【統計学】二項分布の正規近似　ド・モアブル=ラプラスの定理

\begin{align}n! \approx \sqrt{2\pi n }(n / e)^n\end{align}

\begin{align} \lim_{n\to \infty} \cfrac{k}{n} = p\end{align}

\begin{align} \mathrm{Pr}\{X= k\} &= \begin{pmatrix} n \\ k\end{pmatrix} p^k (1 -p)^{n - k}\\ &= \cfrac{n!}{k! (n- k )!}p^k (1 -p)^{n - k}\\ &\approx \cfrac{ \sqrt{2\pi n }(n / e)^n}{ \sqrt{2\pi k }(k / e)^k \sqrt{2\pi (n -k) }\bigl\{(n- k) / e\bigr\}^{n - k} } p^k (1 -p)^{n - k}\\ &= \sqrt{\cfrac{1}{2\pi n( k/ n) (1- k / n) }} \left(\cfrac{np}{k}\right)^{k} \left\{\cfrac{n(1-p)}{n -k}\right\}^{n -k } \\ &\approx \sqrt{\cfrac{1}{2\pi np (1-p) }} \left(\cfrac{np}{k}\right)^{k} \left\{\cfrac{n(1-p)}{n -k}\right\}^{n -k }. \end{align}

\begin{align}&\sqrt{\cfrac{1}{2\pi np (1-p) }} \left(\cfrac{np}{k}\right)^{k} \left\{\cfrac{n(1-p)}{n -k}\right\}^{n -k }\\ &=\sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{k \log\cfrac{np}{k} \right\} \cdot \exp\left\{ (n - k) \log \cfrac{n(1-p)}{n -k}\right\} \\&= \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ - k \log\cfrac{k}{np}  - (n - k) \log \cfrac{n -k}{n( 1 - p)} \right\}  \\&= \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ - k \log\cfrac{np + z\sqrt{np(1 - p)}}{np}  - (n - k) \log \cfrac{n - np - z\sqrt{np(1 - p)}}{n( 1 - p)} \right\}\\ &= \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ - k \log \left(1 + z \sqrt{\cfrac{1 - p}{np}} \right)  - (n - k) \log \left( 1 - z \sqrt{\cfrac{ p}{n(1 - p)}} \right) \right\} \end{align}

\begin{align}\log (1 + x) \approx x - \cfrac{x^2}{2} + \cdots \end{align}

\begin{align}& \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ - k \log \left(1 + z\sqrt{\cfrac{(1 - p)}{np}} \right)  - (n - k) \log \left( 1 - z\sqrt{\cfrac{ p}{n( 1 - p)}} \right) \right\} \\ &=  \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ - k \left( z\sqrt{\cfrac{(1 - p)}{np}}  - \cfrac{1}{2} z^2\cfrac{(1 - p)}{np}  + O(n^{-3/2}) \right) \right. \\ &\qquad\qquad\qquad\qquad\qquad\  \left. - (n - k) \left( - z\sqrt{\cfrac{ p}{n( 1 - p)}} + \cfrac{1}{2} z^2\cfrac{ p}{n( 1 - p)} + O(n^{- 3/2})\right) \right\}\\ &= \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ - \left\{ np + z\sqrt{np(1-p)} \right\} \left( z\sqrt{\cfrac{(1 - p)}{np}}  - \cfrac{1}{2} z^2\cfrac{(1 - p)}{np}  + O(n^{-3/2}) \right) \right. \\ &\qquad\qquad\qquad\qquad\qquad\ \left. - \left\{n (1 - p) - z\sqrt{np(1-p)} \right\} \left( - z\sqrt{\cfrac{ p}{n( 1 - p)}} + \cfrac{1}{2} z^2\cfrac{ p}{n( 1 - p)} + O(n^{- 3/2})\right) \right\} \\ &= \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ - z\sqrt{np(1 - p)} + \cfrac{z^2}{2}(1 - p) - z^2 (1 - p)  + O(n^{-1/2}) \right. \\ &\qquad\qquad\qquad\qquad\qquad\ \left. + z\sqrt{np(1 - p)} - \cfrac{z^2}{2}p - z^2p + O(n^{-1/2})  \right\} \\ &= \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ -  \cfrac{z^2}{2} (1 - p)  -\cfrac{z^2}{2}p + O(n^{-1/2})  \right\}\\ &=  \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ -  \cfrac{z^2}{2}  + O(n^{-1/2})  \right\}. \\&=  \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ -  \cfrac{(k - np)^2}{2np(1-p)}  + O(n^{-1/2})  \right\} \end{align}

\begin{align}\mathrm{Pr}\{X = k\} &\approx \sqrt{\cfrac{1}{2\pi np (1-p) }}  \exp\left\{ -\cfrac{(k - np)^2}{2np(1-p)}  \right\} .\end{align}

\begin{align}\mu = \mathrm{E}[X_i] = p, \quad \sigma^2  = \mathrm{Var}[X_i] = p( 1 - p), \quad i = 1,2, \ldots, n\end{align}

\begin{align}Z &= \sqrt{n}(\bar{X} - \mu) / \sigma\\ \label{eq1} &= \cfrac{\bar{X} - p }{\sqrt{p(1-p)/ n}} \sim N(0, 1). \tag{1}\end{align}