In Gluts's "Intermediate Probability" one question says "The random variables $X$ and $Y$ are independent and $N(0,1)$-distributed. Determine $E(X|X>Y)$."
So it seems like the conditional distribution of both $X$ and $Y$ ought to be: $$f_{X,Y|X>Y}(x,y)=\frac{f_X(x)f_Y(y)}{1-F(Y)},-\infty<y<x<\infty$$ And then I can integrate $y$ out to get $f_{X|X>Y}$: $$f_{X|X>Y}=\int_{-\infty}^x\frac{f_X(x)f_Y(y)}{1-F(Y)}dy=f_X(x)\ln(1-F(x)),-\infty<x<\infty$$ Which means: $$E(X|X>Y)=\int_{-\infty}^\infty xf_X(x)\ln(1-F(X))dx$$ And I have no idea how to solve this integral. Is there another way to approach the problem?
$$ \mathbb{E}[X|X>Y]=\frac{\mathbb{E}[X1\{X>Y\}]}{\mathbb{P}\{X>Y\}}, $$ where $$ \mathbb{E}[X1\{X>Y\}]=\mathbb{E}[\mathbb{E}[X1\{X>Y\}\mid Y]] \\ =\int_{-\infty}^{\infty}\left(\int_y^\infty x\phi(x)dx\right)\phi(y)dy \\ =\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\right)\phi(y)dy=\frac{1}{2\sqrt{\pi}} $$
and $\mathbb{P}\{X>Y\}=\frac{1}{2}$.