Variance of Bernoulli Distribution/Proof 1

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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 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:

\(\ds \expect {\paren {X - \expect X}^2}\) \(=\) \(\ds \paren {1 - p}^2 \times p + \paren {0 - p}^2 \times \paren {1 - p}\)
\(\ds \) \(=\) \(\ds p - 2 p^2 + p^3 + p^2 - p^3\)
\(\ds \) \(=\) \(\ds p - p^2\)
\(\ds \) \(=\) \(\ds p \paren {1 - p}\)

$\blacksquare$