まずは，
\(\Large \displaystyle Y_i = a_0 + a_1 X_i + u_i \)
\(\Large \displaystyle Y_i = \hat{a_0} + \hat{a_1} X_i + \hat{u_i} \)
この二つの式から始めましょう．
ここで，不偏推定量，s2，を定義します．
\(\Large \displaystyle s^2 = \frac{ \sum_{i=1}^n \hat{u_i}^2 }{n-2} \)
上の二つ目の式を代入すれば，
\(\Large \displaystyle s^2 = \frac{ \sum_{i=1}^n (Y_i - \hat{a_0} - \hat{a_1} X_i )^2 }{n-2} \)
つぎに，誤差，ui，を書き直します．
\(\Large \displaystyle u_i = Y_i - a_0 - a_1 X_i = ( \hat{a_0} + \hat{a_1} X_i + \hat{u_i}) - a_0 - a_1 X_i\)
\(\Large \displaystyle = ( \hat{a_0} - a_0 ) +( \hat{a_1} - a_1 ) X_i+ \hat{u_i} \)
両辺を二乗すると，
\(\Large \displaystyle u_i^2= [ ( \hat{a_0} - a_0 ) +( \hat{a_1} - a_1 ) X_i+ \hat{u_i}]^2 \)
\(\Large \displaystyle \hspace{20 pt} = ( \hat{a_0} - a_0 )^2 +( \hat{a_1} - a_1 )^2 X_i^2+ \hat{u_i}^2 \)
\(\Large \displaystyle \hspace{40 pt}+ 2 ( \hat{a_0} - a_0 )( \hat{a_1} - a_1 )X_i +2( \hat{a_0} - a_0 ) \hat{u_i} + 2( \hat{a_1} - a_1 ) X_i \hat{u_i}\)
総和をとると，
\(\Large \displaystyle \sum_{i=1}^n u_i^2= n( \hat{a_0} - a_0 )^2 +( \hat{a_1} - a_1 )^2 \sum_{i=1}^n X_i^2+ \sum_{i=1}^n \hat{u_i}^2 \)
\(\Large \displaystyle \hspace{50 pt} + 2
( \hat{a_0} - a_0 )( \hat{a_1} - a_1 )\sum_{i=1}^n X_i +2( \hat{a_0} - a_0 ) \sum_{i=1}^n \hat{u_i} + 2( \hat{a_1} - a_1 ) \sum_{i=1}^n X_i \hat{u_i}\)
ここで，ここ，を参考にして，
\(\Large \displaystyle \sum_{i=1}^{n} \left( Y_i - \hat{a_0} - \hat{a_1}X_i \right) = \sum_{i=1}^{n} \hat{u_i} = 0 \)
\(\Large \displaystyle \sum_{i=1}^{n} X_i \left( Y_i - \hat{a_0} - \hat{a_1}X_i \right) = \sum_{i=1}^{n} X_i \hat{u_i} = 0 \)
となりますので，先ほどの総和の式のうち，第4，５項は０となります．したがって，
\(\Large \displaystyle \sum_{i=1}^n u_i^2= n( \hat{a_0} - a_0 )^2 +( \hat{a_1} - a_1 )^2 \sum_{i=1}^n X_i^2+ \sum_{i=1}^n \hat{u_i}^2 + 2 n ( \hat{a_0} - a_0 )( \hat{a_1} - a_1 )\bar{ X} \)
期待値をとると，
\(\Large \displaystyle E \left[ \sum_{i=1}^n u_i^2 \right] = n E \left[ ( \hat{a_0} - a_0 )^2 \right] + E \left[ ( \hat{a_1} - a_1 )^2 \right] \sum_{i=1}^n X_i^2+ E \left[ \sum_{i=1}^n \hat{u_i}^2 \right] + 2 n E \left[ ( \hat{a_0} - a_0 )( \hat{a_1} - a_1 ) \right] \bar{ X} \)
と，4つの項の式となります．一つ一つ考えていきましょう．
・左辺
\(\Large \displaystyle E \left[ \hat{u_i}^2 \right] = \sigma^2 \)，なので，
\(\Large \displaystyle E \left[ \sum_{i=1}^n \hat{u_i}^2 \right] = n \sigma^2 \)
・第2項
\(\Large \displaystyle V[X] = E[X^2] - ( E[X])^2 \)，より，
\(\Large \displaystyle V\left[ \hat{ a_1} - a_1 \right] = E\left[ (\hat{ a_1} - a_1)^2 \right] - \left( E \left[ \hat{ a_1} - a_1 \right] \right)^2 \)
\(\Large \displaystyle E\left[ (\hat{ a_1} - a_1)^2 \right] = V\left[ \hat{ a_1} - a_1 \right] + \left( E \left[ \hat{ a_1} - a_1 \right] \right)^2 \)
右辺第1項は，
\(\Large \displaystyle V\left[ \hat{ a_1} - a_1 \right] = V \left[\hat{a_1} \right] = \frac{\sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2}\)
右辺第2項は，
\(\Large \displaystyle \hat{a_1} = a_1 + \sum_{i=1}^{n} \omega_i u_i \)，なので，
\(\Large \displaystyle \left( E \left[ \hat{ a_1} - a_1 \right] \right)^2 = \left( E \left[ \sum_{i=1}^{n} \omega_i u_i \right] \right)^2 = \left( \sum_{i=1}^{n} \omega_i E \left[ u_i \right] \right)^2 =0 \)
，となるので
\(\Large \displaystyle E\left[ (\hat{ a_1} - a_1)^2 \right] =\frac{\sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \)
となります．
・第1項
ここから，
\(\Large \displaystyle \hat{a_0} - a_0 = \left( \hat{a_1} - a_1 \right) \bar{X} + \bar{u} \)，なので，
\(\Large \displaystyle E \left[ ( \hat{a_0} - a_0 )^2 \right] = E \left[ \left\{ \left( \hat{a_1} - a_1 \right) \bar{X} + \bar{u} \right\}^2 \right] \)
\(\Large \displaystyle = \bar{X}^2 E \left[ \left( \hat{a_1} - a_1 \right)^2 \right] + E \left[ 2 \left( \hat{ a_1} - a_1 \right) \bar{X} \bar{u} \right] + E \left[ \bar{u}^2 \right] \)
右辺第2項は，
