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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?

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

$$\begin{align}f_{X^3}(u) &=\dfrac{\mathrm d~~~}{\mathrm d~u}F_{X^3}(u)\\[1ex] &=\dfrac{\mathrm d~~~}{\mathrm d~u}\mathsf P(X\leq u^{1/3})\\[1ex] &=\dfrac{\mathrm d~\tfrac 13u^{1/3}}{\mathrm d~u~~~}\mathbf 1_{u^{1/3}\in[0..3]}\\[1ex]&=\tfrac 19 u^{-2/3}\mathbf 1_{u\in[0..27]}\end{align}$$

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  • $\begingroup$ The factor $\frac13$ is necessary because the original RV is uniform in the range $[0,3]$ $\endgroup$ – dfnu Oct 16 at 5:16
  • $\begingroup$ Now it is correct. $\endgroup$ – dfnu Oct 16 at 5:18

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