Distribution of U+V if (U,V) is i.i.d. exponential I have the following independent exponentials:
$f_U = f_V = e^{-x}, x>o$
And I want to determine Z=U+V. It seems jacobian would work just fine (I define the auxiliary transformation W=U) but my joint distribution ends up being an exponentail on Z only ($f_z = e^{-z})$, which seems weird. The transformation seem one to one...
Any hints (before a full answer)?
EDIT: My approach:
Define $Z=U+V$ and $W=U$. These are my transformations. Now the inverses:
$$
U(z,w) = w ~ V(z,w) = z - w
$$
Which gives jacobian 1. Hence:
$f_{Z,W} = f_{U,V}(U(z,w),V(z,w)) = e^{-z}$
Now to find marginal Z, I would have to integrate from $0$ to $w$? Alternative approaches are welcome aswell.
Thanks.
 A: To sum up, $f_U(u)=\mathrm e^{-u}\mathbf 1_{u\gt0}$, $f_V(v)=\mathrm e^{-v}\mathbf 1_{v\gt0}$ and $(U,V)$ is independent hence the density $f_{(U,V)}$ of $(U,V)$ is such that $f_{(U,V)}(u,v)=\mathrm e^{-u-v}\mathbf 1_{u\gt0,v\gt0}$. Now the Jacobian of the transformation $(u,v)\mapsto(w,z)=(u,u+v)$ is $1$ and the inverse of this transformation is $(w,z)\mapsto(u,v)=(w,z-w)$ hence $(W,Z)=(U,U+V)$ has density
$$
f_{(W,Z)}(w,z)=f_{(U,V)}(w,z-w)=\mathrm e^{-w-(z-w)}\mathbf 1_{w\gt0,z-w\gt0}=\mathrm e^{-z}\mathbf 1_{z\gt w\gt0}.$$ Finally, the density $f_Z$ of $Z$ is the marginal of $f_{(W,Z)}$, that is,
$$
f_Z(z)=\int_\mathbb Rf_{(W,Z)}(w,z)\mathrm dw=\mathbf 1_{z\gt0}\int_0^z\mathrm e^{-z}\mathrm dw=z\mathrm e^{-z}\mathbf 1_{z\gt0}.
$$
As usual (and as already explained several times on the site), to write fully the densities, that is, including the relevant indicator functions, makes these computations mechanical hence trivial.
A: Since alternative answers are accepted as well......
$U$ and $V$ are IID exponential random variables with mean 1.
The Moment Generating Function for the exponential random variable is 
$$M_U(t)=M_V(t)=\frac{1}{1-t}$$
Now if $Z=U+V$, since they are IID, by the properties of MGF's,
$$M_Z(t)=M_{U+V}(t)=M_U(t)M_V(t)=\left(\frac{1}{1-t}\right)\left(\frac{1}{1-t}\right)=\left(\frac{1}{1-t}\right)^2$$
This is precisely the Moment Generating Function for the Gamma Distribution
$$f(x)=\frac{x^{k-1}e^{\frac{-x}{\theta}}}{\theta^k\Gamma{(k)}}, x\gt0$$ with parameters $\theta=1$, and $k=2$.  Thus
$$f(z)=\frac{z^{2-1}e^{\frac{-z}{1}}}{1^2\Gamma{(2)}}=\frac{ze^{-z}}{(2-1)!}=ze^{-z},z\gt0$$
