(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...) |
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(같은 사용자의 중간 판 12개는 보이지 않습니다) | |||
1번째 줄: | 1번째 줄: | ||
;[[회귀분석]] [[제곱합]] 분할 증명 | |||
;회귀분석 자승합 분할 증명 | |||
==문제== | ==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>\sum (Y_i-\overline{Y})^2 = \sum (\hat{Y_i}-\overline{Y})^2 + \sum (Y_i-\hat{Y_i})^2</math>임을 증명하라. | ||
* 다시 말해서 <math>SS_{total} = SS_{regression} + SS_{error}</math>를 증명하라는 문제. | |||
==증명== | ==증명== | ||
<math> | <math>\sum\limits_{i = 1}^n (Y_i - \overline{Y})^2</math> | ||
\ | |||
</math> | <math>= \sum\limits_{i = 1}^n (Y_i - \overline{Y} + \hat{Y}_i - \hat{Y}_i)^2</math> | ||
<math> | <math>= \sum\limits_{i = 1}^n ((\hat{Y}_i - \overline{Y}) + (Y_i - \hat{Y}_i))^2</math> | ||
= \ | |||
</math> | |||
<math> | <math>= \sum\limits_{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> | <math>= \sum\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2 + 2 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i) (\hat{Y}_i - \overline{Y})</math> | ||
= \ | |||
</math> | |||
<math> | <math>= \sum\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2</math> | ||
= \ | <math>+ \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2</math> | ||
</math> | <math>+ 2 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)(\beta_0 + \beta_1 X_i - \overline{Y})</math> | ||
<math> | <math>= \sum\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2 + 2 ( \beta_0 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)</math> | ||
= \ | <math>+ \beta_1 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i) X_i - \overline{Y} \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i))</math> | ||
+ \ | |||
</math> | |||
<math> | |||
+ | |||
</math> | |||
<math> | :여기서 마지막 항은 0이므로 소거된다.<ref><math>\because \sum\limits_{i = 1}^n (Y_i - \hat{Y_i}) = \sum\limits_{i=1}^n (Y_i - \beta_0 - \beta_1 X_i) = 0</math></ref><ref><math>\sum\limits_{i = 1}^n (Y_i - \hat{Y_i}) X_i = \sum\limits_{i=1}^n (Y_i - \beta_0 - \beta_1 X_i) X_i = 0</math></ref> | ||
</math> | |||
<math> | |||
</math> | |||
<math> | <math>= \sum\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2</math> | ||
= \ | |||
</math> | |||
증명 끝. | |||
==주석== | |||
<references/> | |||
</ | |||
[[분류:통계]] | [[분류:통계]] | ||
[[분류:증명]] | [[분류:증명]] |
2020년 9월 10일 (목) 01:33 기준 최신판
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.
- 다시 말해서 [math]\displaystyle{ SS_{total} = SS_{regression} + SS_{error} }[/math]를 증명하라는 문제.
2 증명[ | ]
[math]\displaystyle{ \sum\limits_{i = 1}^n (Y_i - \overline{Y})^2 }[/math]
[math]\displaystyle{ = \sum\limits_{i = 1}^n (Y_i - \overline{Y} + \hat{Y}_i - \hat{Y}_i)^2 }[/math]
[math]\displaystyle{ = \sum\limits_{i = 1}^n ((\hat{Y}_i - \overline{Y}) + (Y_i - \hat{Y}_i))^2 }[/math]
[math]\displaystyle{ = \sum\limits_{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\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2 + 2 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i) (\hat{Y}_i - \overline{Y}) }[/math]
[math]\displaystyle{ = \sum\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 }[/math] [math]\displaystyle{ + \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2 }[/math] [math]\displaystyle{ + 2 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)(\beta_0 + \beta_1 X_i - \overline{Y}) }[/math]
[math]\displaystyle{ = \sum\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2 + 2 ( \beta_0 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i) }[/math] [math]\displaystyle{ + \beta_1 \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i) X_i - \overline{Y} \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)) }[/math]
[math]\displaystyle{ = \sum\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum\limits_{i = 1}^n (Y_i - \hat{Y}_i)^2 }[/math]
증명 끝.