Re: [R] Large determinant problem

Ravi Varadhan wrote: > 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 i

Re: [R] Large determinant problem

What you say about mixture models is true in general, however this fit was the best of 100 random EM starts. Unbounded likelihoods I believe are only a problem for continuous data mixture models and mine was discrete. Anyway it's nearly midnight now here so I'd better sleep. Before I go, here are t

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 an

[R] Large determinant problem

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 Mean3rd Qu. Max. -1.160e+04 0.000e

[R] Calculating large determinants

I apologise for not including a reproducible example with this query but I hope that I can make things clear without one. I am fitting some finite mixture models to data. Each mixture component has p parameters (p=29 in my application) and there are q components to the mixture. The number of data

[R] Dimension of a vector

Consider the following: > A <- 1:10 > A [1] 1 2 3 4 5 6 7 8 9 10 > dim(A) NULL > dim(A) <- c(2,5) > A [,1] [,2] [,3] [,4] [,5] [1,]13579 [2,]2468 10 > dim(A) [1] 2 5 > dim(A) <- 10 > A [1] 1 2 3 4 5 6 7 8 9 10 > dim(A) [1] 10 Would it

Re: [R] Re-ordering factors

Thanks to James and Phil and Peter for their helpful suggestions. I think that I should also point out one way *not* to do the job: > xtabs(Count ~ Education + Age_Group, data=educ) Age_Group Education>64 25-34 35-44 45-54 55-64 CompletedHS 7558 16431 1855 9435 8795

[R] Re-ordering factors

A small example before I begin my query: > educ <- read.table(efile, header=TRUE) > educ Education Age_Group Count 1 IncompleteHS 25-34 5416 2 IncompleteHS 35-44 5030 3 IncompleteHS 45-54 5777 4 IncompleteHS 55-64 7606 5 IncompleteHS >64 13746 6 CompletedHS