tag:blogger.com,1999:blog-2198942534740642384.post7191442627325761942..comments2023-10-24T03:16:41.009-07:00Comments on Econometrics Beat: Dave Giles' Blog: An Exercise With the SURE ModelDave Gileshttp://www.blogger.com/profile/05389606956062019445noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2198942534740642384.post-22555313486305073192014-04-19T09:17:08.441-07:002014-04-19T09:17:08.441-07:00A+ (squared!)A+ (squared!)Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-18375743439717564852014-04-19T09:14:29.798-07:002014-04-19T09:14:29.798-07:00I have two ways to demonstrate this result:
A. Kr...I have two ways to demonstrate this result:<br />A. Kruskal's condition says that GLS=OLS if I can find a matrix G such that WX=XG, where W is the covariance matrix of the disturbances (which I'll denote S@I(n,n) since I can't figure out how to write Sigma or the Kronecker product in this comment). The condition says X2=X1*C (just post multiply the left and rhs of the condition by X2). Therefore we can write X=diag(X1,X2)=diag(X1,X1)A =(I(2,2)@X1)A where A'=(I(n,n) C'). Then use the property of Kronecker products to show that (S@I(n,n))*(I(2,2)@X1)=(I(2,2)@X1)@(S@I(n,n)), so G=(S@I(n,n))A. If you don't want to use Kruskal's theorem, you can follow the reasoning above and add a few lines to show that GLS is just OLS on each equation.<br /><br />B. Subract sigma(1,2)/sigma(1,1) times the first equation from the second. The transformed error u2-(sigma(1,2)/sigma(1,1))*u1 is uncorrelated with u1. So GLS amounts to minimizing a weighted sum of the squared errors from the first equation and the second equation. Given the GLS estimate b1, the n.s. condition for b2 is that it satisfies<br />X2'(y2-X2*b2-(sigma(1,2)/sigma(1,1))(y1-X1*b1))=0. <br />By symmetry, we also see that the n.s. condition for b1 given b2 is that it satisfies<br />X1'(y1-X1*b1-(sigma(1,2)/sigma(2,2))(y2-X2*b2))=0. <br /><br />The ols estimators satisfy<br />X2'(y2-X2*b2ols)=0 and X1'(y1-X1*b1ols)=0<br /><br />Your condition says that a vector is orthogonal to span(X1) iff it is orthogonal to span(X2). So we also get<br />X1'(y2-X2*b2ols)=0. and X2'(y1-X1*b1ols)=0<br /><br />From this we quickly see that under your condition, the OLS estimators satisfies the n.s. condition for the GLS estimators, so they'll be the same.Anonymousnoreply@blogger.com