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).
Best Answer
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.