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I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.

Consider the binomial distribution:

$$ \begin{equation}\begin{aligned} P(X=k) &={\binom n k} p^k (1-p)^{n-k}\\ &=\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} \end{aligned}\end{equation} $$
Substitute $m=np$ , or $p=\frac{m}{n}$ : $$ \begin{equation}\begin{aligned} P(X=k) &=\frac{n!}{k!(n-k)!} \left(\frac{m}{n}\right)^k \left(1-\frac{m}{n}\right)^{n-k}\\ &=\frac{n!}{k!(n-k)!} \frac{m^k}{n^k} \left(1-\frac{m}{n}\right)^{n}\left(1-\frac{m}{n}\right)^{-k} \end{aligned}\end{equation} $$
Slightly rearrange $$ \begin{equation}\begin{aligned} &=\frac{n!}{n^k(n-k)!} \left(1-\frac{m}{n}\right)^{-k}\frac{m^k}{k!}\left(1-\frac{m}{n}\right)^{n} \end{aligned}\end{equation} $$

Note that $$ \begin{equation}\begin{aligned} & \lim_{n\rightarrow \infty} \frac{n!}{n^{k}(n-k)! } =1,\quad\lim_{n\rightarrow \infty} \left(1-\frac{m}{n}\right)^{-k} =1,\quad \lim_{n\rightarrow \infty} \left(1-\frac{m}{n}\right)^{n} =e^{-m} \end{aligned}\end{equation} $$

Thus, we have the final result which is equal to the formula for the Poisson distribution.

$$ =\frac{m^k e^{-m}}{k!} $$

In all these steps, what I don't understand is the following limit: $$ \lim_{n\rightarrow \infty} \frac{n!}{n^{k}(n-k)! } =1 $$

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It is rather obvious if you cancel the factorials:

$$\frac{n!}{n^{k}(n-k)! } =\frac{\overbrace{n(n-1)\cdots (n-k+1)}^{k\; factors}}{n^k}= 1\cdot \left(1-\frac{1}{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\stackrel{n \to \infty}{\longrightarrow} 1$$

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  • $\begingroup$ Thank you so much. I didn't know it was right there under my eyes. $\endgroup$ – billyandriam Apr 13 at 5:01
  • $\begingroup$ You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-) $\endgroup$ – trancelocation Apr 13 at 5:03
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$$a_n=\frac{n!}{n^{k}(n-k)! }\implies \log(a_n)=\log(n!)-k \log(n)-\log((n-k)!)$$

Use Stirling approximation and continue with Taylor series to get $$\log(a_n)=\frac{k(1-k)}{2 n}+O\left(\frac{1}{n^2}\right)$$ Continue with Taylor $$a_n=e^{\log(a_n)}=1+\frac{k(1-k)}{2 n}+O\left(\frac{1}{n^2}\right)$$

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