Prove the mean of the weibull distribution

1. Nov 2, 2012

Chantry

1. The problem statement, all variables and given/known data
This is the full question, but we're only asked to do b.

A random variable x has a Weibull Distribution if and only if its probability density is given by f(x) = {kx^(b-a) *exp(-ax^b) for x > 0 }
where a and b > 0.

a) Express k in terms of a and b.
b) Show that u = a^(-1/b) * gamma(1 + 1/b)

2. Relevant equations

All of the following could be of use.
Mx(t) = ∫exp(xt)f(x) dx
and the limit as t → 0 of M'x(t) = u = mean
u = Ex(x) = ∫xf(x)dx
gamma(a) = ∫x^(1-a)exp(-x)dx

Also the integral of f(x) = 1.

3. The attempt at a solution

I tried making a substitution for x, so that I could get the gamma function out of the integral, but that x^b really throws me.

I tried integrating by parts, but that just gives an even more complicated expression. I wanted to get rid of part of the equation by setting it to one, but no matter how you treat it you'll have a nasty exp(-ax^b) to deal with.

There's obviously some trick I'm supposed to use to figure it out.

Can anyone provide any hints? I've spent a good couple hours both trying to solve it and googling similar solutions.

2. Nov 3, 2012

Chantry

Bumping back to first page.

3. Nov 4, 2012

haruspex

What substitution did you try? What is the obvious one to get rid of the power within the exponential?

4. Nov 4, 2012

Chantry

Not sure why, but it's not letting me edit the first post.

It should be kx^(b-1) *exp(-ax^b) not ∫kx^(b-a) *exp(-ax^b), but anyway, I took another crack at it and I solved it!

u = ax^b
du = abx^(b-1) dx

∫kx^(b-1) *exp(-ax^b) dx
= k/(ab)∫exp(-u) du
= k/ab[1] = 1, so k = ab.

∫kx^(b) *exp(-ax^b) dx
= k∫x^(b) *exp(-ax^b) dx
= k/(ab)∫(u/a)^(1/b) exp(-u)du
= a^(-1/b)k/(ab) ∫u^(1/b) exp(-u) du
= a^(-1/b-1)k/b Γ(1/b + 1)
sub k = ab in and you get
=a^(-1/b)Γ(1/b + 1)

Thanks a lot! I guess I just needed to know that substitution was the method to solve it for me to be able to do the math.

Last edited: Nov 4, 2012
5. Nov 4, 2012

haruspex

Good show.

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