies agree with the superrational version, not the game-theoretic rational one.<br />
<br />
Unlike the standard prisoner's dilemma, in the iterated prisoner's dilemma the defection strategy is counter-intuitive and fails badly to predict the behavior of human players. Within standard economic theory, though, this is the only correct answer. The superrational strategy in the iterated prisoner's dilemma with fixed N is to cooperate against a superrational opponent, and in the limit of large N, experimental results on strategies agree with the superrational version, not the game-theoretic rational one.<br />
<br />
与标准的囚徒困境不同，在迭代囚徒困境中，叛变策略是严重违反直觉的，以至于不能很好地预测人类玩家的行为。然而，在标准的经济理论中，这是唯一正确的答案。具有固定次数 N的迭代囚徒困境中的<font color="#ff8000">超理性 superrational</font>策略是与超理性对手进行合作，在N很大的限制下，实验结果的策略与超理性结果的策略一致，而不是博弈论的理性结果。<br />
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<br />
<br />
For [[cooperation]] to emerge between game theoretic rational players, the total number of rounds ''N'' must be unknown to the players. In this case "always defect" may no longer be a strictly dominant strategy, only a Nash equilibrium. Amongst results shown by [[Robert Aumann]] in a 1959 paper, rational players repeatedly interacting for indefinitely long games can sustain the cooperative outcome.<br />
<br />
For cooperation to emerge between game theoretic rational players, the total number of rounds N must be unknown to the players. In this case "always defect" may no longer be a strictly dominant strategy, only a Nash equilibrium. Amongst results shown by Robert Aumann in a 1959 paper, rational players repeatedly interacting for indefinitely long games can sustain the cooperative outcome.<br />
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为了使合作在博弈论的理性参与者之间出现，参与者必须不知道回合总数N。在这种情况下，“总是叛变”可能不再是一个严格占优策略，而只是一个纳什均衡。罗伯特·奥曼 Robert Aumann在1959年的一篇论文中表明，理性参与者在无限多次的博弈中通过反复互动可以维持合作的结果。<br />
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<br />
According to a 2019 experimental study in the ''American Economic Review'' which tested what strategies real-life subjects used in iterated prisoners' dilemma situations with perfect monitoring, the majority of chosen strategies were always defect, [[Tit for tat|tit-for-tat]], and [[Grim trigger]]. Which strategy the subjects chose depended on the parameters of the game.<ref>{{Cite journal|last=Dal Bó|first=Pedro|last2=Fréchette|first2=Guillaume R.|date=2019|title=Strategy Choice in the Infinitely Repeated Prisoner's Dilemma|journal=American Economic Review|language=en|volume=109|issue=11|pages=3929–3952|doi=10.1257/aer.20181480|issn=0002-8282}}</ref><br />
<br />
According to a 2019 experimental study in the American Economic Review which tested what strategies real-life subjects used in iterated prisoners' dilemma situations with perfect monitoring, the majority of chosen strategies were always defect, tit-for-tat, and Grim trigger. Which strategy the subjects chose depended on the parameters of the game.<br />
<br />
根据《美国经济评论》于2019年进行的一项实验研究，该实验中通过完美的监控测试了现实中被用在迭代囚徒困境情况下的策略，监测选择的策略总是背叛，针锋相对的和 <font color="#ff8000"> 冷酷触发策略 Grim trigger</font>。受试者选择的策略取决于博弈的参数。<ref>{{Cite journal|last=Dal Bó|first=Pedro|last2=Fréchette|first2=Guillaume R.|date=2019|title=Strategy Choice in the Infinitely Repeated Prisoner's Dilemma|journal=American Economic Review|language=en|volume=109|issue=11|pages=3929–3952|doi=10.1257/aer.20181480|issn=0002-8282}}</ref><br />
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<br />
<br />
===Strategy for the iterated prisoner's dilemma===<br />
迭代囚徒困境下的策略<br />
Interest in the iterated prisoner's dilemma (IPD) was kindled by [[Robert Axelrod]] in his book ''[[The Evolution of Cooperation]]'' (1984). In it he reports on a tournament he organized of the ''N'' step prisoner's dilemma (with ''N'' fixed) in which participants have to choose their mutual strategy again and again, and have memory of their previous encounters. Axelrod invited academic colleagues all over the world to devise computer strategies to compete in an IPD tournament. The programs that were entered varied widely in algorithmic complexity, initial hostility, capacity for forgiveness, and so forth.<br />
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Interest in the iterated prisoner's dilemma (IPD) was kindled by Robert Axelrod in his book The Evolution of Cooperation (1984). In it he reports on a tournament he organized of the N step prisoner's dilemma (with N fixed) in which participants have to choose their mutual strategy again and again, and have memory of their previous encounters. Axelrod invited academic colleagues all over the world to devise computer strategies to compete in an IPD tournament. The programs that were entered varied widely in algorithmic complexity, initial hostility, capacity for forgiveness, and so forth.<br />
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罗伯特·阿克塞尔罗德 Robert Axelrod在他的著作《合作的进化》(1984)中激起了人们对迭代囚徒困境(IPD)的兴趣。在这篇文章中，他报道了自己组织的固定N次囚徒困境的比赛，参与者必须一次又一次地选择他们的共同策略，并且要记住他们之前的遭遇。阿克塞尔罗德邀请世界各地的学术界同仁设计计算机策略来参加IPD锦标赛。输入的程序在算法复杂性、最初敌意、宽恕能力等方面有很大差异。<br />
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Axelrod discovered that when these encounters were repeated over a long period of time with many players, each with different strategies, greedy strategies tended to do very poorly in the long run while more [[altruism|altruistic]] strategies did better, as judged purely by self-interest. He used this to show a possible mechanism for the evolution of altruistic behaviour from mechanisms that are initially purely selfish, by [[natural selection]].<br />
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Axelrod discovered that when these encounters were repeated over a long period of time with many players, each with different strategies, greedy strategies tended to do very poorly in the long run while more altruistic strategies did better, as judged purely by self-interest. He used this to show a possible mechanism for the evolution of altruistic behaviour from mechanisms that are initially purely selfish, by natural selection.<br />
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阿克塞尔罗德发现，当这些遭遇长时间在许多玩家身上重复发生时，每个玩家都有不同的策略，从长远来看，贪婪策略往往表现得非常糟糕，而更加利他的策略表现得更好，这完全是根据自身利益来判断的。他利用这一结果揭示了通过自然选择，从最初纯粹自私行为向利他行为进化的可能机制。<br />
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The winning [[deterministic algorithm|deterministic]] strategy was tit for tat, which [[Anatol Rapoport]] developed and entered into the tournament. It was the simplest of any program entered, containing only four lines of [[BASIC]], and won the contest. The strategy is simply to cooperate on the first iteration of the game; after that, the player does what his or her opponent did on the previous move. Depending on the situation, a slightly better strategy can be "tit for tat with forgiveness". When the opponent defects, on the next move, the player sometimes cooperates anyway, with a small probability (around 1–5%). This allows for occasional recovery from getting trapped in a cycle of defections. The exact probability depends on the line-up of opponents.<br />
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The winning deterministic strategy was tit for tat, which Anatol Rapoport developed and entered into the tournament. It was the simplest of any program entered, containing only four lines of BASIC, and won the contest. The strategy is simply to cooperate on the first iteration of the game; after that, the player does what his or her opponent did on the previous move. Depending on the situation, a slightly better strategy can be "tit for tat with forgiveness". When the opponent defects, on the next move, the player sometimes cooperates anyway, with a small probability (around 1–5%). This allows for occasional recovery from getting trapped in a cycle of defections. The exact probability depends on the line-up of opponents.<br />
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最终获胜的决定性策略是针锋相对策略，这是阿纳托尔·拉波波特 Anatol Rapoport开发并参加比赛的策略。这是所有参赛程序中最简单的一个，只有四行 BASIC 语言，并且赢得了比赛。策略很简单，就是在游戏的第一次迭代中进行合作; 在此之后，玩家将执行做他的对手在前一步中所做的事情。根据具体情况，一个稍微好一点的策略可以是“带着宽恕之心针锋相对”。当对手叛变时，在下一次博弈中，玩家有时还是会合作，但概率很小(大约1-5%)。这允许博弈偶尔能从陷入叛变循环中恢复过来。确切的概率取决于对手的安排。<br />
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By analysing the top-scoring strategies, Axelrod stated several conditions necessary for a strategy to be successful.<br />
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By analysing the top-scoring strategies, Axelrod stated several conditions necessary for a strategy to be successful.<br />
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通过分析得分最高的战略，阿克塞尔罗德阐述了战略成功的几个必要条件。<br />
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; Nice: The most important condition is that the strategy must be "nice", that is, it will not defect before its opponent does (this is sometimes referred to as an "optimistic" algorithm). Almost all of the top-scoring strategies were nice; therefore, a purely selfish strategy will not "cheat" on its opponent, for purely self-interested reasons first.<br />
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Nice: The most important condition is that the strategy must be "nice", that is, it will not defect before its opponent does (this is sometimes referred to as an "optimistic" algorithm). Almost all of the top-scoring strategies were nice; therefore, a purely selfish strategy will not "cheat" on its opponent, for purely self-interested reasons first.<br />
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;友好:最重要的条件是策略必须是好的，也就是说，它不会在对手之前叛变(这有时被称为“乐观”算法)。几乎所有得分最高的策略都是友好的; 因此，一个纯粹的自私策略不会为了纯粹自身的利益而“欺骗”对手。<br />
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; Retaliating: However, Axelrod contended, the successful strategy must not be a blind optimist. It must sometimes retaliate. An example of a non-retaliating strategy is Always Cooperate. This is a very bad choice, as "nasty" strategies will ruthlessly exploit such players.<br />
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Retaliating: However, Axelrod contended, the successful strategy must not be a blind optimist. It must sometimes retaliate. An example of a non-retaliating strategy is Always Cooperate. This is a very bad choice, as "nasty" strategies will ruthlessly exploit such players.<br />
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;报复:然而，阿克塞尔罗德认为，成功的战略决不能是盲目的乐观主义。它有时必须进行报复。非报复策略的一个例子就是永远合作。这是一个非常糟糕的选择，因为“肮脏”的策略会无情地利用这些玩家。<br />
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; Forgiving: Successful strategies must also be forgiving. Though players will retaliate, they will once again fall back to cooperating if the opponent does not continue to defect. This stops long runs of revenge and counter-revenge, maximizing points.<br />
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Forgiving: Successful strategies must also be forgiving. Though players will retaliate, they will once again fall back to cooperating if the opponent does not continue to defect. This stops long runs of revenge and counter-revenge, maximizing points.<br />
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;宽容: 成功的策略也必须是宽容的。虽然玩家会报复，但如果对手不继续叛变，他们将再次回到合作的状态。这阻止了长时间的报复和反报复，最大限度地提高积分。<br />
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; Non-envious: The last quality is being non-envious, that is not striving to score more than the opponent.<br />
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Non-envious: The last quality is being non-envious, that is not striving to score more than the opponent.<br />
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;不嫉妒: 最后一个品质是不嫉妒，不强求比对手得分更多。<br />
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The optimal (points-maximizing) strategy for the one-time PD game is simply defection; as explained above, this is true whatever the composition of opponents may be. However, in the iterated-PD game the optimal strategy depends upon the strategies of likely opponents, and how they will react to defections and cooperations. For example, consider a population where everyone defects every time, except for a single individual following the tit for tat strategy. That individual is at a slight disadvantage because of the loss on the first turn. In such a population, the optimal strategy for that individual is to defect every time. In a population with a certain percentage of always-defectors and the rest being tit for tat players, the optimal strategy for an individual depends on the percentage, and on the length of the game.<br />
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The optimal (points-maximizing) strategy for the one-time PD game is simply defection; as explained above, this is true whatever the composition of opponents may be. However, in the iterated-PD game the optimal strategy depends upon the strategies of likely opponents, and how they will react to defections and cooperations. For example, consider a population where everyone defects every time, except for a single individual following the tit for tat strategy. That individual is at a slight disadvantage because of the loss on the first turn. In such a population, the optimal strategy for that individual is to defect every time. In a population with a certain percentage of always-defectors and the rest being tit for tat players, the optimal strategy for an individual depends on the percentage, and on the length of the game.<br />
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对于一次性的囚徒困境博弈，最优(点数最大化)策略就是简单的叛变; 正如上面所说，无论对手的构成如何，这都是正确的。然而，在迭代囚徒困境博弈中，最优策略取决于可能的对手的策略，以及他们对叛变和合作的反应。例如，考虑一个群体，其中每个人每次都会叛变，只有一个人遵循针锋相对的策略。那个人就会由于第一回合的失利而处于轻微的不利地位。在这样一个群体中，个体的最佳策略是每次都叛变。在一定比例的总是选择背叛的玩家和其余组成选择针锋相对策略的玩家的人群中，个人的最佳策略取决于这一比例和博弈的次数。<br />
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In the strategy called Pavlov, [[win-stay, lose-switch]], faced with a failure to cooperate, the player switches strategy the next turn.<ref>http://www.pnas.org/content/pnas/93/7/2686.full.pdf</ref> In certain circumstances,{{specify|date=November 2012}} Pavlov beats all other strategies by giving preferential treatment to co-players using a similar strategy.<br />
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In the strategy called Pavlov, win-stay, lose-switch, faced with a failure to cooperate, the player switches strategy the next turn. In certain circumstances, Pavlov beats all other strategies by giving preferential treatment to co-players using a similar strategy.<br />
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在所谓的<font color="#ff8000">巴甫洛夫策略 Pavlov strategy</font>中，赢-保持，输-变换，面对一次合作失败，玩家将在下一次变换策略。<ref>http://www.pnas.org/content/pnas/93/7/2686.full.pdf</ref>在某些情况下，{{specify|date=November 2012}}巴甫洛夫通过使用类似策略给与合作者优惠待遇打败了其他所有策略。<br />
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Deriving the optimal strategy is generally done in two ways:<br />
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Deriving the optimal strategy is generally done in two ways:<br />
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得出最佳策略通常有两种方法:<br />
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* [[Bayesian Nash equilibrium]]: If the statistical distribution of opposing strategies can be determined (e.g. 50% tit for tat, 50% always cooperate) an optimal counter-strategy can be derived analytically.{{efn|1=For example see the 2003 study<ref>{{cite web|url= http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|title=Bayesian Nash equilibrium; a statistical test of the hypothesis|url-status=dead|archive-url= https://web.archive.org/web/20051002195142/http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|archive-date=2005-10-02|publisher=[[Tel Aviv University]]}}</ref> for discussion of the concept and whether it can apply in real [[economic]] or strategic situations.}}<br />
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<font color="#ff8000">贝叶斯纳什均衡 Bayesian Nash equilibrium</font>：如果可以确定对立策略的统计分布（例如，50%针锋相对，50%总是合作），那么，可以通过分析得出最佳的反策略{{efn|1=例如2003年的研究<ref>{{cite web|url= http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|title=Bayesian Nash equilibrium; a statistical test of the hypothesis|url-status=dead|archive-url= https://web.archive.org/web/20051002195142/http://econ.hevra.haifa.ac.il/~mbengad/seminars/whole1.pdf|archive-date=2005-10-02|publisher=[[Tel Aviv University]]}}</ref>讨论这一概念以及它是否可以应用于实际经济或战略情况。}}<br />
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* [[Monte Carlo method|Monte Carlo]] simulations of populations have been made, where individuals with low scores die off, and those with high scores reproduce (a [[genetic algorithm]] for finding an optimal strategy). The mix of algorithms in the final population generally depends on the mix in the initial population. The introduction of mutation (random variation during reproduction) lessens the dependency on the initial population; empirical experiments with such systems tend to produce tit for tat players (see for instance Chess 1988),{{Clarify|date=August 2016}} but no analytic proof exists that this will always occur.<ref>{{Citation|last=Wu|first=Jiadong|title=Cooperation on the Monte Carlo Rule: Prisoner's Dilemma Game on the Grid|date=2019|work=Theoretical Computer Science|volume=1069|pages=3–15|editor-last=Sun|editor-first=Xiaoming|publisher=Springer Singapore|language=en|doi=10.1007/978-981-15-0105-0_1|isbn=978-981-15-0104-3|last2=Zhao|first2=Chengye|editor2-last=He|editor2-first=Kun|editor3-last=Chen|editor3-first=Xiaoyun}}</ref><br />
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<font color="#ff8000">蒙特卡洛方法 Monte Carlo method </font>已经对种群进行了模拟，分数低的个体死亡，分数高的个体繁殖（<font color="#ff8000">遗传算法 genetic algorithm </font>用于寻找一个最佳策略）。最终群体中的算法组合通常取决于初始总体的组合。引入突变（繁殖过程中的随机变异）可以减少对初始种群的依赖性。使用这种系统进行经验性实验往往会为针锋相对的玩家带来麻烦（见Chess 1988），{{Clarify|date=August 2016}}，但是没有分析证据表明这种情况会一直发生。<ref>{{Citation|last=Wu|first=Jiadong|title=Cooperation on the Monte Carlo Rule: Prisoner's Dilemma Game on the Grid|date=2019|work=Theoretical Computer Science|volume=1069|pages=3–15|editor-last=Sun|editor-first=Xiaoming|publisher=Springer Singapore|language=en|doi=10.1007/978-981-15-0105-0_1|isbn=978-981-15-0104-3|last2=Zhao|first2=Chengye|editor2-last=He|editor2-first=Kun|editor3-last=Chen|editor3-first=Xiaoyun}}</ref><br />
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Although tit for tat is considered to be the most [[robust]] basic strategy, a team from [[Southampton University]] in England introduced a new strategy at the 20th-anniversary iterated prisoner's dilemma competition, which proved to be more successful than tit for tat. This strategy relied on collusion between programs to achieve the highest number of points for a single program. The university submitted 60 programs to the competition, which were designed to recognize each other through a series of five to ten moves at the start.<ref>{{cite press release|url= http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|publisher=University of Southampton|title=University of Southampton team wins Prisoner's Dilemma competition|date=7 October 2004|url-status=dead|archive-url= https://web.archive.org/web/20140421055745/http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|archive-date=2014-04-21}}</ref> Once this recognition was made, one program would always cooperate and the other would always defect, assuring the maximum number of points for the defector. If the program realized that it was playing a non-Southampton player, it would continuously defect in an attempt to minimize the score of the competing program. As a result, the 2004 Prisoners' Dilemma Tournament results show [[University of Southampton]]'s strategies in the first three places, despite having fewer wins and many more losses than the GRIM strategy. (In a PD tournament, the aim of the game is not to "win" matches&nbsp;– that can easily be achieved by frequent defection). Also, even without implicit collusion between [[computer program|software strategies]] (exploited by the Southampton team) tit for tat is not always the absolute winner of any given tournament; it would be more precise to say that its long run results over a series of tournaments outperform its rivals. (In any one event a given strategy can be slightly better adjusted to the competition than tit for tat, but tit for tat is more robust). The same applies for the tit for tat with forgiveness variant, and other optimal strategies: on any given day they might not "win" against a specific mix of counter-strategies. An alternative way of putting it is using the Darwinian [[Evolutionarily stable strategy|ESS]] simulation. In such a simulation, tit for tat will almost always come to dominate, though nasty strategies will drift in and out of the population because a tit for tat population is penetrable by non-retaliating nice strategies, which in turn are easy prey for the nasty strategies. [[Richard Dawkins]] showed that here, no static mix of strategies form a stable equilibrium and the system will always oscillate between bounds.}} this strategy ended up taking the top three positions in the competition, as well as a number of positions towards the bottom.<br />
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Although tit for tat is considered to be the most robust basic strategy, a team from Southampton University in England introduced a new strategy at the 20th-anniversary iterated prisoner's dilemma competition, which proved to be more successful than tit for tat. This strategy relied on collusion between programs to achieve the highest number of points for a single program. The university submitted 60 programs to the competition, which were designed to recognize each other through a series of five to ten moves at the start. Once this recognition was made, one program would always cooperate and the other would always defect, assuring the maximum number of points for the defector. If the program realized that it was playing a non-Southampton player, it would continuously defect in an attempt to minimize the score of the competing program. As a result, the 2004 Prisoners' Dilemma Tournament results show University of Southampton's strategies in the first three places, despite having fewer wins and many more losses than the GRIM strategy. (In a PD tournament, the aim of the game is not to "win" matches&nbsp;– that can easily be achieved by frequent defection). Also, even without implicit collusion between software strategies (exploited by the Southampton team) tit for tat is not always the absolute winner of any given tournament; it would be more precise to say that its long run results over a series of tournaments outperform its rivals. (In any one event a given strategy can be slightly better adjusted to the competition than tit for tat, but tit for tat is more robust). The same applies for the tit for tat with forgiveness variant, and other optimal strategies: on any given day they might not "win" against a specific mix of counter-strategies. An alternative way of putting it is using the Darwinian ESS simulation. In such a simulation, tit for tat will almost always come to dominate, though nasty strategies will drift in and out of the population because a tit for tat population is penetrable by non-retaliating nice strategies, which in turn are easy prey for the nasty strategies. Richard Dawkins showed that here, no static mix of strategies form a stable equilibrium and the system will always oscillate between bounds.}} this strategy ended up taking the top three positions in the competition, as well as a number of positions towards the bottom.<br />
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尽管针锋相对被认为是最有力的基本策略，来自英格兰南安普敦大学的一个团队在20周年的迭代囚徒困境竞赛中提出了一个新策略，这个策略被证明比针锋相对更为成功。这种策略依赖于程序之间的串通，以获得单个程序的最高分数。这所大学提交了60个程序，这些程序的设计目的是在比赛开始时通过一系列的5到10个动作来互相认识。<ref>{{cite press release|url= http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|publisher=University of Southampton|title=University of Southampton team wins Prisoner's Dilemma competition|date=7 October 2004|url-status=dead|archive-url= https://web.archive.org/web/20140421055745/http://www.southampton.ac.uk/mediacentre/news/2004/oct/04_151.shtml|archive-date=2014-04-21}}</ref>一旦认识建立，一个程序总是合作，另一个程序总是叛变，保证叛变者得到最多的分数。如果这个程序意识到它正在和一个非南安普顿的球员比赛，它会不断地叛变，试图最小化与之竞争的程序的得分。因此，2004年囚徒困境锦标赛的结果显示了南安普敦大学战略位居前三名，尽管它比冷酷战略赢得更少，输的更多。(在囚徒困境锦标赛中，比赛的目的不是“赢”比赛——这一点频繁叛变很容易实现)。此外，即使没有软件策略之间的暗中串通(南安普顿队利用了这一点) ，针锋相对并不总是任何特定锦标赛的绝对赢家; 更准确地说，它是在一系列锦标赛中的长期结果超过了它的竞争对手。(在任何一个事件中，一个给定的策略可以比针锋相对稍微更好地适应竞争，但是针锋相对更稳健)。这同样适用于带有宽恕变量的针锋相对，和其他最佳策略: 在任何特定的一天，他们可能不会“赢得”一个特定的混合反战略。另一种方法是使用达尔文 Darwinian的<font color="#ff8000"> ESS模拟 ESS simulation</font>。在这样的模拟中，针锋相对几乎总是占主导地位，尽管讨厌的策略会在人群中漂移，因为使用针锋相对策略的人群可以通过非报复性的好策略进行渗透，这反过来使他们容易成为讨厌策略的猎物。理查德·道金斯 Richard Dawkins指出，在这里，没有静态的混合策略会形成一个稳定的平衡，系统将始终在边界之间振荡。这种策略最终在比赛中获得了前三名的成绩，或者是接近垫底的成绩。<br />
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This strategy takes advantage of the fact that multiple entries were allowed in this particular competition and that the performance of a team was measured by that of the highest-scoring player (meaning that the use of self-sacrificing players was a form of [[minmaxing]]). In a competition where one has control of only a single player, tit for tat is certainly a better strategy. Because of this new rule, this competition also has little theoretical significance when analyzing single agent strategies as compared to Axelrod's seminal tournament. However, it provided a basis for analysing how to achieve cooperative strategies in multi-agent frameworks, especially in the presence of noise. In fact, long before this new-rules tournament was played, Dawkins, in his book ''[[The Selfish Gene]]'', pointed out the possibility of such strategies winning if multiple entries were allowed, but he remarked that most probably Axelrod would not have allowed them if they had been submitted. It also relies on circumventing rules about the prisoner's dilemma in that there is no communication allowed between the two players, which the Southampton programs arguably did with their opening "ten move dance" to recognize one another; this only reinforces just how valuable communication can be in shifting the balance of the game.<br />
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This strategy takes advantage of the fact that multiple entries were allowed in this particular competition and that the performance of a team was measured by that of the highest-scoring player (meaning that the use of self-sacrificing players was a form of minmaxing). In a competition where one has control of only a single player, tit for tat is certainly a better strategy. Because of this new rule, this competition also has little theoretical significance when analyzing single agent strategies as compared to Axelrod's seminal tournament. However, it provided a basis for analysing how to achieve cooperative strategies in multi-agent frameworks, especially in the presence of noise. In fact, long before this new-rules tournament was played, Dawkins, in his book The Selfish Gene, pointed out the possibility of such strategies winning if multiple entries were allowed, but he remarked that most probably Axelrod would not have allowed them if they had been submitted. It also relies on circumventing rules about the prisoner's dilemma in that there is no communication allowed between the two players, which the Southampton programs arguably did with their opening "ten move dance" to recognize one another; this only reinforces just how valuable communication can be in shifting the balance of the game.<br />
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这种策略利用了这样一个事实，即在这场特殊的比赛中允许多个参赛项目，并且团队的表现由得分最高的项目来衡量(这意味着使用自我牺牲的项目是一种分数最大化的形式)。在一个只能控制一个玩家的比赛中，针锋相对当然是一个更好的策略。由于这一新规则的存在，与阿克塞尔罗德的具有深远影响的竞赛相比，这种竞赛在分析单个主体策略时也就没有什么理论意义。然而，它为在分析多主体框架下，特别是在存在干扰的情况下，如何实现协作策略提供了基础。事实上，早在这场新规则锦标赛开始之前，道金斯就在他的《自私的基因》一书中指出，如果允许多次参赛，这种策略就有可能获胜，但他说，如果提交这种策略的话，阿克塞尔罗德很可能不会允许。因为它依赖于规避囚徒困境的规则，即两个参与者之间不允许交流，南安普顿的项目可以说在开场的“十步舞”中就是这样做以认识对方的; 这只是强调了交流在改变游戏平衡方面的价值。<br />
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===Stochastic iterated prisoner's dilemma===<br />
随机的迭代囚徒困境<br />
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In a stochastic iterated prisoner's dilemma game, strategies are specified by in terms of "cooperation probabilities".<ref name=Press2012>{{cite journal|last1=Press|first1=WH|last2=Dyson|first2=FJ|title=Iterated Prisoner's Dilemma contains strategies that dominate any evolutionary opponent|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|date=26 June 2012|volume=109|issue=26|pages=10409–13|doi=10.1073/pnas.1206569109|pmid=22615375|pmc=3387070|bibcode=2012PNAS..10910409P}}</ref> In an encounter between player ''X'' and player ''Y'', ''X'' 's strategy is specified by a set of probabilities ''P'' of cooperating with ''Y''. ''P'' is a function of the outcomes of their previous encounters or some subset thereof. If ''P'' is a function of only their most recent ''n'' encounters, it is called a "memory-n" strategy. A memory-1 strategy is then specified by four cooperation probabilities: <math>P=\{P_{cc},P_{cd},P_{dc},P_{dd}\}</math>, where <math>P_{ab}</math> is the probability that ''X'' will cooperate in the present encounter given that the previous encounter was characterized by (ab). For example, if the previous encounter was one in which ''X'' cooperated and ''Y'' defected, then <math>P_{cd}</math> is the probability that ''X'' will cooperate in the present encounter. If each of the probabilities are either 1 or 0, the strategy is called deterministic. An example of a deterministic strategy is the tit for tat strategy written as ''P''={1,0,1,0}, in which ''X'' responds as ''Y'' did in the previous encounter. Another is the [[win–stay, lose–switch]] strategy written as ''P''={1,0,0,1}, in which ''X'' responds as in the previous encounter, if it was a "win" (i.e. cc or dc) but changes strategy if it was a loss (i.e. cd or dd). It has been shown that for any memory-n strategy there is a corresponding memory-1 strategy which gives the same statistical results, so that only memory-1 strategies need be considered.<ref name="Press2012"/><br />
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In a stochastic iterated prisoner's dilemma game, strategies are specified by in terms of "cooperation probabilities". In an encounter between player X and player Y, X 's strategy is specified by a set of probabilities P of cooperating with Y. P is a function of the outcomes of their previous encounters or some subset thereof. If P is a function of only their most recent n encounters, it is called a "memory-n" strategy. A memory-1 strategy is then specified by four cooperation probabilities: <math>P=\{P_{cc},P_{cd},P_{dc},P_{dd}\}</math>, where <math>P_{ab}</math> is the probability that X will cooperate in the present encounter given that the previous encounter was characterized by (ab). For example, if the previous encounter was one in which X cooperated and Y defected, then <math>P_{cd}</math> is the probability that X will cooperate in the present encounter. If each of the probabilities are either 1 or 0, the strategy is called deterministic. An example of a deterministic strategy is the tit for tat strategy written as P={1,0,1,0}, in which X responds as Y did in the previous encounter. Another is the win–stay, lose–switch strategy written as P={1,0,0,1}, in which X responds as in the previous encounter, if it was a "win" (i.e. cc or dc) but changes strategy if it was a loss (i.e. cd or dd). It has been shown that for any memory-n strategy there is a corresponding memory-1 strategy which gives the same statistical results, so that only memory-1 strategies need be considered.<br />
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在随机迭代<font color="#ff8000"> 囚徒困境prisoner's dilemma</font>博弈中，策略由“合作概率”来确定。<ref name=Press2012>{{cite journal|last1=Press|first1=WH|last2=Dyson|first2=FJ|title=Iterated Prisoner's Dilemma contains strategies that dominate any evolutionary opponent|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|date=26 June 2012|volume=109|issue=26|pages=10409–13|doi=10.1073/pnas.1206569109|pmid=22615375|pmc=3387070|bibcode=2012PNAS..10910409P}}</ref>在玩家''X''和玩家''Y''之间的遭遇中，''X''‘s的策略由一组与''Y''合作的概率''P''确定，''P''是他们之前遭遇的结果的函数，或者是其中的一些子集。如果''P''只是它们最近遇到次数 ''n''的函数，那么它被称为“记忆-n”策略。我们可以由四个联合概率指定一个记忆-1策略: <math>P=\{P_{cc},P_{cd},P_{dc},P_{dd}\}</math>，其中<math>P_{ab}</math>是在当前遭遇中基于先前联合的概率。如果每个概率都是1或0，这种策略称为确定性策略。确定性策略的一个例子是针锋相对策略，写成 p {1,0,1,0} ，其中 x 的反应和 y 在前一次遭遇中的反应一样。另一种是胜-保持-败-转换策略，它被写成 p {1,0,0,1} ，在这种策略中，如果 x 获得胜利(即:cc 或 dc)，x会做出与上一次遭遇一样的反应 ，但如果失败，x会改变策略(即cd 或 dd)。研究表明，对于任何一种记忆-n 策略，存在一个相应的记忆-1策略，这个策略给出相同的统计结果，因此只需要考虑记忆-1策略。<ref name="Press2012"/><br />
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If we define ''P'' as the above 4-element strategy vector of ''X'' and <math>Q=\{Q_{cc},Q_{cd},Q_{dc},Q_{dd}\}</math> as the 4-element strategy vector of ''Y'', a transition matrix ''M'' may be defined for ''X'' whose ''ij'' th entry is the probability that the outcome of a particular encounter between ''X'' and ''Y'' will be ''j'' given that the previous encounter was ''i'', where ''i'' and ''j'' are one of the four outcome indices: ''cc'', ''cd'', ''dc'', or ''dd''. For example, from ''X'' 's point of view, the probability that the outcome of the present encounter is ''cd'' given that the previous encounter was ''cd'' is equal to <math>M_{cd,cd}=P_{cd}(1-Q_{dc})</math>. (The indices for ''Q'' are from ''Y'' 's point of view: a ''cd'' outcome for ''X'' is a ''dc'' outcome for ''Y''.) Under these definitions, the iterated prisoner's dilemma qualifies as a [[stochastic process]] and ''M'' is a [[stochastic matrix]], allowing all of the theory of stochastic processes to be applied.<ref name="Press2012"/><br />
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If we define P as the above 4-element strategy vector of X and <math>Q=\{Q_{cc},Q_{cd},Q_{dc},Q_{dd}\}</math> as the 4-element strategy vector of Y, a transition matrix M may be defined for X whose ij th entry is the probability that the outcome of a particular encounter between X and Y will be j given that the previous encounter was i, where i and j are one of the four outcome indices: cc, cd, dc, or dd. For example, from X 's point of view, the probability that the outcome of the present encounter is cd given that the previous encounter was cd is equal to <math>M_{cd,cd}=P_{cd}(1-Q_{dc})</math>. (The indices for Q are from Y 's point of view: a cd outcome for X is a dc outcome for Y.) Under these definitions, the iterated prisoner's dilemma qualifies as a stochastic process and M is a stochastic matrix, allowing all of the theory of stochastic processes to be applied.<br />
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如果我们将''P''定义为''X''的上述4元策略向量，并将<math>Q=\{Q_{cc},Q_{cd},Q_{dc},Q_{dd}\}</math>定义为''Y''的4元策略向量，则对于''X''可以定义一个转移矩阵''M''，其第ij项是''X''和''Y''之间特定相遇的结果为j的概率，给定i，其中i和j是cc、cd、dc或dd 四个结果索引中的一个。例如，从''X''的角度来看，如果给定''cd''，那么这次的结果是''cd''的概率等于<math>M_{cd,cd}=P_{cd}(1-Q_{dc})</math>。(''Q''的指标是 从''Y''的角度: ''X''的''cd''结果是''Y''的''dc''结果)在这些定义下，重复的囚徒困境被定义为一个随机过程，''M''是一个随机矩阵，允许应用所有的随机过程理论。<ref name="Press2012"/><br />
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One result of stochastic theory is that there exists a stationary vector ''v'' for the matrix ''M'' such that <math>v\cdot M=v</math>. Without loss of generality, it may be specified that ''v'' is normalized so that the sum of its four components is unity. The ''ij'' th entry in <math>M^n</math> will give the probability that the outcome of an encounter between ''X'' and ''Y'' will be ''j'' given that the encounter ''n'' steps previous is ''i''. In the limit as ''n'' approaches infinity, ''M'' will converge to a matrix with fixed values, giving the long-term probabilities of an encounter producing ''j'' which will be independent of ''i''. In other words, the rows of <math>M^\infty</math> will be identical, giving the long-term equilibrium result probabilities of the iterated prisoners dilemma without the need to explicitly evaluate a large number of interactions. It can be seen that ''v'' is a stationary vector for <math>M^n</math> and particularly <math>M^\infty</math>, so that each row of <math>M^\infty</math> will be equal to ''v''. Thus the stationary vector specifies the equilibrium outcome probabilities for ''X''. Defining <math>S_x=\{R,S,T,P\}</math> and <math>S_y=\{R,T,S,P\}</math> as the short-term payoff vectors for the {cc,cd,dc,dd} outcomes (From ''X'' 's point of view), the equilibrium payoffs for ''X'' and ''Y'' can now be specified as <math>s_x=v\cdot S_x</math> and <math>s_y=v\cdot S_y</math>, allowing the two strategies ''P'' and ''Q'' to be compared for their long term payoffs.<br />
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One result of stochastic theory is that there exists a stationary vector v for the matrix M such that <math>v\cdot M=v</math>. Without loss of generality, it may be specified that v is normalized so that the sum of its four components is unity. The ij th entry in <math>M^n</math> will give the probability that the outcome of an encounter between X and Y will be j given that the encounter n steps previous is i. In the limit as n approaches infinity, M will converge to a matrix with fixed values, giving the long-term probabilities of an encounter producing j which will be independent of i. In other words, the rows of <math>M^\infty</math> will be identical, giving the long-term equilibrium result probabilities of the iterated prisoners dilemma without the need to explicitly evaluate a large number of interactions. It can be seen that v is a stationary vector for <math>M^n</math> and particularly <math>M^\infty</math>, so that each row of <math>M^\infty</math> will be equal to v. Thus the stationary vector specifies the equilibrium outcome probabilities for X. Defining <math>S_x=\{R,S,T,P\}</math> and <math>S_y=\{R,T,S,P\}</math> as the short-term payoff vectors for the {cc,cd,dc,dd} outcomes (From X 's point of view), the equilibrium payoffs for X and Y can now be specified as <math>s_x=v\cdot S_x</math> and <math>s_y=v\cdot S_y</math>, allowing the two strategies P and Q to be compared for their long term payoffs.<br />
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随机理论的一个结果是，矩阵''M''存在一个平稳向量''v''使得<math>v\cdot M=v</math>成立。一般地，我们可以指定''v''是标准化的，因此它的4个组成部分之和为1。the equilibrium payoffs for and can now be specified as and, allowing the two strategies ''P'' and ''Q'' to be compared for their long term payoffs.第''ij''项<math>M^n</math>给出了''X''和''Y''相遇的结果的概率为''j''，给定前面相遇''n''步的概率是''i''。当''n''趋于无穷时，''M''收敛于一个具有固定值的矩阵，并且''j''趋向一个长期概率，与''i''独立。换句话说， <math>M^\infty</math>的行将是相同的，从而给出了重复囚徒困境的长期均衡结果概率，而不需要明确地计算大量的相互作用。可以看出，''v''是<math>M^n</math>特别是<math>M^\infty</math>, 的平稳向量，因此<math>M^\infty</math>的每一行都等于''v''。因此平稳向量指定了''X''的均衡结果概率。定义<math>S_x=\{R,S,T,P\}</math>和<math>S_y=\{R,T,S,P\}</math>作为{cc,cd,dc,dd}结果的短期收益向量(从''X''的角度来看) ，现在可以将''X''和''Y''的均衡收益指定为<math>s_x=v\cdot S_x</math>和<math>s_y=v\cdot S_y</math>，使得''P''、''Q''两种策略的长期收益可以比较。<br />
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====Zero-determinant strategies====<br />
<font color="#ff8000">零决定策略 Zero-determinant strategies</font><br />
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[[File:IPD Venn.svg|right|thumb|upright=2.5|The relationship between zero-determinant (ZD), cooperating and defecting strategies in the iterated prisoner's dilemma (IPD) illustrated in a [[Venn diagram]]. Cooperating strategies always cooperate with other cooperating strategies, and defecting strategies always defect against other defecting strategies. Both contain subsets of strategies that are robust under strong selection, meaning no other memory-1 strategy is selected to invade such strategies when they are resident in a population. Only cooperating strategies contain a subset that are always robust, meaning that no other memory-1 strategy is selected to invade and replace such strategies, under both strong and [[weak selection]]. The intersection between ZD and good cooperating strategies is the set of generous ZD strategies. Extortion strategies are the intersection between ZD and non-robust defecting strategies. Tit-for-tat lies at the intersection of cooperating, defecting and ZD strategies.]]<br />
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The relationship between zero-determinant (ZD), cooperating and defecting strategies in the iterated prisoner's dilemma (IPD) illustrated in a [[Venn diagram. Cooperating strategies always cooperate with other cooperating strategies, and defecting strategies always defect against other defecting strategies. Both contain subsets of strategies that are robust under strong selection, meaning no other memory-1 strategy is selected to invade such strategies when they are resident in a population. Only cooperating strategies contain a subset that are always robust, meaning that no other memory-1 strategy is selected to invade and replace such strategies, under both strong and weak selection. The intersection between ZD and good cooperating strategies is the set of generous ZD strategies. Extortion strategies are the intersection between ZD and non-robust defecting strategies. Tit-for-tat lies at the intersection of cooperating, defecting and ZD strategies.]]<br />
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<font color="#ff8000">维恩图 Venn diagram</font>中讨论了<font color="#ff8000">迭代囚徒困境 iterated prisoner's dilemma</font>(IPD)中零决定策略(ZD)、合作策略和背叛策略之间的关系。合作策略总是与其他合作策略相互配合，而背叛策略总是与其他背叛策略相抵触。这两种策略都包都含在强选择下稳健的策略子集，这意味着当它们驻留在一个种群中时，没有选择其他的记忆-1策略来入侵此策略。只有合作策略包含在始终稳健的策略子集，意味着无论选择强项还是弱项，都不会选择其他任何记忆-1策略来入侵和替换此策略。零决定策略和良好的合作策略之间的交集是一组宽松的零决定策略。勒索策略是零决定策略和非稳健背叛策略的交集。针锋相对是合作、背叛和零决定策略的交集。<br />
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In 2012, [[William H. Press]] and [[Freeman Dyson]] published a new class of strategies for the stochastic iterated prisoner's dilemma called "zero-determinant" (ZD) strategies.<ref name="Press2012"/> The long term payoffs for encounters between ''X'' and ''Y'' can be expressed as the determinant of a matrix which is a function of the two strategies and the short term payoff vectors: <math>s_x=D(P,Q,S_x)</math> and <math>s_y=D(P,Q,S_y)</math>, which do not involve the stationary vector ''v''. Since the determinant function <math>s_y=D(P,Q,f)</math> is linear in ''f'', it follows that <math>\alpha s_x+\beta s_y+\gamma=D(P,Q,\alpha S_x+\beta S_y+\gamma U)</math> (where ''U''={1,1,1,1}). Any strategies for which <math>D(P,Q,\alpha S_x+\beta S_y+\gamma U)=0</math> is by definition a ZD strategy, and the long term payoffs obey the relation <math>\alpha s_x+\beta s_y+\gamma=0</math>.<br />
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In 2012, William H. Press and Freeman Dyson published a new class of strategies for the stochastic iterated prisoner's dilemma called "zero-determinant" (ZD) strategies. The long term payoffs for encounters between X and Y can be expressed as the determinant of a matrix which is a function of the two strategies and the short term payoff vectors: <math>s_x=D(P,Q,S_x)</math> and <math>s_y=D(P,Q,S_y)</math>, which do not involve the stationary vector v. Since the determinant function <math>s_y=D(P,Q,f)</math> is linear in f, it follows that <math>\alpha s_x+\beta s_y+\gamma=D(P,Q,\alpha S_x+\beta S_y+\gamma U)</math> (where U={1,1,1,1}). Any strategies for which <math>D(P,Q,\alpha S_x+\beta S_y+\gamma U)=0</math> is by definition a ZD strategy, and the long term payoffs obey the relation <math>\alpha s_x+\beta s_y+\gamma=0</math>.<br />
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2012年，威廉·H·普莱斯 William H. Press和弗里曼·戴森 Freeman Dyson针对随机迭代囚徒困境提出了一类新的策略，称为“零决定”策略。<ref name="Press2012"/>''X''和''Y''之间的长期收益可以表示为一个矩阵的决定因素，它是两个策略和短期收益向量的函数: <math>s_x=D(P,Q,S_x)</math>和<math>s_y=D(P,Q,S_y)</math>，不涉及平稳向量''v''。 由于行列式函数<math>s_y=D(P,Q,f)</math>在''f''中是线性的，因此可以推出<math>\alpha s_x+\beta s_y+\gamma=D(P,Q,\alpha S_x+\beta S_y+\gamma U)</math>(其中''U''={1,1,1,1})。任何策略的<math>D(P,Q,\alpha S_x+\beta S_y+\gamma U)=0</math>被定义为零决定策略，长期收益服从关系式<math>\alpha s_x+\beta s_y+\gamma=0</math>。<br />
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Tit-for-tat is a ZD strategy which is "fair" in the sense of not gaining advantage over the other player. However, the ZD space also contains strategies that, in the case of two players, can allow one player to unilaterally set the other player's score or alternatively, force an evolutionary player to achieve a payoff some percentage lower than his own. The extorted player could defect but would thereby hurt himself by getting a lower payoff. Thus, extortion solutions turn the iterated prisoner's dilemma into a sort of [[ultimatum game]]. Specifically, ''X'' is able to choose a strategy for which <math>D(P,Q,\beta S_y+\gamma U)=0</math>, unilaterally setting <math>s_y</math> to a specific value within a particular range of values, independent of ''Y'' 's strategy, offering an opportunity for ''X'' to "extort" player ''Y'' (and vice versa). (It turns out that if ''X'' tries to set <math>s_x</math> to a particular value, the range of possibilities is much smaller, only consisting of complete cooperation or complete defection.<ref name="Press2012"/>)<br />
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Tit-for-tat is a ZD strategy which is "fair" in the sense of not gaining advantage over the other player. However, the ZD space also contains strategies that, in the case of two players, can allow one player to unilaterally set the other player's score or alternatively, force an evolutionary player to achieve a payoff some percentage lower than his own. The extorted player could defect but would thereby hurt himself by getting a lower payoff. Thus, extortion solutions turn the iterated prisoner's dilemma into a sort of ultimatum game. Specifically, X is able to choose a strategy for which <math>D(P,Q,\beta S_y+\gamma U)=0</math>, unilaterally setting <math>s_y</math> to a specific value within a particular range of values, independent of Y 's strategy, offering an opportunity for X to "extort" player Y (and vice versa). (It turns out that if X tries to set <math>s_x</math> to a particular value, the range of possibilities is much smaller, only consisting of complete cooperation or complete defection.)<br />
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针锋相对是一种零决定策略，在不获得超越其他玩家优势的意义下是“公平”的。然而，零决定策略空间还包含这样的策略：在两个玩家的情况下，可以允许一个玩家单方面设置另一个玩家的分数，或者强迫进化的玩家获得比他自己的分数低一些的收益。被勒索的玩家可能会背叛，但会因此获得较低的回报并且受到伤害。因此，勒索的解决方案将迭代囚徒困境转化为一种<font color="#ff8000">最后通牒博弈 ultimatum game </font>。具体来说，''X''能够选择一种策略，对于这种策略，<math>D(P,Q,\beta S_y+\gamma U)=0</math>单方面地将<math>s_y</math>设置为一个特定值范围内的特定值，与''Y''的策略无关，为''X''提供了“勒索”玩家''Y''的机会(反之亦然)。(事实证明，如果''X''试图将<math>s_x</math>设置为一个特定的值，那么可能的范围要小得多，只包括完全合作或完全叛变。<ref name="Press2012"/>)<br />
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An extension of the IPD is an evolutionary stochastic IPD, in which the relative abundance of particular strategies is allowed to change, with more successful strategies relatively increasing. This process may be accomplished by having less successful players imitate the more successful strategies, or by eliminating less successful players from the game, while multiplying the more successful ones. It has been shown that unfair ZD strategies are not [[evolutionarily stable strategy|evolutionarily stable]]. The key intuition is that an evolutionarily stable strategy must not only be able to invade another population (which extortionary ZD strategies can do) but must also perform well against other players of the same type (which extortionary ZD players do poorly, because they reduce each other's surplus).<ref>{{cite journal|last=Adami|first=Christoph|author2=Arend Hintze|title=Evolutionary instability of Zero Determinant strategies demonstrates that winning isn't everything|journal=Nature Communications|volume=4|year=2013|page=3|arxiv=1208.2666|doi=10.1038/ncomms3193|pmid=23903782|pmc=3741637|bibcode=2013NatCo...4.2193A}}</ref><br />
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An extension of the IPD is an evolutionary stochastic IPD, in which the relative abundance of particular strategies is allowed to change, with more successful strategies relatively increasing. This process may be accomplished by having less successful players imitate the more successful strategies, or by eliminating less successful players from the game, while multiplying the more successful ones. It has been shown that unfair ZD strategies are not evolutionarily stable. The key intuition is that an evolutionarily stable strategy must not only be able to invade another population (which extortionary ZD strategies can do) but must also perform well against other players of the same type (which extortionary ZD players do poorly, because they reduce each other's surplus).<br />
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迭代囚徒困境的一个扩展是进化的随机迭代囚徒困境，其中允许特定策略的相对丰度改变，更成功的策略相对增加。这个过程可以通过让不太成功的玩家模仿更成功的策略，或者通过从游戏中淘汰不太成功的玩家，同时让更成功的玩家成倍增加。研究表明，不公平的零决定策略不是进化稳定策略。关键的直觉告诉我们，进化稳定策略不仅要能够入侵另一个群体(这是勒索零决定策略可以做到的) ，而且还要在同类型的其他玩家面前表现良好(勒索零决定策略玩家表现不佳，因为他们减少了彼此的盈余)。<ref>{{cite journal|last=Adami|first=Christoph|author2=Arend Hintze|title=Evolutionary instability of Zero Determinant strategies demonstrates that winning isn't everything|journal=Nature Communications|volume=4|year=2013|page=3|arxiv=1208.2666|doi=10.1038/ncomms3193|pmid=23903782|pmc=3741637|bibcode=2013NatCo...4.2193A}}</ref><br />
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Theory and simulations confirm that beyond a critical population size, ZD extortion loses out in evolutionary competition against more cooperative strategies, and as a result, the average payoff in the population increases when the population is larger. In addition, there are some cases in which extortioners may even catalyze cooperation by helping to break out of a face-off between uniform defectors and [[win–stay, lose–switch]] agents.<ref name=Hilbe2013 /><br />
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Theory and simulations confirm that beyond a critical population size, ZD extortion loses out in evolutionary competition against more cooperative strategies, and as a result, the average payoff in the population increases when the population is larger. In addition, there are some cases in which extortioners may even catalyze cooperation by helping to break out of a face-off between uniform defectors and win–stay, lose–switch agents.<br />
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理论和模拟证实，超过一个临界种群规模，零决定勒索在与更多合作策略的进化竞争中会失败，因此，种群越大，种群的平均收益就越大。此外，在某些情况下，勒索者甚至可能通过帮助打破统一的背叛者与使用“赢-保持-输”策略的转换玩家之间的对峙而促进合作。<ref name=Hilbe2013 /><br />
