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

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

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

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

    The book has the answer in the back, saying it should be $\displaystyle Gamma(\alpha = n, \beta = \frac{\theta}{n})$, 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 $\displaystyle Y=\sum_{k=1}^nX_k$, where each $\displaystyle X_k\sim\Gamma(1,\theta)$.
    Then by indep $\displaystyle Y\sim\Gamma(n,\theta)$.
    Then let $\displaystyle Z=Y/n$ and do a calc one change of variables.

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