二項分布(binomial distribution)からポアソン分布(Poisson distribution)を導く

二項分布 $B(n, p)$

$$B(n,p)=f_X(x)=\{\begin{array}{ll}{\displaystyle {}_n{C}_x{p}^x(1-p{)}^{n-x}}& x\in \{0,1,2,\dots ,n\}\\ {\displaystyle 0}& x\notin \{0,1,2,\dots ,n\}\end{array}$$

$_nC_x$の展開と変形

$$\begin{array}{rcl}{\displaystyle {}_n{C}_x}& =& {\displaystyle \frac{n!}{x!(n-x)!}}\\ & =& {\displaystyle \frac{1}{x!}\frac{n!}{(n-x)!}}\\ & =& {\displaystyle \frac{1}{x!}n(n-1)(n-2)\cdots (n-x+1)}\\ & =& {\displaystyle \frac{1}{x!}\{\frac{nn}{n}\frac{n(n-1)}{n}\frac{n(n-2)}{n}\cdots \frac{n(n-x+1)}{n}\}}\\ & =& {\displaystyle \frac{1}{x!}{n}^x\{\frac{n}{n}\frac{(n-1)}{n}\frac{(n-2)}{n}\cdots \frac{(n-x+1)}{n}\}}\\ & =& {\displaystyle \frac{n^x}{x!}\{\frac{n}{n}(\frac{n}{n}-\frac{1}{n})(\frac{n}{n}-\frac{2}{n})\cdots (\frac{n}{n}-\frac{x-1}{n})\}}\\ & =& {\displaystyle \frac{n^x}{x!}\{1(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{x-1}{n})\}}\\ & =& {\displaystyle \frac{n^x}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{x-1}{n})\}}\end{array}$$

$(1-p)^{n-x}$の展開と変形

$$\begin{array}{rcl}{\displaystyle (1-p{)}^{n-x}}& =& {\displaystyle (1-p{)}^n(1-p{)}^{-x}}\\ & =& {\displaystyle {\left[{[\{1+(-p){\}}^{\frac{1}{-p}}]}^{-p}\right]}^n(1-p{)}^{-x}\dots a=(a^{\frac{1}{x}}{)}^x}\\ & =& {\displaystyle \{{\left(\mathrm{e}\right)}^{-p}{\}}^n(1-p{)}^{-x}\dots {(1+x)}^{\frac{1}{x}}={(1+\frac{1}{y})}^y=\mathrm{e}}\\ & =& {\displaystyle {\mathrm{e}}^{-np}(1-p{)}^{-x}\dots {\left(a^x\right)}^y=a^{xy}}\end{array}$$

展開と変形の結果を当てはめる

$$\begin{array}{rcl}{\displaystyle {}_n{C}_x{p}^x(1-p{)}^{n-x}}& =& {\displaystyle \frac{n^x}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{k-1}{n})\}{p}^x{\mathrm{e}}^{-np}(1-p{)}^{-x}}\\ & =& {\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{k-1}{n})\}{n}^x{p}^x{\mathrm{e}}^{-np}(1-p{)}^{-x}}\\ & =& {\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{k-1}{n})\}(np{)}^x{\mathrm{e}}^{-np}(1-p{)}^{-x}}\end{array}$$

$\lambda=np$，$\lambda$を一定に$n$を十分大きくとると, $p$は極めて小さくなる

$$\begin{array}{rcl}{\displaystyle f_X(x)}& =& {\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{k-1}{n})\}(np{)}^x{\mathrm{e}}^{-np}(1-p{)}^{-x}}\\ & =& {\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{k-1}{n})\}(\lambda {)}^x{\mathrm{e}}^{-\lambda }(1-p{)}^{-x}\dots \lambda =np}\\ & =& {\displaystyle \frac{1}{x!}\{(1)(1)\cdots (1)\}(\lambda {)}^x{\mathrm{e}}^{-\lambda }(1{)}^{-x}\dots \frac{C}{n}\to 0,p\to 0}\\ & =& {\displaystyle \frac{1}{x!}{\lambda }^x{\mathrm{e}}^{-\lambda }}\\ & =& {\displaystyle \frac{{\lambda }^x}{x!}{\mathrm{e}}^{-\lambda }}\\ & =& {\displaystyle Po(\lambda )}\end{array}$$

