Talking about goodness-of-fit measures, my professor mentioned both centered and uncentered $R^2$ but I am not sure I have understood the difference between them, in terms of their practical use.

In formulas, he defined $R^2 = \frac{y'P_{[X]}M_{[1]}P_{[X]}y}{y'M_{[1]}y}$ and $R^2_u = \frac{y'P_{[X]}y}{y'y}$, where $P_{[X]}$ is the orthogonal projector of the matrix $X$ of regressors, $M_{[1]}$ is the "residual maker" for a matrix $X$ which contains the unity vector as a column; and $\boldsymbol{y}$ is the vector of dependent variables.

My question is about the main differences between these two measures of fit and about the cases in which I should use the one or the other.


I don't know much about econometric. But I think your question is a statistical one in essence. Consider an OLS model $$\boldsymbol{y}=\boldsymbol{X\beta}+\boldsymbol{\varepsilon}.$$ Let $V=\operatorname{col}(\boldsymbol{X})$, also take $V_0\subset V$ to be the "intercept subspace". $R^2$ can be defined as a ratio of two "sum of squares": $$R^2=\frac{\lVert\hat{\boldsymbol{y}}-\hat{\boldsymbol{y}}_0\rVert^2}{\lVert\boldsymbol{y}-\hat{\boldsymbol{y}}_0\rVert^2}.$$ Using projection matrices, which are idempotent and symmetric, this is equivalently saying: $$R^2=\frac{\lVert(\boldsymbol{P}_V-\boldsymbol{P}_{V_0})\boldsymbol{y}\rVert^2}{\lVert(\boldsymbol{I}-\boldsymbol{P}_{V_0})\boldsymbol{y}\rVert^2}=\frac{\lVert(\boldsymbol{I}-\boldsymbol{P}_{V_0})\boldsymbol{P}_V\boldsymbol{y}\rVert^2}{\lVert(\boldsymbol{I}-\boldsymbol{P}_{V_0})\boldsymbol{y}\rVert^2}=\frac{\boldsymbol{y}'\boldsymbol{P}_V(\boldsymbol{I}-\boldsymbol{P}_{V_0})\boldsymbol{P}_V\boldsymbol{y}}{\boldsymbol{y}'(\boldsymbol{I}-\boldsymbol{P}_{V_0})\boldsymbol{y}}.$$ In the usual OLS we take $\boldsymbol{X}=\begin{pmatrix}\boldsymbol{1}_n&\boldsymbol{x}\end{pmatrix}$ and $V_0=\operatorname{span}\{\boldsymbol{1}_n\}$. Then $$\boldsymbol{I}-\boldsymbol{P}_{V_0}=\boldsymbol{I}-\frac{1}{n}\boldsymbol{1}_n\boldsymbol{1}_n'=\boldsymbol{M}_1$$ (which implies the "residual maker" $\boldsymbol{M}_1$ is a projection matrix onto $V_0^\perp$). If we force the intercept term to be $0$ by choosing $\boldsymbol{X}=\begin{pmatrix}\boldsymbol{0}_n&\boldsymbol{x}\end{pmatrix}$ and $V_0=\operatorname{span}\{\boldsymbol{0}_n\}=\{\boldsymbol{0}_n\}$. Then $\boldsymbol{I}-\boldsymbol{P}_{V_0}=\boldsymbol{I}$, so $$R^2=\frac{\boldsymbol{y}'\boldsymbol{P}_V(\boldsymbol{I}-\boldsymbol{P}_{V_0})\boldsymbol{P}_V\boldsymbol{y}}{\boldsymbol{y}'(\boldsymbol{I}-\boldsymbol{P}_{V_0})\boldsymbol{y}}=\frac{\boldsymbol{y}'\boldsymbol{P}_V\boldsymbol{y}}{\boldsymbol{y}'\boldsymbol{y}}.$$ In short, the "centered" $R^2$ is the usual $R^2$, and the "uncentered" $R^2$ is the $R^2$ when the model does not contain an intercept term. The word "centered", I think, comes from the fact that $$\boldsymbol{P}_{V_0}\boldsymbol{y}=\frac{1}{n}\boldsymbol{1}_n\boldsymbol{1}_n'\boldsymbol{y}=\bar{y}\boldsymbol{1}_n.$$

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  • $\begingroup$ Thanks for your explanation. Thus, the only practical difference is that we cannot compute the centered $R^2$ when the intercept is not in the model and so we should use the uncentered one? $\endgroup$ – PhDing Feb 28 '16 at 10:41
  • $\begingroup$ @Alessandro: I guess you can say that. In general it's not a good practice to drop the intercept, see when is it ok to remove the intercept in lm()? $\endgroup$ – Francis Feb 29 '16 at 10:41

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