Why does the Poisson distribution $\!f(k; \lambda)= \Pr(X=k)= \frac{\lambda^k \exp{(-\lambda})}{k!}$ contain the exponential function $\exp$, while its relation to the binomial distribution would suggest it's all about powers of $2$?
Going from binomial distribution to Poisson distribution
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2This is no more and no less odd than the convergence of $\left(1+\frac{a}n\right)^n$ to $\mathrm e^a$, for every value of $a$. – 2012-07-24
1 Answers
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The Binomial $(n,p)$ distribution puts weight $p_n(k)={n\choose k}p^k(1-p)^{n-k}$ on each integer $k$ such that $0\leqslant k\leqslant n$. Fix $k\geqslant0$ and let $n\to\infty$ and $p\to0$ in such a way that $pn\to a$.
Then, ${n\choose k}\sim\frac{n^k}{k!}$, $p^k\sim \left(\frac{a}n\right)^k$, $(1-p)^n\to\mathrm e^{-a}$ and $(1-p)^{-k}\to1$. Hence $p_n(k)\to\mathrm e^{-a}\frac{a^k}{k!}$ the weight at $k$ of the Poisson distribution of parameter $a$.
Finally, the appearance of $\mathrm e^{-a}$ is due to the fact that $(1-p)^n\to\mathrm e^{-a}$ when $n\to\infty$ and $p\to0$ in the regime $pn\to a$, that is, to the fact that $\left(1-\frac{a}n\right)^n\to\mathrm e^{-a}$ when $n\to\infty$.