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