# Let $Z=Z\cdot 1_S+Z\cdot 1_{S^c}$ where $Z$ is standard normal. What is the distribution of $Z$ give $Z\cdot 1_S$

Let $S \subset \mathbb{R}$ and let $Z$ be a standard normal random variable. Let \begin{align} Z_1&=Z \cdot 1_{S},\\ Z_2&=Z \cdot 1_{S^c}, \end{align} where $1_S$ is an indicator function and $S=[-a,a]$ for some $a>0$.

Can we find the distribution of $Z$ given $Z_1$ or $Z_2$?

This question was raised here.

Thanks.

Sure, you can find either. Let's say $Z_2.$ If we have $Z_2=0,$ this means that $Z$ is inside the range $(-a,a).$ Thus, conditional on $Z_2=0,$ $Z$ is distributed as a truncated Gaussian $$f(z|Z_2=0) = \frac{e^{-z^2/2}}{\int_{-a}^ae^{-x^2/2}dx}.$$ On the other hand, if $Z_2\ne 0,$ then it must be that $Z$ is outside of the range and we have $Z = Z_2.$ So, conditional on $Z_2=z\ne0,$ $Z$ is an atom at $z$ $$P(Z=z'|Z_2=z\ne0) = \delta_{z,z'}$$
It's worth noting that there's a potential issue if $Z=0,$ but that happens with probability zero, so there's no need to handle that case.