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Math Help - moment generating function for poisson disttribution

  1. #1
    Member Danshader's Avatar
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    moment generating function for poisson disttribution

    Hi guys i am new. I have a problem in finding moment generating functions. The forum helped me in solving binomial, geometric and other few random equations but i still couldn't understand how the moment generating function for a poisson distribution is derived out from its probability density function.

    Any reply will be appreciated.
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  2. #2
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    Quote Originally Posted by Danshader View Post
    Hi guys i am new. I have a problem in finding moment generating functions. The forum helped me in solving binomial, geometric and other few random equations but i still couldn't understand how the moment generating function for a poisson distribution is derived out from its probability density function.

    Any reply will be appreciated.
    Well, I guess this is about the last one in the list!

    m_X(t) = \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y e^{-\lambda}}{y!} = e^{-\lambda} \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y}{y!}


    = e^{-\lambda} \sum_{y=0}^{\infty} \frac{(\lambda e^t)^y}{y!}


    using the standard series \sum_{y=0}^{\infty} \frac{(x)^y}{y!} = e^x and substituting x = \lambda e^t *


    = e^{-\lambda} \, e^{\lambda e^t} = e^{\lambda(e^t - 1)}.


    * If you don't like doing this there is another clever way of doing it.
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    Member Danshader's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Well, I guess this is about the last one in the list!

    m_X(t) = \sum_{n=0}^{\infty} e^{nt} \frac{\lambda^n e^{-\lambda}}{n!} = e^{-\lambda} \sum_{n=0}^{\infty} e^{nt} \frac{\lambda^n}{n!}


    = e^{-\lambda} \sum_{n=0}^{\infty} \frac{(\lambda e^t)^n}{n!}


    using the standard series \sum_{n=0}^{\infty} \frac{(y)^n}{n!} = e^y and substituting y = \lambda e^t *


    = e^{-\lambda} \, e^{\lambda e^t} = e^{\lambda(e^t - 1)}.


    * If you don't like doing this there is another clever way of doing it.
    ahhh... ic i didnt know there was that standard series thingy xD thanks for the quick reply ^-^ btw what is the other method O.o? no harm in knowing more.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Well, I guess this is about the last one in the list!

    m_X(t) = \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y e^{-\lambda}}{y!} = e^{-\lambda} \sum_{y=0}^{\infty} e^{yt} \frac{\lambda^y}{y!}


    = e^{-\lambda} \sum_{y=0}^{\infty} \frac{(\lambda e^t)^y}{y!}

    [snip]
    Let \mu = \lambda e^t:

    = e^{-\lambda} \sum_{y=0}^{\infty} \frac{\mu^y}{y!}


    = e^{-\lambda} \, e^{\mu} \sum_{y=0}^{\infty} \frac{e^{-\mu}\, \mu^y}{y!}


    \frac{e^{-\mu}\, \mu^y}{y!} is recognised as the pdf of a random variable following a Poisson distribution with mean \mu. Therefore \sum_{y=0}^{\infty} \frac{e^{-\mu}\, \mu^y}{y!} = 1:


    = e^{-\lambda} \, e^{\mu} (1)

    Substitute back that \mu = \lambda e^t:

    = e^{-\lambda} \, e^{\lambda e^t} = e^{\lambda(e^t - 1)}.
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  5. #5
    Member Danshader's Avatar
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    ahhhh..... now thats a better way to understand ^-^ thanks again
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