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Math Help - finding the mgf function

  1. #1
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    finding the mgf function

    Can someone help me with this problem? I don't understand how to do it, I know that the mgf is definied as the integral from negative to positive infinity e^(tx)f(x) but I don't understand how to do it for this question
    Last edited by halldo; November 11th 2009 at 10:56 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    This is discrete, so it's the sum instead

     E(e^{tX})=.2\sum_{x=1}^{\infty} e^{tx}(.8)^{x-1}

    = .25\sum_{x=1}^{\infty} e^{tx}(.8)^x

     =.25\sum_{x=1}^{\infty} (.8e^t)^x

    Now use the geometric sum from calc2, that's why this is a geometric random variable

    You can prove directly that the mean is 1/p.
    So with p=.2 so the mean is indeed 5.
    Last edited by matheagle; November 11th 2009 at 09:11 PM.
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