Prove the mean of the weibull distribution

[Chantry]

Homework Statement

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)

Homework 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.

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.

 

[Chantry]

Bumping back to first page.

 

[haruspex]

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

 

[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.

 

[haruspex]

Good show.