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Thread: Limiting Distribution of Poisson RV

  1. #1
    Pur
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    Limiting Distribution of Poisson RV

    Let $\displaystyle Z_n\sim poisson(\mu=n)$. Show that the limiting distribution of $\displaystyle Y_n=\frac{Z_n-n}{\sqrt{n}}$ is a standard normal distribution.

    If anyone can point me in the right direction it would be appreciated. So far I have already attempted to use the mgf method:
    $\displaystyle \mathbb{E}(\exp(tY_n))= \exp\left\{\frac{-tn}{\sqrt{n}}\right\}\mathbb{E}\left(\frac{tZ_n}{\ sqrt{n}} \right)$
    and using the fact that $\displaystyle \mathbb{E}\left( tZ_n\right)=e^{n(e^{t}-1)}$, we have $\displaystyle \exp\left\{\frac{-tn}{\sqrt{n}}\right\}e^{n(e^{t/\sqrt{n}}-1)} $.However, I do not see how the limiting distribution of this will converge to the mgf for a standard normal.
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  2. #2
    MHF Contributor

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    You could say that the Poisson distribution of parameter $\displaystyle n$ is the distribution of the sum of $\displaystyle n$ i.i.d. Poisson(1) random variables, and therefore simply apply the CLT.

    Or you compute the limit of the mgf: the exponent in the mgf is $\displaystyle -t\sqrt{n}+n(e^{t/\sqrt{n}}-1)=-t\sqrt{n}+n(\frac{t}{\sqrt{n}}+\frac{t^2}{2n}+o(\f rac{1}{n}))$ $\displaystyle =\frac{t^2}{2}+o(1)$, i.e. its limit is $\displaystyle \frac{t^2}{2}$ therefore you get that the limit of the mgf is $\displaystyle e^{t^2/2}$.
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