回帰係数の確率分布(線形単回帰)

$\hat{β} = {({}^{t}X X)}^{- 1} {}^{t}X y$ ･･････(1)

$y = (\begin{matrix} y_{1} \\ y_{2} \\ ⋮ \\ y_{n} \end{matrix})$ ･･･(2) ， $X = (\begin{matrix} 1 & x_{1} \\ 1 & x_{2} \\ ⋮ & ⋮ \\ 1 & x_{n} \end{matrix})$ ･･･(3)， $\hat{β} = (\begin{matrix} {\hat{β}}_{0} \\ {\hat{β}}_{1} \end{matrix})$ ･･･(4)

とおく．

${}^{t}X X$ $= (\begin{matrix} 1 & 1 & \dots & 1 \\ x_{1} & x_{2} & \dots & x_{n} \end{matrix}) (\begin{matrix} 1 & x_{1} \\ 1 & x_{2} \\ ⋮ & ⋮ \\ 1 & x_{n} \end{matrix})$ $= (\begin{matrix} n & \sum_{i = 1}^{n} x_{i} \\ \sum_{j = 1}^{n} x_{j} & \sum_{k = 1}^{n} x_{k}^{2} \end{matrix})$ $= (\begin{matrix} n & n \bar{x} \\ n \bar{x} & \sum_{k = 1}^{n} x_{k}^{2} \end{matrix})$

${({}^{t}X X)}^{- 1} = {(\begin{matrix} n & n \bar{x} \\ n \bar{x} & \sum_{k = 1}^{n} x_{k}^{2} \end{matrix})}^{- 1}$ $= \frac{1}{n \sum_{k = 1}^{n} x_{k}^{2} - n^{2} {\bar{x}}^{2}} (\begin{matrix} \sum_{k = 1}^{n} x_{k}^{2} & - n \bar{x} \\ - n \bar{x} & n \end{matrix})$

${({}^{t}X X)}^{- 1} {}^{t}X$ $= \frac{1}{n \sum_{k = 1}^{n} x_{k}^{2} - n^{2} {\bar{x}}^{2}} (\begin{matrix} \sum_{k = 1}^{n} x_{k}^{2} & - n \bar{x} \\ - n \bar{x} & n \end{matrix}) (\begin{matrix} 1 & 1 & \dots & 1 \\ x_{1} & x_{2} & \dots & x_{n} \end{matrix})$

分散の関係式より

$\frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - {\bar{x}}^{2}$

$\sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} = \sum_{i = 1}^{n} x_{i}^{2} - n {\bar{x}}^{2}$ ･･････(5)

よって

$n \sum_{k = 1}^{n} x_{k}^{2} - n^{2} {\bar{x}}^{2} = n (\sum_{k = 1}^{n} x_{k}^{2} - n {\bar{x}}^{2})$ $= n \sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}$

$= \frac{1}{n \sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} (\begin{matrix} \sum_{k = 1}^{n} x_{k}^{2} - n x_{1} \bar{x} & \sum_{k = 1}^{n} x_{k}^{2} - n x_{2} \bar{x} & \dots & \sum_{k = 1}^{n} x_{k}^{2} - n x_{n} \bar{x} \\ n x_{1} - n \bar{x} & n x_{2} - n \bar{x} & \dots & n x_{n} - n \bar{x} \end{matrix})$

$= \frac{1}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} (\begin{matrix} \frac{1}{n} \sum_{k = 1}^{n} x_{k}^{2} - x_{1} \bar{x} & \frac{1}{n} \sum_{k = 1}^{n} x_{k}^{2} - x_{2} \bar{x} & \dots & \frac{1}{n} \sum_{k = 1}^{n} x_{k}^{2} - x_{n} \bar{x} \\ x_{1} - \bar{x} & x_{2} - \bar{x} & \dots & x_{n} - \bar{x} \end{matrix})$

