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{{short description|Probability distribution}}
{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}
{{see also|Negative binomial distribution}}
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{{Probability distribution
{{Probability distribution
{概率分布
| name = Binomial distribution
| name = Binomial distribution
二项分布
| type = mass
| type = mass
类型 = 质量
| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]
| pdf_image = Probability mass function for the binomial distribution
2012年3月24日 | pdf 图片 = 概率质量函数二项分布
| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]
| cdf_image = Cumulative distribution function for the binomial distribution
2012年3月24日 | cdf 图像 = 累积分布函数二项分布
| notation = <math>B(n,p)</math>
| notation = B(n,p)
| 符号 = b (n，p)
| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> – number of trials<br /><math>p \in [0,1]</math> – success probability for each trial<br /><math>q = 1 - p</math>
| parameters = n \in \{0, 1, 2, \ldots\} – number of trials<br />p \in [0,1] – success probability for each trial<br />q = 1 - p
| parameters = 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
| support = <math>k \in \{0, 1, \ldots, n\}</math> – number of successes
| support = k \in \{0, 1, \ldots, n\} – number of successes
| support = k in {0,1，ldots，n }——成功的数量
| pdf = <math>\binom{n}{k} p^k q^{n-k}</math>
| pdf = \binom{n}{k} p^k q^{n-k}
| pdf = binom { n }{ k } p ^ k ^ { n-k }
| cdf = <math>I_{q}(n - k, 1 + k)</math>
| cdf = I_{q}(n - k, 1 + k)
| cdf = i _ { q }(n-k，1 + k)
| mean = <math>np</math>
| mean = np
平均值 = np
| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math>
| median = \lfloor np \rfloor or \lceil np \rceil
中位数 = lfloor np rfloor 或 lceil np rceil
| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math>
| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1
| mode = lfloor (n + 1) p rfloor 或 lceil (n + 1) p rceil-1
| variance = <math>npq</math>
| variance = npq
| variance = npq
| skewness = <math>\frac{q-p}{\sqrt{npq}}</math>
| skewness = \frac{q-p}{\sqrt{npq}}
| skewness = frac { q-p }{ sqrt { npq }}
| kurtosis = <math>\frac{1-6pq}{npq}</math>
| kurtosis = \frac{1-6pq}{npq}
| 峭度 = frac {1-6pq }{ npq }
| 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.
| 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.
| 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。
| mgf = <math>(q + pe^t)^n</math>
| mgf = (q + pe^t)^n
| mgf = (q + pe ^ t) ^ n
| char = <math>(q + pe^{it})^n</math>
| char = (q + pe^{it})^n
| char = (q + pe ^ { it }) ^ n
| pgf = <math>G(z) = [q + pz]^n</math>
| pgf = G(z) = [q + pz]^n
| pgf = g (z) = [ q + pz ] ^ n
| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)
| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)
| fisher = g _ n (p) = frac { n }{ pq } < br/> (对于固定 n)
}}
}}
}}
[[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'' = 8) ends up in the central bin (''k'' = 4) is <math>70/256</math>.]]
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 = 8) ends up in the central bin (k = 4) is 70/256.]]
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]
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'' = 1 − ''p'').
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 = 1 − p).
在概率论和统计学中，参数 n 和 p 的二项分布是 n 个独立实验序列中成功次数的离散概率分布，每个实验询问一个 yes-no 问题，每个实验都有自己的布尔值结果: success/yes/true/one (带概率 p)或 failure/no/false/zero (带概率 q = 1-p)。
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'' = 1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].
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 = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.
一个单一的成功/失败的实验也被称为伯努利实验或伯努利实验，一系列的结果被称为伯努利过程; 对于一个单一的实验，例如，n = 1，二项分布是一个伯努利分布。二项分布是流行的统计显著性二项检验的基础。
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.
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.
二项分布经常被用来模拟 n 大小的样本中的成功数量，这些样本是用 n 大小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，所以得到的分布是超几何分布，而不是二项式。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。
==Definitions==
===Probability mass function===
In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X'' ~ B(''n'', ''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:
In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X ~ B(n, p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:
一般来说，如果随机变量 x 服从参数 n ∈ n 和 p ∈[0,1]的二项分布，我们写 x ~ b (n，p)。在 n 个独立的 Bernoulli 试验中获得 k 成功的概率是由美国概率质量函数协会给出的:
:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math>
f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}
F (k，n，p) = Pr (k; n，p) = Pr (x = k) = binom { n }{ k } p ^ k (1-p) ^ { n-k }
for ''k'' = 0, 1, 2, ..., ''n'', where
for k = 0, 1, 2, ..., n, where
对于 k = 0,1,2，... ，n，其中
:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math>
\binom{n}{k} =\frac{n!}{k!(n-k)!}
(n-k)
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'' − ''k'' failures occur with probability (1 − ''p'')<sup>''n'' − ''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.
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 − k failures occur with probability (1 − p)n − 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.
是二项式系数，因而得名。这个公式可以理解为。K 的成功发生在概率 pk 和 n-k 的失败发生在概率(1-p) n-k 的情况下。然而，k 的成功可以发生在 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 的成功分配方法。
In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k'' > ''n''/2, the probability can be calculated by its complement as
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k > n/2, the probability can be calculated by its complement as
在为二项分布概率创建参考表时，通常表中最多填充 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
:<math>f(k,n,p)=f(n-k,n,1-p). </math>
f(k,n,p)=f(n-k,n,1-p).
F (k，n，p) = f (n-k，n，1-p).
Looking at the expression ''f''(''k'', ''n'', ''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating
Looking at the expression f(k, n, p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating
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
:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math>
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)}
Frac { f (k + 1，n，p)}{ f (k，n，p)} = frac {(n-k) p }{(k + 1)(1-p)}
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>
and comparing it to 1. There is always an integer M that satisfies
并与1相比较。总有一个整数 m 满足
:<math>(n+1)p-1 \leq M < (n+1)p.</math>
(n+1)p-1 \leq M < (n+1)p.
(n + 1) p-1 leq m < (n + 1) p.
''f''(''k'', ''n'', ''p'') is monotone increasing for ''k'' < ''M'' and monotone decreasing for ''k'' > ''M'', with the exception of the case where (''n'' + 1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n'' + 1)''p'' and (''n'' + 1)''p'' − 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]].
f(k, n, p) is monotone increasing for k < M and monotone decreasing for k > M, with the exception of the case where (n + 1)p is an integer. In this case, there are two values for which f is maximal: (n + 1)p and (n + 1)p − 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.
