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

二項分布 \(B(n, p)\)

$$B(n, p) = f_X(x) = \begin{cases} \displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotso ,n\right\}\\ \displaystyle　0 & \quad x \notin \left\{0,1,2, \dotso ,n\right\} \end{cases} $$

\(f_X(x)\)の対数をとる

$$\begin{array}{rcl} \displaystyle \log\left( f_X(x) \right) \displaystyle &=& \log\left( _nC_x\,p^x(1-p)^{n-x} \right)\\ \displaystyle &=& \log\left(\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\right)\\ \displaystyle &=& \log n!-\log x!-\log(n-x)!+\log p^x+\log(1-p)^{n-x}\\ \displaystyle &=& \log n!-\log x!-\log(n-x)!+x\log p+(n-x)\log(1-p)\\ \end{array}$$

一階微分

$$\begin{array}{rcl} \displaystyle \frac{ \mathrm{d} }{ \mathrm{d}x } \log\left(f_X(x) \right) &=& \displaystyle \frac{ \mathrm{d} }{ \mathrm{d}x }\left( \log n!-\log x!-\log(n-x)!+x\log p+(n-x)\log(1-p) \right) \displaystyle \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/06/blog-post_21.html}{n \to \infty場合, \frac{ \mathrm{d} }{ \mathrm{d}x } \log\left(x!\right) \simeq \log\left(x\right)}\\ &=& \displaystyle \frac{ \mathrm{d} }{ \mathrm{d}x }\log n! \displaystyle -\frac{ \mathrm{d} }{ \mathrm{d}x }\log x! \displaystyle -\frac{ \mathrm{d} }{ \mathrm{d}x }\log(n-x)! \displaystyle +\frac{ \mathrm{d} }{ \mathrm{d}x }x\log p \displaystyle +\frac{ \mathrm{d} }{ \mathrm{d}x }(n-x)\log(1-p) \\ &=& \displaystyle 0-\log x+\log(n-x)+\log p-\log(1-p) \\ &=& \displaystyle \log \frac{(n-x)p}{x(1-p)}\\ \end{array}$$

二階微分

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

\(\log\left(f_X(x) \right)\)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)

$$\begin{array}{rcl} \displaystyle \frac{ \mathrm{d} }{ \mathrm{d}x } \log\left(f_X(x) \right)&=&0\\ \displaystyle \log \frac{(n-x)p}{x(1-p)}&=&0\\ \displaystyle \frac{(n-x)p}{x(1-p)}&=&1\,\dotso\,\log\left(1\right)=0\\ \displaystyle (n-x)p&=&x(1-p)\\ \displaystyle np-xp&=&x(1-p)\\ \displaystyle np&=&x(1-p)+xp\\ \displaystyle np&=&x(1-p+p)\\ \displaystyle np&=&x\\ x&=&np\,\dotso\,\log\left(f_X(x)\right)が極値の際のx，二項分布の期待値はnp(=\mu)\\ \displaystyle \frac{ \mathrm{d} }{ \mathrm{d}x } \log\left(f_X(np) \right)&=&0\\ \end{array}$$

二階微分における\(x=np\)場合の値

$$\begin{array}{rcl} \displaystyle \frac{ \mathrm{d}^2 }{ \mathrm{d}x^2 } \log\left(f_X(np) \right) &=& \displaystyle \frac{-n}{np(n-np)}\,\dotso\,xに期待値npを代入 \\ &=& \displaystyle \frac{-n}{n^2p(1-p)} \\ &=& \displaystyle \frac{-1}{np(1-p)}\,\dotso\,二項分布の分散はnp(1-p)(=\sigma^2)なので\frac{-1}{np(1-p)}=\frac{-1}{\sigma^2}\\ \end{array}$$

\(x=\mu\)周りのテイラー展開

$$\begin{array}{rcl} \displaystyle g(x) &=& \frac{1}{0!}g^{(0)}(\mu)(x-\mu)^0 +\frac{1}{1!}g^{(1)}(\mu)(x-\mu)^1 +\frac{1}{2!}g^{(2)}(\mu)(x-\mu)^2+\,\dotso\\ \log\left(f_X(x) \right) &=& \frac{1}{0!}(f_X(\mu))(x-\mu)^0 +\frac{1}{1!}(f_X'(\mu))(x-\mu)^1 +\frac{1}{2!}(f_X''(\mu))(x-\mu)^2+\,\dotso\\ &=& \frac{1}{1}f_X(\mu)1 +\frac{1}{1}0(x-\mu) +\frac{1}{2}\frac{-1}{\sigma^2}(x-\mu)^2+\,\dotso\\ &=& f_X(\mu)+0+\frac{-(x-\mu)^2}{2\sigma^2}\\ &=& f_X(\mu)+\frac{-(x-\mu)^2}{2\sigma^2}\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle \mathrm{e}^{\log\left(f_X(x) \right)}=f_X(x)&=&\mathrm{e}^{\left(f_X(\mu)+\frac{-(x-\mu)^2}{2\sigma^2}\right)}\\ \displaystyle &=&\mathrm{e}^{f_X(\mu)}\mathrm{e}^{\frac{-(x-\mu)^2}{2\sigma^2}}\\ \displaystyle &=&C\mathrm{e}^{\frac{-(x-\mu)^2}{2\sigma^2}}\,\dotso\,\mathrm{C}=\mathrm{e}^{f_X(\mu)}\\ \end{array}$$

定数\(C\)を求める

$$\begin{array}{rcl} \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{\frac{-(x-\mu)^2}{2\sigma^2}}\mathrm{d}x &=& \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{\frac{-1}{2}y^2}\mathrm{d}x \,\dotso\,y=\frac{x-\mu}{\sigma},\frac{ \mathrm{d}y }{ \mathrm{d}x }=\frac{1}{\sigma},\mathrm{d}x=\sigma \mathrm{d}y\\ &=& \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{\frac{-y^2}{2}}\sigma\mathrm{d}y\\ &=& \displaystyle \sigma\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-y^2}{2}}\mathrm{d}y\\ &=& \displaystyle \sigma\sqrt{2\pi} \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/06/gaussian-integral.html}{\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-y^2}{2}}\mathrm{d}y=\sqrt{2\pi}}\\ &=& \sqrt{2\pi\sigma^2}\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle \int_{-\infty}^{\infty}f_X(x)\mathrm{d}x &=&\displaystyle C\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-(x-\mu)^2}{2\sigma^2}}\mathrm{d}x\\ 1&=&\displaystyle C\sqrt{2\pi \sigma^2}\\ C&=&\frac{1}{\sqrt{2\pi \sigma^2}} \end{array}$$

定数\(C\)に値を代入する

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