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Note : this question is a repost, as my previous question had to be deleted for legal reasons.


While comparing PROC MIXED from SAS with the function lme from the nlme package in R, I stumbled upon some rather confusing differences. More specifically, the degrees of freedom in the different tests differ between PROC MIXED and lme, and I wondered why.

Start from the following dataset (R code given below) :

  • ind : factor indicating the individual where the measurement is taken
  • fac : organ where measurement is taken
  • trt : factor indicating the treatment
  • y : some continuous response variable

The idea is to build the following simple models :

y ~ trt + (ind) : ind as a random factor y ~ trt + (fac(ind)) : fac nested in ind as a random factor

Note that the last model should cause singularities, as there's only 1 value of y for every combination of ind and fac.

First Model

In SAS, I build the following model :

PROC MIXED data=Data;
    CLASS ind fac trt;
    MODEL y = trt /s;
    RANDOM ind /s;
run;

According to tutorials, the same model in R using nlme should be :

> require(nlme)
> options(contrasts=c(factor="contr.SAS",ordered="contr.poly"))
> m2<-lme(y~trt,random=~1|ind,data=Data)

Both models give the same estimates for the coefficients and their SE, but when carrying out an F test for the effect of trt, they use a different amount of degrees of freedom :

SAS : 
Type 3 Tests of Fixed Effects 
Effect Num DF Den DF     F  Value Pr > F 
trt         1      8  0.89        0.3724 

R : 
> anova(m2)
            numDF denDF  F-value p-value
(Intercept)     1     8 70.96836  <.0001
trt             1     6  0.89272  0.3812

Question1: What is the difference between both tests? Both are fitted using REML, and use the same contrasts.

NOTE: I tried different values for the DDFM= option (including BETWITHIN, which theoretically should give the same results as lme)

Second Model

In SAS :

PROC MIXED data=Data;
    CLASS ind fac trt;
    MODEL y = trt /s;
    RANDOM fac(ind) /s;
run;

The equivalent model in R should be :

> m4<-lme(y~trt,random=~1|ind/fac,data=Data)

In this case, there are some very odd differences :

  • R fits without complaining, whereas SAS notes that the final hessian is not positive definite (which doesn't surprise me a bit, see above)
  • The SE on the coefficients differ (is smaller in SAS)
  • Again, the F test used a different amount of DF (in fact, in SAS that amount = 0)

SAS output :

Effect     trt Estimate Std Error  DF t Value Pr > |t| 
Intercept        0.8863    0.1192  14    7.43 <.0001 
trt       Cont  -0.1788    0.1686   0   -1.06 . 

R Output :

> summary(m4)
...
Fixed effects: y ~ trt 
               Value Std.Error DF   t-value p-value
(Intercept)  0.88625 0.1337743  8  6.624963  0.0002
trtCont     -0.17875 0.1891855  6 -0.944840  0.3812
...

(Note that in this case, the F and T test are equivalent and use the same DF.)

Interestingly, when using lme4 in R the model doesn't even fit :

> require(lme4)
> m4r <- lmer(y~trt+(1|ind/fac),data=Data)
Error in function (fr, FL, start, REML, verbose)  : 
  Number of levels of a grouping factor for the random effects
must be less than the number of observations

Question 2: What is the difference between these models with nested factors? Are they specified correctly and if so, how comes the results are so different?


Simulated Data in R :

Data <- structure(list(y = c(1.05, 0.86, 1.02, 1.14, 0.68, 1.05, 0.22, 
1.07, 0.46, 0.65, 0.41, 0.82, 0.6, 0.49, 0.68, 1.55), ind = structure(c(1L, 
2L, 3L, 1L, 3L, 4L, 4L, 2L, 5L, 6L, 7L, 8L, 6L, 5L, 7L, 8L), .Label = c("1", 
"2", "3", "4", "5", "6", "7", "8"), class = "factor"), fac = structure(c(1L, 
1L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 2L), .Label = c("l", 
"r"), class = "factor"), trt = structure(c(2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("Cont", 
"Treat"), class = "factor")), .Names = c("y", "ind", "fac", "trt"
), row.names = c(NA, -16L), class = "data.frame")

