I've been working on this problem for 3 hours now, and my complete lack of progress is getting disheartening. I've looked up definitions, proofs, and have even seen a solution for this particular problem. However, I did not understand the steps the solution used, which means I have no business reading it. I'm hoping that if I show my attempt, y'all can give me some pointers. Thanks in advance!

Here is my work:

$$\mathrm{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])] =E[XY-X*E[Y]-Y*E[X]+E[X]*E[Y]]$$

By linearity of expected value:

=E[XY]-E[X*E[Y]]-E[Y*E[X]]+E[E[X]*E[Y]], as E[Y] and E[X] are constants, E[X*E[Y]]=E[Y]*E[X]










I really have no idea where to go with this problem. If E[X*E[Y]]=E[X]E[Y], I cannot figure out why E[XE[Y|X]]$\neq$E[X]E[Y|X]. I believe I am making a mistake when I go from the bold line to the next, but I can't figure out why.

  • $\begingroup$ First observe that it suffices to prove $Cov(X,Y)=Cov(X,E[Y|X])$ for $X,Y$ with $E(X)=E(Y)=0$. Now, can you prove this? $\endgroup$ – Miheer Nov 24 '13 at 6:14
  • $\begingroup$ Also observe that $E(XE[X|Y])$ is a constant while $E(X)E[Y|X]$ is a random variable, so they may not always be equal. $\endgroup$ – Miheer Nov 24 '13 at 6:24

If E[X*E[Y]]=E[X]E[Y], I cannot figure out why E[XE[Y|X]]≠E[X]E[Y|X].

Indeed, for every integrable random variables $X$ and $Y$, $E[XE[Y]]=E[X]E[Y]$ since $E[Y]$ is simply a constant and one knows that $E[\lambda X]=\lambda E[X]$ for every constant $\lambda$.

But $Z=E[Y\mid X]$ is a random variable and, for some general random variable $Z$, $E[XZ]\ne E[X]Z$ simply because the LHS is a number and the RHS a random variable.

What happens here is that $Z$ is a specific random variable, namely, the conditional expectation of some random variable $Y$ with respect to $X$. And it is a fact (actually, almost part of the definition) that $E[u(X)E[Y\mid X]]=E[u(X)Y]$ for every measurable function $u$ (technically, such that $u(X)Y$ is integrable). You might want to apply this to your problem (if you try and still fail to solve it, I might add some further hints).


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