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 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，尽管总的来说仍然不太可能) ，被称为模式。
===Example===
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
Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is
假设抛出一枚有偏差的硬币时，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
:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>
f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.
F (4,6,0.3) = binom {6}{4}0.3 ^ 4(1-0.3) ^ {6-4} = 0.059535.
===Cumulative distribution function===
The [[cumulative distribution function]] can be expressed as:
The cumulative distribution function can be expressed as:
累积分布函数可以表达为:
:<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>
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},
F (k; n，p) = Pr (x le k) = sum _ { i = 0} ^ { lfloor k rfloor }{ n choose i } p ^ i (1-p) ^ { n-i } ,
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.
楼层是 k 下面的”楼层” ，也就是。小于或等于 k 的最大整数。
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>
It can also be represented in terms of the regularized incomplete beta function, as follows:
它也可以用正规化的不完全 beta 函数来表示，如下:
:<math>\begin{align}
<math>\begin{align}
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
F(k;n,p) & = \Pr(X \le k) \\
F(k;n,p) & = \Pr(X \le k) \\
F (k; n，p) & = Pr (x le k)
&= I_{1-p}(n-k, k+1) \\
&= I_{1-p}(n-k, k+1) \\
& = i _ {1-p }(n-k，k + 1)
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.
& = (n-k){ n choose k } int _ 0 ^ {1-p } t ^ { n-k-1}(1-t) ^ k，dt.
\end{align}</math>
\end{align}</math>
结束{ align } </math >
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>
which is equivalent to the cumulative distribution function of the -distribution:
这相当于-分布的累积分布函数:
:<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>
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).
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)。
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].
Some closed-form bounds for the cumulative distribution function are given below.
下面给出了累积分布函数的一些封闭形式界。
== Properties ==
===Expected value and variance===
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>
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:
如果 x ~ b (n，p) ，即 x 是一个二元分布的随机变量，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 的期望值是:
:<math> \operatorname{E}[X] = np.</math>
\operatorname{E}[X] = np.
操作员名称{ e }[ x ] = np。
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
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
这是从线性的期望值，以及事实上是相同的伯努利随机变量的和，每个与期望值。换句话说，如果 x1，ldots，xn 是带参数的伯努利随机变量，那么 x = x1 + cdots + xn
:<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>
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.
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。
The [[variance]] is:
The variance is:
方差是:
:<math> \operatorname{Var}(X) = np(1 - p).</math>
\operatorname{Var}(X) = np(1 - p).
操作符名称{ Var }(x) = np (1-p)。
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.
这同样也是从这样一个事实出发: 一个独立随机变量之和的方差是方差之和。
===Higher moments===
The first 6 central moments are given by
The first 6 central moments are given by
前6个中心时刻由
:<math>\begin{align}
<math>\begin{align}
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
\mu_1 &= 0, \\
\mu_1 &= 0, \\
1 & = 0,
\mu_2 &= np(1-p),\\
\mu_2 &= np(1-p),\\
mu _ 2 & = np (1-p) ,
\mu_3 &= np(1-p)(1-2p),\\
\mu_3 &= np(1-p)(1-2p),\\
mu _ 3 & = np (1-p)(1-2p) ,
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\
mu _ 4 & = np (1-p)(1 + (3n-6) p (1-p)) ,
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\
mu _ 5 & = np (1-p)(1-2p)(1 + (10n-12) p (1-p)) ,
\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).
\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).
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).
\end{align}</math>
\end{align}</math>
结束{ align } </math >
===Mode===
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'' + 1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n'' + 1)''p'' and (''n'' + 1)''p'' − 1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:
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 + 1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n + 1)p and (n + 1)p − 1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:
通常二项式 b (n，p)分布的模式等于 lfloor (n + 1) p 楼层，其中 lfloor cdot 楼层是楼层函数。然而，当(n + 1) p 是整数且 p 既不是0也不是1时，分布有两种模式: (n + 1) p 和(n + 1) p-1。当 p 等于0或1时，模态将相应地为0和 n。这些情况可概述如下:
: <math>\text{mode} =
<math>\text{mode} =
1.1.1.2.1
\begin{cases}
\begin{cases}
开始{ cases }
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\
文本{ if }(n + 1) p text { is 0 or a noninteger } ,
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\
(n + 1) ，p text { and }(n + 1) ，p-1 & text { if }(n + 1) p in {1，dots，n } ,
n & \text{if }(n+1)p = n + 1.
n & \text{if }(n+1)p = n + 1.
n & text { if }(n + 1) p = n + 1.
\end{cases}</math>
\end{cases}</math>
结束{ cases } </math >
'''Proof:''' Let
Proof: Let
证据: 让
:<math>f(k)=\binom nk p^k q^{n-k}.</math>
f(k)=\binom nk p^k q^{n-k}.
F (k) = binom nk p ^ k q ^ { n-k }.
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>.
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.
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和 f (k) = 0对于 k neq n。这证明了 p = 0时模式为0，p = 1时模式为 n。
Let <math>0 < p < 1</math>. We find
Let 0 < p < 1. We find
让0 < p < 1。我们发现
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.
Frac { f (k + 1)}{ f (k)} = frac {(n-k) p }{(k + 1)(1-p)}.
From this follows
From this follows
由此可见
:<math>\begin{align}
<math>\begin{align}
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
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\
K > (n + 1) p-1 right tarrow f (k + 1) < f (k)
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\
K = (n + 1) p-1 right tarrow f (k + 1) = f (k)
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)
K < (n + 1) p-1 right tarrow f (k + 1) > f (k)
\end{align}</math>
\end{align}</math>
结束{ align } </math >
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>
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.
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。
===Median===
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:
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:
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，甚至可能是非唯一的。然而，已经确立了若干特别成果:
* 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>
* Any median ''m'' must lie within the interval ⌊''np''⌋ ≤ ''m'' ≤ ⌈''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>
* A median ''m'' cannot lie too far away from the mean: {{nowrap||''m'' − ''np''| ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal
| last1 = Hamza | first1 = K.
| doi = 10.1016/0167-7152(94)00090-U
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right)
F (k; n，p) leq exp left (- nD left (frac { k }{ n } parallel p right)))
| journal = Statistics & Probability Letters
| volume = 23
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):
其中 d (a | | p)是一枚硬币和一枚 p 相对熵之间的距离。在伯努利(a)和伯努利(p)分布之间:
| pages = 21–25
| year = 1995
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!
