PDF of $Y = WX$ There are two independent random variables, $X \sim \mathcal{N}(0, 1)$ and $W$ whose PMF is given by
$$
P(W = w) = \begin{cases} \frac{1}{2} \hspace{3mm} \text{if} \hspace{3mm} w = \pm1 \\
0 \hspace{3mm} \text{otherwise}.
\end{cases}
$$
A third random variable is defined as $Y = WX$. I want to find the density of $Y$.
\begin{align}
P(Y \leq y) &= P(WX \leq y)\\ &= P(X \leq \frac{y}{W}) \\&= \sum_{w \in \{1, -1\}}P(X \leq \frac{y}{w})P(W = w)\\ &= \frac{1}{2}\int_{-\infty}^{-y}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx + \frac{1}{2}\int_{-\infty}^{y}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx
\end{align}
When I differentiate the CDF to get PDF of $Y$, both the terms cancel out due to sign of $y$ in one of the integrals. I know that $Y \sim \mathcal{N}(0, 1)$. What am I doing wrong? Thanks.
 A: You are on the right track. The mistake is $X\leq \frac{y}{W}$. Since, W can take both values +1, -1 you cannot take it directly to the denominator without changing inequality.
A: $P(WX \le y) \ne P(X \le y/W)$ because dividing by $W$ when $W$ is negative will change the direction of the inequality.
A: [EDIT]
$$\eqalign{P(Y \le y) = P(WX \le y) &= P(W = 1, \; X \le y) + P(W = -1,\; -X \le y)\cr &= P(W = 1, \; X \le y) + P(W = -1, \; X \ge -y)\cr
&= \frac{1}{2 \sqrt{2\pi}} \int_{-\infty}^y e^{-x^2/2}\; dx + 
 \frac{1}{2 \sqrt{2\pi}} \int_{-y}^\infty e^{-x^2/2}\; dx}$$
Applying the change of variables $x \to -x$, you should get $$ \int_{-y}^{\infty} e^{-x^2/2}\; dx = \int_{-\infty}^y e^{-x^2/2}\; dx $$
so the CDF of $Y$ is exactly the CDF of the standard normal distribution.
A: In this cases I think it is better to solve the exercise using the Total Probability Theorem. Just for another example find the density of
$Y=X+W$
And note that
$(X+W|W=-1)\sim N(-1;1)$
and
$(X+W|W=1)\sim N(1;1)$
Thus
$f_Y(y)=\frac{1}{2}f_1+\frac{1}{2}f_2$
Where $f_1$ and $f_2$ are the densities of  the two conditional gaussian rv's
