# Show that Cov(X,Y)=Cov(X,E(Y|X)).

Let X, Y be independent random variables. I've been working on this for a while and I think this question just requires skillful manipulation of the expectations E(X) and E(Y|X). At one point I got that Cov(X,Y) is 0... which is incorrect. Since then I have started over and here's what I have:

Cov(X, E(Y|X))=E((X-E(X)(E(Y|X)-E(E(Y|X))) =E(X E(Y|X))-X E(E(Y|X))-E(X)E(Y|X)+E(X) E(E(Y|X)) =E(E(XY|X))-E(E(XY|X))

And I'm not so sure what to do from there. Did I just make things more complicated?

• The last = sign is wrong. For example, E(E(Y|X))=E(Y) hence E(X)E(E(Y|X))=E(X)E(Y).
– Did
Oct 14, 2015 at 5:30
• You don't need to assume $X$ and $Y$ are independent. If $X$ and $Y$ are independent, both sides are $0$. Oct 14, 2015 at 6:46
• math.stackexchange.com/q/578919/321264 Feb 21, 2020 at 7:00

Note that $$E[XY|X] = X \cdot E[Y|X] \,\,\,\,\,\,\,\,\,\,\,(1)$$ since $X$ is given and therefore "known".

Also, recall that $$E[E(X|Y)]=E[X]. \,\,\,\,\,\,\,\,\,\,\,\,\, (2)$$

Now, using that fact that $Cov(X,Y) = E[XY]-E[X] E[Y]$ on this problem gives $$\begin{array}{lcl} Cov(X,E[Y|X]) &=& E[X \cdot E(Y|X)] - E[X] \cdot E[E(Y|X)]\\ \\ &=& E[E(XY|X)] - E[X] \cdot E[E(Y|X)] \,\,\,\, \mbox{by (1)}\\ \\ &=& E[XY]-E[X] \cdot E[Y] = Cov(X,Y) \end{array}$$

Rewrite as $$Cov(X,Y-E(Y|X))=0$$ which is true because $E(Y|X)$ of $Y$ is an orthogonal projection onto space of functions measurable with respect to $\sigma(X)$.

Alternatively, expand both sides $$E(XY)-E(X)E(Y)=E(XE(Y|X))-E(X)E(E(Y|X))$$ Second terms on both sides cancel out and first terms are equal per definition of conditional expectation.

Let $Z=E(Y|X)$. Then $$E(Z)=E[E(Y|X)]=E(Y),\quad E(XZ)=E[XE(Y|X)]=E[E(XY|X)]=E(XY).$$ It is then transparent that $$\text{Cov}(X,Z)=E(XZ)-E(X)E(Z)=E(XY)-E(X)E(Y)=\text{Cov}(X,Y).$$ The independence between $X$ and $Y$ are not needed.