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第150行： − + − − 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). − − 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. − − 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. − + − − − − ===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: − − − − :<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} − − − − for ''k'' = 0, 1, 2, ..., ''n'', where − − for k = 0, 1, 2, ..., n, where − − − − :<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math> − − \binom{n}{k} =\frac{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. − + − 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 − − − − − − :<math>f(k,n,p)=f(n-k,n,1-p). </math> − − 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 − − :<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)} − 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 − − − :<math>(n+1)p-1 \leq M < (n+1)p.</math> − − (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. 第279行：
第201行： − + − − − − 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 − − − − :<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. + − − ===Cumulative distribution function=== − − 累积分布函数 − − − − The [[cumulative distribution function]] can be expressed as: − − The cumulative distribution function can be expressed as: − − − :<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}, − − − 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. − − − − 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: − + − :<math>\begin{align} 第346行：
第230行： − − − − 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: − − − :<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). − − − 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. 第373行：
第241行： − + − + − <font color="#ff8000">期望值 Expected value </font>和<font color="#ff8000">方差 variance </font> + − − − 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''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是: 第396行：
第257行： − − 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 − − \operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np. 第411行：
第266行： − The [[variance]] is: − − The variance is: 第424行：
第276行： − − 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. 第433行：
第281行： − ===Higher moments=== − − <font color="#ff8000">高阶矩 Higher moments </font> − The first 6 central moments are given by + − The first 6 central moments are given by 第459行：
第303行： − + − − − − 模 − − − − 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: 第486行：
第320行： − − − '''Proof:''' Let − − Proof: Let − − − :<math>f(k)=\binom nk p^k q^{n-k}.</math> − − 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. − − − Let <math>0 < p < 1</math>. We find − − Let 0 < p < 1. We find − − − :<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)}. − − − From this follows − − From this follows − − 第546行：
第348行： − − − 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. 第556行：
第353行： − ===Median=== − − 中位数 − 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: − * 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> − − *任何中位数''m''都必须满足⌊''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> − + − − − − − − − F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) − − − − − − 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): − − − − − − D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \! − − | pmid = − − | pmc = − − Asymptotically, this bound is reasonably tight; see − − 第641行：
第401行： − * 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 − − * 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. 第663行：
第417行： − + − − − 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''. 第685行：
第436行： − − 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. − − − 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. + − − 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. 第714行：
第456行： − 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})+ − 当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到<math>\widehat{p} = 0</math>，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到:<math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math>\widehat{p_{\text{rule of 3}}} = \frac{3}{n})</math> 第728行：
第468行： − Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.+ − 渐近地，这个边界是相当严格的；详见<ref name="ag"/>。 − − 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. − − 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> − − In the equations for confidence intervals below, the variables have the following meaning: − − which implies the simpler but looser bound 第755行：
