# 二项分布

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{{Probability distribution

{{Probability distribution

 | name       = Binomial distribution

 | name       = Binomial distribution


 | type       = mass

 | type       = mass


 | pdf_image  = Probability mass function for the binomial distribution

 | pdf_image  = Probability mass function for the binomial distribution


| 概率质量函数图像 = 二项分布的概率质量函数 Probability mass function for the binomial distribution

 | cdf_image  = Cumulative distribution function for the binomial distribution

 | cdf_image  = Cumulative distribution function for the binomial distribution


| 累积分布函数图像 = 二项分布的累积分布函数 Cumulative distribution function for the binomial distribution

 | notation   = $\displaystyle{ B(n,p) }$

 | notation   = B(n,p)


| 符号 = $\displaystyle{ B(n,p) }$

 | parameters = $\displaystyle{ n \in \{0, 1, 2, \ldots\} }$ – number of trials$\displaystyle{ p \in [0,1] }$ – success probability for each trial$\displaystyle{ q = 1 - p }$

 | parameters = n \in \{0, 1, 2, \ldots\} – number of trialsp \in [0,1] – success probability for each trialq = 1 - p


| 参数 =
$\displaystyle{ n \in \{0, 1, 2, \ldots\} }$ – --- 试验次数；
$\displaystyle{ p \in [0,1] }$ – -- 每个试验的成功概率；
$\displaystyle{ q = 1 - p }$

 | support    = $\displaystyle{ k \in \{0, 1, \ldots, n\} }$ – number of successes

 | support    = k \in \{0, 1, \ldots, n\} – number of successes


| 支持 =
$\displaystyle{ k \in \{0, 1, \ldots, n\} }$ – --- 成功的数量

 | pdf        = $\displaystyle{ \binom{n}{k} p^k q^{n-k} }$

 | pdf        = \binom{n}{k} p^k q^{n-k}


|概率质量函数 Probability mass function = $\displaystyle{ \binom{n}{k} p^k q^{n-k} }$

 | cdf        = $\displaystyle{ I_{q}(n - k, 1 + k) }$

 | cdf        = I_{q}(n - k, 1 + k)


| 累积分布函数 Cumulative distribution function = $\displaystyle{ I_{q}(n - k, 1 + k) }$

 | mean       = $\displaystyle{ np }$

 | mean       = np


 | median     = $\displaystyle{ \lfloor np \rfloor }$ or $\displaystyle{ \lceil np \rceil }$

 | median     = \lfloor np \rfloor or \lceil np \rceil


 | mode       = $\displaystyle{ \lfloor (n + 1)p \rfloor }$ or $\displaystyle{ \lceil (n + 1)p \rceil - 1 }$

 | mode       = \lfloor (n + 1)p \rfloor or \lceil (n + 1)p \rceil - 1


| 模 mode = $\displaystyle{ \lfloor (n + 1)p \rfloor }$$\displaystyle{ \lceil (n + 1)p \rceil - 1 }$

 | variance   = $\displaystyle{ npq }$

 | variance   = npq


| 方差 variance = $\displaystyle{ npq }$

 | skewness   = $\displaystyle{ \frac{q-p}{\sqrt{npq}} }$

 | skewness   = \frac{q-p}{\sqrt{npq}}


| 偏度 skewness = $\displaystyle{ \frac{q-p}{\sqrt{npq}} }$

 | kurtosis   = $\displaystyle{ \frac{1-6pq}{npq} }$

 | kurtosis   = \frac{1-6pq}{npq}


| 峰度 kurtosis = $\displaystyle{ \frac{1-6pq}{npq} }$

 | entropy    = $\displaystyle{ \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right) }$ in shannons. For nats, use the natural log in the log.

 | entropy    = \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right) in shannons. For nats, use the natural log in the log.


