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I am having serious trouble understanding the results of my Fixed Effects panel regression. I am using two fixed effects (on year and regions) and I get a negative Adjusted R2 (i am using the plm package in R). Why? Please find a screenshot of my output below.

I think I am not prone to the usual mistake - all my variables are fairly time variant. Other than I could not find useful explanations for my problem.

Might it have something to do with the unbalancedness of the panel?

Any suggestions would be wonderful!

Screenshot of Output

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    $\begingroup$ Your explanatory variables do not have much explanatory power (as evidenced by the low standard $R^2$), and you have quite a few of them, so that the penalty for complex models as incorporated into the adjusted $R^2$ leads to a negative value. $\endgroup$ – Christoph Hanck Jan 9 '20 at 17:06
  • $\begingroup$ But the result is the same even if I just use one explanatory variable $\endgroup$ – BeSeLuFri Jan 9 '20 at 19:11
  • $\begingroup$ You may also want to show these results. $\endgroup$ – Christoph Hanck Jan 10 '20 at 5:28
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It is well-known that the relationship between $R^2$ and adjusted $R^2$ in a linear regression (and ultimately, a fixed-effects regression can also be seen as a linear regression, see e.g. Difference between fixed effects dummies and fixed effects estimator?) is (see e.g. Is $R^2_{adjusted}$ both unbiased and consistent under the alternative in simple regression?) $$ R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-k} $$ For a simple linear regression ($k=2$) as discussed in the comments to the original question we obtain $$ R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-2} $$ Hence, $R^2_{adjusted}<0$ in a simple linear regression if $$ R^2<1-\frac{n-2}{n-1} $$ or $$ R^2<\frac{n-1-(n-2)}{n-1}=\frac{1}{n-1} $$ Hence, adjusted $R^2$ is negative when the original $R^2$ is very small. In the general case, we obtain $R^2_{adjusted}<0$ if $$ R^2<\frac{k-1}{n-1} $$ Hence, a somewhat larger $R^2$ is possible to still obtain a negative adjusted $R^2$.

At the same time, $R^2_{adjusted}<0$ can be seen to mostly be a small-sample issue (relative to $k$, of course, as the OP's example nicely illustrates) in that the difference between $R^2$ and $R^2_{adjusted}$ vanishes as $n$ increases and that we always (provided there is a constant in the column space of the regressors) have $R^2\geq0$.

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    $\begingroup$ Thank you very much! Thus, the negative Adj. R2 is a result of the region and time dummies being included in k. E.g. $$ 1 - (1 - 0.080791) * ((2669 - 1) / (2669 - (13 + 311 + 8))) = -0.0494 $$ with n = 2669, R2=0.080791, k = 12 regressors+intercept+311 region dummies + 8 time dummies (with 312 regions and 9 years). Perhaps I should ask this as a new question: But, does it make any sense to include the time and region dummies in the calculation of Adjusted R2? Without it my Adjusted R2 would be $$ 1 - (1 - 0.080791) * ((2669 - 1)/(2669 - 13)) = 0.07663795 $$ $\endgroup$ – BeSeLuFri Jan 10 '20 at 9:03
  • $\begingroup$ Hank - marked your answer as correct (As it answered my question). If I have to ask a new question (on what to include in k in Adj R2) tell me. $\endgroup$ – BeSeLuFri Jan 10 '20 at 9:05
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    $\begingroup$ Yes, I definitely think these coefficients should be included in the calculation, as they are ultimately parameters that are fitted in the model and therefore consume degrees of freedom, even if they do not show up in the output (usually because they are not of key interest). As the first link I give for example explains, the region dummies amount to fitting a unit-specific intercept for each region, so 311 additional parameters. $\endgroup$ – Christoph Hanck Jan 10 '20 at 9:11

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