第5講回帰分析

ガウス・マルコフの定理

E [ϵ_{i}] = 0

$\begin{equation} E[\epsilon_i] = 0 \end{equation}$

V a r [ϵ_{i}] = σ^{2} < \infty

$\begin{equation} Var[\epsilon_i] = \sigma^2 < \infty \end{equation}$

E [ϵ_{i} ϵ_{j}] = 0 (i \neq j)

$\begin{equation} E[\epsilon_i\epsilon_j] = 0~~~~(i \ne j) \end{equation}$

のとき，最小二乗推定量

\hat{β}

$\hat{\bm{\beta}}$ は，最良線形不偏推定量 (BLUE)となる．

補足1: 単回帰分析の場合

$\hat{\beta}_1'$ を任意の線形推定量とし，

{\hat{β}}_{1}^{'} = \sum_{i = 1}^{n} c_{i} y_{i}

$\begin{equation} \hat{\beta}_1' = \sum_{i = 1}^{n}c_i\,y_i \end{equation}$

と書く．

\begin{aligned} {\hat{β}}_{1}^{'} & = \sum_{i = 1}^{n} c_{i} (β_{0} + β_{1} x_{i} + ϵ_{i}) \\ = β_{0} \sum_{i = 1}^{n} c_{i} + β_{1} \sum_{i = 1}^{n} c_{i} x_{i} + \sum_{i = 1}^{n} c_{i} ϵ_{i} \end{aligned}

$\begin{align} \hat{\beta}_1' &= \sum_{i = 1}^{n}c_i\,(\beta_0 + \beta_1 x_i + \epsilon_i) \\ &= \beta_0\sum_{i = 1}^{n}c_i + \beta_1\sum_{i = 1}^{n}c_i\,x_i + \sum_{i = 1}^{n}c_i\,\epsilon_i \\ \end{align}$

\begin{aligned} E [{\hat{β}}_{1}^{'}] & = E [β_{0} \sum_{i = 1}^{n} c_{i}] + E [β_{1} \sum_{i = 1}^{n} c_{i} x_{i}] + E [\sum_{i = 1}^{n} c_{i} ϵ_{i}] \\ = β_{0} E [\sum_{i = 1}^{n} c_{i}] + β_{1} E [\sum_{i = 1}^{n} c_{i} x_{i}] + E [\sum_{i = 1}^{n} c_{i} ϵ_{i}] \\ = β_{0} E [\sum_{i = 1}^{n} c_{i}] + β_{1} E [\sum_{i = 1}^{n} c_{i} x_{i}] + \sum_{i = 1}^{n} c_{i} E [ϵ_{i}] \\ = β_{0} E [\sum_{i = 1}^{n} c_{i}] + β_{1} E [\sum_{i = 1}^{n} c_{i} x_{i}] \end{aligned}

$\begin{align} E[\hat{\beta}_1'] &= E\left[\beta_0\sum_{i = 1}^{n}c_i\right] + E\left[\beta_1\sum_{i = 1}^{n}c_i\,x_i \right] + E\left[\sum_{i = 1}^{n}c_i\,\epsilon_i\right] \\ &= \beta_0\,E\left[\sum_{i = 1}^{n}c_i\right] + \beta_1\,E\left[\sum_{i = 1}^{n}c_i\,x_i \right] + E\left[\sum_{i = 1}^{n}c_i\,\epsilon_i\right] \\ &= \beta_0\,E\left[\sum_{i = 1}^{n}c_i\right] + \beta_1\,E\left[\sum_{i = 1}^{n}c_i\,x_i \right] + \sum_{i = 1}^{n}c_i\,E\left[\epsilon_i\right] \\ &= \beta_0\,E\left[\sum_{i = 1}^{n}c_i\right] + \beta_1\,E\left[\sum_{i = 1}^{n}c_i\,x_i \right] \\ \end{align}$

すべての

β_{0}, β_{1}

$\beta_0,~\beta_1$ に対し，

E [{\hat{β}}_{1}^{'}] = β_{1}

$E[\hat{\beta}_1'] = \beta_1$ であるためには，

\sum_{i = 1}^{n} c_{i} = 0

$\begin{equation} \sum_{i = 1}^{n}c_i = 0 \end{equation}$

\sum_{i = 1}^{n} c_{i} x_{i} = 1

$\begin{equation} \sum_{i = 1}^{n}c_i\,x_i = 1 \end{equation}$

でなければならない．ここで，

\begin{aligned} \sum_{i = 1}^{n} {c_{i} - \frac{x_{i} - \bar{x}}{S_{x x}}}^{2} & = \sum_{i = 1}^{n} {c_{i}^{2} - 2 c_{i} \frac{x_{i} - \bar{x}}{S_{x x}} + \frac{(x_{i} - \bar{x})^{2}}{{S_{x x}}^{2}}} \\ = \sum_{i = 1}^{n} c_{i}^{2} - 2 \sum_{i = 1}^{n} c_{i} \frac{x_{i} - \bar{x}}{S_{x x}} + \sum_{i = 1}^{n} \frac{(x_{i} - \bar{x})^{2}}{{S_{x x}}^{2}} \\ = \sum_{i = 1}^{n} c_{i}^{2} - 2 \sum_{i = 1}^{n} c_{i} \frac{x_{i}}{S_{x x}} + 2 \sum_{i = 1}^{n} c_{i} \frac{\bar{x}}{S_{x x}} + \sum_{i = 1}^{n} \frac{(x_{i} - \bar{x})^{2}}{{S_{x x}}^{2}} \\ = \sum_{i = 1}^{n} c_{i}^{2} - 2 \frac{1}{S_{x x}} \sum_{i = 1}^{n} c_{i} x_{i} + 2 \frac{\bar{x}}{S_{x x}} \sum_{i = 1}^{n} c_{i} + \frac{1}{{S_{x x}}^{2}} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2} \\ = \sum_{i = 1}^{n} c_{i}^{2} - \frac{2}{S_{x x}} + 0 + \frac{1}{{S_{x x}}^{2}} S_{x x} \\ = \sum_{i = 1}^{n} c_{i}^{2} - \frac{1}{S_{x x}} \end{aligned}