積率母凾数(moment-generating function)を考える

$$\begin{array}{rcl}{\displaystyle M_X(t)}& \equiv & {\displaystyle E[{\mathrm{e}}^{tX}]}\\ & =& {\displaystyle \sum _{x=0}^{\mathrm{\infty }}({\mathrm{e}}^{tx})\frac{{\lambda }^x}{x!}{\mathrm{e}}^{-\lambda }}\\ & =& {\displaystyle {\mathrm{e}}^{-\lambda }\sum _{x=0}^{\mathrm{\infty }}({\mathrm{e}}^{tx})\frac{{\lambda }^x}{x!}}\\ & =& {\displaystyle {\mathrm{e}}^{-\lambda }\sum _{x=0}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{tx}{\lambda }^x}{x!}}\\ & =& {\displaystyle {\mathrm{e}}^{-\lambda }\sum _{x=0}^{\mathrm{\infty }}\frac{({\mathrm{e}}^t\lambda {)}^x}{x!}}\\ & =& {\displaystyle {\mathrm{e}}^{-\lambda }{\mathrm{e}}^{{\mathrm{e}}^t\lambda }\dots \sum _{x=0}^{\mathrm{\infty }}\frac{a^x}{x!}={\mathrm{e}}^a}\\ & =& {\displaystyle {\mathrm{e}}^{{\mathrm{e}}^t\lambda -\lambda }={\mathrm{e}}^{\lambda ({\mathrm{e}}^t-1)}}\end{array}$$

期待値，分散

$$\begin{array}{rcl}{\displaystyle M_X^{(m)}(0)}& \equiv & \frac{{\mathrm{d}}^m}{{\mathrm{d}}^{m}t}{M}_x(t){|}_{t=0}\\ & =& {\displaystyle E[X^m{\mathrm{e}}^{tX}]{|}_{t=0}}\\ & =& {\displaystyle E[X^m]}\end{array}$$ $$\begin{array}{rcl}{\displaystyle E[X]}& =& {\displaystyle M_X^1(0)}\\ & =& {\displaystyle \{\frac{\mathrm{d}}{\mathrm{d}t}({\mathrm{e}}^{\lambda ({\mathrm{e}}^t-1)})\}{|}_{t=0}}\\ & =& {\displaystyle \{\frac{\mathrm{d}}{\mathrm{d}s}({\mathrm{e}}^s)\frac{\mathrm{d}s}{\mathrm{d}t}\}{|}_{t=0}\dots s=\lambda ({\mathrm{e}}^t-1),\frac{\mathrm{d}s}{\mathrm{d}t}=\lambda {\mathrm{e}}^t}\\ & =& {\displaystyle \{({\mathrm{e}}^{\lambda ({\mathrm{e}}^t-1)})(\lambda {\mathrm{e}}^t)\}{|}_{t=0}\dots \frac{\mathrm{d}}{\mathrm{d}x}{\mathrm{e}}^x={\mathrm{e}}^x}\\ & =& {\displaystyle \{\lambda ({\mathrm{e}}^{\lambda ({\mathrm{e}}^t-1)+t})\}{|}_{t=0}}\\ & =& {\displaystyle \lambda ({\mathrm{e}}^{\lambda ({\mathrm{e}}^0-1)+0})}\\ & =& {\displaystyle \lambda {\mathrm{e}}^0\dots a^0=1}\\ & =& \lambda \dots a^0=1\end{array}$$ $$\begin{array}{rcl}{\displaystyle E[X^2]}& =& {\displaystyle M_X^2(0)}\\ & =& {\displaystyle \{\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}({\mathrm{e}}^{\lambda ({\mathrm{e}}^t-1)})\}{|}_{t=0}}\\ & =& {\displaystyle \{\frac{\mathrm{d}}{\mathrm{d}t}\lambda ({\mathrm{e}}^{\lambda ({\mathrm{e}}^t-1)+t})\}{|}_{t=0}\dots E[X]\mathrm{の}\mathrm{展}\mathrm{開}\mathrm{か}\mathrm{ら}．}\\ & =& {\displaystyle \{\frac{\mathrm{d}}{\mathrm{d}s}(\lambda {\mathrm{e}}^s)\frac{\mathrm{d}s}{\mathrm{d}t}\}{|}_{t=0}\dots s=\lambda ({\mathrm{e}}^t-1)+t,\frac{\mathrm{d}s}{\mathrm{d}t}=\lambda {\mathrm{e}}^t+1}\\ & =& {\displaystyle \{(\lambda {\mathrm{e}}^{\lambda ({\mathrm{e}}^t-1)+1})(\lambda {\mathrm{e}}^t+1)\}{|}_{t=0}\dots \frac{\mathrm{d}}{\mathrm{d}x}C{\mathrm{e}}^x=C{\mathrm{e}}^x}\\ & =& {\displaystyle (\lambda {\mathrm{e}}^{\lambda ({\mathrm{e}}^0-1)+1})(\lambda {\mathrm{e}}^0+1)}\\ & =& {\displaystyle \lambda {\mathrm{e}}^0(\lambda +1)\dots a^0=1}\\ & =& \lambda (\lambda +1)\dots a^0=1\end{array}$$ $$\begin{array}{rcl}{\displaystyle V[X]}& =& {\displaystyle E[X^2]-E[X{]}^2}\\ & =& \lambda (\lambda +1)-{\lambda }^2\\ & =& {\lambda }^2+\lambda -{\lambda }^2\\ & =& \lambda \end{array}$$