Bernoulli Distribution: Expectation, Variance and Moment Generating Function Proofs

Expectation

• Define success as \(X=1\) with probability \(p\)
• Define failure as \(X=0\) with probability \(1-p\)

\[\begin{align*} E(X) &= \sum_{k \in \text{Img}(X)} k \cdot P(X = k)\\ &= \sum_{0}^1 k P(X = k) \\ &= 0\cdot P(X=0) + 1 \cdot P(X=1)\\ &= 0 \cdot p^0(1-p)^{1-0} + 1 \cdot p^1(1-p)^{1-1} \\ &= 0 + p \\ &= p \end{align*}\]

Variance

\[\begin{align*} E(X^2) &= \sum_{k \in \text{Img}(X)} k^2 \cdot P(X = k)\\ &= \sum_{0}^1 k^2 P(X = k) \\ &= 0^2\cdot P(X=0) + 1^2 \cdot P(X=1)\\ &= 0 \cdot p^0(1-p)^{1-0} + 1 \cdot p^1(1-p)^{1-1} \\ &= 0 + p \\ &= p \\ Var(X) &= E(X^2) - [E(X)]^2 \\ &= p - p^2 \\ &= p(1-p) \end{align*}\]

Moment Generating Function

\[\begin{align*} M_X(t) &= E(e^{tX}) \\ &= \sum_{k=0}^{1} e^{tk} P(X = k)\\ &= e^0 p^0(1-p)^{1-0} + e^t p^1(1-p)^{1-1} \\ &= (1-p) + pe^t\\ &= q + pe^t \end{align*}\]