D (a 并行 p) = (a) log frac { a }{ p } + (1-a) log frac {1-a }{1-p }。\!
| pmid =
| pmc =
Asymptotically, this bound is reasonably tight; see
从渐近的角度来看，这个界限是相当紧密的; 参见
}}</ref>
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),
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)) ,
* The median is unique and equal to ''m'' = [[Rounding|round]](''np'') when |''m'' − ''np''| ≤ min{''p'', 1 − ''p''} (except for the case when ''p'' = {{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/>
which implies the simpler but looser bound
这意味着更简单但更松散的约束
* When ''p'' = 1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n'' − 1) ≤ ''m'' ≤ {{sfrac|1|2}}(''n'' + 1) is a median of the binomial distribution. If ''p'' = 1/2 and ''n'' is even, then ''m'' = ''n''/2 is the unique median.
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).
F (k; n，p) geq frac1{ sqrt {2n } exp left (- nD left (frac { k }{ n } parallel p right))).
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:
对于 p = 1/2和 k ≥3n/8的偶数 n，可以使分母为常数:
===Tail bounds===
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''.
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!
F (k; n，tfrac {1}{2}) geq frac {1}{15} exp left (- 16n left (frac {1}{2}-frac { k }{ n }右) ^ 2 right)。\!
[[Hoeffding's inequality]] yields the simple bound
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math>
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'' < ''n''), but Hoeffding's bound evaluates to a positive constant.
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.
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 Lehmann-scheffé 定理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。
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>
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.
利用 Β分布作为共轭先验分布时，p 的 Bayes 估计也是一个封闭形式。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) ，可容许和一致的概率。
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math>
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.
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)。这种方法被称为继承法则，它是在18世纪由皮埃尔-西蒙·拉普拉斯引进的。
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):
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})
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 })
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math>
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.
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.
即使对于非常大的 n 值，均值的实际分布也是非常非正态的。针对这一问题，提出了几种估计置信区间的方法。
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>
:<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>
In the equations for confidence intervals below, the variables have the following meaning:
在下面的置信区间等式中，这些变量具有以下含义:
which implies the simpler but looser bound
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math>
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>
:<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>
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .
1-widehat { p，})}{ n }.
== Statistical Inference ==
A continuity correction of 0.5/n may be added.
可以加上0.5/n 的连续性修正。
=== Estimation of parameters ===
{{seealso|Beta distribution#Bayesian inference}}
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]].
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .
1-tilde { p })}{ n + z ^ 2}.
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.
Here the estimate of p is modified to
这里 p 的估计被修改为
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]].
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }
1 + frac {1}{2} z2}{ n + z2}
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>)
=== Confidence intervals ===
{{Main|Binomial proportion confidence interval}}
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).
Sin ^ 2 left (arcsin left (sqrt { widehat { p，} right) pm frac { z }{2 sqrt { n } right).
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.
In the equations for confidence intervals below, the variables have the following meaning:
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes
The notation in the formula below differs from the previous formulas in two respects:
下列公式中的符号在两个方面不同于以前的公式:
* <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> = 0.05, so <math>1 - \tfrac{1}{2}\alpha</math> = 0.975 and <math>z</math> = 1.96.
==== Wald method ====
:: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>
<math>\frac{
我不知道
\widehat{p\,} + \frac{z^2}{2n} + z
2n } + z
: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}
\sqrt{
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
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +
1-widehat { p，})}{ n } +
==== Agresti–Coull method ====
\frac{z^2}{4 n^2}
4 n ^ 2}
<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>
}
}
}{
}{
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math>
1 + \frac{z^2}{n}
1 + frac { z ^ 2}{ n }
}</math>
{/math >
: Here the estimate of ''p'' is modified to
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>
The exact (Clopper–Pearson) method is the most conservative.
确切的(克洛佩尔-皮尔森)方法是最保守的。
==== Arcsine method ====
Let X ~ B(n,p1) and Y ~ B(m,p2) be independent. Let T = (X/n)/(Y/m).
设 x ~ b (n，p1)和 y ~ b (m，p2)是独立的。设 t = (X/n)/(Y/m)。
<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>
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m.
然后对数(t)近似呈正态分布，其中平均对数(p1/p2)和方差((1/p1)-1)/n + ((1/p2)-1)/m。
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>
==== Wilson (score) method ====
If X ~ B(n, p) and Y | X ~ B(X, q) (the conditional distribution of Y, given X), then Y is a simple binomial random variable with distribution Y ~ B(n, pq).
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。
{{Main|Binomial proportion confidence interval#Wilson score interval}}
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 ~ B(n, 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 ~ B(X, q) and therefore Y ~ B(n, pq).
例如，想象一下把 n 个球扔到一个篮子里，然后把击中的球扔到另一个篮子里。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 yy 的概率，那么击中 yy 的球数是 y ~ b (x，q) ，因此 y ~ b (n，pq)。
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>
* 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 − ''x'')-th quantile'.
* 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> = 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>.
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,
由于 x sim b (n，p)和 y sim b (x，q) ，由全概率公式,
<math>\begin{align}
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
:: <math>\frac{
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
\widehat{p\,} + \frac{z^2}{2n} + z
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
和 = sum { k = m } n binom { n }{ k }{ m } p ^ k ^ m (1-p) ^ { n-k }(1-q) ^ { k-m }
\sqrt{
\end{align}</math>
结束{ align } </math >
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as
由于 tbinom { n }{ k } tbinom { k }{ m } = tbinom { n }{ m } tbinom { n-m }{ k-m } ，上述方程可表示为
\frac{z^2}{4 n^2}
\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}
Pr [ y = m ] = sum _ { k = m } ^ { n } binom { n-m }{ k-m } p ^ k ^ m (1-p) ^ { n-k }(1-q) ^ { k-m }
}
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
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
}{
<math>\begin{align}
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
1 + \frac{z^2}{n}
\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]
[ 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 ]
}</math><ref>{{cite book
&= \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)
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)
| chapter = Confidence intervals
\end{align}</math>
结束{ align } </math >
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
After substituting i = k - m in the expression above, we get
在上面的表达式中用 i = k-m 代替后，我们得到了
| title = Engineering Statistics Handbook
\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)
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)
| publisher = NIST/Sematech
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields
注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。用最终产量来代替这一点
| year = 2012
<math>\begin{align}
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
| access-date = 2017-07-23
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
}}</ref>
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}
& = binom { n }{ m }(pq) ^ m (1-pq) ^ { n-m }
\end{align}</math>
结束{ align } </math >
==== Comparison ====
and thus Y \sim B(n, pq) as desired.
