Binomial distribution

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Template:Short description Template:Redirect Template:Probability distribution</math> 

 | kurtosis   = <math>\frac{1-6pq}{npq}</math>
 | entropy    = <math>\frac{1}{2} \log_2 (2\pi enpq) + O \left( \frac{1}{n} \right)</math>
 in shannons. For nats, use the natural log in the log.
 | mgf        = <math>(q + pe^t)^n</math>
 | char       = <math>(q + pe^{it})^n</math>
 | pgf        = <math>G(z) = [q + pz]^n</math>
 | fisher     = <math> g_n(p) = \frac{n}{pq} </math>
(for fixed <math>n</math>)

}} Template:Probability fundamentals

Binomial distribution for Template:Math
with Template:Mvar and Template:Mvar as in Pascal's triangle

The probability that a ball in a Galton box with 8 layers (Template:Math) ends up in the central bin (Template:Math) is Template:Math.

In probability theory and statistics, the binomial distribution with parameters Template:Mvar and Template:Mvar is the discrete probability distribution of the number of successes in a sequence of Template:Mvar independent experiments, each asking a yes–no question, and each with its own Boolean-valued outcome: success (with probability Template:Mvar) or failure (with probability Template:Math). 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, that is, when Template:Math, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the binomial test of statistical significance.<ref>Template:Cite book</ref>

The binomial distribution is frequently used to model the number of successes in a sample of size Template:Mvar drawn with replacement from a population of size Template:Mvar. 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 Template:Mvar much larger than Template:Mvar, the binomial distribution remains a good approximation, and is widely used.

Definitions

Probability mass function

If the random variable Template:Mvar follows the binomial distribution with parameters <math>n \isin \mathbb{N}</math> (a natural number) and Template:Math, we write Template:Math. The probability of getting exactly Template:Mvar successes in Template:Mvar independent Bernoulli trials (with the same rate Template:Mvar) is given by the probability mass function: <math display="block">f(k,n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math> for Template:Math, where <math display="block">\binom{n}{k} =\frac{n!}{k!(n-k)!}</math> is the binomial coefficient. The formula can be understood as follows: Template:Math is the probability of obtaining the sequence of Template:Mvar independent Bernoulli trials in which Template:Mvar trials are "successes" and the remaining Template:Math trials are "failures". Since the trials are independent with probabilities remaining constant between them, any sequence of Template:Mvar trials with Template:Mvar successes (and Template:Math failures) has the same probability of being achieved (regardless of positions of successes within the sequence). There are <math display="inline">\binom{n}{k}</math> such sequences, since the binomial coefficient <math display="inline">\binom{n}{k}</math> counts the number of ways to choose the positions of the Template:Mvar successes among the Template:Mvar trials. The binomial distribution is concerned with the probability of obtaining any of these sequences, meaning the probability of obtaining one of them (Template:Math) must be added <math display="inline">\binom{n}{k}</math> times, hence <math display="inline">\Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k}</math>.

In creating reference tables for binomial distribution probability, usually, the table is filled in up to Template:Math values. This is because for Template:Math, the probability can be calculated by its complement as <math display="block">f(k,n,p)=f(n-k,n,1-p). </math>

Looking at the expression Template:Math as a function of Template:Mvar, there is a Template:Mvar value that maximizes it. This Template:Mvar value can be found by calculating <math display="block"> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math> and comparing it to 1. There is always an integer Template:Mvar that satisfies<ref>Template:Cite book</ref> <math display="block">(n+1)p-1 \leq M < (n+1)p.</math>

Template:Math is monotone increasing for Template:Math and monotone decreasing for Template:Math, with the exception of the case where Template:Math is an integer. In this case, there are two values for which Template:Mvar is maximal: Template:Math and Template:Math. Template:Mvar 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.

Equivalently, Template:Math. Taking the floor function, we obtain Template:Math.Template:NoteTag

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 <math display="block">f(4,6,0.3) = \binom{6}{4} 0.3^4 (1-0.3)^{6-4}= 0.059535.</math>

Cumulative distribution function

The cumulative distribution function can be expressed as: <math display="block">F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math> where <math>\lfloor k\rfloor</math> is the "floor" under Template:Mvar; that is, the greatest integer less than or equal to Template:Mvar.

