X is Uniform $[-\theta,\theta]$ what is the distribution of $Y=\frac{1}{x^{2}}$? 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! 
 A: 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|+f_{X}\left(-\frac{1}{\sqrt y}\right)\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|
\\&=2 f_{X}\left(\frac{1}{\sqrt y}\right)\left|\frac{\mathrm{d}x}{\mathrm{d}y}\right|
\\&=\frac{1}{2\theta y^{3/2}}\mathbf1_{y>1/\theta^2}\qquad,\,\theta>0
\end{align}
A: 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.