因此 y sim b (n，pq)为所需。
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}
The Bernoulli distribution is a special case of the binomial distribution, where n = 1. Symbolically, X ~ B(1, p) has the same meaning as X ~ Bernoulli(p). Conversely, any binomial distribution, B(n, p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.
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。
==Related distributions==
===Sums of binomials===
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).
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。
If ''X'' ~ B(''n'', ''p'') and ''Y'' ~ B(''m'', ''p'') are independent binomial variables with the same probability ''p'', then ''X'' + ''Y'' is again a binomial variable; its distribution is ''Z=X+Y'' ~ B(''n+m'', ''p''):
:<math>\begin{align}
\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]\\
Binomial [[probability mass function and normal probability density function approximation for n = 6 and p = 0.5]]
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]
&= \binom{n+m}k p^k (1-p)^{n+m-k}
\end{align}</math>
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n, p) is given by the normal distribution
如果 n 足够大，那么分布的倾斜就不会太大。在这种情况下，通过正态分布给出 b (n，p)的合理近似
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>
\mathcal{N}(np,\,np(1-p)),
数学{ n }(np，np (1-p))
===Ratio of two binomial distributions===
and this basic approximation can be improved in a simple way by using a suitable continuity correction.
通过适当的连续性修正，可以简单地改进这种基本近似。
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:
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:
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>
Let ''X'' ~ B(''n'',''p''<sub>1</sub>) and ''Y'' ~ B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').
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}}
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 }
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>) − 1)/''n'' + ((1/''p''<sub>2</sub>) − 1)/''m''.
===Conditional binomials===
If ''X'' ~ B(''n'', ''p'') and ''Y'' | ''X'' ~ B(''X'', ''q'') (the conditional distribution of ''Y'', given ''X''), then ''Y'' is a simple binomial random variable with distribution ''Y'' ~ B(''n'', ''pq'').
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 ≥ 20 and p ≤ 0.05, or if n ≥ 100 and np ≤ 10.
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 的泊松分佈可以作为二项分布 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。
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'' ~ B(''n'', ''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'' ~ B(''X'', ''q'') and therefore ''Y'' ~ B(''n'', ''pq'').
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.
关于泊松近似的准确性，参见 Novak，ch。4，及其中的参考资料。
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],
:<math>\begin{align}
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.
给定一个统一的先验，给定具有观察到的成功事件的独立事件的成功概率的后验概率是一个 Β分布。
\end{align}</math>
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as
:<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>
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
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.
随机数产生的方法，其中的边缘分布是一个二项分布。
:<math>\begin{align}
\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]
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.
从二项分布中产生随机样本的一种方法是使用反演算法。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0到1之间的样本，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。
&= \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)
\end{align}</math>
After substituting <math> i = k - m </math> in the expression above, we get
:<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>
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r + s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p = 1/2.
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。
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
:<math>\begin{align}
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}
\end{align}</math>
and thus <math> Y \sim B(n, pq) </math> as desired.
{{hidden end}}
===Bernoulli distribution===
The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n'' = 1. Symbolically, ''X'' ~ B(1, ''p'') has the same meaning as ''X'' ~ Bernoulli(''p''). Conversely, any binomial distribution, B(''n'', ''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>
===Poisson binomial distribution===
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>
{{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>
Category:Discrete distributions
类别: 离散分布
===Normal approximation===
Category:Factorial and binomial topics
类别: 阶乘和二项式主题
Category:Conjugate prior distributions
范畴: 共轭先验分布
[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n'' = 6 and ''p'' = 0.5]]
Category:Exponential family distributions
类别: 指数族分布
<noinclude>
<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude>
[[Category:待整理页面]]
{{short description|Probability distribution}}
{{Redirect|Binomial model|the binomial model in options pricing|Binomial options pricing model}}
{{see also|Negative binomial distribution}}
<!-- EDITORS! Please see [[Wikipedia:WikiProject Probability#Standards]] for a discussion of standards used for probability distribution articles such as this one.
<!-- EDITORS! Please see Wikipedia:WikiProject Probability#Standards for a discussion of standards used for probability distribution articles such as this one.
< ! -- 编辑！参见 Wikipedia: WikiProject Probability # standards for a discussion for the standards used for the 概率分布文章使用的标准，如本文。
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{{Probability distribution
{{Probability distribution
{概率分布
| name = Binomial distribution
| name = Binomial distribution
二项分布
| type = mass
| type = mass
类型 = 质量
| pdf_image = [[File:Binomial distribution pmf.svg|300px|Probability mass function for the binomial distribution]]
| pdf_image = Probability mass function for the binomial distribution
2012年3月24日 | pdf 图片 = 概率质量函数二项分布
| cdf_image = [[File:Binomial distribution cdf.svg|300px|Cumulative distribution function for the binomial distribution]]
| cdf_image = Cumulative distribution function for the binomial distribution
2012年3月24日 | cdf 图像 = 累积分布函数二项分布
| notation = <math>B(n,p)</math>
| notation = B(n,p)
| 符号 = b (n，p)
| parameters = <math>n \in \{0, 1, 2, \ldots\}</math> – number of trials<br /><math>p \in [0,1]</math> – success probability for each trial<br /><math>q = 1 - p</math>
| parameters = n \in \{0, 1, 2, \ldots\} – number of trials<br />p \in [0,1] – success probability for each trial<br />q = 1 - p
| parameters = 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
| support = <math>k \in \{0, 1, \ldots, n\}</math> – number of successes
| support = k \in \{0, 1, \ldots, n\} – number of successes
| support = k in {0,1，ldots，n }——成功的数量
| pdf = <math>\binom{n}{k} p^k q^{n-k}</math>
| pdf = \binom{n}{k} p^k q^{n-k}
| pdf = binom { n }{ k } p ^ k ^ { n-k }
| cdf = <math>I_{q}(n - k, 1 + k)</math>
| cdf = I_{q}(n - k, 1 + k)
| cdf = i _ { q }(n-k，1 + k)
| mean = <math>np</math>
| mean = np
平均值 = np
| median = <math>\lfloor np \rfloor</math> or <math>\lceil np \rceil</math>
| median = \lfloor np \rfloor or \lceil np \rceil
中位数 = lfloor np rfloor 或 lceil np rceil
| mode = <math>\lfloor (n + 1)p \rfloor</math> or <math>\lceil (n + 1)p \rceil - 1</math>
| mode = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1
| mode = lfloor (n + 1) p rfloor 或 lceil (n + 1) p rceil-1
| variance = <math>npq</math>
| variance = npq
| variance = npq
| skewness = <math>\frac{q-p}{\sqrt{npq}}</math>
| skewness = \frac{q-p}{\sqrt{npq}}
| skewness = frac { q-p }{ sqrt { npq }}
| kurtosis = <math>\frac{1-6pq}{npq}</math>
| kurtosis = \frac{1-6pq}{npq}
| 峭度 = frac {1-6pq }{ npq }
| 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.
