Sorry that I forgot to include R Help in the addressee list. Here is one of
my earlier follow-up emails.


---------- Forwarded message ----------
From: Xiaogang Su <xiaogangsu@gmail.com>
Date: Wed, Feb 26, 2014 at 8:48 AM
Subject: Re: [R] Fitting glm with maxit=0?
To: Prof Brian Ripley <ripley@stats.ox.ac.uk>


Thanks Prof. Ripley for the reply. Certainly, the theorem is for MLE. But
MLE is not the only best asyptotical normal estimator.  In particular, i am
interested in a class of estimators that are close to MLE in an order
smaller than 1/sqrt(n). By Slutsky theorem, they also follow the same
asymptotic normal dist as MLE. In computing the information matrix, one
could use the estimate itself instead of plugging in MLE.

Anyhow, i think it would be a good feature to have with glm.

One last note, forcing at least one more iteration, as in the current
version of glm, is not wrong as that essentially gives a one step estimator.

=============================
Xiaogang Su, Ph.D.
Associate Professor
Department of Mathematical Sciences
University of Texas at El Paso
500 W. University Ave.
El Paso, Texas 79968-0514
xsu@utep.edu
xiaogangsu@gmail.com
https://sites.google.com/site/xgsu00/
On Feb 26, 2014 1:36 AM, "Prof Brian Ripley" <ripley@stats.ox.ac.uk> wrote:

> The theory used assumes that the estimates are MLEs (of the linear
> predictor).
>
> One could say that
>
> the variance-covariance matrix for any arbitrarily given estimates
>
> is zero: there is no variability.
>
> On 26/02/2014 08:24, Xiaogang Su wrote:
>
>> Dear All,
>>
>> Does anyone know if there is any way to obtain the variance-covariance
>> matrix for any arbitrarily given estimates with the glm() function?
>>
>> Here is what I really want. Given an arbitrary estimate (i.e., as starting
>> points with the start= argument), the glm() function could return the
>> corresponding variance-covariance matrix (or Hessian) and other quantities
>> with no Netwon-Raphson iteration? This could have been done by setting
>> maxit=0, but very unfortunately that is not an option in glm.control(). To
>> illustrate the problem,
>>
>> mydata <- read.csv("http://www.ats.ucla.edu/stat/data/binary.csv")
>> beta0 <- 1:3
>> control0 <- glm.control(epsilon = 1e10, maxit = 0, trace = FALSE)
>> fit <- glm(admit ~ gre + gpa, data = mydata, family = "binomial",
>> start=beta0, control=control0)
>> summary(fit)$"cov.scaled"
>>
>> By the way, I am aware that I could directly compute the covariance matrix
>> using the formula. I also know that I could extract the corresponding
>> deviance by using the offset option.
>>
>> Any suggestion is greatly appreicated.
>>
>> Thanks,
>> Xiaogang Su
>>
>> =============================
>> Xiaogang Su, Ph.D.
>> Associate Professor
>> Department of Mathematical Sciences
>> University of Texas at El Paso
>> 500 W. University Ave.
>> El Paso, Texas 79968-0514
>> xsu@utep.edu
>> xiaogangsu@gmail.com
>> https://sites.google.com/site/xgsu00/
>>
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>>
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>> PLEASE do read the posting guide http://www.R-project.org/
>> posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
> --
> Brian D. Ripley,                  ripley@stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>

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