Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Covariance ▷ Covariance under independence

Theorem: Let $X$ and $Y$ be independent random variables. Then, the covariance of $X$ and $Y$ is zero:

$\label{eq:cov-ind} X, Y \; \text{independent} \quad \Rightarrow \quad \mathrm{Cov}(X,Y) = 0 \; .$ $\label{eq:cov-mean} \mathrm{Cov}(X,Y) = \mathrm{E}(X\,Y) - \mathrm{E}(X) \, \mathrm{E}(Y) \; .$

For independent random variables, the expected value of the product is equal to the product of the expected values:

$\label{eq:mean-mult} \mathrm{E}(X\,Y) = \mathrm{E}(X) \, \mathrm{E}(Y) \; .$

Taking \eqref{eq:cov-mean} and \eqref{eq:mean-mult} together, we have

$\label{eq:cov-ind-qed} \begin{split} \mathrm{Cov}(X,Y) &\overset{\eqref{eq:cov-mean}}{=} \mathrm{E}(X\,Y) - \mathrm{E}(X) \, \mathrm{E}(Y) \\ &\overset{\eqref{eq:mean-mult}}{=} \mathrm{E}(X) \, \mathrm{E}(Y) - \mathrm{E}(X) \, \mathrm{E}(Y) \\ &= 0 \; . \end{split}$
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Metadata: ID: P158 | shortcut: cov-ind | author: JoramSoch | date: 2020-09-03, 06:05.