第486行： − − 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> 第767行：
第496行： − + − <font color="#ff8000">统计推断 Statistical Inference</font>+ − A continuity correction of 0.5/n may be added.+ − 可以加上0.5/n 的连续校正。 − − === Estimation of parameters === − − <font color="#ff8000">参数估计 Estimation of parameters</font> − − {{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]]. 第792行：
第511行： − − <math>\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }.</math> − − 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. − − − 第806行：
第518行： − 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]].+ − 对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。 第814行：
第525行： − − 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 === − <font color="#ff8000">置信区间 Confidence intervals </font>+ 第832行：
第540行： − 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: 第849行：
第554行： − − The notation in the formula below differs from the previous formulas in two respects: 第860行：
第563行： − + − − <font color="#ff8000">Wald 法</font> − − :<math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math> − − : A [[continuity correction]] of 0.5/''n'' may be added. {{clarify|date=July 2012}}; 第874行：
第571行： − + − − 阿格里斯蒂-库尔方法 Agresti–Coull method 第887行：
第582行： − : Here the estimate of ''p'' is modified to − − The exact (Clopper–Pearson) method is the most conservative. 第899行：
第591行： − ==== Arcsine method ==== − − <font color="#ff8000">弧线法 Arcsine method </font> − Let X ~ B(n,p1) and Y ~ B(m,p2) be independent. Let T = (X/n)/(Y/m).+ 第909行：
第598行： − − − Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m. 第919行：
第605行： − + − − <font color="#ff8000">威尔逊法 Wilson (score) method </font> 第930行：
第614行： − − 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). 第938行：
第620行： − − * 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, + 第957行：
第634行： − :: <math>\frac{}+ 第965行：
第642行： − − Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as 第973行：
第648行： − − 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 第1,013行：
第686行： − + − − and thus Y \sim B(n, pq) as desired. − 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}} 第1,029行：
第696行： − 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. 第1,039行：
第705行： − + − 二项式之和+ − − 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 个独立的不同的伯努利试验B(pi)和的分布。 第1,052行：
第715行： − <math>\operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\</math> − − Binomial [[probability mass function and normal probability density function approximation for n = 6 and p = 0.5]] 第1,061行：
第721行： − 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 − − − 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)), + − − − ===Ratio of two binomial distributions=== − 两个二项分布的比值+ − − and this basic approximation can be improved in a simple way by using a suitable continuity correction. − − − 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: 第1,095行：
第743行： − − 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}} − 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''.+ − 则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>) - 1)/''n'' + ((1/''p''<sub>2</sub>) - 1)/''m''。 − + − <font color="#ff8000">条件二项式 Conditional binomials </font> − − − 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. − − − 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. 第1,138行：
第771行： − − Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]], 第1,150行：
第781行： − − Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution. − − − Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as − − 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. − − 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. − − After substituting <math> i = k - m </math> in the expression above, we get − − 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. − − 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. − − − − 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> 第1,216行：
第816行： − + − − <font color="#ff8000">泊松二项分布 Poisson binomial distribution </font> − 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> 第1,254行：
第823行： − Category:Discrete distributions − + − − <font color="#ff8000">正态逼近 Normal approximation </font> − − Category:Factorial and binomial topics − − − Category:Conjugate prior distributions − + − Category: Exponential family distributions
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[[File:Pascal's triangle; binomial distribution.svg|thumb|280px|Binomial distribution for <math>p=0.5</math><br />with ''n'' and ''k'' as in [[Pascal's triangle]]<br /><br />The probability that a ball in a [[Bean machine|Galton box]] with 8 layers (''n'' = 8) ends up in the central bin (''k'' = 4) is <math>70/256</math>.]]