| 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.
| 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。
| mgf = <math>(q + pe^t)^n</math>
| mgf = (q + pe^t)^n
| mgf = (q + pe ^ t) ^ n
| char = <math>(q + pe^{it})^n</math>
| char = (q + pe^{it})^n
| char = (q + pe ^ { it }) ^ n
| pgf = <math>G(z) = [q + pz]^n</math>
| pgf = G(z) = [q + pz]^n
| pgf = g (z) = [ q + pz ] ^ n
| fisher = <math> g_n(p) = \frac{n}{pq} </math><br />(for fixed <math>n</math>)
| fisher = g_n(p) = \frac{n}{pq} <br />(for fixed n)
| fisher = g _ n (p) = frac { n }{ pq } < br/> (对于固定 n)
}}
}}
}}
[[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'' = 8) ends up in the central bin (''k'' = 4) is <math>70/256</math>.]]
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 = 8) ends up in the central bin (k = 4) is 70/256.]]
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]
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'' = 1 − ''p'').
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 = 1 − p).
在概率论和统计学中，参数 n 和 p 的二项分布是 n 个独立实验序列中成功次数的离散概率分布，每个实验询问一个 yes-no 问题，每个实验都有自己的布尔值结果: success/yes/true/one (带概率 p)或 failure/no/false/zero (带概率 q = 1-p)。
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'' = 1, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the popular [[binomial test]] of [[statistical significance]].
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 = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.
一个单一的成功/失败的实验也被称为伯努利实验或伯努利实验，一系列的结果被称为伯努利过程; 对于一个单一的实验，例如，n = 1，二项分布是一个伯努利分布。二项分布是流行的统计显著性二项检验的基础。
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.
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.
二项分布经常被用来模拟 n 大小的样本中的成功数量，这些样本是用 n 大小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，所以得到的分布是超几何分布，而不是二项式。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。
==Definitions==
===Probability mass function===
In general, if the [[random variable]] ''X'' follows the binomial distribution with parameters ''n'' [[∈]] [[natural number|ℕ]] and ''p'' ∈ [0,1], we write ''X'' ~ B(''n'', ''p''). The probability of getting exactly ''k'' successes in ''n'' independent Bernoulli trials is given by the [[probability mass function]]:
In general, if the random variable X follows the binomial distribution with parameters n ∈ ℕ and p ∈ [0,1], we write X ~ B(n, p). The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function:
一般来说，如果随机变量 x 服从参数 n ∈ n 和 p ∈[0,1]的二项分布，我们写 x ~ b (n，p)。在 n 个独立的 Bernoulli 试验中获得 k 成功的概率是由美国概率质量函数协会给出的:
:<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math>
f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}
F (k，n，p) = Pr (k; n，p) = Pr (x = k) = binom { n }{ k } p ^ k (1-p) ^ { n-k }
for ''k'' = 0, 1, 2, ..., ''n'', where
for k = 0, 1, 2, ..., n, where
对于 k = 0,1,2，... ，n，其中
:<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math>
\binom{n}{k} =\frac{n!}{k!(n-k)!}
(n-k)
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'' − ''k'' failures occur with probability (1 − ''p'')<sup>''n'' − ''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.
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 − k failures occur with probability (1 − p)n − 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.
是二项式系数，因而得名。这个公式可以理解为。K 的成功发生在概率 pk 和 n-k 的失败发生在概率(1-p) n-k 的情况下。然而，k 的成功可以发生在 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 的成功分配方法。
In creating reference tables for binomial distribution probability, usually the table is filled in up to ''n''/2 values. This is because for ''k'' > ''n''/2, the probability can be calculated by its complement as
In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k > n/2, the probability can be calculated by its complement as
在为二项分布概率创建参考表时，通常表中最多填充 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
:<math>f(k,n,p)=f(n-k,n,1-p). </math>
f(k,n,p)=f(n-k,n,1-p).
F (k，n，p) = f (n-k，n，1-p).
Looking at the expression ''f''(''k'', ''n'', ''p'') as a function of ''k'', there is a ''k'' value that maximizes it. This ''k'' value can be found by calculating
Looking at the expression f(k, n, p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating
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
:<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math>
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)}
Frac { f (k + 1，n，p)}{ f (k，n，p)} = frac {(n-k) p }{(k + 1)(1-p)}
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>
and comparing it to 1. There is always an integer M that satisfies
并与1相比较。总有一个整数 m 满足
:<math>(n+1)p-1 \leq M < (n+1)p.</math>
(n+1)p-1 \leq M < (n+1)p.
(n + 1) p-1 leq m < (n + 1) p.
''f''(''k'', ''n'', ''p'') is monotone increasing for ''k'' < ''M'' and monotone decreasing for ''k'' > ''M'', with the exception of the case where (''n'' + 1)''p'' is an integer. In this case, there are two values for which ''f'' is maximal: (''n'' + 1)''p'' and (''n'' + 1)''p'' − 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]].
f(k, n, p) is monotone increasing for k < M and monotone decreasing for k > M, with the exception of the case where (n + 1)p is an integer. In this case, there are two values for which f is maximal: (n + 1)p and (n + 1)p − 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.
