Are these transformations of the $\beta^\prime$ distribution from $\beta$ and to $F$ correct? Motivation
I have a prior on a random variable $X\sim \beta(\alpha,\beta)$ but I need to transform the variable to $Y=\frac{X}{1-X}$, for use in an analysis and I would like to know the distribution of $Y$.
Wikipedia states:

if $X\sim\beta(\alpha,\beta)$ then $\frac{X}{1-X} \sim\beta^\prime(\alpha,\beta)$

Thus, the distribution is $Y\sim\beta^\prime(\alpha,\beta)$. The software that I am using, JAGS, does not support the $\beta^\prime$ distribution. So I would like to find an equivalent of a distribution that is supported by JAGS, such as the $F$ or $\beta$.
In addition to the above relationship between the $\beta$ and $\beta^\prime$,
Wikipedia states:

if $X\sim\beta^\prime(\alpha,\beta)$ then $\frac{X\beta}{\alpha}\sim F(2\alpha, 2\beta)$

Unfortunately, neither of these statements are referenced. 
Questions


*

*1) Can I find $c$, $d$ for $Y\sim\beta^\prime(\alpha,\beta)$ where $Y\sim\beta(c,d)$

*2) Are these transformations correct? If so, are there limitations to using them, or a reason to use one versus the other (I presume $\beta$ is a more direct transformation, but why)? 

*3) Where can I find such a proof or how would one demonstrate the validity of these relatively simple transformations?


2 and 3) Both of these transformations are correct; you can prove them with the cdf (cumulative distribution function) technique.  I don't see any limitations on using them. 
Here is the derivation for the first transformation using the cdf technique.  The derivation for the other will be similar.  Let $X \sim \beta(\alpha, \beta)$.  Let $Y = \frac{X}{1-X}.$  Then
$$P\left(Y \leq y\right) = P\left(\frac{X}{1-X} \leq y\right) = P\left(X \leq y(1-X)\right) = P\left(X(1+y) \leq y\right) = P\left(X \leq \frac{y}{1+y}\right)$$
$$= \int_0^{\frac{y}{1+y}} \frac{x^{\alpha - 1} (1-x)^{\beta-1}}{B(\alpha,\beta)} dx.$$
Differentiating both sides of this equation then yields the pdf (probability density function) of $Y$.  We have 
$$f_Y(y)=\frac{\left(\frac{y}{1+y}\right)^{\alpha - 1} \left(1-\frac{y}{1+y}\right)^{\beta-1}}{B(\alpha,\beta)} \frac{d}{dy} \left(\frac{y}{1+y}\right)$$
$$= \frac{\left(\frac{y}{1+y}\right)^{\alpha - 1} \left(\frac{1}{1+y}\right)^{\beta-1}}{B(\alpha,\beta)} \left(\frac{1}{1+y}\right)^2$$
$$= \frac{y^{\alpha - 1} \left(1+y\right)^{-\alpha-\beta}}{B(\alpha,\beta)},$$
which is the pdf of a $\beta'(\alpha,\beta)$ random variable.
