Binomial distribution

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 (with probability p) or failure (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, that is, when n = 1, 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 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.

If the random variable X follows the binomial distribution with parameters \(n \isin \mathbb{N}\) (a natural number) and p ∈ [0, 1], we write X ~ B(n, p). The probability of getting exactly k successes in n independent Bernoulli trials (with the same rate p) is given by the probability mass function: \[f(k,n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\] for k = 0, 1, 2, ..., n, where \[\binom{n}{k} =\frac{n!}{k!(n-k)!}\] is the binomial coefficient. The formula can be understood as follows: p q is the probability of obtaining the sequence of n independent Bernoulli trials in which k trials are "successes" and the remaining n − k trials are "failures". Since the trials are independent with probabilities remaining constant between them, any sequence of n trials with k successes (and n − k failures) has the same probability of being achieved (regardless of positions of successes within the sequence). There are \(\binom{n}{k}\) such sequences, since the binomial coefficient \(\binom{n}{k}\) counts the number of ways to choose the positions of the k successes among the n trials. The binomial distribution is concerned with the probability of obtaining any of these sequences, meaning the probability of obtaining one of them (p q) must be added \(\binom{n}{k}\) times, hence ⁠\(\textstyle \Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)⁠.

In creating reference tables for binomial distribution probability, usually, the table is filled in up to ⁠\(\textstyle \tfrac{n}{2}\)⁠ values. This is because for ⁠\(\textstyle k > \tfrac{n}{2}\)⁠, the probability can be calculated by its complement as \[f(k,n,p)=f(n-k,n,1-p).\]

Looking at the expression f(k, n, p) as a function of k, there is a k value that maximizes it. This k value can be found by calculating \[\frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)}\] and comparing it to 1. There is always an integer M that satisfies \[(n+1)p-1 \leq M < (n+1)p.\]

f(k, n, p) is monotone increasing for k < M and monotone decreasing for k > M, with the exception of the case where (n + 1)p is an integer. In this case, there are two values for which f is maximal: (n + 1)p and (n + 1)p − 1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.

Suppose a biased coin comes up heads with probability 0.3 when tossed. The probability of seeing exactly 4 heads in 6 tosses is \[f(4,6,0.3) = \binom{6}{4} 0.3^4 (1-0.3)^{6-4}= 0.059535.\]

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

It can also be represented in terms of the regularized incomplete beta function, as follows: \[\begin{aligned} F(k;n,p) & = \Pr(X \le k) \\ &= I_{1-p}(n-k, k+1) \\ & = (n-k) {n \choose k} \int_0^{1-p} t^{n-k-1} (1-t)^k \, dt , \end{aligned}\] which is equivalent to the cumulative distribution functions of the beta distribution and of the F-distribution: \[F(k;n,p) = F_{\text{beta-distribution}}\left(x=1-p;\alpha=n-k,\beta=k+1\right)\] \[F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).\]

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

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: \[\operatorname{E}[X] = np.\]

This follows from the linearity of the expected value along with the fact that X is the sum of n identical Bernoulli random variables, each with expected value p. In other words, if \(X_1, \ldots, X_n\) are identical (and independent) Bernoulli random variables with parameter p, then X = X₁ + ... + X_n and \[\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.\]

The variance is: \[\operatorname{Var}(X) = npq = np(1 - p).\]

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

The first 6 central moments, defined as \(\mu _{c}=\operatorname {E} \left[(X-\operatorname {E} [X])^{c}\right]\), are given by \[\begin{aligned} \mu_1 &= 0, \\ \mu_2 &= np \left(1-p\right), \\ \mu_3 &= np \left(1-p\right) \left(1-2p\right), \\ \mu_4 &= np \left(1-p\right) \left[1 + \left(3n-6\right) p \left(1-p\right)\right],\\ \mu_5 &= np \left(1-p\right) \left(1-2p\right) \left[1 + \left(10n-12\right) p \left(1-p\right)\right],\\ \mu_6 &= np \left(1-p\right) \left[1 - 30p\left(1-p\right)\left[1-4p(1-p)\right] + 5np \left(1-p\right)\left[5 - 26p\left(1-p\right)\right] + 15n^2 p^2 \left(1-p\right)^2\right]. \end{aligned}\]

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

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

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

Proof: Let \[f(k)=\binom nk p^k q^{n-k}.\]

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 \(k\neq n\). 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 \[\begin{aligned} k > (n+1)p-1 \Rightarrow f(k+1) < f(k) \\ k = (n+1)p-1 \Rightarrow f(k+1) = f(k) \\ k < (n+1)p-1 \Rightarrow f(k+1) > f(k) \end{aligned}\]

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.

