Independence of a Gaussian random variable and the product of another Gaussian random variable and a Bernoulli random variable

Question

Let $X$ and $Y$ be two independent Gaussian random variables with mean $0$ and variance $σ^2_X$ and $σ^2_Y$ respectively. Let $Z$ be a random variable measurable with respect to $σ(Y)$ and suppose that $Z$ assumes only value $1$ or $−1$. Show that $(ZX, Y)$ is a 2-dimensional Gaussian random vector and determine its variance and covariance matrix. Say also if $ZX$ and $Z$ are independent.

In an attempt to solve this, I first decided to find the distribution of $ZX$ by using the law of Total Probability. I assumed independence for $Z$ and $X$ since they come from different sigma algebras. As I seek the distribution of $ZX$, I asked my about the chance that $ZX \le t$ for some arbitrary real value $t$. To handle the discreteness of $Z$, consider enumerating its possible values:

$$\Pr[ZX \le t] = \Pr[ X \le t \text{ and }Z=1] + \Pr[X \le t \text{ and } Z=-1].$$ Now, can you please show me how to continue from here?

Answer 1

First, $Z$ and $X$ are independent since they come from independent sigma-algebras. Next, let $\Pr[Z=1]=p$. You missed minus sign in equality: $$ \Pr[ZX \le t] = \Pr[ X \le t \text{ and }Z=1] + \Pr[-X \le t \text{ and } Z=-1] $$ continue using independence $$ \Pr[ZX \le t] =p\Pr[X\le t]+(1-p)\Pr[-X\le t]. $$ Note that $X$ is centered Gaussian random variable and then $-X$ has the same distribution as $X$, so both probabilities above coincide: $\Pr[-X\le t]=\Pr[X\le t]$ for any $t$. Then $$ \Pr[ZX \le t] =\Pr[X\le t]\cdot(p+(1-p))=\Pr[X\le t]. $$ It follows that $ZX$ is Gaussian random variable with mean $0$ and variance $\sigma_X^2$.