回帰係数の推定量の分布
\(\boldsymbol{E} \sim \mathcal{N}(0,\mathbf{I}_n \sigma^2)\)とすると,\(\hat{\boldsymbol{\beta}},\hat{\sigma^2}\)の推定量の分布は以下のようになる, \[ \hat{\boldsymbol{\beta}} \sim \mathcal{N}_{k+1}(\boldsymbol{\beta}, \sigma^2(\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}) \\ \frac{(n-(k+1))\hat{\sigma^2}}{\sigma^2} \sim \chi^2_{n-(k+1)} \\ \hat{\boldsymbol{\beta}}と\hat{\sigma^2}が独立 \]
- 証明
\(\hat{\boldsymbol{\beta}}(\boldsymbol{Y}) = (\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}}\boldsymbol{Y}\)
\(\boldsymbol{Y} = \mathbf{X}\boldsymbol{\beta} + \boldsymbol{E} \sim \mathcal{N}_{n}(\mathbf{X}\boldsymbol{\beta},\mathbf{I}_n \sigma^2)\)から \[ \begin{align} \hat{\boldsymbol{\beta}} = (\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}}\boldsymbol{Y} &\sim \mathcal{N}_{k+1}((\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}}\mathbf{X}\boldsymbol{\beta}, ((\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}})(\mathbf{I}_n \sigma^2)((\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}})^{\mathsf{T}}) \\ &= \mathcal{N}_{k+1}(\boldsymbol{\beta},\sigma^2(\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}) \end{align} \\ \hat{\sigma}^2 = \frac{1}{n-k-1}RSS_{k+1}= \frac{1}{n-k-1}(\boldsymbol{Y}-\mathbf{X}\hat{\boldsymbol{\beta}})^{\mathsf{T}}(\boldsymbol{Y}-\mathbf{X}\hat{\boldsymbol{\beta}}) \]
\[ \begin{align} \frac{(n-(k+1))\hat{\sigma^2}}{\sigma^2} &= \frac{(\boldsymbol{Y}-\mathbf{X}\hat{\boldsymbol{\beta}})^{\mathsf{T}}}{\sigma}\frac{(\boldsymbol{Y}-\mathbf{X}\hat{\boldsymbol{\beta}})}{\sigma} \end{align} \]
\[ \mathbf{X}\hat{\boldsymbol{\beta}} \sim \mathcal{N}_n(\mathbf{X}\boldsymbol{\beta},\sigma^2\mathbf{X}(\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}}) \\ \boldsymbol{Y} - \mathbf{X}\hat{\boldsymbol{\beta}} \sim \mathcal{N}_n(\mathbf{X}\boldsymbol{\beta}-\mathbf{X}\boldsymbol{\beta}, \sigma^2(\mathbf{X}(\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}}-\mathbf{I}_n)) =\mathcal{N}_n(\mathbf{0},\sigma^2(\mathbf{X}(\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}}-\mathbf{I}_n)) \\ \frac{(\boldsymbol{Y}-\mathbf{X}\hat{\boldsymbol{\beta}})}{\sigma} \sim \mathcal{N}_n(\mathbf{0},\mathbf{X}(\mathbf{X}^{\mathsf{T}}\mathbf{X})^{-1}\mathbf{X}^{\mathsf{T}}-\mathbf{I}_n) \]