I found my problem:
The following function gave llGp1, llGp2 and ll22:
logLik_lm <- function(object){
res <- resid(object)
n <- length(res)
s2MLE <- sum(res^2)/n
lglk <- (-n/2)*(log(2*pi*s2MLE)+1)
lglk
}
logLik(fitGp1)
logLik(fitGp1)-logLik_lm(fitGp1)
llGp1 - logLik_lm(fitGp1)
logLik(fitGp2)
logLik(fitGp2)-logLik_lm(fitGp2)
llGp2 - logLik_lm(fitGp2)
logLik(fit22)
logLik(fit22)-logLik_lm(fit22)
ll22 -logLik_lm(fit22)
These differences were all 0 to within roundoff error. That
confirmed for me that I could safely compare logLik.lm and logLik.lme.
What I thought should have been a linear operation wasn't. Please
excuse the waste of your time.
Thanks,
Spencer Graves
On 8/29/23 11:15 AM, Spencer Graves wrote:
Hello, all:
I have a dataset with 2 groups. I want to estimate 2 means and 2
standard deviations. I naively think I should be able to use lme to do
that, e.g., lme(y~gp, random=y~1|gp, method='ML'). I think I should get
the same answer as from lm(y~1, ...) within each level of group. I can
get the same means, but I don't know how to extract the within-gp
standard deviations, and the sum of logLik for the latter two does not
equal the former.
TOY EXAMPLE:
library(nlme)
set.seed(1)
lmePblm <- data.frame(y=c(rnorm(5, 1, 2), rnorm(5,3,5)),
gp=factor(rep(1:2, each=5)))
fit22 <- lme(y~gp, lmePblm, random=~1|gp, method='ML')
fitGp1 <- lm(y~1, lmePblm[lmePblm$gp==1, ])
fitGp2 <- lm(y~1, lmePblm[lmePblm$gp==2, ])
(ll22 <- logLik(fit22))
(llGp1 <- logLik(fitGp1))
(llGp2 <- logLik(fitGp2))
# Why isn't (ll22 = llGp1+llGp2)?
(ll22 - llGp1-llGp2)
# And secondarily, how can I get the residual standard deviations
# within each gp from fit22?
Thanks,
Spencer Graves
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