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Thread: Distribution function question

  1. #1
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    Distribution function question

    I'm not quite sure if my lower and upper bound for the following function are correct.

    $\displaystyle f(y) = \left\{ \begin{array}{rcl}
    2(1-y) & \mbox{for} & 0 \leq y \leq 1 \\
    0 & \mbox{for} & \mbox{other}
    \end{array}\right.$

    Where $\displaystyle U_1 = 2Y-1$

    $\displaystyle P(Y \leq \frac{u+1}{2}$

    where $\displaystyle -1 \leq u \leq 1$

    would it be:

    $\displaystyle \int_0^{\frac{u+1}{2}}f(y) \ dy?$
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  2. #2
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    then this function:

    $\displaystyle =\int^{\frac{u+1}{2}}_0 2(1-y) \ dy $

    $\displaystyle =2\left(y-\frac{y^2}{2}\right) \bigg{|}^{\frac{u+1}{2}}_0$

    $\displaystyle =2\left(\frac{u+1}{2} - \frac{\left(\frac{u+1}{2}\right)^2}{2}\right)$

    $\displaystyle =u+1-(u^2+2u+1) = -u^2-u$

    taking the derivative I get: $\displaystyle -u-1$

    from the answer in the book of the book it's supposed to be:

    $\displaystyle \frac{1-u}{2}$

    I can't seem to find where I went wrong...
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lllll View Post
    then this function:

    $\displaystyle =\int^{\frac{u+1}{2}}_0 2(1-y) \ dy $

    $\displaystyle =2\left(y-\frac{y^2}{2}\right) \bigg{|}^{\frac{u+1}{2}}_0$

    $\displaystyle =2\left(\frac{u+1}{2} - \frac{\left(\frac{u+1}{2}\right)^2}{2}\right)$

    $\displaystyle =u+1- {\color{red} \frac {(u^2+2u+1)}4} = ...$
    see the red
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