Index: The Book of Statistical ProofsProbability DistributionsUnivariate discrete distributionsBernoulli distribution ▷ Mean

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

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

Then, the mean or expected value of $X$ is

\[\label{eq:bern-mean} \mathrm{E}(X) = p \; .\]

Proof: The expected value is the probability-weighted average of all possible values:

\[\label{eq:mean} \mathrm{E}(X) = \sum_{x \in \mathcal{X}} x \cdot \mathrm{Pr}(X = x) \; .\]

Since there are only two possible outcomes for a Bernoulli random variable, we have:

\[\label{eq:bern-mean-qed} \begin{split} \mathrm{E}(X) &= 0 \cdot \mathrm{Pr}(X = 0) + 1 \cdot \mathrm{Pr}(X = 1) \\ &= 0 \cdot (1-p) + 1 \cdot p \\ &= p \; . \\ \end{split}\]
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Metadata: ID: P22 | shortcut: bern-mean | author: JoramSoch | date: 2020-01-16, 10:58.