The question is
Let U and V be two independent stochastic variables with $U\sim e(1)$ and with $P(V=1)=P(V=-1)=1/2$
Find the density function of W=UV
First i want to find the distribution funktion as follows
$F_W(w)=P(W\leq w)=P(V\cdot U \leq w)=P\bigg(\{ V=1\}\cap\{U \leq w\}\bigg)+P\bigg(\{V=-1\}\cap \{U\leq -w\} \bigg) $
And this is where I´m stuck, how do i find $F_W(w)$ for $w\geq 0$ and $w<0$
Edit: Figured it out
If $w\geq 0$
$F_W(w)=P(U\cdot V \leq w)=P\bigg(\{ V=1\}\cap\{U \leq w\}\bigg)+P\bigg(\{ V=-1\}\cap\{-U \leq w\}\bigg) =\frac{1}{2}(1-e^{-w})+\frac{1}{2}$
If $w<0$
$F_W(w)=P(U\cdot V \leq w)=P\bigg(\{ V=-1\}\cap\{U \geq -w\}\bigg) +P\bigg(\{ V=1\}\cap\{U \leq w\}\bigg)= \frac{1}{2}(1-P(U\leq -w))+\frac{1}{2}(P(U\leq w))=\frac{1}{2}(1-1+e^w)+0=\frac{1}{2}e^{w}$
so $F_W(w)=\begin{cases}\frac{1}{2}e^{-w} & \ \ w<0 \\ \frac{1}{2}(1-e^{-w})+\frac{1}{2} & \ \ w\geq 0 \end{cases}$
This leads to $f_X(x)=\frac{1}{2}e^{-|w|}$ $\forall w$