Find PDF of Y=1/X²

Question

$f(x)=1/(2θ) , -θ≤x≤θ$ then Pdf of $Y=1/X^2$. I have tried this question in cdf method as $P(Y≤y)= P(1/X^2≤y) = P(-1/√y≤X≤1/√y) = F(1/√y)-F(-1/√y)$. Therefore PDF of $Y= (-1/2y^{3/2})f(1/√y)+(-1/2y^{3/2})f(-1/√y)=(1/2θ)y^{3/2}$. But I don't know I have get it correct or not and I also don't know how to get y's limit

Answer 1

Since $X^2 \in [0,\theta^2]$ then $1/X^2 \in [1/\theta^2,\infty)$. The distribution of $X^2$ is $$P(X^2\leq z)=P(-\sqrt{z}\leq X \leq \sqrt{z})=\frac{1}{2\theta}\int_{[-\sqrt{z},\sqrt{z}]}dx=\frac{\sqrt{z}}{\theta}$$ with density $$f_{X^2}(z)=\frac{1}{2\theta}z^{-\frac{1}{2}}$$ The inverse distribution of $X^2$ is $$P(X^{-2}\leq y)=P(X^2\geq y^{-1})=1-P(X^2\leq y^{-1})=1-\frac{1}{\theta \sqrt{y}}$$ and its density is $$f_{X^{-2}}(y)=\frac{1}{2\theta }y^{-3/2}$$ where $y \in [1/\theta^2,\infty)$.