Let $X\sim N(\mu,\Sigma)$, $t\in\mathbb{R}$, and $a$ be a non-zero vector of the same dimension as $X$. Define a random vector $Y=X\mathbb{1}(a^\top X\ge t)$, where $\mathbb{1}$ denotes the indicator function. Is there a clean formula for the covariance matrix $Cov(Y)$?

  • $\begingroup$ @PeterLeopold Modified. I am looking for a neat answer to this question other than brute-force integration. $\endgroup$ – O. Richard Apr 10 '19 at 15:58
  • $\begingroup$ The vector of interest is $a-\mu$, which can be written in the $\Sigma$ basis. Then the question becomes whether $X$ will have an expected projection onto $a-\mu$ that is greater than the same threshold $t$. I think you can marginalize onto $a$, and the expected projection should be available in terms of $\Phi$, the cdf of a one dimensional normal distribution. Is that the kind of clean formula you are looking for? My thinking is based on the probability that that projection is greater than t. The covariance of the vector $Y$ by itself doesn't mean much to me. $\endgroup$ – Peter Leopold Apr 10 '19 at 16:22
  • $\begingroup$ @PeterLeopold Not sure why $a-\mu$ plays an important role here. Note that if $Y$ is defined instead as $Y=X\mathbb{1}(a^\top X=t)$, then there is a formula using the conditional distribution of normal. $\endgroup$ – O. Richard Apr 10 '19 at 16:29
  • $\begingroup$ Consider $|a|$ small & $|\mu|$ large. Then $P(a^TX>t)\approx 0.5)$ since $X$ is almost certainly vector long compared to $a$ & half of the time it will point in the half-space defined by $a$'s direction. However if $|a|>>|\mu|$, then $P(a^TX>t)\approx 0$ since $X$'s projection onto $a$ may never be greater than $t$ even in the best scenario. We can choose $t$ large enough in the 1st case or small enough in the 2nd case to change the results, but that isn't' interesting. Really we either commit to $\mu=0$ or failing that, transform the entire space (including $a$) by $-\mu$ to accomplish this. $\endgroup$ – Peter Leopold Apr 10 '19 at 16:38

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