3

I'd like to come up with a probabilistic argument to evaluate the following limit: $$\lim_{n\to\infty} e^{-n} \sum_{k = 0}^n \frac{n^k}{k!}$$

The reason is that, if $X \sim \mathrm{Poisson}(n)$, the limit translates to $$\lim_{n\to\infty} F_X(n)$$ where $F_X$ is the cumulative distribution function of $X$. I know that it can be expressed by resorting to the gamma function and its incomplete counterpart, but I don't think I'll be able to evaluate the limit if I substitute that expression.

Is there a simpler probabilistic argument in this case? In the meantime, I confirmed the limit indeed exists, and it's $1/2$:

limit graph

import scipy as sp, matplotlib.pyplot as plt

F_X = lambda n: (sp.stats.distributions.poisson.rvs(n, size=100000) <= n).mean()
ns = np.logspace(0, 6, 30)
fig, ax = plt.subplots(figsize=(10, 6))
ax.semilogx(ns, list(map(F_X, ns)), marker='o')
ax.set_xlabel('$n$', fontsize=16)
ax.set_ylabel('$F_X$', fontsize=16)
Graham Kemp
  • 129,094
rubik
  • 9,344
  • 3
    One of the answers here gives a probabilistic method: https://math.stackexchange.com/questions/160248/evaluating-lim-limits-n-to-infty-e-n-sum-limits-k-0n-fracnkk – Travis Willse Jul 04 '17 at 11:07
  • Related: https://math.stackexchange.com/questions/2091338/rigorous-real-analysis-proof-of-de-moivre-laplace-theorem – Jack D'Aurizio Jul 04 '17 at 11:10
  • 1
    Hints: 1. Sum of $n$ possion random $Po(1)$ variables is a possion random variable $Po(n)$. 2. Central limit theorem. 3. For $X \sim \mathcal{N}(0, \sigma^2), P(X>0) = \frac{1}{2}$ – Dhruv Kohli Jul 04 '17 at 11:34
  • You can use the central limit theorem. – Angina Seng Jul 04 '17 at 11:51

0 Answers0