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Definition of moment generating function

  1. May 13, 2010 #1
    [tex]M(t)=E(e^{ty})=\sum_{y=0}^{n} e^{ty}p(y)[/tex]

    Is this correct?
     
  2. jcsd
  3. May 13, 2010 #2
    No. I'm not going to try to play with TeX here too much, so I'll point you to http://en.wikipedia.org/wiki/Moment-generating_function which is fairly complete. What you really need, in is probably

    [tex]\[M(t)=\sum_{n=0}^{\infty}\frac{t^nm_n}{n!}\][/tex]

    In other words, differentiating M(x) n times gives you the nth moment of the distribution.
     
  4. May 14, 2010 #3

    statdad

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    Homework Helper

    Yes, for discrete distributions. For continuous distributions replace the sum with an integral.

    Added: the upper limit of [itex] n [/itex] only if the distribution takes on finitely many values (example: binomial). if the distribution takes on infinitely many values, the mgf is

    [tex]
    m_Y(t) = \sum_{y=1}^\infty {e^{ty} p(y)}
    [/tex]
     
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