Binomial distribution

File:Pascal's triangle; binomial distribution.svg

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.

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 $n\in \mathbb{\mathbb{N}}$ (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: $$f(k,n,p)=\mathrm{Pr}(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$$ for Template:Math, where $$(\genfrac{}{}{0ex}{}{n}{k})=\frac{n!}{k!(n-k)!}$$ 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 $(\genfrac{}{}{0ex}{}{n}{k})$ such sequences, since the binomial coefficient $(\genfrac{}{}{0ex}{}{n}{k})$ 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 $(\genfrac{}{}{0ex}{}{n}{k})$ times, hence $\mathrm{Pr}(X=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$.

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

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 $$\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)}$$ and comparing it to 1. There is always an integer Template:Mvar that satisfies $$(n+1)p-1\le M<(n+1)p.$$

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 $$f(4,6,0.3)=(\genfrac{}{}{0ex}{}{6}{4})0.3^4(1-0.3{)}^{6-4}=0.059535.$$

Cumulative distribution function

The cumulative distribution function can be expressed as: $$F(k;n,p)=\mathrm{Pr}(X\le k)=\sum _{i=0}^{\lfloor k\rfloor }(\genfrac{}{}{0ex}{}{n}{i})p^i(1-p{)}^{n-i},$$ where $\lfloor k\rfloor$ 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: $$\begin{array}{rl}F(k;n,p)& =\mathrm{Pr}(X\le k)\\ & =I_{1-p}(n-k,k+1)\\ & =(n-k)(\genfrac{}{}{0ex}{}{n}{k})\underset{0}{\overset{1-p}{\int }}{t}^{n-k-1}(1-t{)}^{k}dt,\end{array}$$ which is equivalent to the cumulative distribution functions of the beta distribution and of the [[F-distribution|Template:Mvar-distribution]]: $$F(k;n,p)=F_{\text{beta-distribution}}(x=1-p;\alpha =n-k,\beta =k+1)$$ $$F(k;n,p)=F_{F\text{-distribution}}(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)).$$

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: $$\mathrm{E}[X]=np.$$

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 $X_1,\dots ,X_n$ are identical (and independent) Bernoulli random variables with parameter Template:Mvar, then Template:Math and $$\mathrm{E}[X]=\mathrm{E}[X_1+\cdots +X_n]=\mathrm{E}[X_1]+\cdots +\mathrm{E}[X_n]=p+\cdots +p=np.$$

The variance is: $$\mathrm{Var}(X)=npq=np(1-p).$$

This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.

Higher moments

The first 6 central moments, defined as ${\mu }_c=\mathrm{E}[(X-\mathrm{E}[X]{)}^c]$, are given by $$\begin{array}{rl}{\mu }_1& =0,\\ {\mu }_2& =np(1-p),\\ {\mu }_3& =np(1-p)(1-2p),\\ {\mu }_4& =np(1-p)[1+(3n-6)p(1-p)],\\ {\mu }_5& =np(1-p)(1-2p)[1+(10n-12)p(1-p)],\\ {\mu }_6& =np(1-p)[1-30p(1-p)[1-4p(1-p)]+5np(1-p)[5-26p(1-p)]+15n^2{p}^2{(1-p)}^2].\end{array}$$

The non-central moments satisfy $$\begin{array}{rl}\mathrm{E}[X]& =np,\\ \mathrm{E}[X^2]& =np(1-p)+n^2{p}^2,\end{array}$$ and in general $$\mathrm{E}[X^c]=\sum _{k=0}^c\left\{\genfrac{}{}{0ex}{}{c}{k}\right\}{n}^{\underset{\_}{k}}{p}^k,$$ where $\left\{\genfrac{}{}{0ex}{}{c}{k}\right\}$ are the Stirling numbers of the second kind, and $n^{\underset{\_}{k}}=n(n-1)\cdots (n-k+1)$ is the $k$-th falling power of $n$. A simple bound follows by bounding the Binomial moments via the higher Poisson moments: $$\mathrm{E}[X^c]\le {\left[\frac{c}{\ln (1+\frac{c}{np})}\right]}^c\le (np{)}^c\exp \left(\frac{c^2}{2np}\right).$$ This shows that if $c=O(\sqrt{np})$, then $\mathrm{E}[X^c]$ is at most a constant factor away from $\mathrm{E}[X{]}^c$.

