二項分布 $B(n, p)$
$$B(n, p) = f_X(x) =
\begin{cases}
\displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotsc ,n\right\}\\
\displaystyle 0 & \quad x \notin \left\{0,1,2, \dotsc ,n\right\}
\end{cases}
$$
$_nC_x$の展開と変形
$$\begin{array}{rcl}
\displaystyle _nC_x&=&\displaystyle \frac{n!}{x!(n-x)!}\\
&=&\displaystyle \frac{1}{x!}\frac{n!}{(n-x)!}\\
&=&\displaystyle \frac{1}{x!}n(n-1)(n-2)\dotsm(n-x+1)\\
&=&\displaystyle \frac{1}{x!}\{\frac{n\,n}{n}\frac{n\,(n-1)}{n}\frac{n\,(n-2)}{n}\dotsm\frac{n\,(n-x+1)}{n}\}\\
&=&\displaystyle \frac{1}{x!}n^x\{\frac{n}{n}\frac{(n-1)}{n}\frac{(n-2)}{n}\dotsm\frac{(n-x+1)}{n}\}\\
&=&\displaystyle \frac{n^x}{x!}\{\frac{n}{n}(\frac{n}{n}-\frac{1}{n})(\frac{n}{n}-\frac{2}{n})\dotsm(\frac{n}{n}-\frac{x-1}{n})\}\\
&=&\displaystyle \frac{n^x}{x!}\{1(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{x-1}{n})\}\\
&=&\displaystyle \frac{n^x}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{x-1}{n})\}\\
\end{array}$$
$(1-p)^{n-x}$の展開と変形
$$\begin{array}{rcl}
\displaystyle (1-p)^{n-x}&=&\displaystyle (1-p)^n(1-p)^{-x}\\
&=&\displaystyle \left[\left[\{1+(-p)\}^{\frac{1}{-p}}\right]^{-p}\right]^n(1-p)^{-x}
\,\dotso\,a=(a^\frac{1}{x})^{x}\\
&=&\displaystyle \{\left(\mathrm{e}\right)^{-p}\}^n(1-p)^{-x}
\,\dotso\,\left(1+x\right)^{\frac{1}{x}}=\left(1+\frac{1}{y}\right)^{y}=\mathrm{e}\\
&=&\displaystyle \mathrm{e}^{-np}(1-p)^{-x}\,\dotso\,\left(a^x\right)^y=a^{xy}\\
\end{array}$$
展開と変形の結果を当てはめる
$$\begin{array}{rcl}
\displaystyle _nC_x\,p^x(1-p)^{n-x}&=&\displaystyle \frac{n^x}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}p^x\mathrm{e}^{-np}(1-p)^{-x}\\
&=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}n^{x}p^x\mathrm{e}^{-np}(1-p)^{-x}\\
&=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}(np)^{x}\mathrm{e}^{-np}(1-p)^{-x}\\
\end{array}$$
$\lambda=np$，$\lambda$を一定に$n$を十分大きくとると, $p$は極めて小さくなる
$$\begin{array}{rcl}
\displaystyle f_X(x)&=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}(np)^{x}\mathrm{e}^{-np}(1-p)^{-x}\\
&=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}(\lambda)^{x}\mathrm{e}^{-\lambda}(1-p)^{-x}\,\dotso\,\lambda=np\\
&=&\displaystyle \frac{1}{x!}\{(1)(1)\dotsm(1)\}(\lambda)^{x}\mathrm{e}^{-\lambda}(1)^{-x}\,\dotso\,\frac{C}{n}\to 0, p\to 0\\
&=&\displaystyle \frac{1}{x!}\lambda^{x}\mathrm{e}^{-\lambda}\\
&=&\displaystyle \frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\
&=&\displaystyle Po(\lambda)\\
\end{array}$$
積率母凾数(moment-generating function)を考える
$$\begin{array}{rcl}
\displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\
&=&\displaystyle \sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\
&=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\\
&=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{\mathrm{e}^{tx}\lambda^{x}}{x!}\\
&=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{(\mathrm{e}^{t}\lambda)^{x}}{x!}\\
&=&\displaystyle \mathrm{e}^{-\lambda}\mathrm{e}^{\mathrm{e}^{t}\lambda}
\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{x=0}^{\infty}\frac{a^{x}}{x!}=\mathrm{e}^a}\\
&=&\displaystyle \mathrm{e}^{\mathrm{e}^{t}\lambda-\lambda}=\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}
\end{array}$$
期待値，分散
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
$$\begin{array}{rcl}
\displaystyle E[X]&=&\displaystyle M_X^1(0)\\
&=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\
&=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0}
\,\dotso\,s=\lambda(\mathrm{e}^{t}-1),\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}\\
&=&\displaystyle \left\{ (\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)})(\lambda\mathrm{e}^{t}) \right\}|_{t=0}
\,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }\mathrm{e}^{x}=\mathrm{e}^{x}\\
&=&\displaystyle \left\{ \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0}\\
&=&\displaystyle \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+0})\\
&=&\displaystyle \lambda\mathrm{e}^0
\,\dotso\,a^0=1\\
&=&\lambda
\,\dotso\,a^0=1\\
\end{array}$$
$$\begin{array}{rcl}
\displaystyle E[X^2]&=&\displaystyle M_X^2(0)\\
&=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\
&=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0}
\,\dotso\,E[X]の展開から．\\
&=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\lambda\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0}
\,\dotso\,s=\lambda(\mathrm{e}^{t}-1)+t,\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}+1\\
&=&\displaystyle \left\{ (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+1})(\lambda\mathrm{e}^{t}+1) \right\}|_{t=0}
\,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }C\mathrm{e}^{x}=C\mathrm{e}^{x}\\
&=&\displaystyle (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+1})(\lambda\mathrm{e}^{0}+1)\\
&=&\displaystyle \lambda\mathrm{e}^0(\lambda+1)
\,\dotso\,a^0=1\\
&=&\lambda(\lambda+1)
\,\dotso\,a^0=1\\
\end{array}$$
$$\begin{array}{rcl}
\displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\
&=&\lambda(\lambda+1)-\lambda^2\\
&=&\lambda^2+\lambda-\lambda^2\\
&=&\lambda\\
\end{array}$$
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