证明二项分布的逼近

$\phantom{\rule{.85em}{0ex}}\text{P}\left(X=k\right)\text{≈}\frac{{\lambda }^{k}}{k!}{e}^{-\lambda }\left(k\text{= 0, 1, 2, ...}\right)$

$\phantom{\rule{.85em}{0ex}}{\text{C}}_{n}^{k}\phantom{\rule{.2em}{0ex}}{p}^{k}\phantom{\rule{.2em}{0ex}}{\left(1-p\right)}^{n-k}$

$\text{=}\phantom{\rule{.2em}{0ex}}\frac{n!}{k!\left(n-k\right)!}\phantom{\rule{.2em}{0ex}}{p}^{k}\phantom{\rule{.2em}{0ex}}{\left(1-p\right)}^{n-k}$

$\text{=}\phantom{\rule{.2em}{0ex}}\frac{n\left(n-1\right)\text{...}\left(n-k+1\right)}{k!}\phantom{\rule{.2em}{0ex}}{p}^{k}\phantom{\rule{.2em}{0ex}}{\left(1-p\right)}^{n-k}$

$\phantom{\rule{.85em}{0ex}}\underset{p\to 0+}{\underset{n\to \infty }{\text{lim}}}\frac{n\left(n-1\right)\text{...}\left(n-k+1\right)}{k!}\phantom{\rule{.2em}{0ex}}{p}^{k}\phantom{\rule{.2em}{0ex}}{\left(1-p\right)}^{n-k}$

$\text{=}\phantom{\rule{.2em}{0ex}}\underset{p\to 0+}{\underset{n\to \infty }{\text{lim}}}\frac{n\left(n-1\right)\text{...}\left(n-k+1\right){p}^{k}}{k!}\phantom{\rule{.2em}{0ex}}{\left(1-p\right)}^{-\frac{1}{p}\left(-pn\right)}$

$\underset{x\to 0}{\text{lim}}{\left(1+x\right)}^{\frac{1}{x}}$

$\text{=}\phantom{\rule{.2em}{0ex}}\underset{p\to 0+}{\underset{n\to \infty }{\text{lim}}}\frac{{\left(np\right)}^{k}}{k!}{e}^{\left(-np\right)}$

$\text{=}\phantom{\rule{.2em}{0ex}}\frac{{\lambda }^{k}}{k!}{e}^{-\lambda }$