Deriving the conditional distribution of a multivariate normal, for inequalities

Question

This question is slightly related to Deriving the conditional distributions of a multivariate normal distribution. In that question, the following situation was given.

If $Y$ follows a multivariate normal distribution, $Y \sim N(\mu,\Sigma)$ and you partition $Y$ into $Y = [y_1,y_2]$, how can you derive the resulting conditional distribution of $y_2|y_1=a$? In words, if you start from a multivariate normal distribution, and you fix some of the values($y_1=a$), what is the conditional distribution of the remaining elements of $Y$?

My question is, how can you arrive at the conditional distribution $y_2|y_1<a$, again for $Y \sim N(\mu,\Sigma)$ and $Y = [y_1,y_2]$? It seems to me that the approach used in the case $y_2|y_1=a$ does not work here.

Answer 1

We find $f_{Y_1,Y_2}(y_1|y_2)$, which is normal, with mean as a function of $y_2$, $\mu_{y_1|y_2}(y_2)$ and deviation, $\sigma$, as shown in the wiki page. Then, we can calculate $P(Y_1<a|Y_2=y_2)=\Phi\left(\frac{a-\mu(y_2)}{\sigma}\right)=\phi(y_2)$. Using Bayes Rule, we'd have $$f_{Y_2|Y_1<a}(y_2)=\frac{P(Y_1<a|Y_2=y_2)f_{Y_2}(y_2)}{P(Y_1<a)}\propto \phi(y_2)f_{Y_2}(y_2)$$ which isn't normal any more (in general) due to the term $\phi(y_2)$. This was for the bivariate case (your variables seem to be), but can be generalized into multivariate similarly if you like. However, $y_1$ is always univariate since you compare it with a constant.