| 熵 entropy = $\displaystyle{ \frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right) }$香农熵 Shannon entropy测量。对于分布式消息队列系统 NATS ，使用日志中的自然日志。

 | mgf        = $\displaystyle{ (q + pe^t)^n }$

 | mgf        = (q + pe^t)^n


| 矩量母函数 Moment Generating Function = $\displaystyle{ (q + pe^t)^n }$

 | char       = $\displaystyle{ (q + pe^{it})^n }$

 | char       = (q + pe^{it})^n


| 特征函数 characteristic function = $\displaystyle{ (q + pe^{it})^n }$

 | pgf        = $\displaystyle{ G(z) = [q + pz]^n }$

 | pgf        = G(z) = [q + pz]^n


| 概率母函数 probability generating function = $\displaystyle{ G(z) = [q + pz]^n }$

 | fisher     = $\displaystyle{ g_n(p) = \frac{n}{pq} }$(for fixed $\displaystyle{ n }$)

 | fisher     =  g_n(p) = \frac{n}{pq} (for fixed n)


| 费雪信息量 fisher information = $\displaystyle{ g_n(p) = \frac{n}{pq} }$
(对于固定的 $\displaystyle{ n }$)

}}

}}

}}

Binomial distribution for $\displaystyle{ p=0.5 }$
with n and k as in Pascal's triangle

The probability that a ball in a Galton box with 8 layers (n = 8) ends up in the central bin (k = 4) is $\displaystyle{ 70/256 }$.

Binomial distribution for p=0.5
with n and k as in [[Pascal's triangle

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.]]

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).

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 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.

## Definitions

### Probability mass function

In general, if the random variable X follows the binomial distribution with parameters n and p ∈ [0,1], we write X ~ B(np). 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:

$\displaystyle{ f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k} }$

f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}

$\displaystyle{ f(k,n,p) = \Pr(k;n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k} }$

for k = 0, 1, 2, ..., n, where

for k = 0, 1, 2, ..., n, where

$\displaystyle{ \binom{n}{k} =\frac{n!}{k!(n-k)!} }$

\binom{n}{k} =\frac{n!}{k!(n-k)!}

$\displaystyle{ \binom{n}{k} =\frac{n!}{k!(n-k)!} }$

is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability 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 $\displaystyle{ \binom{n}{k} }$ 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

$\displaystyle{ f(k,n,p)=f(n-k,n,1-p). }$

f(k,n,p)=f(n-k,n,1-p).

$\displaystyle{ f(k,n,p)=f(n-k,n,1-p). }$.

Looking at the expression f(knp) 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

$\displaystyle{ \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} }$
\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)}


$\displaystyle{ \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} }$

and comparing it to 1. There is always an integer M that satisfies[1]

and comparing it to 1. There is always an integer M that satisfies

$\displaystyle{ (n+1)p-1 \leq M \lt (n+1)p. }$

(n+1)p-1 \leq M < (n+1)p.

$\displaystyle{ (n+1)p-1 \leq M \lt (n+1)p. }$.

f(knp) 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) 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(knp)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

Suppose a 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

$\displaystyle{ f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535. }$

f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.

$\displaystyle{ f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535. }$.

### Cumulative distribution function

The cumulative distribution function can be expressed as:

The cumulative distribution function can be expressed as:

$\displaystyle{ F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i}, }$

F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},

$\displaystyle{ F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i}, }$ ,

where $\displaystyle{ \lfloor k\rfloor }$ 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.

$\displaystyle{ \lfloor k\rfloor }$是k的向下取整 round down，即小于或等于k的最大整数。

It can also be represented in terms of the regularized incomplete beta function, as follows:[3]

It can also be represented in terms of the regularized incomplete beta function, as follows:

\displaystyle{ \begin{align} F(k;n,p) & = \Pr(X \le k) \\ &= I_{1-p}(n-k, k+1) \\ & = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt. \end{align} }

which is equivalent to the cumulative distribution function of the F-distribution:[5]

which is equivalent to the cumulative distribution function of the -distribution:

$\displaystyle{ F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right). }$

F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).

$\displaystyle{ F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right). }$

Some closed-form bounds for the cumulative distribution function are given below.

Some closed-form bounds for the cumulative distribution function are given below.