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While extortionary ZD strategies are not stable in large populations, another ZD class called "generous" strategies ''is'' both stable and robust. In fact, when the population is not too small, these strategies can supplant any other ZD strategy and even perform well against a broad array of generic strategies for iterated prisoner's dilemma, including win–stay, lose–switch. This was proven specifically for the [[Prisoner's dilemma#Special case: Donation game|donation game]] by Alexander Stewart and Joshua Plotkin in 2013.<ref name=Stewart2013>{{cite journal|last=Stewart|first=Alexander J.|author2=Joshua B. Plotkin|title=From extortion to generosity, evolution in the Iterated Prisoner's Dilemma|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|year=2013|doi=10.1073/pnas.1306246110|pmid=24003115|volume=110|issue=38|pages=15348–53|bibcode=2013PNAS..11015348S|pmc=3780848}}</ref> Generous strategies will cooperate with other cooperative players, and in the face of defection, the generous player loses more utility than its rival. Generous strategies are the intersection of ZD strategies and so-called "good" strategies, which were defined by Akin (2013)<ref name=Akin2013>{{cite arxiv|last=Akin|first=Ethan|title=Stable Cooperative Solutions for the Iterated Prisoner's Dilemma|year=2013|page=9|class=math.DS|eprint=1211.0969}} {{bibcode|2012arXiv1211.0969A}}</ref> to be those for which the player responds to past mutual cooperation with future cooperation and splits expected payoffs equally if he receives at least the cooperative expected payoff. Among good strategies, the generous (ZD) subset performs well when the population is not too small. If the population is very small, defection strategies tend to dominate.<ref name=Stewart2013 /><br />
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While extortionary ZD strategies are not stable in large populations, another ZD class called "generous" strategies is both stable and robust. In fact, when the population is not too small, these strategies can supplant any other ZD strategy and even perform well against a broad array of generic strategies for iterated prisoner's dilemma, including win–stay, lose–switch. This was proven specifically for the donation game by Alexander Stewart and Joshua Plotkin in 2013. Generous strategies will cooperate with other cooperative players, and in the face of defection, the generous player loses more utility than its rival. Generous strategies are the intersection of ZD strategies and so-called "good" strategies, which were defined by Akin (2013) to be those for which the player responds to past mutual cooperation with future cooperation and splits expected payoffs equally if he receives at least the cooperative expected payoff. Among good strategies, the generous (ZD) subset performs well when the population is not too small. If the population is very small, defection strategies tend to dominate.<br />
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虽然勒索零决定策略在人口众多的情况下并不稳定，但另一种宽松的零决定策略既稳定又稳健。事实上，当人口不算太少的时候，这些策略可以取代任何其他零决定策略，甚至在一系列针对迭代囚徒困境的广泛通用策略（包括“获胜-保持-输”的转换策略）中表现良好。亚历山大·斯图尔特 Alexander Stewart和约书亚·普洛特金 Joshua Plotkin在2013年的捐赠博弈中证明了这一点。<ref name=Stewart2013>{{cite journal|last=Stewart|first=Alexander J.|author2=Joshua B. Plotkin|title=From extortion to generosity, evolution in the Iterated Prisoner's Dilemma|journal=[[Proceedings of the National Academy of Sciences of the United States of America]]|year=2013|doi=10.1073/pnas.1306246110|pmid=24003115|volume=110|issue=38|pages=15348–53|bibcode=2013PNAS..11015348S|pmc=3780848}}</ref>宽松的策略会与其他合作的玩家合作，面对背叛，慷慨的玩家比他的对手失去更多的效用。宽松策略是零决定策略和所谓的“好”策略的交集，阿金(2013) <ref name=Akin2013>{{cite arxiv|last=Akin|first=Ethan|title=Stable Cooperative Solutions for the Iterated Prisoner's Dilemma|year=2013|page=9|class=math.DS|eprint=1211.0969}} {{bibcode|2012arXiv1211.0969A}}</ref> Among good strategies, the generous (ZD) subset performs well when the population is not too small. If the population is very small, defection strategies tend to dominate.将这两种策略定义为玩家对过去的相互合作作出回应，并在至少获得合作预期收益的情况下平均分配预期收益的策略。在好的策略中，当总体不太小时，宽松(零决定)子集表现良好。如果总体很少，背叛策略往往占主导地位。<ref name=Stewart2013 /><br />
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===Continuous iterated prisoner's dilemma===<br />
<font color="#ff8000">连续迭代囚徒困境 Continuous iterated prisoner's dilemma </font> <br />
Most work on the iterated prisoner's dilemma has focused on the discrete case, in which players either cooperate or defect, because this model is relatively simple to analyze. However, some researchers have looked at models of the continuous iterated prisoner's dilemma, in which players are able to make a variable contribution to the other player. Le and Boyd<ref>{{cite journal | last1 = Le | first1 = S. | last2 = Boyd | first2 = R. |name-list-format=vanc| year = 2007 | title = Evolutionary Dynamics of the Continuous Iterated Prisoner's Dilemma | url = | journal = Journal of Theoretical Biology | volume = 245 | issue = 2| pages = 258–67 | doi = 10.1016/j.jtbi.2006.09.016 | pmid = 17125798 }}</ref> found that in such situations, cooperation is much harder to evolve than in the discrete iterated prisoner's dilemma. The basic intuition for this result is straightforward: in a continuous prisoner's dilemma, if a population starts off in a non-cooperative equilibrium, players who are only marginally more cooperative than non-cooperators get little benefit from [[Assortative mating|assorting]] with one another. By contrast, in a discrete prisoner's dilemma, tit for tat cooperators get a big payoff boost from assorting with one another in a non-cooperative equilibrium, relative to non-cooperators. Since nature arguably offers more opportunities for variable cooperation rather than a strict dichotomy of cooperation or defection, the continuous prisoner's dilemma may help explain why real-life examples of tit for tat-like cooperation are extremely rare in nature (ex. Hammerstein<ref>Hammerstein, P. (2003). Why is reciprocity so rare in social animals? A protestant appeal. In: P. Hammerstein, Editor, Genetic and Cultural Evolution of Cooperation, MIT Press. pp. 83–94. </ref>)<br />
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Most work on the iterated prisoner's dilemma has focused on the discrete case, in which players either cooperate or defect, because this model is relatively simple to analyze. However, some researchers have looked at models of the continuous iterated prisoner's dilemma, in which players are able to make a variable contribution to the other player. Le and Boyd found that in such situations, cooperation is much harder to evolve than in the discrete iterated prisoner's dilemma. The basic intuition for this result is straightforward: in a continuous prisoner's dilemma, if a population starts off in a non-cooperative equilibrium, players who are only marginally more cooperative than non-cooperators get little benefit from assorting with one another. By contrast, in a discrete prisoner's dilemma, tit for tat cooperators get a big payoff boost from assorting with one another in a non-cooperative equilibrium, relative to non-cooperators. Since nature arguably offers more opportunities for variable cooperation rather than a strict dichotomy of cooperation or defection, the continuous prisoner's dilemma may help explain why real-life examples of tit for tat-like cooperation are extremely rare in nature (ex. Hammerstein<ref>Hammerstein, P. (2003). Why is reciprocity so rare in social animals? A protestant appeal. In: P. Hammerstein, Editor, Genetic and Cultural Evolution of Cooperation, MIT Press. pp. 83–94. </ref>)<br />
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关于迭代囚徒困境的研究大多集中在离散情况下，在这种情况下，参与者要么合作，要么背叛，因为这个模型分析起来比较简单。然而，一些研究人员已经研究了连续迭代囚徒困境模型，在这个模型中，玩家能够对另一个玩家做出可变的贡献。乐 Le和博伊德 Boyd<ref>{{cite journal | last1 = Le | first1 = S. | last2 = Boyd | first2 = R. |name-list-format=vanc| year = 2007 | title = Evolutionary Dynamics of the Continuous Iterated Prisoner's Dilemma | url = | journal = Journal of Theoretical Biology | volume = 245 | issue = 2| pages = 258–67 | doi = 10.1016/j.jtbi.2006.09.016 | pmid = 17125798 }}</ref>发现，在这种情况下，合作比离散迭代的囚徒困境更难发展。这个结果的基本直觉很简单: 在一个持续的囚徒困境中，如果一个人群开始处于非合作均衡状态，那么与非合作者相比，合作程度稍高的玩家不会从相互配合中获益。相比之下，在离散的囚徒困境中，相对于非合作者，针锋相对的合作者在非合作均衡中相互配合会获得巨大的回报。由于自然界可以提供更多的机会来进行各种各样的合作，而不是严格地将合作或背叛分为两类，因此连续的囚徒困境可以帮助解释为什么现实生活中针锋相对的合作的例子在自然界中极其罕见。（例如，哈默斯坦 Hammerstein <ref>Hammerstein, P. (2003). Why is reciprocity so rare in social animals? A protestant appeal. In: P. Hammerstein, Editor, Genetic and Cultural Evolution of Cooperation, MIT Press. pp. 83–94. </ref>）。<br />
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even though tit for tat seems robust in theoretical models.<br />
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even though tit for tat seems robust in theoretical models.<br />
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尽管在理论模型中，针锋相对策略似乎是稳健的。<br />
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===Emergence of stable strategies===<br />
<font color="#ff8000">稳定策略的出现 Emergence of stable strategies </font> <br />
Players cannot seem to coordinate mutual cooperation, thus often get locked into the inferior yet stable strategy of defection. In this way, iterated rounds facilitate the evolution of stable strategies.<ref>{{cite book|last=Spaniel|first=William|title=Game Theory 101: The Complete Textbook|year=2011}}</ref> Iterated rounds often produce novel strategies, which have implications to complex social interaction. One such strategy is win-stay lose-shift. This strategy outperforms a simple Tit-For-Tat strategy&nbsp;– that is, if you can get away with cheating, repeat that behavior, however if you get caught, switch.<ref>{{cite journal|last=Nowak|first=Martin|author2=Karl Sigmund|title=A strategy of win-stay, lose-shift that outperforms tit-for-tat in the Prisoner's Dilemma game|journal=Nature|year=1993|volume=364|issue=6432|doi=10.1038/364056a0|pages=56–58|pmid=8316296|bibcode=1993Natur.364...56N}}</ref><br />
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Players cannot seem to coordinate mutual cooperation, thus often get locked into the inferior yet stable strategy of defection. In this way, iterated rounds facilitate the evolution of stable strategies. Iterated rounds often produce novel strategies, which have implications to complex social interaction. One such strategy is win-stay lose-shift. This strategy outperforms a simple Tit-For-Tat strategy&nbsp;– that is, if you can get away with cheating, repeat that behavior, however if you get caught, switch.<br />
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玩家似乎不能协调相互合作，因此常常陷入劣等而稳定的背叛策略。这样，迭代回合可以促进稳定策略的发展。<ref>{{cite book|last=Spaniel|first=William|title=Game Theory 101: The Complete Textbook|year=2011}}</ref>迭代回合往往产生新颖的策略，这对复杂的社会互动有影响。其中一个策略就是“赢-保持-输”的转变。这个策略比一个简单的针锋相对策略要好&nbsp;–也就是说，如果你能逃脱作弊的惩罚，就重复这个行为，如果你被抓住了，就改变策略。<ref>{{cite journal|last=Nowak|first=Martin|author2=Karl Sigmund|title=A strategy of win-stay, lose-shift that outperforms tit-for-tat in the Prisoner's Dilemma game|journal=Nature|year=1993|volume=364|issue=6432|doi=10.1038/364056a0|pages=56–58|pmid=8316296|bibcode=1993Natur.364...56N}}</ref><br />
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The only problem of this tit-for-tat strategy is that they are vulnerable to signal error. The problem arises when one individual cheats in retaliation but the other interprets it as cheating. As a result of this, the second individual now cheats and then it starts a see-saw pattern of cheating in a chain reaction.<br />
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The only problem of this tit-for-tat strategy is that they are vulnerable to signal error. The problem arises when one individual cheats in retaliation but the other interprets it as cheating. As a result of this, the second individual now cheats and then it starts a see-saw pattern of cheating in a chain reaction.<br />
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这种针锋相对策略的唯一问题是它们很容易出现信号错误。当一个人因报复而作弊，而另一个人将其单纯解释为欺骗时，就会出现问题。结果，第二个人现在作弊，然后在接下来的连锁反应中开始了反复交替的作弊模式。<br />
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==Real-life examples==<br />
现实生活的例子<br />
The prisoner setting may seem contrived, but there are in fact many examples in human interaction as well as interactions in nature that have the same payoff matrix. The prisoner's dilemma is therefore of interest to the [[social science]]s such as [[economics]], [[politics]], and [[sociology]], as well as to the biological sciences such as [[ethology]] and [[evolutionary biology]]. Many natural processes have been abstracted into models in which living beings are engaged in endless games of prisoner's dilemma. This wide applicability of the PD gives the game its substantial importance.<br />
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The prisoner setting may seem contrived, but there are in fact many examples in human interaction as well as interactions in nature that have the same payoff matrix. The prisoner's dilemma is therefore of interest to the social sciences such as economics, politics, and sociology, as well as to the biological sciences such as ethology and evolutionary biology. Many natural processes have been abstracted into models in which living beings are engaged in endless games of prisoner's dilemma. This wide applicability of the PD gives the game its substantial importance.<br />
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囚犯的环境似乎是人为的，但实际上，在人类交往以及自然界的交互中有许多具有相同收益矩阵的例子。因此，囚徒困境是经济学、政治学、社会学等社会科学以及动物行为学、进化生物学等生物学研究的热点问题。许多自然过程都被抽象为生物进行无休止的囚徒困境博弈的模型。囚徒困境这种广泛的适用性让博弈变得非常重要。<br />
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===Environmental studies===<br />
环境研究<br />
In [[environmental studies]], the PD is evident in crises such as global [[climate change|climate-change]]. It is argued all countries will benefit from a stable climate, but any single country is often hesitant to curb [[Carbon dioxide|{{Co2}}]] emissions. The immediate benefit to any one country from maintaining current behavior is wrongly perceived to be greater than the purported eventual benefit to that country if all countries' behavior was changed, therefore explaining the impasse concerning climate-change in 2007.<ref>{{cite news|newspaper=[[The Economist]]|url=http://www.economist.com/finance/displaystory.cfm?story_id=9867020|title=Markets & Data|date=2007-09-27}}</ref><br />
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In environmental studies, the PD is evident in crises such as global climate-change. It is argued all countries will benefit from a stable climate, but any single country is often hesitant to curb Carbon dioxide| emissions. The immediate benefit to any one country from maintaining current behavior is wrongly perceived to be greater than the purported eventual benefit to that country if all countries' behavior was changed, therefore explaining the impasse concerning climate-change in 2007.<br />
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在环境研究中，囚徒困境在诸如全球气候变化等危机中显而易见。有人认为，所有国家都将从稳定的气候中受益，但是每一个国家通常都在限制二氧化碳排放方面犹豫不决。人们错误地认为，如果所有国家的行为都改变，任何一个国家保持目前的行为所带来的直接好处都会大于所谓的最终好处，这就解释了2007年气候变化方面的僵局。<ref>{{cite news|newspaper=[[The Economist]]|url=http://www.economist.com/finance/displaystory.cfm?story_id=9867020|title=Markets & Data|date=2007-09-27}}</ref><br />
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An important difference between climate-change politics and the prisoner's dilemma is uncertainty; the extent and pace at which pollution can change climate is not known. The dilemma faced by government is therefore different from the prisoner's dilemma in that the payoffs of cooperation are unknown. This difference suggests that states will cooperate much less than in a real iterated prisoner's dilemma, so that the probability of avoiding a possible climate catastrophe is much smaller than that suggested by a game-theoretical analysis of the situation using a real iterated prisoner's dilemma.<ref>{{cite web|last=Rehmeyer|first=Julie|title=Game theory suggests current climate negotiations won't avert catastrophe|url=https://www.sciencenews.org/article/game-theory-suggests-current-climate-negotiations-won%E2%80%99t-avert-catastrophe|work=Science News|publisher=Society for Science & the Public|date=2012-10-29}}</ref><br />
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An important difference between climate-change politics and the prisoner's dilemma is uncertainty; the extent and pace at which pollution can change climate is not known. The dilemma faced by government is therefore different from the prisoner's dilemma in that the payoffs of cooperation are unknown. This difference suggests that states will cooperate much less than in a real iterated prisoner's dilemma, so that the probability of avoiding a possible climate catastrophe is much smaller than that suggested by a game-theoretical analysis of the situation using a real iterated prisoner's dilemma.<br />
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气候变化政治与囚徒困境之间的一个重要区别是不确定性; 污染对气候变化的影响程度和速度尚不清楚。因此，政府面临的困境不同于囚徒困境，因为合作的回报是未知的。这种差异表明，各国之间的合作远远少于真正的迭代囚徒困境中的合作，因此避免可能发生的气候灾难的可能性远远小于使用真正的迭代囚徒困境博弈论情景分析<ref>{{cite web|last=Rehmeyer|first=Julie|title=Game theory suggests current climate negotiations won't avert catastrophe|url=https://www.sciencenews.org/article/game-theory-suggests-current-climate-negotiations-won%E2%80%99t-avert-catastrophe|work=Science News|publisher=Society for Science & the Public|date=2012-10-29}}</ref><br />
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Osang and Nandy (2003) provide a theoretical explanation with proofs for a regulation-driven win-win situation along the lines of [[Michael Porter]]'s hypothesis, in which government regulation of competing firms is substantial.<ref>{{cite thesis|type=paper|url= http://faculty.smu.edu/tosang/pdf/regln0803.pdf|first=Thomas|last=Osang|first2=Arundhati|last2=Nandyyz|date=August 2003|title=Environmental Regulation of Polluting Firms: Porter's Hypothesis Revisited}}</ref><br />
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Osang and Nandy (2003) provide a theoretical explanation with proofs for a regulation-driven win-win situation along the lines of Michael Porter's hypothesis, in which government regulation of competing firms is substantial.<br />
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欧桑 Osang和南迪 Nandy (2003)提供了一个理论解释，并根据迈克尔·波特 Michael Porter的假设，即政府对竞争企业的监管是实质性的，证明了监管驱动的双赢局面。<ref>{{cite thesis|type=paper|url= http://faculty.smu.edu/tosang/pdf/regln0803.pdf|first=Thomas|last=Osang|first2=Arundhati|last2=Nandyyz|date=August 2003|title=Environmental Regulation of Polluting Firms: Porter's Hypothesis Revisited}}</ref><br />
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===Animals===<br />
动物<br />
Cooperative behavior of many animals can be understood as an example of the prisoner's dilemma. Often animals engage in long term partnerships, which can be more specifically modeled as iterated prisoner's dilemma. For example, [[guppy|guppies]] inspect predators cooperatively in groups, and they are thought to punish non-cooperative inspectors.<br />
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Cooperative behavior of many animals can be understood as an example of the prisoner's dilemma. Often animals engage in long term partnerships, which can be more specifically modeled as iterated prisoner's dilemma. For example, guppies inspect predators cooperatively in groups, and they are thought to punish non-cooperative inspectors.<br />
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许多动物的合作行为可以理解为囚徒困境的一个例子。通常动物会建立长期的伙伴关系，这种关系可以更具体地模拟为迭代囚徒困境。例如，孔雀鱼成群结队地合作监察捕食者，它们被认为是在惩罚不合作的监察者。<br />
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[[Vampire bats]] are social animals that engage in reciprocal food exchange. Applying the payoffs from the prisoner's dilemma can help explain this behavior:<ref>{{cite book|last=Dawkins|first=Richard|title=The Selfish Gene|year=1976|publisher=Oxford University Press}}</ref><br />
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Vampire bats are social animals that engage in reciprocal food exchange. Applying the payoffs from the prisoner's dilemma can help explain this behavior:<br />
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吸血蝙蝠是从事相互的食物交换的群居动物。应用囚徒困境收益可以帮助解释这种行为: <ref>{{cite book|last=Dawkins|first=Richard|title=The Selfish Gene|year=1976|publisher=Oxford University Press}}</ref><br />
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* C/C: "Reward: I get blood on my unlucky nights, which saves me from starving. I have to give blood on my lucky nights, which doesn't cost me too much."<br />
* 合作/合作："回报：我在不幸运的晚上得到了能让我果腹的血，那在幸运的晚上我也应该分出点血，那不会花费多少。"<br />
* D/C: "Temptation: You save my life on my poor night. But then I get the added benefit of not having to pay the slight cost of feeding you on my good night."<br />
* 背叛/合作："诱惑：你在我的不幸的夜里救了我，但在我的幸运夜我不会给你血，那样我会活的更好。"<br />
* C/D: "Sucker's Payoff: I pay the cost of saving your life on my good night. But on my bad night you don't feed me and I run a real risk of starving to death."<br />
* 合作/叛变："可怜者的回报：在我的幸运夜我救了你的命，但在我的不幸夜里你没有救我，我有饿死的风险。"<br />
* D/D: "Punishment: I don't have to pay the slight costs of feeding you on my good nights. But I run a real risk of starving on my poor nights."<br />
* 叛变/叛变："惩罚：我在我的幸运夜里不必付出代价来救你，但我在我的不幸夜里有挨饿的风险。"<br />
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===Psychology===<br />
心理学<br />
In [[addiction]] research / [[behavioral economics]], [[George Ainslie (psychologist)|George Ainslie]] points out<ref>{{cite book |first=George|last=Ainslie |title=Breakdown of Will |year=2001 |isbn=978-0-521-59694-7}}</ref> that addiction can be cast as an intertemporal PD problem between the present and future selves of the addict. In this case, ''defecting'' means ''relapsing'', and it is easy to see that not defecting both today and in the future is by far the best outcome. The case where one abstains today but relapses in the future is the worst outcome&nbsp;– in some sense the discipline and self-sacrifice involved in abstaining today have been "wasted" because the future relapse means that the addict is right back where he started and will have to start over (which is quite demoralizing, and makes starting over more difficult). Relapsing today and tomorrow is a slightly "better" outcome, because while the addict is still addicted, they haven't put the effort in to trying to stop. The final case, where one engages in the addictive behavior today while abstaining "tomorrow" will be familiar to anyone who has struggled with an addiction. The problem here is that (as in other PDs) there is an obvious benefit to defecting "today", but tomorrow one will face the same PD, and the same obvious benefit will be present then, ultimately leading to an endless string of defections.<br />
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In addiction research / behavioral economics, George Ainslie points out that addiction can be cast as an intertemporal PD problem between the present and future selves of the addict. In this case, defecting means relapsing, and it is easy to see that not defecting both today and in the future is by far the best outcome. The case where one abstains today but relapses in the future is the worst outcome&nbsp;– in some sense the discipline and self-sacrifice involved in abstaining today have been "wasted" because the future relapse means that the addict is right back where he started and will have to start over (which is quite demoralizing, and makes starting over more difficult). Relapsing today and tomorrow is a slightly "better" outcome, because while the addict is still addicted, they haven't put the effort in to trying to stop. The final case, where one engages in the addictive behavior today while abstaining "tomorrow" will be familiar to anyone who has struggled with an addiction. The problem here is that (as in other PDs) there is an obvious benefit to defecting "today", but tomorrow one will face the same PD, and the same obvious benefit will be present then, ultimately leading to an endless string of defections.<br />
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在成瘾研究/行为经济学中，乔治·安斯利 George Ainslie指出<ref>{{cite book |first=George|last=Ainslie |title=Breakdown of Will |year=2001 |isbn=978-0-521-59694-7}}</ref>，可以将成瘾视为成瘾者现在和未来自我之间的跨期囚徒困境问题。在这种情况下，背叛意味着复发，很容易看出，目前和未来都没有背叛是迄今为止最好的结果。如果一个人今天戒了，但在将来又复吸，这是最糟糕的结果&nbsp;–从某种意义上来说，今天戒瘾所包含的纪律和自我牺牲已经被“浪费”了，因为未来的复吸意味着瘾君子又回到了他开始的地方，他将被迫重新开始(这相当令人沮丧，也使得重新开始更加困难)。今天和明天复发是一个稍微“好一点”的结果，因为尽管瘾君子仍然上瘾，但他们没有努力去尝试停止。最后一种情况是，现在与成瘾斗争的任何人都会熟悉现在的成瘾行为，而在明天放弃。这里的问题是(和其他囚徒困境问题一样)，背叛“今天”有一个明显的好处，但明天这个人将面临同样的囚徒困境问题，同样明显的好处是背叛，最终导致一连串无休止的背叛。<br />
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[[John Gottman]] in his research described in "the science of trust" defines good relationships as those where partners know not to enter the (D,D) cell or at least not to get dynamically stuck there in a loop.<br />
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John Gottman in his research described in "the science of trust" defines good relationships as those where partners know not to enter the (D,D) cell or at least not to get dynamically stuck there in a loop.<br />
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约翰·高特曼 John Gottman在他的研究《信任的科学》中将良好的关系定义为伙伴知道不进入(背叛，背叛)牢房中或者至少不要陷入这样的动态循环关系中。<br />
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===Economics===<br />
经济学<br />
The prisoner's dilemma has been called the ''[[Escherichia coli|E. coli]]'' of social psychology, and it has been used widely to research various topics such as [[Oligopoly|oligopolistic]] competition and collective action to produce a collective good.<ref>{{Cite journal|last=Axelrod|first=Robert|date=1980|title=Effective Choice in the Prisoner's Dilemma|journal=The Journal of Conflict Resolution|volume=24|issue=1|pages=3–25|issn=0022-0027|jstor=173932|doi=10.1177/002200278002400101|url=https://semanticscholar.org/paper/fd1ab82470446bfb12c39f0c577644291027cf76}}</ref> <br />
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The prisoner's dilemma has been called the E. coli of social psychology, and it has been used widely to research various topics such as oligopolistic competition and collective action to produce a collective good. <br />
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囚徒困境被称为社会心理学中的“大肠杆菌”，它被广泛用于研究寡头垄断竞争和集体行动来产生集体利益等问题。<ref>{{Cite journal|last=Axelrod|first=Robert|date=1980|title=Effective Choice in the Prisoner's Dilemma|journal=The Journal of Conflict Resolution|volume=24|issue=1|pages=3–25|issn=0022-0027|jstor=173932|doi=10.1177/002200278002400101|url=https://semanticscholar.org/paper/fd1ab82470446bfb12c39f0c577644291027cf76}}</ref><br />
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Advertising is sometimes cited as a real-example of the prisoner's dilemma. When [[cigarette advertising]] was legal in the United States, competing cigarette manufacturers had to decide how much money to spend on advertising. The effectiveness of Firm A's advertising was partially determined by the advertising conducted by Firm B. Likewise, the profit derived from advertising for Firm B is affected by the advertising conducted by Firm A. If both Firm A and Firm B chose to advertise during a given period, then the advertisement from each firm negates the other's, receipts remain constant, and expenses increase due to the cost of advertising. Both firms would benefit from a reduction in advertising. However, should Firm B choose not to advertise, Firm A could benefit greatly by advertising. Nevertheless, the optimal amount of advertising by one firm depends on how much advertising the other undertakes. As the best strategy is dependent on what the other firm chooses there is no dominant strategy, which makes it slightly different from a prisoner's dilemma. The outcome is similar, though, in that both firms would be better off were they to advertise less than in the equilibrium. Sometimes cooperative behaviors do emerge in business situations. For instance, cigarette manufacturers endorsed the making of laws banning cigarette advertising, understanding that this would reduce costs and increase profits across the industry. <ref>{{Citation needed|reason=This reference doesn't mention or support the claimed historical account.|date=December 2012}}</ref><ref>{{efn|1=This argument for the development of cooperation through trust is given in ''[[The Wisdom of Crowds]]'', where it is argued that long-distance [[capitalism]] was able to form around a nucleus of [[Religious Society of Friends|Quakers]], who always dealt honourably with their business partners. (Rather than defecting and reneging on promises&nbsp;– a phenomenon that had discouraged earlier long-term unenforceable overseas contracts). It is argued that dealings with reliable merchants allowed the [[meme]] for cooperation to spread to other traders, who spread it further until a high degree of cooperation became a profitable strategy in general [[commerce]]}} </ref> This analysis is likely to be pertinent in many other business situations involving advertising. <ref>{{Citation needed|reason=This doesn't sound like cooperation|date=November 2012}}</ref><br />
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Advertising is sometimes cited as a real-example of the prisoner's dilemma. When cigarette advertising was legal in the United States, competing cigarette manufacturers had to decide how much money to spend on advertising. The effectiveness of Firm A's advertising was partially determined by the advertising conducted by Firm B. Likewise, the profit derived from advertising for Firm B is affected by the advertising conducted by Firm A. If both Firm A and Firm B chose to advertise during a given period, then the advertisement from each firm negates the other's, receipts remain constant, and expenses increase due to the cost of advertising. Both firms would benefit from a reduction in advertising. However, should Firm B choose not to advertise, Firm A could benefit greatly by advertising. Nevertheless, the optimal amount of advertising by one firm depends on how much advertising the other undertakes. As the best strategy is dependent on what the other firm chooses there is no dominant strategy, which makes it slightly different from a prisoner's dilemma. The outcome is similar, though, in that both firms would be better off were they to advertise less than in the equilibrium. Sometimes cooperative behaviors do emerge in business situations. For instance, cigarette manufacturers endorsed the making of laws banning cigarette advertising, understanding that this would reduce costs and increase profits across the industry. This analysis is likely to be pertinent in many other business situations involving advertising.<br />
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广告有时被认为是囚徒困境的一个真实例子。当香烟广告在美国是合法的时候，相互竞争的香烟制造商必须决定在广告上花多少钱。公司A的广告效果部分取决于公司B的广告效果。同样，公司B的广告带来的利润也受到公司A的广告影响。如果公司A和公司B都选择在给定的时间段内做广告，那么一家公司的广告就会抵消另一方的广告，倘若收入保持不变，费用就会因广告成本而增加。两家公司都将从广告减少中获益。然而，如果B公司选择不做广告，A公司就可以通过广告获得巨大的利益。尽管如此，一家公司的最佳广告数量仍取决于另一家公司的广告投放量。由于最佳策略取决于其他公司的选择，因此这里没有占主导地位的策略，这使得它与囚徒困境略有不同。但结果是相似的，如果两家公司的广告都少于均衡状态，他们的处境会更好。有时合作行为确实会在商业环境中出现。例如，香烟制造商支持立法禁止香烟广告，因为这将降低成本并增加整个行业的利润。<ref>{{Citation needed|reason=This reference doesn't mention or support the claimed historical account.|date=December 2012}}</ref><ref>{{efn|1=This argument for the development of cooperation through trust is given in ''[[The Wisdom of Crowds]]'', where it is argued that long-distance [[capitalism]] was able to form around a nucleus of [[Religious Society of Friends|Quakers]], who always dealt honourably with their business partners. (Rather than defecting and reneging on promises&nbsp;– a phenomenon that had discouraged earlier long-term unenforceable overseas contracts). It is argued that dealings with reliable merchants allowed the [[meme]] for cooperation to spread to other traders, who spread it further until a high degree of cooperation became a profitable strategy in general [[commerce]]}} </ref>这种分析可能适用于许多其他涉及广告的商业情况。<ref>{{Citation needed|reason=This doesn't sound like cooperation|date=November 2012}}</ref> <br />
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Without enforceable agreements, members of a [[cartel]] are also involved in a (multi-player) prisoner's dilemma.<ref>{{Cite book|last1=Nicholson|first=Walter|year=2000|title=Intermediate microeconomics and its application|edition=8th|location=Fort Worth, TX|publisher=Dryden Press : Harcourt College Publishers|isbn=978-0-030-25916-6}}</ref> 'Cooperating' typically means keeping prices at a pre-agreed minimum level. 'Defecting' means selling under this minimum level, instantly taking business (and profits) from other cartel members. [[Anti-trust]] authorities want potential cartel members to mutually defect, ensuring the lowest possible prices for [[consumer]]s.<br />
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Without enforceable agreements, members of a cartel are also involved in a (multi-player) prisoner's dilemma. 'Cooperating' typically means keeping prices at a pre-agreed minimum level. 'Defecting' means selling under this minimum level, instantly taking business (and profits) from other cartel members. Anti-trust authorities want potential cartel members to mutually defect, ensuring the lowest possible prices for consumers.<br />
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没有可强制执行的协议，卡特尔 cartel的成员国也会陷入(多玩家)囚徒困境。<ref>{{Cite book|last1=Nicholson|first=Walter|year=2000|title=Intermediate microeconomics and its application|edition=8th|location=Fort Worth, TX|publisher=Dryden Press : Harcourt College Publishers|isbn=978-0-030-25916-6}}</ref> “合作”通常意味着将价格保持在预先商定的最低水平。“背叛”意味着低于最低价格水平销售，并立即从其他卡特尔成员那里获得业务(和利润)。反垄断机构希望潜在的卡特尔成员相互背叛，确保消费者获得尽可能低的价格。<br />
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===Sport===<br />
运动<br />
[[Doping in sport]] has been cited as an example of a prisoner's dilemma.<ref name="wired">{{cite journal|last=Schneier |first=Bruce |url=https://www.wired.com/opinion/2012/10/lance-armstrong-and-the-prisoners-dilemma-of-doping-in-professional-sports/ |title=Lance Armstrong and the Prisoners' Dilemma of Doping in Professional Sports &#124; Wired Opinion |journal=Wired |publisher=Wired.com |date=2012-10-26 |accessdate=2012-10-29}}</ref><br />
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Doping in sport has been cited as an example of a prisoner's dilemma.<br />
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体育运动中的兴奋剂被认为是囚徒困境的一个例子。<ref name="wired">{{cite journal|last=Schneier |first=Bruce |url=https://www.wired.com/opinion/2012/10/lance-armstrong-and-the-prisoners-dilemma-of-doping-in-professional-sports/ |title=Lance Armstrong and the Prisoners' Dilemma of Doping in Professional Sports &#124; Wired Opinion |journal=Wired |publisher=Wired.com |date=2012-10-26 |accessdate=2012-10-29}}</ref><br />
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Two competing athletes have the option to use an illegal and/or dangerous drug to boost their performance. If neither athlete takes the drug, then neither gains an advantage. If only one does, then that athlete gains a significant advantage over their competitor, reduced by the legal and/or medical dangers of having taken the drug. If both athletes take the drug, however, the benefits cancel out and only the dangers remain, putting them both in a worse position than if neither had used doping.<ref name="wired" /><br />
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Two competing athletes have the option to use an illegal and/or dangerous drug to boost their performance. If neither athlete takes the drug, then neither gains an advantage. If only one does, then that athlete gains a significant advantage over their competitor, reduced by the legal and/or medical dangers of having taken the drug. If both athletes take the drug, however, the benefits cancel out and only the dangers remain, putting them both in a worse position than if neither had used doping.<br />
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两名参赛运动员可以选择使用非法或危险药物来提高成绩。如果两个运动员都没有服用这种药物，那么他们都不会获得优势。如果只有一个人这样做，那么这个运动员就比他们的竞争对手获得了明显的优势，但由于法律或服用药物的医疗风险，这种优势会减少。然而，如果两名运动员都服用了这种药物，那么好处就被抵消了，只剩下风险，这使得他们的处境比没有服用兴奋剂的情况更加糟糕。<ref name="wired" /><br />
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===International politics===<br />
国际政治<br />
In [[international politics|international political theory]], the Prisoner's Dilemma is often used to demonstrate the coherence of [[strategic realism]], which holds that in international relations, all states (regardless of their internal policies or professed ideology), will act in their rational self-interest given [[anarchy (international relations)|international anarchy]]. A classic example is an arms race like the [[Cold War]] and similar conflicts.<ref>{{cite journal| title = Arms races as iterated prisoner's dilemma games | author = Stephen J. Majeski | journal = Mathematical and Social Sciences | volume = 7 | issue = 3 | pages = 253–66 | year = 1984 | doi=10.1016/0165-4896(84)90022-2}}</ref> During the Cold War the opposing alliances of [[NATO]] and the [[Warsaw Pact]] both had the choice to arm or disarm. From each side's point of view, disarming whilst their opponent continued to arm would have led to military inferiority and possible annihilation. Conversely, arming whilst their opponent disarmed would have led to superiority. If both sides chose to arm, neither could afford to attack the other, but both incurred the high cost of developing and maintaining a nuclear arsenal. If both sides chose to disarm, war would be avoided and there would be no costs.<br />
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In international political theory, the Prisoner's Dilemma is often used to demonstrate the coherence of strategic realism, which holds that in international relations, all states (regardless of their internal policies or professed ideology), will act in their rational self-interest given international anarchy. A classic example is an arms race like the Cold War and similar conflicts. During the Cold War the opposing alliances of NATO and the Warsaw Pact both had the choice to arm or disarm. From each side's point of view, disarming whilst their opponent continued to arm would have led to military inferiority and possible annihilation. Conversely, arming whilst their opponent disarmed would have led to superiority. If both sides chose to arm, neither could afford to attack the other, but both incurred the high cost of developing and maintaining a nuclear arsenal. If both sides chose to disarm, war would be avoided and there would be no costs.<br />
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在国际政治理论中，囚徒困境经常被用来证明战略现实主义的一致性，这种战略现实主义认为，在国际关系中，由于国际无政府状态，所有国家(无论其国内政策或公开宣称的意识形态如何)都会为了自身的理性利益来行动。一个典型的例子是类似冷战和类似冲突的军备竞赛。<ref>{{cite journal| title = Arms races as iterated prisoner's dilemma games | author = Stephen J. Majeski | journal = Mathematical and Social Sciences | volume = 7 | issue = 3 | pages = 253–66 | year = 1984 | doi=10.1016/0165-4896(84)90022-2}}</ref>在冷战期间，北约和华约组织的对立联盟都可以选择武装或解除武装。从双方的观点来看，解除武装而对手继续武装可能会导致军事劣势和被歼灭。相反，如果选择武装而对手已经解除了武装，那么就会获得优势。如果双方都选择武装自己，那么任何一方都承担不起攻击对方的代价，但是双方都为发展和维持核武库付出了高昂的代价。如果双方都选择裁军，战争就可以避免，也不会有任何代价。<br />
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Although the 'best' overall outcome is for both sides to disarm, the rational course for both sides is to arm, and this is indeed what happened. Both sides poured enormous resources into military research and armament in a [[War of attrition (game)|war of attrition]] for the next thirty years until the Soviet Union could not withstand the economic cost.<ref>{{Citation|last=Kuhn|first=Steven|title=Prisoner's Dilemma|date=2019|url=https://plato.stanford.edu/archives/win2019/entries/prisoner-dilemma/|encyclopedia=The Stanford Encyclopedia of Philosophy|editor-last=Zalta|editor-first=Edward N.|edition=Winter 2019|publisher=Metaphysics Research Lab, Stanford University|access-date=2020-04-12}}</ref> The same logic could be applied in any similar scenario, be it economic or technological competition between sovereign states.<br />
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Although the 'best' overall outcome is for both sides to disarm, the rational course for both sides is to arm, and this is indeed what happened. Both sides poured enormous resources into military research and armament in a war of attrition for the next thirty years until the Soviet Union could not withstand the economic cost. The same logic could be applied in any similar scenario, be it economic or technological competition between sovereign states.<br />
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虽然最好的结果是双方解除武装，但是双方的理性选择是武装起来，事实也的确如此。在接下来的三十年里，双方都在军事研究和武器装备的消耗战上投入了大量的资源，直到苏联无法承受经济损失。<ref>{{Citation|last=Kuhn|first=Steven|title=Prisoner's Dilemma|date=2019|url=https://plato.stanford.edu/archives/win2019/entries/prisoner-dilemma/|encyclopedia=The Stanford Encyclopedia of Philosophy|editor-last=Zalta|editor-first=Edward N.|edition=Winter 2019|publisher=Metaphysics Research Lab, Stanford University|access-date=2020-04-12}}</ref>同样的逻辑也适用于任何类似的情况，无论是主权国家之间的经济竞争还是技术竞争。<br />
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===Multiplayer dilemmas===<br />
<font color="#ff8000">多玩家困境 Multiplayer dilemmas</font><br />
Many real-life dilemmas involve multiple players.<ref>Gokhale CS, Traulsen A. Evolutionary games in the multiverse. Proceedings of the National Academy of Sciences. 2010 Mar 23. 107(12):5500–04.</ref> Although metaphorical, [[Garrett Hardin|Hardin's]] [[tragedy of the commons]] may be viewed as an example of a multi-player generalization of the PD: Each villager makes a choice for personal gain or restraint. The collective reward for unanimous (or even frequent) defection is very low payoffs (representing the destruction of the "commons"). A commons dilemma most people can relate to is washing the dishes in a shared house. By not washing dishes an individual can gain by saving his time, but if that behavior is adopted by every resident the collective cost is no clean plates for anyone.<br />
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Many real-life dilemmas involve multiple players. Although metaphorical, Hardin's tragedy of the commons may be viewed as an example of a multi-player generalization of the PD: Each villager makes a choice for personal gain or restraint. The collective reward for unanimous (or even frequent) defection is very low payoffs (representing the destruction of the "commons"). A commons dilemma most people can relate to is washing the dishes in a shared house. By not washing dishes an individual can gain by saving his time, but if that behavior is adopted by every resident the collective cost is no clean plates for anyone.<br />
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许多现实生活中的困境牵涉到多个参与者。<ref>Gokhale CS, Traulsen A. Evolutionary games in the multiverse. Proceedings of the National Academy of Sciences. 2010 Mar 23. 107(12):5500–04.</ref>尽管具有隐喻性，但哈丁的<font color="#ff8000">公地悲剧 tragedy of the commons</font>可以看作是囚徒困境多个参与者的一个例子: 每个村民做出选择是为了个人利益还是克制。对于一致(甚至频繁)叛变的集体回报是非常低的(代表了对“公共资源”的破坏)。大多数人可能会遇到的公地困境是在一个共用的房子里洗碗。通过不洗碗，个人可以节省时间，但如果每个居民都选择这种行为，那么集体的代价是任何人都没有干净的盘子。<br />
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The commons are not always exploited: [[William Poundstone]], in a book about the prisoner's dilemma (see References below), describes a situation in New Zealand where newspaper boxes are left unlocked. It is possible for people to [[Excludability|take a paper without paying]] (''defecting'') but very few do, feeling that if they do not pay then neither will others, destroying the system. Subsequent research by [[Elinor Ostrom]], winner of the 2009 [[Nobel Memorial Prize in Economic Sciences]], hypothesized that the tragedy of the commons is oversimplified, with the negative outcome influenced by outside influences. Without complicating pressures, groups communicate and manage the commons among themselves for their mutual benefit, enforcing social norms to preserve the resource and achieve the maximum good for the group, an example of effecting the best case outcome for PD.<ref>{{cite web|url=http://volokh.com/2009/10/12/elinor-ostrom-and-the-tragedy-of-the-commons/ |title=The Volokh Conspiracy " Elinor Ostrom and the Tragedy of the Commons |publisher=Volokh.com |date=2009-10-12 |accessdate=2011-12-17}}</ref><br />
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The commons are not always exploited: William Poundstone, in a book about the prisoner's dilemma (see References below), describes a situation in New Zealand where newspaper boxes are left unlocked. It is possible for people to take a paper without paying (defecting) but very few do, feeling that if they do not pay then neither will others, destroying the system. Subsequent research by Elinor Ostrom, winner of the 2009 Nobel Memorial Prize in Economic Sciences, hypothesized that the tragedy of the commons is oversimplified, with the negative outcome influenced by outside influences. Without complicating pressures, groups communicate and manage the commons among themselves for their mutual benefit, enforcing social norms to preserve the resource and achieve the maximum good for the group, an example of effecting the best case outcome for PD.<br />
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公共资源并不总是被利用: 威廉·庞德斯通 William Poundstone在一本关于囚徒困境的书(见下文参考文献)中描述了新西兰的一种情况，信箱没有上锁。人们可以不付钱就拿报纸(背叛) ，但很少有人这样做，他们觉得如果他们不付钱，那么其他人也不会付钱，这会摧毁整个系统。2009年诺贝尔经济学奖获得者埃莉诺·奥斯特罗姆 Elinor Ostrom随后的研究认为公地悲剧过于简单化，其负面结果会受到外部影响。在没有复杂压力的情况下，团体之间为了共同利益进行沟通和管理，执行社会规范以保护资源并为团体实现最大利益，这是实现囚徒困境最佳结果的一个例子。<ref>{{cite web|url=http://volokh.com/2009/10/12/elinor-ostrom-and-the-tragedy-of-the-commons/ |title=The Volokh Conspiracy " Elinor Ostrom and the Tragedy of the Commons |publisher=Volokh.com |date=2009-10-12 |accessdate=2011-12-17}}</ref><br />
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==Related games==<br />
相关博弈<br />
===Closed-bag exchange===<br />
<font color="#ff8000">封闭袋子交换 Closed-bag exchange </font><br />
[[File:Prisoner's Dilemma briefcase exchange (colorized).svg|thumb|The prisoner's dilemma as a briefcase exchange]]<br />
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The prisoner's dilemma as a briefcase exchange<br />
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囚徒困境是一个公文包式的交换<br />
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[[Douglas Hofstadter]]<ref name="dh">{{cite book | first=Douglas R. | last=Hofstadter| authorlink=Douglas Hofstadter | title= Metamagical Themas: questing for the essence of mind and pattern | publisher= Bantam Dell Pub Group| year=1985 | isbn=978-0-465-04566-2|chapter= Ch.29 ''The Prisoner's Dilemma Computer Tournaments and the Evolution of Cooperation''.| title-link=Metamagical Themas}}</ref> once suggested that people often find problems such as the PD problem easier to understand when it is illustrated in the form of a simple game, or trade-off. One of several examples he used was "closed bag exchange":<br />
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Douglas Hofstadter once suggested that people often find problems such as the PD problem easier to understand when it is illustrated in the form of a simple game, or trade-off. One of several examples he used was "closed bag exchange":<br />
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侯世达 Douglas Hofstadter <ref name="dh">{{cite book | first=Douglas R. | last=Hofstadter| authorlink=Douglas Hofstadter | title= Metamagical Themas: questing for the essence of mind and pattern | publisher= Bantam Dell Pub Group| year=1985 | isbn=978-0-465-04566-2|chapter= Ch.29 ''The Prisoner's Dilemma Computer Tournaments and the Evolution of Cooperation''.| title-link=Metamagical Themas}}</ref>曾经指出，人们通常会发现诸如囚徒困境的问题，比如，当它以一个简单囚徒困境博弈的形式，或者以权衡的方式表现出来时，会更容易理解。他使用的几个例子之一是“封闭袋子交换” :<br />
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{{quote|Two people meet and exchange closed bags, with the understanding that one of them contains money, and the other contains a purchase. Either player can choose to honor the deal by putting into his or her bag what he or she agreed, or he or she can defect by handing over an empty bag.}}<br />
两人相遇并交换包裹，事先知道一个包里装着钱，一个装着订单。任一玩家都可选择尊重交易，放入事先约定的东西；也可以选择背叛，交换空的公文包。<br />
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Defection always gives a game-theoretically preferable outcome.<ref>{{Cite web|url=https://users.auth.gr/kehagiat/Research/GameTheory/06GamesToPlay/Prisoner%27s_dilemma.htm#Closed_Bag_Exchange|title=Prisoner's dilemma - Wikipedia, the free encyclopedia|website=users.auth.gr|access-date=2020-04-12}}</ref><br />
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Defection always gives a game-theoretically preferable outcome.<br />
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背叛总是会带来一个理论上更可取的结果。<ref>{{Cite web|url=https://users.auth.gr/kehagiat/Research/GameTheory/06GamesToPlay/Prisoner%27s_dilemma.htm#Closed_Bag_Exchange|title=Prisoner's dilemma - Wikipedia, the free encyclopedia|website=users.auth.gr|access-date=2020-04-12}}</ref><br />
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===''Friend or Foe?''===<br />
朋友还是敌人？<br />
''[[Friend or Foe? (TV series)|Friend or Foe?]]'' is a game show that aired from 2002 to 2005 on the [[Game Show Network]] in the US. It is an example of the prisoner's dilemma game tested on real people, but in an artificial setting. On the game show, three pairs of people compete. When a pair is eliminated, they play a game similar to the prisoner's dilemma to determine how the winnings are split. If they both cooperate (Friend), they share the winnings 50–50. If one cooperates and the other defects (Foe), the defector gets all the winnings and the cooperator gets nothing. If both defect, both leave with nothing. Notice that the reward matrix is slightly different from the standard one given above, as the rewards for the "both defect" and the "cooperate while the opponent defects" cases are identical. This makes the "both defect" case a weak equilibrium, compared with being a strict equilibrium in the standard prisoner's dilemma. If a contestant knows that their opponent is going to vote "Foe", then their own choice does not affect their own winnings. In a specific sense, ''Friend or Foe'' has a rewards model between prisoner's dilemma and the [[Chicken (game)|game of Chicken]].<br />
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Friend or Foe? is a game show that aired from 2002 to 2005 on the Game Show Network in the US. It is an example of the prisoner's dilemma game tested on real people, but in an artificial setting. On the game show, three pairs of people compete. When a pair is eliminated, they play a game similar to the prisoner's dilemma to determine how the winnings are split. If they both cooperate (Friend), they share the winnings 50–50. If one cooperates and the other defects (Foe), the defector gets all the winnings and the cooperator gets nothing. If both defect, both leave with nothing. Notice that the reward matrix is slightly different from the standard one given above, as the rewards for the "both defect" and the "cooperate while the opponent defects" cases are identical. This makes the "both defect" case a weak equilibrium, compared with being a strict equilibrium in the standard prisoner's dilemma. If a contestant knows that their opponent is going to vote "Foe", then their own choice does not affect their own winnings. In a specific sense, Friend or Foe has a rewards model between prisoner's dilemma and the game of Chicken.<br />
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朋友还是敌人？是一个竞赛节目，从2002年至2005年在美国的Game show Network播出。这是囚徒困境博弈在真人身上测试的一个例子，只是在人为的环境中。在游戏节目中，有三对选手参加比赛。当一对被淘汰时，他们会玩一个类似囚徒困境的游戏来决定奖金如何分配。如果他们都合作(朋友) ，他们分享奖金50-50。如果一方合作而另一方背叛(敌人) ，那么叛变者将得到所有的奖金，而合作者将一无所获。如果双方都背叛，那么双方都将一无所有。请注意，奖励矩阵与上面给出的标准矩阵略有不同，因为“双方都背叛”和“合作而对方背叛”情况下的奖励是相同的。与标准囚徒困境中的严格均衡相比，这使得“两个都背叛”情况成为一个弱均衡。如果一个参赛者知道他们的对手将投票给“敌人” ，那么他们自己的选择不会影响他们自己的奖金。从特定意义上讲，“朋友还是敌人”节目在囚徒困境和“胆小鬼”博弈之间有一个奖励模型。<br />
