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Thread: using moment generating function to find expectation

  1. #1
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    using moment generating function to find expectation

    A discrete random variable X is pmf p(x)=(1-p)^(x-1)p for x=1,2,...
    where 0<p<1

    (a) Find the moment generating function of X
    (which is pe^t / (1-qe^t)

    (b) Hence, or otherwise, find E(X(X-1)...(X-r+1))

    So what i am trying to do is:
    E(X(X-1)...(X-r+1)
    =E(e^(lnx)+e^ln(x-1)+...+e^ln(x-r+1))
    =pe^(lnx)/(1-qe^(lnx))*...*pe^ln(x-r+1)/(1-qe^ln(x-r+1)
    then I don't know how to proceed, can somebody help ?
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  2. #2
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    Re: using moment generating function to find expectation

    Your first difficulty is that ab is not equal to e^ln(a)+ e^ln(b).

    It is true that ab= e^(ln(ab))= e^(ln(a)+ ln(b))= (e^ln(a))(e^ln(b)). The exponentials are multiplied, not added.
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  3. #3
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    Re: using moment generating function to find expectation

    E(X(X-1)...(X-r+1)
    =E(e^lnx*e^ln(x-1)*...*e^ln(x-r+1))
    =E(e^lnx)E(e^ln(x-1))...E(e^ln(x-r+1))
    =E(X)E(X-1)...E(X-r+1)
    =1/p*(1/p-1)*...*(1/p(r-1))

    so the answer is r!q^(r-1)/p^r
    how can I proceed?
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  4. #4
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    Re: using moment generating function to find expectation

    There is no "q" in your original formulation but there is a "q" in the answer. Is q= 1- p?
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  5. #5
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    Re: using moment generating function to find expectation

    Quote Originally Posted by HallsofIvy View Post
    There is no "q" in your original formulation but there is a "q" in the answer. Is q= 1- p?
    yes, this is a common if annoying abbreviation.
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