SST, SSR, SSE 관계 유도

정의 $SST=\sum^n_{i=1}(y_ i-\bar{y})^2 \tag{1}$
$SSR=\sum^n_ {i=1}(\hat{y}_ i-\bar{y})^2 \tag{2}$
$SSE=\sum^n_ {i=1}(y_ i-\hat{y}_ i)^2 \tag{3}$

$SST=SSE+SSR$ 유도

$SST=\sum^n_{i=1}(y_ i-\bar{y})^2 \tag{4}$
$\sum^n_{i=1}(y_ i-\bar{y})^2=\sum^n_{i=1}(y_ i-\hat{y}_ i+\hat{y}_ i-\bar{y})^2 \tag{5}$ $\sum^n_{i=1}(y_ i-\hat{y}_ i+\hat{y}_ i-\bar{y})^2=\sum^n_{i=1}[(y_ i-\hat{y}_ i)+(\hat{y}_ i-\bar{y})]^2 \tag{6}$
$\sum^n_{i=1}[(y_ i-\hat{y}_ i)+(\hat{y}_ i-\bar{y})]^2=\sum^n_{i=1}[(y_ i-\hat{y}_ i)^2+2(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})+(\hat{y}_ i-\bar{y})^2] \tag{7}$
$\sum^n_{i=1}[(y_ i-\hat{y}_ i)^2+2(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})+(\hat{y}_ i-\bar{y})^2]=\sum^n_{i=1}(y_ i-\hat{y}_ i)^2+\sum^n_{i=1}2(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})+\sum^n_{i=1}(\hat{y}_ i-\bar{y})^2 \tag{8}$

(2), (3)을 (8)에 대입하면, $\sum^n_{i=1}(y_ i-\hat{y}_ i)^2+\sum^n_{i=1}2(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})+\sum^n_{i=1}(\hat{y}_ i-\bar{y})^2=SSE+\sum^n_{i=1}2(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})+SSR \tag{9}$

$SST=SSE+SSR$이 되기 위해서는, $\sum^n_{i=1}2(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})=0$가 성립해야 함.
최소제곱량 추정 (46)에 의해, $\bar{y}=\hat{\beta}_ 0+\hat{\beta}_ 1\bar{x} \tag{10}$

$\hat{y}_ i=\hat{\beta}_ 0+\hat{\beta}_ 1x_ i \tag{11}$

$\sum^n_{i=1}(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})\tag{12}$
$\sum^n_{i=1}(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})=\sum^n_{i=1}(y_ i-\hat{\beta}_ 0+\hat{\beta}_ 1x_ i)(\hat{\beta}_ 0+\hat{\beta}_ 1x_ i-\hat{\beta}_ 0+\hat{\beta}_ 1\bar{x})\tag{13}$
$\sum^n_{i=1}(y_ i-\hat{\beta}_ 0+\hat{\beta}_ 1x_ i)(\hat{\beta}_ 0+\hat{\beta}_ 1x_ i-\hat{\beta}_ 0+\hat{\beta}_ 1\bar{x})=\sum^n_{i=1}(y_ i-\hat{\beta}_ 0+\hat{\beta}_ 1x_ i)(\hat{\beta}_ 1x_ i+\hat{\beta}_ 1\bar{x}) \tag{13}$
$\sum^n_{i=1}(y_ i-\hat{\beta}_ 0+\hat{\beta}_ 1x_ i)(\hat{\beta}_ 1x_ i+\hat{\beta}_ 1\bar{x})=\hat{\beta}_ 1\sum^n_{i=1}(y_ i-\hat{\beta}_ 0+\hat{\beta}_ 1x_ i)(x_ i+\bar{x}) \tag{13}$
$\hat{\beta}_ 1\sum^n_{i=1}(y_ i-\hat{\beta}_ 0+\hat{\beta}_ 1x_ i)(x_ i+\bar{x})=\hat{\beta}_ 1\sum^n_{i=1}y_ i(x_ i+\bar{x})-\hat{\beta}_ 0\hat{\beta}_ 1\sum^n_{i=1}(x_ i+\bar{x})+\hat{\beta}_ 1^2\sum^n_{i=1}x_ i(x_ i+\bar{x}) \tag{14}$
최소제곱량 추정 (36)에 의해, $\sum^n_{i=1}(x_ i+\bar{x})=0 \tag{15}$

(15)를 (14)에 대입하면, $\hat{\beta}_ 1\sum^n_{i=1}y_ i(x_ i+\bar{x})-\hat{\beta}_ 0\hat{\beta}_ 1\sum^n_{i=1}(x_ i+\bar{x})+\hat{\beta}_ 1^2\sum^n_{i=1}x_ i(x_ i+\bar{x})=\hat{\beta}_ 1\sum^n_{i=1}y_ i(x_ i+\bar{x})+\hat{\beta}_ 1^2\sum^n_{i=1}x_ i(x_ i+\bar{x}) \tag{16}$

최소제곱량 추정 (43)에 의해, $\sum^n_{i=1}x_ i(x_ i+\bar{x})=\sum^n_{i=1}(x_ i+\bar{x})^2 \tag{17}$
$\sum^n_{i=1}y_ i(x_ i+\bar{x})=\sum^n_{i=1}(x_ i+\bar{x})(y_ i+\bar{y}) \tag{18}$
(17), (18)을 (16)에 대입하면,

$\hat{\beta}_ 1\sum^n_{i=1}y_ i(x_ i+\bar{x})+\hat{\beta}_ 1^2\sum^n_{i=1}x_ i(x_ i+\bar{x})=\hat{\beta}_ 1\sum^n_{i=1}(x_ i+\bar{x})(y_ i+\bar{y})+\hat{\beta}_ 1^2\sum^n_{i=1}(x_ i+\bar{x})^2 \tag{19}$
$\hat{\beta}_ 1\sum^n_{i=1}(x_ i+\bar{x})(y_ i+\bar{y})+\hat{\beta}_ 1^2\sum^n_{i=1}(x_ i+\bar{x})^2=\hat{\beta}_ 1 S_{XY}+\hat{\beta}_ 1^2S_{XX} \tag{20}$
$\hat{\beta}_ 1 S_{XY}+\hat{\beta}_ 1^2S_{XX}=\hat{\beta}_ 1 (S_{XY}+\hat{\beta}_ 1S_{XX}) \tag{20}$

최소제곱량 추정 (47)에 의해, $\hat{\beta}_ 1=\frac{S_{XY}}{S_{XX}} \tag{21}$

(21)을 (20)에 대입하면, $\hat{\beta}_ 1(S_{XY}+\hat{\beta}_ 1S_{XX})=\hat{\beta}_ 1 (S_{XY}+\frac{S_{XY}}{S_{XX}}S_{XX}) \tag{22}$
$\hat{\beta}_ 1(S_{XY}+\frac{S_{XY}}{S_{XX}}S_{XX})=\hat{\beta}_ 1(S_{XY}+S_{XY})=0 \tag{23}$

(23)을 (9)에 대입하면, $SSE+\sum^n_{i=1}2(y_ i-\hat{y}_ i)(\hat{y}_ i-\bar{y})+SSR=SSE+SSR \tag{24}$
따라서 $SST=SSE+SSR$ ■