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Bernoulli distribution
Probability mass function Three examples of Bernoulli distribution:   $P(x=0)=0.2$ and $P(x=1)=0.8$   $P(x=0)=0.8$ and $P(x=1)=0.2$   $P(x=0)=0.5$ and $P(x=1)=0.5$
Notation	$Bernoulli\left(p\right)$
Parameters	$0\le p\le 1$ $q=1-p$
Support	$k\in \{0,1\}$
PMF	$\{\begin{array}{cc}q=1-p& \text{if }k=0\\ p& \text{if }k=1\end{array}$
CDF	$\{\begin{array}{cc}0& \text{if }k<0\\ 1-p& \text{if }0\le k<1\\ 1& \text{if }k\ge 1\end{array}$
Mean	$p$
Median	$\{\begin{array}{cc}0& \text{if }p<1/2\\ [0,1]& \text{if }p=1/2\\ 1& \text{if }p>1/2\end{array}$
Mode	$\{\begin{array}{cc}0& \text{if }p<1/2\\ 0,1& \text{if }p=1/2\\ 1& \text{if }p>1/2\end{array}$
Variance	$p(1-p)=pq$
MAD	$2p(1-p)=2pq$
Skewness	$\frac{q-p}{\sqrt{pq}}$
Excess kurtosis	$\frac{1-6pq}{pq}$
Entropy	$-q\ln q-p\ln p$
MGF	$q+p{e}^t$
CF	$q+p{e}^{it}$
PGF	$q+pz$
Fisher information	$\frac{1}{pq}$

In probability theory and statistics, the Bernoulli distribution, named after Swiss mathematician Jacob Bernoulli,^[1] is the discrete probability distribution of a random variable which takes the value 1 with probability $p$ and the value 0 with probability $q=1-p$. Less formally, it can be thought of as a model for the set of possible outcomes of any single experiment that asks a yes–no question. Such questions lead to outcomes that are Boolean-valued: a single bit whose value is success/yes/true/one with probability p and failure/no/false/zero with probability q. It can be used to represent a (possibly biased) coin toss where 1 and 0 would represent "heads" and "tails", respectively, and p would be the probability of the coin landing on heads (or vice versa where 1 would represent tails and p would be the probability of tails). In particular, unfair coins would have $p\ne 1/2.$ The Bernoulli distribution is a special case of the binomial distribution where a single trial is conducted (so n would be 1 for such a binomial distribution).

If $X$ is a random variable with a Bernoulli distribution, then:

$\begin{array}{cc}Pr(X=1)& =p,\\ Pr(X=0)& =q=1-p.\end{array}$

The probability mass function $f$ of this distribution, over possible outcomes k, is^[2]

$f(k;p)=\{\begin{array}{cc}p& \text{if }k=1,\\ q=1-p& \text{if }k=0.\end{array}$

This can also be expressed as

$f(k;p)=p^k(1-p{)}^{1-k}\text{for }k\in \{0,1\}$

or as

$f(k;p)=pk+(1-p)(1-k)\text{for }k\in \{0,1\}.$

The Bernoulli distribution is a special case of the binomial distribution with $n=1.$^[3]

The kurtosis goes to infinity for high and low values of $p,$ but for $p=1/2$ the two-point distributions including the Bernoulli distribution have a lower excess kurtosis, namely −2, than any other probability distribution.

The Bernoulli distributions for $0\le p\le 1$ form an exponential family.

The maximum likelihood estimator of $p$ based on a random sample is the sample mean.

The expected value of a Bernoulli random variable $X$ is

$E\left[X\right]=p$

This is because for a Bernoulli distributed random variable $X$ with $Pr(X=1)=p$ and $Pr(X=0)=q$ we find^[2]

$\begin{array}{cc}E\left[X\right]& =Pr(X=1)\cdot 1+Pr(X=0)\cdot 0\\ & =p\cdot 1+q\cdot 0\\ & =p.\end{array}$

The variance of a Bernoulli distributed $X$ is

$\mathrm{Var}\left[X\right]=pq=p(1-p)$

We first find

$\begin{array}{cc}E\left[X^2\right]& =Pr(X=1)\cdot 1^2+Pr(X=0)\cdot 0^2\\ & =p\cdot 1^2+q\cdot 0^2\\ & =p=E\left[X\right]\end{array}$

From this follows^[2]

$\begin{array}{cc}\mathrm{Var}\left[X\right]& =E\left[X^2\right]-E[X{]}^2=E[X]-E[X{]}^2\\ & =p-p^2=p(1-p)=pq\end{array}$

With this result it is easy to prove that, for any Bernoulli distribution, its variance will have a value inside $[0,1/4]$.