[[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>.|链接=Special:FilePath/Pascal's_triangle;_binomial_distribution.svg]]
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.]]
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.]]
文章File:Pascal's triangle; binomial distribution.svg是<math>p=0.5</math><br />与n和k相关的二项分布。一个8层(''n'' = 8)的高尔顿盒子中的一个球最终进入中央箱子(''k'' = 4)的概率是<math>70/256</math>。
文章File:Pascal's triangle; binomial distribution.svg是<math>p=0.5</math><br />与n和k相关的二项分布。一个8层(''n'' = 8)的高尔顿盒子中的一个球最终进入中央箱子(''k'' = 4)的概率是<math>70/256</math>。
在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为 p)或失败/否/错误/0 (概率为 q = 1 − p)。
在概率论和统计学中，参数为n和p的二项分布是''n''个独立实验序列中成功次数的<font color="#ff8000">离散概率分布 discrete probability distribution </font>，每个实验结果是一个 是/否问题，每个实验都有布尔值结果: 成功/是/正确/1 (概率为 p)或失败/否/错误/0 (概率为 q = 1 − p)。
一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n'' = 1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是<font color="#ff8000">统计显著性 statistical significance </font>的<font color="#ff8000">二项检验 binomial test </font>的基础。
一个单一的结果为成功或失败的实验也被称为<font color="#ff8000">伯努利试验 Bernoulli trial</font>或<font color="#ff8000">伯努利实验 Bernoulli experiment </font>，一系列伯努利实验结果被称为<font color="#ff8000">伯努利过程 Bernoulli process </font>; 对于一个单一的实验，即''n'' = 1，这个二项分布是一个<font color="#ff8000">伯努利分布 Bernoulli distribution</font>。二项分布是<font color="#ff8000">统计显著性 statistical significance </font>的<font color="#ff8000">二项检验 binomial test </font>的基础。
二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中有放回地抽取的。如果抽样没有把抽取的个体放回总体中，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。
二项分布经常被用来模拟大小为n的样本中的成功数量，这些样本是从大小为N的种群中有放回地抽取的。如果抽样没有把抽取的个体放回总体中，抽样就不是独立的，所以得到的分布是一个<font color="#ff8000">超几何分布 hypergeometric distribution </font>，而不是二项分布。然而，对于N远大于n的情况，二项分布仍然是一个很好的近似，并且被广泛使用。
==Definitions==
==定义==
概率质量函数
概率质量函数
一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X'' ~ B(''n'', ''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:
一般来说，如果<font color="#ff8000">随机变量 random variable </font>X服从参数''n'' [[∈]] [[natural number|ℕ]]且 ''p'' ∈ [0,1]的二项分布，记作''X'' ~ B(''n'', ''p'')。在n个独立的伯努利试验中获得k次成功的概率由概率质量函数给出:
<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math>
<math>f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math>
对于''k'' = 0, 1, 2, ..., ''n''，其中
对于''k'' = 0, 1, 2, ..., ''n''，其中
<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math>
<math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math>
是<font color="#ff8000">二项式系数 binomial coefficient</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n'' − ''k''次失败发生在概率为(1 − ''p'')<sup>''n'' − ''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。
是<font color="#ff8000">二项式系数 binomial coefficient</font>，因此有了分布的名字。这个公式可以理解为，K次成功发生在概率为''p''<sup>''k''</sup>的情况下，''n'' − ''k''次失败发生在概率为(1 − ''p'')<sup>''n'' − ''k''</sup>的情况下。然而，''k''次成功可以发生在''n''个试验中的任何一个，并且在''n''个试验序列中有<math>\binom{n}{k}</math>种''k''次试验成功的不同分配方法。
在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k'' > ''n''/2，概率可以通过它的补来计算。
在创建二项分布概率的参考表时，通常表中最多填充到''n''/2的值。这是因为对于''k'' > ''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>.
<math>f(k,n,p)=f(n-k,n,1-p). </math>.
把表达式''f''(''k'', ''n'', ''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。
把表达式''f''(''k'', ''n'', ''p'')看作''k''的函数，存在一个''k''值使它达到最大。这个''k'' 值可以通过计算得到。
<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math>
<math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math>
并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref>
并且与1相比较。总有一个整数M满足<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref>
<math>(n+1)p-1 \leq M < (n+1)p.</math>.
<math>(n+1)p-1 \leq M < (n+1)p.</math>.
''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(''n'' + 1)''p'' 和 (''n'' + 1)''p'' −1 两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。
''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(''n'' + 1)''p'' 和 (''n'' + 1)''p'' −1 两个值使''f''达到最大。''M'' 是伯努利试验最有可能的结果(也就是说，发生的可能性最大，尽管仍然存在不发生的情况) ，被称为模。
===Example===
===例子===
假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是
假设抛出一枚<font color="#ff8000">有偏硬币 biased coin </font>时，正面朝上的概率为0.3。在6次抛掷中恰好看到4个正面的概率是
<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.
<math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>.
'''累积分布函数'''
累积分布函数可以表达为:
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> ,
<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> ,
<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。
<math>\lfloor k\rfloor</math>是k的<font color="#ff8000">向下取整 round down</font>，即小于或等于''k''的最大整数。
