Let $X$ have the p.d.f $f(x)= \frac{x^2}9$ , $0 < x < 3$, $0$ otherwise, find the pdf of $Y = X^3$
I have this exercise, but I do not know how to start, how do I know if it is a one to one transformation?
Once I have the inverse, how do I obtain the jacobian of the inverse of Y? I am lost help
The cdf of $Y$ is
$$F_Y(y)=P(Y<y)=P(X^3<y)=P(X<\sqrt[3]y)= \begin{cases} 0,&\text{ if } y<0\\ \int_0^{\sqrt[3]y}\frac{x^2}{9} \ dx&\text{ if } 0\le y \le 27 \\ 1&\text{ otherwise} \end{cases}.$$
Between $0$ and $27$
$$F_Y(y)=\int_0^{\sqrt[3]y}\frac{x^2}{9} \ dx=\frac1{27}\left[x^3\right]_0^{\sqrt[3]y}=\frac1{27}y.$$
It follows that $Y$ is uniformly distributed over $[0,27]$.