It can also be represented in terms of the regularized incomplete beta function, as follows:<ref>Template:Cite book</ref> <math display="block">\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}</math> which is equivalent to the cumulative distribution functions of the beta distribution and of the [[F-distribution|Template:Mvar-distribution]]:<ref>Template:Cite journal</ref> <math display="block">F(k;n,p) = F_{\text{beta-distribution}}\left(x=1-p;\alpha=n-k,\beta=k+1\right)</math> <math display="block">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>

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

Properties

Expected value and variance

If Template:Math, that is, Template:Mvar is a binomially distributed random variable, Template:Mvar being the total number of experiments and Template:Mvar the probability of each experiment yielding a successful result, then the expected value of Template:Mvar is:<ref>See Proof Wiki</ref> <math display="block"> \operatorname{E}[X] = np.</math>

This follows from the linearity of the expected value along with the fact that Template:Mvar is the sum of Template:Mvar identical Bernoulli random variables, each with expected value Template:Mvar. In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter Template:Mvar, then Template:Math and <math display="block">\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math>

The variance is: <math display="block"> \operatorname{Var}(X) = npq = np(1 - p).</math>

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, defined as <math> \mu _{c}=\operatorname {E} \left[(X-\operatorname {E} [X])^{c}\right] </math>, are given by <math display="block">\begin{align} \mu_1 &= 0, \\ \mu_2 &= np \left(1-p\right), \\ \mu_3 &= np \left(1-p\right) \left(1-2p\right), \\ \mu_4 &= np \left(1-p\right) \left[1 + \left(3n-6\right) p \left(1-p\right)\right],\\ \mu_5 &= np \left(1-p\right) \left(1-2p\right) \left[1 + \left(10n-12\right) p \left(1-p\right)\right],\\ \mu_6 &= np \left(1-p\right) \left[1 - 30p\left(1-p\right)\left[1-4p(1-p)\right] + 5np \left(1-p\right)\left[5 - 26p\left(1-p\right)\right] + 15n^2 p^2 \left(1-p\right)^2\right]. \end{align}</math>

The non-central moments satisfy <math display="block">\begin{align} \operatorname {E}[X] &= np, \\ \operatorname {E}[X^2] &= np(1-p)+n^2p^2, \end{align}</math> and in general<ref name="Andreas2008"> Template:Citation</ref><ref name="Nguyen2021"> Template:Citation</ref> <math display="block"> \operatorname {E}[X^c] = \sum_{k=0}^c \left\{ {c \atop k} \right\} n^{\underline{k}} p^k, </math> where <math display="inline"> \left\{{c\atop k}\right\}</math> are the Stirling numbers of the second kind, and <math>n^{\underline{k}} = n(n-1)\cdots(n-k+1)</math> is the <math>k</math>-th falling power of <math>n</math>. A simple bound <ref>Template:Citation</ref> follows by bounding the Binomial moments via the higher Poisson moments: <math display="block"> \operatorname {E}[X^c] \le \left[\frac{c}{\ln\left(1+\frac{c}{np}\right)}\right]^c \le (np)^c \exp\left(\frac{c^2}{2np}\right). </math> This shows that if <math>c=O(\sqrt{np})</math>, then <math>\operatorname {E}[X^c]</math> is at most a constant factor away from <math>\operatorname {E}[X]^c</math>.

The moment-generating function is <math>M_X(t)=\mathbb E[e^{tX}] = (1-p+p e^t)^n</math>.

Mode

Usually the mode of a binomial Template:Math distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the floor function. However, when Template:Math is an integer and Template:Mvar is neither 0 nor 1, then the distribution has two modes: Template:Math and Template:Math. When Template:Mvar is equal to 0 or 1, the mode will be 0 and Template:Mvar correspondingly. These cases can be summarized as follows: <math display="block">\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}</math>

Proof: Let <math display="block">f(k)=\binom nk p^k q^{n-k}.</math>

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>.

