Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate discrete distributions ▷ Bernoulli distribution ▷ Variance

Theorem: Let $X$ be a random variable following a Bernoulli distribution:

$\label{eq:bern} X \sim \mathrm{Bern}(p) \; .$

Then, the variance of $X$ is

$\label{eq:bern-var} \mathrm{Var}(X) = p \, (1-p) \; .$

Proof: The variance is the probability-weighted average of the squared deviation from the expected value across all possible values

$\label{eq:var} \mathrm{Var}(X) = \sum_{x \in \mathcal{X}} (x - \mathrm{E}(X))^2 \cdot \mathrm{Pr}(X = x)$ $\label{eq:var-mean} \mathrm{Var}(X) = \mathrm{E}\left( X^2 \right) - \mathrm{E}(X)^2 \; .$ $\label{eq:bern-mean} X \sim \mathrm{Bern}(p) \quad \Rightarrow \quad \mathrm{E}(X) = p$

and the mean of a squared Bernoulli random variable is

$\label{eq:bern-sqr-mean} \mathrm{E}\left( X^2 \right) = 0^2 \cdot \mathrm{Pr}(X = 0) + 1^2 \cdot \mathrm{Pr}(X = 1) = 0 \cdot (1-p) + 1 \cdot p = p \; .$

Combining \eqref{eq:var-mean}, \eqref{eq:bern-mean} and \eqref{eq:bern-sqr-mean}, we have:

$\label{eq:bern-var-qed} \mathrm{Var}(X) = p - p^2 = p \, (1-p) \; .$
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Metadata: ID: P301 | shortcut: bern-var | author: JoramSoch | date: 2022-01-20, 15:06.