Variance of Bernoulli Distribution/Proof 4

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