对变换变量的推断
R 中的线性回归推断
Jo Hardin
Professor, Pomona College
系数解读——线性
$Y = \beta_0 + \beta_1 \cdot X + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[Y_X] = \beta_0 + \beta_1 \cdot X$
$E[Y_{X+1}] = \beta_0 + \beta_1 \cdot (X+1)$
$\beta_1 = E[Y_{X+1}] - E[Y_X]$
系数解读——非线性 X
$Y = \beta_0 + \beta_1 \cdot \ln(X) + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[Y_{\ln(X)}] = \beta_0 + \beta_1 \cdot \ln(X)$
$E[Y_{\ln(X)+1}] = \beta_0 + \beta_1 \cdot (\ln(X)+1)$
$\beta_1 = E[Y_{\ln(X)+1}] - E[Y_{\ln(X)}]$
系数解读——非线性 Y
$\ln(Y) = \beta_0 + \beta_1 \cdot X + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[\ln(Y)_X] = \beta_0 + \beta_1 \cdot X$
$E[\ln(Y)_{X+1}] = \beta_0 + \beta_1 \cdot (X+1)$
$\beta_1 = E[\ln(Y)_{X+1}] - E[\ln(Y)_X]$
系数解读——X 与 Y 均非线性
$\ln(Y) = \beta_0 + \beta_1 \cdot \ln(X) + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[\ln(Y)_{\ln(X)}] = \beta_0 + \beta_1 \cdot \ln(X)$
$E[\ln(Y)_{\ln(X)+1}] = \beta_0 + \beta_1 \cdot (\ln(X)+1)$
$\beta_1 = E[\ln(Y)_{\ln(X)+1}] - E[\ln(Y)_{\ln{X}}]$
系数解读——自然对数（特殊情形）
$\ln(Y) = \beta_0 + \beta_1 \cdot \ln(X) + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[\ln(Y)_{\ln(X)}] = \beta_0 + \beta_1 \cdot \ln(X)$
$E[\ln(Y)_{\ln(X)+1}] = \beta_0 + \beta_1 \cdot (\ln(X)+1)$
$\beta_1 = E[\ln(Y)_{\ln(X)+1}] - E[\ln(Y)_{\ln{X}}]$
OR (when $X$ and $Y$ are both transformed using natural log):
$\beta_1 = $ $X$ 每增加 1%，$Y$ 的百分比变化
Passons à la pratique !
R 中的线性回归推断
