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No estimated scale value given

2 messages · Ben Bolker, Douglas Bates

#
The not-necessarily-obvious thing is that none of the
parameter estimates change when you go from poisson,
binomial, etc. to their quasi- equivalents -- the variance
on all points is inflated by the same amount, so the
point estimates don't change.  So you can go ahead and
use the quasi- variant to get the estimated scale.

  I'm still struggling with exactly what "sigma" is in
the quasi- case (as is Doug Bates) -- it seems
NOT equal to the Pearson residuals^2/(resid df) ?

  It would be worth exploring this some more ...

  cheers
   Ben Bolker

set.seed(1001)
ntot <- 1000
x <- runif(ntot)
f <- rep(1:20,each=50)
reff <- rnorm(20,mean=0,sd=0.5)
y <- rnbinom(ntot,mu=exp(2*x-1+reff[f]),size=1)
library(lme4)

m1 <- glmer(y~x+(1|f),family=poisson)
m2 <- update(m1,family=quasipoisson)

lme4:::sigma(m1) ## 1
lme4:::sigma(m2) ## 1.04 (??)

## residuals returned are Pearson residuals
##  (uncorrected for overdispersion)

myres <- (y-fitted(m1))/sqrt(fitted(m1))
all.equal(myres,residuals(m1))
## not quite identical, but nearly
all(abs(myres-residuals(m1))<1e-4)

all.equal(residuals(m1),residuals(m2)) ## TRUE

sum(residuals(m1)^2)/ntot ## NOT sigmaML
m1 at deviance["sigmaML"]
m2 at deviance["sigmaML"]
Amy Wade wrote:

  
    
1 day later
#
On Tue, Feb 17, 2009 at 2:23 PM, Ben Bolker <bolker at ufl.edu> wrote:
The first one is 1 by definition.  The second is the penalized
weighted residual sum of squares at the parameter estimates divided by
the number of observations.  You did not include the penalty term in
your calculations.

As we have discussed, it is not clear if this is a meaningful quantity
in the quasi-Poisson case.