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The CDF of a Poisson distribution with rate parameter $\lambda$ is $$ P(n;\lambda)=\sum_{k=0}^n \frac{\lambda^ke^{-\lambda}}{k!}. $$ As $n$ goes to infinity, the CDF would certainly approach 1.

Now, consider the case when the rate parameter is $\xi n$ with $\xi\in(0,1)$ being a given constant. According to the last answer in Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$, the following limit should be 1: $$ \lim_{n\to\infty}P(n;\xi n)=\lim_{n\to\infty} \sum_{k=0}^n\frac{(n\xi)^ke^{-n\xi}}{k!}=1. $$ Is it possible to check the convergence rate of the above limit? More specifically, I wonder whether the convergence rate is faster than $n$, i.e., whether $n[1-P(n;\xi n)]=O(1)$?

I checked using software that the convergence rate is faster than $n$, but I don't know how to show it rigorously. Can anyone provide some hints and insights? Thanks!

ydwang
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1 Answers1

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We have $$S_n= \sum_{k=0}^n\frac{(n\xi)^ke^{-n\xi}}{k!}=\frac{\Gamma (n+1,n \xi )}{\Gamma (n+1)}$$

If you consider the "easier" case $$T_n=\frac{\Gamma (n+1,(n+1) \xi )}{\Gamma (n+1)}$$

Now, consider Nemes's paper (have a look here at formula $8.11.7$, truncating the summation to $k=2$, taking logarithms, using Stirling expansion, it seems that at the first order (hoping no mistake) $$\log(T_n) \sim n (1-\xi +\log (\xi ))$$

Hoping that this could help.

Claude Leibovici
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