1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$$f_{|X|}(x)= \begin{cases} \frac{2}{3} \,\, 0 \leq x \leq 1 \\ \frac{1}{3}\,\, 1 < x \leq 2\\ 0 \,\,\text{otherwise} \end{cases}$$

$\endgroup$
  • $\begingroup$ How did you know that it is -x to x for the values 0 $\leq x \leq 1 $ and -1 to x for the other case? $\endgroup$ – user3325193 Aug 13 '14 at 4:51
  • 1
    $\begingroup$ Because $|X|$ cannot be less than 0. If $x$ is between $0$ and $1$ then $|X| \leq x$ iff $-x \leq X \leq x$ $\endgroup$ – RRL Aug 13 '14 at 4:55
  • 1
    $\begingroup$ If $x$ is between $1$ and $2$ then $|X| \leq x$ iff $-1 \leq X \leq x$. $X$ cannot be less than $-1$. $\endgroup$ – RRL Aug 13 '14 at 4:57
  • $\begingroup$ Thank you! That really helped. $\endgroup$ – user3325193 Aug 13 '14 at 4:58
  • $\begingroup$ @user3325193: You're welcome. $\endgroup$ – RRL Aug 13 '14 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.