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I have a PDF $$ f_X(x) = \frac{1}{4}(x+1), \ \ \ \ \ \ \ \ \ 0 < x < 2 $$ I would like to use $$Z = \frac{9}{(X+1)^2}$$ and find a new PDF.

Since, $z = g(x) = 9/(x+1)^2$ is monotone increasing, I thought I can use the following formula ($X$ has pdf $f_X(x)$ and let $Y=g(X)$) $$ f_Y(y) =\begin{cases} f_X(g^{-1}(y)) \left| \frac{d}{dy} g^{-1}(y) \right| & y \in \mathcal{Y} \\ 0 & otherwise \end{cases} $$

What I got is: $$ f_Z(z) = \frac{1}{4} \left( \frac{9}{(z+1)^2} + 1 \right) \cdot \left| \frac{-9\cdot2(x+1)}{\left\{ ((z+1)^2 \right\}^2} \right| $$ But it seems it's quite different from the answer, $f_Z(z) = 9/(8z^2),\ 1

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    The inverse should be $g^{-1}(y)=\frac{3}{\sqrt{y}}-1$. – 2017-01-27

1 Answers 1

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Thanks to the @user375366 's help, I could get the correct answer. I spell out my calculation for future reference.

  • Calculate $g^{-1}(z)$ $$ z = \frac{9}{(x+1)^2} \\ \Longleftrightarrow \frac{9}{z} = (x + 1)^2\\ \Longleftrightarrow x = \frac{3}{\sqrt{z}}-1 $$ Hence, $$f_X(g^{-1}(z)) = \frac{1}{4}\frac{3}{\sqrt{z}}.$$

  • Calculate $\left|\frac{d}{dz}g^{-1}(z) \right|$ $$ \left| \left(\frac{3}{z^{\frac{1}{2}}}-1 \right)^{'} \right| = \left| - \frac{3}{2} \cdot \frac{1}{\sqrt{z^3}}\right| $$

These results give $$f_Z(z) = \frac{9}{8z^2}$$