Pdf of $ Y = X^2 $ when $X$ is uniform on $[-1,1]$

Question

Suppose you have $X\sim $ Uniform $[-1,1] $. $f_X(x)$= $\frac12, -1 \lt x \lt 1$, and $0$ otherwise.

$ Y = X^2 $. Find the pdf of $Y$.

We know that $Y = X^2 $ will always be $\geq$ to $0$. I also know that $F_Y(y)=P(Y\leq y)=P(X^2 \leq y) = P(\vert X \rvert \leq \sqrt{y}) = P(-\sqrt{y} \leq X \leq \sqrt{y}). $ I'm confused on how to trial certain values of Y to obtain the different ranges of the distribution function/ pdf.

Answer 1

You are right. You found

$F_Y(y)=F_X(\sqrt{y})-F_X(-\sqrt{y})=\sqrt{y}$

You have only to remember that $F_X(x)=\frac{x+1}{2}$.

Now deriving you get

$f_Y(y)=\frac{1}{2\sqrt{y}}\mathbb{1}_{(0;1]}(y)$