Dear Udaya Wagle,
Lets think about what a Hausman test does: It compares estimates from a more robust but less efficient model with estimates from a less robust but more efficient model. The last model is only more efficient if its model assumptions are met. Efficient means that the standard errors and confidence intervals are smaller (and this is true for the entire variance covariance matrix). So we are trying to choose between a model that has smaller standard errors but strict assumptions, and a model with larger standard errors and less strict assumptions. If the model assumptions of the efficient model are met, all elements of the variance covariance matrix of the robust model will be larger than the elements of the variance covariance matrix of the efficient model. Subtracting the efficient variance covariance matrix from the robust variance covariance matrix should than yield a matrix containing only positive entries (i.e. a positive definite matrix). If one of variances or covariances
from the robust model is smaller than the efficient model, than subtracting the two matrices will yield a matrix with at least one entry that is negative (a matrix that is not positive definite). The advantage from using an efficient model (smaller standard errors) is thus not completely realized. So this would suggest you should go for the robust model.
However, all elements of the variance covariance matrix of the efficient model are only assured to be smaller than the elements from the robust model in "the wonderful kingdom of Asymptotia" (i.e. if you have an infinitely large sample). If you have a smaller sample than a standard error or covariance from the robust model could be smaller just by sampling variance. This is not uncommon since the number of elements in a variance covariance can get very large.
So the fact that you find a non-positive definite matrix means that the efficient model (RE) is not more efficient than your robust model (FE). This can either be the result of not meeting the model assumptions of the efficient model, or because the benefit of using an efficient model in terms of efficiency is not large enough to outrun random sampling error. This last point may be a slightly too harsh if you have many variables (i.e. a large variance covariance matrix).
HTH,
Maarten
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Maarten L. Buis
Department of Social Research Methodology
Vrije Universiteit Amsterdam
Boelelaan 1081
1081 HV Amsterdam
The Netherlands
visiting adress:
Buitenveldertselaan 3 (Metropolitan), room Z214
+31 20 5986715
http://home.fsw.vu.nl/m.buis/
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At vrijdag 23 december 2005 4:16 Udaya Wagle wrote:
> I am trying to figure out whether FE or RE is more appropriate
> for my model that I am estimating with panel data. Obviously,
> the regular 'hausman test' reports, "V_b-V_B is not positive
> definite." Even the 'sigmaless' and 'sigmamore' options that
> are supposed to minimize the likelihood of getting non-positive
> differences have failed to resolve the problem.
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