# Bernoulli distribution

Short description: Probability distribution modeling a coin toss which need not be fair
Parameters 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 $\displaystyle{ P(x=1) = 0{.}5 }$ $\displaystyle{ 0 \leq p \leq 1 }$ $\displaystyle{ q = 1 - p }$ $\displaystyle{ k \in \{0,1\} }$ $\displaystyle{ \begin{cases} q=1-p & \text{if }k=0 \\ p & \text{if }k=1 \end{cases} }$ $\displaystyle{ \begin{cases} 0 & \text{if } k \lt 0 \\ 1 - p & \text{if } 0 \leq k \lt 1 \\ 1 & \text{if } k \geq 1 \end{cases} }$ $\displaystyle{ p }$ $\displaystyle{ \begin{cases} 0 & \text{if } p \lt 1/2\\ \left[0, 1\right] & \text{if } p = 1/2\\ 1 & \text{if } p \gt 1/2 \end{cases} }$ $\displaystyle{ \begin{cases} 0 & \text{if } p \lt 1/2\\ 0, 1 & \text{if } p = 1/2\\ 1 & \text{if } p \gt 1/2 \end{cases} }$ $\displaystyle{ p(1-p) = pq }$ $\displaystyle{ \frac{q - p}{\sqrt{pq}} }$ $\displaystyle{ \frac{1 - 6pq}{pq} }$ $\displaystyle{ -q\ln q - p\ln p }$ $\displaystyle{ q+pe^t }$ $\displaystyle{ q+pe^{it} }$ $\displaystyle{ q+pz }$ $\displaystyle{ \frac{1}{pq} }$

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 $\displaystyle{ p }$ 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.

## Properties

If $\displaystyle{ X }$ is a random variable with this 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

$\displaystyle{ f(k;p) = \begin{cases} p & \text{if }k=1, \\ q = 1-p & \text {if } k = 0. \end{cases} }$[2]

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 $\displaystyle{ n = 1. }$[3]

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 than any other probability distribution, namely −2.

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

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

## Mean

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

$\displaystyle{ \operatorname{E}\left(X\right)=p }$

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

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

## Variance

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

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

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

## Skewness

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{align} \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{align} }

## Higher moments and cumulants

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{align} \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{align} }

The higher central moments can be expressed more compactly in terms of $\displaystyle{ \mu_2 }$ and $\displaystyle{ \mu_3 }$

\displaystyle{ \begin{align} \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{align} }

The first six cumulants are

\displaystyle{ \begin{align} \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{align} }

## Related distributions

• 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).[2]
The Bernoulli distribution is simply $\displaystyle{ \operatorname{B}(1, p) }$, also written as $\displaystyle{ \mathrm{Bernoulli} (p). }$