Thank you ... $\displaystyle \frac{1}{2}$ indeed.

Further related question:

I'm told:

"If $\displaystyle Y=g(X,Z)$ holds and $\displaystyle g(x,z)$ is a "well behaved" function, monotonous for each $\displaystyle x$ in $\displaystyle z$ (

*in* $\displaystyle z$?!),

then $\displaystyle p_Z(z)=\int_{-\infty}^{\infty}p_{X,Y}\big(x,g(x,z)\big)\left|\fr ac{\partial g}{\partial z}(x,z)\right|dx$."

How do I find what function $\displaystyle g$ is and then use the given formula?

The solution given is $\displaystyle e^{-z/2}-e^{-z} \text{ ; }z>0$.