Let <math>0 < p < 1</math>. We find <math display="block">\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}.</math>

From this follows <math display="block">\begin{align} k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\ k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\ k < (n+1)p-1 \Rightarrow f(k+1) > f(k) \end{align}</math>

So when <math>(n+1)p-1</math> is an integer, then <math>(n+1)p-1</math> and <math>(n+1)p</math> is a mode. In the case that <math>(n+1)p-1\notin \Z</math>, then only <math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math> is a mode.<ref>See also Template:Cite web</ref>

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:

Template:Cite journal</ref>

Template:Cite journal</ref>

  • When <math display="inline">p = \tfrac{1}{2} </math> and Template:Mvar is odd, any number Template:Mvar in the interval <math display="inline"> \frac{1}{2} \left(n-1\right) \leq m \leq \frac{1}{2} \left(n+1\right)</math> is a median of the binomial distribution. If <math display="inline">p = \tfrac{1}{2} </math> and Template:Mvar is even, then <math display="inline">m = \tfrac{n}{2} </math> is the unique median.

Tail bounds

For Template:Math, 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 Template:Mvar 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 Template:Math.

Hoeffding's inequality yields the simple bound <math display="block"> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math> which is however not very tight. In particular, for Template:Math, we have that Template:Math (for fixed Template:Mvar, Template:Mvar with Template:Math), but Hoeffding's bound evaluates to a positive constant.

A sharper bound can be obtained from the Chernoff bound:<ref name="ag">Template:Cite journal</ref> <math display="block"> F(k;n,p) \leq \exp\left(-n D{\left(\frac{k}{n}\parallel p\right)}\right) </math> where Template:Math is the relative entropy (or Kullback-Leibler divergence) between an Template:Mvar-coin and a Template:Mvar-coin (that is, between the Template:Math and Template:Math distribution): <math display="block"> D(a\parallel p)=(a)\ln\frac{a}{p}+(1-a)\ln\frac{1-a}{1-p}. \!</math>

Asymptotically, this bound is reasonably tight; see <ref name="ag"/> for details.

One can also obtain lower bounds on the tail Template:Math, known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that<ref>Template:Cite book</ref> <math display="block"> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-n D{\left(\frac{k}{n}\parallel p\right)}\right),</math> which implies the simpler but looser bound <math display="block"> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math>

For Template:Math and Template:Math for even Template:Mvar, it is possible to make the denominator constant:<ref>Template:Cite web</ref> <math display="block"> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math>

Statistical inference

Estimation of parameters

Template:See also

When Template:Mvar is known, the parameter Template:Mvar can be estimated using the proportion of successes: <math display="block"> \widehat{p} = \frac{x}{n}.</math> 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 (that is, Template:Mvar). It is also consistent both in probability and in MSE. This statistic is asymptotically normal thanks to the central limit theorem, because it is the same as taking the mean over Bernoulli samples. It has a variance of <math> \operatorname{Var}(\hat{p}) = \frac{p(1-p)}{n}</math>, a property which is used in various ways, such as in Wald's confidence intervals.

A closed form Bayes estimator for Template:Mvar also exists when using the Beta distribution as a conjugate prior distribution. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the posterior mean estimator is: <math display="block"> \widehat{p}_b = \frac{x+\alpha}{n+\alpha+\beta}.</math> The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (Template:Math), it approaches the MLE solution.<ref>Template:Cite journal</ref> The Bayes estimator is biased (how much depends on the priors), admissible and consistent in probability. Using the Bayesian estimator with the Beta distribution can be used with Thompson sampling.

For the special case of using the 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 display="block"> \widehat{p}_b = \frac{x+1}{n+2}.</math> (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.

When relying on Jeffreys prior, the prior is <math display="inline">\operatorname{Beta}(\alpha{=}\tfrac{1}{2}, \, \beta{=}\tfrac{1}{2})</math>,<ref>Marko Lalovic (https://stats.stackexchange.com/users/105848/marko-lalovic), Jeffreys prior for binomial likelihood, URL (version: 2019-03-04): https://stats.stackexchange.com/q/275608</ref> which leads to the estimator: <math display="block"> \widehat{p}_{\mathrm{Jeffreys}} = \frac{x+\frac{1}{2}}{n+1}.</math>