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The rewards matrix is<br />
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The rewards matrix is<br />
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奖励矩阵是<br />
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{| class="wikitable"<br />
<br />
! {{diagonal split header|{{color|#009|Pair 1}}|{{color|#900|Pair 2}}}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"Friend"<br />(cooperate)}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"Foe"<br />(defect)}}<br />
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|-<br />
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! scope="row" style="width:6em;" | {{color|#009|"Friend"<br />(cooperate)}}<br />
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| {{diagonal split header|{{color|#009|1}}|{{color|#900|1}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|2}}|transparent}}<br />
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|-<br />
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! scope="row" | {{color|#009|"Foe"<br />(defect)}}<br />
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| {{diagonal split header|{{color|#009|2}}|{{color|#900|0}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
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|}<br />
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{| class="wikitable"<br />
<br />
! {{diagonal split header|{{color|#009|Pair 1}}|{{color|#900|Pair 2}}}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"朋友"<br />(合作)}}<br />
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! scope="col" style="width:6em;" | {{color|#900|"敌人"<br />(背叛)}}<br />
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|-<br />
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! scope="row" style="width:6em;" | {{color|#009|"朋友"<br />(合作)}}<br />
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| {{diagonal split header|{{color|#009|1}}|{{color|#900|1}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|2}}|transparent}}<br />
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|-<br />
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! scope="row" | {{color|#009|"敌人"<br />(背叛)}}<br />
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| {{diagonal split header|{{color|#009|2}}|{{color|#900|0}}|transparent}}<br />
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| {{diagonal split header|{{color|#009|0}}|{{color|#900|0}}|transparent}}<br />
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|}<br />
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This payoff matrix has also been used on the [[United Kingdom|British]] [[television]] programmes ''Trust Me'', ''[[Shafted]]'', ''[[The Bank Job (TV series)|The Bank Job]]'' and ''[[Golden Balls]]'', and on the [[United States|American]] shows ''[[Bachelor Pad]]'' and ''[[Take It All (game show)|Take It All]]''. Game data from the ''[[Golden Balls]]'' series has been analyzed by a team of economists, who found that cooperation was "surprisingly high" for amounts of money that would seem consequential in the real world, but were comparatively low in the context of the game.<ref>{{cite journal | ssrn=1592456 | title=Split or Steal? Cooperative Behavior When the Stakes Are Large | author=Van den Assem, Martijn J. | journal=Management Science |date=January 2012 | volume=58 | issue=1 | pages=2–20 | doi=10.1287/mnsc.1110.1413| url=http://faculty.chicagobooth.edu/richard.thaler/research/pdf/Split%20or%20Steal%20Cooperative%20Behavior%20When%20the%20Stakes%20Are%20Large.pdf }}</ref><br />
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This payoff matrix has also been used on the British television programmes Trust Me, Shafted, The Bank Job and Golden Balls, and on the American shows Bachelor Pad and Take It All. Game data from the Golden Balls series has been analyzed by a team of economists, who found that cooperation was "surprisingly high" for amounts of money that would seem consequential in the real world, but were comparatively low in the context of the game.<br />
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英国电视节目《相信我》、《阴影》、《银行工作》和《黄金球》以及美国电视节目《单身公寓》和《全部拿走》也采用了这种奖励矩阵。一个经济学家团队分析了“金球奖”系列的游戏数据，他们发现，现实生活中，合作对于金额而言“惊人地高” ，但在游戏的背景下，相对较低。<ref>{{cite journal | ssrn=1592456 | title=Split or Steal? Cooperative Behavior When the Stakes Are Large | author=Van den Assem, Martijn J. | journal=Management Science |date=January 2012 | volume=58 | issue=1 | pages=2–20 | doi=10.1287/mnsc.1110.1413| url=http://faculty.chicagobooth.edu/richard.thaler/research/pdf/Split%20or%20Steal%20Cooperative%20Behavior%20When%20the%20Stakes%20Are%20Large.pdf }}</ref><br />
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===Iterated snowdrift===<br />
<font color="#ff8000">迭代雪堆 Iterated snowdrift </font><br />
{{main|snowdrift game}}<br />
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Researchers from the [[University of Lausanne]] and the [[University of Edinburgh]] have suggested that the "Iterated Snowdrift Game" may more closely reflect real-world social situations. Although this model is actually a [[chicken game]], it will be described here. In this model, the risk of being exploited through defection is lower, and individuals always gain from taking the cooperative choice. The snowdrift game imagines two drivers who are stuck on opposite sides of a [[snowdrift]], each of whom is given the option of shoveling snow to clear a path, or remaining in their car. A player's highest payoff comes from leaving the opponent to clear all the snow by themselves, but the opponent is still nominally rewarded for their work.<br />
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Researchers from the University of Lausanne and the University of Edinburgh have suggested that the "Iterated Snowdrift Game" may more closely reflect real-world social situations. Although this model is actually a chicken game, it will be described here. In this model, the risk of being exploited through defection is lower, and individuals always gain from taking the cooperative choice. The snowdrift game imagines two drivers who are stuck on opposite sides of a snowdrift, each of whom is given the option of shoveling snow to clear a path, or remaining in their car. A player's highest payoff comes from leaving the opponent to clear all the snow by themselves, but the opponent is still nominally rewarded for their work.<br />
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来自洛桑大学和爱丁堡大学的研究人员认为，“迭代雪堆游戏”可能更能反映现实世界的社会状况。虽然这个模型实际上是一个胆小鬼博弈。在这个模型中，由于背叛可以降低被剥削的风险，个体总是从合作选择中获益。这个雪堆游戏可以设想两个司机被困在雪堆的两侧，每个司机都可以选择铲雪清理道路，或者留在自己的车里。一个玩家的最高回报来自于让对手清除所有的积雪，但是仍然可以从对手的工作中得到回报。<br />
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This may better reflect real world scenarios, the researchers giving the example of two scientists collaborating on a report, both of whom would benefit if the other worked harder. "But when your collaborator doesn’t do any work, it’s probably better for you to do all the work yourself. You’ll still end up with a completed project."<ref>{{cite web|last=Kümmerli|first=Rolf|title='Snowdrift' game tops 'Prisoner's Dilemma' in explaining cooperation|url=http://phys.org/news111145481.html|accessdate=11 April 2012}}</ref><br />
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This may better reflect real world scenarios, the researchers giving the example of two scientists collaborating on a report, both of whom would benefit if the other worked harder. "But when your collaborator doesn’t do any work, it’s probably better for you to do all the work yourself. You’ll still end up with a completed project."<br />
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这可能更好地反映了现实世界的情景，研究人员举了两位科学家合作完成一份报告的例子，如果另一位科学家更加努力地工作，这两位科学家都会受益。“但当你的合作者不做任何工作时，你自己完成所有的工作可能会更好。你最终还是会完成一个项目。” <ref>{{cite web|last=Kümmerli|first=Rolf|title='Snowdrift' game tops 'Prisoner's Dilemma' in explaining cooperation|url=http://phys.org/news111145481.html|accessdate=11 April 2012}}</ref><br />
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{|<br />
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|-<br />
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|<br />
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{| class="wikitable" style="text-align: center;"<br />
<br />
|+ Example snowdrift payouts (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! Cooperates !! Defects<br />
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|-<br />
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! Cooperates<br />
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| 200, 200 || 100, 300<br />
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|-<br />
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! Defects<br />
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| 300, 100 || 0, 0<br />
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|}<br />
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||<br />
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{| class="wikitable" style="text-align: center;margin-left:2em;"<br />
<br />
|+ Example PD payouts (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! Cooperates !! Defects<br />
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|-<br />
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! Cooperates<br />
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| 200, 200 || -100, 300<br />
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|-<br />
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! Defects<br />
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| 300, -100 || 0, 0<br />
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|}<br />
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{|<br />
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|-<br />
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|<br />
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{| class="wikitable" style="text-align: center;"<br />
<br />
|+ 迭代雪堆的支出示例 (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! 合作 !! 背叛<br />
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|-<br />
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! 合作<br />
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| 200, 200 || 100, 300<br />
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|-<br />
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! 背叛<br />
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| 300, 100 || 0, 0<br />
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|}<br />
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{| class="wikitable" style="text-align: center;margin-left:2em;"<br />
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|+ 囚徒困境支出示例 (A, B)<br />
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! {{diagonal split header|&nbsp;A|B&nbsp;}} !! 合作 !! 背叛<br />
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|-<br />
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! 合作<br />
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| 200, 200 || -100, 300<br />
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|-<br />
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! 背叛<br />
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| 300, -100 || 0, 0<br />
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|}<br />
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===Coordination games===<br />
<font color="#ff8000">协调博弈 Coordination games</font><br />
{{main|coordination games}}<br />
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In coordination games, players must coordinate their strategies for a good outcome. An example is two cars that abruptly meet in a blizzard; each must choose whether to swerve left or right. If both swerve left, or both right, the cars do not collide. The local [[left- and right-hand traffic]] convention helps to co-ordinate their actions.<br />
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In coordination games, players must coordinate their strategies for a good outcome. An example is two cars that abruptly meet in a blizzard; each must choose whether to swerve left or right. If both swerve left, or both right, the cars do not collide. The local left- and right-hand traffic convention helps to co-ordinate their actions.<br />
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在协调博弈中，参与者必须协调自己的策略以获得一个好的结果。一个例子是两辆车在暴风雪中突然相遇，每辆车必须选择是左转还是右转。如果两辆车都向左转弯，或者都向右转弯，那么两辆车就不会相撞。当地的左右向交通惯例有助于协调他们的行动。<br />
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Symmetrical co-ordination games include [[Stag hunt]] and [[Bach or Stravinsky]].<br />
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Symmetrical co-ordination games include Stag hunt and Bach or Stravinsky.<br />
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对称的协调游戏包括猎鹿 Stag hunt和巴赫 Bach或斯特拉文斯基 Stravinsky。<br />
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===Asymmetric prisoner's dilemmas===<br />
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<font color="#ff8000">不对称的囚徒困境 Asymmetric prisoner's dilemmas</font><br />
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A more general set of games are asymmetric. As in the prisoner's dilemma, the best outcome is co-operation, and there are motives for defection. Unlike the symmetric prisoner's dilemma, though, one player has more to lose and/or more to gain than the other. Some such games have been described as a prisoner's dilemma in which one prisoner has an [[alibi]], whence the term "alibi game".<ref>{{cite conference|last1=Robinson |first1=D.R. |last2=Goforth |first2=D.J. |title=Alibi games: the Asymmetric Prisoner' s Dilemmas |date=May 5, 2004 |url=https://economics.ca/2004/papers/0359.pdf |conference=Meetings of the Canadian Economics Association, Toronto, June 4-6, 2004}}</ref><br />
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A more general set of games are asymmetric. As in the prisoner's dilemma, the best outcome is co-operation, and there are motives for defection. Unlike the symmetric prisoner's dilemma, though, one player has more to lose and/or more to gain than the other. Some such games have been described as a prisoner's dilemma in which one prisoner has an alibi, whence the term "alibi game".<br />
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一个更一般的博弈集是不对称的。就像在囚徒困境中一样，最好的结果是合作，而背叛是有动机的。与对称的囚徒困境不同的是一个玩家比另一个玩家有更多的损失或收获。这样的博弈被描述为囚徒困境，其中一个囚徒有不在场证明，这就是术语“不在场证明游戏”的由来。<ref>{{cite conference|last1=Robinson |first1=D.R. |last2=Goforth |first2=D.J. |title=Alibi games: the Asymmetric Prisoner' s Dilemmas |date=May 5, 2004 |url=https://economics.ca/2004/papers/0359.pdf |conference=Meetings of the Canadian Economics Association, Toronto, June 4-6, 2004}}</ref><br />
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In experiments, players getting unequal payoffs in repeated games may seek to maximize profits, but only under the condition that both players receive equal payoffs; this may lead to a stable equilibrium strategy in which the disadvantaged player defects every X games, while the other always co-operates. Such behaviour may depend on the experiment's social norms around fairness.<ref>{{cite chapter|last1=Beckenkamp |first1=Martin |last2=Hennig-Schmidt |first2=Heike |last3=Maier-Rigaud |first3=Frank P. |chapter=Cooperation in Symmetric and Asymmetric Prisoner's Dilemma Games |date=March 4, 2007 |chapter-url=http://homepage.coll.mpg.de/pdf_dat/2006_25online.pdf |title=[[Max Planck Institute for Research on Collective Goods]]}}</ref><br />
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In experiments, players getting unequal payoffs in repeated games may seek to maximize profits, but only under the condition that both players receive equal payoffs; this may lead to a stable equilibrium strategy in which the disadvantaged player defects every X games, while the other always co-operates. Such behaviour may depend on the experiment's social norms around fairness.<br />
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在实验中，在重复博弈中获得不均等收益的参与者可能会寻求利润最大化，但是前提是两个玩家都必须获得均等的收益。这可能会导致一个稳定的均衡策略，即弱势参与者在每隔X场博弈中都会背叛，而另一个参与者总是保持合作。这种行为可能取决于实验围绕公平的社会规范。<ref>{{cite chapter|last1=Beckenkamp |first1=Martin |last2=Hennig-Schmidt |first2=Heike |last3=Maier-Rigaud |first3=Frank P. |chapter=Cooperation in Symmetric and Asymmetric Prisoner's Dilemma Games |date=March 4, 2007 |chapter-url=http://homepage.coll.mpg.de/pdf_dat/2006_25online.pdf |title=[[Max Planck Institute for Research on Collective Goods]]}}</ref><br />
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==Software==<br />
软件<br />
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Several software packages have been created to run prisoner's dilemma simulations and tournaments, some of which have available source code.<br />
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Several software packages have been created to run prisoner's dilemma simulations and tournaments, some of which have available source code.<br />
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已经有一些可以用来运行囚徒困境模拟和比赛的软件包，其中一些有可用的源代码。<br />
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* The source code for the [[The Evolution of Cooperation|second tournament]] run by Robert Axelrod (written by Axelrod and many contributors in [[Fortran]]) is available [http://www-personal.umich.edu/~axe/research/Software/CC/CC2.html online]<br />
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* [https://web.archive.org/web/19991010053242/http://www.lifl.fr/IPD/ipd.frame.html Prison], a library written in [[Java (programming language)|Java]], last updated in 1998<br />
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* [https://github.com/Axelrod-Python/Axelrod Axelrod-Python], written in [[Python (programming language)|Python]]<br />
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* [http://selborne.nl/ipd/ play the Iterative Prisoner's Dilemma in the browser], play against strategies or let strategies play against other strategies<br />
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==In fiction==<br />
在小说中<br />
[[Hannu Rajaniemi]] set the opening scene of his ''[[The Quantum Thief]]'' trilogy in a "dilemma prison". The main theme of the series has been described as the "inadequacy of a binary universe" and the ultimate antagonist is a character called the All-Defector. Rajaniemi is particularly interesting as an artist treating this subject in that he is a Cambridge-trained mathematician and holds a PhD in [[mathematical physics]]&nbsp;– the interchangeability of matter and information is a major feature of the books, which take place in a "post-singularity" future. The first book in the series was published in 2010, with the two sequels, ''[[The Fractal Prince]]'' and ''[[The Causal Angel]]'', published in 2012 and 2014, respectively.<br />
<br />
Hannu Rajaniemi set the opening scene of his The Quantum Thief trilogy in a "dilemma prison". The main theme of the series has been described as the "inadequacy of a binary universe" and the ultimate antagonist is a character called the All-Defector. Rajaniemi is particularly interesting as an artist treating this subject in that he is a Cambridge-trained mathematician and holds a PhD in mathematical physics&nbsp;– the interchangeability of matter and information is a major feature of the books, which take place in a "post-singularity" future. The first book in the series was published in 2010, with the two sequels, The Fractal Prince and The Causal Angel, published in 2012 and 2014, respectively.<br />
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汉努·拉贾尼埃米 Hannu Rajaniemi将他的《量子窃贼》三部曲的开场场景设置在一个“囚徒困境”中。该系列的主题被描述为“双重宇宙的不足” ，最终的对手是一个叫做全面背叛者的角色。作为一个处理这个问题的艺术家，拉贾尼埃米尤其有趣，因为他是剑桥大学培养的数学家，拥有数学物理学博士学位——物质和信息的可互换性是这本书的一个主要特征，它发生在<font color="#ff8000">后奇点post-singularity</font>的未来。该系列的第一本书于2010年出版，其续集《分形王子》和《因果天使》分别于2012年和2014年出版。<br />
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A game modeled after the (iterated) prisoner's dilemma is a central focus of the 2012 video game ''[[Zero Escape: Virtue's Last Reward]]'' and a minor part in its 2016 sequel ''[[Zero Escape: Zero Time Dilemma]]''.<br />
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A game modeled after the (iterated) prisoner's dilemma is a central focus of the 2012 video game Zero Escape: Virtue's Last Reward and a minor part in its 2016 sequel Zero Escape: Zero Time Dilemma.<br />
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一个（迭代）囚徒困境博弈的模型是2012年电子游戏《零度逃脱: 美德的最后奖励》的重点，也是2016年续集《零度逃脱: 极限脱出刻之困境》的一个次要部分。<br />
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In ''The Mysterious Benedict Society and the Prisoner's Dilemma'' by [[Trenton Lee Stewart]], the main characters start by playing a version of the game and escaping from the "prison" altogether. Later they become actual prisoners and escape once again.<br />
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In The Mysterious Benedict Society and the Prisoner's Dilemma by Trenton Lee Stewart, the main characters start by playing a version of the game and escaping from the "prison" altogether. Later they become actual prisoners and escape once again.<br />
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在特伦顿·李·斯图尔特 Trenton Lee Stewart的《神秘的本尼迪克特社会和囚徒困境》中，主要角色从玩一个版本的游戏开始，然后一起逃离“监狱”。后来他们变成了真正的囚犯，再次越狱。<br />
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In ''[[The Adventure Zone]]: Balance'' during ''The Suffering Game'' subarc, the player characters are twice presented with the prisoner's dilemma during their time in two liches' domain, once cooperating and once defecting.<br />
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In The Adventure Zone: Balance during The Suffering Game subarc, the player characters are twice presented with the prisoner's dilemma during their time in two liches' domain, once cooperating and once defecting.<br />
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在冒险区: 苦难游戏的平衡中，玩家角色在他们在两个领域的时间内两次呈现出囚徒困境，一次是合作，一次是背叛。<br />
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In the 8th novel from the author James S. A. Corey [[Tiamat's Wrath]] . Winston Duarte explains the prisoners dilemma in his 14-year-old daughter, Teresa, to train her in strategic thinking. {{cn|date=April 2020}}<br />
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In the 8th novel from the author James S. A. Corey Tiamat's Wrath . Winston Duarte explains the prisoners dilemma in his 14-year-old daughter, Teresa, to train her in strategic thinking. <br />
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在作者詹姆斯·S·A·科里·提亚玛特 James S. A. Corey Tiamat的《愤怒》中的第八部小说中，温斯顿•杜阿尔特 Winston Duarte向他14岁的女儿特蕾莎 Teresa解释了她面临的囚徒困境，来训练她的战略思维能力。<br />
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==See also==<br />
请参阅<br />
{{div col|colwidth=18em}}<br />
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* [[Abilene paradox]]<br />
<font color="#ff8000"> 阿背伦悖论</font><br />
* [[Centipede game]]<br />
<font color="#ff8000"> 蜈蚣博弈</font><br />
* [[Christmas truce]]<br />
<font color="#ff8000"> 圣诞休战</font><br />
* [[Folk theorem (game theory)]]<br />
<font color="#ff8000"> 无名氏定理（博弈论）/font><br />
* [[Free-rider problem]]<br />
<font color="#ff8000"> 搭便车问题</font><br />
* [[Hobbesian trap]]<br />
]<font color="#ff8000"> 霍布斯主义陷阱</font><br />
* [[Innocent prisoner's dilemma]]<br />
<font color="#ff8000"> 无辜囚徒困局</font><br />
* [[Liar Game]]<br />
<font color="#ff8000">说谎者博弈</font><br />
* [[Optional prisoner's dilemma]]<br />
<font color="#ff8000"> 可选择囚徒困境</font><br />
* [[Robert H. Frank#Prisoner's dilemma and cooperation|Prisoner's dilemma and cooperation]]<br />
罗伯特·H·弗兰克囚徒的困境和合作|<font color="#ff8000"> 囚徒困境</font>和合作<br />
* [[Public goods game]]<br />
<font color="#ff8000"> 公共商品博弈</font><br />
* [[Gift-exchange game]]<br />
<font color="#ff8000"> 互利博弈</font><br />
* [[Reciprocal altruism]]<br />
<font color="#ff8000"> 相互利他行为</font><br />
* [[Social preferences]]<br />
<font color="#ff8000"> 社会偏好</font><br />
* [[Swift trust theory]]<br />
<font color="#ff8000"> 快速信任理论</font><br />
* [[Unscrupulous diner's dilemma]]<br />
]<font color="#ff8000"> 无道德食客困境</font><br />
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==References==<br />
==参考==<br />
{{notelist}}<br />
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{{reflist|colwidth=30em}}<br />
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==Further reading==<br />
==拓展阅读==<br />
{{refbegin|30em}}<br />
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* [[S.M. Amadae|Amadae, S.]] (2016). 'Prisoner's Dilemma,' ''Prisoners of Reason.'' [[Cambridge University Press]], NY, pp.&nbsp;24–61.<br />
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* {{cite book |first1=Robert |last1=Aumann |authorlink=Robert Aumann |chapter=Acceptable points in general cooperative ''n''-person games |editor1-first=R. D. |editor1-last=Luce |editor2-first=A. W. |editor2-last=Tucker |title=Contributions to the Theory 23 of Games IV |series=Annals of Mathematics Study |volume=40 |pages=287–324 |publisher=Princeton University Press |location=Princeton NJ |year=1959 |mr=0104521}}<br />
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* [[Robert Axelrod|Axelrod, R.]] (1984). ''[[The Evolution of Cooperation]]''. {{isbn|0-465-02121-2}}<br />
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* [[Cristina Bicchieri|Bicchieri, Cristina]] (1993). Rationality and Coordination. [[Cambridge University Press]].<br />
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* {{cite journal |first1=David M. |last1=Chess |date=December 1988 |title=Simulating the evolution of behavior: the iterated prisoners' dilemma problem |url=http://www.complex-systems.com/pdf/02-6-4.pdf |journal=Complex Systems |volume=2 |issue=6 |pages=663–70}}<br />
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* [[Melvin Dresher|Dresher, M.]] (1961). ''The Mathematics of Games of Strategy: Theory and Applications'' [[Prentice-Hall]], Englewood Cliffs, NJ.<br />
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* Greif, A. (2006). ''Institutions and the Path to the Modern Economy: Lessons from Medieval Trade.'' Cambridge University Press, [[Cambridge]], UK.<br />
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* [[Anatol Rapoport|Rapoport, Anatol]] and Albert M. Chammah (1965). ''Prisoner's Dilemma''. [[University of Michigan Press]].<br />
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{{refend}}<br />
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==External links==<br />
外部链接<br />
*{{Commonscat-inline}}<br />
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* [http://plato.stanford.edu/entries/prisoner-dilemma/ Prisoner's Dilemma (''Stanford Encyclopedia of Philosophy'')]<br />
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* [http://www.msri.org/ext/larryg/pages/15.htm The Bowerbird's Dilemma] The Prisoner's Dilemma in ornithology&nbsp;– mathematical cartoon by Larry Gonick.<br />
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* [https://www.youtube.com/watch?v=_1SEXTVsxjk The Prisoner's Dilemma] The Prisoner's Dilemma with Lego minifigures.<br />
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* {{cite encyclopedia |last1=Dixit |first1=Avinash |authorlink1=Avinash Dixit |last2= Nalebuff |first2=Barry |authorlink2=Barry Nalebuff |editor=[[David R. Henderson]]|encyclopedia=[[Concise Encyclopedia of Economics]] |title=Prisoner's Dilemma |url=http://www.econlib.org/library/Enc/PrisonersDilemma.html |year=2008 |edition= 2nd |publisher=[[Library of Economics and Liberty]] |location=Indianapolis |isbn=978-0865976658 |oclc=237794267}}<br />
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* [http://gametheory101.com/The_Prisoner_s_Dilemma.html Game Theory 101: Prisoner's Dilemma]<br />
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* [https://www.youtube.com/watch?v=I71mjZefg8g Dawkins: Nice Guys Finish First]<br />
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* [https://axelrod.readthedocs.io/en/stable/ Axelrod] Iterated Prisoner's Dilemma [[Python (programming language)|Python]] library<br />
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* [http://gametheorygames.nl/index.html Play the Iterated Prisoner's Dilemma on gametheorygames.nl]<br />
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* [https://web.archive.org/web/20141011014608/http://demo.otree.org/demo/Prisoner%27s+Dilemma/ Play Prisoner's Dilemma on ''oTree''] (N/A 11-5-17)<br />
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* Nicky Case's [https://web.archive.org/web/20181229222135/https://ncase.me/trust/ Evolution of Trust], an example of the donation game<br />
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* [http://iterated-prisoners-dilemma.info Iterated Prisoner's Dilemma online game] by Wayne Davis<br />
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{{Decision theory paradoxes}}<br />
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{{Game theory}}<br />
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{{Authority control}}<br />
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[[Category:Non-cooperative games]]<br />
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Category:Non-cooperative games<br />
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类别: 非合作性游戏<br />
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[[Category:Thought experiments]]<br />
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Category:Thought experiments<br />
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类别: 思维实验<br />
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[[Category:Dilemmas]]<br />
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Category:Dilemmas<br />
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类别: 困境<br />
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[[Category:Environmental studies]]<br />
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Category:Environmental studies<br />
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类别: 环境研究<br />
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[[Category:Social psychology]]<br />
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Category:Social psychology<br />
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类别: 社会心理学<br />
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[[Category:Moral psychology]]<br />
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Category:Moral psychology<br />
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范畴: 道德心理学<br />
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<noinclude><br />
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<small>This page was moved from [[wikipedia:en:Prisoner's dilemma]]. Its edit history can be viewed at [[囚徒困境/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19340二项分布2020-11-28T17:38:19Z<p>Smile：</p>
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<div>此词条暂由南风翻译。已由Smile审校<br />
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
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{{Probability distribution<br />
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<font color="#ff8000">概率分布 Probability distribution </font> <br />
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| name = Binomial distribution<br />
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| name = Binomial distribution<br />
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名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
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| type = mass<br />
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| type = mass<br />
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类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
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| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
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| pdf_image = Probability mass function for the binomial distribution<br />
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| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
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| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
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| cdf_image = Cumulative distribution function for the binomial distribution<br />
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| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
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| notation = B(n,p)<br />
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| 符号 = <math>B(n,p)</math><br />
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| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
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| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
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| 参数 = <br /><math>n \in \{0, 1, 2, \ldots\}</math> &ndash; --- 试验次数； <br /><math>p \in [0,1]</math> &ndash; -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
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| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
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| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
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| 支持 = <br /><math>k \in \{0, 1, \ldots, n\}</math> &ndash; --- 成功的数量<br />
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| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| pdf = \binom{n}{k} p^k q^{n-k}<br />
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|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
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| cdf = I_{q}(n - k, 1 + k)<br />
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| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
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| mean = <math>np</math><br />
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| mean = np<br />
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<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
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| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
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| median = \lfloor np \rfloor or \lceil np \rceil<br />
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<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
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| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
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| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
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| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
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| variance = <math>npq</math><br />
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| variance = npq<br />
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| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
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| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| skewness = \frac{q-p}{\sqrt{npq}}<br />
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| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
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| kurtosis = \frac{1-6pq}{npq}<br />
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| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
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| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
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| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
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| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math>用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
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| mgf = <math>(q + pe^t)^n</math><br />
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| mgf = (q + pe^t)^n<br />
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| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
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| char = <math>(q + pe^{it})^n</math><br />
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| char = (q + pe^{it})^n<br />
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| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
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| pgf = <math>G(z) = [q + pz]^n</math><br />
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| pgf = G(z) = [q + pz]^n<br />
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| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
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| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
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| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
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| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
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}}<br />
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}}<br />
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[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
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Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
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文章File:Pascal's triangle; binomial distribution.svg是<math>p=0.5</math><br />与n和k相关的二项分布。一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。<br />
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In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
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In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
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在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
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A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
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A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
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一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
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二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
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==Definitions==<br />
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===Probability mass function===<br />
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概率质量函数<br />
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In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
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In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
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一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
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:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
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<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
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for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
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对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
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:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
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<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
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is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
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是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
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在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
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:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
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f(k,n,p)=f(n-k,n,1-p). <br />
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<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
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Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
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Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
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把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
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:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
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<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<br />
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并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
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(n+1)p-1 \leq M < (n+1)p.<br />
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<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
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f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;< ''M''&nbsp;是单调递增的，对''k''&nbsp;> ''M''&nbsp;是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−1&nbsp;两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
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===Example===<br />
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Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
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:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
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f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
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<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
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===Cumulative distribution function===<br />
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累积分布函数<br />
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The [[cumulative distribution function]] can be expressed as:<br />
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The cumulative distribution function can be expressed as:<br />
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累积分布函数可以表达为:<br />
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:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
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F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
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<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
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<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
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where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
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<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
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It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
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It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
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在<font color="#ff8000">正则化不完全的beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
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:<math>\begin{align}<br />
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F(k;n,p) & = \Pr(X \le k) \\<br />
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&= I_{1-p}(n-k, k+1) \\<br />
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& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
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\end{align}</math><br />
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which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
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which is equivalent to the cumulative distribution function of the -distribution: <br />
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这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
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:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
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F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
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<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
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Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
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Some closed-form bounds for the cumulative distribution function are given below.<br />
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下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
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== Properties ==<br />
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===Expected value and variance===<br />
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<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
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If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
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If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
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如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
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:<math> \operatorname{E}[X] = np.</math><br />
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\operatorname{E}[X] = np.<br />
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<math> \operatorname{E}[X] = np.</math>。<br />
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This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
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This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
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这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
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:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
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\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
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<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
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The [[variance]] is:<br />
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The variance is:<br />
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方差是:<br />
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:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
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\operatorname{Var}(X) = np(1 - p).<br />
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<math> \operatorname{Var}(X) = np(1 - p).</math><br />
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This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
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This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
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这也是因为独立随机变量和的方差是方差之和。<br />
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===Higher moments===<br />
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<font color="#ff8000">高阶矩 Higher moments </font> <br />
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The first 6 central moments are given by <br />
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The first 6 central moments are given by <br />
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前6个中心矩由<br />
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:<math>\begin{align}<br />
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\mu_1 &= 0, \\<br />
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\mu_2 &= np(1-p),\\<br />
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\mu_3 &= np(1-p)(1-2p),\\<br />
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\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
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\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
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\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\end{align}</math><br />
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===Mode===<br />
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模<br />
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Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
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: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
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\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
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n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
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<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
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证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
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<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
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<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
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当<math>0 < p < 1</math>。我们发现<br />
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:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
}}</ref><br />
<br />
<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（<math>''p''&nbsp;=&nbsp;{{sfrac|1|2}}</math>和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足{{sfrac|1|2}} (''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp; {{sfrac|1|2}} (''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
<br />
<math>F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right);</math><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<math>F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计:<math> \widehat{p} = \frac{x}{n}</math>。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta)</math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到<math>\widehat{p} = 0</math>，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到:<math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math>\widehat{p_{\text{rule of 3}}} = \frac{3}{n})</math><br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
<math>\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }</math><br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
<math>\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.</math><br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta) </math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
<math>\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }</math><br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
<font color="#ff8000">置信区间 Confidence intervals </font><br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
:<math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个地方不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
==== Wald method ====<br />
<br />
<font color="#ff8000">Wald 法</font><br />
<br />
:<math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{p}{z^2}{2n}\widehat{p\,} + \frac{z^2}{2n} + z</math><br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added. {{clarify|date=July 2012}};<br />
<br />