\(\Large \displaystyle = E \left[ \left( \hat{ a_1} - a_1 \right) \right] = E \left[ \left( \hat{ a_1} \right) \right] -a_1 = 0 \)
となり，右辺第3項も０となるので，
\(\Large \displaystyle E \left[ ( \hat{a_0} - a_0 )^2 \right] = \bar{X}^2 E \left[ \left( \hat{a_1} - a_1 \right)^2 \right] \)
上の第2項の式から，
\(\Large \displaystyle E \left[ \left( \hat{a_1} - a_1 \right)^2 \right] =\frac{\sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2}\)
となるので，
\(\Large \displaystyle n E \left[ \left( \hat{a_1} - a_1 \right)^2 \right] = \frac{n \sigma^2 \bar{X}^2}{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2}
= \frac{n \sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \frac{1}{n} \sum_{i=1}^{n} X_i^2
= \frac{ \sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \sum_{i=1}^{n} X_i^2\)
・第4項
ここから，
\(\Large \displaystyle \hat{a_0} - a_0 = - \left( \hat{a_1} - a_1 \right) \bar{X} + \bar{u} \)，なので，
\(\Large \displaystyle \left( \hat{a_0} - a_0 \right) \left( \hat{a_1} - a_1 \right) = \left\{ - \left( \hat{a_1} - a_1 \right) \bar{X} + \bar{u} \right\} \left( \hat{a_1} - a_1 \right) \)
\(\Large \displaystyle = \left( \hat{a_1} - a_1 \right)^2 \bar{X} + \left( \hat{a_1} - a_1 \right)\bar{u} \)
期待値は，
\(\Large \displaystyle E \left[ \left( \hat{a_0} - a_0 \right) \left( \hat{a_1} - a_1 \right) \right] = E \left[ - \left( \hat{a_1} - a_1 \right)^2 \bar{X} \right] + E \left[\left( \hat{a_1} - a_1 \right)\bar{u} \right] \)
第1項，右辺第2項と同様に，
\(\Large \displaystyle E \left[ \left( \hat{a_0} - a_0 \right) \left( \hat{a_1} - a_1 \right) \right] =- \frac{\sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \bar{X}+ 0 =- \frac{\sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \bar{X}\)
となります．
したがって，まとめると，
\(\Large \displaystyle E \left[ \sum_{i=1}^n u_i^2 \right] = n E \left[ ( \hat{a_0} - a_0 )^2 \right] + E \left[ ( \hat{a_1} - a_1 )^2 \right] \sum_{i=1}^n X_i^2+ E \left[ \sum_{i=1}^n \hat{u_i}^2 \right] + 2 n E \left[ ( \hat{a_0} - a_0 )( \hat{a_1} - a_1 ) \right] \bar{ X} \)
は，
\(\Large \displaystyle n \sigma^2 = \frac{ \sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \sum_{i=1}^{n} X_i^2
+ \frac{\sigma^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \sum_{i=1}^{n} X_i^2
+ E \left[ \sum_{i=1}^n \hat{u_i}^2 \right]
- 2 n \frac{\sigma^2 \bar{X} }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \bar{X} \)
となります．
\(\Large \displaystyle = \frac{ \sigma^2 \sum_{i=1}^{n} X_i^2}{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2}
+ \frac{\sigma^2 \sum_{i=1}^{n} X_i^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2}
+ E \left[ \sum_{i=1}^n \hat{u_i}^2 \right]
- \frac{2 n \sigma^2 \bar{X}^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2} \)
\(\Large \displaystyle = 2 \sigma^2 \frac{ \sum_{i=1}^{n} X_i^2 -n \bar{X}^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2}
+ E \left[ \sum_{i=1}^n \hat{u_i}^2 \right]
\)
ここで，右辺第一項の分数に注目すると
\(\Large \displaystyle \frac{ \sum_{i=1}^{n} X_i^2 -n \bar{X}^2 }{\sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2}
= \frac{ \sum_{i=1}^{n} X_i^2 -n \bar{X}^2 }{\sum_{i=1}^{n} \left( X_i^2 - 2 X_i \bar{X} + \bar{X}^2 \right)}
\)
\(\Large \displaystyle = \frac{ \sum_{i=1}^{n} X_i^2 -n \bar{X}^2 }{\sum_{i=1}^{n} X_i^2 - 2 \bar{X} \sum_{i=1}^{n} X_i + n \bar{X}^2 } \)
\(\Large \displaystyle = \frac{ \sum_{i=1}^{n} X_i^2 -n \bar{X}^2 }{\sum_{i=1}^{n} X_i^2 - n \bar{X}^2 } \)
\(\Large \displaystyle = 1 \)
と簡単になるので，
\(\Large \displaystyle n \sigma^2 = 2 \sigma^2 + E \left[ \sum_{i=1}^n \hat{u_i}^2 \right] \)
つまり，
\(\Large \displaystyle \sigma^2 = \frac{ E \left[ \sum_{i=1}^n \hat{u_i}^2 \right]}{n-2} = s^2 \)
と，前頁の，ｓ2は母分散， σ2の不偏推定量との関係を導き出すことができました．
それにしても．．．．面倒くさい．．．