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I would like do create a mixed linear model for an unbalanced dataset (different number of events per subject and a few missing values for some time points). I am using R version 3.2.1 (2015-06-18), package: nlme_3.1-120.

Here comes simulated data:

library(nlme)
set.seed(1)
subject    <- factor(rep(c(1, 1, 2, 3, 4, 4, 4, 5, 6, 7, 7, 8, 9, 9, 10, 
                           11, 11, 11, 12, 13), 10))
event      <- factor(rep(1:20, 10))
timepoint  <- rep(1:10, each = 20)
measure    <- rnorm(length(timepoint)) + timepoint*0.3
timepoint  <- factor(timepoint)
measure[sample(1:length(measure), rpois(5,4))] <- NA
data       <- data.frame(subject=subject, event=event, timepoint=timepoint, 
                         measure=measure)
str(data)

The model should predict the variable “measure” over different time points as fixed effect and for subjects and events as random effects.

base      <- lme(measure ~ 1,         data=data, random= ~ 1|subject, 
                 na.action=na.exclude, method="ML")
intercept <- lme(measure ~ timepoint, data=data, random= ~ 1|subject, 
                 na.action=na.exclude, method="ML")
nested    <- lme(measure ~ timepoint, data=data, random= ~ 1|subject/event, 
                 na.action=na.exclude, method="ML")
anova(base, intercept, nested)

I would like to fit random intercept and slope, because intercept and slope can vary among subjects and events. However when I add the random slope effect, the model does not converge. It does not through any error message, but it runs to infinity. What can I do create a model with random slope that converges?

cave model runs endless

slope <- lme(measure ~ timepoint, data=data, random= ~ timepoint|subject, 
             na.action=na.exclude, method="ML")

I tried also this

cave model runs endless

slope2 <- lme(measure ~ timepoint, data=data, random= ~ timepoint|subject, 
              na.action=na.exclude, method="ML", control=list(opt="optim"))

cave some models may run endless

slope3      <- lme(measure ~ timepoint, data=data, random= ~ timepoint|subject/event, 
                   na.action=na.exclude, method="ML", control = list(opt="optim"))
covariance  <- lme(measure ~ timepoint, data=data, random= ~ timepoint|subject, 
                   correlation=corAR1(),na.action = na.exclude, method="ML")
covariance2 <- lme(measure ~ timepoint, data=data, random= ~ timepoint|subject, 
                   correlation=corAR1(0), na.action=na.exclude, method="ML", 
                   control=list(opt="optim"))
covariance3 <- lme(measure ~ timepoint, data=data, random= ~ timepoint|subject, 
                   correlation=corAR1(0), na.action=na.exclude, method="ML", 
                   control=list(maxlter=1000))
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  • 1
    $\begingroup$ What are the "cave" bits for? $\endgroup$ – Glen_b -Reinstate Monica Jul 22 '15 at 4:07
  • $\begingroup$ Thank you for the remark Glen_b, it is a warning, because at least on my computer some models run endless and can crash R. I have added a comment to make that clearer. $\endgroup$ – Kev Jul 22 '15 at 18:56
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    $\begingroup$ The main issue here is treating timepoint as categorical. That means that fitting a term like timepoint|subject will require fitting a full 10x10 variance-covariance matrix (55 distinct parameters) across 13 subjects. It would be much more practical to fit timepoint as a continuous covariate ... $\endgroup$ – Ben Bolker Jul 22 '15 at 21:02
  • $\begingroup$ Ah, with the edit and the explanation, it makes more sense. I couldn't see what underground hollows had to do with it. $\endgroup$ – Glen_b -Reinstate Monica Jul 22 '15 at 23:24
  • $\begingroup$ Dear Ben Bolker, Dear @AdamO, Thank your for the very helpful comments and explanations. I see your points and from the simulated data it was not clear, why timepoint is a categorical variable, but in fact the measurements were taken at different time points for example at the beginning of the recording, 2 minutes before the event, during the event, 1 minute, 2 minutes, 3 minutes, 5 minutes, 10 minutes after the event and at the end of the recording. That's why I treated it like a categorical and not like a continuous variable. $\endgroup$ – Kev Jul 24 '15 at 17:15
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@AdamO has done a good job identifying the specific error in your code. Let me address the question more generally. Here is how I simulate a linear mixed effects model:

Mixed effects models assume each unit has random effects drawn from a multivariate normal distribution. (When a model is estimated, it is the variances and covariances of that multivariate normal that are being estimated for the random effects.) I start by specifying this distribution and generating (pseudo-)random values to serve as the random effects. It is often convenient to specify the variances as $1$, so that the covariance is the correlation between slopes and intercepts (which is easier for me to conceptualize).

library(MASS)
ni = 13                                                 # number of subjects
RE = mvrnorm(ni, mu=c(0,0), Sigma=rbind(c(1.0, 0.3),
                                        c(0.3, 1.0) ))
colnames(RE) = c("ints","slopes");  t(round(RE,2))
#         [,1]  [,2]  [,3] [,4]  [,5]  [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
# ints    0.81 -0.52 -0.65 1.30 -0.29 -1.15 0.04 0.05 0.00 -0.29  2.40 -0.05 -0.47
# slopes -1.82  0.81 -0.70 1.28  0.82 -0.18 0.74 1.14 0.93 -0.20  0.04  0.68 -0.53

Next, I would generate my $X$ variables. I can't really follow the logic of your example, so I will use time as my only regressor.

nj   = 10                              # number of timepoints
data = data.frame(ID   = rep(1:ni,   each=nj), 
                  time = rep(1:nj,   times=ni),
                  RE.i = rep(RE[,1], each=nj),
                  RE.s = rep(RE[,2], each=nj),
                  y    = NA                    )
head(data, 14)
#    ID time       RE.i       RE.s  y
# 1   1    1  0.8051709 -1.8152973 NA
# 2   1    2  0.8051709 -1.8152973 NA
# 3   1    3  0.8051709 -1.8152973 NA
# 4   1    4  0.8051709 -1.8152973 NA
# 5   1    5  0.8051709 -1.8152973 NA
# 6   1    6  0.8051709 -1.8152973 NA
# 7   1    7  0.8051709 -1.8152973 NA
# 8   1    8  0.8051709 -1.8152973 NA
# 9   1    9  0.8051709 -1.8152973 NA
# 10  1   10  0.8051709 -1.8152973 NA
# 11  2    1 -0.5174601  0.8135761 NA
# 12  2    2 -0.5174601  0.8135761 NA
# 13  2    3 -0.5174601  0.8135761 NA
# 14  2    4 -0.5174601  0.8135761 NA

Having generated your random effects and your regressors, you can specify the data generating process. Since you want some randomly missed timepoints, there is a level of additional complexity here. (Note that these data are missing completely at random; for more on simulating missing data, see: How to simulate the different types of missing data.)

y       = with(data, (0 + RE.i) + (.3 + RE.s)*time + rnorm(n=ni*nj, mean=0, sd=1))
m       = rbinom(n=ni*nj, size=1, prob=.1)  
y[m==1] = NA
data$y  = y
head(data, 14)
#    ID time       RE.i       RE.s           y
# 1   1    1  0.8051709 -1.8152973  -0.8659219
# 2   1    2  0.8051709 -1.8152973  -3.6961761
# 3   1    3  0.8051709 -1.8152973  -4.2188711
# 4   1    4  0.8051709 -1.8152973  -4.8380769
# 5   1    5  0.8051709 -1.8152973  -5.4126362
# 6   1    6  0.8051709 -1.8152973  -8.3894008
# 7   1    7  0.8051709 -1.8152973          NA
# 8   1    8  0.8051709 -1.8152973 -11.3710128
# 9   1    9  0.8051709 -1.8152973 -14.2095646
# 10  1   10  0.8051709 -1.8152973 -14.7627970
# 11  2    1 -0.5174601  0.8135761   0.2018260
# 12  2    2 -0.5174601  0.8135761          NA
# 13  2    3 -0.5174601  0.8135761   3.9232935
# 14  2    4 -0.5174601  0.8135761          NA