The skewness is $\frac{q-p}{\sqrt{pq}}=\frac{1-2p}{\sqrt{pq}}$. When we take the standardized Bernoulli distributed random variable $\frac{X-E\left[X\right]}{\sqrt{\mathrm{Var}\left[X\right]}}$ we find that this random variable attains $\frac{q}{\sqrt{pq}}$ with probability $p$ and attains $-\frac{p}{\sqrt{pq}}$ with probability $q$. Thus we get

$\begin{array}{cc}{\gamma }_1& =E\left[{\left(\frac{X-E\left[X\right]}{\sqrt{\mathrm{Var}\left[X\right]}}\right)}^3\right]\\ & =p\cdot {\left(\frac{q}{\sqrt{pq}}\right)}^3+q\cdot {(-\frac{p}{\sqrt{pq}})}^3\\ & =\frac{1}{{\sqrt{pq}}^3}(p{q}^3-q{p}^3)\\ & =\frac{pq}{{\sqrt{pq}}^3}(q^2-p^2)\\ & =\frac{(1-p{)}^2-p^2}{\sqrt{pq}}\\ & =\frac{1-2p}{\sqrt{pq}}=\frac{q-p}{\sqrt{pq}}.\end{array}$

The raw moments are all equal because $1^k=1$ and $0^k=0$.

$E\left[X^k\right]=Pr(X=1)\cdot 1^k+Pr(X=0)\cdot 0^k=p\cdot 1+q\cdot 0=p=E\left[X\right].$

The central moment of order $k$ is given by ${\mu }_k=(1-p)(-p{)}^k+p(1-p{)}^k.$ The first six central moments are $\begin{array}{cc}{\mu }_1& =0,\\ {\mu }_2& =p(1-p),\\ {\mu }_3& =p(1-p)(1-2p),\\ {\mu }_4& =p(1-p)(1-3p(1-p\left)\right),\\ {\mu }_5& =p(1-p)(1-2p)(1-2p(1-p\left)\right),\\ {\mu }_6& =p(1-p)(1-5p(1-p\left)\right(1-p(1-p)\left)\right).\end{array}$ The higher central moments can be expressed more compactly in terms of ${\mu }_2$ and ${\mu }_3$ $\begin{array}{cc}{\mu }_4& ={\mu }_2(1-3{\mu }_2),\\ {\mu }_5& ={\mu }_3(1-2{\mu }_2),\\ {\mu }_6& ={\mu }_2(1-5{\mu }_2(1-{\mu }_2\left)\right).\end{array}$ The first six cumulants are $\begin{array}{cc}{\kappa }_1& =p,\\ {\kappa }_2& ={\mu }_2,\\ {\kappa }_3& ={\mu }_3,\\ {\kappa }_4& ={\mu }_2(1-6{\mu }_2),\\ {\kappa }_5& ={\mu }_3(1-12{\mu }_2),\\ {\kappa }_6& ={\mu }_2(1-30{\mu }_2(1-4{\mu }_2\left)\right).\end{array}$

• If $X_1,\dots ,X_n$ are independent, identically distributed (i.i.d.) random variables, all Bernoulli trials with success probability p, then their sum is distributed according to a binomial distribution with parameters n and p:

  $\sum _{k=1}^n{X}_k\sim B(n,p)$ (binomial distribution).^[2]

The Bernoulli distribution is simply $B(1,p)$, also written as $Bernoulli\left(p\right).$

• The categorical distribution is the generalization of the Bernoulli distribution for variables with any constant number of discrete values.
• The Beta distribution is the conjugate prior of the Bernoulli distribution.^[4]
• The geometric distribution models the number of independent and identical Bernoulli trials needed to get one success.
• If $Y\sim Bernoulli\left(\frac{1}{2}\right)$, then $2Y-1$ has a Rademacher distribution.
• The Bernoulli distribution is a special case of the two-point distribution, for which the possible outcomes need not be 0 and 1.^[5]

• Bernoulli process, a random process consisting of a sequence of independent Bernoulli trials
• Bernoulli sampling
• Binary entropy function
• Binary decision diagram

• Abhirath, dwivedi; Kotz, Samuel; Kemp, Adrienne W. (1993). Univariate Discrete Distributions (2nd ed.). Wiley. ISBN 0-471-54897-9.
• Peatman, John G. (1963). Introduction to Applied Statistics. New York: Harper & Row. pp. 162–171.

Wikimedia Commons has media related to Bernoulli distribution.

• "Binomial distribution", Encyclopedia of Mathematics, EMS Press, 2001 [1994].
• Weisstein, Eric W. "Bernoulli Distribution". MathWorld.
• Interactive graphic: Univariate Distribution Relationships.