# [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
------------------------------------------------------------

```