回帰係数の確率分布（線形単回帰）

【前提条件】

誤差はランダムに生じ，平均 $0$ の正規分布に従うと仮定する．式で表すと以下のようになる．

$ε_{i} \sim N (0, σ^{2})$ ：誤差は正規分布に従い，その平均は $0$ （ $E (ε_{i}) = 0$ ），その分散は $σ^{2}$ （ $E (ε_{i}^{2}) = V (ε_{i}) = σ^{2}$ ），である．
$i \neq j$ のとき， $ε_{i}$ と $ε_{j}$ は互いに独立である（独立ならば無相関であるから， $i \neq j$ のとき共分散 $Cov (ε_{i}, ε_{j}) = 0$ である（ここを参照））．

$V ({\hat{β}}_{1}) = V (β_{1} + \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x}) ε_{i}}{\sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}})$

$ε_{i}$ を確率変数とし， $V (a X + b) = a^{2} V (X)$ の関係を使って計算すると

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

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

前提条件と $X$ と $Y$ が独立のとき $V (X + Y) = V (X) + V (Y)$ が成り立つことより

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

$V (ε_{i}) = σ^{2}$ と仮定していることより

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

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

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

$V ({\hat{β}}_{0})$ $= V (β_{0} + \sum_{j = 1}^{n} \{\frac{1}{n} - \frac{(x_{j} - \bar{x}) \bar{x}}{\sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}}\} ε_{j})$

$ε_{i}$ を確率変数とし， $V (a X + b) = a^{2} V (X)$ の関係を使って計算すると

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

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

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

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

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

前提条件と $X$ と $Y$ が独立のとき $V (X + Y) = V (X) + V (Y)$ が成り立つことより

$= \frac{1}{{\{n \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}\}}^{2}} [{\{\sum_{i = 1}^{n} x_{i}^{2} - n x_{1} \bar{x}\}}^{2} V (ε_{1}) + {\{\sum_{i = 1}^{n} x_{i}^{2} - n x_{2} \bar{x}\}}^{2} V (ε_{2}) +$ $\dots + {\{\sum_{i = 1}^{n} x_{i}^{2} - n x_{n} \bar{x}\}}^{2} V (ε_{n})]$

$V (ε_{i}) = σ^{2}$ と仮定していることより

$= \frac{1}{{\{n \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}\}}^{2}} [{\{\sum_{i = 1}^{n} x_{i}^{2} - n x_{1} \bar{x}\}}^{2} σ^{2} + {\{\sum_{i = 1}^{n} x_{i}^{2} - n x_{2} \bar{x}\}}^{2} σ^{2} +$ $\dots + {\{\sum_{i = 1}^{n} x_{i}^{2} - n x_{n} \bar{x}\}}^{2} σ^{2}]$

$= \frac{σ^{2}}{{\{n \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}\}}^{2}} [\{{(\sum_{i = 1}^{n} x_{i}^{2})}^{2} - 2 n x_{1} \bar{x} (\sum_{i = 1}^{n} x_{i}^{2}) + n^{2} x_{1}^{2} {\bar{x}}^{2}\} +$ $\{{(\sum_{i = 1}^{n} x_{i}^{2})}^{2} - 2 n x_{2} \bar{x} (\sum_{i = 1}^{n} x_{i}^{2}) + n^{2} x_{2}^{2} {\bar{x}}^{2}\} +$ $\dots + \{{(\sum_{i = 1}^{n} x_{i}^{2})}^{2} - 2 n x_{n} \bar{x} (\sum_{i = 1}^{n} x_{i}^{2}) + n^{2} x_{n}^{2} {\bar{x}}^{2}\}]$

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

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

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

分散の関係式より

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

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

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

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

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　最終更新日： 2026年7月22日