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In showing us that Binomial distribution: $$B_{N,p}(n) := \binom {N}{n} p^n(1-p)^{N-n}$$ tends to Poisson's: $$P_ \lambda (n) = \dfrac {\lambda ^n}{n!}e^{-\lambda}$$where I guess lambda should be defined as $\lambda:=\lim _N Np$ (it is the limit of the expected value of $\cal B$), my (mechanics) teacher did something i don't understand: he substituted $p=\frac {\lambda}{N}$ before evaluating the limit as $N$ goes to infty. Is this correct? If $\lambda$ is defined as I said above, then I think it isn't, for it must be calculated inside the limit substituting $p=\frac {Np}{N}$. Also, the simplification makes the computation of the limit much easier, so maybe that's why he did it. Note: I'm not looking for a full demonstration, I just want to know if the procedure of my teacher is correct. Related question: How would you prove $\lim _N (1-p)^N=e^{-\lambda}$, given $ \lim _N Np =\lambda \in \mathbb R $? Using the result: $$\lim _{N\rightarrow +\infty} (1+\frac{1}{N})^N = e$$ my attempt was:

$$(1-p)^N=\left[\left(1-\dfrac{Np}{N}\right)^{-\frac{N}{Np}}\right]^{-Np}=\exp\left\{-Np\cdot\ln\left[\left(1-\dfrac{Np}{N}\right)^{-\frac{N}{Np}}\right]\right\}$$

$$\ell = \exp\left\{-\lim_{N\rightarrow+\infty}Np\cdot\ln\left[\lim_{N\rightarrow+\infty}\left(1-\dfrac{Np}{N}\right)^{-\frac{N}{Np}}\right]\right\}=e^{-\lambda}.$$

Notice that this is not immediate since $p=f(n)$.

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  • $\begingroup$ What you call a Bernoulli is actually a Binomial, no? $\endgroup$ – leonbloy Apr 17 '13 at 15:05
  • $\begingroup$ Yes, isn't it the same? $\endgroup$ – pppqqq Apr 17 '13 at 17:00
  • $\begingroup$ AFAIK, a Bernoulli variable takes values 0/1, and the Binomial is the sum of $n$ Bernoulli. $\endgroup$ – leonbloy Apr 17 '13 at 18:04
  • $\begingroup$ Thank you, i'm really talking about binomial distribution. $\endgroup$ – pppqqq Apr 17 '13 at 18:07
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What he's proving is a weaker result that says if $p=\lambda/N$ then $\displaystyle\binom N n p^n(1-p)^{N-n}\to\frac{\lambda^n e^{-\lambda}}{n!}$ as $N\to\infty$. That is of interest in its own right, but what is proved is less than what would have been proved if the assumption had been that $Np\to\lambda$ rather than $Np=\lambda$.

The other result you ask about is that $$ \left(1-\frac\lambda N\right)^N \to e^{-\lambda}\text{ as }N\to\infty. $$ That's usually stated in calculus texts in something like the following form $$ \lim_{N\to\infty}\left(1+\frac x N\right)^N = e^x.\tag{1} $$ And at this point I'm realizing that the really short and simple proof of this that I might have posted is not sufficiently organized in my mind to make it look as short and simple as it is without some further thought. Maybe later. However, $(1)$ is a standard result in calculus that you should be aware of, and that is probably considered part of the prerequisite material for the course you're in.

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  • $\begingroup$ Thank you for the first part, i got it. About second part, my question was another one, I must have stated it not very well. It was: given $Np \rightarrow \lambda $ and the classic limit $ (1+1/n)^n \rightarrow e $, to show $ (1-p)^n \rightarrow e^{-\lambda} $. Since $p=f(n)$ you can't immediately apply the classic result, right? $\endgroup$ – pppqqq Apr 17 '13 at 16:59
  • $\begingroup$ In the "classic result" in the form in which I stated it, the thing in the role of $p$ does depend on $N$. In the notation of your original posting, it should be called $N$ rather than $n$. $\endgroup$ – Michael Hardy Apr 17 '13 at 22:27

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