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 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，尽管总的来说仍然不太可能) ，被称为模式。
===Example===
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
Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is
假设抛出一枚有偏差的硬币时，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
:<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>
f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.
F (4,6,0.3) = binom {6}{4}0.3 ^ 4(1-0.3) ^ {6-4} = 0.059535.
===Cumulative distribution function===
The [[cumulative distribution function]] can be expressed as:
The cumulative distribution function can be expressed as:
累积分布函数可以表达为:
:<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>
F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},
F (k; n，p) = Pr (x le k) = sum _ { i = 0} ^ { lfloor k rfloor }{ n choose i } p ^ i (1-p) ^ { n-i } ,
where <math>\lfloor k\rfloor</math> is the "floor" under ''k'', i.e. the [[greatest integer]] less than or equal to ''k''.
where \lfloor k\rfloor is the "floor" under k, i.e. the greatest integer less than or equal to k.
楼层是 k 下面的”楼层” ，也就是。小于或等于 k 的最大整数。
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>
It can also be represented in terms of the regularized incomplete beta function, as follows:
它也可以用正规化的不完全 beta 函数来表示，如下:
:<math>\begin{align}
<math>\begin{align}
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
F(k;n,p) & = \Pr(X \le k) \\
F(k;n,p) & = \Pr(X \le k) \\
F (k; n，p) & = Pr (x le k)
&= I_{1-p}(n-k, k+1) \\
&= I_{1-p}(n-k, k+1) \\
& = i _ {1-p }(n-k，k + 1)
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.
& = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt.
& = (n-k){ n choose k } int _ 0 ^ {1-p } t ^ { n-k-1}(1-t) ^ k，dt.
\end{align}</math>
\end{align}</math>
结束{ align } </math >
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>
which is equivalent to the cumulative distribution function of the -distribution:
这相当于-分布的累积分布函数:
:<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>
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).
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)。
Some closed-form bounds for the cumulative distribution function are given [[#Tail bounds|below]].
Some closed-form bounds for the cumulative distribution function are given below.
下面给出了累积分布函数的一些封闭形式界。
== Properties ==
===Expected value and variance===
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>
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:
如果 x ~ b (n，p) ，即 x 是一个二元分布的随机变量，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 的期望值是:
:<math> \operatorname{E}[X] = np.</math>
\operatorname{E}[X] = np.
操作员名称{ e }[ x ] = np。
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
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
这是从线性的期望值，以及事实上是相同的伯努利随机变量的和，每个与期望值。换句话说，如果 x1，ldots，xn 是带参数的伯努利随机变量，那么 x = x1 + cdots + xn
:<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>
\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.
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。
The [[variance]] is:
The variance is:
方差是:
:<math> \operatorname{Var}(X) = np(1 - p).</math>
\operatorname{Var}(X) = np(1 - p).
操作符名称{ Var }(x) = np (1-p)。
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.
这同样也是从这样一个事实出发: 一个独立随机变量之和的方差是方差之和。
===Higher moments===
The first 6 central moments are given by
The first 6 central moments are given by
前6个中心时刻由
:<math>\begin{align}
<math>\begin{align}
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
\mu_1 &= 0, \\
\mu_1 &= 0, \\
1 & = 0,
\mu_2 &= np(1-p),\\
\mu_2 &= np(1-p),\\
mu _ 2 & = np (1-p) ,
\mu_3 &= np(1-p)(1-2p),\\
\mu_3 &= np(1-p)(1-2p),\\
mu _ 3 & = np (1-p)(1-2p) ,
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\
\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\
mu _ 4 & = np (1-p)(1 + (3n-6) p (1-p)) ,
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\
\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\
mu _ 5 & = np (1-p)(1-2p)(1 + (10n-12) p (1-p)) ,
\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).
\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).
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).
\end{align}</math>
\end{align}</math>
结束{ align } </math >
===Mode===
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'' + 1)''p'' is an integer and ''p'' is neither 0 nor 1, then the distribution has two modes: (''n'' + 1)''p'' and (''n'' + 1)''p'' − 1. When ''p'' is equal to 0 or 1, the mode will be 0 and ''n'' correspondingly. These cases can be summarized as follows:
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 + 1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n + 1)p and (n + 1)p − 1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:
通常二项式 b (n，p)分布的模式等于 lfloor (n + 1) p 楼层，其中 lfloor cdot 楼层是楼层函数。然而，当(n + 1) p 是整数且 p 既不是0也不是1时，分布有两种模式: (n + 1) p 和(n + 1) p-1。当 p 等于0或1时，模态将相应地为0和 n。这些情况可概述如下:
: <math>\text{mode} =
<math>\text{mode} =
1.1.1.2.1
\begin{cases}
\begin{cases}
开始{ cases }
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\
\lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\
文本{ if }(n + 1) p text { is 0 or a noninteger } ,
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\
(n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\
(n + 1) ，p text { and }(n + 1) ，p-1 & text { if }(n + 1) p in {1，dots，n } ,
n & \text{if }(n+1)p = n + 1.
n & \text{if }(n+1)p = n + 1.
n & text { if }(n + 1) p = n + 1.
\end{cases}</math>
\end{cases}</math>
结束{ cases } </math >
'''Proof:''' Let
Proof: Let
证据: 让
:<math>f(k)=\binom nk p^k q^{n-k}.</math>
f(k)=\binom nk p^k q^{n-k}.
F (k) = binom nk p ^ k q ^ { n-k }.
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>.
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.
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和 f (k) = 0对于 k neq n。这证明了 p = 0时模式为0，p = 1时模式为 n。
Let <math>0 < p < 1</math>. We find
Let 0 < p < 1. We find
让0 < p < 1。我们发现
:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.
\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.
Frac { f (k + 1)}{ f (k)} = frac {(n-k) p }{(k + 1)(1-p)}.
From this follows
From this follows
由此可见
:<math>\begin{align}
<math>\begin{align}
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
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\
k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\
K > (n + 1) p-1 right tarrow f (k + 1) < f (k)
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\
k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\
K = (n + 1) p-1 right tarrow f (k + 1) = f (k)
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)
k < (n+1)p-1 \Rightarrow f(k+1) > f(k)
K < (n + 1) p-1 right tarrow f (k + 1) > f (k)
\end{align}</math>
\end{align}</math>
结束{ align } </math >
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>
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.
