第5講回帰分析

回帰係数の期待値

$\hat{\beta}_0$ と $\hat{\beta}_1$ が不偏推定量であることを示す．

\begin{aligned} E [{\hat{β}}_{1}] & = E [β_{1} + \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i}] \\ = E [β_{1}] + E [\sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) ϵ_{i}] \\ = β_{1} + \sum_{i = 1}^{n} (\frac{x_{i} - \bar{x}}{S_{x x}}) E [ϵ_{i}] \\ = β_{1} \end{aligned}

$\begin{align} E[\hat{\beta}_1] &= E\left[\beta_1 + \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) \epsilon_i\right] \\ &= E\left[\beta_1\right] +E\left[ \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right) \epsilon_i\right] \\ &= \beta_1 + \sum_{i=1}^{n}\left(\frac{x_i - \bar{x}}{S_{xx}}\right)\,E\left[\epsilon_i\right] \\ &= \beta_1 \end{align}$

\begin{aligned} E [{\hat{β}}_{0}] & = E [\bar{y} - {\hat{β}}_{1} \bar{x}] \\ = E [\bar{y}] - E [{\hat{β}}_{1}] \bar{x} \\ = β_{0} + β_{1} \bar{x} - β_{1} \bar{x} \\ = β_{0} \end{aligned}

$\begin{align} E[\hat{\beta}_0] &= E[\bar{y} - \hat{\beta}_1\bar{x}] \\ &= E[\bar{y}] - E[\hat{\beta}_1]\bar{x} \\ &= \beta_0 + \beta_1\bar{x} - \beta_1\bar{x} \\ &= \beta_0 \end{align}$

第5講 回帰分析

回帰係数の期待値

第5講回帰分析