第5講回帰分析

単回帰モデル

次の単回帰モデルを考える．

y_{i} = β_{0} + β_{1} x_{i} + ϵ_{i}

$\begin{equation} y_{i} = \beta_0 + \beta_1 x_{i} + \epsilon_{i} \end{equation}$

ϵ_{i} \sim N (0, σ^{2})

$\begin{equation} \epsilon_{i} \sim N(0,~\sigma^2) \end{equation}$

誤差項 $\epsilon_{i}$ は次の条件を満たす確率変数である．

$E[\epsilon_{i}] = 0~~~~i=1,2,\ldots,n$
$V[\epsilon_{i}] = \sigma^{2}~~~~i=1,2,\ldots,n$
$Cov[\epsilon_{i}, \epsilon_{j}] = E(\epsilon_{i}\epsilon_{j}) = 0~~~~i \ne j$

ここで，

\begin{aligned} \bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i} \\ \bar{y} = \frac{1}{n} \sum_{i = 1}^{n} y_{i} \end{aligned}

$\begin{align} \bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_i \\ \bar{y} = \frac{1}{n}\sum_{i=1}^{n}y_i \end{align}$

\begin{aligned} S_{x x} & = \sum_{i = 1}^{n} (x - \bar{x})^{2} \\ S_{y y} & = \sum_{i = 1}^{n} (y - \bar{y})^{2} \\ S_{y x} & = \sum_{i = 1}^{n} (x - \bar{x}) (y - \bar{y}) \end{aligned}

$\begin{align} S_{xx} &= \sum_{i=1}^{n}(x - \bar{x})^{2} \\ S_{yy} &= \sum_{i=1}^{n}(y - \bar{y})^{2} \\ S_{yx} &= \sum_{i=1}^{n}(x - \bar{x})(y - \bar{y}) \end{align}$

e_{i} = y_{i} - (β_{0} + β_{1} x_{i})

$\begin{equation} e_{i} = y_{i} - (\beta_0 + \beta_1 x_{i}) \end{equation}$

とおく．残差

e_{i}

$e_{i}$ の平方和

S_{e} = \sum_{i = 1}^{n} {e_{i}}^{2} = \sum_{i = 1}^{n} (y_{i} - β_{0} - β_{1} x_{i})^{2}

$\begin{equation} S_{e} = \sum_{i=1}^{n}{e_{i}}^{2} = \sum_{i=1}^{n}(y_{i} - \beta_0 - \beta_1 x_{i})^{2} \end{equation}$

を最小にする

β_{0}, β_{1}

$\beta_0,~\beta_1$ の値

\hat{β_{0}}, \hat{β_{1}}

$\hat{\beta_0},~\hat{\beta_1}$ は，

\begin{aligned} \hat{β_{1}} & = \frac{S_{x y}}{S_{x x}} \\ \hat{β_{0}} & = \bar{y} - \hat{β_{1}} \bar{x} \end{aligned}

$\begin{align} \hat{\beta_1} &= \frac{S_{xy}}{S_{xx}} \\ \hat{\beta_0} &= \bar{y} - \hat{\beta_1}\bar{x} \end{align}$

よって，推定式は，

\hat{y} = {\hat{β}}_{0} + {\hat{β}}_{1} x = \bar{y} + {\bar{β}}_{1} (x - \bar{x})

$\begin{equation} \hat{y} = \hat{\beta}_0 + \hat{\beta}_1 x = \bar{y} + \bar{\beta}_1(x - \bar{x}) \end{equation}$

第5講 回帰分析

単回帰モデル

第5講回帰分析