Prove Cov(X, Y) = Cov(X , E(Y|X) ) I try to solve it from  Cov(X,Y) = E(XY) - E(X)E(Y).
However, I get some problems evaluating E(X*E(Y|X)).
Any hint would be appreciated.
 A: We know: 
\begin{align}Cov(X,Y) = E(XY) - E(X)E(Y)\end{align}
Thus, 
\begin{align}Cov(X,E[Y|X]) = E[X \cdot E(Y|X)] - E[X]E[E(Y|X)]\end{align}
As such, to solve the problem, we need to show that : 
\begin{align}E[X\cdot E(Y|X)]= E[XY]\end{align} 
as well as:
\begin{align}E[E(Y|X)] = E[Y]\end{align}
We want to prove for any function $r: S\rightarrow\mathbb{R}$:
\begin{align} E[r(X) \cdot E(Y|X)] = E[r(X)Y]\end{align}
Proof:
\begin{align}
E[r(X)E(Y|X)] &= \int_S r(x)E(Y|X=x)g(x)dx \\
&= \int_S r(x) \left[\int_T y\cdot h(y|x)dy\right]g(x)dx\\
&=\int_S \int_T r(x)\cdot y\cdot h(y|x)\cdot g(x) dydx\\
&= \int_{S\times T} r(x)\cdot y\cdot f(x,y) d(x,y)\\
&= E[r(X)Y] \text{, as wanted.}
\end{align}
Additionally, this implies that $E[E(Y|X)] = E[Y]$, by letting $r(x)=1$.
A: I believe you could say this:
Given $X$, $X$ is constant and $E[\alpha\cdot Y] = \alpha\cdot E[Y]$ therefore $X\cdot E[Y|X] = E[X\cdot Y|X]$
$\implies E[X\cdot E[Y|X]] = E[E[X\cdot Y|X]] = E[X\cdot Y]$ which will get you to the result you want. I'm sure about the last step though...
