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Math Help - Distribution for |X|

  1. #1
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    Distribution for |X|

    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!
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  2. #2
    Senior Member
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    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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