Is (great) enhancement of $R^2$ sufficient to justify the use of another model ? other desciptives variables ?

  • $\begingroup$ For an extreme example, set $\hat y_i = y_i$. Then you'll get $R^2 = 1$, but clearly this is a terrible model. $R^2_{adj}$ is better but still not perfect. See stats.stackexchange.com/questions/13314/…, for example $\endgroup$
    – jld
    Apr 12, 2016 at 15:16
  • $\begingroup$ You're getting lots of negative signals here, which are in part I think a reaction to your sufficient. But it's also true that of similar models (e.g. similar number of predictors) that with highest $R^2$ may well be the best. But there are no guarantees: look at graphs, look at residuals, consider whether the functional form matches the data, think about what makes scientific or practical sense. None of these is easy for beginners, or even very experienced people, but they are all more important than $R^2$. $\endgroup$
    – Nick Cox
    Apr 12, 2016 at 15:31
  • $\begingroup$ I excluded overfitting because of the model used (use of kernel, lot of 'overlapping data', this tend to give smooth surfaces that does not make detours to perfectly fit a point), this exclusion seems to be confirmed by plots, i have yet to look at residuals. The first model is quite simple and justified but the second is not. The use of the second model is justified ex post by its complexity (:/) and the enhancement of $R^2$. $\endgroup$ Apr 12, 2016 at 16:45

1 Answer 1


The short answer would be no. You could , per example, overfit your model and get a greater R squared. Try comparing your model by AIC (https://en.wikipedia.org/wiki/Akaike_information_criterion ).

  • $\begingroup$ Is there comparison between two models already fitted? If so, there must be an afdition or deletion of variables and or ... $\endgroup$
    – user10619
    Apr 12, 2016 at 15:20
  • $\begingroup$ As @subhashc.davar is pointing out, if you compare two models with the same dimension via AIC then you are just comparing the likelihoods. For iid normal errors linear regression, this is exactly equivalent to comparing $R^2$s. If the models have different dimension then there is value to using AIC over $R^2$. $\endgroup$
    – jld
    Apr 12, 2016 at 15:27
  • $\begingroup$ If the y variable is the same but he is using different x variable then AIC is more valuable to identify the most parcimonious model. $\endgroup$ Apr 12, 2016 at 15:53
  • $\begingroup$ What about the use of x,y for the first model and x,y,z for the second ? $\endgroup$ Apr 12, 2016 at 16:47
  • $\begingroup$ As long as you don't change the y variable you can compare the model $\endgroup$ Apr 12, 2016 at 16:59

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