[R] Large determinant problem
Ravi Varadhan
rvaradhan at jhmi.edu
Mon Dec 10 16:08:32 CET 2007
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.
> Hmm, S'S is numerically singular. crossprod(S) would be a better way to
> compute it than crossprod(S,S) (it does use a different algorithm), but
> look at the singular values of S, which I suspect will show that S is
> numerically singular.
>
> Looks like the answer is 0.
>
>
> On Sun, 9 Dec 2007, maj at stats.waikato.ac.nz wrote:
>
>> I thought I would have another try at explaining my problem. I think
>> that
>> last time I may have buried it in irrelevant detail.
>>
>> This output should explain my dilemma:
>>
>>> dim(S)
>> [1] 1455 269
>>> summary(as.vector(S))
>> Min. 1st Qu. Median Mean 3rd Qu. Max.
>> -1.160e+04 0.000e+00 0.000e+00 -4.132e-08 0.000e+00 8.636e+03
>>> sum(as.vector(S)==0)/(1455*269)
>> [1] 0.8451794
>> # S is a large moderately sparse matrix with some large elements
>>> SS <- crossprod(S,S)
>>> (eigen(SS,only.values = TRUE)$values)[250:269]
>> [1] 9.264883e+04 5.819672e+04 5.695073e+04 1.948626e+04
>> 1.500891e+04
>> [6] 1.177034e+04 9.696327e+03 8.037049e+03 7.134058e+03
>> 1.316449e-07
>> [11] 9.077244e-08 6.417276e-08 5.046411e-08 1.998775e-08
>> -1.268081e-09
>> [16] -3.140881e-08 -4.478184e-08 -5.370730e-08 -8.507492e-08
>> -9.496699e-08
>> # S'S fails to be non-negative definite.
>>
>> I can't show you how to produce S easily but below I attempt at a
>> reproducible version of the problem:
>>
>>> set.seed(091207)
>>> X <- runif(1455*269,-1e4,1e4)
>>> p <- rbinom(1455*269,1,0.845)
>>> Y <- p*X
>>> dim(Y) <- c(1455,269)
>>> YY <- crossprod(Y,Y)
>>> (eigen(YY,only.values = TRUE)$values)[250:269]
>> [1] 17951634238 17928076223 17725528630 17647734206 17218470634
>> 16947982383
>> [7] 16728385887 16569501198 16498812174 16211312750 16127786747
>> 16006841514
>> [13] 15641955527 15472400630 15433931889 15083894866 14794357643
>> 14586969350
>> [19] 14297854542 13986819627
>> # No sign of negative eigenvalues; phenomenon must be due
>> # to special structure of S.
>> # S is a matrix of empirical parameter scores at an approximate
>> # mle for a model with 269 paramters fitted to 1455 observations.
>> # Thus, for example, its column sums are approximately zero:
>>> summary(apply(S,2,sum))
>> Min. 1st Qu. Median Mean 3rd Qu. Max.
>> -1.148e-03 -2.227e-04 -7.496e-06 -6.011e-05 7.967e-05 8.254e-04
>>
>> I'm starting to think that it may not be a good idea to attempt to
>> compute
>> large information matrices and their determinants!
>>
>> Murray Jorgensen
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> --
> Brian D. Ripley, ripley at stats.ox.ac.uk
> Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
> University of Oxford, Tel: +44 1865 272861 (self)
> 1 South Parks Road, +44 1865 272866 (PA)
> Oxford OX1 3TG, UK Fax: +44 1865 272595
>
>
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