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 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 $\sigma_X^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.