在<font color="#ff8000">正则化不完全的beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref>
在<font color="#ff8000">正则化不完全的beta函数 regularized incomplete beta function </font>下，它也可以表示如下: <ref>{{cite book |last=Wadsworth |first=G. P. |title=Introduction to Probability and Random Variables |year=1960 |publisher=McGraw-Hill |location=New York |page=[https://archive.org/details/introductiontopr0000wads/page/52 52] |url=https://archive.org/details/introductiontopr0000wads |url-access=registration }}</ref>
<math>\begin{align}
F(k;n,p) & = \Pr(X \le k) \\
F(k;n,p) & = \Pr(X \le k) \\
\end{align}</math>
\end{align}</math>
这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref>
这相当于<font color="#ff8000">F分布 F-distribution</font>的累积分布函数: <ref>{{cite journal |last=Jowett |first=G. H. |year=1963 |title=The Relationship Between the Binomial and F Distributions |journal=Journal of the Royal Statistical Society D |volume=13 |issue=1 |pages=55–57 |doi=10.2307/2986663 |jstor=2986663 }}</ref>
<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>
<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>
下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。
下面给出了累积分布函数的一些<font color="#ff8000">闭式界 closed-form bounds </font>。
== Properties ==
== 属性 ==
===Expected value and variance===
===期望值和方差 ===
如果''X'' ~ ''B''(''n'', ''p'')，即''X''是一个服从二项分布的随机变量，n 是实验的总数，p 是每个实验得到成功结果的概率，那么''X''的期望值是:
这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math>
这是由于期望值的<font color="#ff8000">线性性 linearity</font>，以及{{mvar|X}}是{{mvar|n}}个相同的伯努利随机变量的线性组合，每个变量都有期望值{{mvar|p}}。换句话说，如果<math>X_1, \ldots, X_n</math>是参数{{mvar|p}}的相同的（且独立的）伯努利随机变量，那么<math>X = X_1 + \cdots + X_n</math>
:<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>
:<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>
<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>
<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>
方差是:
方差是:
这也是因为独立随机变量和的方差是方差之和。
这也是因为独立随机变量和的方差是方差之和。
===高阶矩===
前6个中心矩由
前6个中心矩由
\end{align}</math>
\end{align}</math>
===模===
===Mode===
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n'' + 1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n'' + 1)''p''和(''n'' + 1)''p'' − 1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:
通常二项式''B''(''n'', ''p'')分布的模等于<math>\lfloor (n+1)p\rfloor</math>，其中<math>\lfloor\cdot\rfloor</math>是<font color="#ff8000">向下取整函数 floor function </font>。然而，当(''n'' + 1)''p''是整数且''p''不为0或1时，二项分布有两种模: (''n'' + 1)''p''和(''n'' + 1)''p'' − 1。当''p''等于0或1时，对应的模为0或n。这些情况可总结如下:
\end{cases}</math>
\end{cases}</math>
证明: 让
证明: 让
<math>f(k)=\binom nk p^k q^{n-k}.</math>
<math>f(k)=\binom nk p^k q^{n-k}.</math>
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。
当<math>p=0</math>，只有<math>f(0)</math>有一个非零值，<math>f(0)=1</math>。当<math>p=1</math>，我们发现当<math>k\neq n</math>，<math>f(n)=1</math>且<math>f(k)=0</math>。这证明了<math>p=0</math>时模为0，<math>p=1</math>时模为<math>n</math>。
当<math>0 < p < 1</math>。我们发现
当<math>0 < p < 1</math>。我们发现
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.
<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.
由此可见
由此可见
:<math>\begin{align}
:<math>\begin{align}
\end{align}</math>
\end{align}</math>
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref>
所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref>
===中位数===
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:
一般来说，没有单一的公式可以找到一个二项分布的中位数，甚至可能不是唯一的。然而，几个特殊的结果是已经确定的:
* 如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref>
*如果''np''是一个整数，那么它的均值，中位数和模相同且等于''np''。<ref>{{cite journal|last=Neumann|first=P.|year=1966|title=Über den Median der Binomial- and Poissonverteilung|journal=Wissenschaftliche Zeitschrift der Technischen Universität Dresden|volume=19|pages=29–33|language=German}}</ref><ref>Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", [[The Mathematical Gazette]] 94, 331-332.</ref>
* 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
* 任何中位数''m''都必须满足⌊''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>
| 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
| journal = Statistics & Probability Letters
| volume = 23
| pages = 21–25
| year = 1995
}}</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)\right),
* 中位数是唯一的并且等于''m'' = [[Rounding|round]](''np'')，此时|''m'' − ''np''| ≤ min{''p'', 1 − ''p''}（<math>''p'' = {{sfrac|1|2}}</math>和 ''n'' 是奇数的情况除外）
*中位数是唯一的并且等于''m'' = [[Rounding|round]](''np'')，此时|''m'' − ''np''| ≤ min{''p'', 1 − ''p''}（<math>''p'' = {{sfrac|1|2}}</math>和 ''n'' 是奇数的情况除外）
这意味着更简单但更宽松的界限
这意味着更简单但更宽松的界限
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)\right).
