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

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?

## 1 Answer

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 $$\sigma_X^2$$.