(새 문서: ==문제== <math>\sum (Y_i-\overline{Y})^2 = \sum (\hat{Y_i}-\overline{Y})^2 + \sum (Y_i-\hat{Y_i})^2</math>임을 증명하라. :(<math>SS_{total} = SS_{regression} + SS_{error}</mat...) |
(→증명) |
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4번째 줄: | 4번째 줄: | ||
==증명== | ==증명== | ||
<math> | <math>\sum_{i = 1}^n (Y_i - \overline{Y})^2</math> | ||
\sum_{i = 1}^n (Y_i - \overline{Y})^2 | |||
</math> | |||
<math> | <math>= \sum_{i = 1}^n (Y_i - \overline{Y} + \hat{Y}_i - \hat{Y}_i)^2</math> | ||
= \sum_{i = 1}^n (Y_i - \overline{Y} + \hat{Y}_i - \hat{Y}_i)^2 | |||
</math> | |||
<math> | <math>= \sum_{i = 1}^n ((\hat{Y}_i - \overline{Y}) + (Y_i - \hat{Y}_i))^2</math> | ||
= \sum_{i = 1}^n ((\hat{Y}_i - \overline{Y}) + (Y_i - \hat{Y}_i))^2 | |||
</math> | |||
<math> | <math>= \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> | ||
= \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> | <math>= \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> | ||
= \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> | <math>= \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2</math> | ||
= \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 | <math>+ \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2</math> | ||
</math> | <math>+ 2 \sum_{i = 1}^n (Y_i - \hat{Y}_i)(\beta_0 + \beta_1 X_i - \overline{Y})</math> | ||
<math> | |||
+ \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2 | |||
</math> | |||
<math> | |||
+ 2 \sum_{i = 1}^n (Y_i - \hat{Y}_i)(\beta_0 + \beta_1 X_i - \overline{Y}) | |||
</math> | |||
<math> | <math>= \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> | ||
= \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>+ \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> | |||
<math> | |||
+ \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> | <math>= \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2</math> | ||
= \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum_{i = 1}^n (Y_i - \hat{Y}_i)^2 | |||
</math> | |||
<math> | <math>\because \sum_{i = 1}^n (Y_i - \hat{Y_i}) = \sum_{i=1}^n (Y_i - \beta_0 - \beta_1 X_i) = 0</math> | ||
\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 | and | ||
<math> | <math>\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> | ||
\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> | |||
[[분류:통계]] | [[분류:통계]] | ||
[[분류:증명]] | [[분류:증명]] |
2012년 2월 5일 (일) 01:35 판
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]