# Non linear transformation of uniform distribution

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 $$F_X(b)=\mathbb P(3+U^{1/3} $$=\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}$$ From here you can derive pdf by differentiating cdf.
• With pleasure. Note also that for calculation of expectation you need not cdf or pdf of $X$. $$\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$$ – NCh Nov 24 '19 at 13:11