[R-sig-ME] Can AIC be approximated by -2ln(L) i.e. the deviance, at very large sample size?

[Steve Taylor]

I agree that it is changes in AIC that matter, not its absolute value.  

My understanding is that AIC is only useful for comparing two models fitted on the same data set, i.e. with the same sample size.  So the question of how AIC changes with sample size is of little use beyond curiosity.  

The change in AIC caused by adding a term to the model formula would be of interest.  But the change in AIC caused by adding cases to the sample size is pretty meaningless.

The 2K part is important because it provides a penalty for the change in the number of parameters between a simpler model and a more complex model.

I would advise against making any approximations when calculating AIC, especially considering its main use is in taking the difference between two close large numbers.

cheers,
    Steve


-----Original Message-----
From: r-sig-mixed-models-bounces at r-project.org [mailto:r-sig-mixed-models-bounces at r-project.org] On Behalf Of Emmanuel Curis
Sent: Friday, 1 March 2013 9:18p
To: Chris Howden
Cc: r-sig-mixed-models
Subject: Re: [R-sig-ME] Can AIC be approximated by -2ln(L) i.e. the deviance, at very large sample size?

Hi,

I may be wrong, but I understood that AIC in itself is not as
important as changes in AIC between models, and some authors says that
changes in AIC in the order of more than 10 are enough to favor a
model on another.

And changes in the 2*k term should be in this order of magnitude when
comparing different models.

So my guess would be that it remains important.

On the other hand, if a set of parameters will remain in all models,
it probably can be safely ignored in the 2*k term for all models.

Hope this helps,

On Fri, Mar 01, 2013 at 06:30:53PM +1100, Chris Howden wrote:
< Hi everyone,
< 
< Although not strictly an R issue there often seems to be discussions along
< these lines on this list, so I hope no one minds me posting this. If U do
< please let me know. (and just for the record I am applying this in R)
< 
< I'm trying to get my head around AIC and sample size.
< 
< Now if AIC = -2ln(L) + 2K = Deviance + 2K
< 
< Am I right in thinking that as the Likelihood is the product of
< probabilities then (all else being equal) the larger the sample size the
< smaller the Likelihood?
< Which means that if we have very large sample sizes we expect the -2ln(L)
< term to be a very large number?
< Which would reduce the effect of the parameter correction term 2K?
< 
< 
< Chris Howden B.Sc. (Hons) GStat.
< Founding Partner
< Evidence Based Strategic Development, IP Commercialisation and Innovation,
< Data Analysis, Modelling and Training
< (mobile) 0410 689 945
< (fax) +612 4782 9023
< chris at trickysolutions.com.au
< 
< _______________________________________________
< R-sig-mixed-models at r-project.org mailing list
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-- 
                                Emmanuel CURIS
                                emmanuel.curis at parisdescartes.fr

Page WWW: http://emmanuel.curis.online.fr/index.html

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