Pourquoi

Remarque: $SST$ = somme des carrés au total, $SSE$ = somme des erreurs au carré et $SSR$ = somme des carrés de régression. L'équation dans le titre est souvent écrite comme suit:

\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}

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

Question assez simple, mais je cherche une explication intuitive. Intuitivement, il me semble que $SST\geq SSE+SSR$ aurait plus de sens. Par exemple, supposons que le point $x_i$ n'a correspondant valeur y $y_i=5$ et , où est le point correspondant sur la droite de régression. Supposons également que la valeur moyenne y pour l'ensemble de données est . Ensuite, pour ce point particulier i, $\hat y_i=3$ $\hat y_i$ $\bar y=0$ $SST=(5-0)^2=5^2=25$ , tandis que $SSE=(5-3)^2=2^2=4$ et $SSR=(3-0)^2=3^2=9$ . Évidemment, $9+4<25$ . Ce résultat ne se généraliserait-il pas à l'ensemble des données? Je ne comprends pas.

regression least-squares r-squared Came
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Les discussions très proches ont également de bonnes réponses: stats.stackexchange.com/questions/1447 , stats.stackexchange.com/questions/118 , stats.stackexchange.com/questions/123651 , stats.stackexchange.com/questions/204930 et stats.stackexchange.com/questions/127598 .

whuber

Réponses:

Additionner et soustraire donne Nous devons donc montrer que

\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}

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

. Écriture

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

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

Donc, (a) les résidus

besoin d'être orthogonale aux valeurs ajustées,

, et (b) la somme des besoinsmatière de valeurs ajustées pour être égale à la somme de la variable dépendante,

\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

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

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

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

En fait, je pense que (a) est plus facile de montrer en notation matricielle pour la régression multiple générale de laquelle le boîtier de variable unique est un cas particulier:

\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}

$\begin{eqnarray*} e'X\hat\beta &=&(y-X\hat\beta)'X\hat\beta\\ &=&(y-X(X'X)^{-1}X'y)'X\hat\beta\\ &=&y'(X-X(X'X)^{-1}X'X)\hat\beta\\ &=&y'(X-X)\hat\beta=0 \end{eqnarray*}$

\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
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(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
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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
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This is just the Pythagorean theorem! enter image description here

user0
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stats.stackexchange.com/q/71620/171583, stats.stackexchange.com/a/256532/171583.

ayorgo