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RE: st: gradient and the inverse of the information matrix


From   Jun Xu <mystata@hotmail.com>
To   Listserv STATA <statalist@hsphsun2.harvard.edu>
Subject   RE: st: gradient and the inverse of the information matrix
Date   Wed, 1 May 2013 17:10:55 -0500

Yes, that's another direction that I could take, and I have a straightforward model setup and I just want to test coefficient equivalence. The lr, score, and Wald test should give asymptotically equivalent results. But the score test has some advantage in some aspect. Although I don't have to bother the substantive conclusion with the SAS output, I am interested in the Stata codes behind the operation. I appreciate your help!

Jun
 
----------------------------------------
> Date: Wed, 1 May 2013 23:46:27 +0200
> From: John.Antonakis@unil.ch
> To: statalist@hsphsun2.harvard.edu
> Subject: Re: st: gradient and the inverse of the information matrix
>
> Well, if your model is that straightforward just do a likelihood ratio
> test, by estimating the constrained model, then storing the estimates
>
> est store constrained
>
> Then estimate the unconstrained model and store the estimates:
>
> est store unconstrained
>
> Then do the lrtest:
>
> lrtest unconstrained constrained
>
> If you have not used a -robust vce- then you can use this test.
>
> HTH,
> J.
>
> __________________________________________
>
> John Antonakis
> Professor of Organizational Behavior
> Director, Ph.D. Program in Management
>
> Faculty of Business and Economics
> University of Lausanne
> Internef #618
> CH-1015 Lausanne-Dorigny
> Switzerland
> Tel ++41 (0)21 692-3438
> Fax ++41 (0)21 692-3305
> http://www.hec.unil.ch/people/jantonakis
>
> Associate Editor
> The Leadership Quarterly
> __________________________________________
>
> On 01.05.2013 23:40, Jun Xu wrote:
> > Dear Professor Antonakis,
> >
> > Thanks a lot for those useful leads. I am working on a single-equation categorical dependent variable model. I estimated the model with constraints imposed. Then if I understand the score test correctly, it would be a simple matrix operation of
> >
> > gradient * inv(information matrix) * gradient'
> >
> > where gradient = e(gradient) and inv(information matrix) = e(V)
> >
> > But the results do not match those from SAS, at least not to the point of having rounding errors. Either e(gradient) does not mean what it's named or the inverse of information matrix != e(V), I couldn't think of other possibilities.
> >
> > Jun
> >
> > ----------------------------------------
> >> Date: Wed, 1 May 2013 23:22:03 +0200
> >> From: John.Antonakis@unil.ch
> >> To: statalist@hsphsun2.harvard.edu
> >> Subject: Re: st: gradient and the inverse of the information matrix
> >>
> >> Hi:
> >>
> >> If you estimate your model with -sem- score tests are possible by using
> >> -estat mindices-; see also -estat scoretests.
> >>
> >> Also see the userwritten command -scoregrp- (available through ssc).
> >>
> >> Best,
> >> J.
> >>
> >> __________________________________________
> >>
> >> John Antonakis
> >> Professor of Organizational Behavior
> >> Director, Ph.D. Program in Management
> >>
> >> Faculty of Business and Economics
> >> University of Lausanne
> >> Internef #618
> >> CH-1015 Lausanne-Dorigny
> >> Switzerland
> >> Tel ++41 (0)21 692-3438
> >> Fax ++41 (0)21 692-3305
> >> http://www.hec.unil.ch/people/jantonakis
> >>
> >> Associate Editor
> >> The Leadership Quarterly
> >> __________________________________________
> >>
> >> On 01.05.2013 23:01, Jon Mu wrote:
> >>> Hi Statalisters,
> >>>
> >>> I am trying to check into the (Rao's) score (or commonly known as the Lagrange Multiplier) test for a model that I am working on. I got results from SAS already, and I want to see if those from SAS would square with the one produced from my own Stata codes.
> >>>
> >>> They don't match, and looks like I probably made some mistakes in my Stata codes. For the generalized formula to get the Chi-Square statistic, I need to get the gradient and the inverse of the information matrix. For the inverse of the information matrix, I can grab from e(V) directly without any further calculation.
> >>>
> >>> So I might've made some mistake in the gradient. I've searched through the voluminous Stata pdf documentation using gradient as the key word, and I was not able to find useful information. But I vaguely remember a while back ago when I was also checking into related issues, I read somewhere that the e(gradient) matrix is a gradient with respect to xb, not b, so I suspect that might be the cause. I am wondering if that's the case. If I am right on this, then a follow-up question is how to recover the gradient with respect to b since I feel there might not be a linear transformation that I can use to get it directly. Any input/suggestion would be appreciated.
> >>>
> >>> Jun Xu, PhD
> >>> Associate Professor
> >>> Department of Sociology
> >>> Ball State University
> >>> Muncie, IN 46037
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