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
] On
Behalf Of [EMAIL PROTECTED]
Sent: Sunday, December 09, 2007 2:45 AM
To: Prof Brian Ripley
Cc: r-help@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
So S is numerically singular: the last 10 or 11 values are effectively 0.
As Peter Dalgaard said, scaling may help but only if the parameters are on
drastically different scales.
On Sun, 9 Dec 2007, [EMAIL PROTECTED] wrote:
> What you say about mixture models is true in general, however this fit
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
Prof Brian Ripley wrote:
> 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
On Sun, 9 Dec 2007, [EMAIL PROTECTED] 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
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
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, [EMAIL P
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
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