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The MGF of Order Statistics

  1. Jun 8, 2010 #1

    Suppose we have the following set of independent and identically distributed exponential random variables: [tex]\gamma_1,\,\gamma_2,\ldots ,\,\gamma_N[/tex]. If we arrange them in ascending order we get the following order statistics: [tex]\gamma^{(1)}\leq\gamma^{(2)}\leq\cdots\leq\gamma^{(N)}[/tex].

    I need to find the moment generating function (MGF) of the highest order statistics, i.e.: [tex]\mathcal{M}_{\gamma^{(N)}}(s)=E_{\gamma^{(N)}}[\text{e}^{s\,\gamma}][/tex] in terms of the MGFs of the exponential RVs. Is there any way to connect these MGFs?

    Thanks in advance
  2. jcsd
  3. Jun 9, 2010 #2
    Have you already calculated the MGF for the maximum order statistic? This can be done just from knowing that you have identically distributed, independent exponential RVs. And since the MGF for an exponential RV is pretty simple, maybe there's some way to rewrite the MGF for the max order statistic in terms of them.
  4. Jun 10, 2010 #3
    Thanks for replying. I already derive the MGF of the maximum order statistic, which is as the following:

    [tex]\mathcal{M}_{\gamma^{(N)}}(s)=\,N\,\sum_{k=0}^{N-1}\frac{(-1)^k\,{N-1\choose k}}{k+1-\overline{\gamma}\,s}[/tex]

    but I thought there may be more direct way.

  5. Jun 10, 2010 #4
    Yeah, that looks right. I am assuming that

    [tex]\overline{\gamma} = \frac{1}{\lambda}[/tex] is the mean of the exponential distribution.

    The MGF of the exponential distribution is

    [tex]\phi(t) = \frac{1}{1-\overline{\gamma}t}[/tex]

    The denominator of this looks pretty similar to the denominator of your expression. If you evaluate [tex]\phi(t)[/tex] for a value of t that depends on s, the mean, and k, you can make the denominators match.

    I don't know if there's a more direct way. I'm thinking it is only because of the simple form of the exponential distribution that the MGF of the maximum order statistic can be written in terms of the MGF of the distribution of the [tex]\gamma_i[/tex].
  6. Jun 11, 2010 #5
    You know, I began with the exponential distribution because it is simple, but in my work I am using very complicated distributions like the order statistics of Chi-square random variables. So, I need a general rule that can be used in all cases. Any way, we can derive this general form from matching as you said, but then the question is: is there any matching possible?

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