Distribution for |X|

**tttcomrader** · April 9th 2010, 06:35 AM

Let the distribution function for X be:

$F_X(t)=0 \ \ \ t < -1$

$F_X(t)= \frac {(t+1)^2}{8} \ \ \ -1 \leq t < 1$

$F_X(t)= \frac {-t^2+6t-1}{8} \ \ \ 1 \leq t < 3$

Assume that $Y = | X |$ , find $F_Y(t)$

My solution:

For $-1 \leq t < 1$ , we have $F_Y(t)=P(Y \leq t )=P(|X| \leq t ) = P(-t \leq X \leq t)$

$=P(X \leq t)-P(X \leq -t) =F_X(t)-F_X(-t)$

$= \frac {(t+1)^2}{8}- \frac {(-t+1)^2}{8}= \frac {t}{2} \ \ \ -1 \leq t \leq 1$

But the answer in the back for this portion is 1, so what am I doing wrong?

I'm basically reading and learning from the Statistic text I found in my office to prepare for a grad course in the Fall. I thought it would have been a walk in the park, but it turns it can get hairy here. Thank you!

**hatsoff** · April 9th 2010, 08:05 AM

Either the text is mistaken or there has been a miscommunication.

First of all, remember that if $t<0$ , then $P(|X|\leq t)=0$ . For $0\leq t\leq 1$ , your logic is essentially correct, and $F_{|X|}(t)=t/2$ . And then for $1\leq t\leq 3$ we have $F_X(-t)=0$ , and so $F_{|X|}(t)=F_X(t)$ .

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