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.


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

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    $\begingroup$ You can find answers in questions about ordinary least squares regression and conditional Normal distributions. Here's a focused search with many useful hits: stats.stackexchange.com/… The duplicate is one of those hits: it's a generalization of your question with a directly applicable answer. $\endgroup$ – whuber Dec 7 '20 at 17:31
  • $\begingroup$ Thank you, there it was. $\endgroup$ – Jesper for President Dec 7 '20 at 17:38
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    $\begingroup$ The expectation of $X$ given that $Y$ is doubly truncated is not the same as the expectation of $X | X>Y$. $\endgroup$ – wolfies Dec 7 '20 at 17:49
  • $\begingroup$ On second thought I guess you are right @wolfies. $\endgroup$ – Jesper for President Dec 7 '20 at 18:27
  • $\begingroup$ @Wolfies Apply the duplicate to the case $X=X_1$ and $Y=X_1-X_2$ truncated at $0$ and $\infty.$ That is, all singly truncated random variables are special cases of double truncation where one limit is $\pm \infty.$ $\endgroup$ – whuber Dec 7 '20 at 22:42