1. ## Distribution for |X|

Let the distribution function for X be:

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

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

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

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

My solution:

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

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

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

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!

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

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