From: <maj_at_stats.waikato.ac.nz>

Date: Sun, 9 Dec 2007 18:28:53 +1300 (NZDT)

> summary(apply(S,2,sum))

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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. Received on Sun 09 Dec 2007 - 05:36:54 GMT

Date: Sun, 9 Dec 2007 18:28:53 +1300 (NZDT)

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

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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. Received on Sun 09 Dec 2007 - 05:36:54 GMT

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