Variance of Bernoulli Distribution/Proof 2

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 as Expectation of Square minus Square of Expectation:

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

From Expectation of Function of Discrete Random Variable:

$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$

So:

\(\ds \expect {X^2}\)

\(=\)

\(\ds 1^2 \times p + 0^2 \times \paren {1 - p}\)

\(\ds \)

\(=\)

\(\ds p\)

Then:

\(\ds \var X\)

\(=\)

\(\ds \expect {X^2} - \paren {\expect X}^2\)

\(\ds \)

\(=\)

\(\ds p - p^2\)

Expectation of Bernoulli Distribution

\(\ds \)

\(=\)

\(\ds p \paren {1 - p}\)

$\blacksquare$