## 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:[7]

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:

$\displaystyle{ \operatorname{E}[X] = np. }$
\operatorname{E}[X] = np.


$\displaystyle{ \operatorname{E}[X] = np. }$

This follows from the linearity of the expected value along with fact that X is the sum of n identical Bernoulli random variables, each with expected value p. In other words, if $\displaystyle{ X_1, \ldots, X_n }$ are identical (and independent) Bernoulli random variables with parameter p, then $\displaystyle{ X = X_1 + \cdots + X_n }$ 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

$\displaystyle{ \operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np. }$

\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.

$\displaystyle{ \operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np. }$

The variance is:

The variance is:

$\displaystyle{ \operatorname{Var}(X) = np(1 - p). }$
\operatorname{Var}(X) = np(1 - p).


$\displaystyle{ \operatorname{Var}(X) = np(1 - p). }$

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.

### Higher moments

The first 6 central moments are given by

The first 6 central moments are given by

\displaystyle{ \begin{align} \mu_1 &= 0, \\ \mu_2 &= np(1-p),\\ \mu_3 &= np(1-p)(1-2p),\\ \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_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). \end{align} }

### Mode

Usually the mode of a binomial B(n, p) distribution is equal to $\displaystyle{ \lfloor (n+1)p\rfloor }$, where $\displaystyle{ \lfloor\cdot\rfloor }$ 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:

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:

$\displaystyle{ \text{mode} = \begin{cases} \lfloor (n+1)\,p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger}, \\ (n+1)\,p\ \text{ and }\ (n+1)\,p - 1 &\text{if }(n+1)p\in\{1,\dots,n\}, \\ n & \text{if }(n+1)p = n + 1. \end{cases} }$

Proof: Let

Proof: Let

$\displaystyle{ f(k)=\binom nk p^k q^{n-k}. }$

f(k)=\binom nk p^k q^{n-k}.

$\displaystyle{ f(k)=\binom nk p^k q^{n-k}. }$

For $\displaystyle{ p=0 }$ only $\displaystyle{ f(0) }$ has a nonzero value with $\displaystyle{ f(0)=1 }$. For $\displaystyle{ p=1 }$ we find $\displaystyle{ f(n)=1 }$ and $\displaystyle{ f(k)=0 }$ for $\displaystyle{ k\neq n }$. This proves that the mode is 0 for $\displaystyle{ p=0 }$ and $\displaystyle{ n }$ for $\displaystyle{ p=1 }$.

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.

$\displaystyle{ p=0 }$，只有$\displaystyle{ f(0) }$有一个非零值，$\displaystyle{ f(0)=1 }$。当$\displaystyle{ p=1 }$，我们发现当$\displaystyle{ k\neq n }$$\displaystyle{ f(n)=1 }$$\displaystyle{ f(k)=0 }$。这证明了$\displaystyle{ p=0 }$时模为0，$\displaystyle{ p=1 }$时模为$\displaystyle{ n }$

Let $\displaystyle{ 0 \lt p \lt 1 }$. We find

Let 0 < p < 1. We find

$\displaystyle{ 0 \lt p \lt 1 }$。我们发现

$\displaystyle{ \frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)} }$.

\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.

$\displaystyle{ \frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)} }$.

From this follows

From this follows

\displaystyle{ \begin{align} k \gt (n+1)p-1 \Rightarrow f(k+1) \lt f(k) \\ k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\ k \lt (n+1)p-1 \Rightarrow f(k+1) \gt f(k) \end{align} }

So when $\displaystyle{ (n+1)p-1 }$ is an integer, then $\displaystyle{ (n+1)p-1 }$ and $\displaystyle{ (n+1)p }$ is a mode. In the case that $\displaystyle{ (n+1)p-1\notin \Z }$, then only $\displaystyle{ \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor }$ is a mode.[8]

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.

