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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.

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2 Answers 2

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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, 2016 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, 2016 at 10:41
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The above answer is not correct based on my experience with econometrics. I hope this adds some additional flavor and intuition to the post linked in the first comment of the OP's question above.

Centered R2 is the usual measure and it effectively assesses the improvement in accuracy that your linear model (with a constant/intercept or not) has over just using the mean. If the model is worse than the mean, R2 is negative (this can't happen with a regression that includes a constant/intercept term). Centered R2 is the same as Nash Sutcliffe Efficiency for y and yhat.

Uncentered R2 is uncommon and just tells you how much of y (rather than variation in y about it's mean) has been explained.

Uncentered R2 is a measure that gives a trophy to the loser for participation, which in this case is explaining the non-varying part of y. Centered R2 gives no points for explaining a non-varying quantity and the score starts at 0 when accuracy is equivalent to the mean.

Following [https://stats.stackexchange.com/a/26205/297006], it seems disingenious that R would provide uncentered R2 when the regression lacks a mean. That positive uncentered R2 value very well may be for predictions (yhat) that are worse than the mean (but there's a trophy for you regardless).

I would say never (for econometric analysis, machine learning, and other standard statistical applications) use uncentered R2, but if you do, make sure you don't compare it to centered R2 and assume you have a better fit because the score is higher.

if you center y by subtracting it's mean before the regression and then exclude an intercept term from your regression, then centered R2 and uncentered R2 are identical.

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