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.
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?
