Beta Distribution to F Distribution If $X$ is $Beta\left(\dfrac{ \alpha_1}{ 2 }, \dfrac{\alpha_2}{2}\right)$ then $\dfrac{\alpha_2 X}{\alpha_1(1-X)}$ is $F(\alpha_1, \alpha_2)$? 
Any help is appreciated I don't know where to start. I'm assuming I need the pdf's of each distribution?
 A: Yes, I use the pdf.
$f(x)=\frac{\Gamma(\frac{m}{2}+\frac{n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}x^{\frac{m}{2}-1}(1-x)^{\frac{n}{2}-1}$
and the pdf of function $g(x)$ is $l(y)=f(h(y))|h'(y)|$, where $h$ is the inverse of $g$. 
We get $h(y)=\frac{my}{my+n}$ and $|h'(y)|=\frac{mn}{(my+n)^2}$, then after patient insertion and simplification we get the result of 
$m^{m/2}n^{n/2}\frac{\Gamma(\frac{m}{2}+\frac{n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}y^{m/2-1}(my+n)^{-(m+n)/2}$ , which is indeed $F(m,n)$.
A: I'm not 100% sure this way is valid, but I'm gonna give it a try using the CDF's:
$Y =\dfrac{\alpha_2 X}{\alpha_1(1-X)}$
CDF of $Y = F_{Y}(y) = \mathbb {P}(Y < y) = \mathbb {P}(\dfrac{\alpha_2 X}{\alpha_1(1-X)} < y) = \mathbb {P}(X <\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}) $
The last stage has some limitation, mainly that $ X \neq 1 $, which depends on how you define the support of the Beta distribution. Assuming it does not include 1, we're ok. Note that $\alpha_1 y + \alpha_2 > 0$, so we can divide by it without changing signs. Continuing:
$ = F_{X}(\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}) = $ (according to the CDF definition of a Beta distribution)
$ = I_{\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}}(\dfrac{\alpha_1}{2}, \dfrac{\alpha_2}{2})  $
But this is exactly the definition of the CDF for the F distribution.
