Tại sao

14

Lưu ý: $SST$ = Sum of Squares Total, $SSE$ = Sum of Squared Error và $SSR$ = Regression Sum of Squares. Phương trình trong tiêu đề thường được viết là:

\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}

$\sum_{i=1}^n (y_i-\bar y)^2=\sum_{i=1}^n (y_i-\hat y_i)^2+\sum_{i=1}^n (\hat y_i-\bar y)^2$

Câu hỏi khá đơn giản, nhưng tôi đang tìm kiếm một lời giải thích trực quan. Theo trực giác, dường như với tôi, $SST\geq SSE+SSR$ sẽ có ý nghĩa hơn. Ví dụ, giả sử điểm $x_i$ đã tương ứng với giá trị y $y_i=5$ và , nơi là điểm tương ứng trên đường hồi quy. Cũng giả sử rằng giá trị y trung bình của tập dữ liệu là . Sau đó, cho điểm đặc biệt này i, $\hat y_i=3$ $\hat y_i$ $\bar y=0$ $SST=(5-0)^2=5^2=25$ , trong khi $SSE=(5-3)^2=2^2=4$ và $SSR=(3-0)^2=3^2=9$ . Rõ ràng, $9+4<25$ . Kết quả này sẽ không khái quát cho toàn bộ dữ liệu? Tôi không hiểu

regression least-squares r-squared

— Cam
nguồn

1

Các chủ đề liên quan rất chặt chẽ cũng có câu trả lời tốt: stats.stackexchange.com/questions/1447 , stats.stackexchange.com/questions/118 , stats.stackexchange.com/questions/123651 , stats.stackexchange.com/questions/204930 và stats.stackexchange.com/questions/127598 .

— whuber

15

Cộng và trừ sẽ cho Vì vậy, chúng ta cần chỉ ra rằng

\begin{array}{rcl} \sum_{i = 1}^{n} (y_{i} - \bar{y})^{2} & = & \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i} + {\hat{y}}_{i} - \bar{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} - \bar{y}) + \sum_{i = 1}^{n} ({\hat{y}}_{i} - \bar{y})^{2} \end{array}

$\begin{eqnarray*} \sum_{i=1}^n (y_i-\bar y)^2&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\ &=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2 \end{eqnarray*}$

\sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) ({\hat{y}}_{i} - \bar{y}) = 0

$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=0$

\sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) ({\hat{y}}_{i} - \bar{y}) = \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) {\hat{y}}_{i} - \bar{y} \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i})

$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=\sum_{i=1}^n(y_i-\hat y_i)\hat y_i-\bar y\sum_{i=1}^n(y_i-\hat y_i)$

e_{i} = y_{i} - {\hat{y}}_{i}

$e_i=y_i-\hat y_i$

\sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) {\hat{y}}_{i} = 0

$\sum_{i=1}^n(y_i-\hat y_i)\hat y_i=0$ , and (b) the sum of the fitted values needs to be equal to the sum of the dependent variable,

\sum_{i = 1}^{n} y_{i} = \sum_{i = 1}^{n} {\hat{y}}_{i}

$\sum_{i=1}^ny_i=\sum_{i=1}^n\hat y_i$ .

Actually, I think (a) is easier to show in matrix notation for general multiple regression of which the single variable case is a special case:

\begin{array}{rcl} e^{'} X \hat{β} & = & (y - X \hat{β})^{'} X \hat{β} \\ = & (y - X (X^{'} X)^{- 1} X^{'} y)^{'} X \hat{β} \\ = & y^{'} (X - X (X^{'} X)^{- 1} X^{'} X) \hat{β} \\ = & y^{'} (X - X) \hat{β} = 0 \end{array}

$\begin{eqnarray*} e'X\hat\beta &=&(y-X\hat\beta)'X\hat\beta\\ &=&(y-X(X'X)^{-1}X'y)'X\hat\beta\\ &=&y'(X-X(X'X)^{-1}X'X)\hat\beta\\ &=&y'(X-X)\hat\beta=0 \end{eqnarray*}$ As for (b), the derivative of the OLS criterion function with respect to the constant (so you need one in the regression for this to be true!), aka the normal equation, is

\frac{\partial S S R}{\partial \hat{α}} = - 2 \sum_{i} (y_{i} - \hat{α} - \hat{β} x_{i}) = 0,

$\frac{\partial SSR}{\partial\hat\alpha}=-2\sum_i(y_i-\hat\alpha-\hat\beta x_i)=0,$ which can be rearranged to

\sum_{i} y_{i} = n \hat{α} + \hat{β} \sum_{i} x_{i}

$\sum_i y_i=n\hat\alpha+\hat\beta\sum_ix_i$ The right hand side of this equation evidently also is

\sum_{i = 1}^{n} {\hat{y}}_{i}

$\sum_{i=1}^n\hat y_i$ , as

{\hat{y}}_{i} = \hat{α} + \hat{β} x_{i}

$\hat y_i=\hat\alpha+\hat\beta x_i$ .

