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I am considering this in the sense that I know according to the central limit theorem, for an i.i.d. process $X_n$ (with mean $m$ and variance $σ^2$), the corresponding normalized sum process is: $$ Z_n = \frac{S_n-nm}{σ\sqrt{n}} $$ with $S_n = X_1+X_2+ . . . + X_n$. I know that this does indeed converge in distribution to a zero-mean unit-variance Gaussian. My question is why does this not happen for the Poisson Process. I am speaking of the Poisson process derived as a limit of the Binomial counting process, where $n$, the number of infinitesimal intervals, went to $∞$, and $p$, the success probability, went to $0$, while their product $np$ stayed constant at $\lambda t$. I believe if CLT had worked here, we would have obtained a Gaussian $N(t)$ instead of a discrete $N(t)$.

For the purposes of exploring this problem, if we consider the Taylor expansion: $$E\left[\exp\left(-\frac{j \omega}{\sigma \sqrt{n}}(X_1-m)\right)\right]=\displaystyle\sum\limits_{k=0}^∞ \frac{1}{k!}\left(-\frac{j \omega}{\sigma \sqrt{n}}\right)^k E \left[(X_1-m)^k\right]$$

I desire to examine higher-order the terms when $X_1$ is $$\mathrm{Bernoulli}\left(\frac{\lambda t}{n}\right)$$ as in the Poisson Process. Can we say these term are really neglible when compared to the terms for $k=0,1,2$?

Any help would be greatly appreciated. Thanks!

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    When you say: Does not work for a poisson process, could you clarify as to what is a poisson process here. Do you instead mean $X_n$ are iid poisson random variables?2012-12-14
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    Indeed the question needs clarification--the CLT works all right for the Poisson process.2012-12-14
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    Agree with Did. It does not work with Cauchy distribution for instance as it does not satisfy the requirements for CLT.2012-12-14
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    @GautamShenoy Question updated to hopefully clarify what it is I'm wondering about2012-12-14

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