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.
• Any median m must lie within the interval \(\lfloor np \rfloor\leq m \leq \lceil np \rceil\).
• A median m cannot lie too far away from the mean:\(|m-np|\leq \min\{{\ln2}, \max\{p,1-p\}\}\).
• The median is unique and equal to m = round(np) when |m − np| ≤ min{p, 1 − p} (except for the case when p = 1/2 and n is odd).
• When p is a rational number (with the exception of p = 1/2 and n odd), the median is unique.
• When \(p = \tfrac{1}{2}\) and n is odd, any number m in the interval \(\frac{1}{2} \left(n-1\right) \leq m \leq \frac{1}{2} \left(n+1\right)\) is a median of the binomial distribution. If \(p = \tfrac{1}{2}\) and n is even, then \(m = \tfrac{n}{2}\) is the unique median.

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

Hoeffding's inequality yields the simple bound \[F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!\] 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.

A sharper bound can be obtained from the Chernoff bound: \[F(k;n,p) \leq \exp\left(-n D{\left(\frac{k}{n}\parallel p\right)}\right)\] where D(a ∥ p) is the relative entropy (or Kullback-Leibler divergence) between an a-coin and a p-coin (that is, between the Bernoulli(a) and Bernoulli(p) 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 F(k; n, p), known as anti-concentration bounds. By approximating the binomial coefficient with Stirling's formula it can be shown that \[F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-n D{\left(\frac{k}{n}\parallel p\right)}\right),\] which implies the simpler but looser bound \[F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).\]

For p = 1/2 and k ≥ 3n/8 for even n, it is possible to make the denominator constant: \[F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!\]

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 (that is, x). 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 \(\operatorname{Var}(\hat{p}) = \frac{p(1-p)}{n}\), a property which is used in various ways, such as in Wald's confidence intervals.

A closed form Bayes estimator for 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. 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, \(\operatorname{Beta}(\alpha{=}1,\, \beta{=}1) = U(0,1)\), 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.

When relying on Jeffreys prior, the prior is \(\operatorname{Beta}(\alpha{=}\tfrac{1}{2}, \, \beta{=}\tfrac{1}{2})\), which leads to the estimator: \[\widehat{p}_{\mathrm{Jeffreys}} = \frac{x+\frac{1}{2}}{n+1}.\]

When estimating p with very rare events and a small n (for example, 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 \(\widehat{p}_b\), 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: \[\widehat{p}_{\text{rule of 3}} = \frac{3}{n}.\]

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.

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
• \(\widehat{p\,} = \frac{n_1}{n}\) is the proportion of successes
• \(z\) is the \(1 - \tfrac{1}{2}\alpha\) 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 \(1 - \tfrac{1}{2}\alpha=0.975\) and \(z=1.96\).

If X ~ B(n, p) and Y ~ B(m, p) are independent binomial variables with the same probability p, then X + Y is again a binomial variable; its distribution is Z = X + Y ~ B(n + m, p): \[\begin{aligned} \operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\ &= \binom{n+m}k p^k (1-p)^{n+m-k} \end{aligned}\]

A Binomial distributed random variable X ~ B(n, p) can be considered as the sum of n Bernoulli distributed random variables. So the sum of two Binomial distributed random variables X ~ B(n, p) and Y ~ B(m, p) is equivalent to the sum of n + m Bernoulli distributed random variables, which means Z = X + Y ~ B(n + m, p). This can also be proven directly using the addition rule.

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 B(n + m, p).

The binomial distribution is a special case of the Poisson binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(p_i).

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

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.