When estimating Template:Mvar with very rare events and a small Template:Mvar (for example, if Template:Math), 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>Template:Cite journal</ref> One way is to use the Bayes estimator <math> \widehat{p}_b</math>, leading to: <math display="block"> \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: <math display="block"> \widehat{p}_{\text{rule of 3}} = \frac{3}{n}.</math>

Confidence intervals for the parameter p

Template:Main Template:See also

Even for quite large values of Template:Mvar, the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>Template:Citation</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:

  • Template:Math is the number of successes out of Template:Mvar, the total number of trials
  • <math> \widehat{p\,} = \frac{n_1}{n}</math> is the proportion of successes
  • <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> quantile of a standard normal distribution (that is, probit) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% confidence level the error <math>\alpha=0.05</math>, so <math>1 - \tfrac{1}{2}\alpha=0.975</math> and <math>z=1.96</math>.

Wald method

Template:Main <math display="block"> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>

A continuity correction of Template:Math may be added.Template:Clarify

Agresti–Coull method

Template:Main <ref name=Agresti1988>Template:Citation</ref> <math display="block"> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }</math>

Here the estimate of Template:Mvar is modified to <math display="block"> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>

This method works well for Template:Math and Template:Math.<ref>Template:Cite web</ref> See here for <math>n\leq 10</math>.<ref>Template:Cite web</ref> For Template:Math use the Wilson (score) method below.

Arcsine method

Template:Main <ref name="Pires00">Template:Cite book</ref> <math display="block">\sin^2 \left(\arcsin \left(\sqrt{\hat{p}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>

Wilson (score) method

Template:Main

The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">Template:Citation</ref>

  • Firstly, Template:Math has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the Template:Mvarth quantile of the standard normal distribution', rather than being a shorthand for 'the Template:Mathth 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=0.05</math>, 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>.

<math display="block">\frac{ 

   \hat{p} + \frac{z^2}{2n} + z
   \sqrt{
       \frac{\hat{p} \left(1 - \hat{p}\right)}{n} +
       \frac{z^2}{4 n^2}
   }

}{ 

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

}</math><ref> Template:Cite book</ref>

Comparison

The so-called "exact" (Clopper–Pearson) method is the most conservative.<ref name="Brown2001" /> (Exact does not mean perfectly accurate; rather, it indicates that the estimates will not be less conservative than the true value.)

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

Sums of binomials

If Template:Math and Template:Math are independent binomial variables with the same probability Template:Mvar, then Template:Math is again a binomial variable; its distribution is Template:Math:<ref>Template:Cite book</ref> <math display="block">\begin{align} 

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

\end{align}</math>

A Binomial distributed random variable Template:Math can be considered as the sum of Template:Mvar Bernoulli distributed random variables. So the sum of two Binomial distributed random variables Template:Math and Template:Math is equivalent to the sum of Template:Math Bernoulli distributed random variables, which means Template:Math. This can also be proven directly using the addition rule.

However, if Template:Mvar and Template:Mvar do not have the same probability Template:Mvar, then the variance of the sum will be smaller than the variance of a binomial variable distributed as Template:Math.

Poisson binomial distribution

The binomial distribution is a special case of the Poisson binomial distribution, which is the distribution of a sum of Template:Mvar independent non-identical Bernoulli trials Template:Math.<ref> Template:Cite journal </ref>

Ratio of two binomial distributions

This result was first derived by Katz and coauthors in 1978.<ref name="Katz1978">Template:Cite journal</ref>

Let Template:Math and Template:Math be independent. Let Template:Math.

Then Template:Math is approximately normally distributed with mean Template:Math and variance Template:Math.

Conditional binomials

If Template:Math and Template:Math (the conditional distribution of Template:Mvar, given Template:Mvar), then Template:Mvar is a simple binomial random variable with distribution Template:Math.

For example, imagine throwing Template:Mvar balls to a basket Template:Math and taking the balls that hit and throwing them to another basket Template:Math. If Template:Mvar is the probability to hit Template:Math then Template:Math is the number of balls that hit Template:Math. If Template:Mvar is the probability to hit Template:Math then the number of balls that hit Template:Math is Template:Math and therefore Template:Math.

