Okay so I have been trying to solve a problem in statistics for a while, but I end up getting the wrong answer all the time..
the problem is
$f(y) = 2(1-y), \ 0 \leq y \leq 1$ and $0$ elsewhere
I have to find the density function for $U_{1} = 2Y-1$
Firstly I get that $F_{U}(u) = P(U \leq u) = P ( -\frac{u+1}{2} \leq Y \leq \frac{u+1}{2}) $
to find $f_{y}dy$ I integrate so we have that
$\int_{-\frac{u+1}{2}}^{\frac{u+1}{2}}fydy = F_{Y}(\frac{u+1}{2}) - F_{y} (- \frac{u+1}{2})$
So we have that
$f_{U}(u) = \frac{1}{2} ( f_{Y} ( \frac{u+1}{2})+f_{Y}(-\frac{u+1}{2})$
since $f(y) = 2(1-y)$
$f_{U}(u) = \frac{1}{2} ( 2 ( - \frac{u+1}{2} + 2) + ( 1 + \frac{u+1}{2})) = 2$
But according to my textbook 2 is not the right answer, and I cant really see where my mistake is, any help will be appreciated!
Note that $F_{U}(u) = P(U \leq u) = P ( -\frac{u+1}{2} \leq Y \leq \frac{u+1}{2}) $
should be
$F_{U}(u) = P(U \leq u) = P ( Y \leq \frac{u+1}{2})=F_Y\left(\frac{u+1}2 \right) $
$$f_U(u) = \frac12f_Y\left( \frac{u+1}2\right)$$