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Find the MGF of geometric,neg binomial dist.

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the MGF (Moment generating function) of the
    a. geometric distribution
    b. negative binomial distribution

    2. Relevant equations

    geometric distribution: [tex] f(x)=p^x(1-p)^{x-1} [/tex] where x=1,2,3...

    negative binomial distribution: [tex] f(x)= \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r} [/tex] where x=r, r+1, r+2...

    MGF= [tex] E(e^{tx}) [/tex]

    3. The attempt at a solution

    a. [tex] \sum_{x=1}^{\infty}e^{tx}p^x(1-p)^{x-1} [/tex]
    let [itex] q=1-p [/itex]
    [tex] \sum_{x=1}^{\infty}e^{tx}p^xq^{x-1} [/tex]
    [tex] \sum_{x=0}^{\infty}(pe^t)q^x [/tex]
    [tex] =\frac{pe^t}{1-q} [/tex]
    that's as close as I can get to approximating the solution,
    but the book says the answer is [tex] \frac{pe^t}{1-qe^t} [/tex]

    b. [tex] \sum_{x=r}^{\infty}\frac{(x-1)!}{(x-r)!(r-1)!}e^{tx}p^rq^{x-r} [/tex] where q=1-p
  2. jcsd
  3. Nov 6, 2011 #2

    I like Serena

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  4. Nov 6, 2011 #3
    Those aren't the distributions I'm used to. Are you sure these formula's are correct??

    For the geometric distribution, I got


    and the negative binomial is


    Could you recheck this first?
  5. Nov 6, 2011 #4
    oh good god. I was looking at point binomial.
    so here's what I get now,
    [tex] \sum_{x=1}^{\infty}e^{tx}p(1-p)^{x-1} [/tex] letting q= (1-p)
    [tex] = \frac{p}{q} \sum_{x=1}^{\infty}e^{tx}q^{x} [/tex]
    [tex] = \frac{p}{q} \sum_{x=1}^{\infty}(e^{t}q)^x [/tex]
  6. Nov 6, 2011 #5
    Yes, go on... You have a geometric series right now.
  7. Nov 6, 2011 #6
    the negative binomial I have as being : [tex] \binom{x-1}{r-1} p^r(1-p)^{x-r} [/tex]
    which I figured equals : [tex] \frac{(x-1)!}{(x-1-(r-1))!(r-1)!}p^r(1-p)^{x-r} [/tex]
    [tex] = \frac{(x-1)!}{(x-r)!(r-1)!}p^r(1-p)^{x-r} [/tex]
  8. Nov 6, 2011 #7
    so, [itex] \frac{1}{1-r} [/itex] ?
    [tex] ( \frac{p}{q} ) \frac{1}{1-e^{t}q} [/tex] ?
  9. Nov 6, 2011 #8
    That's the sum of the geometrci series when you start summing from 0. But you start summing from 1 here.
  10. Nov 6, 2011 #9
    but if I sum from zero, won't I have [tex] \frac{p}{q} \sum_{x=0}^{\infty}(e^{t}q)^{x+1} [/tex] oh wait. I see. hold on
  11. Nov 6, 2011 #10
    nevermind. I don't see. am I suppose to have x-1 at x=0 to do the sum?
  12. Nov 6, 2011 #11
    oop, there it is! found the answer to the geometric. thanks for the support on that.
  13. Nov 6, 2011 #12

    Ray Vickson

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    The negative binomial with parameters p and r is the distribution of a sum of r independent geometric random variables with parameter p. What do you know about the MGF of a sum of independent random variables?

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