# Distribution of Product of Random Variables with one being the normal distribution.

Let $X$ and $Z$ be independent, with $X\sim N(0,1)$, and with $\textbf{P}(Z=1)=\textbf{P}(Z=-1)=\frac{1}{2}$. Let $Y=XZ$ (i.e., $Y$ is the product of $X$ and $Z$).

(a) Prove that $Y\sim N(0,1)$.

(b) Prove that $\textbf{P}(|X|=|Y|)=1$.

(c) Prove that $X$ and $Y$ are not independent.

(d) Prove that $\textbf{Cov}(X,Y)=0$.

(e) It is sometimes claimed that if $X$ and $Y$ are normally distributed random variables with $\textbf{Cov}(X,Y)=0$, then $X$ and $Y$ must be independent. Is that claim correct?

My Work: (a) Is really where I am a little stuck. In order to show that $Y\sim N(0,1)$, I want to show that they have the same distribution laws. Thus, I want to show that $\mathcal{L}(Y)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$ given that $\mathcal{L}(X)=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\textbf{1}_B(t)\lambda(dt)$.

However, I don't know how to go about that. Another approach I have considered is to show that for any Borel measurable function $f:\mathbb{R}\to\mathbb{R}$,$\space$ $\textbf{E}(f(X))=\textbf{E}(f(Y))$ for which each is well defined. Which approach do you think is better and How should I start. Any help will be appreciated.

## 2 Answers

\begin{align} P(Y\le y)&=P(XZ\le y)\\ &=0.5P(X\le y)+0.5P(-X\le y)\\ &=P(X\le y)\text{ (using distribution's symmetry)} \end{align}

Part b: \begin{align} \mathbb{P}(|X|=|Y|) &= \mathbb{P}(X=Y\ {\rm or}\ X=-Y) = \mathbb{P}(X=Y \cup X=-Y) = \mathbb{P}(X=XZ \cup X=-XZ) \\ & \hspace{5mm} = \mathbb{P}(X=XZ) + \mathbb{P}(X=-Y)-\mathbb{P}(X=XZ \cap X=-XZ) \\ & \hspace{10mm} = \mathbb{P}(Z=1)+\mathbb{P}(Z=-1)+0=\frac{1}{2}+\frac{1}{2} = 1 \end{align}.