# Is there any way that adjusted R squared would be greater than R squared?

Is there any way that adjusted $R^2$ would be greater than $R^2$? Including cases of extreme values of n and p and negative values of $R^2$.

The formula is $$R^2_{adj} = 1 - \frac{(N-1)}{N-p-1}(1-R^2)$$ where N = sample size, p = number of predictors, and $$R^2$$ is, well, $$R^2$$. So at best with an enormous number of samples and a small number of predictors, it can approach the original $$R^2$$ as $$\frac{(N-1)}{N-p-1}$$ approaches 1.
• The argument is correct--but it's not quite general enough to address the situation implied by the question. The suggestion that $R^2$ could be negative hints that a regression through the origin is contemplated, in which case the correct formula for adjusted $R^2$ is slightly different than the one given here. The conclusion remains true--in fact, the adjustment is even greater in that case--but it might be worth pointing out that the absolute value of the adjusted $R^2$ easily could exceed that of $R^2$ (fwiw). In the most extreme cases $R^2=0$ whereas the adjusted value is $-1$.