Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How would you find the distribution function for the following density functions (Weibull function):

$$f_{X}(x) = c\tau x^{\tau−1}e^{− cx^{\tau}} $$

for $0< x < \infty$, $\tau > 0$ and $c>0$.

share|cite|improve this question
Is the edit by johnny OK? I'm guessing it should actually be $$f_{X}(x) = c\tau x^{\tau−1}e^{− cx^{\tau}} $$ Am I right? – Pedro Tamaroff Dec 8 '12 at 15:44
Certainly looks more Weibullish your way, @PeterTamaroff. :) – johnny Dec 8 '12 at 15:50
Are you looking to evaluate $$\int_0^\infty f_X(x)\text{ ? }$$ – Pedro Tamaroff Dec 8 '12 at 16:08
up vote 2 down vote accepted

So $$F_X(x)=\int_0^x c\tau s^{\tau−1}e^{− cs^{\tau}} ds\\ = c\tau \int_0^x s^{\tau−1}e^{− cs^{\tau}} ds $$

If we use the substitution $s^{\tau}=u$, and $\frac{du}{ds}=\tau s^{\tau-1}$ this simplifies to

$$c\int_0^x e^{− cu} du\\ =\left[-e^{-cu}\right]_0^x\\ =1-e^{-cx}.$$

I hope that I've not given this to you too easily and that this is useful to you.

$\textbf{EDIT}$: I have assumed you were asking for the c.d.f. but the other commenters are correct your question is not entirely clear on its terminology. Also fixed my $\LaTeX$.

share|cite|improve this answer
thank you for editing and answering the question simon. I am new to this forum and didn't know how to write the question in the correct form – Sharingan Dec 8 '12 at 16:57
That's ok. Is this answer what you wanted? Don't forget you can accept the answer as correct by clicking the tick icon on the left side of it. – Simon Hayward Dec 8 '12 at 16:59
Ah that special feeling of rejection when you put up the right answer and you don't get a tick or even an upvote. – Simon Hayward Dec 8 '12 at 18:52
Your welcome. Thanks again for your help :-) – Sharingan Dec 11 '12 at 2:50
@Simon Hayward If you take $u=s^{\tau}$ then you change also the limits, and the result then becomes, $1-e^{-cx^{\tau}}$ or not ? – OBDA Jun 23 '14 at 13:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.