$\hat{β} = {({}^{t}X X)}^{- 1} {}^{t}X y$

$= \frac{1}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} (\begin{matrix} (\frac{1}{n} \sum_{k = 1}^{n} x_{k}^{2} - x_{1} \bar{x}) y_{1} + (\frac{1}{n} \sum_{k = 1}^{n} x_{k}^{2} - x_{2} \bar{x}) y_{2} + \dots + (\frac{1}{n} \sum_{k = 1}^{n} x_{k}^{2} - x_{n} \bar{x}) y_{n} \\ (x_{1} - \bar{x}) y_{1} + (x_{2} - \bar{x}) y_{2} + \dots + (x_{n} - \bar{x}) y_{n} \end{matrix})$

$= \frac{1}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} (\begin{matrix} \frac{1}{n} (\sum_{k = 1}^{n} x_{k}^{2}) (\sum_{i = 1}^{n} y_{i}) - (\sum_{i = 1}^{n} x_{i} y_{i}) \bar{x} \\ \sum_{i = 1}^{n} x_{i} y_{i} - \bar{x} \sum_{i = 1}^{n} y_{i} \end{matrix})$

$= \frac{1}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} (\begin{matrix} (\sum_{k = 1}^{n} x_{k}^{2}) (\frac{1}{n} \sum_{i = 1}^{n} y_{i}) - (\sum_{i = 1}^{n} x_{i} y_{i}) \bar{x} \\ \sum_{i = 1}^{n} x_{i} y_{i} - n \bar{x} \frac{1}{n} \sum_{i = 1}^{n} y_{i} \end{matrix})$

$= \frac{1}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} (\begin{matrix} (\sum_{k = 1}^{n} x_{k}^{2}) \bar{y} - (\sum_{i = 1}^{n} x_{i} y_{i}) \bar{x} \\ \sum_{i = 1}^{n} x_{i} y_{i} - n \bar{x} \bar{y} \end{matrix})$

$= (\begin{matrix} \frac{(\sum_{k = 1}^{n} x_{k}^{2}) \bar{y} - (\sum_{i = 1}^{n} x_{i} y_{i}) \bar{x}}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} \\ \frac{\sum_{i = 1}^{n} x_{i} y_{i} - n \bar{x} \bar{y}}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} \end{matrix})$ ･･････(6)

共分散の関係式より

$\frac{1}{n} \sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})$ $= \frac{1}{n} \sum_{i = 1}^{n} x_{i} y_{i} - \bar{x} \bar{y}$

$\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y}) = \sum_{i = 1}^{n} x_{i} y_{i} - n \bar{x} \bar{y}$ ･･････(7)

ここで

$(\sum_{k = 1}^{n} x_{k}^{2}) \bar{y} - (\sum_{i = 1}^{n} x_{i} y_{i}) \bar{x}$ $= (\sum_{k = 1}^{n} x_{k}^{2}) \bar{y} - n {\bar{x}}^{2} \bar{y}$ $- (\sum_{i = 1}^{n} x_{i} y_{i}) \bar{x} + n {\bar{x}}^{2} \bar{y}$

$= (\sum_{k = 1}^{n} x_{k}^{2} - n {\bar{x}}^{2}) \bar{y}$ $- (\sum_{i = 1}^{n} x_{i} y_{i} - n \bar{x} \bar{y}) \bar{x}$

(5)，(7)の関係式を用いて式を書き換える

$= (\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}) \bar{y}$ $- (\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})) \bar{x}$ ･･････(8)

$= (\begin{matrix} \frac{(\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}) \bar{y} - (\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})) \bar{x}}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} \\ \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} \end{matrix})$

$= (\begin{matrix} \bar{y} - \frac{(\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})) \bar{x}}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} \\ \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}} \end{matrix})$

よって

${\hat{β}}_{0} = \bar{y} - \frac{(\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})) \bar{x}}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}}$ ･･････(9)

${\hat{β}}_{1} = \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) (y_{i} - \bar{y})}{\sum_{k = 1}^{n} {(x_{k} - \bar{x})}^{2}}$ ･･････(10)

(9)と(10)より

${\hat{β}}_{0} = \bar{y} - {\hat{β}}_{1} \bar{x}$

⇒元のページに戻る

ホーム>>カテゴリー分類>>確率>>統計>>回帰係数の確率分布(線形単回帰)

　最終更新日： 2026年7月13日