当''p'' = 1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。
当''p'' = 1/2并且''n''为偶数，k ≥ 3n/8时, 可以使分母为常数。
===Tail bounds===
===<font color="#ff8000">尾部边界</font>===
<font color="#ff8000">尾部边界 Tail bounds </font>
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。
对于''k''≤''np''，可以得出累积分布函数左尾的上界<math>F(k;n,p)=Pr(X \le k)</math>，即最多存在''k''次成功的概率。由于<math>Pr(X \ge k) = F(n-k;n,1-p) </math>，这些界限也可以看作是''k''≥''np''的累积分布函数右尾的边界。
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k'' < ''n'')，但是Hoeffding的约束评价为一个正的常数。
然而，这并不是很严格。特别是，当''p''=1时，有''F''(''k'';''n''，''p'') = 0(对于固定的''k''，''n''与''k'' < ''n'')，但是Hoeffding的约束评价为一个正的常数。
当 n 已知时，参数 p 可以使用成功的比例来估计:<math> \widehat{p} = \frac{x}{n}</math>。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。
当 n 已知时，参数 p 可以使用成功的比例来估计:<math> \widehat{p} = \frac{x}{n}</math>。可以利用<font color="#ff8000">极大似然估计 maximum likelihood estimator </font>和<font color="#ff8000"> 矩方法 method of moments</font>来求出该估计量。<font color="#ff8000">Lehmann-scheffé 定理</font>证明了该估计量是无偏的一致的且方差最小的，因为该估计量是基于一个极小<font color="#ff8000">充分完备统计量 sufficient and complete statistic</font>(即: x).它在概率和<font color="#ff8000">均方误差 MSE</font>方面也是一致的。
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref>
可以从<font color="#ff8000">切尔诺夫界 Chernoff bound</font>中得到一个更清晰的边界。<ref name="ag">{{cite journal |first1=R. |last1=Arratia |first2=L. |last2=Gordon |title=Tutorial on large deviations for the binomial distribution |journal=Bulletin of Mathematical Biology |volume=51 |issue=1 |year=1989 |pages=125–131 |doi=10.1007/BF02458840 |pmid=2706397 |s2cid=189884382 }}</ref>
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta)</math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。
利用 Beta分布作为<font color="#ff8000">共轭先验分布 conjugate prior distribution </font>时，也存在p的封闭形式的<font color="#ff8000">贝叶斯估计 Bayes estimator </font>。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta)</math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math>
:<math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) </math>
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为<font color="#ff8000">继承法则 the rule of succession </font>，它是18世纪皮埃尔-西蒙·拉普拉斯 Pierre-Simon Laplace提出的。
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：
其中''D''(''a'' || ''p'')是参数为a和p的相对熵，即Bernoulli(a)和Bernoulli(p)概率分布的差值：
当估计用非常罕见的事件和一个小的n (例如，如果x = 0) ，那么使用标准估计会得到<math>\widehat{p} = 0</math>，这有时是不现实的和我们不希望看到的。在这种情况下，有各种可供选择的估计值。一种方法是使用贝叶斯估计，得到:<math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math>\widehat{p_{\text{rule of 3}}} = \frac{3}{n})</math>
渐近地，这个边界是相当严格的；详见<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>。
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。
即使对于非常大的 n 值，均值的实际分布是非正态的。针对这一问题，提出了几种估计置信区间的方法。
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref>
我们还可以得到尾部<math>F(k;n,p) </math>的下界，即<font color="#ff8000">反集中界anti-concentration bounds </font>。通过用<font color="#ff8000">斯特林公式 Stirling's formula</font>对二项式系数进行近似，可以看出：<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]}}</ref>
:<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>
:<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>
在下面的置信区间等式中，这些变量具有以下含义:
在下面的置信区间等式中，这些变量具有以下含义:
这意味着更简单但更松散的约束。
这意味着更简单但更松散的约束。
当''p'' = 1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数
当''p'' = 1/2并且''n''为偶数，''k'' ≥ 3''n''/8时, 可以使分母为常数
== Statistical Inference ==
== 统计推断 ==
可以加上0.5/n 的连续校正。
=== 参数估计 ===
Beta分布 贝叶斯推断
Beta分布 贝叶斯推断
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。
当''n''已知时，参数''p''可以用成功的比例来估计：<math> \widehat{p} = \frac{x}{n}.</math>。这个估计是用极大似然估计法和矩估计方法来计算的。这个估计是无偏的、一致的且有最小的方差，由Lehmann-Scheffé定理证明，因为它是基于最小充分完备统计量（即：''x''）。它的概率和均方误差（MSE）也是一致估计。
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .
\tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } .
