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

Chris Howden chris at trickysolutions.com.au
Fri Mar 1 08:30:53 CET 2013


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



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