### 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:

• If np is an integer, then the mean, median, and mode coincide and equal np.[10][11]
• 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[12][13]
• Any median m must lie within the interval ⌊np⌋ ≤ m ≤ ⌈np⌉.[14]
• 任何中位数m都必须满足⌊np⌋ ≤ m ≤ ⌈np⌉。[14]
• A median m cannot lie too far away from the mean: |mnp| ≤ min{ ln 2, max{p, 1 − p} }.[15]

• 中位数m不能离均值太远。|mnp| ≤ min{ ln 2, max{p, 1 − p} }[15]
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),

• The median is unique and equal to m = round(np) when |m − np| ≤ min{p, 1 − p} (except for the case when p = 模板:Sfrac and n is odd).[14]
• 中位数是唯一的并且等于m = round(np)，此时|m − np| ≤ min{p, 1 − p}（$\displaystyle{ ''p'' = {{sfrac|1|2}} }$n 是奇数的情况除外）

which implies the simpler but looser bound

• When p = 1/2 and n is odd, any number m in the interval 模板:Sfrac(n − 1) ≤ m ≤ 模板:Sfrac(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).


For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:

$\displaystyle{ F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right); }$

p = 1/2并且n为偶数，k ≥ 3n/8时, 可以使分母为常数。

### Tail bounds

For knp, upper bounds can be derived for the lower tail of the cumulative distribution function $\displaystyle{ F(k;n,p) = \Pr(X \le k) }$, the probability that there are at most k successes. Since $\displaystyle{ \Pr(X \ge k) = F(n-k;n,1-p) }$, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for knp.

F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!


$\displaystyle{ F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \! }$

Hoeffding's inequality yields the simple bound

$\displaystyle{ F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \! }$

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:[16]

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.

$\displaystyle{ F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right) }$

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.

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):

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})

$\displaystyle{ D(a\parallel p)=(a)\log\frac{a}{p}+(1-a)\log\frac{1-a}{1-p}. \! }$

Asymptotically, this bound is reasonably tight; see [16] 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.

One can also obtain lower bounds on the tail $\displaystyle{ F(k;n,p) }$, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that[17]

$\displaystyle{ F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right), }$

In the equations for confidence intervals below, the variables have the following meaning:

which implies the simpler but looser bound

$\displaystyle{ F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right). }$

For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant:[19]

p = 1/2并且n为偶数，k ≥ 3n/8时, 可以使分母为常数

$\displaystyle{ F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \! }$
 \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .


$\displaystyle{ \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } }$

## Statistical Inference

A continuity correction of 0.5/n may be added.


### Estimation of parameters

Beta分布 贝叶斯推断

When n is known, the parameter p can be estimated using the proportion of successes: $\displaystyle{ \widehat{p} = \frac{x}{n}. }$。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.

n已知时，参数p可以用成功的比例来估计：$\displaystyle{ \widehat{p} = \frac{x}{n}. }$。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 } } .


$\displaystyle{ \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }. }$

A closed form Bayes estimator for p also exists when using the Beta distribution as a conjugate prior distribution. When using a general $\displaystyle{ \operatorname{Beta}(\alpha, \beta) }$ as a prior, the posterior mean estimator is: $\displaystyle{ \widehat{p_b} = \frac{x+\alpha}{n+\alpha+\beta} }$. 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.

Here the estimate of p is modified to


For the special case of using the standard uniform distribution as a non-informative prior ($\displaystyle{ \operatorname{Beta}(\alpha=1, \beta=1) = U(0,1) }$), the posterior mean estimator becomes $\displaystyle{ \widehat{p_b} = \frac{x+1}{n+2} }$ (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.

 \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 }


$\displaystyle{ \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } }$

When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to $\displaystyle{ \widehat{p} = 0, }$ which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.[20] One way is to use the Bayes estimator, leading to: $\displaystyle{ \widehat{p_b} = \frac{1}{n+2} }$). Another method is to use the upper bound of the confidence interval obtained using the rule of three: $\displaystyle{ \widehat{p_{\text{rule of 3}}} = \frac{3}{n} }$)

### Confidence intervals

\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).

