"회귀분석 제곱합 분할 증명"의 두 판 사이의 차이

(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...)
 
 
(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.
:(<math>SS_{total} = SS_{regression} + SS_{error}</math>)
* 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>
\sum_{i = 1}^n (Y_i - \overline{Y})^2
 
</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>
= \sum_{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})^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}) + (Y_i - \hat{Y}_i))^2
</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>
= \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\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2</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\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>
= \sum_{i = 1}^n (\hat{Y}_i - \overline{Y})^2
<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>
+ \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>
: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>
= \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>
+ \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\limits_{i = 1}^n (\hat{Y}_i - \overline{Y})^2 + \sum\limits_{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>
증명 끝.
\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
==주석==
<math>
<references/>
\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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2020년 9월 10일 (목) 01:33 기준 최신판

회귀분석 제곱합 분할 증명
회귀분석 자승합 분할 증명

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]

여기서 마지막 항은 0이므로 소거된다.[1][2]

[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.

3 주석[ | ]

  1. [math]\displaystyle{ \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]
  2. [math]\displaystyle{ \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]