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta) </math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。
利用 Beta分布作为共轭先验分布时，也存在p的封闭形式的贝叶斯估计。当使用一个通用<math>\operatorname{Beta}(\alpha, \beta) </math>作为先验时，后验均值估计量为: <math>\widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta}</math>。贝叶斯估计是渐近有效的，当样本容量趋近无穷大(n →∞)时，它趋近极大似然估计（MLE）解。贝叶斯估计是有偏的(偏多少取决于先验) ，可接受的且一致的概率。
Here the estimate of p is modified to
Here the estimate of p is modified to
这里p的估计被修改为
这里p的估计被修改为
对于使用标准均匀分布作为非信息性的先验概率的特殊情况(<math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>)，后验均值估计变为<math>\widehat{p_b} = \frac{x+1}{n+2}</math> (后验模式应只能得出标准估计量)。这种方法被称为继承法则，它是18世纪 Pierre-Simon Laplace提出的。
<math>\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }</math>
<math>\tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }</math>
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)
当估计值''p''时非常罕见，而且很小（例如：如果x=0），那么使用标准估计器会得到<math> \widehat{p} = 0,</math>，这有时是不现实的，也是不可取的。在这种情况下，有几种不同的可替代的估计方法。<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref>一种方法是使用贝叶斯估计，得到: <math> \widehat{p_b} = \frac{1}{n+2}</math>)。另一种方法是利用从3个规则获得的置信区间的上界: <math> \widehat{p_{\text{rule of 3}}} = \frac{3}{n}</math>)
=== 值信区间 ===
{{Main|Binomial proportion confidence interval}}
{{Main|Binomial proportion confidence interval}}
:<math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>
:<math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name="Brown2001">{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。
即使对于相当大的''n''值，平均数的实际分布是显著非正态的，<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref>由于这个问题，人们提出了几种估计置信区间的方法。
在下面的置信区间公式中，变量具有以下含义
在下面的置信区间公式中，变量具有以下含义
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。
*<math> \widehat{p\,} = \frac{n_1}{n}</math>是成功的比例。
下列公式中的符号在两个地方不同于以前的公式:
下列公式中的符号在两个地方不同于以前的公式:
==== Wald method ====
==== Wald 法 ====
<math>\frac{p}{z^2}{2n}\widehat{p\,} + \frac{z^2}{2n} + z</math>
<math>\frac{p}{z^2}{2n}\widehat{p\,} + \frac{z^2}{2n} + z</math>
可以添加一个0.5/''n''连续调整。（2012年7月更新）
可以添加一个0.5/''n''连续调整。（2012年7月更新）
<math> \sqrt{\frac{p}{n}\widehat{p\,}(1 - \widehat{p\,}){n} </math>
<math> \sqrt{\frac{p}{n}\widehat{p\,}(1 - \widehat{p\,}){n} </math>
==== Agresti–Coull method ====
==== 阿格里斯蒂-库尔方法 ====
<math> \frac{z^2}{4 n^2}</math>
<math> \frac{z^2}{4 n^2}</math>
<math>1 + \frac{z^2}{n}</math>
<math>1 + \frac{z^2}{n}</math>
这里''p''的估计量被修改为
这里''p''的估计量被修改为
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>
:: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>
确切的(克洛佩尔-皮尔森)方法是最保守的。
确切的(克洛佩尔-皮尔森)方法是最保守的。
==== 弧线法 ====
设X ~ B(n,p1)和Y ~ B(m,p2)是独立的。设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>
<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>
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为<math>((1/p1) − 1)/n + ((1/p2) − 1)/m</math>。
然后log(T)近似服从正态分布，均值为log(p1/p2)和方差为<math>((1/p1) − 1)/n + ((1/p2) − 1)/m</math>。
==== Wilson (score) method ====
==== 威尔逊法 ====
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).
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).
