Suppose $U\sim \mathrm{Uniform}(-1, 1)$. Let $Y =1/{U^2}$. What is the distribution of Y?
Here is what I have: $$ \begin{aligned} Y \in [1,\infty)\\ P(Y <y) = P\Big(\dfrac{1}{U^2} < y \Big)\\ = 1 - P(U^2 <y)\\ = 1- P(-\sqrt{y} < U < \sqrt{y})\\ = 1 - [F_u(\sqrt{y}) - F_u(-\sqrt{y})]\\ = 1 -\sqrt{y}\\ \end{aligned} $$ Is that correct?
Edited: Based on comments with correct proof: $$ \begin{aligned} Y \in [1,\infty)\\ P(Y <y) = P\Big(\dfrac{1}{U^2} < y \Big)\\ = 1 - P(U^2 <\dfrac{1}{y})\\ = 1- P(-{\dfrac{1}{\sqrt y}} < U <\dfrac{1}{\sqrt y})\\ = 1 - [F_u(\dfrac{1}{\sqrt y}) - F_u(-\dfrac{1}{\sqrt y})]\\ = 1 -\dfrac{1}{\sqrt{y}}\\ \end{aligned} $$