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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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  • $\lambda$ is the mean, and you are asking the limit of the probability of being at the mean or below. The median is close to the mean, so $\frac12$ would be a natural guess for the limit – Henry Mar 26 '21 at 14:58
  • Empirically for large $n$ it seems something like $\frac12 + \frac{0.266}{\sqrt{n}}$, confirmed in the answers an comments to the linked question – Henry Mar 26 '21 at 15:19
  • Let $X_n\sim \text{Poisson}(n)$. If $n$ is large then $X_n$ is approximately normal with mean $n$ and variance $n$. – Matthew H. Mar 26 '21 at 15:21
  • Also https://math.stackexchange.com/q/3000437/321264. – StubbornAtom Mar 26 '21 at 15:25

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$$\sum_{k=0}^n\frac{n^k\,e^{-n}}{k!}=\frac{\Gamma (n+1,n)}{\Gamma (n+1)} \to \frac 12$$

Claude Leibovici
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