可以添加一个0.5/''n''连续调整。（2012年7月更新）<br />
<br />
<math> \sqrt{\frac{p}{n}\widehat{p\,}(1 - \widehat{p\,}){n} </math><br />
<br />
==== Agresti–Coull method ====<br />
<br />
阿格里斯蒂-库尔方法 Agresti–Coull method<br />
<br />
<math> \frac{z^2}{4 n^2}</math><br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
<math>1 + \frac{z^2}{n}</math><br />
<br />
: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
<font color="#ff8000">弧线法 Arcsine method </font><br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为<math>((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m</math>。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq)。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-''th''分位数的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}</math><br />
<br />
:: <math>\frac{}<br />
<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math><br />
<br />
<math> \widehat{p\,} + \frac{z^2}{2n} + z</math><br />
<br />
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math><br />
<br />
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math><br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为<br />
<br />
<math>\frac{z^2}{4 n^2}</math><br />
<br />
<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对<math>p ^ k = p ^ m p ^ { k-m }</math>进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math> 1 + \frac{z^2}{n}</math><br />
<br />
<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]}</math><ref>{{cite book<br />
<br />
<math> &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math><br />
<br />
| chapter = Confidence intervals<br />
<br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
| year = 2012<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
<math> \Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]</math><br />
<br />
}}</ref><br />
<br />
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此<math>Y \sim B(n, pq)</math>为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<math>\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\</math><br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
<math> &= \binom{n+m}k p^k (1-p)^{n+m-k}</math><br />
<br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
<math>\mathcal{N}(np,\,np(1-p))</math><br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差<math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math><br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
<br />
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分布。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
<br />
<br />
<br />
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
<br />
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
<br />
<br />
<br />
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
<br />
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
<br />
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
<br />
:<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math><br />
<br />
<math>P(p;\alpha,\beta) =\frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.</math><br />
<br />
<math>P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.</math><br />
<br />
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math></math><br />
<br />
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
<br />
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
<br />
<br />
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
<br />
:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
<br />
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
<br />
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
<br />
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
<br />
<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math><br />
<br />
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
<br />
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
:<math>&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math><br />
<br />
After substituting <math> i = k - m </math> in the expression above, we get<br />
<br />
将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
<br />
:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
<br />
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
<br />
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
<br />
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
<br />
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
\end{align}</math><br />
<br />
and thus <math> Y \sim B(n, pq) </math> as desired.<br />
<br />
{{hidden end}}<br />
<br />
<br />
<br />
===Bernoulli distribution===<br />
<br />
伯努利分布<br />
<br />
The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
<br />
<br />
===Poisson binomial distribution===<br />
<br />
<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
<br />
The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
{{Cite journal<br />
<br />
| volume = 3<br />
<br />
| issue = 2<br />
<br />
| pages = 295–312<br />
<br />
| last = Wang<br />
<br />
| first = Y. H.<br />
<br />
| title = On the number of successes in independent trials<br />
<br />
| journal = Statistica Sinica<br />
<br />
| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| url-status = dead<br />
<br />
| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
<br />
}}<br />
</ref><br />
<br />
二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref>{{Cite journal | volume = 3 | issue = 2 | pages = 295–312 | last = Wang | first = Y. H. | title = On the number of successes in independent trials | journal = Statistica Sinica | year = 1993 | url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | url-status = dead | archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | archivedate = 2016-03-03}}</ref><br />
<br />
<br />
<br />
Category:Discrete distributions<br />
<br />
类别: 离散分布<br />
<br />
===Normal approximation===<br />
<br />
<font color="#ff8000">正态逼近 Normal approximation </font><br />
<br />
Category:Factorial and binomial topics<br />
<br />
类别: 阶乘和二项式主题<br />
<br />
<br />
<br />
Category:Conjugate prior distributions<br />
<br />
类别: 共轭先验分布<br />
<br />
[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
<br />
Category: Exponential family distributions<br />
<br />
类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
<br />
<noinclude><br />
<br />
<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
<br />
[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19339二项分布2020-11-28T17:25:54Z<p>Smile：</p>
<hr />
<div>此词条暂由南风翻译。已由Smile审校<br />
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
<br />
{{Probability distribution<br />
<br />
<font color="#ff8000">概率分布 Probability distribution </font> <br />
<br />
| name = Binomial distribution<br />
<br />
| name = Binomial distribution<br />
<br />
名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
<br />
| type = mass<br />
<br />
| type = mass<br />
<br />
类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
<br />
| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
<br />
| pdf_image = Probability mass function for the binomial distribution<br />
<br />
| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
<br />
| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
<br />
| cdf_image = Cumulative distribution function for the binomial distribution<br />
<br />
| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
<br />
| notation = B(n,p)<br />
<br />
| 符号 = <math>B(n,p)</math><br />
<br />
| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
<br />
| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
<br />
| 参数 = <br /><math>n \in \{0, 1, 2, \ldots\}</math> &ndash; --- 试验次数； <br /><math>p \in [0,1]</math> &ndash; -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
<br />
| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
<br />
| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
<br />
| 支持 = <br /><math>k \in \{0, 1, \ldots, n\}</math> &ndash; --- 成功的数量<br />
<br />
| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
<br />
| pdf = \binom{n}{k} p^k q^{n-k}<br />
<br />
|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
<br />
| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
<br />
| cdf = I_{q}(n - k, 1 + k)<br />
<br />
| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
<br />
| mean = <math>np</math><br />
<br />
| mean = np<br />
<br />
<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
<br />
| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
<br />
| median = \lfloor np \rfloor or \lceil np \rceil<br />
<br />
<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
<br />
| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
<br />
| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
<br />
| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
<br />
| variance = <math>npq</math><br />
<br />
| variance = npq<br />
<br />
| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
<br />
| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
<br />
| skewness = \frac{q-p}{\sqrt{npq}}<br />
<br />
| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
<br />
| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
<br />
| kurtosis = \frac{1-6pq}{npq}<br />
<br />
| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
<br />
| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
<br />
| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
<br />
| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math>用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
<br />
| mgf = <math>(q + pe^t)^n</math><br />
<br />
| mgf = (q + pe^t)^n<br />
<br />
| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
<br />
| char = <math>(q + pe^{it})^n</math><br />
<br />
| char = (q + pe^{it})^n<br />
<br />
| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
<br />
| pgf = <math>G(z) = [q + pz]^n</math><br />
<br />
| pgf = G(z) = [q + pz]^n<br />
<br />
| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
<br />
| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
<br />
| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
<br />
| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
<br />
}}<br />
<br />
}}<br />
<br />
}}<br />
<br />
[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
<br />
Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
<br />
文章File:Pascal's triangle; binomial distribution.svg是<math>p=0.5</math><br />与n和k相关的二项分布。一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。<br />
<br />
In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
<br />
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
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在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
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<br />
A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
<br />
A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
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一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
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<br />
The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
<br />
The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
<br />
二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
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<br />
==Definitions==<br />
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<br />
<br />
===Probability mass function===<br />
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概率质量函数<br />
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<br />
<br />
In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
<br />
In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
<br />
一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
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<br />
<br />
:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
<br />
<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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<br />
<br />
for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
<br />
for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
<br />
对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
<br />
<br />
<br />
:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
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<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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<br />
<br />
is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
<br />
is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
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是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
<br />
在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
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:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
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f(k,n,p)=f(n-k,n,1-p). <br />
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<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
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Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
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Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
<br />
把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
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:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
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<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<br />
<br />
并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
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(n+1)p-1 \leq M < (n+1)p.<br />
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<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
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f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;< ''M''&nbsp;是单调递增的，对''k''&nbsp;> ''M''&nbsp;是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−1&nbsp;两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
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===Example===<br />
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Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
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:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
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f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
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<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
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===Cumulative distribution function===<br />
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累积分布函数<br />
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The [[cumulative distribution function]] can be expressed as:<br />
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The cumulative distribution function can be expressed as:<br />
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累积分布函数可以表达为:<br />
<br />
<br />
<br />
:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
<br />
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
<br />
<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
<br />
<br />
<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
<br />
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
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<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
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<br />
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
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It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
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在<font color="#ff8000">正则化不完全的beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
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<br />
:<math>\begin{align}<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
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& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
\end{align}</math><br />
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<br />
<br />
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
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which is equivalent to the cumulative distribution function of the -distribution: <br />
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这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
<br />
<br />
<br />
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
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F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
<br />
<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
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Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
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Some closed-form bounds for the cumulative distribution function are given below.<br />
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下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
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== Properties ==<br />
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===Expected value and variance===<br />
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<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
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If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
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If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
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如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
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<br />
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:<math> \operatorname{E}[X] = np.</math><br />
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\operatorname{E}[X] = np.<br />
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<math> \operatorname{E}[X] = np.</math>。<br />
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This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
<br />
This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
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这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
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:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
<br />
<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
<br />
<br />
The [[variance]] is:<br />
<br />
The variance is:<br />
<br />
方差是:<br />
<br />
:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
\operatorname{Var}(X) = np(1 - p).<br />
<br />
<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
这也是因为独立随机变量和的方差是方差之和。<br />
<br />
<br />
<br />
===Higher moments===<br />
<br />
<font color="#ff8000">高阶矩 Higher moments </font> <br />
<br />
The first 6 central moments are given by <br />
<br />
The first 6 central moments are given by <br />
<br />
前6个中心矩由<br />
<br />
:<math>\begin{align}<br />
<br />
\mu_1 &= 0, \\<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
===Mode===<br />
<br />
模<br />
<br />
<br />
<br />
Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
<br />
<br />
: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
当<math>0 < p < 1</math>。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
}}</ref><br />
<br />
<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（<math>''p''&nbsp;=&nbsp;{{sfrac|1|2}}</math>和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足{{sfrac|1|2}} (''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp; {{sfrac|1|2}} (''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
<br />
<math>F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right);</math><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<math>F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计:<math> \widehat{p} = \frac{x}{n}</math>。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta)</math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到<math>\widehat{p} = 0</math>，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到:<math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math>\widehat{p_{\text{rule of 3}}} = \frac{3}{n})</math><br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
<math>\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }</math><br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
<math>\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.</math><br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta) </math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
<math>\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }</math><br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
<font color="#ff8000">置信区间 Confidence intervals </font><br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
:<math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个地方不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
==== Wald method ====<br />
<br />
<font color="#ff8000">Wald 法</font><br />
<br />
: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{}\widehat{p\,} + \frac{z^2}{2n} + z</math><br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added. {{clarify|date=July 2012}};<br />
<br />
可以添加一个0.5/''n''连续调整。（2012年7月更新）<br />
<br />
<math> \sqrt{\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math><br />
<br />
==== Agresti–Coull method ====<br />
<br />
阿格里斯蒂-库尔方法 Agresti–Coull method<br />
<br />
<math> \frac{z^2}{4 n^2}</math><br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
<math>1 + \frac{z^2}{n}</math><br />
<br />
: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
<font color="#ff8000">弧线法 Arcsine method </font><br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为<math>((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m</math>。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq)。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-''th''分位数的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}</math><br />
<br />
:: <math>\frac{}<br />
<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math><br />
<br />
<math> \widehat{p\,} + \frac{z^2}{2n} + z</math><br />
<br />
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math><br />
<br />
\sqrt{}<br />
<br />
<math> \end{align}</math><br />
<br />
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math><br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为<br />
<br />
<math> \frac{z^2}{4 n^2}</math><br />
<br />
<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对<math>p ^ k = p ^ m p ^ { k-m }</math>进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}</math><br />
<br />
<math> 1 + \frac{z^2}{n}</math><br />
<br />
<math> \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]}</math><ref>{{cite book<br />
<br />
<math> &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math><br />
<br />
| chapter = Confidence intervals<br />
<br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
| year = 2012<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
<math> \Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]</math><br />
<br />
}}</ref><br />
<br />
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此Y \sim B(n, pq)为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\</math><br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
<math> &= \binom{n+m}k p^k (1-p)^{n+m-k}</math><br />
<br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
<math>\mathcal{N}(np,\,np(1-p))</math><br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差<math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math><br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
<br />
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分布。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
<br />
<br />
<br />
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
<br />
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
<br />
<br />
<br />
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
<br />
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
<br />
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
<br />
:<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math><br />
<br />
<math>P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.</math><br />
<br />
<math>P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math><br />
<br />
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
<br />
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
<br />
<br />
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
<br />
:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
<br />
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
<br />
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
<br />
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
<br />
:<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math><br />
<br />
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
<br />
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
\end{align}</math><br />
<br />
After substituting <math> i = k - m </math> in the expression above, we get<br />
<br />
将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
<br />
:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
<br />
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
<br />
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
<br />
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
<br />
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
\end{align}</math><br />
<br />
and thus <math> Y \sim B(n, pq) </math> as desired.<br />
<br />
{{hidden end}}<br />
<br />
<br />
<br />
===Bernoulli distribution===<br />
<br />
伯努利分布<br />
<br />
The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
<br />
<br />
===Poisson binomial distribution===<br />
<br />
<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
<br />
The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
{{Cite journal<br />
<br />
| volume = 3<br />
<br />
| issue = 2<br />
<br />
| pages = 295–312<br />
<br />
| last = Wang<br />
<br />
| first = Y. H.<br />
<br />
| title = On the number of successes in independent trials<br />
<br />
| journal = Statistica Sinica<br />
<br />
| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| url-status = dead<br />
<br />
| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
<br />
}}<br />
</ref><br />
<br />
二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref>{{Cite journal | volume = 3 | issue = 2 | pages = 295–312 | last = Wang | first = Y. H. | title = On the number of successes in independent trials | journal = Statistica Sinica | year = 1993 | url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | url-status = dead | archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | archivedate = 2016-03-03}}</ref><br />
<br />
<br />
<br />
Category:Discrete distributions<br />
<br />
类别: 离散分布<br />
<br />
===Normal approximation===<br />
<br />
<font color="#ff8000">正态逼近 Normal approximation </font><br />
<br />
Category:Factorial and binomial topics<br />
<br />
类别: 阶乘和二项式主题<br />
<br />
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Category:Conjugate prior distributions<br />
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类别: 共轭先验分布<br />
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[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
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Category: Exponential family distributions<br />
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类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
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<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19338二项分布2020-11-28T17:23:46Z<p>Smile：</p>
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<div>此词条暂由南风翻译。已由Smile审校<br />
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
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{{Probability distribution<br />
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<font color="#ff8000">概率分布 Probability distribution </font> <br />
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| name = Binomial distribution<br />
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| name = Binomial distribution<br />
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名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
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| type = mass<br />
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| type = mass<br />
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类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
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| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
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| pdf_image = Probability mass function for the binomial distribution<br />
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| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
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| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
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| cdf_image = Cumulative distribution function for the binomial distribution<br />
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| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
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| notation = B(n,p)<br />
<br />
| 符号 = <math>B(n,p)</math><br />
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| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
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| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
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| 参数 = <br /><math>n \in \{0, 1, 2, \ldots\}</math> &ndash; --- 试验次数； <br /><math>p \in [0,1]</math> &ndash; -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
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| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
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| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
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| 支持 = <br /><math>k \in \{0, 1, \ldots, n\}</math> &ndash; --- 成功的数量<br />
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| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| pdf = \binom{n}{k} p^k q^{n-k}<br />
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|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
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| cdf = I_{q}(n - k, 1 + k)<br />
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| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
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| mean = <math>np</math><br />
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| mean = np<br />
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<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
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| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
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| median = \lfloor np \rfloor or \lceil np \rceil<br />
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<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
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| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
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| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
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| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
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| variance = <math>npq</math><br />
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| variance = npq<br />
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| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
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| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| skewness = \frac{q-p}{\sqrt{npq}}<br />
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| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
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| kurtosis = \frac{1-6pq}{npq}<br />
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| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
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| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
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| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
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| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math>用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
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| mgf = <math>(q + pe^t)^n</math><br />
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| mgf = (q + pe^t)^n<br />
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| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
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| char = <math>(q + pe^{it})^n</math><br />
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| char = (q + pe^{it})^n<br />
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| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
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| pgf = <math>G(z) = [q + pz]^n</math><br />
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| pgf = G(z) = [q + pz]^n<br />
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| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
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| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
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| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
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| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
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}}<br />
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}}<br />
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}}<br />
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[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
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Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
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文章File:Pascal's triangle; binomial distribution.svg是<math>p=0.5</math><br />与n和k相关的二项分布。一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。<br />
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In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
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In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
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在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
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A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
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A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
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一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
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二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
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==Definitions==<br />
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===Probability mass function===<br />
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概率质量函数<br />
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In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
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In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
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一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
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:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
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<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
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for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
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对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
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:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
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<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
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is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
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是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
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在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
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:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
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f(k,n,p)=f(n-k,n,1-p). <br />
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<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
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Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
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Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
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把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
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:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
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<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<br />
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并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
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(n+1)p-1 \leq M < (n+1)p.<br />
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<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
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f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;< ''M''&nbsp是单调递增的，对''k''&nbsp;> ''M''&nbsp是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−1&nbsp两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
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===Example===<br />
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Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
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:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
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f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
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<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
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===Cumulative distribution function===<br />
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累积分布函数<br />
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The [[cumulative distribution function]] can be expressed as:<br />
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The cumulative distribution function can be expressed as:<br />
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累积分布函数可以表达为:<br />
<br />
<br />
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:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
<br />
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
<br />
<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
<br />
<br />
<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
<br />
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
<br />
<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
<br />
<br />
<br />
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
<br />
在<font color="#ff8000">正则化不完全的\beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
<br />
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
<br />
which is equivalent to the cumulative distribution function of the -distribution: <br />
<br />
这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
<br />
<br />
<br />
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
<br />
<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
<br />
<br />
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
<br />
Some closed-form bounds for the cumulative distribution function are given below.<br />
<br />
下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
<br />
<br />
<br />
== Properties ==<br />
<br />
===Expected value and variance===<br />
<br />
<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
<br />
<br />
<br />
If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
<br />
If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
<br />
如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
<br />
<br />
<br />
:<math> \operatorname{E}[X] = np.</math><br />
<br />
\operatorname{E}[X] = np.<br />
<br />
<math> \operatorname{E}[X] = np.</math>。<br />
<br />
<br />
<br />
This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
<br />
This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
<br />
这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
<br />
:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
<br />
<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
<br />
<br />
The [[variance]] is:<br />
<br />
The variance is:<br />
<br />
方差是:<br />
<br />
:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
\operatorname{Var}(X) = np(1 - p).<br />
<br />
<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
这也是因为独立随机变量和的方差是方差之和。<br />
<br />
<br />
<br />
===Higher moments===<br />
<br />
<font color="#ff8000">高阶矩 Higher moments </font> <br />
<br />
The first 6 central moments are given by <br />
<br />
The first 6 central moments are given by <br />
<br />
前6个中心矩由<br />
<br />
:<math>\begin{align}<br />
<br />
\mu_1 &= 0, \\<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
===Mode===<br />
<br />
模<br />
<br />
<br />
<br />
Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
<br />
<br />
: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
当<math>0 < p < 1</math>。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
}}</ref><br />
<br />
<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（''p''&nbsp;= \\sfrac{1}{2}&nbsp; 和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足\\sfrac{1}{2}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp; \\sfrac{1}{2} (''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
<br />
<math>F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right);</math><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>/Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<math>F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计:<math> \widehat{p} = \frac{x}{n}</math>。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta)</math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到<math>\widehat{p} = 0</math>，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到:<math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math>\widehat{p_{\text{rule of 3}}} = \frac{3}{n})</math><br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
<math>\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }</math><br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
<math>\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.</math><br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta) </math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
<math>\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }</math><br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
<font color="#ff8000">置信区间 Confidence intervals </font><br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
:<math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个地方不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
==== Wald method ====<br />
<br />
<font color="#ff8000">Wald 法</font><br />
<br />
: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{<br />
<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z}</math><br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}<br />
\sqrt{<br />
<br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
==== Agresti–Coull method ====<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
<br />
}<br />
<br />
<br />
}{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
<math>1 + \frac{z^2}{n}</math><br />
<br />
: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
<font color="#ff8000">弧线法 Arcsine method </font><br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq)。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-th分位数的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>/alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}<br />
<br />
:: <math>\frac{}<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
\sqrt{<br />
<br />
\end{align}</math><br />
<br />
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math><br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为<br />
<br />
<math> \frac{z^2}{4 n^2}<br />
<br />
\Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对<math>p ^ k = p ^ m p ^ { k-m }</math>进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}<br />
<br />
1 + \frac{z^2}{n}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
}</math><ref>{{cite book}}<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
| chapter = Confidence intervals<br />
<br />
\end{align}</math><br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
在上面的表达式中用 i = k-m 代替后，我们得到了<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。最终将其替换为<br />
<br />
| year = 2012<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
}}</ref><br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
<br />
<br />
\end{align}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此Y \sim B(n, pq)为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\<br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
&= \binom{n+m}k p^k (1-p)^{n+m-k}<br />
<br />
\end{align}</math><br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
<math>\mathcal{N}(np,\,np(1-p))</math><br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差<math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math><br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
<br />
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分布。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
<br />
<br />
<br />
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
<br />
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
<br />
<br />
<br />
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
<br />
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
<br />
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.<br />
<br />
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
<br />
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
<br />
\end{align}</math><br />
<br />
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
<br />
:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
<br />
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
<br />
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
<br />
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
<br />
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
\end{align}</math><br />
<br />
After substituting <math> i = k - m </math> in the expression above, we get<br />
<br />
将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
<br />
:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
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This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
<br />
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
<br />
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
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请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
\end{align}</math><br />
<br />
and thus <math> Y \sim B(n, pq) </math> as desired.<br />
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{{hidden end}}<br />
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===Bernoulli distribution===<br />
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伯努利分布<br />
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The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
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===Poisson binomial distribution===<br />
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<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
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The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
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二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref><br />
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{{Cite journal<br />
<br />
| volume = 3<br />
<br />
| issue = 2<br />
<br />
| pages = 295–312<br />
<br />
| last = Wang<br />
<br />
| first = Y. H.<br />
<br />
| title = On the number of successes in independent trials<br />
<br />
| journal = Statistica Sinica<br />
<br />
| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| url-status = dead<br />
<br />
| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
<br />
}}<br />
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</ref><br />
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Category:Discrete distributions<br />
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类别: 离散分布<br />
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===Normal approximation===<br />
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<font color="#ff8000">正态逼近 Normal approximation </font><br />
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Category:Factorial and binomial topics<br />
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类别: 阶乘和二项式主题<br />
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Category:Conjugate prior distributions<br />
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类别: 共轭先验分布<br />
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[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
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Category: Exponential family distributions<br />
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类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
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<noinclude><br />
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<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19337二项分布2020-11-28T17:21:45Z<p>Smile：</p>
<hr />
<div>此词条暂由南风翻译。已由Smile审校<br />
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
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{{Probability distribution<br />
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<font color="#ff8000">概率分布 Probability distribution </font> <br />
<br />
| name = Binomial distribution<br />
<br />
| name = Binomial distribution<br />
<br />
名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
<br />
| type = mass<br />
<br />
| type = mass<br />
<br />
类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
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| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
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| pdf_image = Probability mass function for the binomial distribution<br />
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| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
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| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
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| cdf_image = Cumulative distribution function for the binomial distribution<br />
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| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
<br />
| notation = B(n,p)<br />
<br />
| 符号 = <math>B(n,p)</math><br />
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| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
