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Math Help - 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.

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

    Where U_1 = 2Y-1

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

    where -1 \leq u \leq 1

    would it be:

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

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

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

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

    =u+1-(u^2+2u+1) = -u^2-u

    taking the derivative I get: -u-1

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

    \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:

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

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

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

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