TITLE: Disprove independence of vector of Gaussians by independence of marginals
QUESTION [5 upvotes]: If we have three random variables $X,Y,Z$, then if $X$ and $Z$ are independent, and $Y$ and $Z$ are independent, it doesn't follow that $Z$ is independent of the vector $(X,Y)$.
There is a simple counter example for this. But, I can't find a counter example in the case where all three are Normal, ie $X,Y,Z$ are Gaussian variables, and not multivariate Gaussian.
Appreciate any help.

REPLY [3 votes]: Example: Let $X$ and $Z$ be iid standard Gaussian (zero mean) variables. And let 
$$
Y =
\left\{
 \begin{array}{ll}
  X  & \mbox{if } |Z| \leq 1 \\
  -X & \mbox{elsewhere} 
 \end{array}
\right.
$$
Then, $Y$ is marginally gaussian, and independent of $Z$: $P(Z|Y) = P(Z)$, as $P(Z|X)=P(Z)$. However, $P(Z|X Y) \ne P(Z)$, because knowing $X$ and $Y$ we know whether $|Z|<1$ or not.