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| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
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| 参数 = <br /><math>n \in \{0, 1, 2, \ldots\}</math> &ndash; --- 试验次数； <br /><math>p \in [0,1]</math> &ndash; -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
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| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
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| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
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| 支持 = <br /><math>k \in \{0, 1, \ldots, n\}</math> &ndash; --- 成功的数量<br />
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| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| pdf = \binom{n}{k} p^k q^{n-k}<br />
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|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
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| cdf = I_{q}(n - k, 1 + k)<br />
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| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
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| mean = <math>np</math><br />
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| mean = np<br />
<br />
<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
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| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
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| median = \lfloor np \rfloor or \lceil np \rceil<br />
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<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
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| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
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| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
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| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
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| variance = <math>npq</math><br />
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| variance = npq<br />
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| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
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| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| skewness = \frac{q-p}{\sqrt{npq}}<br />
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| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
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| kurtosis = \frac{1-6pq}{npq}<br />
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| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
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| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
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| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
<br />
| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math>用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
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| mgf = <math>(q + pe^t)^n</math><br />
<br />
| mgf = (q + pe^t)^n<br />
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| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
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| char = <math>(q + pe^{it})^n</math><br />
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| char = (q + pe^{it})^n<br />
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| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
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| pgf = <math>G(z) = [q + pz]^n</math><br />
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| pgf = G(z) = [q + pz]^n<br />
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| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
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| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
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| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
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| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
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}}<br />
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}}<br />
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}}<br />
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[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
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Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
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文章File:Pascal's triangle; binomial distribution.svg是<math>p=0.5</math><br />与n和k相关的二项分布。一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。<br />
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In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
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In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
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在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
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A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
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A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
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一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
<br />
The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
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二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
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==Definitions==<br />
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<br />
<br />
===Probability mass function===<br />
<br />
概率质量函数<br />
<br />
<br />
<br />
In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
<br />
In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
<br />
一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
<br />
<br />
<br />
:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
<br />
<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
<br />
<br />
for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
<br />
for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
<br />
对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
<br />
<br />
<br />
:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
<br />
\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
<br />
<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
<br />
<br />
<br />
is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
<br />
is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
<br />
是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
<br />
在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
<br />
<br />
<br />
:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
<br />
f(k,n,p)=f(n-k,n,1-p). <br />
<br />
<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
<br />
<br />
<br />
Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
<br />
Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
<br />
把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
<br />
:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
<br />
<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<br />
<br />
并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
<br />
<br />
:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
<br />
(n+1)p-1 \leq M < (n+1)p.<br />
<br />
<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
<br />
<br />
<br />
''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
<br />
f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
<br />
''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;< ''M''&nbsp;是单调递增的，对''k''&nbsp;> ''M''&nbsp;是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−1&nbsp;两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
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<br />
===Example===<br />
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<br />
<br />
Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
<br />
<br />
<br />
:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
<br />
f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
<br />
<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
<br />
<br />
<br />
===Cumulative distribution function===<br />
<br />
累积分布函数<br />
<br />
<br />
<br />
The [[cumulative distribution function]] can be expressed as:<br />
<br />
The cumulative distribution function can be expressed as:<br />
<br />
累积分布函数可以表达为:<br />
<br />
<br />
<br />
:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
<br />
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
<br />
<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
<br />
<br />
<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
<br />
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
<br />
<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
<br />
<br />
<br />
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
<br />
在<font color="#ff8000">正则化不完全的beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
<br />
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
<br />
which is equivalent to the cumulative distribution function of the -distribution: <br />
<br />
这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
<br />
<br />
<br />
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
<br />
<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
<br />
<br />
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
<br />
Some closed-form bounds for the cumulative distribution function are given below.<br />
<br />
下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
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<br />
<br />
== Properties ==<br />
<br />
===Expected value and variance===<br />
<br />
<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
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<br />
If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
<br />
If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
<br />
如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
<br />
<br />
<br />
:<math> \operatorname{E}[X] = np.</math><br />
<br />
\operatorname{E}[X] = np.<br />
<br />
<math> \operatorname{E}[X] = np.</math>。<br />
<br />
<br />
<br />
This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
<br />
This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
<br />
这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
<br />
:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
<br />
<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
<br />
<br />
The [[variance]] is:<br />
<br />
The variance is:<br />
<br />
方差是:<br />
<br />
:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
\operatorname{Var}(X) = np(1 - p).<br />
<br />
<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
这也是因为独立随机变量和的方差是方差之和。<br />
<br />
<br />
<br />
===Higher moments===<br />
<br />
<font color="#ff8000">高阶矩 Higher moments </font> <br />
<br />
The first 6 central moments are given by <br />
<br />
The first 6 central moments are given by <br />
<br />
前6个中心矩由<br />
<br />
:<math>\begin{align}<br />
<br />
\mu_1 &= 0, \\<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
===Mode===<br />
<br />
模<br />
<br />
<br />
<br />
Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
<br />
<br />
: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
当<math>0 < p < 1</math>。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
}}</ref><br />
<br />
<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（<math>''p''&nbsp;=&nbsp;{{sfrac|1|2}}</math>和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足{{sfrac|1|2}} (''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp; {{sfrac|1|2}} (''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
<br />
<math>F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right);</math><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<math>F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计:<math> \widehat{p} = \frac{x}{n}</math>。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta)</math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到<math>\widehat{p} = 0</math>，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到:<math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math>\widehat{p_{\text{rule of 3}}} = \frac{3}{n})</math><br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
<math>\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }</math><br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
<math>\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.</math><br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta) </math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
<math>\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }</math><br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
<font color="#ff8000">置信区间 Confidence intervals </font><br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
:<math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个地方不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
==== Wald method ====<br />
<br />
<font color="#ff8000">Wald 法</font><br />
<br />
: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{}\widehat{p\,} + \frac{z^2}{2n} + z</math><br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added. {{clarify|date=July 2012}};<br />
<br />
可以添加一个0.5/''n''连续调整。（2012年7月更新）<br />
<br />
<math> \sqrt{\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math><br />
<br />
==== Agresti–Coull method ====<br />
<br />
阿格里斯蒂-库尔方法 Agresti–Coull method<br />
<br />
<math> \frac{z^2}{4 n^2}</math><br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
<math>1 + \frac{z^2}{n}</math><br />
<br />
: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
<font color="#ff8000">弧线法 Arcsine method </font><br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为<math>((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m</math>。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq)。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-''th''分位数的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}</math><br />
<br />
:: <math>\frac{}<br />
<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math><br />
<br />
<math> \widehat{p\,} + \frac{z^2}{2n} + z</math><br />
<br />
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math><br />
<br />
\sqrt{}<br />
<br />
<math> \end{align}</math><br />
<br />
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math><br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为<br />
<br />
<math> \frac{z^2}{4 n^2}</math><br />
<br />
<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对<math>p ^ k = p ^ m p ^ { k-m }</math>进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}</math><br />
<br />
<math> 1 + \frac{z^2}{n}</math><br />
<br />
<math> \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]}</math><ref>{{cite book<br />
<br />
<math> &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math><br />
<br />
| chapter = Confidence intervals<br />
<br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
| year = 2012<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
<math> \Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]</math><br />
<br />
}}</ref><br />
<br />
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此Y \sim B(n, pq)为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\</math><br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
<math> &= \binom{n+m}k p^k (1-p)^{n+m-k}</math><br />
<br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
<math>\mathcal{N}(np,\,np(1-p))</math><br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差<math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math><br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
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If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
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如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
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The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
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当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分布。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
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For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
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例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
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Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
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关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
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{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
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Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
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由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
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:<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math><br />
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<math>P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.</math><br />
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<math>P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
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&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math><br />
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Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
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给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
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Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
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由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
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:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
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Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
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将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
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Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
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<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
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:<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math><br />
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One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
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一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
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&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
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\end{align}</math><br />
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After substituting <math> i = k - m </math> in the expression above, we get<br />
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将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
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:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
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This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
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这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
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Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
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请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
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:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
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&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
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\end{align}</math><br />
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and thus <math> Y \sim B(n, pq) </math> as desired.<br />
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{{hidden end}}<br />
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===Bernoulli distribution===<br />
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伯努利分布<br />
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The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
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伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
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===Poisson binomial distribution===<br />
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<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
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The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
{{Cite journal<br />
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| volume = 3<br />
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| issue = 2<br />
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| pages = 295–312<br />
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| last = Wang<br />
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| first = Y. H.<br />
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| title = On the number of successes in independent trials<br />
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| journal = Statistica Sinica<br />
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| year = 1993<br />
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| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
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| url-status = dead<br />
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| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
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| archivedate = 2016-03-03<br />
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}}<br />
</ref><br />
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二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref>{{Cite journal | volume = 3 | issue = 2 | pages = 295–312 | last = Wang | first = Y. H. | title = On the number of successes in independent trials | journal = Statistica Sinica | year = 1993 | url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | url-status = dead | archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | archivedate = 2016-03-03}}</ref><br />
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Category:Discrete distributions<br />
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类别: 离散分布<br />
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===Normal approximation===<br />
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<font color="#ff8000">正态逼近 Normal approximation </font><br />
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Category:Factorial and binomial topics<br />
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类别: 阶乘和二项式主题<br />
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Category:Conjugate prior distributions<br />
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类别: 共轭先验分布<br />
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[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
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Category: Exponential family distributions<br />
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类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
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<noinclude><br />
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<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19336二项分布2020-11-28T16:25:08Z<p>Smile：</p>
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<div>此词条暂由南风翻译。已由Smile审校<br />
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
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{{Probability distribution<br />
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<font color="#ff8000">概率分布 Probability distribution </font> <br />
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| name = Binomial distribution<br />
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| name = Binomial distribution<br />
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名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
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| type = mass<br />
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| type = mass<br />
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类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
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| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
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| pdf_image = Probability mass function for the binomial distribution<br />
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| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
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| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
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| cdf_image = Cumulative distribution function for the binomial distribution<br />
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| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
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| notation = B(n,p)<br />
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| 符号 = <math>B(n,p)</math><br />
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| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
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| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
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| 参数 = <br /><math>n \in \{0, 1, 2, \ldots\}</math> &ndash; --- 试验次数； <br /><math>p \in [0,1]</math> &ndash; -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
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| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
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| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
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| 支持 = <br /><math>k \in \{0, 1, \ldots, n\}</math> &ndash; --- 成功的数量<br />
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| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| pdf = \binom{n}{k} p^k q^{n-k}<br />
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|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
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| cdf = I_{q}(n - k, 1 + k)<br />
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| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
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| mean = <math>np</math><br />
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| mean = np<br />
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<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
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| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
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| median = \lfloor np \rfloor or \lceil np \rceil<br />
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<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
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| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
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| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
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| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
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| variance = <math>npq</math><br />
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| variance = npq<br />
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| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
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| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| skewness = \frac{q-p}{\sqrt{npq}}<br />
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| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
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| kurtosis = \frac{1-6pq}{npq}<br />
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| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
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| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
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| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
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| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math>用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
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| mgf = <math>(q + pe^t)^n</math><br />
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| mgf = (q + pe^t)^n<br />
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| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
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| char = <math>(q + pe^{it})^n</math><br />
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| char = (q + pe^{it})^n<br />
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| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
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| pgf = <math>G(z) = [q + pz]^n</math><br />
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| pgf = G(z) = [q + pz]^n<br />
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| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
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| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
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| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
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| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
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}}<br />
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}}<br />
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}}<br />
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[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
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Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
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文章File:Pascal's triangle; binomial distribution.svg是<math>p=0.5</math><br />与n和k相关的二项分布。一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。<br />
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In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
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In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
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在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
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A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
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A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
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一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
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二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
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==Definitions==<br />
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===Probability mass function===<br />
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概率质量函数<br />
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In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
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In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
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一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
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:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
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<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
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for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
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对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
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:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
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<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
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is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
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是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
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在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
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:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
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f(k,n,p)=f(n-k,n,1-p). <br />
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<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
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Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
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Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
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把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
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:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
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<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<br />
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并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
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(n+1)p-1 \leq M < (n+1)p.<br />
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<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
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f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;< ''M''&nbsp是单调递增的，对''k''&nbsp;> ''M''&nbsp是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−1&nbsp两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
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===Example===<br />
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Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
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:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
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f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
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<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
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===Cumulative distribution function===<br />
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累积分布函数<br />
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The [[cumulative distribution function]] can be expressed as:<br />
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The cumulative distribution function can be expressed as:<br />
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累积分布函数可以表达为:<br />
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:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
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F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
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<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
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where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
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where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
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<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
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It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
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It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
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在<font color="#ff8000">正则化不完全的\beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
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:<math>\begin{align}<br />
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F(k;n,p) & = \Pr(X \le k) \\<br />
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&= I_{1-p}(n-k, k+1) \\<br />
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& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
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\end{align}</math><br />
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which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
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which is equivalent to the cumulative distribution function of the -distribution: <br />
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这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
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:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
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F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
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<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
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Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
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Some closed-form bounds for the cumulative distribution function are given below.<br />
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下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
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== Properties ==<br />
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===Expected value and variance===<br />
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<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
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If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
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If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
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如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
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:<math> \operatorname{E}[X] = np.</math><br />
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\operatorname{E}[X] = np.<br />
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<math> \operatorname{E}[X] = np.</math>。<br />
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This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
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This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
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这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
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:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
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\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
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<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
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The [[variance]] is:<br />
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The variance is:<br />
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方差是:<br />
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:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
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\operatorname{Var}(X) = np(1 - p).<br />
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<math> \operatorname{Var}(X) = np(1 - p).</math><br />
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This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
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This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
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这也是因为独立随机变量和的方差是方差之和。<br />
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===Higher moments===<br />
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<font color="#ff8000">高阶矩 Higher moments </font> <br />
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The first 6 central moments are given by <br />
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The first 6 central moments are given by <br />
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前6个中心矩由<br />
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:<math>\begin{align}<br />
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\mu_1 &= 0, \\<br />
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\mu_2 &= np(1-p),\\<br />
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\mu_3 &= np(1-p)(1-2p),\\<br />
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\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
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\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
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\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
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\end{align}</math><br />
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===Mode===<br />
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模<br />
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Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
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Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
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通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
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: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
当<math>0 < p < 1</math>。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
}}</ref><br />
<br />
<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（''p''&nbsp;= \\sfrac{1}{2}&nbsp; 和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足\\sfrac{1}{2}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp; \\sfrac{1}{2} (''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
<br />
<math>F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right);</math><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>/Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<math>F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计:<math> \widehat{p} = \frac{x}{n}</math>。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta)</math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到<math>\widehat{p} = 0</math>，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到:<math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math>\widehat{p_{\text{rule of 3}}} = \frac{3}{n})</math><br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
<math>\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }</math><br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
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When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
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当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
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<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
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<math>\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.</math><br />
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A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
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利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta) </math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
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Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
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For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
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对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
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\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
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<math>\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }</math><br />
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When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
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当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
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=== Confidence intervals ===<br />
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<font color="#ff8000">置信区间 Confidence intervals </font><br />
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{{Main|Binomial proportion confidence interval}}<br />
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<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
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:<math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
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Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
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即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
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In the equations for confidence intervals below, the variables have the following meaning:<br />
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在下面的置信区间公式中，变量具有以下含义<br />
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* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
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* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
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* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
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*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
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The notation in the formula below differs from the previous formulas in two respects:<br />
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下列公式中的符号在两个地方不同于以前的公式:<br />
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* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
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* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
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<br />
==== Wald method ====<br />
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<font color="#ff8000">Wald 法</font><br />
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: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
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<math>\frac{<br />
<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z}</math><br />
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: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}<br />
\sqrt{<br />
<br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
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==== Agresti–Coull method ====<br />
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\frac{z^2}{4 n^2}<br />
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<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
<br />
}<br />
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<br />
}{<br />
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:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
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<math>1 + \frac{z^2}{n}</math><br />
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: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
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The exact (Clopper–Pearson) method is the most conservative.<br />
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确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
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==== Arcsine method ====<br />
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<font color="#ff8000">弧线法 Arcsine method </font><br />
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Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
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设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
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<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
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<br />
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Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
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然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m。<br />
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: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
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==== Wilson (score) method ====<br />
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<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
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If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
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如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
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{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
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For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
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例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq)。<br />
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The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
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下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
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* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
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首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-th分位数的简写。<br />
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* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>/alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
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<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}<br />
<br />
:: <math>\frac{}<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
\sqrt{<br />
<br />
\end{align}</math><br />
<br />
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math><br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
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由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为<br />
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<math> \frac{z^2}{4 n^2}<br />
<br />
\Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
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Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
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对<math>p ^ k = p ^ m p ^ { k-m }</math>进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}<br />
<br />
1 + \frac{z^2}{n}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
}</math><ref>{{cite book}}<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
| chapter = Confidence intervals<br />
<br />
\end{align}</math><br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
在上面的表达式中用 i = k-m 代替后，我们得到了<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