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

Mode

Usually the mode of a binomial Template:Math distribution is equal to $\lfloor (n+1)p\rfloor$, where $\lfloor \cdot \rfloor$ 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: $$\text{mode}=\{\begin{array}{ll}\lfloor (n+1)p\rfloor & \text{if }(n+1)p\text{ is 0 or a noninteger},\\ (n+1)p\text{ and }(n+1)p-1& \text{if }(n+1)p\in \{1,\dots ,n\},\\ n& \text{if }(n+1)p=n+1.\end{array}$$

Proof: Let $$f(k)=(\genfrac{}{}{0ex}{}{n}{k})p^k{q}^{n-k}.$$

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 $k\ne n$. This proves that the mode is 0 for $p=0$ and $n$ for $p=1$.

Let $0<p<1$. We find $$\frac{f(k+1)}{f(k)}=\frac{(n-k)p}{(k+1)(1-p)}.$$

From this follows $$\begin{array}{r}k>(n+1)p-1\Rightarrow f(k+1)<f(k)\\ k=(n+1)p-1\Rightarrow f(k+1)=f(k)\\ k<(n+1)p-1\Rightarrow f(k+1)>f(k)\end{array}$$

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 $(n+1)p-1\notin \mathbb{\mathbb{Z}}$, 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:

• If Template:Math is an integer, then the mean, median, and mode coincide and equal Template:Math.
• Any median Template:Mvar must lie within the interval $\lfloor np\rfloor \le m\le \lceil np\rceil$.
• A median Template:Mvar cannot lie too far away from the mean:$|m-np|\le \min \{\ln 2,\max \{p,1-p\}\}$.
• The median is unique and equal to Template:Math when Template:Math (except for the case when Template:Math and Template:Mvar is odd).
• When Template:Mvar is a rational number (with the exception of Template:Math and Template:Mvar odd), the median is unique.
• When $p=\frac{1}{2}$ and Template:Mvar is odd, any number Template:Mvar in the interval $\frac{1}{2}(n-1)\le m\le \frac{1}{2}(n+1)$ is a median of the binomial distribution. If $p=\frac{1}{2}$ and Template:Mvar is even, then $m=\frac{n}{2}$ is the unique median.

Tail bounds

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

Hoeffding's inequality yields the simple bound $$F(k;n,p)\le \exp (-2n{(p-\frac{k}{n})}^2),$$ 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: $$F(k;n,p)\le \exp (-nD(\frac{k}{n}\parallel p))$$ 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): $$D(a\parallel p)=(a)\ln \frac{a}{p}+(1-a)\ln \frac{1-a}{1-p}.$$

Asymptotically, this bound is reasonably tight; see 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 $$F(k;n,p)\ge \frac{1}{\sqrt{8n\frac{k}{n}(1-\frac{k}{n})}}\exp (-nD(\frac{k}{n}\parallel p)),$$ which implies the simpler but looser bound $$F(k;n,p)\ge \frac{1}{\sqrt{2n}}\exp (-nD(\frac{k}{n}\parallel p)).$$

For Template:Math and Template:Math for even Template:Mvar, it is possible to make the denominator constant: $$F(k;n,\frac{1}{2})\ge \frac{1}{15}\exp (-16n{(\frac{1}{2}-\frac{k}{n})}^2).$$

Statistical inference

Estimation of parameters

When Template:Mvar is known, the parameter Template:Mvar can be estimated using the proportion of successes: $$\hat{p}=\frac{x}{n}.$$ 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 (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 $\mathrm{Var}(\hat{p})=\frac{p(1-p)}{n}$, 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 $\mathrm{Beta}(\alpha ,\beta )$ as a prior, the posterior mean estimator is: $${\hat{p}}_b=\frac{x+\alpha }{n+\alpha +\beta }.$$ The Bayes estimator is asymptotically efficient and as the sample size approaches infinity (Template:Math), it approaches the MLE solution. 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, $\mathrm{Beta}(\alpha =1,\beta =1)=U(0,1)$, the posterior mean estimator becomes: $${\hat{p}}_b=\frac{x+1}{n+2}.$$ (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.

