泊松分布概率公式的推导

泊松分布概率公式的推导

前言：重在记录，可能出错。

在二项式分布中，假设λ=np,n→∞,p→0,得p=λ/n,对二项式分布的概率公式求极限，即有以下推导

$$\begin{array}{l}\underset{\mathit{n}\to \infty }{\text{l}\text{i}\text{m}}{\mathit{C}}_{\mathit{k}}^{\mathit{n}}\mathit{p}{}^{\mathit{k}}(1-\mathit{p}){}^{\mathit{n}-\mathit{k}}\\ =\underset{\mathit{n}\to \infty }{\text{l}\text{i}\text{m}}\frac{\mathit{n}!}{\mathit{k}!(\mathit{n}-\mathit{k})!}\frac{\mathit{\lambda }{}^{\mathit{k}}}{\mathit{n}{}^{\mathit{k}}}(1-\frac{\mathit{\lambda }}{\mathit{n}}){}^{\mathit{n}}(1-\frac{\mathit{\lambda }}{\mathit{n}}){}^{-\mathit{k}}\text{最}\text{后}\text{一}\text{项}\text{的}\text{极}\text{限}\text{为}1\text{，}\text{因}\text{此}\text{省}\text{略}\\ =\underset{\mathit{n}\to \infty }{\text{l}\text{i}\text{m}}\frac{\mathit{n}⸳(\mathit{n}-1)(\mathit{n}-2)\bullet \bullet \bullet (\mathit{n}-\mathit{k}+1)}{\mathit{n}{}^{\mathit{k}}}\frac{\mathit{\lambda }{}^{\mathit{k}}}{\mathit{k}!}(1-\frac{\mathit{\lambda }}{\mathit{n}}){}^{\mathit{n}}\text{第}\text{一}\text{项}\text{的}\text{极}\text{限}\text{为}1\text{，}\text{因}\text{此}\text{省}\text{略}\\ =\underset{\mathit{n}\to \infty }{\text{l}\text{i}\text{m}}\frac{\mathit{\lambda }{}^{\mathit{k}}}{\mathit{k}!}\left[\right(1+\frac{1}{-\frac{\mathit{n}}{\mathit{\lambda }}}\left){}^{-\mathit{n}/\mathit{\lambda }}\right]{}^{-\mathit{\lambda }}\text{最}\text{后}\text{一}\text{项}\text{中}\text{括}\text{号}\text{里}\text{符}\text{合}(1+\frac{1}{\mathit{n}}){}^{\mathit{n}}\text{的}\text{形}\text{式}\text{，}\text{所}\text{以}\text{极}\text{限}\text{为}\mathit{e}\\ =\frac{\mathit{\lambda }{}^{\mathit{k}}\mathit{e}{}^{-\mathit{\lambda }}}{\mathit{k}!}\end{array}$$

最终推导出来的结果，就是泊松分布的概率公式了——$$\frac{\mathit{\lambda }{}^{\mathit{k}}\mathit{e}{}^{-\mathit{\lambda }}}{\mathit{k}!}$$