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?
 A: 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
$$
\begin{array}{lcl}
Cov(X,E[Y|X]) &=& E[X \cdot E(Y|X)] - E[X] \cdot E[E(Y|X)]\\
\\
&=& E[E(XY|X)] - E[X] \cdot E[E(Y|X)] \,\,\,\, \mbox{by (1)}\\
\\
&=& E[XY]-E[X] \cdot  E[Y] = Cov(X,Y) 
\end{array}
$$
A: 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.
A: 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.
