0
$\begingroup$

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!

  • 0
    What do you mean by "U mean independent of $X$"?2012-11-04
  • 0
    It's just E[U|X] = 02012-11-04
  • 0
    Don't you mean $E[U|X] = E[U]$?2012-11-04

1 Answers 1

2

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.