Find W has the beta distribution

Question

Assume that $X$ has the Fisher distribution with degrees of freedom $m$ and $n$. Prove that $$W = \frac{(m/n)\cdot X}{1+(m/n)\cdot X}$$ has the Beta distribution.

Answer 1

\begin{align} W & = \frac{(m/n)\cdot X}{1+(m/n)\cdot X} \\[8pt] & = \frac{mX}{n+mX} = 1- \frac n {n+mX} \\[15pt] \text{So } & f_W(w) = \frac d {dw} \Pr(W\le w) \\[8pt] & = \frac d {dw} \Pr\left( 1 - \frac n {n+mX} \le w \right) \\[8pt] & = \frac d {dw} \Pr\left( X \le \frac {\frac n {1-w} - n} m \right) \\[8pt] & = \frac d {dw} \Pr\left( X \le \frac n m \cdot \frac w {1-w} \right) \\[8pt] & = \frac d {dw} F\left( \frac n m \cdot \frac w {1-w} \right) \\ & {} \qquad \text{where $F$ is the c.d.f. of} \\ & {} \qquad \quad \text{the Fisher distribution} \\[8pt] & = f\left( \frac n m \cdot \frac w {1-w} \right) \cdot \frac n m \cdot \frac d {dw}\, \frac w {1-w} \end{align} where $f$ is the p.d.f. of the Fisher distribution.

Then simplify.

Answer 2

HINTS

1. Let $g(x) = \frac{ax}{1+ax}$. Can you find $g^{-1}(x)$, the inverse of $g(x)$?
2. Let $U$ be a continuous random variable with pdf and cdf $f_U(\cdot)$ and $F_U(\cdot)$, respectively, and consider the new random variable $Y = g(U)$ for some invertible function $g(\cdot)$. Then, under some conditions on $g(\cdot)$, $$ F_Y(y) = \mathbb{P}[g(U) \le y] = \mathbb{P}\left[U \le g^{-1}(y) \right] = F_U \left(g^{-1}(y)\right) $$ and if you need, you can obtain the pdf $f_Y(y)$ by differentiating the cdf.