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。
===Median===
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:
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:
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，甚至可能是非唯一的。然而，已经确立了若干特别成果:
* 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>
* Any median ''m'' must lie within the interval ⌊''np''⌋ ≤ ''m'' ≤ ⌈''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>
* A median ''m'' cannot lie too far away from the mean: {{nowrap||''m'' − ''np''| ≤ min{ ln 2, max{''p'', 1 − ''p''} }}}.<ref name="Hamza">{{Cite journal
| last1 = Hamza | first1 = K.
| doi = 10.1016/0167-7152(94)00090-U
| title = The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions
F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right)
F (k; n，p) leq exp left (- nD left (frac { k }{ n } parallel p right)))
| journal = Statistics & Probability Letters
| volume = 23
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):
其中 d (a | | p)是一枚硬币和一枚 p 相对熵之间的距离。在伯努利(a)和伯努利(p)分布之间:
| pages = 21–25
| year = 1995
D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!
D (a 并行 p) = (a) log frac { a }{ p } + (1-a) log frac {1-a }{1-p }。\!
| pmid =
| pmc =
Asymptotically, this bound is reasonably tight; see
从渐近的角度来看，这个界限是相当紧密的; 参见
}}</ref>
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),
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)) ,
* The median is unique and equal to ''m'' = [[Rounding|round]](''np'') when |''m'' − ''np''| ≤ min{''p'', 1 − ''p''} (except for the case when ''p'' = {{sfrac|1|2}} and ''n'' is odd).<ref name="KaasBuhrman"/>
which implies the simpler but looser bound
这意味着更简单但更松散的约束
* When ''p'' = 1/2 and ''n'' is odd, any number ''m'' in the interval {{sfrac|1|2}}(''n'' − 1) ≤ ''m'' ≤ {{sfrac|1|2}}(''n'' + 1) is a median of the binomial distribution. If ''p'' = 1/2 and ''n'' is even, then ''m'' = ''n''/2 is the unique median.
F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).
F (k; n，p) geq frac1{ sqrt {2n } exp left (- nD left (frac { k }{ n } parallel p right))).
For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:
对于 p = 1/2和 k ≥3n/8的偶数 n，可以使分母为常数:
===Tail bounds===
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''.
F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!
F (k; n，tfrac {1}{2}) geq frac {1}{15} exp left (- 16n left (frac {1}{2}-frac { k }{ n }右) ^ 2 right)。\!
[[Hoeffding's inequality]] yields the simple bound
:<math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math>
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'' < ''n''), but Hoeffding's bound evaluates to a positive constant.
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.
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 Lehmann-scheffé 定理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。
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>
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.
利用 Β分布作为共轭先验分布时，p 的 Bayes 估计也是一个封闭形式。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) ，可容许和一致的概率。
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math>
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.
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)。这种方法被称为继承法则，它是在18世纪由皮埃尔-西蒙·拉普拉斯引进的。
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):
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})
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 })
:<math> D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \!</math>
Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.
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.
即使对于非常大的 n 值，均值的实际分布也是非常非正态的。针对这一问题，提出了几种估计置信区间的方法。
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>
:<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>
In the equations for confidence intervals below, the variables have the following meaning:
在下面的置信区间等式中，这些变量具有以下含义:
which implies the simpler but looser bound
:<math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math>
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>
:<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>
\widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .
1-widehat { p，})}{ n }.
== Statistical Inference ==
A continuity correction of 0.5/n may be added.
可以加上0.5/n 的连续性修正。
=== Estimation of parameters ===
{{seealso|Beta distribution#Bayesian inference}}
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]].
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .
1-tilde { p })}{ n + z ^ 2}.
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.
Here the estimate of p is modified to
这里 p 的估计被修改为
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]].
\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }
1 + frac {1}{2} z2}{ n + z2}
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>)
=== Confidence intervals ===
{{Main|Binomial proportion confidence interval}}
\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).
Sin ^ 2 left (arcsin left (sqrt { widehat { p，} right) pm frac { z }{2 sqrt { n } right).
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.
In the equations for confidence intervals below, the variables have the following meaning:
* ''n''<sub>1</sub> is the number of successes out of ''n'', the total number of trials
* <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes
The notation in the formula below differs from the previous formulas in two respects:
下列公式中的符号在两个方面不同于以前的公式:
* <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> = 0.05, so <math>1 - \tfrac{1}{2}\alpha</math> = 0.975 and <math>z</math> = 1.96.
==== Wald method ====
:: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>
<math>\frac{
我不知道
\widehat{p\,} + \frac{z^2}{2n} + z
2n } + z
: A [[continuity correction]] of 0.5/''n'' may be added.{{clarify|date=July 2012}}
\sqrt{
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
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +
1-widehat { p，})}{ n } +
==== Agresti–Coull method ====
\frac{z^2}{4 n^2}
4 n ^ 2}
<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>
}
}
}{
}{
:: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .</math>
1 + \frac{z^2}{n}
1 + frac { z ^ 2}{ n }
}</math>
{/math >
: Here the estimate of ''p'' is modified to
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>
The exact (Clopper–Pearson) method is the most conservative.
确切的(克洛佩尔-皮尔森)方法是最保守的。
==== Arcsine method ====
Let X ~ B(n,p1) and Y ~ B(m,p2) be independent. Let T = (X/n)/(Y/m).
设 x ~ b (n，p1)和 y ~ b (m，p2)是独立的。设 t = (X/n)/(Y/m)。
<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>
Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m.
然后对数(t)近似呈正态分布，其中平均对数(p1/p2)和方差((1/p1)-1)/n + ((1/p2)-1)/m。
: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>
==== Wilson (score) method ====
If X ~ B(n, p) and Y | X ~ B(X, q) (the conditional distribution of Y, given X), then Y is a simple binomial random variable with distribution Y ~ B(n, pq).
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。
{{Main|Binomial proportion confidence interval#Wilson score interval}}
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 ~ B(n, 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 ~ B(X, q) and therefore Y ~ B(n, pq).
例如，想象一下把 n 个球扔到一个篮子里，然后把击中的球扔到另一个篮子里。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 yy 的概率，那么击中 yy 的球数是 y ~ b (x，q) ，因此 y ~ b (n，pq)。
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>
* 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 − ''x'')-th quantile'.
