Basically the title. if $X$, $Y$ are independent $N(0,1)$ random variables, find the distribution of $$E(XY|X+2Y)$$
I attempted the following:
Since $\text{Cov}(X-\frac{1}{2}Y,X+2Y)=0$, they are independent. So \begin{align} E((X-\frac{1}{2}Y)(X+2Y)|X+2Y) &=(X+2Y)E(X-\frac{1}{2}Y|X+2Y)\\ &=(X+2Y)E(X-\frac{1}{2}Y)\\ &=(X+2Y)(0-0)=0\\ \end{align} then \begin{align} E(XY|X+2Y)&=\frac{2}{3}(E((X-\frac{1}{2}Y)(X+2Y)|X+2Y)-E(X^2-Y^2|X+2Y))\\ &=\frac{2}{3}(E(Y^2|X+2Y)-E(X^2|X+2Y)) \end{align} Now it is possible to find the densities of $E(X^2|X+2Y)$ and $E(Y^2|X+2Y)$, I could then convolve them to find an answer. But the computations involved are extremely tedious, and given that this is a question from the qualifying exam I assume there is an easier way?
Let $U = X+2Y$ and $V=X-\frac{1}{2}Y$. Then $Y=\frac{2}{5}(U-V)$ and $X=\frac{1}{5}(U+4V)$.
$U \sim N(0, 5)$ and $V \sim N(0, 5/4)$ are independent.
$$E[XY \mid X+2Y] = \frac{2}{25} E[(U-V)(U+4V) \mid U] = \frac{2}{25}(U^2 + 3UE[V] - 4 E[V^2]) = \frac{2}{25}(U^2 - 5).$$