# 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?
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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.

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Thanks for your answer, I am still stuck trying to figure out the answer to part 1; if $Y\sim\beta^prime(\alpha_1,\beta_1)$, can I find $\alpha_2, \beta_2$ such that $Y\sim\beta(\alpha_2,\beta_2$. I am stuck at solving $Y=\frac{X}{1-X}$ for $X$. –  David Dec 10 '10 at 21:28
@David: The solution is (sort of) embedded in my answer; it's $X = \frac{Y}{1+Y}.$ But why would you want to undo the transformation you just did (from $X$ to $Y$ in the first place)? What's wrong with just using the $F$ distribution expression? –  Mike Spivey Dec 10 '10 at 21:31

There is no $c$, $d$ for $Y~\beta^\prime(\alpha,\beta)$ where $Y~\beta(c,d)$

This is because $Y>0$ but $X\in(0; 1)$

Use the $F$ distribution which has support for $Y$ which is defined for positive real numbers.

although this is not a mathematical proof, it provides an example of how to sufficiently support the postulate that $X\frac{\beta}{\alpha}\sim F(2*\alpha,2*\beta)$. Feedback appreciated.

Choose arbitrary parameter values

alpha <- 2
beta  <- 4


Create a vector of a $\beta^\prime(\alpha, \beta)$ variate

X <- rbeta(1000000,alpha, beta)
Y1 <- X/(1-X)


Create a vector of an $F(2*\alpha, 2*\beta)$ variate and divide by $\frac{\beta}{\alpha}$

Y2 <- rf(1000000, 2*alpha, 2*beta) / (beta / alpha)


Compare the variables at decile intervals across their range.

testY1 <- quantile(Y1, seq(0.1,0.9,0.1))
testY2 <- quantile(Y2, seq(0.1,0.9,0.1))

signif(testY1,2) == signif(testY2,2)