$\displaystyle{ \sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right). }$

Even for quite large values of n, the actual distribution of the mean is significantly nonnormal.[22] 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:

• n1 is the number of successes out of n, the total number of trials
• n1n中的成功次数，即试验的总次数。
• $\displaystyle{ \widehat{p\,} = \frac{n_1}{n} }$ is the proportion of successes
• $\displaystyle{ \widehat{p\,} = \frac{n_1}{n} }$是成功的比例。

The notation in the formula below differs from the previous formulas in two respects:

• $\displaystyle{ z }$ is the $\displaystyle{ 1 - \tfrac{1}{2}\alpha }$ quantile of a standard normal distribution (i.e., probit) corresponding to the target error rate $\displaystyle{ \alpha }$. For example, for a 95% confidence level the error $\displaystyle{ \alpha }$ = 0.05, so $\displaystyle{ 1 - \tfrac{1}{2}\alpha }$ = 0.975 and $\displaystyle{ z }$ = 1.96.

• $\displaystyle{ z }$标准正态分布 standard normal distribution $\displaystyle{ 1 - \tfrac{1}{2}\alpha }$分位数(即概率)对应的目标错误率 $\displaystyle{ \alpha }$。例如，95%的置信度 confidence level 的错误率为$\displaystyle{ \alpha }$ = 0.05，因此 $\displaystyle{ 1 - \tfrac{1}{2}\alpha }$ = 0.975 并且$\displaystyle{ z }$ = 1.96.

#### Wald method

Wald 法

$\displaystyle{ \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } . }$
$\displaystyle{ \frac{p}{z^2}{2n}\widehat{p\,} + \frac{z^2}{2n} + z }$

A continuity correction of 0.5/n may be added. 模板:Clarify;

$\displaystyle{ \sqrt{\frac{p}{n}\widehat{p\,}(1 - \widehat{p\,}){n} }$

#### Agresti–Coull method

$\displaystyle{ \frac{z^2}{4 n^2} }$

[23] {

$\displaystyle{ \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } } . }$

$\displaystyle{ 1 + \frac{z^2}{n} }$

Here the estimate of p is modified to

$\displaystyle{ \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } }$

The exact (Clopper–Pearson) method is the most conservative.

#### Arcsine method

Let X ~ B(n,p1) and Y ~ B(m,p2) be independent. Let T = (X/n)/(Y/m).

Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m.

$\displaystyle{ \sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right). }$

#### 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).

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).

The notation in the formula below differs from the previous formulas in two respects:[25]

• Firstly, zx has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the xth 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 $\displaystyle{ z = z_{\alpha / 2} }$ to get the lower bound, or use $\displaystyle{ z = z_{1 - \alpha/2} }$ to get the upper bound. For example: for a 95% confidence level the error $\displaystyle{ \alpha }$ = 0.05, so one gets the lower bound by using $\displaystyle{ z = z_{\alpha/2} = z_{0.025} = - 1.96 }$, and one gets the upper bound by using $\displaystyle{ z = z_{1 - \alpha/2} = z_{0.975} = 1.96 }$.
• 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$\displaystyle{ z = z_{/alpha / 2} }$得到下限，或者使用$\displaystyle{ z = z_{1 - \alpha/2} }$得到上限。例如：对于95%的置信度，误差为$\displaystyle{ alpha }$ = 0.05，所以用$\displaystyle{ z = z_{/alpha/2} = z_{0.025} = - 1.96 }$得到下限，用$\displaystyle{ z = z_{1 - \alpha/2} = z_{0.975} = 1.96 }$得到上限。

Since X \sim B(n, p) and Y \sim B(X, q) , by the law of total probability,

\displaystyle{ \begin{align} }

$\displaystyle{ \frac{} \lt math\gt \Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt] }$

$\displaystyle{ \widehat{p\,} + \frac{z^2}{2n} + z }$

$\displaystyle{ &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} }$

$\displaystyle{ \frac{\widehat{p\,}(1 - \widehat{p\,})}{n} }$

Since \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, the equation above can be expressed as

$\displaystyle{ \frac{z^2}{4 n^2} }$

$\displaystyle{ \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} }$

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

$\displaystyle{ p ^ k = p ^ m p ^ { k-m } }$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

}{

$\displaystyle{ 1 + \frac{z^2}{n} }$

$\displaystyle{ \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]} }$[26]

$\displaystyle{ &= \binom{n}{m} (pq)^m (1-pq)^{n-m} }$

#### Comparison

and thus Y \sim B(n, pq) as desired.

