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
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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 tr
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-hel
summary(gnul)
shows the std error of the coefficient estimate
On Mon, Dec 9, 2019 at 5:16 PM Marc Girondot via R-help
wrote:
>
> Let do a simple glm:
>
> > y=rnorm(100)
> > gnul <- glm(y ~ 1)
> > gnul$coefficients
> (Intercept)
>0.1399966
>
> The logLik shows the fit of two parameters (DF
Dear Marc,
For your simple model, the standard deviation of y is the square-root of the
estimated dispersion parameter:
> set.seed(123)
> y <- rnorm(100)
> gnul <- glm(y ~ 1)
> summary(gnul)
Call:
glm(formula = y ~ 1)
Deviance Residuals:
Min1QMedian3Q Max
-2.3
Let do a simple glm:
> y=rnorm(100)
> gnul <- glm(y ~ 1)
> gnul$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 h
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