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Math Help - Converging distributions - Binomial to Poisson

  1. #1
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    Converging distributions - Binomial to Poisson

    Let Yn be a binomial random variable with n trials and with success probability p. Consider that n tends to infinity and p tends to zero in a way that np remains fixed at np=ƛ. Prove that the distribution of Yn converges to a Poisson distribtution with mean ƛ.

    I am supposed to use mgfs but am unsure of how to do so.
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  2. #2
    Super Member Random Variable's Avatar
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    For the binomial distribution,  M_{x}(t) = [pe^{t}+(1-p)]^{n}

    Now let  p = \frac{\lambda}{n}

    then  M_{x}(t) = [\frac{\lambda}{n}e^{t}+(1-\frac{\lambda}{n})]^{n} = [1+ \frac{(e^{t}-1)\lambda}{n}]^{n}

    now let n go to infinity

     \lim_{n \to \infty} M_{x}(t) = \lim_{n \to \infty} [1+ \frac{(e^{t}-1)\lambda}{n}]^{n}= e^{(e^{t}-1)\lambda} which is the MGF of the Poisson distribution
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  3. #3
    MHF Contributor matheagle's Avatar
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    I usually just use the distribution...........

    {n\choose x}p^x(1-p)^{n-x}

    ={n!\over (n-x)!x!} \left({\lambda\over n}\right)^x \left(1-{\lambda\over n}\right)^{n-x}

    ={n(n-1)\cdots (n-x+1)\over n^x} \left({\lambda^x\over x!}\right) \left(1-{\lambda\over n}\right)^n\left(1-{\lambda\over n}\right)^{-x}

    \to\left(1\right) \left({\lambda^x\over x!}\right) e^{-\lambda}\left(1\right)

    ={\lambda^xe^{-\lambda} \over x!}
    Last edited by matheagle; March 18th 2010 at 05:17 PM.
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