How to find $E[X|X>Y]$

Question

Suppose $X$ and $Y$ are independent standard normal variable. I want to find $E[X|X>Y]$. I calculated that $$ f_{X>Y}(x) = 2\Phi(x)\phi(x)$$ However I couldn't find the expected value using this conditional probability. How can I evaluate the integral?

Answer 1

Here is one way to look at the problem. Let $\alpha = P(X > Y)$ (By symmetry, it should be clear what $\alpha$ is.), then $$ 0 = E[X] = \underbrace{E[X|X > Y]}_{=:x} \alpha + \underbrace{E[X| X < Y]}_{=:y} (1-\alpha) $$ By symmetry $y:= E[X| X < Y] = E[Y | Y < X]$. Now, by linearity of expectation, $$ E[X-Y|X > Y] = x - y $$ But $X-Y \sim N(0,2)$ and if we let $Z \sim N(0,1)$, then $E[X-Y|X > Y] = \sqrt{2} E[Z | Z > 0]$. You should be able to easily compute $E[Z|Z>0]$, from which you get $x-y$, and then $x$ by examining the above equalities.