# 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.

## 1 Answer

Hint: Argue in terms of the cumulative distribution function (CDF).

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...

• plz write in a details . Sep 18 '13 at 13:27
• @OchenaPothik: What is not clear? The calculations following the dots should be pretty straightforward. Anyway, I urge you or your cousin to try and solve the exercise yourself based on my hint, and if this still causes you trouble, then let me know. Sep 18 '13 at 13:33
• Thanks Stefan Hansen . I have solved the problem . Many Many thanks . Sep 18 '13 at 13:42