Finding joint pmfs from marginal pmfs Let $a, b > 0$. The random variables $X$ and $Y$ are independent and their densities are :
$f(x)$ = $1 \gamma(a)*x^{x-1}*e^{-x}, x\geq 0$
$f(y)$ = $1 \gamma(b)*y^{b-1}*e^{-y}, y\geq 0$
Let $U=X+Y$ and $V=X/X+Y$
Find the joint density of U and V and show they are independent.
So far I have found $J(u,v)$ to find the determinant is $-u$. I'm confused how to plug this into the marginal pdfs to find the joint pdfs.
 A: Firstly, when working with absolutely continuous distributions (those with densities), you refer to the density as the pdf (probability density function), not the pmf (probability mass function [used when there are "masses" at singletons]). 
The change of variables formula tells us that
$$
f_{(U,V)}(u,v)=f_{(X,Y)}(x,y)|\det \frac{d(x,y)}{d(u,v)}| = f_X(x)f_Y(y)\cdot|-u|.
$$
where he we have used the fact that $f_{(X,Y)}=f_Xf_Y$ by independence of $X,Y$.
Write $x$ and $y$ in terms of $u,v$ and plug them into the $x,y$ in the above formula. You will find that $f_{(U,V)}$ will end up being the product of two familiar looking densities $h_1(u),h_2(v)$. It then immediately follows that $f_U = h_1, f_V = h_2$ and so $f_{(U,V)} = f_U f_V$ which tells us that $U,V$ are independent.
Can you run through the calculation and see what $h_1,h_2$ appear?
Edit: Note that $\frac{d(x,y)}{d(u,v)}=J(u,v)$ is meant to be evaluated at $(u,v)$.
Hint 1: $x=uv, y=u(1-v)$, and you correctly calculated $J(u,v)=-u$. Plugging into the densities I am getting
$$
f_{(U,V)}(u,v) = \left\{\begin{array}{cc}
 & 
\begin{array}{cc}
 \frac{e^{-u} (u v)^{a-1} (u(1-v))^b}{(1-v) \Gamma (a) \Gamma (b)} & 0\leq uv,0\leq u(1-v) \\
 0 & \text{else} \\
\end{array}
 \\
\end{array}\right.
$$
Do you see what $h_1,h_2$ are now?
