Distribution of quotient of random variables If $X$ and $Y$ are independent random variables such that $X\sim \Gamma(a,b)$ and $Y\sim\Gamma(a,c)$. What is the distribution of random variable $\frac{Y}{X+Y}$? Any help with this ?
 A: To find the p.d.f of the ratio $\frac{Y}{X+Y}$, let us first write its c.d.f. 
Since $X$ and $Y$ are always positive, their ratio is also positive and, therefore, for $0\leq t\lt1$ we can write:
$
P\left(\frac{Y}{X+Y}\leq t\right)=P\left(Y\leq \frac{t}{1-t}X\right)=\int_{0}^{\infty }\left(\int_{0}^{\frac{t}{1-t}x}f_{X}(x)f_{Y}(y)dy\right)dx
$
as $f_{X}(x)f_{Y}(y)$ is the joint p.d.f. of $X$ and $Y$ (the variables are indipendent) and $y$ goes from $0$ to $\frac{t}{1-t}x$ when $x$ goes from $0$ to $\infty$.
The p.d.f. is the derivative of the c.d.f. so we can write:
$
\frac{d}{dt}P\left(\frac{Y}{X+Y}\leq t\right)=\frac{d}{dt}\int_{0}^{\infty }\left(\int_{0}^{\frac{t}{1-t}x}f_{X}(x)f_{Y}(y)dy\right)dx=\int_{0}^{\infty }\frac{d}{dt}\left(\int_{0}^{\frac{t}{1-t}x}f_{Y}(y)dy\right)f_{X}(x)dx
$
let $F_{Y}(y)$ be a primitive for $f_{Y}(y)$ 
$
\frac{d}{dt}\left(\int_{0}^{\frac{t}{1-t}x}f_{Y}(y)dy\right)=\frac{d}{dt}F_{Y}(\frac{t}{1-t}x)=\frac{1}{(1-t)^{2}}xf_{Y}(\frac{t}{1-t}x)
$
so
$
\frac{d}{dt}P\left(\frac{Y}{X+Y}\leq t\right)=\frac{1}{(1-t)^{2}}\int_{0}^{\infty }xf_{Y}(\frac{t}{1-t}x)f_{X}(x)dx
$
The last integral is easy to compute but quite long, you just need to substitute the equations for the p.d.f.s, group the exponentials, make a variable change (I have $z=\left(b+\frac{ct}{1-t}\right)x$) and you'll have your p.d.f.
$
\frac{\Gamma(2a)}{\Gamma^{2}(a)}c^{a}b^{a}\frac{t^{a-1}(1-t)^{a-1}}{(b+(c-b)t)^{2a}}
$
If you want I can help with all the steps needed, but now I have little time to control if the results are correct.

EDIT: I changed the final result (I was wrong while grouping a factor). Now the function correctly become a $Beta(a,a)$ when $c=b$.