Simulated Data :

   y ind fac   trt
1.05   1   l Treat
0.86   2   l Treat
1.02   3   l Treat
1.14   1   r Treat
0.68   3   r Treat
1.05   4   l Treat
0.22   4   r Treat
1.07   2   r Treat
0.46   5   r  Cont
0.65   6   l  Cont
0.41   7   l  Cont
0.82   8   l  Cont
0.60   6   r  Cont
0.49   5   l  Cont
0.68   7   r  Cont
1.55   8   r  Cont
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  • $\begingroup$ @Aaron : Please find your answer included in this post. If you could copy and paste that as an answer, I give you the rep for it. It has been very helpful, so I really want to keep it here on crossvalidated. After you've done that, I delete your answer from the question. $\endgroup$ – Joris Meys Jun 1 '12 at 9:58
  • $\begingroup$ I'm trying to make the team revive your original Q with this unfortunate revision wiped out for good -- thus there is a great chance to restore original answers and merge them here. $\endgroup$ – user88 Jun 1 '12 at 12:25
  • $\begingroup$ @mbq : That would be nice, although I simulated some data (which I use here) and edited the answer of Aaron accordingly. For the other answer, that's going to be a bit more complicated, but I can try as well. $\endgroup$ – Joris Meys Jun 1 '12 at 12:58
  • $\begingroup$ Aaron's answer is an incredibly good one. I hope they see it. Unfortunately, your @Aaron won't contact him unless he's participated in this thread. $\endgroup$ – Wayne Jun 1 '12 at 18:45
  • 1
    $\begingroup$ Yes this was a nice answer. Here I gave a link to the deleted post : stats.stackexchange.com/questions/26556/… I'm going to add the link to the present post. $\endgroup$ – Stéphane Laurent Jun 1 '12 at 19:00
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For the first question, the default method in SAS to find the df is not very smart; it looks for terms in the random effect that syntactically include the fixed effect, and uses that. In this case, since trt is not found in ind, it's not doing the right thing. I've never tried BETWITHIN and don't know the details, but either the Satterthwaite option (satterth) or using ind*trt as the random effect give correct results.

PROC MIXED data=Data;
    CLASS ind fac trt;
    MODEL y = trt /s ddfm=satterth;
    RANDOM ind /s;
run;

PROC MIXED data=Data;
    CLASS ind fac trt;
    MODEL y = trt /s;
    RANDOM ind*trt /s;
run;

As for the second question, your SAS code doesn't quite match your R code; it only has a term for fac*ind, while the R code has a term for both ind and fac*ind. (See the Variance Components output to see this.) Adding this gives the same SE for trt in all models in both Q1 and Q2 (0.1892).

As you note, this is an odd model to fit as the fac*ind term has one observation for each level, so is equivalent to the error term. This is reflected in the SAS output, where the fac*ind term has zero variance. This is also what the error message from lme4 is telling you; the reason for the error is that you most likely misspecified something as you're including the error term in the model in two different ways. Interestingly, there is one slight difference in the nlme model; it's somehow finding a variance term for the fac*ind term in addition to the error term, but you will notice that the sum of these two variances equal the error term from both SAS and nlme without the fac*ind term. However, the SE for trt remains the same (0.1892) as trt is nested in ind, so these lower variance terms don't affect it.

Finally, a general note about degrees of freedom in these models: They are computed after the model is fit, and so differences in the degrees of freedom between different programs or options of a program do not necessarily mean that the model is being fit differently. For that, one must look at the estimates of the parameters, both fixed effect parameters and covariance parameters.

Also, using the t and F approximations with a given number of degrees of freedom is fairly controversial. Not only are there several ways to approximate the df, some believe the practice of doing so is not a good idea anyway. A couple words of advice:

  1. If everything is balanced, compare the results with the traditional least squares method, as they should agree. If it's close to balanced, compute them yourself (assuming balance) so that you can make sure the ones you're using are in the right ballpark.

  2. If you have a large sample size, the degrees of freedom don't matter very much as the distributions get close to a normal and chi-squared.

  3. Check out Doug Bates's methods for inference. His older method is based on MCMC simulation; his newer method is based on profiling the likelihood.

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  • $\begingroup$ Indeed a good answer, although I think that profiling the likelihood solves a different question (appropriate CIs on variance parameters where the profile is non-quadratic) than doing MCMC simulation (which handles both finite-size correction and non-quadraticness). I think bootMer (parametric bootstrap) is closer to the equivalent for mcmcsamp than confint(profile(...)) ... $\endgroup$ – Ben Bolker Jun 3 '12 at 15:33
  • $\begingroup$ @BenBolker: Sure could be. Doug Bates gave a talk here last month and he talked about his ideas on profiling the likelihood. That's about all I know about it so far. $\endgroup$ – Aaron - Reinstate Monica Jun 3 '12 at 18:39

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