Bernoulli distribution

Probability mass function Three examples of Bernoulli distribution:   ${\displaystyle P(x=0)=0{.}2}$ and ${\displaystyle P(x=1)=0{.}8}$   ${\displaystyle P(x=0)=0{.}8}$ and ${\displaystyle P(x=1)=0{.}2}$   ${\displaystyle P(x=0)=0{.}5}$ and ${\displaystyle P(x=1)=0{.}5}$
Parameters	${\displaystyle 0\leq p\leq 1}$ ${\displaystyle q=1-p}$
Support	${\displaystyle k\in \{0,1\))$
PMF	${\displaystyle {\begin{cases}q=1-p&{\text{if ))k=0\\p&{\text{if ))k=1\end{cases))}$
CDF	${\displaystyle {\begin{cases}0&{\text{if ))k<0\\1-p&{\text{if ))0\leq k<1\\1&{\text{if ))k\geq 1\end{cases))}$
Mean	${\displaystyle p}$
Median	${\displaystyle {\begin{cases}0&{\text{if ))p<1/2\\\left[0,1\right]&{\text{if ))p=1/2\\1&{\text{if ))p>1/2\end{cases))}$
Mode	${\displaystyle {\begin{cases}0&{\text{if ))p<1/2\\0,1&{\text{if ))p=1/2\\1&{\text{if ))p>1/2\end{cases))}$
Variance	${\displaystyle p(1-p)=pq}$
MAD	${\displaystyle 2p(1-p)=2pq}$
Skewness	${\displaystyle {\frac {q-p}{\sqrt {pq))))$
Excess kurtosis	${\displaystyle {\frac {1-6pq}{pq))}$
Entropy	${\displaystyle -q\ln q-p\ln p}$
MGF	${\displaystyle q+pe^{t))$
CF	${\displaystyle q+pe^{it))$
PGF	${\displaystyle q+pz}$
Fisher information	${\displaystyle {\frac {1}{pq))}$

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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 ${\displaystyle p}$ and the value 0 with probability ${\displaystyle q=1-p}$. 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 ${\displaystyle p\neq 1/2.}$

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). It is also a special case of the two-point distribution, for which the possible outcomes need not be 0 and 1. ^[2]

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.

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.