Product of a Continuous and Discrete Distribution

Question

Let $X \sim \mathcal{N}(0, 1)$ and $Y$ be a random variable independent of $X$ such that \begin{align*} P(Y=y) = \begin{cases} \frac{1}{2} & y = -1\\ \frac{1}{2} & y = 1\\ 0 & otherwise \end{cases} \end{align*} If $ Z = XY$, show that $Z \sim \mathcal{N}(0,1)$.

Here is what I came up with: \begin{align*} P(Z<z) &= P(XY < z)\\ &=P(X < \frac{z}{Y})\\ &= P(X < \frac{z}{Y} \mid Y = -1)P(Y=-1) + P(X < \frac{z}{Y} \mid Y=1) P(Y=1)\\ &= \ldots\,?\\ &= P(X < z) \end{align*}

Unfortunately, I'm not sure what to do, or whether I am even on the right track. I'd greatly appreciate any pointers!

Answer 1

You need to use the symmetry of the normal distribution about the Y axis.

If $Y=-1$, $Z=-X\sim -N(0,1)$. Due to the symmetry of $N(0,1)$, that means $X\sim N(0,1)$. If $Y=1$, $Z=X\sim N(0,1)$. Since the probability of either one is evenly split, it means $Z\sim N(0,1)$. If you want to do this with the conditional probability equations in your answer, $$ \begin{align} P(Z<z) &=P(XY <z)\\ &=P(X<z|Y=1)P(Y=1)+P(-X<z|Y=-1)P(Y=-1)\\ &=1/2(P(X<z)+P(X>-z)) \end{align} $$ By the symmetry of the normal distribution, $$P(X>-z)=1/2\pi\int_{-z}^\infty e^{-x^2}dx=1/2\pi\int_{-\infty}^{z}e^{-x^2}dx=P(X<z),$$ and so $$ P(Z<z)=P(X<z). $$

Answer 2

Your mistake: you can't divide by $Y$ without knowing its sign (if it's negative, the inequality would reverse).

\begin{align*} P(XY<z) &= P(XY<z \mid Y=1) P(Y=1) + P(XY<z \mid Y=-1) P(Y=-1)\\ &=\frac{1}{2} P(X<z) + \frac{1}{2} P(-X<z)\\ &= \frac{1}{2} P(X<z) + \frac{1}{2} P(X>-z)\\ &= P(X<z). \end{align*} The last step is due to symmetry: $P(X<z)=P(X>-z)$