I am trying to solve this question:
Let X have the pdf:
$f(x) = 6(x-x^2)$ $if$ $0\le x \le1$
Define $Y=3X^2-2X^3$
Show that Y ~ U(0,1)
I am having trouble finding the pdf of Y since I don't know how to find $g^-1(y)$ when Y is defined this way. Can anyone help me get on the right track?
Here is a hint: observe that if $y = g(x) = 3x^2 - 2x^3$, then $$\frac{dy}{dx} = 6x - 6x^2 = f(x)$$ on the interval $0 \le x \le 1$. What does this tell you about the CDF $F_X(x)$ of $X$, and what theorem can you apply?