Let $ X $ be exponentially distributed with parameter $\lambda > 0$ .
$a)$ Find the probability density function for $ Y := exp(X) $.
$b)$ Consider $X$ again. Now let $\varepsilon$ be an independent random variable with $P(\varepsilon = 1 ) = \frac{1}{2} $ and $P(\varepsilon = -1 ) = \frac{1}{2} $. Find the probability density function for $\varepsilon \cdot X$.
So a) was no big deal. but I'm stucked in $b)$.
Remark(Update) : I read that $ \varepsilon \cdot X $ has to be $Laplace(0,\frac{1}{\lambda})$ distributed.
We want to compute CDF of $YX$. [ $Y := \varepsilon$ ]. Let us say that $ a \ge 0 $
We start with $ F(a) = P(YX \le a) = P(X > -a | Y = -1)P( Y=-1) + P(X \le a |Y =1)P(Y=1) = P(X > -a)P( Y=-1) + P(X \le a)P(Y=1) = \frac{1}{2} + (1-e^{-\lambda \cdot a}) \cdot \frac{1}{2}. $
Then we take $F'(a)$ and get that $f(a)$ is $\frac{1}{2} \cdot \lambda \cdot e^{-\lambda \cdot a} $. So $YX$ is $Lap(0,\frac{1}{\lambda})$.
Thank you for your help.