Given $f_Y(y)=4y^3$, on $[0, 1]$, (where $Y$ is a continuous random variable), set $W=2Y$ and find $f_W(w)$:
I determined that $Y$ is the identity mapping on $[0, 1]$, thus $W:[0, 1]\rightarrow [0,2]$ is given by $2x$. I have no idea how the W affects the function. I guessed $f_W(w)=32w^3$ on $[0, 1/2]$ but I'm not exactly sure why. Help