Redundant fixed effects and degrees of freedom

[chungrak9]

Hello. I use EViews 8.

I have 40 cross-sections and 53 time periods. In the pool estimation, I include both cross-section and period fixed effects. Thus, I have 93 fixed effects in total. When I click the redundant fixed effects test, EViews shows three sets of tests: (1) Both effects vs. period effects; (2) Both effects vs. cross-section effects; and (3) Both effects vs. no fixed effects.

(1) For the first test of the cross-section effects, df=(40-1)=39.
(2) For the second test of the period effects, df=(53-1)=52.
(3) For the third test of both effects, df=(93-2)=91.

Would you please explain why the df for the third test is not 92 but 91? Also, would you please specify the null hypotheses of those three tests?

Thanks a lot!
Rakkoo

[EViews Glenn]

There are two adding up restrictions relative to the intercept only. One for the period effects, and one for the cross-section effects. Without the adding up restriction, you have the standard dummy variables trap.

The nulls are the statements after your "vs.". For example, for (1) the null is that there are period effects only.

[chungrak9]

Thank you for your answer. I'd like to have more specific information, because I cannot find such details in the manual. Let me explain what I think about the test. Would you please check any error in it?

In my pooled model, C(1) thru C(40) represent the cross-section fixed effects, and C(101) thru C(153) represent the period fixed effects. I suspect that the null hypotheses for the three redundancy tests are:

(1) H0: C(1)=C(2)=C(3)= ... =C(40)
(2) H0: C(101)=C(102)= ... =C(153)
(3) H0: C(1)=C(2)=C(3)= ... =C(40) AND C(101)=C(102)= ... =C(153)

For the redundant fixed effects test, df is the number of restrictions. So, the degrees of freedom for those tests are:

(1) df=(40-1)=39
(2) df=(53-1)=52
(3) df=(40-1)+(53-1)=91

Are these correct?

Thank you.
Rakkoo

[EViews Glenn]

There's an equal 0 at the end of each of your nulls, and you have to drop one of the coefficients from each block of coefficients because the sum adds up to zero. For example, in (1), H0 is C(2)=...=C(40)=0 since the sum of C(1) to C(40) is constrained to be zero.

[chungrak9]

Thanks a lot!