Falko <[email protected]>:
My own feeling is that the adjusted R2 is useless in all cases.
That said, you can get two different answers from two essentially
equivalent estimation commands by using -xtreg, fe i(id)- and then
-areg, a(id)- because they calculate R2 differently. -areg- acts as
though you included dummy variables to sweep out FE while -xtreg-
reports the "within" R2 as the R2 of the model, ignoring the dummy
variables version. They both act as though the dummy variables are
there when computing the adjusted R2, though, since that gives the
correct df.
Let e(r2) be the R-squared, and k=number of regressors, excluding the
constant, and g=#id. Then adj R2 is 1-(1-e(r2))*(200-1)/(200-k-g).
The difference arises in whether you want to treat the FE as
"explaining" variance in the outcome, e.g.
webuse grunfeld, clear
qui xtreg mval inv, fe
di e(r2),1-(1-e(r2))*(200-1)/(200-11),e(r2_a)
qui areg mval inv, a(com)
di e(r2),1-(1-e(r2))*(200-1)/(200-11),e(r2_a)
If there is a big difference between e(r2) and e(r2_a) using -xtreg-
you may have a low within R2 and a lot of FE, which is not necessarily
a problem. The F-test of the model is a better measure of explanatory
power, and is identical for -xtreg- and -areg-.
On 12/7/07, [email protected] <[email protected]> wrote:
> Dear Stata-Users,
>
> I've been searching the archive for quite a while now but somehow
> couldn't find an answer to my following question:
>
>
> How is the adjustment for the adjusted R-squared calculated when using
> xtreg, fe?
>
> The difference between the within R-squared (which I think is most
> interesting for me) and the adjusted R-squared is really high most of
> the time, so that my results seem pretty useless when taking the
> adjusted R-squared after a xtreg, fe regression.
>
> Can anybody help me? Or show me, how to prove that the adjusted
> R-squared is useless when doing xtreg, fe regressions? ;-)
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