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st: Hausman test


From   Christopher Baum <[email protected]>
To   [email protected]
Subject   st: Hausman test
Date   Sat, 24 Dec 2005 09:20:29 -0500

Marcello said

"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."

The original statement should read "when you form the expression V_r - V_e, the resulting matrix is p.d., implying that it has all positive principal minors." That also implies that all of the DIAGONAL elements of that matrix are strictly positive, but it does not restrict the off-diagonal elements from being negative -- it merely means they cannot be 'too large'. 

The original point regarding the sampling error of coeff. estimates should be thought of as "if the estimated variance of b_1 is larger in the "efficient" model than in the "robust" model, the corresponding element on the diagonal of V_r - V_e will be negative, and Stata will complain..."

Kit Baum, Boston College Economics
http://ideas.repec.org/e/pba1.html
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