Variance of Bernoulli Distribution/Proof 4
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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:
- $\var X = 0 - \mu - \mu^2 = p - p^2 = p \paren {1 - p}$
$\blacksquare$