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Show using the Poisson distribution that

$$\lim_{n \to +\infty} e^{-n} \sum_{k=1}^{n}\frac{n^k}{k!} = \frac {1}{2}$$

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marked as duplicate by Eric Naslund, Macavity, Davide Giraudo, Cameron Buie, Micah Oct 1 '13 at 16:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Second hint, to supplement the Poisson hint: central limit theorem. (Is this (homework)?) – Did Mar 16 '12 at 22:28
It is not homework, just personal interest. I picked up the problem here: – wnvl Mar 16 '12 at 22:32
@wnvl : You should be less formal when you ask questions here and show a little what you've tried or where you are stuck (or admit that you don't know where to start, if that is). We're humans too you know =P – Patrick Da Silva Mar 16 '12 at 22:55
The same question was asked here: – Martin Sleziak May 25 '12 at 11:03
up vote 14 down vote accepted

By the definition of Poisson distribution, if in a given interval, the expected number of occurrences of some event is $\lambda$, the probability that there is exactly $k$ such events happening is $$ \frac {\lambda^k e^{-\lambda}}{k!}. $$ Let $\lambda = n$. Then the probability that the Poisson variable $X_n$ with parameter $\lambda$ takes a value between $0$ and $n$ is $$ \mathbb P(X_n \le n) = e^{-n} \sum_{k=0}^n \frac{n^k}{k!}. $$ If $Y_i \sim \mathrm{Poi}(1)$ and the random variables $Y_i$ are independent, then $\sum\limits_{i=1}^n Y_i \sim \mathrm{Poi}(n) \sim X_n$, hence the probability we are looking for is actually $$ \mathbb P\left( \frac{Y_1 + \dots + Y_n - n}{\sqrt n} \le 0 \right) = \mathbb P( Y_1 + \dots + Y_n \le n) = \mathbb P(X_n \le n). $$ By the central limit theorem, the variable $\frac {Y_1 + \dots + Y_n - n}{\sqrt n}$ converges in distribution towards the Gaussian distribution $\mathscr N(0, 1)$. The point is, since the Gaussian has mean $0$ and I want to know when it is less than equal to $0$, the variance doesn't matter, the result is $\frac 12$. Therefore, $$ \lim_{n \to \infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!} = \lim_{n \to \infty} \mathbb P(X_n \le n) = \lim_{n \to \infty} \mathbb P \left( \frac{Y_1 + \dots + Y_n - n}{\sqrt n} \le 0 \right) = \mathbb P(\mathscr N(0, 1) \le 0) = \frac 12. $$

Hope that helps,

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Edited some confusion between $X_1$ and $Y_i$, just revert to the previous version if you disagree. // The end of the argument does not apply because $\sigma$ depends on $n$ hence $P(N(1,\sigma)\leqslant1)$ cannot be a limit when $n\to\infty$. The correct approach is to apply the CLT to the event $[X_n\leqslant n]=[(S_n-n)/\sqrt{n}\leqslant0]$ where $S_n=Y_1+\cdots+Y_n$ hence $(S_n-n)/\sqrt{n}$ converges in distribution to $N(0,a)$ for some positive $a$ whose value is irrelevant. – Did Mar 17 '12 at 11:54
Curious to know how many upvoters understand the answer... :-) – Did Mar 17 '12 at 11:56
@DidierPiau: Not me. Nice avatar! – Tim Mar 17 '12 at 12:04
@BCLC : That's precisely the CLT : that a sum of i.i.d variables (minus the average divided by standard deviation) converges in distribution to the normal distribution, that is, $\lim_{n \to \infty} \mathbb P \left( \frac {Y_1 + \cdots + Y_n - \mu}{\sigma} \le x \right) \to \mathbb P \left(Z \le x \right)$ when $Y_i$ follows some distribution of mean $\mu$ and variance $\sigma^2$ and $Z \sim \mathcal N(0,1)$. So I did not really switch $\lim$ and $\mathbb P$ properly speaking, I just applied the CLT ; that's because the CLT only guarantees convergence in distribution. – Patrick Da Silva Aug 8 '15 at 3:02
@BCLC : Yes, because the normal distribution is given by a smooth (and in particular continuous) density. – Patrick Da Silva Aug 8 '15 at 3:17

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