Prove that, $E[\operatorname{Cov}(X,Y|Z)]+\operatorname{Cov}(E[X|Z],E[Y|Z])=\operatorname{Cov}(X,Y)$

Question

I have tried the problem in the following manner-
We know that $\operatorname{Cov}(X,Y|Z)=E[XY|Z]-E[X|Z]E[Y|Z]$
$E[\operatorname{Cov}(X,Y|Z)]+\operatorname{Cov}(E[X|Z],E[Y|Z])$
$=E[E[XY|Z]-E[X|Z]E[Y|Z]]+E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]$
$=E[E[XY|Z]]-E[E[X|Z]E[Y|Z]]+E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]$
From that step, I cannot proceed further. Can anyone help to complete the proof? Thanks for assistance in advance.

Answer 1

Assuming the statement to be proven is $$E[{\rm Cov}(X,Y|Z)] + {\rm Cov}(E[X|Z],E[Y|Z]) = {\rm Cov}(X,Y),$$ you can use the same calculations you do above to get to $$ E[E[XY|Z]] - E[E[X|Z]E[Y|Z]] + E[E[X|Z]E[Y|Z]] - E[E[X|Z]]E[E[Y|Z]]. $$ The middle terms cancel out. To finish, you just need to use the tower property of conditional expectation (or the law of total expectation):

• For any random variables $X$ and $Z$, $E[E[X|Z]] = E[X]$.

This will give $E[E[XY|Z]] = E[XY]$, $E[E[X|Z]] = E[X]$ and $E[E[Y|Z]] = E[Y]$, and you are done.