Let $X_{(1)}< ... <X_{(n)}$ be size n order statistics from $Beta(\alpha,1) (\alpha >0)$ and let $$R_n = X_{(1)}/X_{(n)}$$
Then I want to derive $pdf$ and $cdf$ of $R_(n)$
First I had found the formula such as $$pdf_{X_{(1)},X_{(n)}}(x,y) = n(n-1)\alpha x^{\alpha-1}[y^\alpha-x^\alpha]^{n-2}\alpha y^{\alpha-1}$$
Then since $R_n = X_{(1)}/X_{(n)}$ $$pdf_{X_{(1)}/X_{(n)},X_{(n)}}(r_n=x/y,y) = n(n-1)\alpha(\dfrac{x}{y})^{\alpha-1}[y^\alpha-(\dfrac{x}{y})^\alpha]^{n-2}\alpha y^{\alpha-1}\cdot y$$
where $\mid \dfrac{\partial x}{\partial r_n}\mid = y$
my question is,
1) It there any problem with above reasoning?
2) Is there any more simpler derive pdf of $R_n$? (since now it becomes to intricate integrate and derive the pdf of $R_n$ from it)
The PDF in your post $$f_{X_{(1)},X_{(n)}}(x,y) = n(n-1)\alpha x^{\alpha-1}[y^\alpha-x^\alpha]^{n-2}\alpha y^{\alpha-1}\mathbf 1_{0<x<y<1}$$ is correct but you mess the change of variable big time since $(r,y)=(x/y,y)$ yields $(x,y)=(ry,y)$ and $dxdy=ydrdy$, thus the PDF of $(R_n,X_{(n)})$ is $$f_{R_n,X_{(n)}}(r,y) = f_{X_{(1)},X_{(n)}}(ry,y)\cdot y$$ that is, $$ f_{R_n,X_{(n)}}(r,y)=n(n-1)\alpha^2(ry)^{\alpha-1}[y^\alpha-(ry)^\alpha]^{n-2}y^{\alpha-1}\cdot y\cdot\mathbf 1_{0<ry<y<1}$$ which can be factored as $$f_{R_n,X_{(n)}}(r,y) = \left((n-1)\alpha r^{\alpha-1}[1-r^\alpha]^{n-2}\mathbf 1_{0<r<1}\right)\cdot\left(n\alpha y^{n\alpha-1}\mathbf 1_{0<y<1}\right)$$ Thus, $R_n$ and $X_{(n)}$ are independent with $$f_{R_n}(r) = (n-1)\alpha r^{\alpha-1}[1-r^\alpha]^{n-2}\mathbf 1_{0<r<1}$$ and $$f_{X_{(n)}}(y) = n\alpha y^{n\alpha-1}\mathbf 1_{0<y<1}$$