1 문제
[math]\displaystyle{ \sum (Y_i-\overline{Y})^2 = \sum (\hat{Y_i}-\overline{Y})^2 + \sum (Y_i-\hat{Y_i})^2 }[/math]임을 증명하라.
- ([math]\displaystyle{ SS_{total} = SS_{regression} + SS_{error} }[/math])
2 증명
[math]\displaystyle{ \sum_{i = 1}^n (Y_i - \overline{Y})^2 }[/math]
[math]\displaystyle{ = \sum_{i = 1}^n (Y_i - \overline{Y} + \hat{Y}_i - \hat{Y}_i)^2 }[/math]
[math]\displaystyle{ = \sum_{i = 1}^n ((\hat{Y}_i - \overline{Y}) + (Y_i - \hat{Y}_i))^2 }[/math]
[math]\displaystyle{ = \sum_{i = 1}^n ((\hat{Y}_i - \overline{Y})^2 + (Y_i - \hat{Y}_i)^2 + 2 (Y_i - \hat{Y}_i) (\hat{Y}_i - \overline{Y})) }[/math]
[math]\displaystyle{ = \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2 + 2 \sum_{i = 1}^n (Y_i - \hat{Y}_i) (\hat{Y}_i - \overline{Y}) }[/math]
[math]\displaystyle{ = \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 }[/math] [math]\displaystyle{ + \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2 }[/math] [math]\displaystyle{ + 2 \sum_{i = 1}^n (Y_i - \hat{Y}_i)(\beta_0 + \beta_1 X_i - \overline{Y}) }[/math]
[math]\displaystyle{ = \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2 + 2 ( \beta_0 \sum_{i = 1}^n (Y_i - \hat{Y}_i) }[/math] [math]\displaystyle{ + \beta_1 \sum_{i = 1}^n (Y_i - \hat{Y}_i) X_i - \overline{Y} \sum_{i = 1}^n (Y_i - \hat{Y}_i)) }[/math]
[math]\displaystyle{ = \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2 }[/math]
[math]\displaystyle{ \because \sum_{i = 1}^n (Y_i - \hat{Y_i}) = \sum_{i=1}^n (Y_i - \beta_0 - \beta_1 X_i) = 0 }[/math]
and [math]\displaystyle{ \sum_{i = 1}^n (Y_i - \hat{Y_i}) X_i = \sum_{i=1}^n (Y_i - \beta_0 - \beta_1 X_i) X_i = 0 }[/math]
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