Let $X$ and $Y$ be random variables with $p$ as correlation coefficient. Prove that: $$E(Var(Y|X))=(1-p^2)Var(Y)$$
So I've tried two different ways but with a dead end for me:
$Var(Y)(1-p^2)= \left(1-\left(\frac{Cov(X,Y)}{\sqrt{Var(X)}\sqrt{Var(Y)}}\right)^2\right)Var(Y)= Var(Y)-\frac{(Cov(X,Y))^2}{Var(X)}$
$E(Var(Y|X))=E(E(Y^2|X)-(E(Y|X))^2)=E(Y^2)-E((E(Y|X))^2)$
It seems that some more detail is missing about $X$ and $Y$ (e.g. if they are jointly normal, or as pointed out in the comments if a linear estimation of $E(Y|X)$ is sought). Under such assumption we can do as follows:
The law of total variance is:
$$Var(Y)=E(Var(Y|X))+Var(E(Y|X))$$ Thus $$E(Var(Y|X))=Var(Y)-Var(E(Y|X))$$ Now it suffices to prove that $Var(E(Y|X))=p^2Var(Y)$.
If $Y$ is normally distributed given $X$, then $E(Y|X)$ would be linear in $X$ and using regression equation is given by $$E(Y|X)=EX+p\frac{\sigma_Y}{\sigma_X}(X-EX)$$ which yields $$Var(E(Y|X))=p^2 \frac{Var(Y)}{Var(X)}Var(X)=p^2Var(Y)$$