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注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。最终将其替换为<br />
<br />
| year = 2012<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
}}</ref><br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
<br />
<br />
\end{align}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此Y \sim B(n, pq)为所需值。<br />
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The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
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Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
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<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
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伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
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==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
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The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
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二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\<br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
&= \binom{n+m}k p^k (1-p)^{n+m-k}<br />
<br />
\end{align}</math><br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
<math>\mathcal{N}(np,\,np(1-p))</math><br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差<math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math><br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
<br />
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分布。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
<br />
<br />
<br />
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
<br />
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
<br />
<br />
<br />
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
<br />
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
<br />
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.<br />
<br />
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
<br />
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
<br />
\end{align}</math><br />
<br />
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
<br />
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
<br />
:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
<br />
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
<br />
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
<br />
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
<br />
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
\end{align}</math><br />
<br />
After substituting <math> i = k - m </math> in the expression above, we get<br />
<br />
将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
<br />
:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
<br />
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
<br />
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
<br />
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
<br />
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
\end{align}</math><br />
<br />
and thus <math> Y \sim B(n, pq) </math> as desired.<br />
<br />
{{hidden end}}<br />
<br />
<br />
<br />
===Bernoulli distribution===<br />
<br />
伯努利分布<br />
<br />
The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
<br />
<br />
===Poisson binomial distribution===<br />
<br />
<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
<br />
The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
<br />
二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref><br />
<br />
<br />
{{Cite journal<br />
<br />
| volume = 3<br />
<br />
| issue = 2<br />
<br />
| pages = 295–312<br />
<br />
| last = Wang<br />
<br />
| first = Y. H.<br />
<br />
| title = On the number of successes in independent trials<br />
<br />
| journal = Statistica Sinica<br />
<br />
| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| url-status = dead<br />
<br />
| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
<br />
}}<br />
<br />
</ref><br />
<br />
<br />
<br />
Category:Discrete distributions<br />
<br />
类别: 离散分布<br />
<br />
===Normal approximation===<br />
<br />
<font color="#ff8000">正态逼近 Normal approximation </font><br />
<br />
Category:Factorial and binomial topics<br />
<br />
类别: 阶乘和二项式主题<br />
<br />
<br />
<br />
Category:Conjugate prior distributions<br />
<br />
类别: 共轭先验分布<br />
<br />
[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
<br />
Category: Exponential family distributions<br />
<br />
类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
<br />
<noinclude><br />
<br />
<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
<br />
[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19335二项分布2020-11-28T13:37:05Z<p>Smile：</p>
<hr />
<div>此词条暂由南风翻译。已由Smile审校<br />
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
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{{Probability distribution<br />
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<font color="#ff8000">概率分布Probability distribution </font> <br />
<br />
| name = Binomial distribution<br />
<br />
| name = Binomial distribution<br />
<br />
名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
<br />
| type = mass<br />
<br />
| type = mass<br />
<br />
类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
<br />
| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
<br />
| pdf_image = Probability mass function for the binomial distribution<br />
<br />
| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
<br />
| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
<br />
| cdf_image = Cumulative distribution function for the binomial distribution<br />
<br />
| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
<br />
| notation = B(n,p)<br />
<br />
| 符号 = <math>B(n,p)</math><br />
<br />
| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
<br />
| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
<br />
| 参数 = <br /><math>n \in \{0, 1, 2, \ldots\}</math> &ndash; --- 试验次数； <br /><math>p \in [0,1]</math> &ndash -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
<br />
| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
<br />
| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
<br />
| 支持 = <math>k \in \{0, 1, \ldots, n\}</math> &ndash --- 成功的数量<br />
<br />
| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
<br />
| pdf = \binom{n}{k} p^k q^{n-k}<br />
<br />
|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
<br />
| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
<br />
| cdf = I_{q}(n - k, 1 + k)<br />
<br />
| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
<br />
| mean = <math>np</math><br />
<br />
| mean = np<br />
<br />
<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
<br />
| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
<br />
| median = \lfloor np \rfloor or \lceil np \rceil<br />
<br />
<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
<br />
| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
<br />
| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
<br />
| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
<br />
| variance = <math>npq</math><br />
<br />
| variance = npq<br />
<br />
| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
<br />
| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
<br />
| skewness = \frac{q-p}{\sqrt{npq}}<br />
<br />
| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
<br />
| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
<br />
| kurtosis = \frac{1-6pq}{npq}<br />
<br />
| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
<br />
| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
<br />
| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
<br />
| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math> < br /> 用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
<br />
| mgf = <math>(q + pe^t)^n</math><br />
<br />
| mgf = (q + pe^t)^n<br />
<br />
| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
<br />
| char = <math>(q + pe^{it})^n</math><br />
<br />
| char = (q + pe^{it})^n<br />
<br />
| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
<br />
| pgf = <math>G(z) = [q + pz]^n</math><br />
<br />
| pgf = G(z) = [q + pz]^n<br />
<br />
| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
<br />
| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
<br />
| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
<br />
| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
<br />
}}<br />
<br />
}}<br />
<br />
}}<br />
<br />
[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
<br />
Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
<br />
[[文章[[File:Pascal's triangle; binomial distribution.svg]]< br/> < br/> < br/> 是<math>p=0.5</math><br />与n和k相关的二项分布。< br/> < br/> < br/> 一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。]]<br />
<br />
In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
<br />
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
<br />
在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
<br />
<br />
A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
<br />
A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
<br />
一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
<br />
<br />
The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
<br />
The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
<br />
二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
<br />
<br />
==Definitions==<br />
<br />
<br />
<br />
===Probability mass function===<br />
<br />
概率质量函数<br />
<br />
<br />
<br />
In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
<br />
In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
<br />
一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
<br />
<br />
<br />
:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
<br />
<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
<br />
<br />
for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
<br />
for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
<br />
对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
<br />
<br />
<br />
:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
<br />
\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
<br />
<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
<br />
<br />
<br />
is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
<br />
is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
<br />
是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
<br />
在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
<br />
<br />
<br />
:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
<br />
f(k,n,p)=f(n-k,n,1-p). <br />
<br />
<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
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<br />
<br />
Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
<br />
Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
<br />
把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
<br />
:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
<br />
<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<br />
<br />
并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
<br />
<br />
:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
<br />
(n+1)p-1 \leq M < (n+1)p.<br />
<br />
<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
<br />
<br />
<br />
''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
<br />
f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
<br />
''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;<&nbsp是单调递增的，对''k''&nbsp;>&nbsp是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
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===Example===<br />
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<br />
Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
<br />
<br />
<br />
:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
<br />
f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
<br />
<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
<br />
<br />
<br />
===Cumulative distribution function===<br />
<br />
累积分布函数<br />
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<br />
<br />
The [[cumulative distribution function]] can be expressed as:<br />
<br />
The cumulative distribution function can be expressed as:<br />
<br />
累积分布函数可以表达为:<br />
<br />
<br />
<br />
:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
<br />
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
<br />
<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
<br />
<br />
<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
<br />
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
<br />
<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
<br />
<br />
<br />
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
<br />
在<font color="#ff8000">正则化不完全的\beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
<br />
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
<br />
which is equivalent to the cumulative distribution function of the -distribution: <br />
<br />
这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
<br />
<br />
<br />
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
<br />
<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
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<br />
<br />
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
<br />
Some closed-form bounds for the cumulative distribution function are given below.<br />
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下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
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<br />
<br />
== Properties ==<br />
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===Expected value and variance===<br />
<br />
<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
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If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
<br />
If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
<br />
如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
<br />
<br />
<br />
:<math> \operatorname{E}[X] = np.</math><br />
<br />
\operatorname{E}[X] = np.<br />
<br />
<math> \operatorname{E}[X] = np.</math>。<br />
<br />
<br />
<br />
This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
<br />
This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
<br />
这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
<br />
:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
<br />
<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
<br />
<br />
The [[variance]] is:<br />
<br />
The variance is:<br />
<br />
方差是:<br />
<br />
:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
\operatorname{Var}(X) = np(1 - p).<br />
<br />
<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
这也是因为独立随机变量和的方差是方差之和。<br />
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<br />
<br />
===Higher moments===<br />
<br />
<font color="#ff8000">高阶矩 Higher moments </font> <br />
<br />
The first 6 central moments are given by <br />
<br />
The first 6 central moments are given by <br />
<br />
前6个中心矩由<br />
<br />
:<math>\begin{align}<br />
<br />
\mu_1 &= 0, \\<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
===Mode===<br />
<br />
模<br />
<br />
<br />
<br />
Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
<br />
<br />
: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
当<math>0 < p < 1</math>。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（''p''&nbsp;=&nbsp;{{sfrac|1|2}} 和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足{{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right)<br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>/Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计: \widehat{p} = \frac{x}{n}。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用\operatorname{Beta}(\alpha, \beta)作为先验时，后验均值估计量为: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1))，后验均值估计变为\widehat{p_b} = \frac{x+1}{n+2} (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到\widehat{p} = 0，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到: \widehat{p_b} = \frac{1}{n+2})。另一种方法是利用从3个规则获得的置信区间的上界: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }<br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.<br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用\operatorname{Beta}(\alpha, \beta)作为先验时，后验均值估计量为: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1))，后验均值估计变为\widehat{p_b} = \frac{x+1}{n+2} (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }<br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
<font color="#ff8000">置信区间 Confidence intervals </font><br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
:\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个地方不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
==== Wald method ====<br />
<br />
<font color="#ff8000">Wald 法</font><br />
<br />
:: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{<br />
<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}<br />
<br />
\sqrt{<br />
<br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
==== Agresti–Coull method ====<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
<br />
}<br />
<br />
<br />
}{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
1 + \frac{z^2}{n}<br />
<br />
<br />
<br />
}</math><br />
<br />
: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
<font color="#ff8000">弧线法 Arcsine method </font><br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-th分位数的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>/alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X \sim B(n, p)和Y \sim B(X, q)，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}<br />
<br />
:: <math>\frac{<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
\sqrt{<br />
<br />
\end{align}</math><br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}，上述方程可表示为<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
\Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} <br />
<br />
}<br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对 p ^ k = p ^ m p ^ { k-m }进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
1 + \frac{z^2}{n}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
}</math><ref>{{cite book<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
| chapter = Confidence intervals<br />
<br />
\end{align}</math><br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
在上面的表达式中用 i = k-m 代替后，我们得到了<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。最终将其替换为<br />
<br />
| year = 2012<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
}}</ref><br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
<br />
<br />
\end{align}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此Y \sim B(n, pq)为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\<br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
&= \binom{n+m}k p^k (1-p)^{n+m-k}<br />
<br />
\end{align}</math><br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
\mathcal{N}(np,\,np(1-p))<br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差\sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
<br />
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分佈。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
<br />
<br />
<br />
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
<br />
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
<br />
<br />
<br />
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
<br />
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
<br />
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.<br />
<br />
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
<br />
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
<br />
\end{align}</math><br />
<br />
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
<br />
由于<math>/tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
<br />
:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
<br />
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
<br />
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
<br />
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
<br />
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
\end{align}</math><br />
<br />
After substituting <math> i = k - m </math> in the expression above, we get<br />
<br />
将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
<br />
:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
<br />
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
<br />
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
<br />
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
<br />
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
\end{align}</math><br />
<br />
and thus <math> Y \sim B(n, pq) </math> as desired.<br />
<br />
{{hidden end}}<br />
<br />
<br />
<br />
===Bernoulli distribution===<br />
<br />
伯努利分布<br />
<br />
The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<br />
<br />
<br />
<br />
===Poisson binomial distribution===<br />
<br />
<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
<br />
The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
<br />
二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref><br />
<br />
<br />
{{Cite journal<br />
<br />
| volume = 3<br />
<br />
| issue = 2<br />
<br />
| pages = 295–312<br />
<br />
| last = Wang<br />
<br />
| first = Y. H.<br />
<br />
| title = On the number of successes in independent trials<br />
<br />
| journal = Statistica Sinica<br />
<br />
| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| url-status = dead<br />
<br />
| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
<br />
}}<br />
<br />
</ref><br />
<br />
<br />
<br />
Category:Discrete distributions<br />
<br />
类别: 离散分布<br />
<br />
===Normal approximation===<br />
<br />
<font color="#ff8000">正态逼近 Normal approximation </font><br />
<br />
Category:Factorial and binomial topics<br />
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类别: 阶乘和二项式主题<br />
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Category:Conjugate prior distributions<br />
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类别: 共轭先验分布<br />
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[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
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Category: Exponential family distributions<br />
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类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
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<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19334二项分布2020-11-28T13:25:17Z<p>Smile：</p>
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
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{{Probability distribution<br />
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<font color="#ff8000">概率分布Probability distribution </font> <br />
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| name = Binomial distribution<br />
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| name = Binomial distribution<br />
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名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
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| type = mass<br />
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| type = mass<br />
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类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
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| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
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| pdf_image = Probability mass function for the binomial distribution<br />
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| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
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| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
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| cdf_image = Cumulative distribution function for the binomial distribution<br />
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| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
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| notation = B(n,p)<br />
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| 符号 = <math>B(n,p)</math><br />
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| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
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| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
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| 参数 = <br /><math>n \in \{0, 1, 2, \ldots\}</math> &ndash; --- 试验次数； <br /><math>p \in [0,1]</math> &ndash -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
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| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
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| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
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| 支持 = <math>k \in \{0, 1, \ldots, n\}</math> &ndash --- 成功的数量<br />
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| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| pdf = \binom{n}{k} p^k q^{n-k}<br />
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|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
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| cdf = I_{q}(n - k, 1 + k)<br />
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| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
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| mean = <math>np</math><br />
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| mean = np<br />
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<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
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| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
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| median = \lfloor np \rfloor or \lceil np \rceil<br />
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<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
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| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
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| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
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| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
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| variance = <math>npq</math><br />
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| variance = npq<br />
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| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
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| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| skewness = \frac{q-p}{\sqrt{npq}}<br />
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| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
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| kurtosis = \frac{1-6pq}{npq}<br />
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| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
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| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
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| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
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| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math> < br /> 用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
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| mgf = <math>(q + pe^t)^n</math><br />
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| mgf = (q + pe^t)^n<br />
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| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
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| char = <math>(q + pe^{it})^n</math><br />
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| char = (q + pe^{it})^n<br />
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| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
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| pgf = <math>G(z) = [q + pz]^n</math><br />
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| pgf = G(z) = [q + pz]^n<br />
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| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
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| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
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| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
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| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
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}}<br />
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}}<br />
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}}<br />
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[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
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Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
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[[文章[[File:Pascal's triangle; binomial distribution.svg]]< br/> < br/> < br/> 是<math>p=0.5</math><br />与n和k相关的二项分布。< br/> < br/> < br/> 一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。]]<br />
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In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
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In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
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在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
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A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
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A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
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一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
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The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
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二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
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==Definitions==<br />
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===Probability mass function===<br />
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概率质量函数<br />
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In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
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In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
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一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
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:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
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<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
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for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
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for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
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对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
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:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
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<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
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is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
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is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
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是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
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In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
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在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
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:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
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f(k,n,p)=f(n-k,n,1-p). <br />
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<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
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Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
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Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
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把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
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:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
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<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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and comparing it to 1. There is always an integer ''M'' that satisfies<br />
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并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
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:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
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(n+1)p-1 \leq M < (n+1)p.<br />
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<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
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f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
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''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;<&nbsp是单调递增的，对''k''&nbsp;>&nbsp是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
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===Example===<br />
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Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
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假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
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:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
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f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
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<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
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===Cumulative distribution function===<br />
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累积分布函数<br />
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The [[cumulative distribution function]] can be expressed as:<br />
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The cumulative distribution function can be expressed as:<br />
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累积分布函数可以表达为:<br />
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<br />
<br />
:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
<br />
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
<br />
<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
<br />
<br />
<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
<br />
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
<br />
<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
<br />
<br />
<br />
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
<br />
在<font color="#ff8000">正则化不完全的\beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
<br />
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
<br />
which is equivalent to the cumulative distribution function of the -distribution: <br />
<br />
这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
<br />
<br />
<br />
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
<br />
<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
<br />
<br />
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
<br />
Some closed-form bounds for the cumulative distribution function are given below.<br />
<br />
下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
<br />
<br />
<br />
== Properties ==<br />
<br />
===Expected value and variance===<br />
<br />
<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
<br />
<br />
<br />
If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
<br />
If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
<br />
如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
<br />
<br />
<br />
:<math> \operatorname{E}[X] = np.</math><br />
<br />
\operatorname{E}[X] = np.<br />
<br />
<math> \operatorname{E}[X] = np.</math>。<br />
<br />
<br />
<br />
This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
<br />
This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
<br />
这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
<br />
:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
<br />
<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
<br />
<br />
The [[variance]] is:<br />
<br />
The variance is:<br />
<br />
方差是:<br />
<br />
:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
\operatorname{Var}(X) = np(1 - p).<br />
<br />
<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
这也是因为独立随机变量和的方差是方差之和。<br />
<br />
<br />
<br />
===Higher moments===<br />
<br />
<font color="#ff8000">高阶矩 Higher moments </font> <br />
<br />
The first 6 central moments are given by <br />
<br />
The first 6 central moments are given by <br />
<br />
前6个中心矩由<br />
<br />
:<math>\begin{align}<br />
<br />
\mu_1 &= 0, \\<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
===Mode===<br />
<br />
模<br />
<br />
<br />
<br />
Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
<br />
<br />
: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
当<math>0 < p < 1</math>。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin \Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（''p''&nbsp;=&nbsp;{{sfrac|1|2}} 和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足{{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right)<br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>/Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计: \widehat{p} = \frac{x}{n}。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用\operatorname{Beta}(\alpha, \beta)作为先验时，后验均值估计量为: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1))，后验均值估计变为\widehat{p_b} = \frac{x+1}{n+2} (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到\widehat{p} = 0，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到: \widehat{p_b} = \frac{1}{n+2})。另一种方法是利用从3个规则获得的置信区间的上界: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }<br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.<br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用\operatorname{Beta}(\alpha, \beta)作为先验时，后验均值估计量为: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1))，后验均值估计变为\widehat{p_b} = \frac{x+1}{n+2} (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }<br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
<font color="#ff8000">置信区间 Confidence intervals </font><br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
:\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个地方不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
==== Wald method ====<br />
<br />
<font color="#ff8000">Wald 法</font><br />
<br />
:: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{<br />
<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}<br />
<br />
\sqrt{<br />
<br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
==== Agresti–Coull method ====<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
<br />
}<br />
<br />
<br />
}{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
1 + \frac{z^2}{n}<br />
<br />
<br />
<br />
}</math><br />
<br />
: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
<font color="#ff8000">弧线法 Arcsine method </font><br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-th分位数的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>/alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X \sim B(n, p)和Y \sim B(X, q)，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}<br />
<br />
:: <math>\frac{<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
\sqrt{<br />
<br />
\end{align}</math><br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}，上述方程可表示为<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
\Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} <br />
<br />
}<br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对 p ^ k = p ^ m p ^ { k-m }进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
1 + \frac{z^2}{n}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
}</math><ref>{{cite book<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
| chapter = Confidence intervals<br />
<br />
\end{align}</math><br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
在上面的表达式中用 i = k-m 代替后，我们得到了<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。最终将其替换为<br />
<br />
| year = 2012<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
}}</ref><br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
<br />
<br />
\end{align}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此Y \sim B(n, pq)为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\<br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
&= \binom{n+m}k p^k (1-p)^{n+m-k}<br />
<br />
\end{align}</math><br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
\mathcal{N}(np,\,np(1-p))<br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差\sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
<br />
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分佈。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
<br />
<br />
<br />
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
<br />
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
<br />
<br />
<br />
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
<br />
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
<br />
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.<br />
<br />
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
<br />
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
<br />
\end{align}</math><br />
<br />
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
<br />
由于<math>/tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
<br />
:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
<br />
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
<br />
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
<br />
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
<br />
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
\end{align}</math><br />
<br />
After substituting <math> i = k - m </math> in the expression above, we get<br />
<br />
将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
<br />
:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
<br />
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
<br />
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
<br />
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
<br />
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
\end{align}</math><br />
<br />
and thus <math> Y \sim B(n, pq) </math> as desired.<br />
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{{hidden end}}<br />
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===Bernoulli distribution===<br />
<br />
伯努利分布<br />
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The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<br />
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<br />
===Poisson binomial distribution===<br />
<br />
<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
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The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
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二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref><br />
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{{Cite journal<br />
<br />
| volume = 3<br />
<br />
| issue = 2<br />
<br />
| pages = 295–312<br />
<br />
| last = Wang<br />
<br />
| first = Y. H.<br />
<br />
| title = On the number of successes in independent trials<br />
<br />
| journal = Statistica Sinica<br />
<br />
| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| url-status = dead<br />
<br />
| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
<br />
}}<br />
<br />
</ref><br />
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Category:Discrete distributions<br />
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类别: 离散分布<br />
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===Normal approximation===<br />
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<font color="#ff8000">正态逼近 Normal approximation </font><br />
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Category:Factorial and binomial topics<br />
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类别: 阶乘和二项式主题<br />
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Category:Conjugate prior distributions<br />
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类别: 共轭先验分布<br />
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[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
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Category: Exponential family distributions<br />
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类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
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<noinclude><br />
<br />
<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19317二项分布2020-11-28T08:18:40Z<p>Smile：</p>
<hr />
<div>此词条暂由南风翻译。已由Smile审校<br />
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{{short description|Probability distribution}}<br />
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{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
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{{Probability distribution<br />
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<font color="#ff8000">概率分布Probability distribution </font> <br />
<br />
| name = Binomial distribution<br />
<br />
| name = Binomial distribution<br />
<br />
名称 = <font color="#ff8000">二项分布 Binomial distribution </font> <br />
<br />
| type = mass<br />
<br />
| type = mass<br />
<br />
类型 = 质量，这里指<font color="#ff8000">离散型 discrete</font> <br />
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| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
<br />
| pdf_image = Probability mass function for the binomial distribution<br />
<br />
| 概率质量函数图像 = '''<font color="#ff8000">二项分布的概率质量函数 Probability mass function for the binomial distribution </font>'''<br />
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| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
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| cdf_image = Cumulative distribution function for the binomial distribution<br />
<br />
| 累积分布函数图像 = '''<font color="#ff8000">二项分布的累积分布函数 Cumulative distribution function for the binomial distribution </font>'''<br />
| notation = <math>B(n,p)</math><br />
<br />
| notation = B(n,p)<br />
<br />
| 符号 = <math>B(n,p)</math><br />
<br />
| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
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| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
<br />
| 参数 = <math>n \= \{0, 1, 2, \ldots\}</math> &ndash -- 试验次数； <br /><math>p \in [0,1]</math> &ndash -- 每个试验的成功概率； <br /><math>q = 1 - p</math><br />
<br />
| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
<br />
| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
<br />
| 支持 = <math>k \in \{0, 1, \ldots, n\}</math> &ndash --- 成功的数量<br />
<br />
| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
<br />
| pdf = \binom{n}{k} p^k q^{n-k}<br />
<br />
|<font color="#ff8000">概率质量函数 Probability mass function </font> = <math>\binom{n}{k} p^k q^{n-k}</math><br />
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| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
<br />
| cdf = I_{q}(n - k, 1 + k)<br />
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| <font color="#ff8000">累积分布函数 Cumulative distribution function </font> = <math>I_{q}(n - k, 1 + k)</math><br />
<br />
| mean = <math>np</math><br />
<br />
| mean = np<br />
<br />
<font color="#ff8000">平均值 mean</font> = <math>np</math><br />
<br />
| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
<br />
| median = \lfloor np \rfloor or \lceil np \rceil<br />
<br />
<font color="#ff8000">中位数 median</font> = <math>\lfloor np \rfloor</math> 或 <math>\lceil np \rceil</math><br />
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| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
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| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
<br />
| <font color="#ff8000">模 mode</font> = <math>\lfloor (n + 1)p \rfloor</math> 或 <math>\lceil (n + 1)p \rceil - 1</math><br />
<br />
| variance = <math>npq</math><br />
<br />
| variance = npq<br />
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| <font color="#ff8000">方差 variance</font> = <math>npq</math><br />
<br />
| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
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| skewness = \frac{q-p}{\sqrt{npq}}<br />
<br />
| <font color="#ff8000">偏度 skewness</font> = <math>\frac{q-p}{\sqrt{npq}}</math><br />
<br />
| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
<br />
| kurtosis = \frac{1-6pq}{npq}<br />
<br />
| <font color="#ff8000">峰度 kurtosis</font> = <math>\frac{1-6pq}{npq}</math><br />
<br />
| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
<br />
| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
<br />
| <font color="#ff8000">熵 entropy</font> = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math> < br /> 用<font color="#ff8000">香农熵 Shannon entropy</font>测量。对于<font color="#ff8000">分布式消息队列系统 NATS </font>，使用日志中的自然日志。<br />
<br />
| mgf = <math>(q + pe^t)^n</math><br />
<br />
| mgf = (q + pe^t)^n<br />
<br />
| <font color="#ff8000">矩量母函数 Moment Generating Function</font> = <math>(q + pe^t)^n</math><br />
<br />
| char = <math>(q + pe^{it})^n</math><br />
<br />
| char = (q + pe^{it})^n<br />
<br />
| <font color="#ff8000">特征函数 characteristic function</font> = <math>(q + pe^{it})^n</math><br />
<br />
| pgf = <math>G(z) = [q + pz]^n</math><br />
<br />
| pgf = G(z) = [q + pz]^n<br />
<br />
| <font color="#ff8000">概率母函数 probability generating function</font> = <math>G(z) = [q + pz]^n</math><br />
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| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
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| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
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| <font color="#ff8000">费雪信息量 fisher information</font> = <math> g_n(p) = \frac{n}{pq} </math><br />(对于固定的 <math>n</math>)<br />
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}}<br />
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}}<br />
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}}<br />
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[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
<br />
Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
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[[文章[[File:Pascal's triangle; binomial distribution.svg]]< br/> < br/> < br/> 是<math>p=0.5</math><br />与n和k相关的二项分布。< br/> < br/> < br/> 一个8层(''n''&nbsp;=&nbsp;8)的高尔顿盒子中的一个球最终进入中央箱子(''k''&nbsp;=&nbsp;4)的概率是<math>70/256</math>。]]<br />
<br />
In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
<br />
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
