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Math Help - Expectation of Poisson ^ -1

  1. #1
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    Expectation of Poisson ^ -1

    I've been able to work through most of the problem so far, so I'm only posting what's relevant:

    X_1, X_2, ..., X_n \sim Poisson(\theta) and Y = \sum{X_i}

    What I need help with is calculating: E(\frac{n}{Y+1})

    It seems like it ought to be easy, but I'm drawing a blank. Is using the mgf the strategy I should be looking at? If so, how?
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  2. #2
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    Hello,

    If they follow a Poisson distribution, it means that \mathbb{P}(X_1=k)=\frac{e^{-\lambda}\lambda^k}{k!}, k\in\mathbb{N}

    By the mgf method, you can see that Y follows a Poisson distribution with parameter n\theta

    Then, recall that for any measurable and bounded function h, \mathbb{E}(h(Z))=\sum_{z\in D} h(z)\mathbb{P}(Z=z), where D is the domain of the rv Z.

    So here, we have :
    E=\mathbb{E}\left(\frac{n}{Y+1}\right)=\sum_{k=0}^  \infty \frac{n}{k+1}\cdot \mathbb{P}(Y=k)

    E=\sum_{k=0}^\infty \frac{n}{k+1}\cdot \frac{(n\theta)^k}{k!} \cdot e^{-n\theta}

    E=e^{-n\theta}\sum_{k=0}^\infty \frac{n^{k+1} \theta^k}{(k+1)!}

    E=e^{-n\theta}\cdot\frac1\theta \sum_{k=0}^\infty \frac{(n\theta)^{k+1}}{(k+1)!}

    Change the indice :
    E=e^{-n\theta}\cdot\frac1\theta \sum_{k=1}^\infty \frac{(n\theta)^k}{k!}



    Now recall this : e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=1+\sum_{k=1}^\infty \frac{x^k}{k!} \Rightarrow \sum_{k=1}^\infty \frac{x^k}{k!}=e^x-1
    Last edited by Moo; May 14th 2009 at 11:09 PM. Reason: typo mentioned below
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  3. #3
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    Quote Originally Posted by Moo View Post
    Now recall this : e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=1+\sum_{k=1}^\infty \frac{x^k}{k!} \Rightarrow \sum_{k=1}^\infty \frac{x^k}{k!}=1-e^x
    Thanks! I got through the rest of the problem, but shouldn't the above come out to: e^x-1?
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  4. #4
    Moo
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    Quote Originally Posted by Zithuan View Post
    Thanks! I got through the rest of the problem, but shouldn't the above come out to: e^x-1?
    Yes, of course !
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