TITLE: Conditional expectation of X given X is greater than Y
QUESTION [5 upvotes]: I wonder if someone can help me with this question:
"Let $X,Y \sim \mathcal{N}(0,1)$ where X and Y are independent. Find $E[X | X>Y]$."
I know that $E[X|Y=y]= \int_{-\infty}^{\infty} x f_{X|Y}(x,y)dx$, but in this case one is conditioning on an inequality, and I don't know how to treat that.
Thanks in advance!

REPLY [2 votes]: Note that
\begin{align*}
E(X \mid X > Y) = \frac{E(X\mathbb{1}_{X>Y})}{P(X>Y)}= 2E(X\mathbb{1}_{X>Y}).
\end{align*}
Moreover,
\begin{align*}
E(X\mathbb{1}_{X>Y}) &= E\big(E(X\mathbb{1}_{X>Y} \mid X) \big)\\
&=E(X\Phi(X))\\
&=\int_{-\infty}^{\infty}x\Phi(x)\phi(x) dx,
\end{align*}
where $\Phi$ is the cumulative distribution of a standard normal and $\phi$ is the density.