<s>[INST] Find the probability density function of the random variable Let $X$ be a random variable with
$$f(x) = \frac{1}{2 \theta} , -\theta < x < \theta$$
Let $Y=\frac{1}{X^2}$.Then what is the probability density function of  $Y$?
Case $1:$ let $y < 0$. Then $P(\{Y \le y\}) = 0$
Case $2$: let $0 \le y < \theta$.
Then $P(Y \le y) = P(\frac{1}{X^2} \le y) = 1 - P(\frac{1}{X^2} \ge y) = 1 - P(X^2 \le y) = 1- P(-\sqrt{y} \le X \le \sqrt{y}) = \int_{-\theta}^{y}f(y)dy - \int_{-\theta}^{-y}f(y)dy$
Case $3:$ $y \ge \theta$
$P(Y \le y) =1$
The  on differentiating the function $F(y)$ we get the probability density function.
Is my approach correct?
 [/INST] $\begin{aligned}
F_Y(y) &= P(Y \le y) \\ 
&= P \left(\frac{1}{X^2} \le y\right) \\
&= 1 - P \left(X^2 \le \frac 1y \right) \\
&= 1- P \left(- \frac 1 {\sqrt{y}} \le X \le \frac 1 {\sqrt{y}}\right) \\
&= 1 - \frac{1}{\theta \cdot \sqrt y} \\
\end{aligned}$
Now differentiating it gives you the density function of $~Y$ and as $ Y = \dfrac 1 {X^2}~, ~$ support of $Y$ is $ ~\displaystyle \frac 1 {\theta^2} \lt y \lt \infty$
Alternatively, applying Jacobian method -
$\begin{aligned} 
f_Y(y) &= \frac{f_X \left( 1 / {\sqrt y}\right)}{|dy / dx|_{x = 1 / \sqrt y}} + \frac{f_X \left(- 1 / {\sqrt y}\right)}{|dy / dx|_{x = - 1 / \sqrt y}} \\
&= \frac {1}{2\theta} \cdot \frac{1}{2 y^{3/2}} + \frac {1}{2\theta} \cdot \frac{1}{2 y^{3/2}} \\
\text {So, } f_Y(y) &= \frac{1}{2 \theta \cdot y^{3/2}}~, ~\frac{1}{\theta^2} \lt \ y \lt \infty
\end{aligned}$
</s>