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X ~ B(n, p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y ~ B(X, q)，那么Y ~ B(n, pq)。
例如，想象一下把 n 个球扔到一个篮子UX里，然后把击中的球扔到另一个篮子UY里。如果 p 是击中 UX 的概率，那么X ~ B(n, p)是击中 UX 的球数。如果 q 是击中 UY 的概率，那么击中 UY的球数是Y ~ B(X, q)，那么Y ~ B(n, pq)。
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref>
下面的公式中的符号与前面的公式有两个不同之处<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref>
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1 − ''x'')-''th''分位数的简写。
首先，''z''<sub>''x''</sub>在下式中的解释略有不同：它的普通含义是标准正态分布''x-th''的分位数，而不是(1 − ''x'')-''th''分位数的简写。
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>alpha</math> = 0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。
*其次，这个公式没有使用加减法来定义两个界限。相反，我们可以使用<math>z = z_{/alpha / 2}</math>得到下限，或者使用<math>z = z_{1 - \alpha/2}</math>得到上限。例如：对于95%的置信度，误差为<math>alpha</math> = 0.05，所以用<math>z = z_{/alpha/2} = z_{0.025} = - 1.96</math>得到下限，用<math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>得到上限。
由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>,
由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>,
<math>\begin{align}</math>
<math>\begin{align}</math>
<math>\frac{}
<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math>
<math>\Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]</math>
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
<math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math>
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}</math>，上述方程可表示为
<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>
<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>
对<math>p ^ k = p ^ m p ^ { k-m }</math>进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果
对<math>p ^ k = p ^ m p ^ { k-m }</math>进行分解，从总和中取出所有不依赖于 k 的项，现在就得到了结果
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math>
<math> &= \binom{n}{m} (pq)^m (1-pq)^{n-m}</math>
==== Comparison ====
==== 比较 ====
因此<math>Y \sim B(n, pq)</math>为所需值。
因此<math>Y \sim B(n, pq)</math>为所需值。
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001">{{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>
最精确的二项式比例置信区间#Clopper–Pearson区间方法是最保守的。<ref name="Brown2001" />
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。
Wald法虽然是教科书上普遍推荐的方法，但却是最偏颇的方法。
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)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。
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)是 n 个伯努利试验和的分布，每个试验的概率 p 相同。
===Sums of binomials===
===二项式之和===
二项分布是泊松二项分布的一个特例，也叫一般二项分布，它是 n 个独立的不同的伯努利试验B(pi)和的分布。
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''):
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''):
二项式n = 6 and p = 0.5的概率质量函数和正态概率密度函数近似
二项式n = 6 and p = 0.5的概率质量函数和正态概率密度函数近似
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n, p)的合理近似
如果 n 足够大，那么分布的偏斜就不会太大。在这种情况下，通过正态分布给出B(n, p)的合理近似
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。
但是，如果''X''和''Y''的概率''p''不一样，那么和的方差将是小于二项式变量的方差的分布为<math>B(n+m, \bar{p}).\,</math>。
<math>\mathcal{N}(np,\,np(1-p))</math>
<math>\mathcal{N}(np,\,np(1-p))</math>
===两个二项式分布的比值===
通过适当的连续性修正，可以简单地改进这种基本近似。
通过适当的连续性修正，可以简单地改进这种基本近似。
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:
基本近似通常随着 n 的增加而改进(至少20) ，当 p 不接近0或1时更好。经验法则可以用来判断 n 是否足够大，p的极值是否远离0或1:
令''X'' ~ B(''n'',''p''<sub>1</sub>)和''Y'' ~ B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。
令''X'' ~ B(''n'',''p''<sub>1</sub>)和''Y'' ~ B(''m'',''p''<sub>2</sub>)独立，''T'' = (''X''/''n'')/(''Y''/''m'')。
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差<math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math>
例如，假设从大群体中随机抽取了 n 个人，然后询问他们是否同意某种说法。同意的人的比例取决于样本。如果 n 组人群被重复随机地取样，其比例将遵循一个近似正态分布，均值等于总体中一致性的真实比例 p，标准差<math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math>
则log(''T'')近似正态分布，均值为log(''p''<sub>1</sub>/''p''<sub>2</sub>)，方差为((1/''p''<sub>1</sub>) - 1)/''n'' + ((1/''p''<sub>2</sub>) - 1)/''m''。
===Conditional binomials===
===条件二项式nditional 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'').
