> [!NOTE] Lemma > If $X_{1}, X_{2},\dots ,X_{n}$ are [[Uncorrelated Square-Integrable Discrete Real-Valued Random Variables|uncorrelated square-integrable discrete real-valued random variables]] with the [[Expectation of Discrete Real-Valued Random Variable|mean]] and [[Variance of Square-Integrable Discrete Real-Valued Random Variable|variance]] $\mu, \sigma^{2}.$ Then $\mathbb{E}\left[ \frac{X_{1}+\dots+X_{n}}{n} \right] = \mu$and $\text{Var}\left( \frac{X_{1}+\dots+X_{n}}{n} \right) = \frac{\sigma^{2}}{n}.$ Proof: By [[Expectation of Discrete Real-Valued Random Variable is Linear|linearity of expectation]] and the fact the $X_{i}$ have the same mean $\mu,$ $\mathbb{E}\left[ \frac{X_{1}+\dots+X_{n}}{n} \right] = \frac{\mathbb{E}[X_{1}]+\dots \mathbb{E}[X_{n}]}{n }= \frac{n\mu}{n} = \mu.$ By [[Variance of Sum of Pairwise Independent Square-Integrable Discrete Real-Valued Random Variables]] and [[Variance of Linear Transformation of Square-Integrable Discrete Real-Valued Random Variable]], $\begin{align} \text{Var}\left( \frac{X_{1}+\dots+X_{n}}{n} \right) &= \frac{1}{n^{2}} \text{Var} (X_{1}+\dots+X_{n}) \\ &= \frac{\text{Var}(X_{1})+\dots+\text{Var}(X_{n})}{n^{2}} \\ &= \frac{n\sigma^{2}}{n^{2}} \\ &= \frac{\sigma^{2}}{n}. \end{align}$