[R] help about how can R compute AIC?
Arnau Mir Torres
arnau.mir at uib.es
Tue Oct 14 18:37:42 CEST 2008
El 14/10/2008, a las 18:05, Martin Maechler escribió:
>>>>>> "AMT" == Arnau Mir Torres <arnau.mir at uib.es>
>>>>>> on Tue, 14 Oct 2008 17:13:01 +0200 writes:
>>>>>> "AMT" == Arnau Mir Torres <arnau.mir at uib.es>
>>>>>> on Tue, 14 Oct 2008 17:13:01 +0200 writes:
>
> AMT> Hello.
>
> AMT> I need to know how can R compute AIC when I study a
> regression model?
> AMT> For example, if I use these data:
> AMT> growth tannin
> AMT> 1 12 0
> AMT> 2 10 1
> AMT> 3 8 2
> AMT> 4 11 3
> AMT> 5 6 4
> AMT> 6 7 5
> AMT> 7 2 6
> AMT> 8 3 7
> AMT> 9 3 8
> AMT> and I do
> AMT> model <- lm (growth ~ tannin)
> AMT> AIC(model)
>
> AMT> R responses:
> AMT> 38.75990
>
> AMT> I know the following formula to compute AIC:
> AMT> AIC= -2*log-likelihood + 2*(p+1)
>
> AMT> In my example, it would be:
> AMT> AIC=-2*log-likelihood + 2*2
> AMT> but I don't know how R computes log-likelihood:
>
> AMT> logLik(model)
> AMT> 'log Lik.' -16.37995 (df=3)
>
> and so?
What is the formula to compute logLik? I don't know how to compute "by
hand" logLik(model) and obtain -16.37995.
Arnau.
>
>
> Hint: Your only problem is that your 'p' is wrongly off by one.
> 2nd Hint: sigma is a parameter, too
>
------------------------------------------------------------
Arnau Mir Torres
Edifici A. Turmeda
Campus UIB
Ctra. Valldemossa, km. 7,5
07122 Palma de Mca.
tel: (+34) 971172987
fax: (+34) 971173003
email: arnau.mir at uib.es
URL: http://dmi.uib.es/~arnau
------------------------------------------------------------
More information about the R-help
mailing list