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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 $2$?

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Where is binomial distribution linked with powers of 2? And $\exp(-\lambda) = 2^{-\lambda \cdot \log_2 e}$, so write it in powers of 2, if you wish :) – martini Jul 24 '12 at 7:58
@martini: Probably powers of two is too much, but yes/no decissions. And the relation you posted is true, but no unnice factor like $\log(e)$ appreas in the Poisson distribution formula. – NikolajK Jul 24 '12 at 8:12
This 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 $a$. – Did Jul 24 '12 at 8:15
up vote 4 down vote accepted

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 $n\to\infty$.

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