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在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为&nbsp;p)或失败/否/错误/0 (概率为 q&nbsp;=&nbsp;1&nbsp;−&nbsp;p)。<br />
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<br />
A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
<br />
A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
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一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n''&nbsp;=&nbsp;1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是流行<font color="#ff8000">统计显著性 statistical significance </font><font color="#ff8000">二项检验 binomial test </font>的基础。<br />
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<br />
The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
<br />
The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
<br />
二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中用替代物抽取的。如果抽样没有用替代物，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。<br />
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<br />
==Definitions==<br />
<br />
<br />
<br />
===Probability mass function===<br />
<br />
概率质量函数<br />
<br />
<br />
<br />
In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
<br />
In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
<br />
一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:<br />
<br />
<br />
<br />
:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
<br />
<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
<br />
<br />
for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
<br />
for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
<br />
对于''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n''，其中<br />
<br />
<br />
<br />
:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
<br />
\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
<br />
<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
<br />
<br />
<br />
is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
<br />
is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
<br />
是<font color="#ff8000">二项式系数</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n''&nbsp;−&nbsp;''k''次失败发生在概率为(1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。<br />
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<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
<br />
在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k''&nbsp;>&nbsp;''n''/2，概率可以通过它的补来计算。<br />
<br />
<br />
<br />
:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
<br />
f(k,n,p)=f(n-k,n,1-p). <br />
<br />
<math>f(k,n,p)=f(n-k,n,1-p). </math>.<br />
<br />
<br />
<br />
Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
<br />
Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
<br />
把表达式''f''(''k'',&nbsp;''n'',&nbsp;''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。<br />
<br />
:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
<br />
<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<br />
<br />
并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
<br />
<br />
:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
<br />
(n+1)p-1 \leq M < (n+1)p.<br />
<br />
<math>(n+1)p-1 \leq M < (n+1)p.</math>.<br />
<br />
<br />
<br />
''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
<br />
f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
<br />
''f''(''k'',&nbsp;''n'',&nbsp;''p'')对''k''&nbsp;<&nbsp是单调递增的，对''k''&nbsp;>&nbsp是单调递减的，但(''n''&nbsp;+&nbsp;1)''p''是整数的情况除外。在这种情况下，有(''n''&nbsp;+&nbsp;1)''p'' 和 (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。<br />
<br />
<br />
<br />
===Example===<br />
<br />
<br />
<br />
Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是<br />
<br />
<br />
<br />
:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
<br />
f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
<br />
<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.<br />
<br />
<br />
<br />
===Cumulative distribution function===<br />
<br />
累积分布函数<br />
<br />
<br />
<br />
The [[cumulative distribution function]] can be expressed as:<br />
<br />
The cumulative distribution function can be expressed as:<br />
<br />
累积分布函数可以表达为:<br />
<br />
<br />
<br />
:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
<br />
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
<br />
<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> ,<br />
<br />
<br />
<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
<br />
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
<br />
<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。<br />
<br />
<br />
<br />
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
<br />
在<font color="#ff8000">正则化不完全的\beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
<br />
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
<br />
which is equivalent to the cumulative distribution function of the -distribution: <br />
<br />
这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref><br />
<br />
<br />
<br />
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
<br />
<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
<br />
<br />
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
<br />
Some closed-form bounds for the cumulative distribution function are given below.<br />
<br />
下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。<br />
<br />
<br />
<br />
== Properties ==<br />
<br />
===Expected value and variance===<br />
<br />
<font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> <br />
<br />
<br />
<br />
If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
<br />
If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
<br />
如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:<br />
<br />
<br />
<br />
:<math> \operatorname{E}[X] = np.</math><br />
<br />
\operatorname{E}[X] = np.<br />
<br />
<math> \operatorname{E}[X] = np.</math>。<br />
<br />
<br />
<br />
This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
<br />
This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
<br />
这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math><br />
<br />
:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
<br />
<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
<br />
<br />
The [[variance]] is:<br />
<br />
The variance is:<br />
<br />
方差是:<br />
<br />
:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
\operatorname{Var}(X) = np(1 - p).<br />
<br />
<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
这也是因为独立随机变量和的方差是方差之和。<br />
<br />
<br />
<br />
===Higher moments===<br />
<br />
<font color="#ff8000">高阶矩 Higher moments </font> <br />
<br />
The first 6 central moments are given by <br />
<br />
The first 6 central moments are given by <br />
<br />
前6个中心矩由<br />
<br />
:<math>\begin{align}<br />
<br />
\mu_1 &= 0, \\<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
===Mode===<br />
<br />
模<br />
<br />
<br />
<br />
Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n''&nbsp;+&nbsp;1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n''&nbsp;+&nbsp;1)''p''和(''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:<br />
<br />
<br />
: <math>\text{mode} =<br />
<br />
\begin{cases}<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
\end{cases}</math> <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
当<math>0 < p < 1</math>。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
\end{align}</math><br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin \Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
*任何中位数''m''都必须满足⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉。<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
<br />
*中位数''m''不能离均值太远。{{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}<ref name="Hamza">{{Cite journal<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中D(a || p)是参数为a和p的<font color="#ff8000">相对熵 relative entropy </font>，即Bernoulli(a)和Bernoulli(p)概率分布的差值:<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限十分严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
*中位数是唯一的并且等于''m''&nbsp;=&nbsp;[[Rounding|round]](''np'')，此时|''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''}（''p''&nbsp;=&nbsp;{{sfrac|1|2}} 和 ''n'' 是奇数的情况除外）<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更宽松的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于''p''&nbsp;=&nbsp;1/2且''n''是奇数，任意''m''满足{{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1)是一个二项分布的中位数。如果''p''&nbsp;=&nbsp;1/2且''n'' 是偶数，那么''m''&nbsp;=&nbsp;''n''/2是唯一的中位数:<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right)<br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。<br />
<br />
===Tail bounds===<br />
<br />
<font color="#ff8000">尾部边界 Tail bounds </font> <br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>/Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
<font color="#ff8000">霍夫丁不等式 Hoeffding's inequality </font>得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但是Hoeffding的约束评价为一个正的常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计: \widehat{p} = \frac{x}{n}。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用\operatorname{Beta}(\alpha, \beta)作为先验时，后验均值估计量为: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1))，后验均值估计变为\widehat{p_b} = \frac{x+1}{n+2} (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到\widehat{p} = 0，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到: \widehat{p_b} = \frac{1}{n+2})。另一种方法是利用从3个规则获得的置信区间的上界: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐近地，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
当''p''&nbsp;=&nbsp;1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } }<br />
<br />
<br />
== Statistical Inference ==<br />
<br />
<font color="#ff8000">统计推断 Statistical Inference</font><br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
<font color="#ff8000">参数估计 Estimation of parameters</font><br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
Beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math>。This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.<br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用\operatorname{Beta}(\alpha, \beta)作为先验时，后验均值估计量为: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。<br />
<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里p的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1))，后验均值估计变为\widehat{p_b} = \frac{x+1}{n+2} (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }<br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
<font color="#ff8000">置信区间 Confidence intervals </font><br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
:\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个地方不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
* <math>z</math>是<font color="#ff8000">标准正态分布 standard normal distribution </font>的<math>1 - \tfrac{1}{2}\alpha</math>分位数(即概率)对应的目标错误率 <math>\alpha</math>。例如，95%的<font color="#ff8000">置信度 confidence level </font>的错误率为<math>\alpha</math>&nbsp;=&nbsp;0.05，因此 <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 并且<math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
==== Wald method ====<br />
<br />
<font color="#ff8000">Wald 法</font><br />
<br />
:: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{<br />
<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}<br />
<br />
\sqrt{<br />
<br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
==== Agresti–Coull method ====<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
<br />
}<br />
<br />
<br />
}{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
1 + \frac{z^2}{n}<br />
<br />
<br />
<br />
}</math><br />
<br />
: Here the estimate of ''p'' is modified to<br />
这里''p''的估计量被修改为<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
<font color="#ff8000">弧线法 Arcsine method </font><br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设X&nbsp;~&nbsp;B(n,p1)和Y&nbsp;~&nbsp;B(m,p2)是独立的。设T = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
<font color="#ff8000">威尔逊法 Wilson (score) method </font><br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果X&nbsp;~&nbsp;B(n,&nbsp;p)和Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (给定Y的条件分布&nbsp;X) ，则Y是服从Y&nbsp;~&nbsp;B(n,&nbsp;pq)的简单二项随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X&nbsp;~&nbsp;B(n,&nbsp;p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y&nbsp;~&nbsp;B(X,&nbsp;q)，那么Y&nbsp;~&nbsp;B(n,&nbsp;pq。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1&nbsp;−&nbsp;''x'')-th分位数的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>/alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于X \sim B(n, p)和Y \sim B(X, q)，由<font color="#ff8000">全概率公式 the law of total probability </font>,<br />
<br />
<br />
<br />
<math>\begin{align}<br />
<br />
:: <math>\frac{<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
\sqrt{<br />
<br />
\end{align}</math><br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}，上述方程可表示为<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
\Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} <br />
<br />
}<br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对 p ^ k = p ^ m p ^ { k-m }进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
1 + \frac{z^2}{n}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
}</math><ref>{{cite book<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
| chapter = Confidence intervals<br />
<br />
\end{align}</math><br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
在上面的表达式中用 i = k-m 代替后，我们得到了<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。最终将其替换为<br />
<br />
| year = 2012<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
}}</ref><br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
<br />
<br />
\end{align}</math> <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此Y \sim B(n, pq)为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" /><br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中n&nbsp;=&nbsp;1。在符号上，X&nbsp;~&nbsp;B(1,&nbsp;p)与X&nbsp;~&nbsp;Bernoulli(p)具有相同的意义。反之，任何二项分布B(n,&nbsp;p)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率相同且为''p''，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\<br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5的概率质量函数和正态概率密度函数近似<br />
<br />
&= \binom{n+m}k p^k (1-p)^{n+m-k}<br />
<br />
\end{align}</math><br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n,&nbsp;p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
\mathcal{N}(np,\,np(1-p))<br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由卡兹 Katz和合著者在1978年得出的。<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
<br />
<br />
Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
<br />
令''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>)和''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。<br />
<br />
For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差\sigma = \sqrt{\frac{p(1-p)}{n}}<br />
<br />
Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
<br />
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
<br />
<br />
<br />
<br />
===Conditional binomials===<br />
<br />
<font color="#ff8000">条件二项式 Conditional binomials </font><br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
<br />
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分佈。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n&nbsp;≥&nbsp;20和p&nbsp;≤&nbsp;0.05,或者如果n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10，则这个近似是好的。<br />
<br />
<br />
<br />
For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
<br />
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
<br />
<br />
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
<br />
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。<br />
<br />
<br />
<br />
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
<br />
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
<br />
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.<br />
<br />
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
<br />
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。<br />
<br />
\end{align}</math><br />
<br />
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
<br />
由于<math>/tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
<br />
:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
<br />
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
<br />
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
<br />
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
<br />
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
<br />
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
\end{align}</math><br />
<br />
After substituting <math> i = k - m </math> in the expression above, we get<br />
<br />
将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
<br />
:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
<br />
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
<br />
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r&nbsp;+&nbsp;s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p&nbsp;=&nbsp;1/2的情况。<br />
<br />
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
<br />
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
\end{align}</math><br />
<br />
and thus <math> Y \sim B(n, pq) </math> as desired.<br />
<br />
{{hidden end}}<br />
<br />
<br />
<br />
===Bernoulli distribution===<br />
<br />
伯努利分布<br />
<br />
The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
<br />
伯努利分布是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<br />
<br />
<br />
<br />
===Poisson binomial distribution===<br />
<br />
<font color="#ff8000">泊松二项分布 Poisson binomial distribution </font><br />
<br />
The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
<br />
二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref><br />
<br />
<br />
{{Cite journal<br />
<br />
| volume = 3<br />
<br />
| issue = 2<br />
<br />
| pages = 295–312<br />
<br />
| last = Wang<br />
<br />
| first = Y. H.<br />
<br />
| title = On the number of successes in independent trials<br />
<br />
| journal = Statistica Sinica<br />
<br />
| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| url-status = dead<br />
<br />
| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
<br />
}}<br />
<br />
</ref><br />
<br />
<br />
<br />
Category:Discrete distributions<br />
<br />
类别: 离散分布<br />
<br />
===Normal approximation===<br />
<br />
<font color="#ff8000">正态逼近 Normal approximation </font><br />
<br />
Category:Factorial and binomial topics<br />
<br />
类别: 阶乘和二项式主题<br />
<br />
<br />
<br />
Category:Conjugate prior distributions<br />
<br />
类别: 共轭先验分布<br />
<br />
[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
<br />
Category: Exponential family distributions<br />
<br />
类别: <font color="#ff8000">指数族分布 Exponential family distributions </font> <br />
<br />
<noinclude><br />
<br />
<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
<br />
[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83&diff=19071二项分布2020-11-24T00:52:35Z<p>Smile：</p>
<hr />
<div>此词条暂由南风翻译<br />
<br />
{{short description|Probability distribution}}<br />
<br />
{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}<br />
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{{see also|Negative binomial distribution}}<br />
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{{Probability distribution<br />
<br />
{{Probability distribution<br />
<br />
{概率分布<br />
<br />
| name = Binomial distribution<br />
<br />
| name = Binomial distribution<br />
<br />
二项分布<br />
<br />
| type = mass<br />
<br />
| type = mass<br />
<br />
类型 = 质量<br />
<br />
| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]<br />
<br />
| pdf_image = Probability mass function for the binomial distribution<br />
<br />
2012年3月24日 | pdf 图片 = '''<font color="#ff8000">概率质量函数二项分布</font>'''<br />
<br />
| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]<br />
<br />
| cdf_image = Cumulative distribution function for the binomial distribution<br />
<br />
2012年3月24日 | cdf 图像 = '''<font color="#ff8000">累积分布函数二项分布</font>'''<br />
| notation = <math>B(n,p)</math><br />
<br />
| notation = B(n,p)<br />
<br />
| 符号 = b (n，p)<br />
<br />
| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> &ndash; number of trials<br /><math>p \in [0,1]</math> &ndash; success probability for each trial<br /><math>q = 1 - p</math><br />
<br />
| parameters = n \in \{0, 1, 2, \ldots\} &ndash; number of trials<br />p \in [0,1] &ndash; success probability for each trial<br />q = 1 - p<br />
<br />
| 参数 = n in {0,1,2，ldots } -- 试验次数 < br/> p in [0,1] -- 每个试验的成功概率 < br/> q = 1-p<br />
<br />
| support = <math>k \in \{0, 1, \ldots, n\}</math> &ndash; number of successes<br />
<br />
| support = k \in \{0, 1, \ldots, n\} &ndash; number of successes<br />
<br />
| support = k in {0,1，ldots，n }——成功的数量<br />
<br />
| pdf = <math>\binom{n}{k} p^k q^{n-k}</math><br />
<br />
| pdf = \binom{n}{k} p^k q^{n-k}<br />
<br />
| pdf = [math]\binom{n}{k} p^k q^{n-k}[/math]<br />
<br />
| cdf = <math>I_{q}(n - k, 1 + k)</math><br />
<br />
| cdf = I_{q}(n - k, 1 + k)<br />
<br />
| cdf = i _ { q }(n-k，1 + k)<br />
<br />
| mean = <math>np</math><br />
<br />
| mean = np<br />
<br />
平均值 = np<br />
<br />
| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math><br />
<br />
| median = \lfloor np \rfloor or \lceil np \rceil<br />
<br />
中位数 = lfloor np rfloor 或 lceil np rceil<br />
<br />
| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math><br />
<br />
| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1<br />
<br />
| 模 = lfloor (n + 1) p rfloor 或 lceil (n + 1) p rceil-1<br />
<br />
| variance = <math>npq</math><br />
<br />
| variance = npq<br />
<br />
| 方差 = npq<br />
<br />
| skewness = <math>\frac{q-p}{\sqrt{npq}}</math><br />
<br />
| skewness = \frac{q-p}{\sqrt{npq}}<br />
<br />
| 偏度 = frac { q-p }{ sqrt { npq }}<br />
<br />
| kurtosis = <math>\frac{1-6pq}{npq}</math><br />
<br />
| kurtosis = \frac{1-6pq}{npq}<br />
<br />
| 峰度 = frac {1-6pq }{ npq }<br />
<br />
| entropy = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math><br /> in [[Shannon (unit)|shannons]]. For [[nat (unit)|nats]], use the natural log in the log.<br />
<br />
| entropy = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)<br /> in shannons. For nats, use the natural log in the log.<br />
<br />
| 熵 = frac {1}{2} log _ 2(2 pi enpq) + o left (frac {1}{ n } right) < br/> in shannons。对于 nats，使用原木中的自然原木。<br />
<br />
| mgf = <math>(q + pe^t)^n</math><br />
<br />
| mgf = (q + pe^t)^n<br />
<br />
| mgf = (q + pe ^ t) ^ n<br />
<br />
| char = <math>(q + pe^{it})^n</math><br />
<br />
| char = (q + pe^{it})^n<br />
<br />
| char = (q + pe ^ { it }) ^ n<br />
<br />
| pgf = <math>G(z) = [q + pz]^n</math><br />
<br />
| pgf = G(z) = [q + pz]^n<br />
<br />
| pgf = g (z) = [ q + pz ] ^ n<br />
<br />
| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)<br />
<br />
| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)<br />
<br />
| fisher = g _ n (p) = frac { n }{ pq } < br/> (对于固定 n)<br />
<br />
}}<br />
<br />
}}<br />
<br />
}}<br />
<br />
[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n''&nbsp;=&nbsp;8) ends up in the central bin (''k''&nbsp;=&nbsp;4) is <math>70/256</math>.]]<br />
<br />
Binomial distribution for p=0.5<br />with n and k as in [[Pascal's triangle<br /><br />The probability that a ball in a Galton box with 8 layers (n&nbsp;=&nbsp;8) ends up in the central bin (k&nbsp;=&nbsp;4) is 70/256.]]<br />
<br />
在[帕斯卡三角形 < br/> < br/> < br/> 中，p = 0.5 < br/> 与 n 和 k 相关的二项分布为[[帕斯卡三角形 < br/> < br/> < br/> 一个8层的高尔顿盒子中的球最终进入中央箱子(k = 4)的概率是70/256]<br />
<br />
In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters ''n'' and ''p'' is the [[discrete probability distribution]] of the number of successes in a sequence of ''n'' [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[boolean-valued function|boolean]]-valued [[outcome (probability)|outcome]]: [[wikt:success|success]]/[[yes and no|yes]]/[[truth value|true]]/[[one]] (with [[probability]] ''p'') or [[failure]]/[[yes and no|no]]/[[false (logic)|false]]/[[zero]] (with [[probability]] ''q''&nbsp;=&nbsp;1&nbsp;−&nbsp;''p''). <br />
<br />
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q&nbsp;=&nbsp;1&nbsp;−&nbsp;p). <br />
<br />
在概率论和统计学中，参数为 n 和 p 的'''<font color="#ff8000">二项分布</font>'''是 n 个独立实验序列中成功次数的'''<font color="#ff8000">离散概率分布</font>'''，每个实验询问一个 是-否 问题，每个实验都有自己的布尔值结果: 成功/是/正确/一 (概率为 p)或 失败/否/错误/零 (概率为 q = 1-p)。<br />
<br />
<br />
A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., ''n''&nbsp;=&nbsp;1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].<br />
<br />
A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n&nbsp;=&nbsp;1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.<br />
<br />
一个单一的结果为成功/失败的实验也被称为'''<font color="#ff8000">伯努利试验</font>'''或伯努利实验，一系列伯努利实验结果被称为'''<font color="#ff8000">伯努利过程</font>'''; 对于一个单一的实验，例如，n = 1，这个二项分布是一个'''<font color="#ff8000">伯努利分布</font>'''。二项分布是流行'''<font color="#ff8000">统计显著性</font>''''''<font color="#ff8000">二项检验</font>'''的基础。<br />
<br />
<br />
The binomial distribution is frequently used to model the number of successes in a sample of size ''n'' drawn [[With replacement|with replacement]] from a population of size ''N''. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one. However, for ''N'' much larger than ''n'', the binomial distribution remains a good approximation, and is widely used.<br />
<br />
The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.<br />
<br />
二项分布经常被用来模拟大小为 n 的样本中的成功数量，这些样本是用 大小为N的种群中的替代物抽取的。如果抽样没有更换，抽样就不是独立的，所以得到的分布是'''<font color="#ff8000">超几何分布</font>'''，而不是二项分布。然而，对于 N 比 n 大得多的情况，二项分布仍然是一个很好的近似值，并且被广泛使用。<br />
<br />
<br />
==Definitions==<br />
<br />
<br />
<br />
===Probability mass function===<br />
<br />
概率质量函数<br />
<br />
<br />
<br />
In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:<br />
<br />
In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X&nbsp;~&nbsp;B(n,&nbsp;p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:<br />
<br />
一般来说，如果'''<font color="#ff8000">随机变量</font>''' x 服从参数 n ∈ N 和 p ∈[0,1]的二项分布，我们写作 x ~ b (n，p)。在 n 个独立的伯努利试验中获得 k 成功的概率是由'''<font color="#ff8000">概率质量函数</font>'''给出的:<br />
<br />
<br />
<br />
:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math><br />
<br />
f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}<br />
<br />
F (k，n，p) = Pr (k; n，p) = Pr (x = k) = binom { n }{ k } p ^ k (1-p) ^ { n-k }<br />
<br />
<br />
<br />
for ''k''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;''n'', where<br />
<br />
for k&nbsp;=&nbsp;0,&nbsp;1,&nbsp;2,&nbsp;...,&nbsp;n, where<br />
<br />
对于 k = 0,1,2，... ，n，其中<br />
<br />
<br />
<br />
:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math><br />
<br />
\binom{n}{k} =\frac{n!}{k!(n-k)!}<br />
<br />
(n-k) <br />
<br />
<br />
<br />
is the [[binomial coefficient]], hence the name of the distribution. The formula can be understood as follows. ''k'' successes occur with probability ''p''<sup>''k''</sup> and ''n''&nbsp;−&nbsp;''k'' failures occur with probability (1&nbsp;−&nbsp;''p'')<sup>''n''&nbsp;−&nbsp;''k''</sup>. However, the ''k'' successes can occur anywhere among the ''n'' trials, and there are <math>\binom{n}{k}</math> different ways of distributing ''k'' successes in a sequence of ''n'' trials.<br />
<br />
is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n&nbsp;−&nbsp;k failures occur with probability (1&nbsp;−&nbsp;p)n&nbsp;−&nbsp;k. However, the k successes can occur anywhere among the n trials, and there are \binom{n}{k} different ways of distributing k successes in a sequence of n trials.<br />
<br />
是'''<font color="#ff8000">二项式系数</font>'''，因而得名。这个公式可以理解为。K次成功发生在概率为 pk 的情况下， n-k 次失败发生在概率为(1-p) n-k 的情况下。然而，k 次成功可以发生在 n 个试验中的任何地方，并且在 n 个试验序列中有不同的 k 次成功的分配方法。<br />
<br />
<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k''&nbsp;>&nbsp;''n''/2, the probability can be calculated by its complement as<br />
<br />
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k&nbsp;>&nbsp;n/2, the probability can be calculated by its complement as<br />
<br />
在为二项分布概率创建参考表时，通常表中最多填充到 n/2的值。这是因为对于 k > n/2，概率可以通过它的补来计算<br />
<br />
<br />
<br />
:<math>f(k,n,p)=f(n-k,n,1-p). </math><br />
<br />
f(k,n,p)=f(n-k,n,1-p). <br />
<br />
F (k，n，p) = f (n-k，n，1-p).<br />
<br />
<br />
<br />
Looking at the expression ''f''(''k'',&nbsp;''n'',&nbsp;''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating<br />
<br />
Looking at the expression f(k,&nbsp;n,&nbsp;p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating<br />
<br />
把表达式 f (k，n，p)看作 k 的函数，有一个 k 值使它最大化。这个 k 值可以通过计算找到<br />
<br />
:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math><br />
<br />
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} <br />
<br />
Frac { f (k + 1，n，p)}{ f (k，n，p)} = frac {(n-k) p }{(k + 1)(1-p)}<br />
<br />
and comparing it to 1. There is always an integer ''M'' that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref><br />
<br />
and comparing it to 1. There is always an integer M that satisfies<br />
<br />
并与1相比较。总有一个整数 M 满足<br />
<br />
<br />
<br />
:<math>(n+1)p-1 \leq M < (n+1)p.</math><br />
<br />
(n+1)p-1 \leq M < (n+1)p.<br />
<br />
(n + 1) p-1 leq m < (n + 1) p.<br />
<br />
<br />
<br />
''f''(''k'',&nbsp;''n'',&nbsp;''p'') is monotone increasing for ''k''&nbsp;<&nbsp;''M'' and monotone decreasing for ''k''&nbsp;>&nbsp;''M'', with the exception of the case where (''n''&nbsp;+&nbsp;1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. ''M'' is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].<br />
<br />
f(k,&nbsp;n,&nbsp;p) is monotone increasing for k&nbsp;<&nbsp;M and monotone decreasing for k&nbsp;>&nbsp;M, with the exception of the case where (n&nbsp;+&nbsp;1)p is an integer. In this case, there are two values for which f is maximal: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.<br />
<br />
F (k，n，p)对 k < m 是单调递增的，对 k > m 是单调递减的，但(n + 1) p 是整数的情况除外。在这种情况下，有两个值使 f 是最大的: (n + 1) p 和(n + 1) p-1。M 是伯努利试验最有可能的结果(也就是说，最有可能的结果，尽管总的来说仍然存在不发生的情况) ，被称为'''<font color="#ff8000">模</font>'''。<br />
<br />
<br />
<br />
===Example===<br />
<br />
<br />
<br />
Suppose a [[fair coin|biased coin]] comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is<br />
<br />
假设抛出一枚'''<font color="#ff8000">有偏硬币</font>'''时，正面朝上的概率为0.3。在6次抛掷中正好看到4个正面的概率是<br />
<br />
<br />
<br />
:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math><br />
<br />
f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.<br />
<br />
F (4,6,0.3) = binom {6}{4}0.3 ^ 4(1-0.3) ^ {6-4} = 0.059535.<br />
<br />
<br />
<br />
===Cumulative distribution function===<br />
<br />
累积分布函数<br />
<br />
<br />
<br />
The [[cumulative distribution function]] can be expressed as:<br />
<br />
The cumulative distribution function can be expressed as:<br />
<br />
'''<font color="#ff8000">累积分布函数</font>'''可以表达为:<br />
<br />
<br />
<br />
:<math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math><br />
<br />
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},<br />
<br />
F (k; n，p) = Pr (x le k) = sum _ { i = 0} ^ { lfloor k rfloor }{ n choose i } p ^ i (1-p) ^ { n-i } ,<br />
<br />
<br />
<br />
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.<br />
<br />
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.<br />
<br />
是 k 下面的”楼层” ，也就是。小于或等于 k 的'''<font color="#ff8000">最大整数</font>'''。<br />
<br />
<br />
<br />
It can also be represented in terms of the [[regularized incomplete beta function]], as follows:<ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref><br />
<br />
It can also be represented in terms of the regularized incomplete beta function, as follows:<br />
<br />
它也可以用'''<font color="#ff8000">正则化不完全 beta 函数</font>'''来表示，如下:<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3.3.3.3.3.3.3.3.3.3.3.4.3.3.3.3.3.3.3.3.3.3<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
F(k;n,p) & = \Pr(X \le k) \\<br />
<br />
F (k; n，p) & = Pr (x le k) <br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
<br />
&= I_{1-p}(n-k, k+1) \\<br />
<br />
& = i _ {1-p }(n-k，k + 1)<br />
<br />
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.<br />
<br />
& = (n-k){ n choose k } int _ 0 ^ {1-p } t ^ { n-k-1}(1-t) ^ k，dt.<br />
<br />
\end{align}</math><br />
<br />
\end{align}</math><br />
<br />
结束{ align } </math > <br />
<br />
<br />
<br />
which is equivalent to the [[cumulative distribution function]] of the [[F-distribution|{{mvar|F}}-distribution]]:<ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref> <br />
<br />
which is equivalent to the cumulative distribution function of the -distribution: <br />
<br />
这相当于分布的'''<font color="#ff8000">累积分布函数</font>''':<br />
<br />
<br />
<br />
:<math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math><br />
<br />
F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).<br />
<br />
F (k; n，p) = f _ { f text {-distribution }}左(x = frac {1-p }{ p } frac { k + 1}{ n-k } ; d _ 1 = 2(n-k) ，d _ 2 = 2(k + 1)右)。<br />
<br />
<br />
<br />
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].<br />
<br />
Some closed-form bounds for the cumulative distribution function are given below.<br />
<br />
下面给出了'''<font color="#ff8000">累积分布函数</font>'''的一些闭式界。<br />
<br />
<br />
<br />
== Properties ==<br />
<br />
===Expected value and variance===<br />
<br />
期望值和方差<br />
<br />
<br />
<br />
If ''X'' ~ ''B''(''n'', ''p''), that is, ''X'' is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the [[expected value]] of ''X'' is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref><br />
<br />
If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:<br />
<br />
如果 x ~ b (n，p) ，即 x 是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验产生成功结果的概率，那么 x 的'''<font color="#ff8000">期望值</font>'''是:<br />
<br />
<br />
<br />
:<math> \operatorname{E}[X] = np.</math><br />
<br />
\operatorname{E}[X] = np.<br />
<br />
操作员名称{ e }[ x ] = np。<br />
<br />
<br />
<br />
This follows from the linearity of the expected value along with fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then <math>X = X_1 + \cdots + X_n</math> and<br />
<br />
This follows from the linearity of the expected value along with fact that is the sum of identical Bernoulli random variables, each with expected value . In other words, if X_1, \ldots, X_n are identical (and independent) Bernoulli random variables with parameter , then X = X_1 + \cdots + X_n and<br />
<br />
这是由于两随机变量线性组合的期望值实际上是两个独立的伯努利随机变量的期望的线性组合。换句话说，如果 x1，ldots，xn 是独立的伯努利随机变量，那么 x = x1 + cdots + xn<br />
<br />
:<math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math><br />
<br />
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.<br />
<br />
操作员名称{ e }[ x ] = 操作员名称{ e }[ x _ 1 + 圆点 + x _ n ] = 操作员名称{ e }[ x _ 1] + 圆点 + 操作员名称{ e }[ x _ n ] = p + 圆点 + p = np。<br />
<br />
<br />
<br />
The [[variance]] is:<br />
<br />
The variance is:<br />
<br />
'''<font color="#ff8000">方差</font>'''是:<br />
<br />
:<math> \operatorname{Var}(X) = np(1 - p).</math><br />
<br />
\operatorname{Var}(X) = np(1 - p).<br />
<br />
操作符名称{ Var }(x) = np (1-p)。<br />
<br />
<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.<br />
<br />
这同样也是从这样一个事实出发: 独立随机变量之和的方差是方差之和。<br />
<br />
<br />
<br />
===Higher moments===<br />
<br />
高阶矩<br />
<br />
The first 6 central moments are given by <br />
<br />
The first 6 central moments are given by <br />
<br />
前6个中心矩由<br />
<br />
:<math>\begin{align}<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3.3.3.3.3.3.3.3<br />
<br />
\mu_1 &= 0, \\<br />
<br />
\mu_1 &= 0, \\<br />
<br />
1 & = 0,<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
\mu_2 &= np(1-p),\\<br />
<br />
mu _ 2 & = np (1-p) ,<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
\mu_3 &= np(1-p)(1-2p),\\<br />
<br />
mu _ 3 & = np (1-p)(1-2p) ,<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\<br />
<br />
mu _ 4 & = np (1-p)(1 + (3n-6) p (1-p)) ,<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\<br />
<br />
mu _ 5 & = np (1-p)(1-2p)(1 + (10n-12) p (1-p)) ,<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).<br />
<br />
mu _ 6 & = np (1-p)(1-30p (1-p)(1-4p (1-p)) + 5 np (1-p)(5-26p (1-p)) + 15 n ^ 2 p ^ 2(1-p) ^ 2).<br />
<br />
\end{align}</math><br />
<br />
\end{align}</math><br />
<br />
结束{ align } </math > <br />
<br />
<br />
<br />
===Mode===<br />
<br />
模<br />
<br />
<br />
<br />
Usually the [[mode (statistics)|mode]] of a binomial ''B''(''n'', ''p'') distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when (''n''&nbsp;+&nbsp;1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n''&nbsp;+&nbsp;1)''p'' and (''n''&nbsp;+&nbsp;1)''p''&nbsp;−&nbsp;1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:<br />
<br />
Usually the mode of a binomial B(n, p) distribution is equal to \lfloor (n+1)p\rfloor, where \lfloor\cdot\rfloor is the floor function. However, when (n&nbsp;+&nbsp;1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n&nbsp;+&nbsp;1)p and (n&nbsp;+&nbsp;1)p&nbsp;−&nbsp;1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:<br />
<br />
通常二项式 b (n，p)分布的'''<font color="#ff8000">模</font>'''等于 lfloor (n + 1) p 楼层，其中 lfloor cdot 楼层是'''<font color="#ff8000">下限函数</font>'''。然而，当(n + 1) p 是整数且 p 既不是0也不是1时，分布有两种模: (n + 1) p 和(n + 1) p-1。当 p 等于0或1时，模将相应地为0和 n。这些情况可概述如下:<br />
<br />
<br />
: <math>\text{mode} =<br />
<br />
<math>\text{mode} =<br />
<br />
1.1.1.2.1<br />
<br />
\begin{cases}<br />
<br />
\begin{cases}<br />
<br />
开始{ cases }<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\<br />
<br />
文本{ if }(n + 1) p text { is 0 or a noninteger } ,<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\<br />
<br />
(n + 1) ，p text { and }(n + 1) ，p-1 & text { if }(n + 1) p in {1，dots，n } ,<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
n & \text{if }(n+1)p = n + 1.<br />
<br />
n & text { if }(n + 1) p = n + 1.<br />
<br />
\end{cases}</math><br />
<br />
\end{cases}</math><br />
<br />
结束{ cases } </math > <br />
<br />
<br />
<br />
'''Proof:''' Let<br />
<br />
Proof: Let<br />
<br />
证明: 让<br />
<br />
<br />
<br />
:<math>f(k)=\binom nk p^k q^{n-k}.</math><br />
<br />
f(k)=\binom nk p^k q^{n-k}.<br />
<br />
F (k) = binom nk p ^ k q ^ { n-k }.<br />
<br />
<br />
<br />
For <math>p=0</math> only <math>f(0)</math> has a nonzero value with <math>f(0)=1</math>. For <math>p=1</math> we find <math>f(n)=1</math> and <math>f(k)=0</math> for <math>k\neq n</math>. This proves that the mode is 0 for <math>p=0</math> and <math>n</math> for <math>p=1</math>.<br />
<br />
For p=0 only f(0) has a nonzero value with f(0)=1. For p=1 we find f(n)=1 and f(k)=0 for k\neq n. This proves that the mode is 0 for p=0 and n for p=1.<br />
<br />
对于 p = 0，只有 f (0)有一个非零值，f (0) = 1。对于 p = 1，我们发现 f (n) = 1，对于 k neq n， f (k) = 0。这证明了 p = 0时模为0，p = 1时模为 n。<br />
<br />
<br />
<br />
Let <math>0 < p < 1</math>. We find<br />
<br />
Let 0 < p < 1. We find<br />
<br />
让0 < p < 1。我们发现<br />
<br />
<br />
<br />
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.<br />
<br />
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.<br />
<br />
Frac { f (k + 1)}{ f (k)} = frac {(n-k) p }{(k + 1)(1-p)}.<br />
<br />
<br />
<br />
From this follows<br />
<br />
From this follows<br />
<br />
由此可见<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3.3<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\<br />
<br />
K > (n + 1) p-1 right tarrow f (k + 1) < f (k)<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\<br />
<br />
K = (n + 1) p-1 right tarrow f (k + 1) = f (k)<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)<br />
<br />
K < (n + 1) p-1 right tarrow f (k + 1) > f (k)<br />
<br />
\end{align}</math><br />
<br />
\end{align}</math><br />
<br />
结束{ align } </math > <br />
<br />
<br />
<br />
So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref><br />
<br />
So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.<br />
<br />
所以当(n + 1) p-1是一个整数时，(n + 1) p-1和(n + 1) p 是一个模。在(n + 1) p-1 notin z 的情况下，只有 lfloor (n + 1) p-1楼 + 1 = lfloor (n + 1) p-1楼是模。<br />
<br />
<br />
<br />
===Median===<br />
<br />
中位数<br />
<br />
In general, there is no single formula to find the [[median]] for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:<br />
<br />
一般来说，没有单一的公式可以找到一个二项分布的'''<font color="#ff8000">中位数</font>'''，甚至可能是非唯一的。然而，已经确立了若干特别成果:<br />
<br />
* If ''np'' is an integer, then the mean, median, and mode coincide and equal ''np''.<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref><br />
<br />
如果“np”是一个整数，那么它的均值，中位数和模相同且等于“np”。<br />
<br />
* Any median ''m'' must lie within the interval ⌊''np''⌋&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;⌈''np''⌉.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref><br />
<br />
任何中位数“ m”都必须在⌊''np''⌋≤≤m''≤⌈''np''⌉范围内。<br />
<br />
* A median ''m'' cannot lie too far away from the mean: {{nowrap|&#124;''m'' − ''np''&#124; ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal<br />
<br />
中位数“m”不能离均值太远。<br />
<br />
| last1 = Hamza | first1 = K.<br />
<br />
| doi = 10.1016/0167-7152(94)00090-U<br />
<br />
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions<br />
<br />
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) <br />
<br />
F (k; n，p) leq exp left (- nD left (frac { k }{ n } parallel p right))) <br />
<br />
| journal = Statistics & Probability Letters<br />
<br />
| volume = 23<br />
<br />
where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):<br />
<br />
其中 D(a | | p)是a-coin，p-coin相对熵之间的距离。在伯努利(a)和伯努利(p)分布之间:<br />
==~~~[翻译]存疑“a-coin，p-coin”<br />
<br />
| pages = 21–25<br />
<br />
| year = 1995<br />
<br />
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!<br />
<br />
D (a 并行 p) = (a) log frac { a }{ p } + (1-a) log frac {1-a }{1-p }。\!<br />
<br />
| pmid = <br />
<br />
| pmc = <br />
<br />
Asymptotically, this bound is reasonably tight; see <br />
<br />
从渐近的角度来看，这个界限相当严格; 参见<br />
<br />
}}</ref><br />
<br />
F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),<br />
<br />
F (k; n，p) geq frac {1}{ sqrt {8n tfrac { k }{ n }(1-tfrac { k }{ n })} exp left (- nD left (frac { k }{ n } parallel p right)) ,<br />
<br />
* The median is unique and equal to ''m''&nbsp;=&nbsp;[[Rounding|round]](''np'') when |''m''&nbsp;−&nbsp;''np''|&nbsp;≤&nbsp;min{''p'',&nbsp;1&nbsp;−&nbsp;''p''} (except for the case when ''p''&nbsp;=&nbsp;{{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/><br />
<br />
中位数是唯一的并且等于“m”<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的界限<br />
<br />
* When ''p''&nbsp;=&nbsp;1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n''&nbsp;−&nbsp;1)&nbsp;≤&nbsp;''m''&nbsp;≤&nbsp;{{sfrac|1|2}}(''n''&nbsp;+&nbsp;1) is a median of the binomial distribution. If ''p''&nbsp;=&nbsp;1/2 and ''n'' is even, then ''m''&nbsp;=&nbsp;''n''/2 is the unique median.<br />
<br />
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).<br />
<br />
F (k; n，p) geq frac1{ sqrt {2n } exp left (- nD left (frac { k }{ n } parallel p right))).<br />
<br />
<br />
<br />
<br />
<br />
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:<br />
<br />
对于 p = 1/2且 k ≥3n/8且n为偶数时，可以使分母为常数:<br />
<br />
===Tail bounds===<br />
<br />
尾部边界<br />
<br />
For ''k'' ≤ ''np'', upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most ''k'' successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for ''k'' ≥ ''np''.<br />
<br />
对于''k''≤''np''，可以得出累积分布函数下尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''成功的概率。由于<math>/Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数上尾的界限。<br />
<br />
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!<br />
<br />
F (k; n，tfrac {1}{2}) geq frac {1}{15} exp left (- 16n left (frac {1}{2}-frac { k }{ n }右) ^ 2 right)。\!<br />
<br />
<br />
<br />
[[Hoeffding's inequality]] yields the simple bound<br />
<br />
[[Hoeffding's不等式]]得到简单的边界<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math><br />
<br />
<br />
<br />
which is however not very tight. In particular, for ''p'' = 1, we have that ''F''(''k'';''n'',''p'') = 0 (for fixed ''k'', ''n'' with ''k''&nbsp;<&nbsp;''n''), but Hoeffding's bound evaluates to a positive constant. <br />
<br />
然而，这并不是很严格。特别是，对于''p''=1，我们有''F''(''k'';''n''，''p'')=0(对于固定的''k''，''n''与''k''&nbsp;<&nbsp;''n'')，但Hoeffding''的约束评价为一个正常数。<br />
<br />
When n is known, the parameter p can be estimated using the proportion of successes: \widehat{p} = \frac{x}{n}. This estimator is found using maximum likelihood estimator and also the method of moments. This estimator is unbiased and uniformly with minimum variance, proven using Lehmann–Scheffé theorem, since it is based on a minimal sufficient and complete statistic (i.e.: x). It is also consistent both in probability and in MSE.<br />
<br />
当 n 已知时，参数 p 可以使用成功的比例来估计: widehat { p } = frac { x }{ n }。利用极大似然估计和矩方法求出了该估计量。利用 '''<font color="#ff8000">Lehmann-scheffé 定理</font>'''证明了该估计量的无偏一致最小方差，因为该估计量是基于一个极小充分完全统计量(即:。: x).它在概率和均方误差方面也是一致的。<br />
<br />
<br />
<br />
A sharper bound can be obtained from the [[Chernoff bound]]:<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref><br />
<br />
可以从[[Chernoff界]]中得到一个更清晰的边界<br />
<br />
A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general \operatorname{Beta}(\alpha, \beta) as a prior, the posterior mean estimator is: \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}. The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (n → ∞), it approaches the MLE solution. The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability.<br />
<br />
利用 beta分布作为'''<font color="#ff8000">共轭先验分布</font>'''时，也存在p的封闭形式 Bayes 估计。当使用一个通用算子名{ Beta }(alpha，Beta)作为先验时，后验平均估计量为: widehat { p _ b } = frac { x + alpha }{ n + alpha + Beta }。贝叶斯估计是渐近有效的，当样本容量接近无穷大(n →∞)时，它逼近最大似然估计解。贝叶斯估计是有偏的(多少取决于先验) ，可接受且一致的概率。<br />
<br />
<br />
<br />
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math><br />
<br />
For the special case of using the standard uniform distribution as a non-informative prior (\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)), the posterior mean estimator becomes \widehat{p_b} = \frac{x+1}{n+2} (a posterior mode should just lead to the standard estimator). This method is called the rule of succession, which was introduced in the 18th century by Pierre-Simon Laplace.<br />
<br />
对于使用标准均匀分布作为先验概率的特殊情况(操作者名{ Beta }(alpha = 1，Beta = 1) = u (0,1)) ，后验均值估计变为广义{ p _ b } = frac { x + 1}{ n + 2}(后验模式应该只导致标准估计)。这种方法被称为'''<font color="#ff8000">继承法则</font>'''，它是在18世纪由皮埃尔-西蒙·拉普拉斯引进的。<br />
<br />
<br />
<br />
where ''D''(''a'' || ''p'') is the [[Kullback–Leibler divergence|relative entropy]] between an ''a''-coin and a ''p''-coin (i.e. between the Bernoulli(''a'') and Bernoulli(''p'') distribution):<br />
<br />
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to \widehat{p} = 0, which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: \widehat{p_b} = \frac{1}{n+2}). Another method is to use the upper bound of the confidence interval obtained using the rule of three: \widehat{p_{\text{rule of 3}}} = \frac{3}{n})<br />
<br />
在估计具有非常罕见事件和小 n (例如，n)的 p 时。: 如果 x = 0) ，那么使用标准估计器会导致广义{ p } = 0，这有时是不现实的，也是不受欢迎的。在这种情况下，有各种可供选择的估计值。一种方法是使用 Bayes 估计，导致: widehat { p _ b } = frac {1}{ n + 2})。另一种方法是使用使用3个规则获得的置信区间的上界: widehat { p { text { rule of 3}}}} = frac {3}{ n })<br />
<br />
<br />
<br />
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math><br />
<br />
<br />
<br />
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.<br />
<br />
渐进地讲，这个边界是相当严格的；详见<ref name="ag"/>。<br />
<br />
<br />
<br />
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于非常大的 n 值，均值的实际分布也是非正态的。针对这一问题，提出了几种估计置信区间的方法。<br />
<br />
One can also obtain ''lower'' bounds on the tail <math>F(k;n,p) </math>, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref><br />
<br />
我们还可以得到尾部<math>F(k;n,p) </math>的''下'界，即'''<font color="#ff8000">反集中界</font>'''。通过用斯特林公式对二项式系数进行近似，可以看出：<math>F(k;n,p)</math>是一个反集中的界限。<br />
<br />
:<math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math><br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间等式中，这些变量具有以下含义:<br />
<br />
which implies the simpler but looser bound<br />
<br />
这意味着更简单但更松散的约束。<br />
<br />
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math><br />
<br />
<br />
<br />
For ''p'' = 1/2 and ''k'' ≥ 3''n''/8 for even ''n'', it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf }}</ref><br />
<br />
<br />
<br />
:<math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math><br />
<br />
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .<br />
<br />
1-widehat { p，})}{ n }.<br />
<br />
<br />
<br />
== Statistical Inference ==<br />
<br />
统计推断<br />
<br />
A continuity correction of 0.5/n may be added.<br />
<br />
可以加上0.5/n 的连续校正。<br />
<br />
=== Estimation of parameters ===<br />
<br />
参数估计<br />
<br />
{{seealso|Beta distribution#Bayesian inference}}<br />
<br />
beta分布 贝叶斯推断<br />
<br />
<br />
<br />
When ''n'' is known, the parameter ''p'' can be estimated using the proportion of successes: <math> \widehat{p} = \frac{x}{n}.</math> This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[Minimal sufficient|minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: ''x''). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]].<br />
<br />
当''n''已知时，参数''p''可以用成功的比例来估计。 <math> \widehat{p} = \frac{x}{n}.</math> 这个估计是用[[最大似然估计法]]和[[矩量法（统计学）|矩量法]]来计算的。这个估计是[[估计的偏倚|无偏估计]]，并与[[最小方差无偏估计|最小方差]]一致，用[[Lehmann-Scheffé定理]]证明，因为它是基于[[最小充分性|最小充分性]]和[[完全性（统计学）|完全性]]统计（即：''x''）。它在概率和[[平均平方误差|MSE]]方面也是[[一致估计|一致]]。<br />
<br />
<br />
<br />
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .<br />
<br />
1-tilde { p })}{ n + z ^ 2}.<br />
<br />
A closed form [[Bayes estimator]] for ''p'' also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is: <math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>. The Bayes estimator is [[Asymptotic efficiency (Bayes)|asymptotically efficient]] and as the sample size approaches infinity (''n'' → ∞), it approaches the [[Maximum likelihood estimation|MLE]] solution. The Bayes estimator is [[Bias of an estimator|biased]] (how much depends on the priors), [[Bayes estimator#Admissibility|admissible]] and [[Consistent estimator|consistent]] in probability.<br />
<br />
当使用[[Beta分布]]作为[[共轭先验|共轭]]时，''p''的闭合形式[[贝叶斯估计]]也存在。[[先验分布]]。当使用一般的<math><operatorname{Beta}(\alpha，\beta)</math>作为先验时，[[贝叶斯估计#后验均值|后验均值]]估计是。<math> \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。Bayes估计是[[渐进效率(Bayes)|asymptotically efficient]]，当样本量接近无穷大(''n'' → ∞)时，它接近[[最大似然估计|MLE]]解。贝叶斯估计是[[估计的偏倚|偏倚]]。(多少取决于前值)，[[贝叶斯估计#可接受性|可接受性]]和[[一致性估计|一致性]]的概率。<br />
<br />
<br />
<br />
Here the estimate of p is modified to<br />
<br />
这里 p 的估计被修改为<br />
<br />
For the special case of using the [[Standard uniform distribution|standard uniform distribution]] as a [[non-informative prior]] (<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>), the posterior mean estimator becomes <math> \widehat{p_b} = \frac{x+1}{n+2}</math> (a [[Bayes estimator#Posterior mode|posterior mode]] should just lead to the standard estimator). This method is called the [[rule of succession]], which was introduced in the 18th century by [[Pierre-Simon Laplace]].<br />
<br />
对于使用[[标准均匀分布|标准均匀分布]]作为[[非信息先验]]的特殊情况（<math>operatorname{Beta}(alpha=1, β=1)=U(0, 1)</math>)，后均值估计变成<math> \widehat{p_b}=\frac{x+1}{n+2}</math>(一个[[贝叶斯估计#后模|后模]]应该只是导致标准估计)。这种方法被称为[[继承规则]]，由[[Pierre-Simon Laplace]]在18世纪引入。<br />
<br />
<br />
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } <br />
<br />
1 + frac {1}{2} z2}{ n + z2}<br />
<br />