At this point, you can fit your model. I typically use the lme4 package.

library(lme4)
summary(lmer(y~time+(time|ID), data))
# Linear mixed model fit by REML ['lmerMod']
# Formula: y ~ time + (time | ID)
#    Data: data
# 
# REML criterion at convergence: 378.3
# 
# Scaled residuals: 
#      Min       1Q   Median       3Q      Max 
# -2.48530 -0.61824 -0.08551  0.59285  2.70687 
# 
# Random effects:
#   Groups   Name        Variance Std.Dev. Corr 
#   ID       (Intercept) 0.9970   0.9985        
#            time        0.8300   0.9110   -0.05
#   Residual             0.7594   0.8715        
# Number of obs: 112, groups:  ID, 13
# 
# Fixed effects:
#             Estimate Std. Error t value
# (Intercept)  0.03499    0.33247   0.105
# time         0.53454    0.25442   2.101
# 
# Correlation of Fixed Effects:
#      (Intr)
# time -0.100
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  • $\begingroup$ Dear Ben Bolker, Dear AdamO, Dear gung, Thank your for the very helpful comments and explanations. I see your points and from the simulated data it was not clear, why timepoint is a categorical variable, but in fact the measurements were taken at different time points for example at the beginning of the recording, 2 minutes before the event, during the event, 1 minute, 2 minutes, 3 minutes, 5 minutes, 10 minutes after the event and at the end of the recording. That's why I treated it like a categorical and not like a continuous variable. $\endgroup$ – Kev Jul 23 '15 at 20:42
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    $\begingroup$ @Kev, I am the only person who is notified of your comment. If you want others to be notified, you need to comment after AdamO's answer, or in the comments under the Q w/ the "@" symbol before BenBolker's user name (as I have done for you here). That said, your comment doesn't make much sense, but you should be able to adapt the code I provide here. If you are focused on some event, you can use -2, 0, 1, ... for the time variable & even add a linear spline term w/ a knot at the event, if that's what you're after. $\endgroup$ – gung - Reinstate Monica Jul 23 '15 at 20:49
  • $\begingroup$ @gung If I interpret your set-up correctly, all random effects would have intercepts and slopes drawn from the same bivariate normal distribution, such that the intercept and slope for any given RE would be correlated. So, here is my follow-up question re: recreating model here: if for instance a RE is the different trees in the sample, there wouldn't be slope - only intercept. So, in that case, would I draw the intercepts from a simple rnorm. And if so, where would there be any correlation? $\endgroup$ – Antoni Parellada Dec 22 '15 at 16:44
  • $\begingroup$ This is where I am. $\endgroup$ – Antoni Parellada Dec 22 '15 at 16:44
  • $\begingroup$ @AntoniParellada, by default mixed effects models assume the random effects are normally distributed (ie multivariate normal), although it is possible to deal with other distributions. You can have a set of independent random effects; that would amount to setting the off diagonal elements of the covariance matrix to 0 or using multiple rnorm() calls. If you only have a random intercept, none of that applies & you can just use rnorm(). I don't have time to try to work through your larger situation right now; I'll try to get to it when I can. $\endgroup$ – gung - Reinstate Monica Dec 22 '15 at 16:52
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There is an obvious mistake in your simulation. However, in general, it's impossible to generate data so that a random slopes model is guaranteed to converge.

The fix you need to apply is to timepoint. Timepoint is a factor. You should not be using a factor level variable in a random slopes model, it is completely aliased with random intercept.

Try

data$timepoint <- as.numeric(data$timepoint)

and

slope <- lme(measure ~ factor(timepoint), data=data, 
  random=~timepoint|subject, na.action=na.exclude, method="ML")

This converges instantly. It's also appropriately nested within other models.

Make good use of the try() command to "capture" simulation output with converge-fails. You can explore interesting behavior with numerical solvers that are "at the boundary" of their capabilities.

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