[R] Denominator Degrees of Freedom in lme() -- Adjusting and Understanding Them

Ken Kelley KKelley at nd.edu
Tue Oct 21 22:42:11 CEST 2003

Hello all.

I was wondering if there is any way to adjust the denominator degrees of 
freedom in lme(). It seems to me that there is only one method that can be 
used. As has been pointed out previously on the list, the denominator 
degrees of freedom given by lme() do not match those given by SAS Proc 
Mixed or HLM5. Proc Mixed, for example, offers five different options for 
computing the denominator degrees of freedom. Is there anyway to make such 
specifications in lme(), so that the degrees of freedom will correspond 
with the output given from Proc Mixed.

I've looked at Pinheiro and Bates' Mixed-Effects Models book (especially p. 
91), but I still don't quite understand the method used for determining the 
degrees of freedom in lme(). When analyzing longitudinal data with the 
straight-line growth model (intercept and slope both have fixed and random 
effects), the degrees of freedom seem to be N*T-N-1, where N is total 
sample size and T is the number of timepoints (at least when data are 
balanced). In the Pinheiro and Bates book (p. 91), the degrees of freedom 
are given as m_i-(m_1-1+pi), where m_i is the number of groups at the ith 
level, m_0=1 if an intercept is included and p_i is the sum of the degrees 
of freedom corresponding to the terms estimated. I'm not sure how the 
N*T-N-1 matches up with the formula given on page 91. It seems to me the 
number of "groups" (i.e., m_i) would be equal to N, the number of 
individuals (note that this is what is given as the "number of groups" in 
the summary of the lme() object.). However, as more occasions of 
measurements are added, the number of degrees of freedom gets larger, 
making it seems as though m_i represents the total number of observations, 
not the "number of groups."

For example, if N=2 and T=3, you end up with 3 degrees of freedom using 
lme(). Increasing T to 10 has not changed the number of groups (i.e., N 
still equals 2), but the degrees of freedom increases to 17. In such a 
situation SAS Proc Mixed would still have 1 degree of freedom (N-1) 
regardless of T, as the number of "groups" have not changed (just the 
number of observations per group have changed).

Any insight into understanding the denominator degrees of freedom for the 
fixed effects would be appreciated. Since the degrees of freedom given by 
lme() can be made to be arbitrarily larger than those given by PROC MIXED 
(i.e., by having an arbitrarily large number of measurement occasions for 
each individual), and since the degrees of freedom affect the standard 
errors, then the hypothesis tests, then the p values, the differences 
between the methods is surprising. It seems one of the methods would be 
better than the other since they can potentially be so much different.

Thanks and have a good one,
P.S. I have posted this to both the R and Multilevel Modeling list.

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