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Prove the mean of the weibull distribution

  1. Nov 2, 2012 #1
    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. jcsd
  3. Nov 3, 2012 #2
    Bumping back to first page.
     
  4. Nov 4, 2012 #3

    haruspex

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    What substitution did you try? What is the obvious one to get rid of the power within the exponential?
     
  5. Nov 4, 2012 #4
    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
  6. Nov 4, 2012 #5

    haruspex

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    Good show.
     
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