Converting Repeated Measures mixed model formula from SAS to R - 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 process as well as I.
So, my question is I would like to formulate a repeated measures ancova in R from sas proc mixed procedure:
proc mixed data=df1;
FitStatistics=akaike
class GROUP person day;
model Y = GROUP X1 / solution alpha=.1 cl;
repeated / type=cs subject=person group=GROUP;
lsmeans GROUP;
run;
Here is the SAS output using the data created in R (below):
. Effect panel Estimate Error DF t Value Pr > |t| Alpha Lower Upper
Intercept -9.8693 251.04 7 -0.04 0.9697 0.1 -485.49 465.75
panel 1 -247.17 112.86 7 -2.19 0.0647 0.1 -460.99 -33.3510
panel 2 0 . . . . . . .
X1 20.4125 10.0228 7 2.04 0.0811 0.1 1.4235 39.4016
Below is how I formulated the model in R using 'nlme' package, but am not getting similar coefficient estimates:
## create reproducible example fake panel data set:
set.seed(94); subject.id = abs(round(rnorm(10)*10000,0))
set.seed(99); sds = rnorm(10,15,5);means = 1:10*runif(10,7,13);trends = runif(10,0.5,2.5)
this = NULL; set.seed(98)
for(i in 1:10) { this = c(this,rnorm(6, mean = means[i], sd = sds[i])*trends[i]*1:6)}
set.seed(97)
that = sort(rep(rnorm(10,mean = 20, sd = 3),6))
df1 = data.frame(day = rep(1:6,10), GROUP = c(rep('TEST',30),rep('CONTROL',30)),
Y = this,
X1 = that,
person = sort(rep(subject.id,6)))
## use package nlme
require(nlme)
## run repeated measures mixed model using compound symmetry covariance structure:
summary(lme(Y ~ GROUP + X1, random = ~ +1 | person,
correlation=corCompSymm(form=~day|person), na.action = na.exclude,
data = df1,method='REML'))
Now, the output from R, which I now realize is similar to the output from lm():
Value Std.Error DF t-value p-value
(Intercept) -626.1622 527.9890 50 -1.1859379 0.2413
GROUPTEST -101.3647 156.2940 7 -0.6485518 0.5373
X1 47.0919 22.6698 7 2.0772934 0.0764
I believe I'm close as to the specification, but not sure what piece I'm missing to make the results match (within reason..). Any help would be appreciated!
UPDATE: Using the code in the answer below, the R output becomes:
> summary(model2)
Scroll to bottom for the parameter estimates -- look! identical to SAS.
Linear mixed-effects model fit by REML
Data: df1
AIC BIC logLik
776.942 793.2864 -380.471
Random effects:
Formula: ~GROUP - 1 | person
Structure: Diagonal
GROUPCONTROL GROUPTEST Residual
StdDev: 184.692 14.56864 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:
TEST CONTROL
1.000000 3.068837
Fixed effects: Y ~ GROUP + X1
Value Std.Error DF t-value p-value
(Intercept) -9.8706 251.04678 50 -0.0393178 0.9688
GROUPTEST -247.1712 112.85945 7 -2.1900795 0.0647
X1 20.4126 10.02292 7 2.0365914 0.0811

Answer 1

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 (σg1 = τg2 + 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 = σk1 + σk2.
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 = τg2 + σ1 + σg2.

Answer 2

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.