Template:Hidden begin Since <math> X \sim \mathrm{B}(n, p) </math> and <math> Y \sim \mathrm{B}(X, q) </math>, by the law of total probability, <math display="block">\begin{align} 

  \Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
  &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
\end{align}</math>

Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as <math display="block"> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math> Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields <math display="block">\begin{align} 

  \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]
  &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k}  \right)
\end{align}</math>

After substituting <math> i = k - m </math> in the expression above, we get <math display="block"> \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> 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 display="block">\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 \mathrm{B}(n, pq) </math> as desired. Template:Hidden end

Bernoulli distribution

The Bernoulli distribution is a special case of the binomial distribution, where Template:Math. Symbolically, Template:Math has the same meaning as Template:Math. Conversely, any binomial distribution, Template:Math, is the distribution of the sum of Template:Mvar independent Bernoulli trials, Template:Math, each with the same probability Template:Mvar.<ref>Template:Cite web</ref>

Normal approximation

Template:See also

Binomial probability mass function and normal probability density function approximation for Template:Math and Template:Math

If Template:Mvar is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to Template:Math is given by the normal distribution <math display="block"> \mathcal{N}(np,\,np(1-p)),</math> and this basic approximation can be improved in a simple way by using a suitable continuity correction. The basic approximation generally improves as Template:Mvar increases (at least 20) and is better when Template:Mvar is not near to 0 or 1.<ref name="bhh">Template:Cite book</ref> Various rules of thumb may be used to decide whether Template:Mvar is large enough, and Template:Mvar is far enough from the extremes of zero or one:

  • One rule<ref name="bhh"/> is that for Template:Math the normal approximation is adequate if the absolute value of the skewness is strictly less than 0.3; that is, if <math display="block">\frac{|1-2p|}{\sqrt{np(1-p)}}=\frac1{\sqrt{n}}\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<0.3.</math>

This can be made precise using the Berry–Esseen theorem.

  • A stronger rule states that the normal approximation is appropriate only if everything within 3 standard deviations of its mean is within the range of possible values; that is, only if <math display="block"> \mu \pm 3\sigma = n p \pm 3 \sqrt{np(1-p)}\in(0,n).</math>
This 3-standard-deviation rule is equivalent to the following conditions, which also imply the first rule above. <math display="block">n>9 \left(\frac{1-p}{p} \right)\quad\text{and}\quad n>9\left(\frac{p}{1-p}\right).</math>

Template:Hidden begin The rule <math> np\pm3\sqrt{np(1-p)}\in(0,n)</math> is totally equivalent to request that <math display="block">np-3\sqrt{np(1-p)}>0\quad\text{and}\quad np+3\sqrt{np(1-p)}<n.</math> Moving terms around yields: <math display="block">np>3\sqrt{np(1-p)}\quad\text{and}\quad n(1-p)>3\sqrt{np(1-p)}.</math> Since <math>0<p<1</math>, we can apply the square power and divide by the respective factors <math>np^2</math> and <math>n(1-p)^2</math>, to obtain the desired conditions: <math display="block">n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math> Notice that these conditions automatically imply that <math>n>9</math>. On the other hand, apply again the square root and divide by 3, <math display="block">\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}>0 \quad \text{and} \quad \frac{\sqrt{n}}3 > \sqrt{\frac{p}{1-p}}>0.</math> Subtracting the second set of inequalities from the first one yields: <math display="block">\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}>-\frac{\sqrt{n}}3;</math> and so, the desired first rule is satisfied, <math display="block">\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<\frac{\sqrt{n}}3.</math> Template:Hidden end

  • Another commonly used rule is that both values Template:Math and Template:Math must be greater than<ref>Template:Cite book</ref><ref>Template:Cite web</ref> or equal to 5. However, the specific number varies from source to source, and depends on how good an approximation one wants. In particular, if one uses 9 instead of 5, the rule implies the results stated in the previous paragraphs.