The exact (Clopper–Pearson) method is the most conservative.[22]

The Wald method, although commonly recommended in textbooks, is the most biased.模板:Clarify

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。

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.

## 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).

If X ~ B(np) and Y ~ B(mp) 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+mp):

$\displaystyle{ \operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\ }$

$\displaystyle{ &= \binom{n+m}k p^k (1-p)^{n+m-k} }$

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 smaller than the variance of a binomial variable distributed as $\displaystyle{ B(n+m, \bar{p}).\, }$

\mathcal{N}(np,\,np(1-p)),


$\displaystyle{ \mathcal{N}(np,\,np(1-p)) }$

### Ratio of two binomial distributions

and this basic approximation can be improved in a simple way by using a suitable continuity correction.

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:

This result was first derived by Katz and coauthors in 1978.[27]

Let X ~ B(n,p1) and Y ~ B(m,p2) be independent. Let T = (X/n)/(Y/m).

X ~ B(n,p1)和Y ~ B(m,p2)独立，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(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m.

### Conditional binomials

If X ~ B(np) and Y | X ~ B(Xq) (the conditional distribution of Y, given X), then Y is a simple binomial random variable with distribution Y ~ B(npq).

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 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(np) 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(Xq) and therefore Y ~ B(npq).

Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.

Since $\displaystyle{ X \sim B(n, p) }$ and $\displaystyle{ Y \sim B(X, q) }$, by the law of total probability,

$\displaystyle{ \Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt] }$

$\displaystyle{ P(p;\alpha,\beta) =\frac{p^{\alpha-1}(1-p)^{\beta-1}}{\mathrm{B}(\alpha,\beta)}. }$

$\displaystyle{ P (p; alpha，beta) = frac { p ^ { alpha-1}(1-p) ^ { beta-1}{ mathrm { b }(alpha，beta)}}. }$

$\displaystyle{ &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} }$[/itex]

Given a uniform prior, the posterior distribution for the probability of success given independent events with observed successes is a beta distribution.

Since $\displaystyle{ \tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m}, }$ the equation above can be expressed as

$\displaystyle{ \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} }$

Factoring $\displaystyle{ p^k = p^m p^{k-m} }$ and pulling all the terms that don't depend on $\displaystyle{ k }$ out of the sum now yields

$\displaystyle{ p^k = p^m p^{k-m} }$ 进行分解，并将所有不依赖于 $\displaystyle{ k }$ 的项从总和中抽出，即可得到

Methods for random number generation where the marginal distribution is a binomial distribution are well-established.

$\displaystyle{ \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt] }$

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.

$\displaystyle{ &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right) }$

After substituting $\displaystyle{ i = k - m }$ in the expression above, we get

$\displaystyle{ i = k - m }$ 代入上述表达式后，我们得到了

$\displaystyle{ \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) }$

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 $\displaystyle{ (p - pq + 1 - p)^{n-m} }$ by the binomial theorem. Substituting this in finally yields

\displaystyle{ \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} }

and thus $\displaystyle{ Y \sim B(n, pq) }$ as desired.

### 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(np), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.[28]

### 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(pi).[30]

Category:Discrete distributions

### Normal approximation

Category:Factorial and binomial topics

Category:Conjugate prior distributions

Binomial probability mass function and normal probability density function approximation for n = 6 and p = 0.5

Category: Exponential family distributions

This page was moved from wikipedia:en:Binomial distribution. Its edit history can be viewed at 二项分布/edithistory

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26. {{cite book $\displaystyle{ &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right) }$ | chapter = Confidence intervals | chapter-url = http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm After substituting i = k - m in the expression above, we get | 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) | 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 | year = 2012 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 $\displaystyle{ \Pr[Y=m] &= \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt] }$ }}
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