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Large determinant problem

It is evident that you do not have enough information in the data to
estimate 9 mixture components.  This is clearly indicated by a positive
semi-definite information matrix, S, that is less than full rank.  You can
monitor the rank of the information matrix, as you increase the number of
components, and stop when you suspect rank-deficiency.

Ravi.


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Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvaradhan at jhmi.edu

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

 

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-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of maj at stats.waikato.ac.nz
Sent: Sunday, December 09, 2007 2:45 AM
To: Prof Brian Ripley
Cc: r-help at r-project.org
Subject: Re: [R] Large determinant problem

I tried crossprod(S) but the results were identical. The term
-0.5*log(det(S)) is  a complexity penalty meant to make it unattractive to
include too many components in a finite mixture model. This case was for a
9-component mixture. At least up to 6 components the determinant behaved
as expected and increased with the number of components.

Thanks for your comments.
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