Joint density with restrain I can't solve this problem...tried several areas for limits of integration but solution just doesn't match with solution from end of textbook (Stirzaker:Probability and random variables).
Exercise 6.6.2
Let X and Y have joint density 
$$f(x,y)=e^{-y}$$ 
$$0<x<y<\infty$$
Find joint density for Z=X+Y.
I just can't figure it up and became really frustrated, especially being self-learner. I haven't write any of my trials because I only ask for someone to please help me with bounds of integration(I know formula for joint density etc.)
Many thanks!
 A: Assuming you mean
$$f(x,y)=e^{-y}$$ for the density and the problem is to verify that this is a distribution, the integral is 
$$f_{X,Y}(x,y)=\int_0^\infty\int_0^ye^{-y}dxdy$$
which after some calculus...
$$f_{X,Y}(x,y)=\int_0^\infty[e^{-y}x|_0^ydy$$
$$=\int_0^\infty ye^{-y}dy$$
which we can be tricky and write as 
$$=\frac{1}{e}\int_0^\infty ye^{-y+1}dy$$
$$=\frac{1}{e}\int_0^\infty \frac{d}{dy} e^{-y+1}dy$$
and using integral of derivative is derivative of integral
$$=\frac{1}{e}e^{-y+1}|_0^\infty$$
$$=\frac{e}{e}$$
EDIT:
To find the joint density $Z=X+Y$ then you can either apply the convolution formula or use standard transformation of variables, let's do the second.
Define $$U=X,V=X+Y$$
so $$g_1(X,Y)=X,g_2(Y)=X+Y$$
so we need to find the inverse transformation which is given by 
$$g_1^{-1}(U,V)=U,g_2^{-1}(U,V) = V-U$$
Okay so we know that the transformation of random variables is given by 
$$f_{U,V}(U,V)=f_{X,Y}(g_1^{-1}(U,V),g_2^{-1}(U,V)|J|$$
where $J$ is the jacobian is the derivative of 
$$g_1^{-1}(U,V),g_2^{-1}(U,V)$$ with respect to $U,V$
This gives $1$ after some computation (ask in comments if you want to see it).
So plugging in for our density we get
$$f_{U,V}(U,V)=f_{X,Y}(U,V-U)$$
$$f_{U,V}(u,v)=e^{u-v}$$
So now we want to integrate out $U$ this is where the support comes in
if $$0<x<y<\infty$$ then we can write this as 
$$\{(x,y) | x<y\}$$
$$\{(u,v) | 2u < v\}$$
so we get the final answer as 
$$\int_0^{\frac{v}{2}}e^{u-v}du$$
Which if you plug into wolfram alpha (or do some hard calculus) as 
integral from 0 to v/2 of e^(u-v) du gives you the desired answer (where I used $v=X+Y$ instead of $Z=X+Y$)
