Testing Panel Data Regression Models with Spatial Error Correlation



2.5 LR Test for Ha: λ = ¾2 = 0

We also compute the Likelihood ratio (LR) test for Ha: A = ¾2l = 0.  Estimation of the

unrestricted log-likelihood function is obtained using the method of scoring. The details of
the estimation procedure are available upon request from the authors. Let
A°, A, b and b
denote the unrestricted maximum likelihood estimators and let Bb = INAW and u = yX0b,
then the unrestricted maximum log-likelihood estimator function is given by

NT         1                                    1

Lu =       ln2¼¾° ~n InljTAIN + (BA0IA)-1 j] + (T — 1) ln B —   .l ti§-1^;    (2.21)

2            2                                        2¾°

see Anselin (1988), where §b is obtained from (2.8) with Bb replacing B and Ab replacing A: But
under the null hypothesis
Ha, the variance-covariance matrix reduces to U = u = ¾°IτN
and the restricted maximum likelihood estimator of β is ~~ols , so that ~ = y X,∕~ols
are the OLS residuals and ¾° = U0U=ΝT. Therefore, the restricted maximum log-likelihood
function under
Haa is given by

NT        1

Lr = —in ln 2¼¾° — ..2 u ~∙                                               (2.22)

2            2¾°

Hence, the likelihood ratio test statistic for Haa : A = ¾2p = 0 is given by

LRJ = 2(LuLr);                                                            (2.23)

and this should be asymptotically distributed as a mixture of Â2 given in (2.20) under the
null hypothesis.

2.6 Conditional LM Test for Had: λ = 0 (assuming ¾2 > 0)

When one uses LM2; given by (2.16), to test Hac : A = 0; one implicitly assumes that the
random region effects do not exist. This may lead to incorrect decisions especially when
¾2
is large. To overcome this problem, this section derives a conditional LM test for spatially
uncorrelated disturbances assuming the possible existence of random regional effects. The
null hypothesis for this model is
Had : A = 0 (assuming ¾2p > 0). Under the null hyphothesis,
the variance-covariance matrix reduces to
a = ¾2pJT IN + ¾2°INT . It is the familiar form
of the one-way error component model, see Baltagi(1995), with
-1 = (¾2)-1(Jτ IN) +
(
¾°)-1(Et   IN), where ¾2 = iμ + ¾°; and Et = ItJt. Using derivations analogous

to those for the joint LM-test, see Appendix A.1, we obtain the following LM test for Had vs
H1d,



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