Variance of Bernoulli Distribution/Proof 4

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

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:

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

$\blacksquare$