PDF of sum of two random variables $Z = X + Y$ where $0 < x < y < \infty$ I have two random variables $X$ and $Y$ with joint PDF
$$
f_{x,y}(x,y) = e^{-y} ; 0<x<y<\infty
$$
and 0 otherwise.
Given $Z=X+Y$, I need to find the PDF of $Z$. This is what I did, but my end result makes no sense; I think it might have to do with the bounds of $X$ and $Y$ but I'm having a lot of trouble wrapping my head around the way $0<x<y<\infty$ interacts with $Z=X+Y$.
We know that $F_z(z)=P(Z\leq z)=P(x \leq z-y)$
From this we get
$$
F_z(z)=\int_{y=0}^\infty\int_{x=0}^{z-y}f_{x,y}(x,y)dxdy
$$
Then, differentiating with respect to z, first we need to move the derivative inside the first integral and we get
$$
\frac{\partial}{\partial z} F_z(z)=\int_{y=0}^\infty \frac{\partial}{\partial z}\int_{x=0}^{z-y}f_{x,y}(x,y)dxdy
$$
since the bounds of the outer integral are not functions of z. Then we do it again and get
$$
\int_{y=0}^\infty \frac{\partial}{\partial z}\int_{x=0}^{z-y}f_{x,y}(x,y)dxdy=\int_{y=0}^\infty(1\cdot f_{x,y}(z-y,y)-0+\int_{x=0}^{z-y}\frac{\partial}{\partial z} f_{x,y}(x,y)dx)dy
$$
But since $f_{x,y}(x,y)$ is not a function of $z$, the inner integral disappears and we're left with
$$
f_z(z)=\int_0^\infty f_{x,y}(z-y,y)dy=\int_{0}^\infty e^{-y}dy=1
$$
since $f_{x,y}$ only depends on $y$, but this answer clearly doesn't make sense - how could the PDF of $Z$ just be 1? I'm really having a lot of trouble understanding this, so any help would be greatly appreciated.
 A: 
Please see the diagram. The way you have set up, you will have to split your integral into two. Instead I would suggest,
$F_z(z)=P(Z\leq z)=P(y \leq z-x)$
$F_Z (z) = \displaystyle \int_0^{z/2}\int_x^{z-x} f(x,y) ~ dy ~ dx = e^{-z} (e^{z/2} - 1)^2$
So, $f_Z(z) = e^{-z/2} - e^{-z}, 0 \lt z \lt \infty$
A: You forgot the $1_{0<x<y}$ in the density $f_{x,y}$.  You should have instead:
\begin{align*}
f_Z(z)&=\int_{-\infty}^\infty f_{x,y}(z-y,y)\,\mathrm{d}y\\
&=\int_{-\infty}^\infty 1_{0<z-y<y}(y) e^{-y}\,\mathrm{d}y\\
&=\int_{-\infty}^\infty 1_{y<z}(y)1_{2y>z}(y) e^{-y}\,\mathrm{d}y\\
&=\int_{z/2}^z e^{-y}\,\mathrm{d}y\\
&=e^{-z/2}-e^z.
\end{align*}
A: First let us use the transformation $U=X+Y$. and $V=Y$.
So the Jacobian is $\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}
1 & 1 \\
1 & 0 
\end{vmatrix}$
So the Joint distribution of $U,V$ is
$$f_{U,V}(u,v) = e^{-v}$$
Now to find the range of $u$ and $v$.
We see that $0<x<y$
So $0<x\implies 0<u-v\implies v<u$.
And we have $x<y\implies u-v<v\implies \frac{u}{2}<v$.
So you see that $v$ has to lie between $\frac{u}{2}<v<u$.
So the joint density is $$f_{U,V}(u,v)= e^{-v}\,\, \\0<u<\infty\,\, ,\frac{u}{2}<v<u $$
The graph is this. $f$ is non zero only in the shaded portion.

So we integrate $f_{U,V}(u,v)$ over $v$ to get the distribution for $u$.
We get $$f_{U}(u) = \int_{\frac{u}{2}}^{u}e^{-v}dv=e^{-\frac{u}{2}}-e^{-u}$$
So the pdf of $U=X+Y$ is
$$f_{U}(u) = e^{-\frac{u}{2}}-e^{-u}\,\,,u>0$$ and $0$ elsewhere