* 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> = 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>.
Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,
由于 x sim b (n，p)和 y sim b (x，q) ，由全概率公式,
<math>\begin{align}
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
:: <math>\frac{
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
\widehat{p\,} + \frac{z^2}{2n} + z
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
和 = sum { k = m } n binom { n }{ k }{ m } p ^ k ^ m (1-p) ^ { n-k }(1-q) ^ { k-m }
\sqrt{
\end{align}</math>
结束{ align } </math >
\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +
Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as
由于 tbinom { n }{ k } tbinom { k }{ m } = tbinom { n }{ m } tbinom { n-m }{ k-m } ，上述方程可表示为
\frac{z^2}{4 n^2}
\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}
Pr [ y = m ] = sum _ { k = m } ^ { n } binom { n-m }{ k-m } p ^ k ^ m (1-p) ^ { n-k }(1-q) ^ { k-m }
}
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
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
}{
<math>\begin{align}
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
1 + \frac{z^2}{n}
\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]
[ 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 ]
}</math><ref>{{cite book
&= \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)
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)
| chapter = Confidence intervals
\end{align}</math>
结束{ align } </math >
| chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
After substituting i = k - m in the expression above, we get
在上面的表达式中用 i = k-m 代替后，我们得到了
| title = Engineering Statistics Handbook
\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)
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)
| publisher = NIST/Sematech
Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields
注意，上面的和(在括号中)等于(p-pq + 1-p) ^ { n-m }二项式定理。用最终产量来代替这一点
| year = 2012
<math>\begin{align}
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
| access-date = 2017-07-23
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
}}</ref>
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}
& = binom { n }{ m }(pq) ^ m (1-pq) ^ { n-m }
\end{align}</math>
结束{ align } </math >
==== Comparison ====
and thus Y \sim B(n, pq) as desired.
因此 y sim b (n，pq)为所需。
The exact ([[Binomial proportion confidence interval#Clopper–Pearson interval|Clopper–Pearson]]) method is the most conservative.<ref name="Brown2001" />
The Wald method, although commonly recommended in textbooks, is the most biased.{{clarify|reason=what sense of bias is this|date=July 2012}}
The Bernoulli distribution is a special case of the binomial distribution, where n = 1. Symbolically, X ~ B(1, p) has the same meaning as X ~ Bernoulli(p). Conversely, any binomial distribution, B(n, p), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.
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。
==Related distributions==
===Sums of binomials===
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).
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。
If ''X'' ~ B(''n'', ''p'') and ''Y'' ~ B(''m'', ''p'') are independent binomial variables with the same probability ''p'', then ''X'' + ''Y'' is again a binomial variable; its distribution is ''Z=X+Y'' ~ B(''n+m'', ''p''):
:<math>\begin{align}
\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]\\
Binomial [[probability mass function and normal probability density function approximation for n = 6 and p = 0.5]]
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]
&= \binom{n+m}k p^k (1-p)^{n+m-k}
\end{align}</math>
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n, p) is given by the normal distribution
如果 n 足够大，那么分布的倾斜就不会太大。在这种情况下，通过正态分布给出 b (n，p)的合理近似
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>
\mathcal{N}(np,\,np(1-p)),
数学{ n }(np，np (1-p))
===Ratio of two binomial distributions===
and this basic approximation can be improved in a simple way by using a suitable continuity correction.
通过适当的连续性修正，可以简单地改进这种基本近似。
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:
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:
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>
Let ''X'' ~ B(''n'',''p''<sub>1</sub>) and ''Y'' ~ B(''m'',''p''<sub>2</sub>) be independent. Let ''T'' = (''X''/''n'')/(''Y''/''m'').
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}}
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 }
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>) − 1)/''n'' + ((1/''p''<sub>2</sub>) − 1)/''m''.
===Conditional binomials===
If ''X'' ~ B(''n'', ''p'') and ''Y'' | ''X'' ~ B(''X'', ''q'') (the conditional distribution of ''Y'', given ''X''), then ''Y'' is a simple binomial random variable with distribution ''Y'' ~ B(''n'', ''pq'').
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 ≥ 20 and p ≤ 0.05, or if n ≥ 100 and np ≤ 10.
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 的泊松分佈可以作为二项分布 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。
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'' ~ B(''n'', ''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'' ~ B(''X'', ''q'') and therefore ''Y'' ~ B(''n'', ''pq'').
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.
关于泊松近似的准确性，参见 Novak，ch。4，及其中的参考资料。
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],
:<math>\begin{align}
\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}.
P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}.
&= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.
给定一个统一的先验，给定具有观察到的成功事件的独立事件的成功概率的后验概率是一个 Β分布。
\end{align}</math>
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as
:<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>
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
Methods for random number generation where the marginal distribution is a binomial distribution are well-established.
随机数产生的方法，其中的边缘分布是一个二项分布。
:<math>\begin{align}
\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]
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.
从二项分布中产生随机样本的一种方法是使用反演算法。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0到1之间的样本，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。
&= \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)
\end{align}</math>
After substituting <math> i = k - m </math> in the expression above, we get
:<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>
This distribution was derived by Jacob Bernoulli. He considered the case where p = r/(r + s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p = 1/2.
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。
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
:<math>\begin{align}
\Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
&= \binom{n}{m} (pq)^m (1-pq)^{n-m}
\end{align}</math>
and thus <math> Y \sim B(n, pq) </math> as desired.
{{hidden end}}
===Bernoulli distribution===
The [[Bernoulli distribution]] is a special case of the binomial distribution, where ''n'' = 1. Symbolically, ''X'' ~ B(1, ''p'') has the same meaning as ''X'' ~ Bernoulli(''p''). Conversely, any binomial distribution, B(''n'', ''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>
===Poisson binomial distribution===
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>
{{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>
Category:Discrete distributions
类别: 离散分布
===Normal approximation===
Category:Factorial and binomial topics
类别: 阶乘和二项式主题
Category:Conjugate prior distributions
范畴: 共轭先验分布
[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n'' = 6 and ''p'' = 0.5]]
Category:Exponential family distributions
类别: 指数族分布
<noinclude>
<small>This page was moved from [[wikipedia:en:Binomial distribution]]. Its edit history can be viewed at [[二项分布/edithistory]]</small></noinclude>
[[Category:待整理页面]]