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What is the significance of "large city" in the definition of the following Poisson variable :

"Number of phone calls placed during a ten second interval in a large city"

I guess either $n \to \infty$ or $p$ very small is not implied by "large city".

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4 Answers 4

A Poisson r.v. $N$ follows a Poisson distribution, which has only one parameter $\lambda$; and takes values in $\mathbb{N}$ — what do you refer to with $n$ and $p$? (are you thinking instead of a binomial distribution $\operatorname{Bin}(n,p)$?)

I assume the "large city" assumption is just so that you don't have to worry about the number of calls being too big wrt the number of people in the city (technically, if you model it with a Poisson-distributed r.v. $N$, it could take any non-negative integer value with non-zero probability, including very large ones).

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This is not quite a definition, just an illustration of a quantity, well modeled by a Poisson random variable. I am not sure what $n,p$ in your question are, but I think that you need a large city because if you have a small population sample, they don't call each other according to the same frequency distribution...

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I assume the OP is referring to this http://en.wikipedia.org/wiki/Poisson_limit_theorem for $n, p$.

$p$ is probability that a person makes a phone call during a particular 10-second interval, which is very small. $n$ is the number of people in the city. The large city assumption guarantees $n$ is large, so this can be modeled by $Poisson(np)$.

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Consider the binomial random variable: $P(X=i)=\frac{n!}{(n-i)!i!}p^i(1-p)^{n-i}$

If you let $n\rightarrow \infty$ and $p \rightarrow 0$ for fix $np$ and consider $p=\frac{\mu}{n}$

Use the following identities:

$\lim_{n \to \infty} (1-\frac{\mu}{n})^n= e^{-\mu}$ and $\frac{n!}{(n-i)!} \approx n^i$ and you arrive at the Poisson random variable:

$P(X=i)=\frac{e^{-\mu}\mu^{i}}{i!}$

Thus, to model a phenomenon with Poisson distribution its important that the population is large ($n\rightarrow \infty$) and events are rare ($p \rightarrow 0$).

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