Suppose I have two continuous random variables $X$ and $Z$, $X$ and $Z$ are not independent. While I know that $Z \sim \mathcal{N}(0, 1)$, I have no information for the distribution of $X$. Now one of my friend said that the conditional distribution $p(Z|X)$ should also be a Gaussian. I am not sure if it's correct or not?

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    $\begingroup$ No this is not correct. I do not know what you mean by $p(Z|X)$, but likely you mean a density $f_{Z|X}(z|x)$ (or perhaps you mean $P[Z\leq z|X=x]$). Anyway, just take $Z$ Gaussian $N(0,1)$ and define $X$ to be uniform over $[0,1]$ if $Z>0$, and $X$ uniform over $[-2,-1]$ if $Z\leq 0$. Then $f_{Z|X}(z|1/2)$ is not a Gaussian PDF since we know it is zero for all $z<0$ (since $X=1/2\implies Z>0$). $\endgroup$ – Michael Apr 11 at 22:27
  • $\begingroup$ I mean the PDF function of $Z$ given $X$. $\endgroup$ – lenhhoxung Apr 11 at 22:56

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