[R] lme: null deviance, deviance due to the random effects, residual deviance

[Patrick Giraudoux]

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