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From |
Marcello Pagano <[email protected]> |

To |
[email protected] |

Subject |
Re: st: RE: Alternative to Hausman test for FE vs RE models |

Date |
Fri, 23 Dec 2005 06:54:39 -0500 |

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:

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.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

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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**Follow-Ups**:**RE: st: RE: Alternative to Hausman test for FE vs RE models***From:*Udaya Wagle <[email protected]>

**References**:**st: RE: Alternative to Hausman test for FE vs RE models***From:*"Maarten Buis" <[email protected]>

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