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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

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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: mymathforum.com/viewtopic.php?f=24&t=28627. –  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: sosmath.com/CBB/viewtopic.php?t=28258 –  Martin Sleziak May 25 '12 at 11:03

1 Answer 1

up vote 12 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
@Didier Piau : $\mathbb P( \mathscr N(1,\sigma_n) \le 1 )$ does not depend on $\sigma$, so I understand I did things wrong because I didn't apply CRT the most natural way, but what I said still stands, does it? –  Patrick Da Silva Mar 17 '12 at 16:09
@Didier Piau : I've given a little bit more thought about your comment and edited my answer. I was actually worried about the $X /Y$ thing, but you made it clear. And yes, usually when we "switch to the normal approximation" we always subtract the mean and divide through by the variance... I shouldn't have lost that reflex. After reading a little bit on the CLT again I got back on track and agreed with you. I edited my answer to reflect that. –  Patrick Da Silva Mar 17 '12 at 16:13

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