Let $X\sim N(0,\sigma^2)$ and Z a Bernoulli random variable with
$\begin{equation}\qquad \Pr(Z=z) = \begin{cases}
p & \mbox{if $z = 1$}, \\
1-p & \mbox{if $z = 0$},\\
0 & \mbox{otherwise}.
\end{cases} \end{equation}$
I'm struggling to work out the distribution of $y=Z\cdot X$. It seems straightforward if $Z$ takes values 1 and -1 (The product of a normal and Bernoulli variables, independent from each other). But if I try the same approach, I get stuck at some stage with
$\Pr(Y\leq y) = \frac{1}{2} \Pr(1\cdot X \leq y) + \frac{1}{2} \Pr(0\cdot X \leq y)$.
Not sure how to go on from here. Any help is really appreciated.
Let $y \in (-\infty,0)$. Then:
$F_{Y}(y) = \mathbb P(ZX \le y) = \mathbb P(ZX \le y | Z = 1)\mathbb P(Z=1) + \mathbb P(ZX \le y | Z=0)\mathbb P(Z=0) =$
$ p\mathbb P(X \le y) + (1-p)\mathbb P(0 \le y) = p\mathbb P(X \le y) = p\int_{-\infty}^y \frac{1}{\sqrt{2}\pi\sigma}e^{-\frac{x^2}{2\sigma^2}}dx $
Similar for $y \in [0,+\infty)$, we get:
$F_Y(y) =p\mathbb P(X \le y) + (1-p)\mathbb P(0 \le y) = 1-p + p\mathbb P(X \le y) = 1-p +p\int_{-\infty}^y \frac{1}{\sqrt{2}\pi\sigma}e^{-\frac{x^2}{2\sigma^2}}dx $
So, when $\mu$ is the distribution of $\mathcal N(0,\sigma^2)$, then $\mu_{Y}$ is given by:
$\mu_Y(A) = (1-p)\cdot \delta_{0}(A) + p\cdot \mu(A)$