"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...)
 
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]