What is the expectation
$$\mathbb E[X_1 \lvert X_1 > X_2]$$
assuming that
$$(X_1,X_2) \sim \mathcal MVN(0,\Sigma),$$
with $\mathcal{MVN}$ being the multivarite normal.
I would expect this to have an answer somewhere, but I cannot seem to find the duplicate. The closest I came was this question.
ANSWER
To arrive at the answer using the question that has be referred to as a duplicate a little work is required. I will sligthly generalize the answer because that is the result I need so instead of solving for $\mathbb E[X_1 \lvert X_1 >X_2]$, I will solve for
$$\mathbb E[X_1 \lvert X_1 >X_2 + u] $$
First note that
$$\mathbb E[X_1 \lvert X_1 >X_2 + u] = \mathbb E[X_1 \lvert U> u],$$
with $U:=X_1-X_2$. From properties of the multivariate normal it follows that
$$\begin{bmatrix}1&0\\1 & -1 \end{bmatrix}\begin{bmatrix}X_1\\X_2 \end{bmatrix} = \begin{bmatrix}X_1\\U \end{bmatrix} \sim \mathcal{MVN}(0,D\Sigma D^\top)$$
where $D$ is matrix premultiplied on $(X_1,X_2)^\top$.
From the duplicate (the equations given in the question of the duplicate and taken from wikipedia, not the answer itself) it follows that
$$\mathbb E[X_1 \lvert U > u] = \mu_{X_1} + \frac{\sigma_{X_1,U}}{\sigma_U} \left( \frac{\phi((u-\mu_u)/\sigma_u)}{1- \Phi\left((u-\mu_u)/\sigma_U \right)} \right) = \frac{\sigma_{X_1,U}}{\sigma_U} \left( \frac{\phi(u/\sigma_U)}{1- \Phi\left(u/\sigma_U \right)} \right),$$ where the last identity follows from $\mathbb E[U] =0$ and $\mathbb E[X_1] = 0$. Furthermore, $$\sigma_{X_1,U} = \sigma^2_{X_1}-\sigma_{X_1,X_2},$$
implying that
$$\frac{\sigma_{X_1,U}}{\sigma_U} = \frac{\sigma_{X_1}\sigma_{X_2}}{\sigma_U}\left( \frac{\sigma_{X_1}}{\sigma_{X_2}}-\rho_{X_1,X_2} \right).$$
Another related post is this post on Roy models
