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If adjusted R squared is superior to R squared, then why do statistical software continue to report the latter? Is there any kind of situation when a researcher may prefer to use R squared instead of adjusted R squared?

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  • $\begingroup$ What kind of regression are you dealing with? If I am not mistaken, for linear regression, there is no difference between the R-squared and the adjusted R-squared. So in this case it is very appropriate to use the plain R-squared value. $\endgroup$
    – alesc
    Apr 22 '15 at 7:38
  • $\begingroup$ A linear one. But statistical packages provides both measures. That's why I wonder why. $\endgroup$
    – Mike Senin
    Apr 22 '15 at 7:42
  • $\begingroup$ Well, according to Wiki, the equation is a bit different even for linear regression (p=1). But the whole point of adjusted R-squared is "The use of an adjusted R2 is an attempt to take account of the phenomenon of the R2 automatically and spuriously increasing when extra explanatory variables are added to the model.". Linear regression doesn't have any additional explanatory variable, because it is the most primitive type of regression. $\endgroup$
    – alesc
    Apr 22 '15 at 7:51
  • $\begingroup$ @alesc, I know that. What I don't know is why to report both values. $\endgroup$
    – Mike Senin
    Apr 22 '15 at 7:52
  • $\begingroup$ What are you trying to prove with your R-squared value? Do you compare different regression models? If you compare linear and non-linear regression models, then it would make sense to use adjusted R-squared, otherwise the plain R-square will be sufficient. But then again, you can also use the adjusted R-squared even for linear regression :) I personally would not report both values. So choose one metric and report only that value (either R-square or adjusted R-square). $\endgroup$
    – alesc
    Apr 22 '15 at 7:56
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Under conditions for instance explained here, $R^2$ measures the proportion of the variance in the dependent variable explained by the regression, which is a natural measure. Adjusted $R^2$ does not have this interpretation, as it modifies the $R^2$ value.

So while adjusted $R^2$ has the indisputable advantage of not increasing automatically when the number of regressors goes up, you pay a price in terms of how you can interpret the measure.

Note I am not advocating the use of one or the other, just giving a possible reason for why people still use the standard $R^2$.

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    $\begingroup$ Quick question: is it perhaps true that $R^2_{adj.}$ is a consistent estimator of the population $R^2$ under some conditions, e.g. a well-specified model? Then it would make sense to report $R^2_{adj.}$ in place of $R^2$. $\endgroup$ Apr 22 '15 at 11:56
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    $\begingroup$ Yes, but as we can write $R_{adj.}^2=1-\frac{n-1}{n-K}+\frac{n-1}{n-K}R^2$ and, obviously, $\frac{n-1}{n-K}\to1$ (at least when, as is mostly assumed, $K$ remains fixed as $n\to\infty$), we have that $R_{adj.}^2-R^2=o_p(1)$, so that does not seem to be a reason to prefer one over the other. $\endgroup$ Apr 22 '15 at 12:02
  • $\begingroup$ $K$ is of course the number of regressors $\endgroup$ Apr 22 '15 at 12:03
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    $\begingroup$ Well...do we define population $R^2$ as $1-\sigma^2/Var(y)$? If so, writing $R^2_{adj.}=1-\frac{s^2}{\sum_i(y-\bar{y})^2/(n-1)}$ ($s^2$ the d.f.-adjusted variance estimate dividing by $n-K$) shows that both the estimator of the error variance in the numerator and that of the variance of $y$ in the denominator are unbiased for the respective population parameters, $E(s^2)=\sigma^2$ and $E[\sum_i(y-\bar{y})^2/(n-1)]=Var(y)$. But that does not make the ratio an unbiased estimator of the ratios of the parameters, as the expectation operator does not pass through nonlinear functions in general. $\endgroup$ Apr 22 '15 at 12:40
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    $\begingroup$ Thanks. Perhaps I should have posted my comments as a separate question, then I could have upvoted your answers. Since I suspected similar things have been asked, I just hoped for a short confirmation/disconfirmation, comment style. You were more explicit than that, I appreciate it! $\endgroup$ Apr 22 '15 at 13:16
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Adjusted R-squared is useful for comparing different regression models. This task cannot be accomplished by R-squared which, as Others have already said, has another informative goal, that is expressing the proportion of variance of the dependent variable that is explained by the regression model under investigation.

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