When relying on Jeffreys prior, the prior is $\mathrm{Beta}(\alpha =\frac{1}{2},\beta =\frac{1}{2})$, which leads to the estimator: $${\hat{p}}_{\mathrm{J}\mathrm{e}\mathrm{f}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{y}\mathrm{s}}=\frac{x+\frac{1}{2}}{n+1}.$$

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 $\hat{p}=0,$ which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator ${\hat{p}}_b$, leading to: $${\hat{p}}_b=\frac{1}{n+2}.$$ Another method is to use the upper bound of the confidence interval obtained using the rule of three: $${\hat{p}}_{\text{rule of 3}}=\frac{3}{n}.$$

Confidence intervals for the parameter p

Even for quite large values of Template:Mvar, the actual distribution of the mean is significantly nonnormal. 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
• $\hat{p}=\frac{n_1}{n}$ is the proportion of successes
• $z$ is the $1-\frac{1}{2}\alpha$ quantile of a standard normal distribution (that is, probit) corresponding to the target error rate $\alpha$. For example, for a 95% confidence level the error $\alpha =0.05$, so $1-\frac{1}{2}\alpha =0.975$ and $z=1.96$.

Wald method

$$\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.$$

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

Agresti–Coull method

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

Here the estimate of Template:Mvar is modified to $$\tilde{p}=\frac{n_1+\frac{1}{2}{z}^2}{n+z^2}$$

This method works well for Template:Math and Template:Math. See here for $n\le 10$. For Template:Math use the Wilson (score) method below.

Arcsine method

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

Wilson (score) method

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

• 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 $z=z_{\alpha /2}$ to get the lower bound, or use $z=z_{1-\alpha /2}$ to get the upper bound. For example: for a 95% confidence level the error $\alpha =0.05$, so one gets the lower bound by using $z=z_{\alpha /2}=z_{0.025}=-1.96$, and one gets the upper bound by using $z=z_{1-\alpha /2}=z_{0.975}=1.96$.

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

Comparison

The so-called "exact" (Clopper–Pearson) method is the most conservative. (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

Related distributions

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: $$\begin{array}{rl}\mathrm{P}(Z=k)& =\sum _{i=0}^k[(\genfrac{}{}{0ex}{}{n}{i})p^i(1-p{)}^{n-i}][(\genfrac{}{}{0ex}{}{m}{k-i})p^{k-i}(1-p{)}^{m-k+i}]\\ & =(\genfrac{}{}{0ex}{}{n+m}{k})p^k(1-p{)}^{n+m-k}\end{array}$$

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.

Ratio of two binomial distributions

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

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.

[Proof]

Since $X\sim \mathrm{B}(n,p)$ and $Y\sim \mathrm{B}(X,q)$, by the law of total probability, $$\begin{array}{rl}\mathrm{Pr}[Y=m]& =\sum _{k=m}^n\mathrm{Pr}[Y=m\mid X=k]\mathrm{Pr}[X=k]\\ & =\sum _{k=m}^n(\genfrac{}{}{0ex}{}{n}{k})(\genfrac{}{}{0ex}{}{k}{m})p^k{q}^m(1-p{)}^{n-k}(1-q{)}^{k-m}\end{array}$$ Since $(\genfrac{}{}{0ex}{}{n}{k})(\genfrac{}{}{0ex}{}{k}{m})=(\genfrac{}{}{0ex}{}{n}{m})(\genfrac{}{}{0ex}{}{n-m}{k-m}),$ the equation above can be expressed as $$\mathrm{Pr}[Y=m]=\sum _{k=m}^n(\genfrac{}{}{0ex}{}{n}{m})(\genfrac{}{}{0ex}{}{n-m}{k-m})p^k{q}^m(1-p{)}^{n-k}(1-q{)}^{k-m}$$ 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 $$\begin{array}{rl}\mathrm{Pr}[Y=m]& =(\genfrac{}{}{0ex}{}{n}{m})p^m{q}^m(\sum _{k=m}^n(\genfrac{}{}{0ex}{}{n-m}{k-m})p^{k-m}(1-p{)}^{n-k}(1-q{)}^{k-m})\\ & =(\genfrac{}{}{0ex}{}{n}{m})(pq{)}^m(\sum _{k=m}^n(\genfrac{}{}{0ex}{}{n-m}{k-m}){(p(1-q))}^{k-m}(1-p{)}^{n-k})\end{array}$$ After substituting $i=k-m$ in the expression above, we get $$\mathrm{Pr}[Y=m]=(\genfrac{}{}{0ex}{}{n}{m})(pq{)}^m(\sum _{i=0}^{n-m}(\genfrac{}{}{0ex}{}{n-m}{i})(p-pq{)}^i(1-p{)}^{n-m-i})$$ Notice that the sum (in the parentheses) above equals $(p-pq+1-p{)}^{n-m}$ by the binomial theorem. Substituting this in finally yields $$\begin{array}{rl}\mathrm{Pr}[Y=m]& =(\genfrac{}{}{0ex}{}{n}{m})(pq{)}^m(p-pq+1-p{)}^{n-m}\\ & =(\genfrac{}{}{0ex}{}{n}{m})(pq{)}^m(1-pq{)}^{n-m}\end{array}$$ and thus $Y\sim \mathrm{B}(n,pq)$ as desired.

