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Can someone help prove this theorem? Many thanks!

If $p\to0$ and $n\to\infty$ in such a way that $\lim np = \lambda > 0$, then for $k=0, 1,\dots$: $$\lim_{n\to\infty}\binom nkp^k (1-p)^{n-k}=\frac{\lambda^k}{k!} e^{-\lambda}$$

kjetil b halvorsen
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cliu
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    One standard way is to find the limit of the logarithm by applying Stirling's approximation to $\log\binom{n}{k}=\log\Gamma(n+1)-\log\Gamma(n-k+1)-\log(k!).$ Another is to analyze the convergence of the characteristic or cumulant generating functions of the Binomial distribution. There also exist perfectly elementary demonstrations (but they tend to be lengthier). If you're doing this for homework or self-study, then, please indicate what facts you are able to use in the proof, for otherwise you may get answers that are not appropriate for this exercise. – whuber Apr 11 '20 at 14:15
  • https://en.wikipedia.org/wiki/Poisson_limit_theorem – StubbornAtom Apr 11 '20 at 16:20
  • See https://stats.stackexchange.com/questions/261119/intuitively-understand-why-the-poisson-distribution-is-the-limiting-case-of-the – kjetil b halvorsen Apr 11 '20 at 16:31

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One of the elementary versions follows as below, (a simplified and common version with $\lambda=np$, instead of limiting): $$\begin{align}\lim_{n\rightarrow\infty}{n\choose k}p^k(1-p)^{n-k}&=\lim_{n\rightarrow\infty }\frac{n!}{(n-k)!k!}\left(\frac{\lambda}{n}\right)^k\left(1-\frac{\lambda}{n}\right)^{n-k}\\&=\frac{\lambda^k}{k!}\lim_{n\rightarrow\infty }\underbrace{\frac{n!}{(n-k)!n^k}}_{\rightarrow1}\underbrace{\left(1-{\lambda\over n}\right)^{n}}_{\rightarrow e^{-\lambda}}\underbrace{\left(1-{\lambda\over n}\right)^{-k}}_{\rightarrow1}\end{align}$$

It's a good exercise to show each of these limits.

gunes
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  • I have a problem with the last term, which is $(1-p)^{-k}$, independent of $n$. I do not see it why it should approach 1. – F. Tusell Apr 11 '20 at 16:11
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    @F.Tusell $p$ is not independent of $n$ since they should satisfy $pn=\lambda$ – gunes Apr 11 '20 at 16:12
  • OK, clear enough. I had missed the first bit of the question "If $p\to 0\ldots$". Thank you. – F. Tusell Apr 11 '20 at 16:16
  • You assume throughout that $\lambda=np$ but it is only assumed that $\lambda=\lim np.$ The substitution of $\lambda$ for $np$ is not justified, even though it does produce the desired answer. – whuber Apr 11 '20 at 18:08
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    @whuber thanks, I've at least noted down in my answer. It's as if the limit sign is hidden in between the words. – gunes Apr 11 '20 at 18:28