Consider random variable $X$ and its distribution $f = \dfrac{3x^2}{2}$ where $-1 < x <1$
What is the distribution of $Y= |X|$? What about $V=X^2$? And $W=X^3$?
I basically did the following $F(y) = P(Y \leq y) = P(|X|\leq y) = P(-y \leq X \leq y) = \int_{-y}^{y}3x^2/2 dx = y^3$
Is the first one even right? I am worried about the $-y$ since the bounds on $x$ is $-1 < x <1$, do I need to substitute the end points? How would I do it if I were to use transformations? Example I would try $Y = X$ and $Y=-X$. Split them up and find their distribution and then add them. I eventually got to $f_x(y)(y)' = \dfrac{3y^2}{2}$. So multiply by 2 and I get $3y^2$, which contradicts my other answer...