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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2 Answers
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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 $R^2$.
answered Apr 22, 2015 at 9:58
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1$\begingroup$ 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 $R^2$. $\endgroup$ Commented Apr 22, 2015 at 11:56
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3$\begingroup$ 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. $\endgroup$ Commented Apr 22, 2015 at 12:02
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$\begingroup$ $K$ is of course the number of regressors $\endgroup$ Commented Apr 22, 2015 at 12:03
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1$\begingroup$ 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. $\endgroup$ Commented Apr 22, 2015 at 12:40
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1$\begingroup$ 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! $\endgroup$ Commented Apr 22, 2015 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.
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. $\endgroup$