모형에 상수 항이있는 경우 은 X 의 열 공간에 있습니다 ( ˉ Y 1 n 과 마찬가지로 나중에 유용함). 피팅 Y는 관측의 정사영 인 Y 그 열 공간에 의해 형성되는 평면 상. 이 수단 잔차 벡터의 E = Y – Y는 평면에 수직이며, 따라서, 행 1 N . 내적을 고려하면 ∑ n i = 1 e i = 0 이므로1n

${\mathbf{1}}_{\mathbf{n}}$

$\mathbf{1_n}$X

$\mathbf{X}$

$\mathbf{X}$Y¯1n

$\overline{Y}{\mathbf{1}}_{\mathbf{n}}$

$\bar{Y}\mathbf{1_n}$Y^

$\hat{\mathbf{Y}}$

$\mathbf{\hat{Y}}$Y

$\mathbf{Y}$

$\mathbf{Y}$e=y−y^

$\mathbf{e}=\mathbf{y}-\hat{\mathbf{y}}$

$\mathbf{e} = \mathbf{y} - \mathbf{\hat{y}}$1n

${\mathbf{1}}_{\mathbf{n}}$

$\mathbf{1_n}$∑ni=1ei=0

$\sum _{i=1}^n{e}_i=0$

$\sum_{i=1}^n e_i = 0$ 는 0으로 합산되어야합니다. 이후 Y 난 = ^ Y I + E 난 우리 결론 Σ N 난 = 1 Y를 난 = Σ N 난 = 1 ^ Y를 제가 너무 모두 끼워 관찰 응답 평균 있는지 ˉ Y를 .e

$\mathbf{e}$

$\mathbf{e}$Yi=Yi^+ei

$Y_i=\widehat{Y_i}+e_i$

$Y_i = \hat{Y_i} + e_i$∑ni=1Yi=∑ni=1Yi^

$\sum _{i=1}^n{Y}_i=\sum _{i=1}^n\widehat{Y_i}$

$\sum_{i=1}^n Y_i = \sum_{i=1}^n \hat{Y_i}$Y¯

$\overline{Y}$

$\bar{Y}$

도면에서 점선 나타내는 과 Y – ˉ Y 1 N 이며, 중심 관찰 끼워 맞춤 응답 벡터. 각도의 코사인 θ 이러한 벡터 사이 그러므로의 상관 것이다 Y 및 Y 정의 다중 상관 계수이고, R . 이들 잔차 벡터의 벡터를 형성하는 삼각형 때문에 직각되는 Y – ˉ Y 1 N 놓여 평면뿐만 Y−Y¯1n

$\mathbf{Y}-\overline{Y}{\mathbf{1}}_{\mathbf{n}}$

$\mathbf{Y} - \bar{Y}\mathbf{1_n}$Y^−Y¯1n

$\hat{\mathbf{Y}}-\overline{Y}{\mathbf{1}}_{\mathbf{n}}$

$\mathbf{\hat{Y}} - \bar{Y}\mathbf{1_n}$θ

$\theta$

$\theta$Y

$Y$

$Y$Y^

$\hat{Y}$

$\hat{Y}$R

$R$

$R$Y^−Y¯1n

$\hat{\mathbf{Y}}-\overline{Y}{\mathbf{1}}_{\mathbf{n}}$

$\mathbf{\hat{Y}} - \bar{Y}\mathbf{1_n}$ 는 그것에 직교합니다. 금후:e

$\mathbf{e}$

$\mathbf{e}$

$$R = \cos(\theta) = \frac{\text{adj}}{\text{hyp}} = \frac{\|\mathbf{\hat{Y}} - \bar{Y}\mathbf{1_n}\|}{\|\mathbf{Y} - \bar{Y}\mathbf{1_n}\|}$$

피타고라스를 삼각형에 적용 할 수도 있습니다.

$$\|\mathbf{Y} - \bar{Y}\mathbf{1_n}\|^2 = \|\mathbf{Y} - \mathbf{\hat{Y}}\|^2 + \|\mathbf{\hat{Y}} - \bar{Y}\mathbf{1_n}\|^2$$

다음과 같이 더 친숙 할 수 있습니다.

$$\sum_{i=1}^{n} (Y_i - \bar{Y})^2 = \sum_{i=1}^{n} (Y_i - \hat{Y}_i)^2 + \sum_{i=1}^{n} (\hat{Y}_i - \bar{Y})^2$$

이것은 제곱의 합, 의 분해입니다 .SStotal=SSresidual+SSregression

$S{S}_{\text{total}}=S{S}_{\text{residual}}+S{S}_{\text{regression}}$

$SS_{\text{total}} = SS_{\text{residual}} + SS_{\text{regression}}$

결정 계수에 대한 표준 정의는 다음과 같습니다.

$$R^2 = 1 - \frac{SS_{\text{residual}}}{SS_{\text{total}}} = 1 - \frac{\sum_{i=1}^n (y_i - \hat{y}_i)^2}{\sum_{i=1}^n (y_i - \bar{y})^2} = 1 - \frac{\|\mathbf{Y} - \mathbf{\hat{Y}}\|^2}{\|\mathbf{Y} - \bar{Y}\mathbf{1_n}\|^2}$$

When the sums of squares can be partitioned, it takes some straightforward algebra to show this is equivalent to the “proportion of variance explained” formulation,

$$R^2 = \frac{SS_{\text{regression}}}{SS_{\text{total}}} = \frac{\sum_{i=1}^n (\hat{y}_i - \bar{y})^2}{\sum_{i=1}^n (y_i - \bar{y})^2} = \frac{\|\mathbf{\hat{Y}} - \bar{Y}\mathbf{1_n}\|^2}{\|\mathbf{Y} - \bar{Y}\mathbf{1_n}\|^2}$$

There is a geometric way of seeing this from the triangle, with minimal algebra. The definitional formula gives R2=1−sin2(θ)

$R^2=1-{\sin }^2(\theta )$

$R^2 = 1 - \sin^2(\theta)$ and with basic trigonometry we can simplify this to cos2(θ)

${\cos }^2(\theta )$

$\cos^2(\theta)$. This is the link between R2

$R^2$

$R^2$ and R

$R$

$R$.

Note how vital it was for this analysis to have fitted an intercept term, so that 1n

${\mathbf{1}}_{\mathbf{n}}$

$\mathbf{1_n}$ was in the column space. Without this, the residuals would not have summed to zero, and the mean of the fitted values would not have coincided with the mean of Y

$Y$

$Y$. In that case we couldn’t have drawn the triangle; the sums of squares would not have decomposed in a Pythagorean manner; R2

$R^2$

$R^2$ would not have had the frequently-quoted form SSreg/SStotal

$S{S}_{\text{reg}}/S{S}_{\text{total}}$

$SS_{\text{reg}}/SS_{\text{total}}$ nor be the square of R

$R$

$R$. In this situation, some software (including R) uses a different formula for R2

$R^2$

$R^2$ altogether.

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