MemexPoisson Lawb(n, p)→dP(λ) if n≫1 & p≪1 & λ=npb(n,~p) \xrightarrow{d} P(\lambda) ~ \text{if} ~ n \gg 1 ~ \& ~ p \ll 1 ~ \& ~ \lambda = npb(n, p)dP(λ) if n≫1 & p≪1 & λ=np Proof limn→∞b(n, p)=limn→∞(nx)px(1−p)n−x=limn→∞n!(n−x)! x!px(1−p)n−x\lim_{n \to \infty} b(n, ~p) = \lim_{n \to \infty} {n \choose x} p^x (1-p) ^{n-x} = \lim_{n \to \infty} {n! \over {(n-x)! ~ x!}} p^x (1-p) ^{n-x}n→∞limb(n, p)=n→∞lim(xn)px(1−p)n−x=n→∞lim(n−x)! x!n!px(1−p)n−x =limn→∞n!x!(n−x)!λxnx(1−λn)n−x=λxx!limn→∞n!(n−x)!1nx(1−λn)n−x= \lim_{n \to \infty} {n! \over {x! (n-x)!}} {\lambda^x \over n^x} (1-{\lambda \over n})^{n-x} = {\lambda^x \over x!} \lim_{n \to \infty} {n! \over {(n-x)!}} {1 \over n^x} (1- {\lambda \over n})^{n-x}=n→∞limx!(n−x)!n!nxλx(1−nλ)n−x=x!λxn→∞lim(n−x)!n!nx1(1−nλ)n−x =λxx!limn→∞nnn−1nn−2n⋯n−x+1n(1−λn)n(1−λn)−x= {\lambda^x \over x!} \lim_{n \to \infty} {n \over n} {{n-1} \over n} {{n-2} \over n} \cdots {{n-x+1} \over n} (1 - {\lambda \over n})^n (1 - {\lambda \over n})^{-x}=x!λxn→∞limnnnn−1nn−2⋯nn−x+1(1−nλ)n(1−nλ)−x =λxx!limn→∞1⋅limn→∞n−1n⋯limn→∞n−x+1nlimn→∞(1−λn)nlimn→∞(1−λn)−x= {\lambda^x \over x!} \lim_{n \to \infty} 1 \cdot \lim_{n \to \infty} {{n-1} \over n} \cdots \lim_{n \to \infty} {{n-x+1} \over n} \lim_{n \to \infty} ({1- {\lambda \over n}})^n \lim_{n \to \infty} ({1- {\lambda \over n}})^{-x}=x!λxn→∞lim1⋅n→∞limnn−1⋯n→∞limnn−x+1n→∞lim(1−nλ)nn→∞lim(1−nλ)−x =λxx!limn→∞(1+−λn)n=λxe−λx!= {\lambda^x \over x!} \lim_{n \to \infty} (1+ {-\lambda \over n})^n = {{\lambda^x e^{-\lambda}} \over x!}=x!λxn→∞lim(1+n−λ)n=x!λxe−λStirling ApproximationPrevious PageCountingNext Page