Dear Bert,
It's perhaps a bit pedantic to point it out, but the dispersion is estimated
from the Pearson statistic (sum of squared residuals or weighted squared
residuals) not from the residual deviance. You can see this in the code for
summary.glm().
Best,
John
-----------------------------
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
Web: http::/socserv.mcmaster.ca/jfox
> On Dec 9, 2019, at 10:45 AM, Bert Gunter <bgunter.4...@gmail.com> wrote:
>
> 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.
>
> Bert
>
> On Mon, Dec 9, 2019 at 7:17 AM Marc Girondot via R-help <
> r-help@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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>>
>
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>
> ______________________________________________
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