對變數轉換後的推論
R 中的線性迴歸推論
Jo Hardin
Professor, Pomona College
解讀係數－線性
$Y = \beta_0 + \beta_1 \cdot X + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[Y_X] = \beta_0 + \beta_1 \cdot X$
$E[Y_{X+1}] = \beta_0 + \beta_1 \cdot (X+1)$
$\beta_1 = E[Y_{X+1}] - E[Y_X]$
解讀係數－X 非線性
$Y = \beta_0 + \beta_1 \cdot \ln(X) + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[Y_{\ln(X)}] = \beta_0 + \beta_1 \cdot \ln(X)$
$E[Y_{\ln(X)+1}] = \beta_0 + \beta_1 \cdot (\ln(X)+1)$
$\beta_1 = E[Y_{\ln(X)+1}] - E[Y_{\ln(X)}]$
解讀係數－Y 非線性
$\ln(Y) = \beta_0 + \beta_1 \cdot X + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[\ln(Y)_X] = \beta_0 + \beta_1 \cdot X$
$E[\ln(Y)_{X+1}] = \beta_0 + \beta_1 \cdot (X+1)$
$\beta_1 = E[\ln(Y)_{X+1}] - E[\ln(Y)_X]$
解讀係數－X 與 Y 皆非線性
$\ln(Y) = \beta_0 + \beta_1 \cdot \ln(X) + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon)$
$E[\ln(Y)_{\ln(X)}] = \beta_0 + \beta_1 \cdot \ln(X)$
$E[\ln(Y)_{\ln(X)+1}] = \beta_0 + \beta_1 \cdot (\ln(X)+1)$
$\beta_1 = E[\ln(Y)_{\ln(X)+1}] - E[\ln(Y)_{\ln{X}}]$
解讀係數－自然對數（特例）
$\ln(Y) = \beta_0 + \beta_1 \cdot \ln(X) + \epsilon$, 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 = $ $X$ 每變動 1% 時，$Y$ 的百分比變動
一起來練習吧！
R 中的線性迴歸推論
