Find distribution of $Z=\frac{X+Y}{2}$ given $f_{X,Y}(x,y)=e^{-(x+y)}$ Excercise
Let $X, Y$ be random variables such that their joint density function is defined by:
$f_{X,Y}(x,y)=e^{-(x+y)},  \enspace x,y>0$.
Find the distribution of Z defined as:
$Z=\frac{X+Y}{2}$.

Attempt of solution
I am using the fact that:
$F_Z(z)=P\{\frac{X+Y}{2}<z\}$, which implies that:
$F_Z(z)=\int_{0}^{\infty}\int_{0}^{2z-y}e^{-(x+y)}dxdy$.
My problem is that this leads to:
$F_Z(z)=\int_{0}^{\infty}e^{-y}-e^{-2z}dy$,
and this integral does not converge.

What am I getting wrong here? Thanks in advance.

Note that marginal densities of $X$ and $Y$ are $f_X(x)=e^{-x}$ and $f_Y(y)=e^{-y}$. 
Then, $Z=X+Y$ has density according to the convolution:
$$f_Z(z)=\int_0^\infty f_X(x)f_Y(z-x)dx = \int_0^z e^{-x}e^{-(z-x)}dx=ze^{-z}.$$
Density of $W=Z/2$ is 
$$f_W(w)=f_Z(2w)|dz/dw| = 4we^{-2w}.$$
