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X is Uniform $[-\theta,\theta], \theta>0$ what is the distribution of $Y=\frac{1}{x^{2}}$

So I've been working on some transformation questions; however, most of them have been one to one. I am a bit stuck on how to proceed with this question.

So far, I have: $G(y) = P(Y\leq y) = P(\frac{1}{x^{2}}\leq y) = P(-\frac{1}{\sqrt{y}}\leq x \leq \frac{1}{\sqrt{y}})$

I'm pretty sure this will be a piecewise distribution for various ranges of $X$, but I'm having a hard time coming up with the ranges, because at 0 $Y$ is undefined. Can any one give some guidance?

Thank you in advance!

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    $\begingroup$ I find it helpful to view $X$ as a 50-50 mixture of a Uniform$(0,\theta)$ and Uniform$(-\theta,0)$ distribution. Therefore $Y$ is a mixture of their reciprocal squares. Notice that the reciprocal squares of these two component distributions must be identically distributed, because one component distribution is merely the negative of the other. This reduces your question to finding, for any $z\gt 0,$ $\Pr(1/Z^2 \le z)$ where $\Pr(Z \le w) = \min(1, w/\theta)$ for all $w\gt 0.$ This eliminates all difficulties in dealing with negative numbers in the inequalities. $\endgroup$ – whuber Dec 19 '18 at 16:22
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    $\begingroup$ Additional advice: since $\theta$ obviously is a scale parameter for $X,$ algebra instantly shows $1/\theta^2$ must be a scale parameter for $Y.$ Therefore you might as well set $\theta=1$ for all your work and then, at the end, you can introduce appropriate multiples of $1/\theta^2$ in the answer. This makes any arithmetic you do very, very simple. $\endgroup$ – whuber Dec 19 '18 at 16:25
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First, $G(0) = 0$. For $y > 0$, $P(x^{-2} \leq y) = P(x^2 \geq y^{-1}) = P(x \geq y^{-0.5}) + P(x \leq - y^{-0.5})$.

You just missed one change from $\leq$ to $\geq$ in the middle.

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Or just directly apply the change of variables formula, somewhat mechanically.

Pdf of $X$ is $$f_X(x)=\frac{1}{2\theta}\mathbf1_{|x|<\theta}\quad,\,\theta>0$$

You are transforming $X\to Y$ such that $Y=\frac{1}{X^2}$.

So, $$x=\begin{cases}\frac{1}{\sqrt y}&,\text{ if }0<x<\theta\\-\frac{1}{\sqrt y}&,\text{ if }-\theta<x<0\end{cases}$$

And $$|x|<\theta \implies x^2<\theta^2\implies y>\frac{1}{\theta^2} $$

Absolute value of jacobian is $$\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|=\frac{1}{2y^{3/2}}$$

Pdf of $Y$ should be

\begin{align} f_Y(y)&=f_X\left(\frac{1}{\sqrt y}\right)\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|\mathbf1_{y>1/\theta^2}+f_{X}\left(-\frac{1}{\sqrt y}\right)\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|\mathbf1_{y>1/\theta^2} \\&=2 f_{X}\left(\frac{1}{\sqrt y}\right)\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|\mathbf1_{y>1/\theta^2} \\&=\frac{1}{2\theta y^{3/2}}\mathbf1_{y>1/\theta^2}\qquad,\,\theta>0 \end{align}

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