Let $U$ be uniformly distributed over $[0,1]$. Let $X:=3+U^{1/3}$. What is the cdf and pdf of $X$?

My approach:

The constant factor results in a shift of $3$ to the right, resulting in $\mathbf{P}(3+U^{1/3} \leq b)$ in $[0,1]$ is equal to $\mathbf{P}(U^{1/3} \leq b)$ in $[3,4]$ is equal to $\mathbf{P}(U \leq b^3)$ in $[3,4]$. From here on I just need to compute the cdf and derive the pdf. Am I on the right track? Please show me the last step(s).


Cdf of $U$ is $$ F_U(t)=\begin{cases}0, & t < 0 \cr t, & 0\leq t <1 \cr 1, & t\geq 1\end{cases} $$ Therefore $$ F_X(b)=\mathbb P(3+U^{1/3}<b)=\mathbb P(U^{1/3}<b-3) = \mathbb P(U<(b-3)^3) = F_U((b-3)^3) $$ $$ =\begin{cases}0,&(b-3)^3 < 0,\cr (b-3)^3, & 0\leq (b-3)^3 < 1, \cr 1, & (b-3)^3 \geq 1.\end{cases} = \begin{cases}0,& b < 3,\cr (b-3)^3, & 3 \leq b < 4, \cr 1, & b \geq 4.\end{cases} $$ From here you can derive pdf by differentiating cdf.

  • $\begingroup$ Thank you! I just computed the pdf, expectation and variance with this. You helped a lot. $\endgroup$ – Marc Nov 24 '19 at 13:08
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    $\begingroup$ With pleasure. Note also that for calculation of expectation you need not cdf or pdf of $X$. $$ \mathbb E[X] = 3+\mathbb E[U^{1/3}] = 3+\int_0^1 u^{1/3} du $$ and the same for 2nd moments and others: $$ \mathbb E[X^2] = \mathbb E[(3+U^{1/3})^2] = \int_0^1 (3+u^{1/3})^2 du $$ $\endgroup$ – NCh Nov 24 '19 at 13:11
  • $\begingroup$ Nice, that makes the calculations I did for expectation and variance a lot easier. Thanks! $\endgroup$ – Marc Nov 24 '19 at 13:18

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