probability distribution of a random variable that is uniformly distributed in [-1, 1] If $X$ is a random variable that is uniformly distributed between $-1$ and $1$, find the PDF of $\sqrt{\vert X\vert}$ and the PDF of $-\ln\vert X\vert$.
Solution or an explantion of approach wooulf work.
thank you in advance for your help
 A: For $\sqrt{|X|}$, take a numbber $t\in [0,1]$,
$$
P(\sqrt{|X|} \leq t)=P({|X|} \leq t^2)= P(-t^2\leq{X} \leq t^2) =t^2.
$$
Differentiate and the density is $2t$ in $[0,1]$ and $0$ otherwise.
For $-\ln{|X|}$ take a number $t \in [0,\infty)$,
$$
P(-\ln{|X|}\leq t) = P(|X|\geq e^{-t}) = P(X\geq e^{-t} \text{  or  }X\leq -e^{-t}) =P(X\geq e^{-t} )+(X\leq -e^{-t}) \\=1-e^{-t}.
$$
Differentiate and the density is $e^{-t}$ on $[0,\infty)$ and $0$ otherwise.
A: $X \sim U[-1, 1]$, $Y = \sqrt{|X|} \Rightarrow$
$$
F_Y(y) = \text{Pr}\left(Y \leq y\right), \quad y \geq 0, 
$$
is a cumulative distribution function (c.d.f.) of $Y$ (by definition).
$$
\begin{aligned}
\text{Pr}\left(Y \leq y\right) &= \text{Pr}\left(\sqrt{|X|} \leq y\right) = \text{Pr}\left(|X| \leq y^2\right) = \text{Pr}\left(-y^2 \leq X \leq y^2\right) = \\
&= \left|\begin{aligned}
&\text{since }X\sim U[-1, 1] \Leftrightarrow \text{c.d.f. of } X\text{ is }\\
&F_X(x) = 
\left\{
\begin{array}{rl}
0, & x \leq -1 \\
\frac{x-(-1)}{1-(-1)} = \frac{x+1}{2}, & -1 < x \leq 1 \\
1, & x > 1 
\end{array}
\right.
\end{aligned}\right| = \\
&= F_X(y^2) - F_X(-y^2) = 
\left\{
\begin{array}{rl}
\frac{y^2+1}{2}-\frac{-y^2+1}{2} = y^2, & 0 \leq y^2 \leq 1 \Leftrightarrow 0 \leq y \leq 1\\
1, & y^2 > 1  \Leftrightarrow y > 1
\end{array}
\right.
\end{aligned}
$$
So, the c.d.f. of $Y$, $F_Y(y) = \text{Pr}(Y \leq y)$, $\quad y \geq 0$, is equal to
$$
F_Y(y) = 
\left\{
\begin{array}{rl}
y^2, & 0 \leq y \leq 1\\
1, & y > 1
\end{array}
\right.
$$
Then, the probability density function (p.d.f.) of $Y$ equals to
$$
f_Y(y) = F'_Y(y) = 
\left\{
\begin{array}{rl}
2y, & 0 \leq y \leq 1\\
0, & y > 1
\end{array}
\right.
$$
