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If I want to compare the goodness-of-fit of two regression models, with and without intercept, is it valid to compare the squared correlation coefficient between the fitted values and the data? Since the squared correlation would get back the $R^2$ for the model with intercept, it seems to make sense to compute the squared correlation for the model without intercept and use that to make comparison. I am not entirely sure whether this is legitimate. Am I missing anything? If this approach is not valid, what methods can I use to make such comparisons?

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  • $\begingroup$ I think this is a good idea. You just lose the interpretation of the $R^2$ as the "proportion of variability explained by the model". $\endgroup$
    – Stéphane Laurent
    Commented Jan 11, 2013 at 21:09
  • $\begingroup$ Sorry I misunderstood the question. I didn't see your goal is to compare the two models. As said by ttnphns below, you would get the same correlation for both models. $\endgroup$
    – Stéphane Laurent
    Commented Jan 12, 2013 at 10:47

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$R^2 = 1-SS_{residuals}/SS_{total}$, where "SS" is "sum of squares".

In the model with intercept included, $SS_{total}$ is the sum of squares about the dependent variable's mean. In the model with intercept suppressed, $SS_{total}$ is the sum of squares about 0, i.e. the sum of squares in the non-centered dependent variable. Therefore, one cannot directly compare the two R-squares.

Update. Well, now for the question. It sounds this: when there is intercept, R-square is equal to the squared correlation between the predicted values and the observed values; when there is no intercept, it is not: the mentioned squared correlation is another thing than R-square. Can we use this correlation in place of R-square to compare a model with intercept with a model without intercept?

If both models are the same (same set of IVs) except for the intercept, both correlations "predicted with observed" will be the same because in both cases the predicted values are just linear combinations of the same terms (only coefficients being different). The correlations are equal despite the fact that no-intercept model a priori fits worse than intercept model. So the answer to the question is no, we shouldn't use that correlation.

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    $\begingroup$ I am aware that $R^2$ for a regression model without intercept does not have the valid interpretation. That's why I am asking whether it's valid to compare squared correlation instead of $R^2$. $\endgroup$
    – user13587
    Commented Jan 11, 2013 at 20:27
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    $\begingroup$ Aren't the two squared correlation values equal? The predicted values being left by both models are linear combinations of the same terms except for a constant term which does not affect a correlation; the correlation between predicted values of the two models is thus 1. $\endgroup$
    – ttnphns
    Commented Jan 11, 2013 at 22:38
  • $\begingroup$ @ttnphns Right. I think you should include this comment in your answer. $\endgroup$
    – Stéphane Laurent
    Commented Jan 12, 2013 at 10:47
  • $\begingroup$ Thanks for pointing that out. What methods should I use to compare which model fits better? $\endgroup$
    – user13587
    Commented Jan 12, 2013 at 21:10