The probability density function of gamma distribution is given like $f(x)=\begin{cases}\frac{\beta^{a}x^{a-1}e^{-\beta x}}{\Gamma({a})}& \text{x>0}\\ 0,\text{elsewhere}\end{cases}$
Let X and Y be two independent random variables with gamma distribution $\Gamma(a_{1},\beta)$ and $\Gamma(a_{2},\beta)$ respectively. Here $\alpha$,$\beta_{1}$,$\beta_{2}$ are all positive constants. Define $U=X+Y$ and $V=\frac{X}{Y}$. Find the joint density of $U$ and $V$. Are $U$ and $V$ independent random variables?
*My attempt was first I used the independence of $X$ and $Y$ hence found the joint probability distribution function of $X$ and $Y$, which is $f_{x}(x)\times f_{y}(y)$. As a result it is something like $f_{x,y}(x,y)=\begin{cases}\frac{\beta^{a_{1}+a_{2}}\times x^{a_{1}-1}\times y^{a_{2}-1}\times e^{-\beta(x+y)}}{\Gamma (a_{1})\times \Gamma (a_{2})} \text{x>0 ,y>0} \\ 0,\text{elsewhere} \end{cases}$. Now I used the transformation rule which is $f_{u,v}(u,v)=f_{x,y}(x,y)\times\frac{1}{|J(x,y)|}$ where $J(x,y)=\frac{1}{\partial(\frac{(u,v)}{(x,y)})}$ Therefore $f_{u,v}(u,v)=\frac{y^{2}}{x+y}\times f_{x,y}(x,y)$ is my conclusion. As final step in compact form I wrote that $f_{u,v}(u,v)=\begin{cases}\frac{(\frac{\beta\times u}{v+1})^{a_{1}+a_{2}}\times e^{-\beta u}\times v^{a_{1}-1}}{u} &\text{u>0, v>0}\\ 0 &\text{otherwise}\end{cases}$
Now I will try to show $f_{u,v}(u,v)=f_{u}(u)\times f_{v}(v)$ for independence. $f_{u}(u)=\int_{0}^{\infty}f_{u,v}(u,v)dv$ and $f_{v}(v)=\int_{0}^{\infty}f_{u,v}(u,v)du$ but I have some issues in the last step in the integration. Is my approach corect to find the independence of $U$ and $V$. Should I identify the region under the transformation? is there an easier way to do that find the independence? Is my joint density of $U$ and $V$ correct?
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