# Relation between Weibull and exponential distributions

The probability distribution function of a Weibull distribution is as follows: $$f(x) = a\cdot b^{-a}x^{a-1}\cdot e^{(-x/b)^a},\quad x>0$$ for parameters $a,b>0$.

I have to show that $X\sim\mathrm{Weibull}(a,b)$ iff $X^a\sim\mathrm{expo}(b^a)$. Please help me to solve this question. This problem is taken from excercise of "Simulation Modeling And Analysis" book. If there is an solution book to you, I will be greatly helpful you can give that to me.

If $X\sim \mathrm{Weibull}(a,b)$, then $X$ has CDF $F_X$ given by $$F_X(x)=P(X\leq x)= \begin{cases} 1-\exp\left(-(x/b)^a\right),\quad &x>0,\\ 0,&x\leq 0. \end{cases}$$ Similarly, if $Y\sim \mathrm{exp}(\lambda)$, then $Y$ has CDF $F_Y$ given by $$F_Y(y)= \begin{cases} 1-\exp(-\lambda x),\quad &y>0,\\ 0,&y\leq 0. \end{cases}$$ So for the first direction, assume that $X\sim \mathrm{Weibull}(a,b)$ for some $a,b>0$. Then we aim at finding the CDF for $X^a$ and confirm that this corresponds to the CDF of an exponential distribution with parameter $b^a$. For $x\leq 0$ we have $P(X^a\leq x)=0$ and for $x>0$ we have $$P(X^a\leq x)=P(X\leq x^{1/a})=\ldots$$ and hence...