$\begin{align} \sum_{i = 1}^{n}\left\{c_i - \frac{x_i - \bar{x}}{S_{xx}}\right\}^2 &= \sum_{i = 1}^{n}\left\{c_i^2 - 2c_i\frac{x_i - \bar{x}}{S_{xx}} + \frac{(x_i - \bar{x})^2}{{S_{xx}}^2}\right\} \\ &= \sum_{i = 1}^{n}c_i^2 - 2\sum_{i = 1}^{n}c_i\frac{x_i - \bar{x}}{S_{xx}} + \sum_{i = 1}^{n}\frac{(x_i - \bar{x})^2}{{S_{xx}}^2} \\ &= \sum_{i = 1}^{n}c_i^2 - 2\sum_{i = 1}^{n}c_i\frac{x_i}{S_{xx}} + 2\sum_{i = 1}^{n}c_i\frac{\bar{x}}{S_{xx}} + \sum_{i = 1}^{n}\frac{(x_i - \bar{x})^2}{{S_{xx}}^2} \\ &= \sum_{i = 1}^{n}c_i^2 - 2\frac{1}{S_{xx}}\sum_{i = 1}^{n}c_i\,x_i + 2\frac{\bar{x}}{S_{xx}}\sum_{i = 1}^{n}c_i + \frac{1}{{S_{xx}}^2}\sum_{i = 1}^{n}(x_i - \bar{x})^2 \\ &= \sum_{i = 1}^{n}c_i^2 - \frac{2}{S_{xx}} + 0+ \frac{1}{{S_{xx}}^2}S_{xx} \\ &= \sum_{i = 1}^{n}c_i^2 - \frac{1}{S_{xx}} \\ \end{align}$

\sum_{i = 1}^{n} c_{i}^{2} = \sum_{i = 1}^{n} {c_{i} - \frac{x_{i} - \bar{x}}{S_{x x}}}^{2} + \frac{1}{S_{x x}}

$\begin{equation} \sum_{i = 1}^{n}c_i^2 = \sum_{i = 1}^{n}\left\{c_i - \frac{x_i - \bar{x}}{S_{xx}}\right\}^2 + \frac{1}{S_{xx}} \end{equation}$

であることを利用すると，

\begin{aligned} V [{\hat{β}}_{1}^{'}] & = V [\sum_{i = 1}^{n} c_{i} y_{i}] \\ = \sum_{i = 1}^{n} {c_{i}}^{2} V [y_{i}] \\ = \sum_{i = 1}^{n} {c_{i}}^{2} σ^{2} \\ = σ^{2} \sum_{i = 1}^{n} {c_{i}}^{2} \\ = σ^{2} {\sum_{i = 1}^{n} {c_{i} - \frac{x_{i} - \bar{x}}{S_{x x}}}^{2} + \frac{1}{S_{x x}}} \\ = σ^{2} \sum_{i = 1}^{n} {c_{i} - \frac{x_{i} - \bar{x}}{S_{x x}}}^{2} + \frac{σ^{2}}{S_{x x}} \end{aligned}

$\begin{align} V[\hat{\beta}_1'] &= V\left[\sum_{i = 1}^{n}c_i\,y_i\right] \\ &= \sum_{i = 1}^{n}{c_i}^2\,V[y_i] \\ &= \sum_{i = 1}^{n}{c_i}^2\,\sigma^2 \\ &= \sigma^2\sum_{i = 1}^{n}{c_i}^2 \\ &= \sigma^2\left\{ \sum_{i = 1}^{n}\left\{c_i - \frac{x_i - \bar{x}}{S_{xx}}\right\}^2 + \frac{1}{S_{xx}} \right\} \\ &= \sigma^2\sum_{i = 1}^{n}\left\{c_i - \frac{x_i - \bar{x}}{S_{xx}}\right\}^2 + \frac{\sigma^2}{S_{xx}} \\ \end{align}$

よって，すべての

i

$i$ について，

c_{i} = \frac{x_{i} - \bar{x}}{S_{x x}}

$\begin{equation} c_i = \frac{x_i - \bar{x}}{S_{xx}} \end{equation}$

のとき，

V [{\hat{β}}_{1}^{'}]

$V[\hat{\beta}_1']$ は最小となる．したがって，

V [{\hat{β}}_{1}^{'}] \geq \frac{σ^{2}}{S_{x x}} = V [{\hat{β}}_{1}]

$\begin{equation} V[\hat{\beta}_1'] \ge \frac{\sigma^2}{S_{xx}} = V[\hat{\beta}_1] \end{equation}$

よって，

V [{\hat{β}}_{1}]

$V[\hat{\beta}_1]$ は最良線形不偏推定量である．

第5講 回帰分析

ガウス・マルコフの定理

補足1: 単回帰分析の場合

第5講回帰分析