1. ## 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:

$\displaystyle \int_0^{\frac{u+1}{2}}f(y) \ dy?$

2. 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...

3. Originally Posted by lllll
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