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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.

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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...

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  • $\begingroup$ plz write in a details . $\endgroup$ Sep 18 '13 at 13:27
  • $\begingroup$ @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. $\endgroup$ Sep 18 '13 at 13:33
  • $\begingroup$ Thanks Stefan Hansen . I have solved the problem . Many Many thanks . $\endgroup$ Sep 18 '13 at 13:42

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