# Another “not bivariate normally distributed” question

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.

## 1 Answer

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).