# How to find the cube of a uniform distribution?

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: $$F_{X^3}(x) = \mathbb{P}(X^3 \leq x) = \mathbb{P}(X \in [0,\sqrt{x}]) = \sqrt{x}$$. But this is completely wrong apparently, can someone please clarify this for me?

• Shouldn't there be a factor $\frac13$? Also: the requirement is to find PDF, while you found CDF, here. – dfnu Oct 16 at 5:03
• @dfnu oh sorry I forgot to add my second part of working. So would the CDF be $(1/3)*\sqrt{x}$ and then the PDF would be the derivative of this? – user35675 Oct 16 at 5:08
• yep, with the right domain, of course. – dfnu Oct 16 at 5:08
• Thank you so much! – user35675 Oct 16 at 5:09

\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}
• The factor $\frac13$ is necessary because the original RV is uniform in the range $[0,3]$ – dfnu Oct 16 at 5:16