I came across this question in a quiz and I was not sure on how to do it since my lecturer didn't clearly teach this. Can anyone assist me with this:
Let $X$ ~ $U(0,3)$, find the density $f(u)$ for $U = X^3$ and calculate its value at $u=2$
I first tried attempting it by doing: $F_{X^3}(x) = \mathbb{P}(X^3 \leq x) = \mathbb{P}(X \in [0,\sqrt[3]{x}]) = \sqrt[3]{x}$. But this is completely wrong apparently, can someone please clarify this for me?
You want to find the probability density function; not the cumulative distribution function.
$$\begin{align}f_{X^3}(u) &=\dfrac{\mathrm d~~~}{\mathrm d~u}F_{X^3}(u)\\[1ex] &=\dfrac{\mathrm d~~~}{\mathrm d~u}\mathsf P(X\leq u^{1/3})\\[1ex] &=\dfrac{\mathrm d~\tfrac 13u^{1/3}}{\mathrm d~u~~~}\mathbf 1_{u^{1/3}\in[0..3]}\\[1ex]&=\tfrac 19 u^{-2/3}\mathbf 1_{u\in[0..27]}\end{align}$$