Template:Hidden begin Assume that both values <math>np</math> and <math>n(1-p)</math> are greater than 9. Since <math>0< p<1</math>, we easily have that <math display="block">np\geq9>9(1-p)\quad\text{and}\quad n(1-p)\geq9>9p.</math> We only have to divide now by the respective factors <math>p</math> and <math>1-p</math>, to deduce the alternative form of the 3-standard-deviation rule: <math display="block">n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math> Template:Hidden end

The following is an example of applying a continuity correction. Suppose one wishes to calculate Template:Math for a binomial random variable Template:Mvar. If Template:Mvar has a distribution given by the normal approximation, then Template:Math is approximated by Template:Math. The addition of 0.5 is the continuity correction; the uncorrected normal approximation gives considerably less accurate results.

This approximation, known as de Moivre–Laplace theorem, is a huge time-saver when undertaking calculations by hand (exact calculations with large Template:Mvar are very onerous); historically, it was the first use of the normal distribution, introduced in Abraham de Moivre's book The Doctrine of Chances in 1738. Nowadays, it can be seen as a consequence of the central limit theorem since Template:Math is a sum of Template:Mvar independent, identically distributed Bernoulli variables with parameter Template:Mvar. This fact is the basis of a hypothesis test, a "proportion z-test", for the value of Template:Mvar using Template:Math, the sample proportion and estimator of Template:Mvar, in a common test statistic.<ref>NIST/SEMATECH, "7.2.4. Does the proportion of defectives meet requirements?" e-Handbook of Statistical Methods.</ref>

For example, suppose one randomly samples Template:Mvar 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 Template:Mvar people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion Template:Mvar of agreement in the population and with standard deviation <math display="block">\sigma = \sqrt{\frac{p(1-p)}{n}}</math>

Poisson approximation

The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product Template:Math converges to a finite limit. Therefore, the Poisson distribution with parameter Template:Math can be used as an approximation to Template:Math of the binomial distribution if Template:Mvar is sufficiently large and Template:Mvar is sufficiently small. According to rules of thumb, this approximation is good if Template:Math and Template:Math<ref>Template:Cite news</ref> such that Template:Math, or if Template:Math and Template:Math such that Template:Math,<ref>Template:Cite book</ref> or if Template:Math and Template:Math.<ref name="nist">NIST/SEMATECH, "6.3.3.1. Counts Control Charts", e-Handbook of Statistical Methods.</ref><ref>Template:Cite web</ref>

Concerning the accuracy of Poisson approximation, see Novak,<ref>Novak S.Y. (2011) Extreme value methods with applications to finance. London: CRC/ Chapman & Hall/Taylor & Francis. Template:ISBN.</ref> ch. 4, and references therein.

Limiting distributions

Beta distribution

The binomial distribution and beta distribution are different views of the same model of repeated Bernoulli trials. The binomial distribution is the PMF of Template:Mvar successes given Template:Mvar independent events each with a probability Template:Mvar of success. Mathematically, when Template:Math and Template:Math, the beta distribution and the binomial distribution are related byTemplate:Clarification needed a factor of Template:Math: <math display="block">\operatorname{Beta}(p;\alpha;\beta) = (n+1)\mathrm{B}(k;n;p)</math>

Beta distributions also provide a family of prior probability distributions for binomial distributions in Bayesian inference:<ref name="MacKay">Template:Cite book</ref> <math display="block">P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\operatorname{Beta}(\alpha,\beta)}.</math> Given a uniform prior, the posterior distribution for the probability of success Template:Mvar given Template:Mvar independent events with Template:Mvar observed successes is a beta distribution.<ref>Template:Cite web</ref>

Computational methods

Random number generation

Template:Further Methods for random number generation where the marginal distribution is a binomial distribution are well-established.<ref>Devroye, Luc (1986) Non-Uniform Random Variate Generation, New York: Springer-Verlag. (See especially Chapter X, Discrete Univariate Distributions)</ref><ref> Template:Cite journal</ref> One way to generate random variates samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that Template:Math for all values Template:Mvar from Template:Math through Template:Mvar. (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.

History

This distribution was derived by Jacob Bernoulli. He considered the case where Template:Math where Template:Mvar is the probability of success and Template:Mvar and Template:Mvar are positive integers. Blaise Pascal had earlier considered the case where Template:Math, tabulating the corresponding binomial coefficients in what is now recognized as Pascal's triangle.<ref name=":1">Template:Cite book</ref>

See also

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Notes

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References

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Further reading

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