Transformation of variables in Beta distribution of first kind If $X$ and $Y$ follow Beta distribution, with parameters $m$ and $n$, of first kind and are independent random variables. Then what is the distribution of $(X+Y)$?
I took $U=X+Y$ and $V=Y$
Then I obtained the Jacobian which is 1.
I've reached up to
$g \left( u,v \right) = \left( \frac{1}{B\left(m,n\right)} \right) \cdot \left(u-v\right)^{\left(m-1\right)} \cdot \left(1-u+v\right)^{\left( n-1\right) } \cdot (\frac{1}{B(m,n)}) \cdot v^{\left( m-1 \right)} \cdot \left( 1-v \right)^{\left(n-1\right)}$
Can it be simplified further.
And i know we have to integrate $g(u,v)$ w.r.t. $v$ to get the pdf of $u$ only. but I ain't getting how.
 A: I think you already did most job. I will write details for my own exercises.
For a Beta distribution with parameter $m,n$
$f(x)=\frac{x^{m-1}(1-x)^{n-1}}{B(m,n)},x\in (0,1)\\ B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$
Let $U=X+Y$ and $V=Y$
Then $X=U-V$ and $Y=V$
$|J|=\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ 
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix}=\begin{vmatrix}
 1& -1\\ 
0 & 1
\end{vmatrix}=1$
Since $X$ and $Y$ are independent random variables with $Beta$ distribution,
$\therefore f(x,y)=\frac{x^{m-1}(1-x)^{n-1}}{B(m,n)}\frac{y^{m-1}(1-y)^{n-1}}{B(m,n)}$
$\therefore g(u,v)=\frac{(u-v)^{m-1}(1-u+v)^{n-1}}{B(m,n)}\frac{v^{m-1}(1-v)^{n-1}}{B(m,n)}$
This is exactly equal to what you got.
Next thing is to integrate out $v$ then we will get the pdf of $u$, then we will get distribution of $X+Y$.
$$g(u)=\int_0^1g(u,v)dv=\int_0^1 \frac{(u-v)^{m-1}(1-u+v)^{n-1}}{B(m,n)}\frac{v^{m-1}(1-v)^{n-1}}{B(m,n)}dv\\=\frac{1}{B^2(m,n)}\int_0^1 (u-v)^{m-1}(1-u+v)^{n-1}v^{m-1}(1-v)^{n-1}dv\\=\frac{1}{B^2(m,n)}\int_0^1 [(u-v)v]^{m-1}[(1-u+v)(1-v)]^{n-1}dv$$
Ok, I have to stop here. just find a solution to this distribution.
Handbook of Beta Distribution and Its Applications
see page 70.
