I have a question regarding the transformation of a gamma distribution. I think I solved the problem, but I am not sure whether it is correct.
Let $X$ and $Y$ be independent and Gamma distributed with paramters $\alpha$ and $\lambda$, i.e. $f_X(x)=f_Y(x)=\frac{\lambda^\alpha\cdot x^{\alpha-1}\cdot e^{-\lambda x}}{\Gamma{(\alpha)}}$ and $\Gamma(\alpha)=\int_0^\infty x^{\alpha -1}e^{-x}dx$
Define $R=\frac{X}{X+Y}, T=X+Y$. What is the joint density of $T$ and $R$?
I would say the inverse of $r$ and $t$ are given by: $x=rt$ and $y=t(1-r)$. The Jacobian $J(r,t)=|t|$.
Following the joint transformation theorem, $f_{RT}(r,t)=f_{XY}(x(r,t),y(r,t))\cdot |J(r,t)|$
Because of independence, $=f_{XY}(x,y)=f_X(x) \cdot f_{Y}(y)=\frac{\lambda^\alpha\cdot x^{\alpha-1}\cdot e^{-\lambda x}}{\Gamma{(\alpha)}}\cdot \frac{\lambda^\alpha\cdot y^{\alpha-1}\cdot e^{-\lambda y}}{\Gamma{(\alpha)}}$
Hence, $f_{RT}(r,t)=\frac{\lambda^\alpha\cdot rt^{\alpha-1}\cdot e^{-\lambda rt}}{\Gamma{(\alpha)}}\cdot \frac{\lambda^\alpha\cdot t(1-r)^{\alpha-1}\cdot e^{-\lambda t(1-r)}}{\Gamma{(\alpha)}}t$
I am not sure how the Gamma-function, if at all, should transform. Could anyone give me a hint here?