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An extra sum of squares $SSR(X_p|X_1,...X_{p-1})$, assuming that no pair of predictor variables are perfectly correlated, measures the marginal reduction in the error sum of squares. Eventually one can view an extra sum of squares as measuring the marginal increase in the regression sum of squares when one or several predictor variables are added to the regression model.

Assuming the number of observational data $n$ is lager than the number of predictor variables $p$.

Can I have a rigourous proof that $SSTO=\sum_{i=1}^n (Y_i-\bar{Y})^2\geq SSR(X_p|X_1,...X_{p-1})\geq0$ ?


Applying LSE

$SSE(X_1,...,X_{p-1})=\sum(Y_i-b_0-b_1X_{1,i}-...-b_{p-1}X_{p-1,i})^2=\sum(Y_i-b_0-b_1X_{1,i}-...-b_{p-1}X_{p-1,i}-0\times X_{p,i})^2\geq\sum(Y_i-b_0'-b_1'X_{1,i}-...-b_{p-1}'X_{p-1,i}-b_p'X_{p,i})^2=SSE(X_1,...,X_{p-1},X_p)$

$SSR(X_p|X_1,...X_{p-1})=SSE(X_1,...,X_{p-1})-SSE(X_1,...,X_{p-1},X_p)\geq 0$

Proved.

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  • $\begingroup$ If this is for self-study or homework, your question should have the self-study tag. See stats.stackexchange.com/tags/self-study/info $\endgroup$ Apr 29, 2014 at 5:10
  • $\begingroup$ Thanks very much for your suggestion. While the text books generally do not talk much about that, it is a question of myself instead of from other sources. I don't know whether it is appropriate to add the self-study tag. @PatrickCoulombe $\endgroup$ Apr 29, 2014 at 5:20

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