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'')('Y''的条件分布，给定 。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y'' ~ B(''n'', ''pq'')。
如果''X'' ~ B(''n'', ''p'')和''Y'' | 'X'' ~ B(''X'', ''q'')('Y''的条件分布，给定 。 ''X''），则''Y''是一个简单的二项式随机变量，其分布为''Y'' ~ B(''n'', ''pq'')。
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分布。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n ≥ 20和p ≤ 0.05,或者如果n ≥ 100 and np ≤ 10，则这个近似是好的。
当试验数量趋于无穷大，而np 保持不变或者至少 p 趋于零时，二项分布收敛到泊松分布。因此，如果 n 是足够大，p 足够小的话，参数为λ = np的泊松分布可以作为二项分布B(n, p)的近似。根据两个经验法则，如果n ≥ 20和p ≤ 0.05,或者如果n ≥ 100 and np ≤ 10，则这个近似是好的。
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X'' ~ B(''n'', ''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y'' ~ B(''X'', ''q'')，因此''Y'' ~ B(''n'', ''pq'')。
例如，想象将''n''个球扔到一个篮子里''U<sub>X</sub>''，然后把击中的球扔到另一个篮子里''U<sub>Y</sub>''。如果''p''是击中''U<sub>X</sub>''的概率，那么''X'' ~ B(''n'', ''p'')就是击中''U<sub>X</sub>''的球数。如果''q''是击中''U<sub>Y</sub>''的概率，那么击中''U<sub>Y</sub>''的球数是''Y'' ~ B(''X'', ''q'')，因此''Y'' ~ B(''n'', ''pq'')。
关于泊松近似的准确性，参见 Novak，ch.4，及其中的参考资料。
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]}}
{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，
由于<math> X \sim B(n, p) </math>和<math> Y \sim B(X, q) </math>，由全概率公式，
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math></math>
<math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math></math>
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。
给定一个一致性先验，给定观察到成功结果的独立事件成功概率的后验分布是一个beta分布。
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为
由于<math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math>上式可表示为
:<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>
:<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>
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到
将 <math> p^k = p^m p^{k-m} </math> 进行分解，并将所有不依赖于 <math> k </math> 的项从总和中抽出，即可得到
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。
<font color="#ff8000">边缘分布 marginal distribution </font>是二项分布较完善的随机数产生方法。
<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math>
<math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]</math>
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。
一种从二项分布中产生随机样本的方法是使用<font color="#ff8000">反演算法 inversion algorithm </font>。要做到这一点，我们必须计算从到的所有值的概率。(为了包含整个样本空间，这些概率的和应该接近于1。)然后，通过使用伪随机数生成器来生成介于0和1之间的样本，可以使用在第一步计算出的概率将计算出的样本转换成离散数。
:<math>&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math>
:<math>&= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math>
将 <math> i = k - m </math> 代入上述表达式后，我们得到了
将 <math> i = k - m </math> 代入上述表达式后，我们得到了
:<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>
:<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>
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r + s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p = 1/2的情况。
这个分布是由雅各布伯努利 Jacob Bernoulli推导出来的。他考虑了p = r/(r + s)的情形，其中 p 是成功的概率，r 和 s 是正整数。早些时候，布莱斯 · 帕斯卡 Blaise Pascal考虑过p = 1/2的情况。
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到
请注意，上述的和（括号内）等于<math> (p - pq + 1 - p)^{n-m} </math>由<font color="#ff8000">二项式定理 binomial theorem</font>得出。将此代入最终得到
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===Bernoulli distribution===
===Bernoulli distribution===
伯努利分布
伯努利分布
伯努利分布是二项分布的特例，其中''n'' = 1.从符号上看，''X'' ~ B(1, ''p'')与''X'' ~ Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'', ''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref>
伯努利分布是二项分布的特例，其中''n'' = 1.从符号上看，''X'' ~ B(1, ''p'')与''X'' ~ Bernoulli(''p'')具有相同的意义。相反，任何二项分布，B(''n'', ''p'')是''n''个伯努利试验的和的分布，每个概率''p''相同。<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|accessdate=18 December 2017}}</ref>
===Poisson binomial distribution===
===泊松二项分布===
二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref>{{Cite journal | volume = 3 | issue = 2 | pages = 295–312 | last = Wang | first = Y. H. | title = On the number of successes in independent trials | journal = Statistica Sinica | year = 1993 | url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | url-status = dead | archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | archivedate = 2016-03-03}}</ref>
二项分布是泊松二项分布或广义二项分布的特例，它是''n''个独立的不相同的伯努利试验之和的分布。B(''p<sub>i</sub>'') <ref>{{Cite journal | volume = 3 | issue = 2 | pages = 295–312 | last = Wang | first = Y. H. | title = On the number of successes in independent trials | journal = Statistica Sinica | year = 1993 | url = http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | url-status = dead | archiveurl = https://web.archive.org/web/20160303182353/http://www3.stat.sinica.edu.tw/statistica/oldpdf/A3n23.pdf | archivedate = 2016-03-03}}</ref>
类别: 离散分布
类别: 离散分布
===Normal approximation===
===正态逼近===
类别: 阶乘和二项式主题
类别: 阶乘和二项式主题
类别: 共轭先验分布
类别: 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]]
[[File:Binomial Distribution.svg|right|250px|thumb|Binomial [[probability mass function]] and normal [[probability density function]] approximation for ''n'' = 6 and ''p'' = 0.5|链接=Special:FilePath/Binomial_Distribution.svg]]
类别: <font color="#ff8000">指数族分布 Exponential family distributions </font>
类别: <font color="#ff8000">指数族分布 Exponential family distributions </font>