Finding the distribution of $xy$ when Distribution of $x$ and $y$ is given If $x$ follows $Beta(u,v)$ and $y$ follows $Ganma(\lambda,u+v)$ then what is the distribution of $xy$?
I could only write the PDF of x and Y but not able to draw PDF of XY. I suppose that there is some relationship that I am not able to catch. Please tell me. Intial Hint also would do for me. 
 A: Assuming $x$ and $y$ are independent, then $xy$ is distributed as $Gamma(\lambda,u)$.  First, write down the joint density, then change variable to $z=xy$ and $w=y-xy$, and then integrate out $w$.
The joint density is $$\frac{x^{u-1}(1-x)^{v-1}}{\Gamma(u)\Gamma(v)}\lambda^{u+v}y^{u+v-1}e^{-\lambda(u+v)}~dx~dy$$ where we've already cancelled the $\Gamma(u+v)$ that appears in the numerator of one factor and the denominator of the other.  
Next compute the Jacobian determinant of $z$ and $w$ with respect to $x$ and $y$, and you get $\frac{\partial z}{\partial x}\frac{\partial w}{\partial y}-\frac{\partial w}{\partial x}\frac{\partial z}{\partial y}=y(1-x)-(-y)x=y$, so $dz~dw = y~dx~dy$.  So you can rewrite the above as  $$\frac{(xy)^{u-1}(y-xy)^{v-1}}{\Gamma(u)\Gamma(v)}\lambda^u\lambda^ve^{-\lambda u}e^{-\lambda v} y~ dx~ dy$$ $$=\frac{z^{u-1}}{\Gamma(u)}\lambda^ue^{-\lambda u}\frac{w^{v-1}}{\Gamma(v)}\lambda^ve^{-\lambda v}~dz~dw$$
And now clearly $z$ and $w$ are independent gammas (since this factors), and you can integrate out $w$ to get the distribution of $z$ as $$=\frac{z^{u-1}}{\Gamma(u)}\lambda^ue^{-\lambda u}~dz$$ so $xy$, which is $z$, is $Gamma(\lambda,u)$.
Note that there is an important related result that if you have independent $R$ which is $Gamma(\lambda,u)$ and $S$ which is $Gamma(\lambda,v)$, then, the distribution of $\frac{R}{R+S}$ is $Beta(u,v)$.  
