PDF of $Z^2$ where $Z$ is uniform $[-1,1]$ (a) Let $Z$ be uniform $[-1,1]$. What is the PDF of $Z^2$.

I would like to fix my previous statement. Using the textbook for uniform $[0,1]$ the PDF of $Z^2$
Say $Y=Z^2$.    

The CDF
$$P(Z^2\le y) = P(-\sqrt {y} \le Z \le \sqrt{y})= F_Y(\sqrt{y})-F_Y(\sqrt{-y})$$ 
Differentiate to obtain the PDF 
$$\frac{1}{2\sqrt y}f_Y(\sqrt y)+\frac{1}{2\sqrt y}f_Y(\sqrt{-y})$$ This is correct if $Z$ ~ $[0,1]$. Does the problem change greatly due to the new interval?
\begin{align}
P(Z^2 \leq y) &amp;= P(-\sqrt{y} \leq Z \leq \sqrt{y}) \\
&amp;= F_Z(\sqrt{y}) - F_Z(-\sqrt{y})
\end{align}

Differentiating

$$f_{Z^2}(y) = \frac{1}{2\sqrt{y}}f_Z(\sqrt{y}) + \frac{1}{2\sqrt{y}}f_Z(-\sqrt{y}) = \frac{1}{\sqrt{y}}f_Z(\sqrt{y})$$

I will leave the evaluation of $f_Z(\sqrt{y})$ as an exercise.
\begin{align}
P(Z^2 \leq y) &amp;= P(-\sqrt{y} \leq Z \leq \sqrt{y}) \\
&amp;= F_Z(\sqrt{y}) - F_Z(-\sqrt{y})
\end{align}

Differentiating

$$f_{Z^2}(y) = \frac{1}{2\sqrt{y}}f_Z(\sqrt{y}) + \frac{1}{2\sqrt{y}}f_Z(-\sqrt{y}) = \frac{1}{\sqrt{y}}f_Z(\sqrt{y})$$

I will leave the evaluation of $f_Z(\sqrt{y})$ as an exercise.