If ${\displaystyle X}$  is a random variable with a Bernoulli distribution, then:

${\displaystyle \Pr(X=1)=p=1-\Pr(X=0)=1-q.}$ 

The probability mass function ${\displaystyle f}$  of this distribution, over possible outcomes k, is

${\displaystyle f(k;p)={\begin{cases}p&{\text{if }}k=1,\\q=1-p&{\text{if }}k=0.\end{cases}}}$ ^[3]

This can also be expressed as

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

or as

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

The Bernoulli distribution is a special case of the binomial distribution with ${\displaystyle n=1.}$ ^[4]

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

The Bernoulli distributions for ${\displaystyle 0\leq p\leq 1}$  form an exponential family.

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

The expected value of a Bernoulli random variable ${\displaystyle X}$  is

${\displaystyle \operatorname {E} [X]=p}$ 

This is due to the fact that for a Bernoulli distributed random variable ${\displaystyle X}$  with ${\displaystyle \Pr(X=1)=p}$  and ${\displaystyle \Pr(X=0)=q}$  we find

${\displaystyle \operatorname {E} [X]=\Pr(X=1)\cdot 1+\Pr(X=0)\cdot 0=p\cdot 1+q\cdot 0=p.}$ ^[3]

The variance of a Bernoulli distributed ${\displaystyle X}$  is

${\displaystyle \operatorname {Var} [X]=pq=p(1-p)}$ 

We first find

${\displaystyle \operatorname {E} [X^{2}]=\Pr(X=1)\cdot 1^{2}+\Pr(X=0)\cdot 0^{2}=p\cdot 1^{2}+q\cdot 0^{2}=p=\operatorname {E} [X]}$ 

From this follows

${\displaystyle \operatorname {Var} [X]=\operatorname {E} [X^{2}]-\operatorname {E} [X]^{2}=\operatorname {E} [X]-\operatorname {E} [X]^{2}=p-p^{2}=p(1-p)=pq}$ ^[3]

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

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

${\displaystyle {\begin{aligned}\gamma _{1}&=\operatorname {E} \left[\left({\frac {X-\operatorname {E} [X]}{\sqrt {\operatorname {Var} [X]}}}\right)^{3}\right]\\&=p\cdot \left({\frac {q}{\sqrt {pq}}}\right)^{3}+q\cdot \left(-{\frac {p}{\sqrt {pq}}}\right)^{3}\\&={\frac {1}{{\sqrt {pq}}^{3}}}\left(pq^{3}-qp^{3}\right)\\&={\frac {pq}{{\sqrt {pq}}^{3}}}(q-p)\\&={\frac {q-p}{\sqrt {pq}}}.\end{aligned}}}$ 

The raw moments are all equal due to the fact that ${\displaystyle 1^{k}=1}$  and ${\displaystyle 0^{k}=0}$ .

${\displaystyle \operatorname {E} [X^{k}]=\Pr(X=1)\cdot 1^{k}+\Pr(X=0)\cdot 0^{k}=p\cdot 1+q\cdot 0=p=\operatorname {E} [X].}$ 

The central moment of order ${\displaystyle k}$  is given by

${\displaystyle \mu _{k}=(1-p)(-p)^{k}+p(1-p)^{k}.}$ 

The first six central moments are

${\displaystyle {\begin{aligned}\mu _{1}&=0,\\\mu _{2}&=p(1-p),\\\mu _{3}&=p(1-p)(1-2p),\\\mu _{4}&=p(1-p)(1-3p(1-p)),\\\mu _{5}&=p(1-p)(1-2p)(1-2p(1-p)),\\\mu _{6}&=p(1-p)(1-5p(1-p)(1-p(1-p))).\end{aligned}}}$ 

The higher central moments can be expressed more compactly in terms of ${\displaystyle \mu _{2}}$  and ${\displaystyle \mu _{3}}$ 

${\displaystyle {\begin{aligned}\mu _{4}&=\mu _{2}(1-3\mu _{2}),\\\mu _{5}&=\mu _{3}(1-2\mu _{2}),\\\mu _{6}&=\mu _{2}(1-5\mu _{2}(1-\mu _{2})).\end{aligned}}}$ 

The first six cumulants are

${\displaystyle {\begin{aligned}\kappa _{1}&=p,\\\kappa _{2}&=\mu _{2},\\\kappa _{3}&=\mu _{3},\\\kappa _{4}&=\mu _{2}(1-6\mu _{2}),\\\kappa _{5}&=\mu _{3}(1-12\mu _{2}),\\\kappa _{6}&=\mu _{2}(1-30\mu _{2}(1-4\mu _{2})).\end{aligned}}}$ 

• If ${\displaystyle X_{1},\dots ,X_{n}}$  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:

  ${\displaystyle \sum _{k=1}^{n}X_{k}\sim \operatorname {B} (n,p)}$  (binomial distribution).^[3]

The Bernoulli distribution is simply ${\displaystyle \operatorname {B} (1,p)}$ , also written as ${\textstyle \mathrm {Bernoulli} (p).}$ 

• 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.^[5]
• The geometric distribution models the number of independent and identical Bernoulli trials needed to get one success.
• If ${\textstyle Y\sim \mathrm {Bernoulli} \left({\frac {1}{2}}\right)}$ , then ${\textstyle 2Y-1}$  has a Rademacher distribution.

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

• Johnson, N. L.; Kotz, S.; Kemp, A. (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.

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