— Christoph Hanck
nguồn

3

(1) Intuition for why $SST = SSR + SSE$

When we try to explain the total variation in Y ( $SST$ ) with one explanatory variable, X, then there are exactly two sources of variability. First, there is the variability captured by X (Sum Square Regression), and second, there is the variability not captured by X (Sum Square Error). Hence, $SST = SSR + SSE$ (exact equality).

(2) Geometric intuition

Please see the first few pictures here (especially the third): https://sites.google.com/site/modernprogramevaluation/variance-and-bias

Some of the total variation in the data (distance from datapoint to $\bar{Y}$ ) is captured by the regression line (the distance from the regression line to $\bar{Y}$ ) and error (distance from the point to the regression line). There's not room left for $SST$ to be greater than $SSE + SSR$ .

(3) The problem with your illustration

You can't look at SSE and SSR in a pointwise fashion. For a particular point, the residual may be large, so that there is more error than explanatory power from X. However, for other points, the residual will be small, so that the regression line explains a lot of the variability. They will balance out and ultimately $SST = SSR + SSE$ . Of course this is not rigorous, but you can find proofs like the above.

Also notice that regression will not be defined for one point: $b_1 = \frac{\sum(X_i -\bar{X})(Y_i-\bar{Y}) }{\sum (X_i -\bar{X})^2}$ , and you can see that the denominator will be zero, making estimation undefined.

Hope this helps.

--Ryan M.

— RMurphy
nguồn

1

When an intercept is included in linear regression(sum of residuals is zero), $SST=SSE+SSR$ .

prove

\begin{array}{rcl} S S T & = & \sum_{i = 1}^{n} (y_{i} - \bar{y})^{2} \\ = & \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i} + {\hat{y}}_{i} - \bar{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} - \bar{y}) + \sum_{i = 1}^{n} ({\hat{y}}_{i} - \bar{y})^{2} \\ = & S S E + S S R + 2 \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) ({\hat{y}}_{i} - \bar{y}) \end{array}

$\begin{eqnarray*} SST&=&\sum_{i=1}^n (y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2\\&=&SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y) \end{eqnarray*}$ Just need to prove last part is equal to 0:

\begin{array}{rcl} \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) ({\hat{y}}_{i} - \bar{y}) & = & \sum_{i = 1}^{n} (y_{i} - β_{0} - β_{1} x_{i}) (β_{0} + β_{1} x_{i} - \bar{y}) \\ = & (β_{0} - \bar{y}) \sum_{i = 1}^{n} (y_{i} - β_{0} - β_{1} x_{i}) + β_{1} \sum_{i = 1}^{n} (y_{i} - β_{0} - β_{1} x_{i}) x_{i} \end{array}

$\begin{eqnarray*} \sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)&=&\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)(\beta_0+\beta_1x_i-\bar y)\\&=&(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i \end{eqnarray*}$ In Least squares regression, the sum of the squares of the errors is minimized.

S S E = \sum_{i = 1}^{n} {(e_{i})}^{2} = \sum_{i = 1}^{n} {(y_{i} - \hat{y_{i}})}^{2} = \sum_{i = 1}^{n} {(y_{i} - β_{0} - β_{1} x_{i})}^{2}

$SSE=\displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum_{i=1}^n\left(y_i - \hat{y_i} \right)^2= \sum_{i=1}^n\left(y_i -\beta_0- \beta_1x_i\right)^2$ Take the partial derivative of SSE with respect to

β_{0}

$\beta_0$ and setting it to zero.

\frac{\partial S S E}{\partial β_{0}} = \sum_{i = 1}^{n} 2 {(y_{i} - β_{0} - β_{1} x_{i})}^{1} = 0

$\frac{\partial{SSE}}{\partial{\beta_0}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$ So

\sum_{i = 1}^{n} {(y_{i} - β_{0} - β_{1} x_{i})}^{1} = 0

$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$ Take the partial derivative of SSE with respect to

β_{1}

$\beta_1$ and setting it to zero.

\frac{\partial S S E}{\partial β_{1}} = \sum_{i = 1}^{n} 2 {(y_{i} - β_{0} - β_{1} x_{i})}^{1} x_{i} = 0

$\frac{\partial{SSE}}{\partial{\beta_1}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$ So

\sum_{i = 1}^{n} {(y_{i} - β_{0} - β_{1} x_{i})}^{1} x_{i} = 0

$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$ Hence,

\sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) ({\hat{y}}_{i} - \bar{y}) = (β_{0} - \bar{y}) \sum_{i = 1}^{n} (y_{i} - β_{0} - β_{1} x_{i}) + β_{1} \sum_{i = 1}^{n} (y_{i} - β_{0} - β_{1} x_{i}) x_{i} = 0

$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i=0$

S S T = S S E + S S R + 2 \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i}) ({\hat{y}}_{i} - \bar{y}) = S S E + S S R

$SST=SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=SSE+SSR$

— DavidCruise
nguồn

1

This is just the Pythagorean theorem! enter image description here

— user0
nguồn

stats.stackexchange.com/q/71620/171583, stats.stackexchange.com/a/256532/171583.

— ayorgo