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This question already has an answer here:

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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    $\begingroup$ Second hint, to supplement the Poisson hint: central limit theorem. (Is this (homework)?) $\endgroup$ – Did Mar 16 '12 at 22:28
  • $\begingroup$ It is not homework, just personal interest. I picked up the problem here: mymathforum.com/viewtopic.php?f=24&t=28627. $\endgroup$ – wnvl Mar 16 '12 at 22:32
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    $\begingroup$ @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 $\endgroup$ – Patrick Da Silva Mar 16 '12 at 22:55
  • $\begingroup$ The same question was asked here: sosmath.com/CBB/viewtopic.php?t=28258 $\endgroup$ – Martin Sleziak May 25 '12 at 11:03
  • $\begingroup$ The Poisson distribution has the properties that, if the mean is an integer, (a) the median is equal to the mean and (b) the modal values are the mean and one less than the mean. Property (a) implies that the sum in this question is at least $\frac12$ and that without its final term the sum would be less than $\frac12$, with the difference reducing towards $0$ as $n$ increases $\endgroup$ – Henry Dec 4 '17 at 0:15
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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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    $\begingroup$ 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. $\endgroup$ – Did Mar 17 '12 at 11:54
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    $\begingroup$ Curious to know how many upvoters understand the answer... :-) $\endgroup$ – Did Mar 17 '12 at 11:56
  • $\begingroup$ @DidierPiau: Not me. Nice avatar! $\endgroup$ – Tim Mar 17 '12 at 12:04
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    $\begingroup$ @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. $\endgroup$ – Patrick Da Silva Aug 8 '15 at 3:02
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    $\begingroup$ @BCLC : Yes, because the normal distribution is given by a smooth (and in particular continuous) density. $\endgroup$ – Patrick Da Silva Aug 8 '15 at 3:17

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