Expectation of a conditional variance I have to prove this $$E(Var(Y|X))=(1-\rho^2)Var(Y)$$ but I got stuck and don't know how to continue.
This is what I've done so far based on this variance formula $Var(Y)=E(Var(Y|X))+ Var(E(Y|X))$
$$Var(Y|X)= E(Y^2|X)- (E(Y|X))^2$$
$$=Var(Y)- Var(E(Y|X))$$
$$=Cov(Y,Y)- E(E(Y|X))^2- E(Y)^2$$
I get to the part when I relate the covariance in order to get the correlation coefficient, but I don't know what to do from there, or maybe what I've done is wrong, so I'll be grateful if any of you can help me out with this.
 A: Let's work from your decomposition formula (the "law of total variance"): $$\mathrm{Var}(Y) = \mathbb{E}(\mathrm{Var}(Y|X)) + \mathrm{Var}(\mathbb{E}(Y|X))$$
Rearranging this, you get
$$\mathrm{Var}(Y)\bigg(1 - \frac{\mathrm{Var}(\mathbb{E}(Y|X))}{\mathrm{Var}(Y)}\bigg) = \mathbb{E}(\mathrm{Var}(Y|X)).$$
So you really just need to show that $\mathrm{Var}(\mathbb{E}(Y|X))\big/\mathrm{Var}(Y) = \rho^2$. Try working this out from here.
A: The identity is false.  I generated a random distribution on $\{\, 0, 1, 2\,\}^2$ to get a case where it is not satisfied. However, we do have the inequality $E[\operatorname{Var} (Y\mid X)] \leqslant (1 -\rho^2) \operatorname{Var}(Y)$.
Since $E[\operatorname{Var}(Y\mid X)] = \operatorname{Var}(Y) -\operatorname{Var}(E[Y\mid X])$, it would suffice to show that
$$
\operatorname{Var}(E[Y\mid X])\geqslant \rho^2 \operatorname{Var}(Y).
$$
This inequality can be rewritten as $\operatorname{Cov}(X, Y)^2\leqslant \operatorname{Var}(X) \operatorname{Var}(E[Y\mid X])$, which follows from Cauchy-Schwarz and the identity $\operatorname{Cov}(X, Y) = \operatorname{Cov}(X, E[Y\mid X])$.
A: We consider the case where $X$ and $Y$ are jointly normal with correlation $\rho$ $(|\rho|<1)$. Let
\begin{align*}
Z = \frac{\frac{Y-E(Y)}{\sqrt{Var(Y)}}-\rho \frac{X-E(X)}{\sqrt{Var(X)}}}{\sqrt{1-\rho^2}}.
\end{align*}
Then $Z$ is also normal and independent of $X-E(X)$, as they are jointly normal with zero correlation. Since 
\begin{align*}
Y=\sqrt{Var(Y)}\left(\sqrt{1-\rho^2} Z + \rho \frac{X-E(X)}{\sqrt{Var(X)}}\right) + E(Y),
\end{align*}
we conclude that
\begin{align*}
E(Y\mid X) = \rho \sqrt{Var(Y)}\frac{X-E(X)}{\sqrt{Var(X)}} + E(Y).
\end{align*}
Therefore,
\begin{align*}
Var\big(E(Y\mid X)\big) = \rho^2\, Var(X)\,Var(Y),
\end{align*}
and, consequently,
\begin{align*}
E\big(Var(Y\mid X)\big)=\left(1-\rho^2\right)Var(Y).
\end{align*}
A: As for the useful identity aforementioned by Herschkorn, namely
$$
\operatorname{Cov}(X, Y) = \operatorname{Cov}(X, E[Y\mid X]),
$$
it follows from the definition of the covariance, by means of the law of total expectation:
\begin{align}
\operatorname{Cov}(X, Y)
&= E[XY] - E[X]\, E[Y]\\
&= E[E[XY\mid X]] - E[X]\cdot E[E[Y\mid X]]\\
&= E[X\cdot E[Y\mid X]] - E[X]\cdot E[E[Y\mid X]]\\
&= \operatorname{Cov}(X, E[Y\mid X]).
\end{align}
