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{\bf Question}

Suppose that $X \sim {\rm beta}(\alpha,\beta)$.  Show that the pdf
of $Y=\frac{1}{X}-1$ is

$$g(y|\alpha,\beta)=\left\{ \begin{array} {ll}
\frac{1}{B(\alpha,\beta)}\frac{y^{\beta-1}}{(1+y)^{\alpha+\beta}},
& {\rm for}\ y \geq 0;\\ 0, & {\rm otherwise} \end{array}
\right.$$

This distribution is called the {\bf beta distribution of the
second type}.

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{\bf Answer}

We have $f(x|\alpha,\beta)\ds
\frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}$

The transformation $Y=\ds\frac{1}{X}-1$ is decreasing and
continuous if $x \in (0,1)$ and also $0<y<\infty$.

$$x=\ds \frac{1}{1+y} \Rightarrow
\ds\frac{dx}{dy}=-\ds\frac{1}{(1+y)^2}$$

Therefore the pdf of $Y$ is \begin{eqnarray*} g(y) & = &
\ds\frac{1}{B(\alpha,\beta)} \cdot
\left(\ds\frac{1}{1+y}\right)^{\alpha-1} \cdot
\left(1-\ds\frac{1}{1+y}\right)^{\beta-1} \cdot
\left(\ds\frac{1}{1+y}\right)^2,\ \ 0 \leq y < \infty\\ & = &
\ds\frac{1}{B(\alpha,\beta)} \cdot
\ds\frac{y^{\beta-1}}{(1+y)^{\alpha+\beta}},\ \ 0 \leq y < \infty
\end{eqnarray*}

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