> [!NOTE] Theorem > Let $X_{1},X_{2},\dots,X_{n}$ be [[Pairwise Independent Set of Discrete Real-Valued Random Variables|pairwise independent discrete real-valued random variables]] that are each [[Square-Integrable Discrete Real-Valued Random Variable|square-integrable]]. Then the [[Variance of Square-Integrable Discrete Real-Valued Random Variable|variance]] of their sum is given by $\text{Var}\left( \sum_{i=1}^{n} X_{i} \right) = \sum_{i=1}^{n} \text{Var}(X_{i})$ **Proof**: For all $i=1,2,\dots,n,$ let $\mu_{i}:=\mathbb{E}[X_{i}]$ (the [[Expectation of Discrete Real-Valued Random Variable|expectation]] of the $X_{i}s). Let $\mu:= \sum_{i=1}^{n} \mu_{i}$ which equals $\mathbb{E}\left[ \sum_{i=1}^{n} X_{i} \right]$ by [[Expectation of Discrete Real-Valued Random Variable is Linear|linearity of expectation]]. Then $\begin{align}\text{Var}\left( \sum_{i=1}^{n} X_{i} \right) &= \mathbb{E}\left[ \left( \left( \sum_{i=1}^{n} X_{i} \right) - \mu \right)^{2} \right] \\&= \mathbb{E} \left[ \left( \sum_{i=1}^{n} (X_{i}-\mu_{i}) \right)^{2} \right] \\ &= \mathbb{E} \left[ \sum_{i=1}^{n} \sum_{j=1}^{n} (X_{i}-\mu_{i}) (X_{j} - \mu_{j}) \right] \\ &= \mathbb{E}\left[ \sum_{i=1}^{n} (X_{i} - \mu_{i })^{2} \right] + \mathbb{E}\left[ \sum_{i \neq j} (X_{j} - \mu_{j}) (X_{i} - \mu_{i}) \right] \\ &= \sum_{i=1}^{n} \mathbb{E}[(X_{i} - \mu)^{2}] + \sum_{i \neq j} \mathbb{E}[(X_{i}-\mu_{i})(X_{j}-\mu_{j})]\\ &= \sum_{i=1}^{n} \text{Var}(X_{i}) + \sum_{i \neq j} (\mathbb{E}[X_{i}X_{j}] - \mu_{i} \mu_{j}) \\ &= \sum_{i=1}^{n} \text{Var} (X_{i}) . \end{align}$since by [[Expectation of Product of Two Independent Discrete Real-Valued Random Variables]], for $i \neq j,$ $\mathbb{E}[X_{i}X_{j}]=\mu_{i}\mu_{j}.$ **Proof**: The result follows directly from [[Variance of Sum of Uncorrelated Square-Integrable Discrete Real-Valued Random Variables]] since [[Pairwise Independent Square-Integrable Discrete Real-Valued Random Variables are Uncorrelated]].