I appreciate the replies. MP's observation on the differential
variance/covariance matrices supports my own observation too, unless I am
wrong in simply subtracting the more efficient variance/covariances from the
less efficient ones. My interest, however, is in figuring out an alternative
way of testing the two models. Even though Stata reports the differential
variance/covariance matrix to be not positive definite, it still yields chi
square statistics and p-values that are significant. Is it simply
disregarding the not positive defining part and then conducting the test and
reporting results? Can I assume this test result to be valid?
Also, if R-squareds are any guide, results indicate that a substantial part
of between effects is captured in the RE model. To me this is helpful to
discern that IDs matter in explaining the variation perhaps more so than
time. Am I correct?
Udaya
-----Original Message-----
From: [email protected]
[mailto:[email protected]]On Behalf Of Marcello Pagano
Sent: Friday, December 23, 2005 6:55 AM
To: [email protected]
Subject: Re: st: RE: Alternative to Hausman test for FE vs RE models
This may be a slow time of year, and I may know nothing about a Hausman
test, but there are some wild statements being made here that sound odd,
and one that are patently false, viz. "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)." The parenthetical remark is false, as a 2x2
matrix with ones on diagonal and twos off-diagonal shows.
"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)." Once again, the parenthetical statement is false,
as the 2x2 matrix with twos on the diagonal and minus ones off-diagonal
proves.
Ordering positive definite matrices can be done, but it ordinarily does
not translate to a simple statement about the signs of the elements of
the matrix obtained as the result of differencing the two matrices.
m.p.
Maarten Buis wrote:
>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 covariance
s 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
>
>-----------------------------------------
>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/
>-----------------------------------------
>
>
>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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>
>
*
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*
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