When estimating ''p'' with very rare events and a small ''n'' (e.g.: if x=0), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator, leading to: <math> \widehat{p_b} = \frac{1}{n+2}</math>). Another method is to use the upper bound of the [[confidence interval]] obtained using the [[Rule of three (statistics)|rule of three]]: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)<br />
<br />
当估计''p''时，如果事件非常罕见，而且''n''很小（例如：如果x=0），那么使用标准估计器会导致<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有各种不同的估计方法。<br />
<br />
<br />
=== Confidence intervals ===<br />
<br />
置信区间<br />
<br />
{{Main|Binomial proportion confidence interval}}<br />
<br />
<br />
<br />
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).<br />
<br />
Sin ^ 2 left (arcsin left (sqrt { widehat { p，} right) pm frac { z }{2 sqrt { n } right).<br />
<br />
Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.<br />
<br />
即使对于相当大的''n''值，平均数的实际分布也是显著非正态的，由于这个问题，人们提出了几种估计置信区间的方法。<br />
<br />
<br />
<br />
In the equations for confidence intervals below, the variables have the following meaning:<br />
<br />
在下面的置信区间公式中，变量具有以下含义<br />
<br />
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials<br />
<br />
* ''n''<sub>1</sub>是''n''中的成功次数，即试验的总次数。<br />
<br />
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes<br />
<br />
* <math> <math> widehat{p,} = \frac{n_1}{n}</math> 是成功的比例。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<br />
<br />
下列公式中的符号在两个方面不同于以前的公式:<br />
<br />
* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.<br />
<br />
<br />
<br />
==== Wald method ====<br />
<br />
Wald法<br />
<br />
:: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math><br />
<br />
<math>\frac{<br />
<br />
我不知道<br />
<br />
<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
2n } + z<br />
<br />
: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}<br />
<br />
\sqrt{<br />
<br />
2.1.1.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3<br />
<br />
<br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
1-widehat { p，})}{ n } + <br />
<br />
==== Agresti–Coull method ====<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
4 n ^ 2}<br />
<br />
<ref name=Agresti1988>{{Citation |last1=Agresti |first1=Alan |last2=Coull |first2=Brent A. |date=May 1998 |title=Approximate is better than 'exact' for interval estimation of binomial proportions |url = http://www.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf |journal=The American Statistician |volume=52 |issue=2 |pages=119–126 |accessdate=2015-01-05 |doi=10.2307/2685469 |jstor=2685469 }}</ref><br />
<br />
}<br />
<br />
}<br />
<br />
<br />
<br />
}{<br />
<br />
}{<br />
<br />
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math><br />
<br />
1 + \frac{z^2}{n}<br />
<br />
1 + frac { z ^ 2}{ n }<br />
<br />
<br />
<br />
}</math><br />
<br />
{/math > <br />
<br />
: Here the estimate of ''p'' is modified to<br />
<br />
<br />
<br />
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math><br />
<br />
The exact (Clopper–Pearson) method is the most conservative.<br />
<br />
确切的(克洛佩尔-皮尔森)方法是最保守的。<br />
<br />
<br />
<br />
==== Arcsine method ====<br />
<br />
弧线法<br />
<br />
Let X&nbsp;~&nbsp;B(n,p1) and Y&nbsp;~&nbsp;B(m,p2) be independent. Let T = (X/n)/(Y/m).<br />
<br />
设 x ~ b (n，p1)和 y ~ b (m，p2)是独立的。设 t = (X/n)/(Y/m)。<br />
<br />
<ref name="Pires00">{{cite book |last=Pires |first=M. A. |chapterurl=https://www.math.tecnico.ulisboa.pt/~apires/PDFs/AP_COMPSTAT02.pdf |chapter=Confidence intervals for a binomial proportion: comparison of methods and software evaluation |editor-last=Klinke |editor-first=S. |editor2-last=Ahrend |editor2-first=P. |editor3-last=Richter |editor3-first=L. |title=Proceedings of the Conference CompStat 2002 |others=Short Communications and Posters |year=2002 }}</ref><br />
<br />
<br />
<br />
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1)&nbsp;−&nbsp;1)/n&nbsp;+&nbsp;((1/p2)&nbsp;−&nbsp;1)/m.<br />
<br />
然后log(t)近似呈正态分布，其中平均log(p1/p2)和方差((1/p1)-1)/n + ((1/p2)-1)/m。<br />
<br />
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math><br />
<br />
<br />
<br />
==== Wilson (score) method ====<br />
<br />
威尔逊法<br />
<br />
If X&nbsp;~&nbsp;B(n,&nbsp;p) and Y&nbsp;|&nbsp;X&nbsp;~&nbsp;B(X,&nbsp;q) (the conditional distribution of Y, given&nbsp;X), then Y is a simple binomial random variable with distribution Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
如果 x ~ b (n，p)和 y | x ~ b (x，q)(给定 x 的条件分布) ，则 y 是具有分布 y ~ b (n，pq)的简单二项式随机变量。<br />
<br />
{{Main|Binomial proportion confidence interval#Wilson score interval}}<br />
<br />
<br />
<br />
For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X&nbsp;~&nbsp;B(n,&nbsp;p) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y&nbsp;~&nbsp;B(X,&nbsp;q) and therefore Y&nbsp;~&nbsp;B(n,&nbsp;pq).<br />
<br />
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是命中 UX 的概率，那么 x ~ b (n，p)是命中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY1 的球数是 y ~ b (x，q) ，因此 y ~ b (n，pq)。<br />
<br />
The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref><br />
<br />
下面的公式中的符号与前面的公式有两个不同之处<br />
<br />
* Firstly, ''z''<sub>''x''</sub> has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the ''x''th quantile of the standard normal distribution', rather than being a shorthand for 'the (1&nbsp;−&nbsp;''x'')-th quantile'.<br />
<br />
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是 "标准正态分布的''x''th分位数"，而不是"(1&nbsp;-&nbsp;''x'')th分位数 "的简写。<br />
<br />
<br />
* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.<br />
<br />
其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差<math>/alpha</math>&nbsp;=&nbsp;0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。<br />
<br />
<br />
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,<br />
<br />
由于 x sim b (n，p)和 y sim b (x，q) ，由全概率公式,<br />
<br />
<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3.3.3.3.3.3.3<br />
<br />
:: <math>\frac{<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
<br />
\widehat{p\,} + \frac{z^2}{2n} + z<br />
<br />
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
<br />
和 = sum { k = m } n binom { n }{ k }{ m } p ^ k ^ m (1-p) ^ { n-k }(1-q) ^ { k-m }<br />
<br />
\sqrt{<br />
<br />
\end{align}</math><br />
<br />
结束{ align } </math > <br />
<br />
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +<br />
<br />
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as<br />
<br />
由于 tbinom { n }{ k } tbinom { k }{ m } = tbinom { n }{ m } tbinom { n-m }{ k-m } ，上述方程可表示为<br />
<br />
\frac{z^2}{4 n^2}<br />
<br />
\Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} <br />
<br />
Pr [ y = m ] = sum _ { k = m } ^ { n } binom { n-m }{ k-m } p ^ k ^ m (1-p) ^ { n-k }(1-q) ^ { k-m }<br />
<br />
}<br />
<br />
Factoring p^k = p^m p^{k-m} and pulling all the terms that don't depend on k out of the sum now yields<br />
<br />
对 p ^ k = p ^ m p ^ { k-m }进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果<br />
<br />
}{<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
1 + \frac{z^2}{n}<br />
<br />
\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
<br />
[ y = m ] & = binom { n }{ m } p ^ m ^ m left (sum { k = m } n binom { n-m }{ k-m } p ^ { k-m }(1-p) ^ { n-k }(1-q) ^ { k-m } right)[2 pt ]<br />
<br />
}</math><ref>{{cite book<br />
<br />
&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
<br />
和 = binom { n }{ m }(pq) ^ m left (sum { k = m } ^ n binom { n-m }{ k-m } left (p (1-q) right) ^ { k-m }(1-p) ^ { n-k } right)<br />
<br />
| chapter = Confidence intervals<br />
<br />
\end{align}</math><br />
<br />
结束{ align } </math > <br />
<br />
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm<br />
<br />
After substituting i = k - m in the expression above, we get<br />
<br />
在上面的表达式中用 i = k-m 代替后，我们得到了<br />
<br />
| title = Engineering Statistics Handbook<br />
<br />
\Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) <br />
<br />
Pr [ y = m ] = binom { n }{ m }(pq) ^ m left (sum _ { i = 0} ^ { n-m } binom { n-m }{ i }(p-pq) ^ i (1-p) ^ { n-m-i }右)<br />
<br />
| publisher = NIST/Sematech<br />
<br />
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields<br />
<br />
注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。最终将其替换为<br />
<br />
| year = 2012<br />
<br />
<math>\begin{align}<br />
<br />
1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3<br />
<br />
| access-date = 2017-07-23<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
<br />
}}</ref><br />
<br />
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
<br />
& = binom { n }{ m }(pq) ^ m (1-pq) ^ { n-m }<br />
<br />
<br />
<br />
\end{align}</math><br />
<br />
结束{ align } </math > <br />
<br />
==== Comparison ====<br />
<br />
and thus Y \sim B(n, pq) as desired.<br />
<br />
因此 y sim b (n，pq)为所需值。<br />
<br />
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" /><br />
<br />
<br />
<br />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}<br />
<br />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。<br />
<br />
<br />
<br />
The Bernoulli distribution is a special case of the binomial distribution, where n&nbsp;=&nbsp;1. Symbolically, X&nbsp;~&nbsp;B(1,&nbsp;p) has the same meaning as X&nbsp;~&nbsp;Bernoulli(p). Conversely, any binomial distribution, B(n,&nbsp;p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.<br />
<br />
伯努利分布是二项分布的一个特例，其中 n = 1。在符号上，x ~ b (1，p)与 x ~ Bernoulli (p)具有相同的意义。反之，任何二项分布 b (n，p)都是 n 个 Bernoulli 试验之和的分布，每个试验的概率 p 相同。<br />
<br />
==Related distributions==<br />
<br />
相关分布<br />
<br />
<br />
<br />
===Sums of binomials===<br />
<br />
二项式之和<br />
<br />
The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).<br />
<br />
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同 Bernoulli 试验 b (pi)的和的分布。<br />
<br />
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'') are independent binomial variables with the same probability ''p'', then ''X''&nbsp;+&nbsp;''Y'' is again a binomial variable; its distribution is ''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p''):<br />
<br />
如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;~&nbsp;B(''m'',&nbsp;''p'')是独立的二项式变量，概率''p''相同，那么''X''&nbsp; +&nbsp;''Y''又是一个二项式变量，其分布是''Z=X+Y''&nbsp;~&nbsp;B(''n+m'',&nbsp;''p'')。<br />
<br />
<br />
<br />
<br />
:<math>\begin{align}<br />
<br />
\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\<br />
<br />
Binomial [[probability mass function and normal probability density function approximation for n&nbsp;=&nbsp;6 and p&nbsp;=&nbsp;0.5]]<br />
<br />
二项式[ n = 6和 p = 0.5的概率质量函数和正态概率密度函数近似]<br />
<br />
&= \binom{n+m}k p^k (1-p)^{n+m-k}<br />
<br />
\end{align}</math><br />
<br />
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n,&nbsp;p) is given by the normal distribution<br />
<br />
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出 b (n，p)的合理近似<br />
<br />
<br />
<br />
However, if ''X'' and ''Y'' do not have the same probability ''p'', then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as <math>B(n+m, \bar{p}).\,</math><br />
<br />
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是[[二项式和方差不等式|小于二项式变量的方差]]分布为<math>B(n+m, \bar{p}).\,</math>。<br />
<br />
\mathcal{N}(np,\,np(1-p)),<br />
<br />
数学{ n }(np，np (1-p))<br />
<br />
<br />
<br />
===Ratio of two binomial distributions===<br />
<br />
两个二项分布的比值<br />
<br />
and this basic approximation can be improved in a simple way by using a suitable continuity correction.<br />
<br />
通过适当的连续性修正，可以简单地改进这种基本近似。<br />
<br />
<br />
<br />
The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1. Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:<br />
<br />
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p 是否远离0或1的极值:<br />
<br />
This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |displayauthors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref><br />
<br />
这个结果最早是由Katz和合作者在1978年得出的。<br />
<br />
<br />
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Let ''X''&nbsp;~&nbsp;B(''n'',''p''<sub>1</sub>) and ''Y''&nbsp;~&nbsp;B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').<br />
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For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation \sigma = \sqrt{\frac{p(1-p)}{n}}<br />
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例如，假设一个人从一个大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例当然取决于样本。如果 n 组人群被重复且真正随机地取样，其比例将遵循一个近似正态分布，其平均值等于总体中一致性的真实比例 p，标准差 σ = sqrt { p (1-p)}{ n }<br />
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Then log(''T'') is approximately normally distributed with mean log(''p''<sub>1</sub>/''p''<sub>2</sub>) and variance ((1/''p''<sub>1</sub>)&nbsp;−&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;−&nbsp;1)/''m''.<br />
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则log(''T'')近似正态分布，均值log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差((1/''p''<sub>1</sub>)&nbsp;-&nbsp;1)/''n''&nbsp;+&nbsp;((1/''p''<sub>2</sub>)&nbsp;-&nbsp;1)/''m''。<br />
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===Conditional binomials===<br />
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条件二项式<br />
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If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
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如果''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')和''Y''&nbsp;|&nbsp;'X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')('Y''的条件分布，给定&nbsp。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
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The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n&nbsp;≥&nbsp;20 and p&nbsp;≤&nbsp;0.05, or if n&nbsp;≥&nbsp;100 and np&nbsp;≤&nbsp;10.<br />
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当试验数量趋于无穷大，而产品 np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分佈。因此，参数 λ = np 的泊松分布可以作为二项分布 b (n，p)的近似，如果 n 是足够大，p 足够小的话。根据两个经验法则，如果 n ≥20和 p ≤0.05，或者如果 n ≥100和 np ≤10，这个近似是好的。<br />
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For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').<br />
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例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'')，因此''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'')。<br />
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Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.<br />
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关于泊松近似的准确性，参见 Novak，ch。4，及其中的参考资料。<br />
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{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}<br />
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Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],<br />
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:<math>\begin{align}<br />
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\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]<br />
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P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.<br />
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P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.<br />
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&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}<br />
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Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.<br />
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给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个 beta分布。<br />
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\end{align}</math><br />
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Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as<br />
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由于<math>/tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为<br />
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:<math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math><br />
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Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields<br />
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将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到<br />
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Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<br />
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'''<font color="#ff8000">边缘分布</font>'''是二项分布的随机数产生方法已经很完善了。<br />
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:<math>\begin{align}<br />
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\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]<br />
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One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that for all values from through . (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.<br />
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一种从二项分布中产生随机样本的方法是使用'''<font color="#ff8000">反演算法</font>'''。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率之和应该接近于一。)然后，通过使用伪随机数生成器来生成数值为0或1的样本，人们可以使用在第一步计算出的概率将计算出的样本转换成离散数。<br />
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&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)<br />
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\end{align}</math><br />
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After substituting <math> i = k - m </math> in the expression above, we get<br />
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将 <math> i = k - m </math> 代入上述表达式后，我们得到了<br />
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:<math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math><br />
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This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r&nbsp;+&nbsp;s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p&nbsp;=&nbsp;1/2.<br />
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这个分布是由雅各布伯努利推导出来的。他考虑了 p = r/(r + s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡考虑过 p = 1/2的情况。<br />
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Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields<br />
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请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m}。</math>由[[二项式定理]]得出。将此代入最终得到<br />
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:<math>\begin{align}<br />
<br />
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]<br />
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&= \binom{n}{m} (pq)^m (1-pq)^{n-m}<br />
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\end{align}</math><br />
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and thus <math> Y \sim B(n, pq) </math> as desired.<br />
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{{hidden end}}<br />
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===Bernoulli distribution===<br />
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伯努利分布<br />
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The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n''&nbsp;=&nbsp;1. Symbolically, ''X''&nbsp;~&nbsp;B(1,&nbsp;''p'') has the same meaning as ''X''&nbsp;~&nbsp;Bernoulli(''p''). Conversely, any binomial distribution, B(''n'',&nbsp;''p''), is the distribution of the sum of ''n'' [[Bernoulli trials]], Bernoulli(''p''), each with the same probability ''p''.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref><br />
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[[伯努利分布]]是二项分布的特例，其中''n''&nbsp;=&nbsp;1.从符号上看，''X''&nbsp;~&nbsp;B(1,&nbsp;''p'')与''X''&nbsp;~&nbsp;伯努利(''p'')具有相同的意义。相反，任何二项分布，B(''n'',&nbsp;''p'')是''n''[[伯努利试验]]，伯努利(''p'')之和的分布，每个概率''p''相同。<br />
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===Poisson binomial distribution===<br />
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泊松二项分布<br />
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The binomial distribution is a special case of the [[Poisson binomial distribution]], or [[general binomial distribution]], which is the distribution of a sum of ''n'' independent non-identical [[Bernoulli trials]] B(''p<sub>i</sub>'').<ref><br />
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二项分布是[[泊松二项分布]]或[[广义二项分布]]的特例，它是''n''独立的非相同[[伯努利试验]]之和的分布。B(''p<sub>i</sub>'')<br />
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{{Cite journal<br />
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| volume = 3<br />
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| issue = 2<br />
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| pages = 295–312<br />
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| last = Wang<br />
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| first = Y. H.<br />
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| title = On the number of successes in independent trials<br />
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| journal = Statistica Sinica<br />
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| year = 1993<br />
<br />
| url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
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| url-status = dead<br />
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| archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf<br />
<br />
| archivedate = 2016-03-03<br />
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}}<br />
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</ref><br />
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Category:Discrete distributions<br />
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类别: 离散分布<br />
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===Normal approximation===<br />
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正态逼近<br />
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Category:Factorial and binomial topics<br />
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类别: 阶乘和二项式主题<br />
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Category:Conjugate prior distributions<br />
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类别: 共轭先验分布<br />
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[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n''&nbsp;=&nbsp;6 and ''p''&nbsp;=&nbsp;0.5]]<br />
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Category:Exponential family distributions<br />
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类别: 指数族分布<br />
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<noinclude><br />
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<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E6%95%B4%E4%BD%93%E8%AE%BA_Holism&diff=19058整体论 Holism2020-11-23T01:43:10Z<p>Smile：</p>
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<div>此词条暂由Henry翻译。已由Smile审校<br />
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'''Holism''' (from [[Ancient Greek|Greek]] {{lang|grc|ὅλος}} ''holos'' "all, whole, entire") is the idea that various [[system]]s (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts.<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> The term "holism" was coined by [[Jan Smuts]] in his 1926 book ''[[Holism and Evolution]]''.<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref><br />
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Holism (from Greek holos "all, whole, entire") is the idea that various systems (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts. The term "holism" was coined by Jan Smuts in his 1926 book Holism and Evolution.<br />
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<font color="#ff8000"> 整体论 Holism</font>（源自希腊holos“all，whole，entire”）是指各种系统（例如物理的、生物的、社会的）应该被视为整体，而不仅仅作为各个部分的集合的观点。<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> “整体论”由扬·斯密茨 Jan Smuts在1926年出版的《整体论与进化》<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref>一书中提出。 <br />
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==Meaning含义==<br />
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The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures.<ref name=oed /> However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts.<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref> In this sense, "holism" may also be spelled "'''wholism'''", and it may be contrasted with [[reductionism]] or [[atomism]].<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
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The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures. However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts. In this sense, "holism" may also be spelled "wholism", and it may be contrasted with reductionism or atomism.<br />
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整体论的确切含义取决于语境。Smuts最初用整体论来指自然界中从单元结构的有序组合中产生整体的倾向。<ref name=oed /> 然而，在整体论的一般用法中，“整体论”通常指整体优于部分之和的观点。<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref>在这个含义下，整体论也被称为<font color="#ff8000">整体主义 wholism </font> ，并与<font color="#ff8000"> 还原论 reductionism</font>或<font color="#ff8000"> 原子论 atomism </font>形成对比。<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
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=== Diet and health饮食与健康 ===<br />
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The term holistic when applied to [[Diet (nutrition)|diet]] or medical [[health]] refers to [[intuitive]] approach to [[food]], [[eating]], or [[Lifestyle (sociology)|lifestyle]].<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref> One example is in the context of [[holistic medicine]], "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms.<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref> In this sense, holism may also be called "'''holiatry'''".<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref> Several approaches are used by [[medical doctors]], [[dietitian]]s, and [[religion|religious]] institutions, usually recommended based on an individual basis.<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref> Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including [[Hinduism]],<ref name="Fenton 2010"></ref> [[Shinto]],<ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref> and the [[Seventh-Day Adventist Church]].<ref name="TODAY.com 2016"></ref><br />
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The term holistic when applied to diet or medical health refers to intuitive approach to food, eating, or lifestyle. One example is in the context of holistic medicine, "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms. In this sense, holism may also be called "holiatry". Several approaches are used by medical doctors, dietitians, and religious institutions, usually recommended based on an individual basis. Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including Hinduism, and the Seventh-Day Adventist Church.<ref name="TODAY.com 2016"></ref><br />
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当用于饮食或医疗健康时，术语“整体”指的是对食物、饮食或生活方式的直觉方法。<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref>举个例子，在整体医学的背景下，整体论是指治疗一个人健康的所有方面，包括心理和社会因素，而不仅仅是他/她的身体状况或症状。<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref>从这个意义上讲，整体论也可以称为<font color="#ff8000"> 整体主义 holiatry </font>。<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref>医生、营养师和宗教机构通常根据个人基础推荐几种方法。<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref>宗教机构的信徒，如<font color="#ff8000">印度教 Hinduism </font> <ref name="Fenton 2010"></ref>、<font color="#ff8000">神道教 Shinto </font><ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref>和<font color="#ff8000">基督复临安息日会 Seventh-Day Adventist Church </font> <ref name="TODAY.com 2016"></ref>的信徒，他们使用一个全面的饮食和健康的方法，事实证明，他们比周围人口的寿命更长。<br />
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== See also参见 ==<br />
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*[[Antireductionism]]<br />
反还原论<br />
*[[Emergentism]]<br />
紧急情况论<br />
*[[Gaia hypothesis]]<br />
盖亚假说<br />
*[[Holistic education]]<br />
整体教育<br />
*[[Holism in science]]<br />
科学中的整体论<br />
* [[Monism]]<br />
一元论<br />
*[[Organicism]]<br />
有机主义<br />
*[[Synergy]]<br />
协同作用 <br />
<br />
<br />
==References参考==<br />
<br />
{{reflist}}<br />
<br />
<br />
<br />
== Further reading 拓展阅读==<br />
<br />
* Fodor, Jerry, and Ernst Lepore, ''Holism: A Shopper's Guide'' Wiley. New York. 1992<br />
<br />
* Phillips, D.C. ''Holistic Thought in Social Science''. Stanford University Press. Stanford. 1976.<br />
<br />
<br />
<br />
== External links外部链接 ==<br />
<br />
{{Wiktionary}}<br />
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*{{Commonscat-inline}}<br />
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Category:Natural philosophy<br />
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[[Category:Philosophy of science]]<br />
<br />
Category:Philosophy of science<br />
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范畴: 科学哲学<br />
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[[Category:Social theories]]<br />
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Category:Social theories<br />
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范畴: 社会理论<br />
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Category:Emergence<br />
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Category:Jan Smuts<br />
<br />
类别: Jan Smuts<br />
<br />
<noinclude><br />
<br />
<small>This page was moved from [[wikipedia:en:Holism]]. Its edit history can be viewed at [[整体论/edithistory]]</small></noinclude><br />
<br />
[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E6%95%B4%E4%BD%93%E8%AE%BA_Holism&diff=19056整体论 Holism2020-11-23T01:16:18Z<p>Smile：</p>
<hr />
<div>此词条暂由Henry翻译。已由Smile审校<br />
<br />
'''Holism''' (from [[Ancient Greek|Greek]] {{lang|grc|ὅλος}} ''holos'' "all, whole, entire") is the idea that various [[system]]s (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts.<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> The term "holism" was coined by [[Jan Smuts]] in his 1926 book ''[[Holism and Evolution]]''.<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref><br />
<br />
Holism (from Greek holos "all, whole, entire") is the idea that various systems (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts. The term "holism" was coined by Jan Smuts in his 1926 book Holism and Evolution.<br />
<br />
<font color="#ff8000"> 整体论 Holism</font>（源自希腊holos“all，whole，entire”）是指各种系统（例如物理的、生物的、社会的）应该被视为整体，而不仅仅作为各个部分的集合的观点。<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> “整体论”这个词是扬·斯密茨 Jan Smuts在1926年出版的《整体论与进化》<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref>一书中提出的。 <br />
<br />
<br />
<br />
==Meaning含义==<br />
<br />
The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures.<ref name=oed /> However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts.<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref> In this sense, "holism" may also be spelled "'''wholism'''", and it may be contrasted with [[reductionism]] or [[atomism]].<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
<br />
The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures. However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts. In this sense, "holism" may also be spelled "wholism", and it may be contrasted with reductionism or atomism.<br />
<br />
<font color="#ff8000"> 整体论</font>的确切含义取决于语境。Smuts最初用<font color="#ff8000"> 整体论</font>来指自然界中从单元结构的有序组合中产生整体的倾向。<ref name=oed /> 然而，在<font color="#ff8000"> 整体论</font>的一般用法中，<font color="#ff8000"> 整体论</font>通常指整体优于部分之和的观点。<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref>在这个含义下，<font color="#ff8000"> 整体论</font>也可以称为<font color="#ff8000">整体主义 wholism </font> ，它可以与<font color="#ff8000"> 还原论 reductionism</font>或<font color="#ff8000"> 原子论 atomism </font>形成对比。<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
<br />
<br />
=== Diet and health饮食与健康 ===<br />
<br />
The term holistic when applied to [[Diet (nutrition)|diet]] or medical [[health]] refers to [[intuitive]] approach to [[food]], [[eating]], or [[Lifestyle (sociology)|lifestyle]].<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref> One example is in the context of [[holistic medicine]], "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms.<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref> In this sense, holism may also be called "'''holiatry'''".<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref> Several approaches are used by [[medical doctors]], [[dietitian]]s, and [[religion|religious]] institutions, usually recommended based on an individual basis.<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref> Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including [[Hinduism]],<ref name="Fenton 2010"></ref> [[Shinto]],<ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref> and the [[Seventh-Day Adventist Church]].<ref name="TODAY.com 2016"></ref><br />
<br />
The term holistic when applied to diet or medical health refers to intuitive approach to food, eating, or lifestyle. One example is in the context of holistic medicine, "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms. In this sense, holism may also be called "holiatry". Several approaches are used by medical doctors, dietitians, and religious institutions, usually recommended based on an individual basis. Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including Hinduism, and the Seventh-Day Adventist Church.<ref name="TODAY.com 2016"></ref><br />
<br />
当用于饮食或医疗健康时，术语“整体”指的是对食物、饮食或生活方式的直觉方法。<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref>举个例子，在整体医学的背景下，<font color="#ff8000"> 整体论</font>是指治疗一个人健康的所有方面，包括心理和社会因素，而不仅仅是他/她的身体状况或症状。<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref>从这个意义上讲，<font color="#ff8000"> 整体论</font>也可以称为<font color="#ff8000"> 整体主义 holiatry </font>。<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref>医生、营养师和宗教机构通常根据个人基础推荐几种方法。<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref>宗教机构的信徒，如<font color="#ff8000">印度教 Hinduism </font> <ref name="Fenton 2010"></ref>、<font color="#ff8000">神道教 Shinto </font><ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref>和<font color="#ff8000">基督复临安息日会 Seventh-Day Adventist Church </font> <ref name="TODAY.com 2016"></ref>的信徒，他们使用一个全面的饮食和健康的方法，事实证明，他们比周围人口的寿命更长。<br />
<br />
<br />
<br />
== See also参见 ==<br />
<br />
*[[Antireductionism]]<br />
反还原论<br />
*[[Emergentism]]<br />
紧急情况论<br />
*[[Gaia hypothesis]]<br />
盖亚假说<br />
*[[Holistic education]]<br />
整体教育<br />
*[[Holism in science]]<br />
科学中的整体论<br />
* [[Monism]]<br />
一元论<br />
*[[Organicism]]<br />
有机主义<br />
*[[Synergy]]<br />
协同作用 <br />
<br />
<br />
==References参考==<br />
<br />
{{reflist}}<br />
<br />
<br />
<br />
== Further reading 拓展阅读==<br />
<br />
* Fodor, Jerry, and Ernst Lepore, ''Holism: A Shopper's Guide'' Wiley. New York. 1992<br />
<br />
* Phillips, D.C. ''Holistic Thought in Social Science''. Stanford University Press. Stanford. 1976.<br />
<br />
<br />
<br />
== External links外部链接 ==<br />
<br />
{{Wiktionary}}<br />
<br />
{{wikiquote}}<br />
<br />
*{{Commonscat-inline}}<br />
<br />
<br />
<br />
{{Positivism}}<br />
<br />
{{JanSmutsFooter}}<br />
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<br />
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[[Category:Holism| ]]<br />
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Category:Philosophical theories<br />
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范畴: 哲学理论<br />
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[[Category:Natural philosophy]]<br />
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Category:Natural philosophy<br />
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[[Category:Philosophy of science]]<br />
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Category:Philosophy of science<br />
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范畴: 科学哲学<br />
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[[Category:Social theories]]<br />
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Category:Social theories<br />
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Category:Jan Smuts<br />
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类别: Jan Smuts<br />
<br />
<noinclude><br />
<br />
<small>This page was moved from [[wikipedia:en:Holism]]. Its edit history can be viewed at [[整体论/edithistory]]</small></noinclude><br />
<br />
[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E6%95%B4%E4%BD%93%E8%AE%BA_Holism&diff=19054整体论 Holism2020-11-22T17:01:16Z<p>Smile：</p>
<hr />
<div>此词条暂由Henry翻译。已由Smile审校<br />
<br />
'''Holism''' (from [[Ancient Greek|Greek]] {{lang|grc|ὅλος}} ''holos'' "all, whole, entire") is the idea that various [[system]]s (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts.<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> The term "holism" was coined by [[Jan Smuts]] in his 1926 book ''[[Holism and Evolution]]''.<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref><br />
<br />
Holism (from Greek holos "all, whole, entire") is the idea that various systems (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts. The term "holism" was coined by Jan Smuts in his 1926 book Holism and Evolution.<br />
<br />
<font color="#ff8000"> 整体论 Holism</font>（源自希腊holos“all，whole，entire”）是指各种系统（例如物理的、生物的、社会的）应该被视为整体，而不仅仅作为各个部分的集合的观点。<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> “整体论”这个词是扬·斯密茨 Jan Smuts在1926年出版的《整体论与进化》<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref>一书中提出的。 <br />
<br />
<br />
<br />
==Meaning含义==<br />
<br />
The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures.<ref name=oed /> However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts.<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref> In this sense, "holism" may also be spelled "'''wholism'''", and it may be contrasted with [[reductionism]] or [[atomism]].<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
<br />
The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures. However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts. In this sense, "holism" may also be spelled "wholism", and it may be contrasted with reductionism or atomism.<br />
<br />
<font color="#ff8000"> 整体论</font>的确切含义取决于语境。Smuts最初用<font color="#ff8000"> 整体论</font>来指自然界中从单元结构的有序组合中产生整体的倾向。<ref name=oed /> 然而，在<font color="#ff8000"> 整体论</font>的一般用法中，<font color="#ff8000"> 整体论</font>通常指整体优于部分之和的观点。<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref>在这个意义上，<font color="#ff8000"> 整体论</font>也可以拼写为<font color="#ff8000">整体主义 wholism </font> ，它可以与<font color="#ff8000"> 还原论 reductionism</font>或<font color="#ff8000"> 原子论 atomism </font>形成对比。<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
<br />
<br />
=== Diet and health饮食与健康 ===<br />
<br />
The term holistic when applied to [[Diet (nutrition)|diet]] or medical [[health]] refers to [[intuitive]] approach to [[food]], [[eating]], or [[Lifestyle (sociology)|lifestyle]].<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref> One example is in the context of [[holistic medicine]], "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms.<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref> In this sense, holism may also be called "'''holiatry'''".<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref> Several approaches are used by [[medical doctors]], [[dietitian]]s, and [[religion|religious]] institutions, usually recommended based on an individual basis.<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref> Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including [[Hinduism]],<ref name="Fenton 2010"></ref> [[Shinto]],<ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref> and the [[Seventh-Day Adventist Church]].<ref name="TODAY.com 2016"></ref><br />
<br />
The term holistic when applied to diet or medical health refers to intuitive approach to food, eating, or lifestyle. One example is in the context of holistic medicine, "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms. In this sense, holism may also be called "holiatry". Several approaches are used by medical doctors, dietitians, and religious institutions, usually recommended based on an individual basis. Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including Hinduism, and the Seventh-Day Adventist Church.<ref name="TODAY.com 2016"></ref><br />
<br />
当用于饮食或医疗健康时，术语“整体”指的是对食物、饮食或生活方式的直觉方法。<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref>举个例子，在整体医学的背景下，<font color="#ff8000"> 整体论</font>是指治疗一个人健康的所有方面，包括心理和社会因素，而不仅仅是他/她的身体状况或症状。<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref>从这个意义上讲，<font color="#ff8000"> 整体论</font>也可以称为<font color="#ff8000"> 整体主义 holiatry </font>。<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref>医生、营养师和宗教机构通常根据个人基础推荐几种方法。<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref>宗教机构的信徒，包括<font color="#ff8000">印度教 Hinduism </font> <ref name="Fenton 2010"></ref>、<font color="#ff8000">神道教 Shinto </font><ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref>和<font color="#ff8000">基督复临安息日会 Seventh-Day Adventist Church </font> <ref name="TODAY.com 2016"></ref>的信徒，尝试了一个全面的饮食和健康的方法，已经得到证明，他们比周围人口的寿命更长。<br />
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== See also参见 ==<br />
<br />
*[[Antireductionism]]<br />
反还原论<br />
*[[Emergentism]]<br />
紧急情况论<br />
*[[Gaia hypothesis]]<br />
盖亚假说<br />
*[[Holistic education]]<br />
整体教育<br />
*[[Holism in science]]<br />
科学中的整体论<br />
* [[Monism]]<br />
一元论<br />
*[[Organicism]]<br />
有机主义<br />
*[[Synergy]]<br />
协同作用 <br />
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==References参考==<br />
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{{reflist}}<br />
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== Further reading 拓展阅读==<br />
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* Fodor, Jerry, and Ernst Lepore, ''Holism: A Shopper's Guide'' Wiley. New York. 1992<br />
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* Phillips, D.C. ''Holistic Thought in Social Science''. Stanford University Press. Stanford. 1976.<br />
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== External links外部链接 ==<br />
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<noinclude><br />
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<small>This page was moved from [[wikipedia:en:Holism]]. Its edit history can be viewed at [[整体论/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smilehttps://wiki.swarma.org/index.php?title=%E6%95%B4%E4%BD%93%E8%AE%BA_Holism&diff=19053整体论 Holism2020-11-22T16:58:07Z<p>Smile：</p>
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<div>此词条暂由Henry翻译。已由Smile审校<br />
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'''Holism''' (from [[Ancient Greek|Greek]] {{lang|grc|ὅλος}} ''holos'' "all, whole, entire") is the idea that various [[system]]s (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts.<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> The term "holism" was coined by [[Jan Smuts]] in his 1926 book ''[[Holism and Evolution]]''.<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref><br />
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Holism (from Greek holos "all, whole, entire") is the idea that various systems (e.g. physical, biological, social) should be viewed as wholes, not merely as a collection of parts. The term "holism" was coined by Jan Smuts in his 1926 book Holism and Evolution.<br />
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<font color="#ff8000"> 整体论 Holism</font>（源自希腊holos“all，whole，entire”）是指各种系统（例如物理的、生物的、社会的）应该被视为整体，而不仅仅作为各个部分的集合的观点。<ref>{{Citation | first = Barry | last = Oshry | title = Seeing Systems: Unlocking the Mysteries of Organizational Life | publisher = Berrett-Koehler | year = 2008}}.</ref><ref>{{Citation | first = Sunny Y | last = Auyang | title = Foundations of Complex-system Theories: in Economics, Evolutionary Biology, and Statistical Physics | publisher = Cambridge University Press | year = 1999}}.</ref> “整体论”这个词是扬·斯密茨 Jan Smuts在1926年出版的《整体论与进化》<ref name=oed>"holism, n." OED Online, [[Oxford University Press]], September 2019, www.oed.com/view/Entry/87726. Accessed 23 October 2019.</ref>一书中提出的。 <br />
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==Meaning含义==<br />
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The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures.<ref name=oed /> However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts.<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref> In this sense, "holism" may also be spelled "'''wholism'''", and it may be contrasted with [[reductionism]] or [[atomism]].<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
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The exact meaning of "holism" depends on context. Smuts originally used "holism" to refer to the tendency in nature to produce wholes from the ordered grouping of unit structures. However, in common usage, "holism" usually refers to the idea that a whole is greater than the sum of its parts. In this sense, "holism" may also be spelled "wholism", and it may be contrasted with reductionism or atomism.<br />
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<font color="#ff8000"> 整体论</font>的确切含义取决于语境。Smuts最初用<font color="#ff8000"> 整体论</font>来指自然界中从单元结构的有序组合中产生整体的倾向。<ref name=oed /> 然而，在<font color="#ff8000"> 整体论</font>的一般用法中，<font color="#ff8000"> 整体论</font>通常指整体优于部分之和的观点。<ref name=poynton>J. C. Poynton (1987) SMUTS'S HOLISM AND EVOLUTION SIXTY YEARS ON, Transactions of the Royal Society of South Africa, 46:3, 181-189, DOI:10.1080/00359198709520121</ref>在这个意义上，<font color="#ff8000"> 整体论</font>也可以拼写为<font color="#ff8000">整体主义 wholism </font> ，它可以与<font color="#ff8000"> 还原论 reductionism</font>或<font color="#ff8000"> 原子论 atomism </font>形成对比。<ref> "wholism, n." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/228738. Accessed 23 October 2019.</ref><br />
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=== Diet and health饮食与健康 ===<br />
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The term holistic when applied to [[Diet (nutrition)|diet]] or medical [[health]] refers to [[intuitive]] approach to [[food]], [[eating]], or [[Lifestyle (sociology)|lifestyle]].<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref> One example is in the context of [[holistic medicine]], "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms.<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref> In this sense, holism may also be called "'''holiatry'''".<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref> Several approaches are used by [[medical doctors]], [[dietitian]]s, and [[religion|religious]] institutions, usually recommended based on an individual basis.<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref> Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including [[Hinduism]],<ref name="Fenton 2010"></ref> [[Shinto]],<ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref> and the [[Seventh-Day Adventist Church]].<ref name="TODAY.com 2016"></ref><br />
<br />
The term holistic when applied to diet or medical health refers to intuitive approach to food, eating, or lifestyle. One example is in the context of holistic medicine, "holism" refers to treating all aspects of a person's health, including psychological and societal factors, rather than only his/her physical conditions or symptoms. In this sense, holism may also be called "holiatry". Several approaches are used by medical doctors, dietitians, and religious institutions, usually recommended based on an individual basis. Adherents of religious institutions, that practice a holistic dietary and health approach, have been shown have longer lifespans than those of surrounding populations, including Hinduism, and the Seventh-Day Adventist Church.<ref name="TODAY.com 2016"></ref><br />
<br />
当用于饮食或医疗健康时，术语“整体”指的是对食物、饮食或生活方式的直觉方法。<ref name="Chesak 2018">{{cite web | last=Chesak | first=Jennifer | title=The No BS Guide to Holistic, Healthier Eating | website=Healthline | date=October 23, 2018 | url=https://www.healthline.com/health/food-nutrition/how-to-start-intuitive-eating | access-date=August 15, 2020}}</ref>举个例子，在整体医学的背景下，<font color="#ff8000"> 整体论</font>是指治疗一个人健康的所有方面，包括心理和社会因素，而不仅仅是他/她的身体状况或症状。<ref>"holistic, adj." OED Online, Oxford University Press, September 2019, www.oed.com/view/Entry/87727. Accessed 23 October 2019.</ref>从这个意义上讲，<font color="#ff8000"> 整体论</font>也可以称为<font color="#ff8000"> 整体论 holiatry </font>。<ref>[https://www.dictionary.com/browse/holism Dictionary.com: holism]</ref>医生、营养师和宗教机构通常根据个人基础推荐几种方法。<ref name="Fenton 2010">{{cite web | last=Fenton | first=Crystal | title=Holistic Diet | website=LIVESTRONG.COM | date=April 16, 2010 | url=https://www.livestrong.com/article/107401-holistic-diet/ | access-date=August 15, 2020}}</ref><ref name="doctoroz.com 2011">{{cite web | title=28-Day Holistic Health Overhaul | website=doctoroz.com | date=January 27, 2011 | url=https://www.doctoroz.com/article/28-day-holistic-health-overhaul | access-date=August 15, 2020}}</ref><ref name="TODAY.com 2016">{{cite web | title=8 foods for a longer, healthier life | website=TODAY.com | date=October 21, 2016 | url=https://www.today.com/health/eat-adventist-8-foods-longer-healthier-life-t13901 | access-date=August 15, 2020}}</ref>宗教机构的信徒，包括<font color="#ff8000">印度教 Hinduism </font> <ref name="Fenton 2010"></ref>、<font color="#ff8000">神道教 Shinto </font><ref name="Eesti Rahvaluule">{{cite web | title=FACING THE SPIRITS: ILLNESS AND HEALING IN A JAPANESE COMMUNITY | website=Eesti Rahvaluule | url=http://www.folklore.ee/rl/pubte/ee/usund/ingl/kalland.html | access-date=October 9, 2020}}</ref>和<font color="#ff8000">基督复临安息日会 Seventh-Day Adventist Church </font> <ref name="TODAY.com 2016"></ref>的信徒，尝试了一个全面的饮食和健康的方法，已经得到证明，他们比周围人口的寿命更长。<br />
<br />
<br />
<br />
== See also参见 ==<br />
<br />
*[[Antireductionism]]<br />
反还原论<br />
*[[Emergentism]]<br />
紧急情况论<br />
*[[Gaia hypothesis]]<br />
盖亚假说<br />
*[[Holistic education]]<br />
整体教育<br />
*[[Holism in science]]<br />
科学中的整体论<br />
* [[Monism]]<br />
一元论<br />
*[[Organicism]]<br />
有机主义<br />
*[[Synergy]]<br />
协同作用 <br />
<br />
<br />
==References参考==<br />
<br />
{{reflist}}<br />
<br />
<br />
<br />
== Further reading 拓展阅读==<br />
<br />
* Fodor, Jerry, and Ernst Lepore, ''Holism: A Shopper's Guide'' Wiley. New York. 1992<br />
<br />
* Phillips, D.C. ''Holistic Thought in Social Science''. Stanford University Press. Stanford. 1976.<br />
<br />
<br />
<br />
== External links外部链接 ==<br />
<br />
{{Wiktionary}}<br />
<br />
{{wikiquote}}<br />
<br />
*{{Commonscat-inline}}<br />
<br />
<br />
<br />
{{Positivism}}<br />
<br />
{{JanSmutsFooter}}<br />
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[[Category:Holism| ]]<br />
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[[Category:Philosophical theories]]<br />
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Category:Philosophical theories<br />
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范畴: 哲学理论<br />
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[[Category:Natural philosophy]]<br />
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Category:Natural philosophy<br />
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范畴: 自然哲学<br />
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[[Category:Philosophy of science]]<br />
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Category:Philosophy of science<br />
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Category:Social theories<br />
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<noinclude><br />
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<small>This page was moved from [[wikipedia:en:Holism]]. Its edit history can be viewed at [[整体论/edithistory]]</small></noinclude><br />
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[[Category:待整理页面]]</div>Smile