Is $E[(X-E[X])U] = E[U]E[X-E[X]]$ for RV $U,X$: U mean independent of X? If so, why? Thanks!
Quick probability question (multiplication of expectations)
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probability
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0Don't you mean $E[U|X] = E[U]$? – 2012-11-04
1 Answers
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This holds for all $X, U$:
$E[(X-E[X])U] = E[XU] - E[X]E[U] = Cov(X,U)$
You are telling us that the first expression equals:
$E[U]E[X-E[X]] = E[U](E[X] - E[E[X]]) = E[U] \cdot 0 = 0$
So this means that the $Cov(X,U)$ is $0$. However that doesn't imply $X$ independent of $U$, because uncorrelatedness does not imply independence.