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I've found questions very similar to the following. But I haven't found any that involve something of the form $X|Z|$.

Let $ Z\sim \mathcal{N}(0,1) $ and $X$ be the discrete random variable such that $$ \Pr(X=-1) = \Pr(X=1) = \frac12. $$

Assume that $X$ and $Z$ are independent.

Now let $Y = X|Z|$. Indeed $Y$ turns out to also be standard normal - this was successfully proven.

The next part is to prove (for fewer marks than the first part) that $(Y,Z)$ is not a bivariate normal vector. I have my own solution for this, presented below. However, I was wondering if there would be any easier approach than the one I have taken? It feels very convoluted (and potentially incorrect...), and I believe it requires way more marks than the amount allocated for this part.


Current solution: With probability 1, \begin{align*} Y+Z &= X|Z| + Z\\ &= \begin{cases} Z(1-X), & Z < 0,\\ Z(1+X), & Z \geq 0. \end{cases} \end{align*} But also with probability 1, $$ Z (1-X) = \begin{cases} 0, & X=1,\\ 2Z, & X=-1. \end{cases} $$ Hence by the law of total probability, or rather considering only one term from it, \begin{align*} \Pr(Y+Z = 0) &\geq \Pr(Y+Z=0 \mid Z<0, X=1) \Pr(Z<0 \mid X=1) \Pr(X = 1) \\ &= 1 \times \frac12 \times \frac12 \tag{$X$, $Z$ indep.}\\ &= \frac14\\ &> 0. \end{align*} So the linear combination $Y+Z$ is certainly not univariate normal, and hence $(Y,Z)$ is not bivariate normal.

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Just consider the conditional distribution of $Y|Z$. This can be characterized as $P(Y = z|Z = z) = 0.5$ and $P(Y = -z|Z = z) = 0.5$. This is no normal distribution and therefore the distribution of $(Y, Z)$ is not bivariate normal (because for a multivariate normal distribution the conditional distributions are also normal).

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