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Thread: Distribution for |X|

  1. #1
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    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!
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  2. #2
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    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 $\displaystyle F_{|X|}(t)=F_X(t)$.
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