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Let $X\sim\text{Uniform} [0,1]$ and let $$Y=\frac{1}{X}-X.$$

Find the pdf of $Y$.

$f(x)=1$ if $0≤x≤1$, $0$ otherwise

$Y$ is a decreasing function. Thus, the pdf should be: $$f_y(y)=-f_x(g^-1(y)).|\frac{d}{dy}(g^-1(y))|$$ But how to find: $g^-1(y)$? ie. how to solve $y=(1/x)-x$ for $x$ to produce the inverse to $g(x)$?

Any advice much appreciated!

A600
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3 Answers3

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Hint. I'm not a fan of using $g^{-1}(y)$ for pedagogical purposes; I prefer $x(y)$ (although this is less "correct"), i.e., meaning $x$ in terms of $y$.

So what is $x$ in terms of $y$? That's your $g^{-1}$.

Clarinetist
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\begin{align*} Y &= \frac{1}{X}-X\\ \implies XY &= 1-X^2\\ X^2+XY &= 1\\ \left(X+Y/2\right)^2&=1+Y/4\quad\text{(Complete the square)}\\ X &= \pm\sqrt{1+Y/4}-Y/2 \end{align*}

Em.
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For completeness, let me attempt to apply the missing part, provided by @probablyme, to this problem:

$f_Y(y)=-f_x(g^-1(y)).|\frac{d}{dy}(g^-1(y))|$ and $ X=\sqrt{1+Y/4}-Y/2$ (right side)

then,

$ f_Y(y)=-f_x(\sqrt{1+Y/4}-Y/2).|\frac{d}{dy}(g^-1(y))|$

As $f_X(x)$ is uniform with 1, the $f_X(g^{-1})$ simplifies to 1. (Is this correct?) (In alternative notation: g(y)=f(x(y)|dx/dy|=|dx/dy|)

$ f_Y(y)=-1.\frac{d}{dy}(\sqrt{1+y/4}-y/2)$

using chain rule for the right derivative term:

$((1+y/4)^{^-2}-y/2))d/dy=$

$=(1/4)(-2(1+y/4)^{-3})-1/2=-1/2(1+y/4)^{-3}-1/2$

Thus,

$f_Y(y)=(-1).(−1/2(1+y/4)^{−3}−1/2)$

Does that look about right? Thanks.

A600
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