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Math Help - moment generating function of geometric distribution

  1. #1
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    moment generating function of geometric distribution

    I seem to be stuck on the moment generating function of a geometric distribution.

    so far

    m(t) = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y=0}^{n} e^{ty} pq^{y-1} = p \sum_{y=0}^{n} e^{ty} q^{y-1}

    how do you go from p \sum_{y=0}^{n} e^{ty} q^{y-1} \ \ \mbox{to} \ \  p \sum_{y=0}^{n} (qe^t)^y where those the -1 in p \sum_{y=0}^{n} e^{ty} q^{y {\color{red}-1}} go?
    Last edited by lllll; March 28th 2008 at 05:27 PM.
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    and is the final solution

    1) \frac{p}{1-qe^t} or

    2) \frac{pe^t}{1-qe^t} ?
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  3. #3
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    Quote Originally Posted by lllll View Post
    I seem to be stuck on the moment generating function of a geometric distribution.

    so far

    m(t) = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y=0}^{n} e^{ty} pq^{y-1} = p \sum_{y=0}^{n} e^{ty} q^{y-1}

    how do you go from p \sum_{y=0}^{n} e^{ty} q^{y-1} \ \ \mbox{to} \ \  p \sum_{y=0}^{n} (qe^t)^y where those the -1 in p \sum_{y=0}^{n} e^{ty} q^{y {\color{red}-1}} go?
    m(t) = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y={\color{red}1}}^{{\color{red}\infty}} e^{ty} pq^{y-1} = p \sum_{y={\color{red}1}}^{{\color{red}\infty}} e^{ty} q^{y-1}


    = p e^{t} \sum_{y={\color{red}1}}^{{\color{red}\infty}} e^{t(y-1)} q^{y-1}


    = p e^{t} \sum_{y={\color{red}1}}^{{\color{red}\infty}} (q e^{t})^{y-1}


    = p e^{t} \sum_{y={\color{blue}0}}^{{\color{red}\infty}} (q e^{t})^{y}

    Now note that the for an infinite geometric series with |r| < 1: \sum_{y={\color{blue}0}}^{{\color{red}\infty}} r^{y} = \frac{1}{1-r}. In the present case, r = q e^t (justification for 0 \leq q e^t < 1 is left to you).


    = p e^{t} \, \frac{1}{1 - q e^t} = \frac{p e^{t}}{1 - q e^t}.
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  4. #4
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    Quote Originally Posted by lllll View Post
    and is the final solution

    1) \frac{p}{1-qe^t} or

    2) \frac{pe^t}{1-qe^t} ?
    My prof uses p(y) = pq^y for y = 0, 1, ..., and that gives me the first result rather than the second one.
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  5. #5
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    Geometric distributions can be defined over {0,1,2,...} or {1,2,3,...}
    It depends on the context...
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    i know that this post is quite old but I can't get in your proof why we have qe^t smaller than one for me e^t is a constant so we should have in some case qe^t bigger than one no ?
    thanks
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  7. #7
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    Quote Originally Posted by morathyl View Post
    i know that this post is quite old but I can't get in your proof why we have qe^t smaller than one for me e^t is a constant so we should have in some case qe^t bigger than one no ?
    thanks
    There is no finite solution for such a case so it is excluded.
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