Product of a exponential and discrete Distribution Let $U \sim e(1)$ and $V$ a discrete random variable independent of $U$ such that
$p_V(v)=1/2$ if $v \in \{-1,1\}$ and $p_V(v)=0$ otherwise.

Problem:
Let $W=UV$. Find the distribution function of $W$ $\forall w\in \mathbb{R}$.

My try:
\begin{align}
P(W\leq w)&amp;=P(UV\leq w)=P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1)\\
&amp;= P(U\leq w)\frac{1}{2}+P(U\geq -w)\frac{1}{2}\\
&amp;=P(U\leq w)\frac{1}{2}+(1-P(U\leq-w)\frac{1}{2}\\
&amp;=\left(\int_0^we^{-u} \, du\right)\frac{1}{2}+\left(1-\int_0^{-w} e^{-u} \, du \right) \frac{1}{2}\\
&amp;=(1-e^{-w})\frac{1}{2}+\frac{1}{2}e^w = F_W(w)
\end{align}

The correct answer in my book is the following
$$F_W(w)=\frac{1}{2}(1-e^{-w})+\frac{1}{2}, w&gt;0$$ and
$$F_W(w)=\frac{1}{2}e^{w}, w\leq 0$$

Can anyone tell me what i am doing wrong?
You wrote:
\begin{align}
&amp; P(W\leq w) = P(UV\leq w) \\[10pt]
= {} &amp;P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1).
\end{align}

What is $P(U\le w\mid V=1)$?

It is $\begin{cases} 1-e^{-w} &amp; \text{if } w\ge0, \\ 0 &amp; \text{if } w&lt;0. \end{cases}$

What is $P(-U\le w\mid V=-1)$?

It is $\begin{cases} e^w &amp; \text{if } w\le0, \\ 1 &amp; \text{if } w&gt;0. \end{cases}$

You need to consider the cases $w\ge0$ and $w&lt;0$ separately.

If you find the density function, you will be able to write it more simply by using the notation $|w|.$
Hint: An exponentially distributed random variable has positive support.