二項分布(binomial distribution)から正規分布(normal distribution)を導く

二項分布 $B(n,p)$

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

$f_X(x)$の対数をとる

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

一階微分

$$\begin{array}{rcl}{\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\log (f_X(x))}& =& {\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(\log n!-\log x!-\log (n-x)!+x\log p+(n-x)\log (1-p)){\displaystyle \dots n\to \mathrm{\infty }\mathrm{場}\mathrm{合},\frac{\mathrm{d}}{\mathrm{d}x}\log (x!)\simeq \log \left(x\right)}}\\ & =& {\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\log n!{\displaystyle -\frac{\mathrm{d}}{\mathrm{d}x}\log x!{\displaystyle -\frac{\mathrm{d}}{\mathrm{d}x}\log (n-x)!{\displaystyle +\frac{\mathrm{d}}{\mathrm{d}x}x\log p{\displaystyle +\frac{\mathrm{d}}{\mathrm{d}x}(n-x)\log (1-p)}}}}}\\ & =& {\displaystyle 0-\log x+\log (n-x)+\log p-\log (1-p)}\\ & =& {\displaystyle \log \frac{(n-x)p}{x(1-p)}}\end{array}$$

二階微分

$$\begin{array}{rcl}{\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\log (f_X(x))}& =& {\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(\log \frac{(n-x)p}{x(1-p)})}\\ & =& {\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\log (n-x)+\frac{\mathrm{d}}{\mathrm{d}x}\log p-\frac{\mathrm{d}}{\mathrm{d}x}\log x-\frac{\mathrm{d}}{\mathrm{d}x}\log (1-p)}\\ & =& {\displaystyle \left(\frac{-1}{n-x}\right)+0-\left(\frac{1}{x}\right)-0}\\ & =& {\displaystyle \frac{-1}{n-x}-\frac{1}{x}}\\ & =& {\displaystyle \frac{-x-(n-x)}{x(n-x)}}\\ & =& {\displaystyle \frac{-n}{x(n-x)}}\end{array}$$

$\log (f_X(x))$の極値を考える(logは単調に増加する凾数なので$f_X(x)$も極値)

$$\begin{array}{rcl}{\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\log (f_X(x))}& =& 0\\ {\displaystyle \log \frac{(n-x)p}{x(1-p)}}& =& 0\\ {\displaystyle \frac{(n-x)p}{x(1-p)}}& =& 1\dots \log \left(1\right)=0\\ {\displaystyle (n-x)p}& =& x(1-p)\\ {\displaystyle np-xp}& =& x(1-p)\\ {\displaystyle np}& =& x(1-p)+xp\\ {\displaystyle np}& =& x(1-p+p)\\ {\displaystyle np}& =& x\\ x& =& np\dots \log (f_X(x))\mathrm{が}\mathrm{極}\mathrm{値}\mathrm{の}\mathrm{際}\mathrm{の}x，\mathrm{二}\mathrm{項}\mathrm{分}\mathrm{布}\mathrm{の}\mathrm{期}\mathrm{待}\mathrm{値}\mathrm{は}np(=\mu )\\ {\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\log (f_X(np))}& =& 0\end{array}$$

二階微分における$x=np$場合の値

$$\begin{array}{rcl}{\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\log (f_X(np))}& =& {\displaystyle \frac{-n}{np(n-np)}\dots x\mathrm{に}\mathrm{期}\mathrm{待}\mathrm{値}np\mathrm{を}\mathrm{代}\mathrm{入}}\\ & =& {\displaystyle \frac{-n}{n^{2}p(1-p)}}\\ & =& {\displaystyle \frac{-1}{np(1-p)}\dots \mathrm{二}\mathrm{項}\mathrm{分}\mathrm{布}\mathrm{の}\mathrm{分}\mathrm{散}\mathrm{は}np(1-p)(={\sigma }^2)\mathrm{な}\mathrm{の}\mathrm{で}\frac{-1}{np(1-p)}=\frac{-1}{{\sigma }^2}}\end{array}$$

$x=\mu$周りのテイラー展開

$$\begin{array}{rcl}{\displaystyle g(x)}& =& \frac{1}{0!}{g}^{(0)}(\mu )(x-\mu {)}^0+\frac{1}{1!}{g}^{(1)}(\mu )(x-\mu {)}^1+\frac{1}{2!}{g}^{(2)}(\mu )(x-\mu {)}^2+\dots \\ \log (f_X(x))& =& \frac{1}{0!}(f_X(\mu ))(x-\mu {)}^0+\frac{1}{1!}(f_X^{\prime }(\mu ))(x-\mu {)}^1+\frac{1}{2!}(f_X^{″}(\mu ))(x-\mu {)}^2+\dots \\ & =& \frac{1}{1}{f}_X(\mu )1+\frac{1}{1}0(x-\mu )+\frac{1}{2}\frac{-1}{{\sigma }^2}(x-\mu {)}^2+\dots \\ & =& f_X(\mu )+0+\frac{-(x-\mu {)}^2}{2{\sigma }^2}\\ & =& f_X(\mu )+\frac{-(x-\mu {)}^2}{2{\sigma }^2}\end{array}$$ $$\begin{array}{rcl}{\displaystyle {\mathrm{e}}^{\log (f_X(x))}=f_X(x)}& =& {\mathrm{e}}^{(f_X(\mu )+\frac{-(x-\mu {)}^2}{2{\sigma }^2})}\\ {\displaystyle }& =& {\mathrm{e}}^{f_X(\mu )}{\mathrm{e}}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}}\\ {\displaystyle }& =& C{\mathrm{e}}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}}\dots \mathrm{C}={\mathrm{e}}^{f_X(\mu )}\end{array}$$

定数$C$を求める

$$\begin{array}{rcl}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}}\mathrm{d}x}& =& {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-1}{2}{y}^2}\mathrm{d}x\dots y=\frac{x-\mu }{\sigma },\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sigma },\mathrm{d}x=\sigma \mathrm{d}y}\\ & =& {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-y^2}{2}}\sigma \mathrm{d}y}\\ & =& {\displaystyle \sigma {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-y^2}{2}}\mathrm{d}y}\\ & =& {\displaystyle \sigma \sqrt{2\pi }\dots {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-y^2}{2}}\mathrm{d}y=\sqrt{2\pi }}\\ & =& \sqrt{2\pi {\sigma }^2}\end{array}$$ $$\begin{array}{rcl}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_X(x)\mathrm{d}x}& =& {\displaystyle C{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}}\mathrm{d}x}\\ 1& =& {\displaystyle C\sqrt{2\pi {\sigma }^2}}\\ C& =& \frac{1}{\sqrt{2\pi {\sigma }^2}}\end{array}$$

定数$C$に値を代入する

$$\begin{array}{rcl}{f}_X(x)& =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{\mathrm{e}}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}}\\ & =& N(\mu ,{\sigma }^2)\end{array}$$