Here's my question: X is uniformly distributed in the interval $[-1,2]$. Find pdf of $|X|$...
So I did P($|X| \le x$) = P($-x \le X \le x$)...
From here I'm not too sure how to proceed. I know the pdf for X is 1/3 because X $\in$ $[-1,2]$ and 1/2-(-1) = 1/3.
So is $|X|$ $\in$ $[0,2]$ and |X|'s pdf = 1/2?
Sorry if this seems rudimentary, thank you for helping.
The distribution for $0 \leq x \leq 2$ is
$$P(|X| \leq x ) = \begin{cases} \int_{-x}^x\frac1{3}dt \,\, 0 \leq x \leq 1 \\ \int_{-1}^x\frac1{3}dt \,\, 1 < x \leq 2 \end{cases}= \begin{cases} \frac{2x}{3} \,\, 0 \leq x \leq 1 \\ \frac{x+1}{3}\,\, 1 < x \leq 2 \end{cases}.$$
Also $P(|X| \leq 0 ) = 0$ and $P(|X| \leq x ) = 1$ for $x > 2$.
Taking the derivative, the PDF is
$$f_{|X|}(x)= \begin{cases} \frac{2}{3} \,\, 0 \leq x \leq 1 \\ \frac{1}{3}\,\, 1 < x \leq 2\\ 0 \,\,\text{otherwise} \end{cases}$$