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

[<kirin_guess@yahoo.com.tw>]

The first version of the ado and help files can be downloaded from:
https://docs.google.com/file/d/0B984NoKuZv46ZDRBWlYzUkNBOFU/edit?usp=sharing
Suggestions and comments are welcome.


Chi-lin




-----Original Message----- 
From: Jon Mu
To Listserv STATA
Subject RE: st: gradient and the inverse of the information matrix
Date Fri, 3 May 2013 16:15:47 -0500

Chi-lin,

Along with suggestions from Maarten, your paper nicely demonstrates how to 
tweak Stata codes to conduct the LM test as well as user-written ado files 
for the same thing. Although there are a few papers floating around out 
there that provide some expository discussions, like the one by Buse (1982) 
and others, none of them has the empirical technicalities in your paper. By 
the way, are those two commands readily downloadable? Thanks a lot!
Jun





-----Original Message----- 
From: kirin_guess@yahoo.com.tw
Sent: Friday, May 03, 2013 1:23 PM
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: gradient and the inverse of the information matrix

I have a manuscript that is exactly related to your question. The article
also briefs some possible reasons why the score test conducted by SAS and
STATA can be different. (See the footnote 6.)
You can download the article from:
https://docs.google.com/file/d/0B984NoKuZv46akZPeGRKcTZHekU/edit?usp=sharing

Chi-lin Tsai





-----Original Message----- 
From: Jon Mu
Sent: Wednesday, May 01, 2013 10:01 PM
To: Listserv STATA
Subject: st: gradient and the inverse of the information matrix

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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