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Math Help - distribution of the mean of an exponential distribution (mgf)

  1. #1
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    distribution of the mean of an exponential distribution (mgf)

    Let \overline{X} be the mean of a random sample from the exponential distribution, Exp( \theta). Using the mgf technique determine the distribution of \overline{X}.

    M_{\overline{X}}(t) = Ee^{t\overline{X}}

    M_{\overline{X}}(t) = Ee^{\frac{t}{n}\sum{X_i}}

    M_{\overline{X}}(t) = e^{\frac{1}{n}}(1-\frac{t}{\theta})^{-n} I'm sure this is wrong, but I don't know why.

    The book has the answer in the back, saying it should be Gamma(\alpha = n, \beta = \frac{\theta}{n}), but I don't know how to get there. MGF problems often seem to give me difficulty.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Let Y=\sum_{k=1}^nX_k, where each X_k\sim\Gamma(1,\theta).
    Then by indep Y\sim\Gamma(n,\theta).
    Then let Z=Y/n and do a calc one change of variables.

    OR if you wish to use MGFs...
    E(e^{\bar X t}) = E(e^{Y(t/n)})=M_Y(t/n).
    Now, I don't know how you're writing your exponential/gamma densities.
    BUT subtitute t/n for t in a \Gamma(n,\theta) MGF and it's over.
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