[R] lme: null deviance, deviance due to the random effects, residual deviance
Patrick Giraudoux
patrick.giraudoux at univ-fcomte.fr
Sun Apr 23 18:08:26 CEST 2006
A maybe trivial and stupid question:
In the case of a lm or glm fit, it is quite informative (to me) to have
a look to the null deviance and the residual deviance of a model. This
is generally provided in the print method or the summary, eg:
Null Deviance: 658.8
Residual Deviance: 507.3
and (a bit simpled minded) I like to think that the proportion of
deviance 'explained' by the model is (658.8-507.3)/658.8 = 23%
In the case of lme models, is it possible and reasonable to try and get the:
- null deviance
- the total deviance due to the the random effect(s)
- the residual deviance?
With the idea that Null deviance = Fixed effects + Random Effects +
Residuals
If yes how to do it ? A lme object provides the following:
> names(glm6)
[1] "modelStruct" "dims" "contrasts" "coefficients"
[5] "varFix" "sigma" "apVar" "logLik"
[9] "numIter" "groups" "call" "method"
[13] "fitted" "residuals" "fixDF" "family"
so no $null.deviance and $deviance elements as in glm objects...
I tried to find out an answer on R-help & Pineihro & Bates (2000).
Partial success only:
- null deviance: Response: possibly yes: see
http://tolstoy.newcastle.edu.au/R/help/05/12/17796.html (Spencer
Graves). The (null?) deviance is -2*logLik(mylme), but a personnal trial
with some glm objects did not led to the same numbers that the one given
by the print.glm method...
- the deviance due to the the random effect(s). I was supposing that the
coefficients given by ranef(mylme) may be an entry... but beyond this, I
guess those coefficients must be weighed in some way... which is a far
beyond my capacities in this matter...
- residual deviance. I was supposing that it may be
sum(residuals(mylme)^2). With some doubts as far as I feel that I am
thinking sum of squares estimation in the context of likelihood and
deviance estimations... So most likely irrelevant. Moreover, in the
case I was exploring, this quantity is much larger than the null
deviance computed as above...
Any hint appreciated,
Patrick Giraudoux
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