Converting Repeated Measures mixed model formula from SAS to R

Question

There are several questions and posts about mixed models for more complex experimental designs, so I thought this more simple model would help other beginners in this proces

Answer 1

Oooh, this is going to be a tricky one, and if it's even possible using standard nlme functions, is going to take some serious study of Pinheiro/Bates.

Before you spend the time doing that though, you should make absolutely sure that this is exact model you need. Perhaps there's something else that might fit the story of your data better. Or maybe there's something R can do more easily that is just as good, but not quite the same.

First, here's my take on what you're doing in SAS with this line:

repeated / type=cs subject=person group=GROUP;

This type=cs subject=person is inducing correlation between all the measurements on the same person, and that correlation is the same for all pairs of days. The group=GROUP is allowing the correlation for each group to be different.

In contrast, here's my take on what your R code is doing:

random = ~ +1 | person,
correlation=corCompSymm(form=~day|person)

This code is actually adding almost the same effect in two different ways; the random line is adding a random effect for each person, and the correlation line is inducing correlation between all the measurements on the same person. However, these two things are almost identical; if the correlation is positive, you get the exact same result by including either of them. I'm not sure what happens when you include both, but I do know that only one is necessary. Regardless, this code has the same correlation for all individuals, it's not allowing each group to have their own correlation.

To let each group have their own correlation, I think you have to build a more complicated correlation structure up out of two different pieces; I've never done this but I'm pretty sure I remember Pinheiro/Bates doing it.

You might consider instead adding a random effect for person and then letting the variance be different for the different groups with weights=varIdent(form=~1|group) (from memory, check my syntax, please). This won't quite be the same but tells a similar story. The story in SAS is that the measurements on some individuals are more correlated than the measurements on other individuals. Thinking about what that means, the measurements for individuals with higher correlation will be closer together than the measurements for individuals with lower correlation. In contrast, the story in R is that the variability of measurements within individuals varies; thinking about that, measurements with higher variability with have lower correlation. So they do tell similar stories, but come at it from opposite sides.

It is even possible (but I would be surprised) that these two models end up being different parameterizations of the same thing. My intuition is that the overall measurement variability will be different in some way. But even if they aren't the same thing, it would be worth writing out the parameterizations just to be sure you understand them and to make sure that they are appropriately describing the story of your data.

Answer 2

Please try below:

model1 <- lme(
  Y ~ GROUP + X1,
  random = ~ GROUP | person,
  correlation = corCompSymm(form = ~ day | person),
  na.action = na.exclude, data = df1, method = "REML"
)
summary(model1)

I think random = ~ groupvar | subjvar option with R lme provides similar result of repeated / subject = subjvar group = groupvar option with SAS/MIXED in this case.

Edit:

SAS/MIXED

R (a revised model2)

model2 <- lme(
  Y ~ GROUP + X1,
  random = list(person = pdDiag(form = ~ GROUP - 1)),
  correlation = corCompSymm(form = ~ day | person),
  weights = varIdent(form = ~ 1 | GROUP),
  na.action = na.exclude, data = df1, method = "REML"
)
summary(model2)

So, I think these covariance structures are very similar (σ_g1 = τ_g² + σ₁).

Edit 2:

Covariate estimates (SAS/MIXED):

Variance            person          GROUP TEST        8789.23
CS                  person          GROUP TEST         125.79
Variance            person          GROUP CONTROL       82775
CS                  person          GROUP CONTROL       33297

So

TEST group diagonal element
  = 125.79 + 8789.23
  = 8915.02
CONTROL group diagonal element
  = 33297 + 82775
  = 116072

where diagonal element = σ_k1 + σ_k².

Covariate estimates (R lme):

Random effects:
 Formula: ~GROUP - 1 | person
 Structure: Diagonal
        GROUP1TEST GROUP2CONTROL Residual
StdDev:   14.56864       184.692 93.28885

Correlation Structure: Compound symmetry
 Formula: ~day | person 
 Parameter estimate(s):
         Rho 
-0.009929987 
Variance function:
 Structure: Different standard deviations per stratum
 Formula: ~1 | GROUP 
 Parameter estimates:
   1TEST 2CONTROL 
1.000000 3.068837 

So

TEST group diagonal element
  = 14.56864^2 + (3.068837^0.5 * 93.28885 * -0.009929987) + 93.28885^2
  = 8913.432
CONTROL group diagonal element
  = 184.692^2  + (3.068837^0.5 * 93.28885 * -0.009929987) + (3.068837 * 93.28885)^2
  = 116070.5

where diagonal element = τ_g² + σ₁ + σ_g².