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.

Normal approximation

File:Binomial Distribution.svg

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 $${𝒩}(np,np(1-p)),$$ 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. 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 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 $$\frac{|1-2p|}{\sqrt{np(1-p)}}=\frac{1}{\sqrt{n}}|\sqrt{\frac{1-p}{p}}-\sqrt{\frac{p}{1-p}}|<0.3.$$

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 $$\mu \pm 3\sigma =np\pm 3\sqrt{np(1-p)}\in (0,n).$$

This 3-standard-deviation rule is equivalent to the following conditions, which also imply the first rule above. $$n>9\left(\frac{1-p}{p}\right)\text{and}n>9\left(\frac{p}{1-p}\right).$$

[Proof]

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

• Another commonly used rule is that both values Template:Math and Template:Math must be greater than 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.

[Proof]

Assume that both values $np$ and $n(1-p)$ are greater than 9. Since $0<p<1$, we easily have that $$np\ge 9>9(1-p)\text{and}n(1-p)\ge 9>9p.$$ We only have to divide now by the respective factors $p$ and $1-p$, to deduce the alternative form of the 3-standard-deviation rule: $$n>9\left(\frac{1-p}{p}\right)\text{and}n>9\left(\frac{p}{1-p}\right).$$

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.

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 $$\sigma =\sqrt{\frac{p(1-p)}{n}}$$

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 such that Template:Math, or if Template:Math and Template:Math such that Template:Math, or if Template:Math and Template:Math.

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

Limiting distributions

• Poisson limit theorem: As Template:Mvar approaches Template:Math and Template:Mvar approaches 0 with the product Template:Math held fixed, the Template:Math distribution approaches the Poisson distribution with expected value Template:Math.
• de Moivre–Laplace theorem: As Template:Mvar approaches Template:Math while Template:Mvar remains fixed, the distribution of $$\frac{X-np}{\sqrt{np(1-p)}}$$ approaches the normal distribution with expected value 0 and variance 1. This result is sometimes loosely stated by saying that the distribution of Template:Mvar is asymptotically normal with expected value 0 and variance 1. This result is a specific case of the central limit theorem.

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: $$\mathrm{Beta}(p;\alpha ;\beta )=(n+1)\mathrm{B}(k;n;p)$$

Beta distributions also provide a family of prior probability distributions for binomial distributions in Bayesian inference: $$P(p;\alpha ,\beta )=\frac{p^{\alpha -1}(1-p{)}^{\beta -1}}{\mathrm{Beta}(\alpha ,\beta )}.$$ 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.

Computational methods

Random number generation

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

See also

• Logistic regression
• Multinomial distribution
• Negative binomial distribution
• Beta-binomial distribution
• Binomial measure, an example of a multifractal measure.
• Statistical mechanics
• Piling-up lemma, the resulting probability when XOR-ing independent Boolean variables

External links

• Interactive graphic: Univariate Distribution Relationships
• Binomial distribution formula calculator
• Difference of two binomial variables: X-Y or |X-Y|
• Querying the binomial probability distribution in WolframAlpha
• Confidence (credible) intervals for binomial probability, p: online calculator available at causaScientia.org