[R] Where is the SD in output of glm with Gaussian distribution
Marc Girondot
m@rc_grt @end|ng |rom y@hoo@|r
Mon Dec 9 17:25:40 CET 2019
Le 09/12/2019 à 16:45, Bert Gunter a écrit :
> In addition, as John's included output shows, only 1 parameter, the
> intercept, is fit. As he also said, the sd is estimated from the
> residual deviance -- it is not a model parameter.
>
> Suggest you spend some time with a glm tutorial/text.
I tried ! But I miss this point. I understand now this point. Thanks a
lot... big progress for me.
But still I don't understand why AIC calculation uses 2 parameters if
the SD is estimated from the residual deviance.
> y=rnorm(100)
> gnul <- glm(y ~ 1)
> logLik(gnul)
'log Lik.' -136.4343 (df=2)
> AIC(gnul)
[1] 276.8687
> -2*logLik(gnul)+2*2
'log Lik.' 276.8687 (df=2)
This is not intuitive when to count SD as a parameter (in AIC) or not in
df.resuidual !
>
> Bert
>
> On Mon, Dec 9, 2019 at 7:17 AM Marc Girondot via R-help
> <r-help using r-project.org <mailto:r-help using r-project.org>> wrote:
>
> Let do a simple glm:
>
> > y=rnorm(100)
> > gnul <- glm(y ~ 1)
> > gnul$coefficients
> (Intercept)
> 0.1399966
>
> The logLik shows the fit of two parameters (DF=2) (intercept) and sd
>
> > logLik(gnul)
> 'log Lik.' -138.7902 (df=2)
>
> But where is the sd term in the glm object?
>
> If I do the same with optim, I can have its value
>
> > dnormx <- function(x, data) {1E9*-sum(dnorm(data, mean=x["mean"],
> sd=x["sd"], log = TRUE))}
> > parg <- c(mean=0, sd=1)
> > o0 <- optim(par = parg, fn=dnormx, data=y, method="BFGS")
> > o0$value/1E9
> [1] 138.7902
> > o0$par
> mean sd
>
> 0.1399966 0.9694405
>
> But I would like have the value in the glm.
>
> (and in the meantime, I don't understand why gnul$df.residual
> returned
> 99... for me it should be 98=100 - number of observations) -1 (for
> mean)
> - 1 (for sd); but it is statistical question... I have asked it in
> crossvalidated [no answer still] !)
>
> Thanks
>
> Marc
>
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