# Variance of Bernoulli Distribution

## Theorem

Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:

$X \sim \Bernoulli p$

Then the variance of $X$ is given by:

$\var X = p \paren {1 - p}$

## Proof 1

From the definition of variance:

$\var X = \expect {\paren {X - \expect X}^2}$

From the Expectation of Bernoulli Distribution, we have $\expect X = p$.

Then by definition of Bernoulli distribution:

 $\displaystyle \expect {\paren {X - \expect X}^2}$ $=$ $\displaystyle \paren {1 - p}^2 \times p + \paren {0 - p}^2 \times \paren {1 - p}$ $\displaystyle$ $=$ $\displaystyle p - 2 p^2 + p^3 + p^2 - p^3$ $\displaystyle$ $=$ $\displaystyle p - p^2$ $\displaystyle$ $=$ $\displaystyle p \paren {1 - p}$

$\blacksquare$

## Proof 2

$\var X = \expect {X^2} - \paren {\expect X}^2$
$\displaystyle \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$

So:

 $\displaystyle \expect {X^2}$ $=$ $\displaystyle 1^2 \times p + 0^2 \times \paren {1 - p}$ $\displaystyle$ $=$ $\displaystyle p$

Then:

 $\displaystyle \var X$ $=$ $\displaystyle \expect {X^2} - \paren {\expect X}^2$ $\displaystyle$ $=$ $\displaystyle p - p^2$ Expectation of Bernoulli Distribution $\displaystyle$ $=$ $\displaystyle p \paren {1 - p}$

$\blacksquare$

## Proof 3

We can simply use the Variance of Binomial Distribution, putting $n = 1$.

$\blacksquare$

## Proof 4

From Variance of Discrete Random Variable from PGF, we have:

$\var X = \map { {\Pi_X}''} 1 + \mu - \mu^2$

where $\mu = \expect X$ is the expectation of $X$.

From the Probability Generating Function of Bernoulli Distribution, we have:

$\map {\Pi_X} s = q + p s$

where $q = 1 - p$.

From Expectation of Bernoulli Distribution, we have $\mu = p$.

We have $\map { {\Pi_X}''} s = 0$ from Derivatives of PGF of Bernoulli Distribution.

Hence:

$\var X = 0 - \mu - \mu^2 = p - p^2 = p \paren {1 - p}$

$\blacksquare$

## Proof 5

From Moment Generating Function of Bernoulli Distribution, the moment generating function of $X$, $M_X$, is given by:

$\displaystyle \map {M_X} t = q + p e^t$
$\displaystyle \var X = \expect {X^2} - \paren {\expect X}^2$
$\displaystyle \expect {X^2} = \map {M_X''} 0$

We have:

 $\displaystyle \map {M_X''} t$ $=$ $\displaystyle \frac {\d^2} {\d t^2} \paren {q + p e^t}$ $\displaystyle$ $=$ $\displaystyle p \frac \d {\d t} \paren {e^t}$ Derivative of Constant, Derivative of Exponential Function $\displaystyle$ $=$ $\displaystyle p e^t$ Derivative of Exponential Function

Setting $t = 0$ gives:

 $\displaystyle \expect {X^2}$ $=$ $\displaystyle p e^0$ $\displaystyle$ $=$ $\displaystyle p$ Exponential of Zero

In Expectation of Bernoulli Distribution, it is shown that:

$\displaystyle \expect X = p$

So:

 $\displaystyle \var X$ $=$ $\displaystyle p - p^2$ $\displaystyle$ $=$ $\displaystyle p \paren {1 - p}$

$\blacksquare$