Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Variance ▷ Variance of a sum

Theorem: The variance of the sum of two random variables equals the sum of the variances of those random variables, plus two times their covariance:

$\label{eq:var-sum} \mathrm{Var}(X+Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2 \, \mathrm{Cov}(X,Y) \; .$

Proof: The variance is defined in terms of the expected value as

$\label{eq:var} \mathrm{Var}(X) = \mathrm{E}\left[ (X-\mathrm{E}(X))^2 \right] \; .$

Using this and the linearity of the expected value, we can derive \eqref{eq:var-sum} as follows:

$\label{eq:var-sum-qed} \begin{split} \mathrm{Var}(X+Y) &\overset{\eqref{eq:var}}{=} \mathrm{E}\left[ ((X+Y)-\mathrm{E}(X+Y))^2 \right] \\ &= \mathrm{E}\left[ ([X-\mathrm{E}(X)] + [Y-\mathrm{E}(Y)])^2 \right] \\ &= \mathrm{E}\left[ (X-\mathrm{E}(X))^2 + (Y-\mathrm{E}(Y))^2 + 2 \, (X-\mathrm{E}(X)) (Y-\mathrm{E}(Y)) \right] \\ &= \mathrm{E}\left[ (X-\mathrm{E}(X))^2 \right] + \mathrm{E}\left[ (Y-\mathrm{E}(Y))^2 \right] + \mathrm{E}\left[ 2 \, (X-\mathrm{E}(X)) (Y-\mathrm{E}(Y)) \right] \\ &\overset{\eqref{eq:var}}{=} \mathrm{Var}(X) + \mathrm{Var}(Y) + 2 \, \mathrm{Cov}(X,Y) \; . \\ \end{split}$
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Metadata: ID: P128 | shortcut: var-sum | author: JoramSoch | date: 2020-07-07, 06:10.