Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
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? –  Gautam Shenoy Dec 14 '12 at 6:27
    
Indeed the question needs clarification--the CLT works all right for the Poisson process. –  Did Dec 14 '12 at 6:33
    
Agree with Did. It does not work with Cauchy distribution for instance as it does not satisfy the requirements for CLT. –  Gautam Shenoy Dec 14 '12 at 6:48
    
@GautamShenoy Question updated to hopefully clarify what it is I'm wondering about –  αδζδζαδζασεβ23τ254634ω5234ησςε Dec 14 '12 at 6:55

1 Answer 1

The frame of the CLT is that one considers sums of i.i.d. random variables with fixed distribution. When the number $n$ of summands becomes large, after centering (by the mean) and scaling (by $1/\sqrt{n}$), the limit in distribution is normal. The frame described in the revised version of the question is quite different, since one considers binomial distributions $B(n,p_n)$ when $n$ becomes large (just like before) and $p_n$ becomes small (quite different from before), and moreover, when $np_n\to\lambda$ for some positive $\lambda$.

In other words, now the distribution of the individual summands changes with $n$. No wonder the results are different! In the latter case, the limit in distribution (without centering nor scaling) is Poisson (with parameter $\lambda$), a result which is sometimes called the law of rare events.

Edit: The new new version of the question concerns the generating function of the random variable $\xi_n=(S_n-m_n)/\sigma_n$ where $S_n$ is $B(n,p_n)$, $m_n=\mathbb E(S_n)$ and $\sigma_n^2=\mathrm{var}(S_n)$. Since the presentation of this context in the question seems to be rather confused, it may be useful to recall that $m_n=np_n$ and $\sigma_n^2=np_n(1-p_n)$ hence, in the asymptotics considered here, $m_n\to\lambda$ and $\sigma_n^2\to\lambda$. In particular, neither $m_n$ nor $\sigma_n^2$ is linear in $n$. Keeping this in mind, you might try to expand anew the generating function of $\xi_n$.

share|improve this answer
    
Ok, understood. Can we also explore this problem with the edit I have made to the question? –  αδζδζαδζασεβ23τ254634ω5234ησςε Dec 14 '12 at 7:42
    
Not sure how this is related to your main concern, but anyhow this seems pretty obvious since $E((X_1-m)^k)$ is bounded and the power of $\sqrt{n}$ increases with $k$. That is, if this is your question... (Note however that one is not supposed to modify ad infinitum one's question like that, especially after it has been answered. If you have another question, post a new question.) –  Did Dec 14 '12 at 8:23
    
I'm sorry I should have posted the edit originally, since this problem from my stochastic processes course asks me to show how the higher-order terms are not negligible to verify that the limit in distribution in not Gaussian. The problem actually also then asks to perform the same analysis for the Wiener process to show that there, it does indeed converge in distribution to a Gaussian. –  αδζδζαδζασεβ23τ254634ω5234ησςε Dec 14 '12 at 9:04
    
See Edit. $ $ $ $ –  Did Dec 14 '12 at 10:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.