Large determinant problem
On Sun, 9 Dec 2007, maj at stats.waikato.ac.nz wrote:
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.
And the singular values were? I am not surprised at this: if you have too many components some of them may not be contributing to the fit or duplicating others: both lead to numerically singular information matrices. In many mixture-fitting problems the log-likelihood is unbounded but with many local maxima: it is very easy to find a poor one.
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
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-- 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
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