[R] random effects with lmer() and lme(), three random factors
Douglas Bates
bates at stat.wisc.edu
Sat Jul 29 17:51:10 CEST 2006
On 7/28/06, Xianqun (Wilson) Wang <xwang at aviaradx.com> wrote:
> Hi, all,
> I have a question about random effects model. I am dealing with a
> three-factor experiment dataset. The response variable y is modeled
> against three factors: Samples, Operators, and Runs. The experimental
> design is as follow:
> 4 samples were randomly chosen from a large pool of test samples. Each
> of the 4 samples was analyzed by 4 operators, randomly selected from a
> group of operators. Each operator independently analyzed same samples
> over 5 runs (runs nested in operator). I would like to know the
> following things:
> (1) the standard deviation within each run;
> (2) the standard deviation between runs;
> (3) the standard deviation within operator
> (4) the standard deviation between operator.
> With this data, I assumed the three factors are all random effects. So
> the model I am looking for is
> Model: y = Samples(random) + Operator(random) + Operator:Run(random) +
> Error(Operator) + Error(Operator:Run) + Residuals
> I am using lme function in nlme package. Here is the R code I have
> 1. using lme:
> First I created a grouped data using
> gx <- groupedData(y ~ 1 | Sample, data=x)
> gx$dummy <- factor(rep(1,nrow(gx)))
> then I run the lme
> fm<- lme(y ~ 1, data=gx,
> random=list(dummy=pdBlocked(list(pdIdent(~Sample-1),
> pdIdent(~Operator-1),
> pdIdent(~Operator:Run-1)))))
> finally, I use VarCorr to extract the variance components
> vc <- VarCorr(fm)
> Variance StdDev
> Operator:Run 1.595713e-10:20 1.263215e-05:20
> Sample 5.035235e+00: 4 2.243933e+00: 4
> Operator 5.483145e-04: 4 2.341612e-02: 4
> Residuals 8.543601e-02: 1 2.922944e-01: 1
> 2. Using lmer in Matrix package:
> fm <- lmer(y ~ (1 | Sample) + (1 | Operator) +
> (1|Operator:Run), data=x)
That model specification can now be written as
fm <- lmer(y ~ (1|Sample) + (1|Operator/Run), x)
> summary(fm)
> Linear mixed-effects model fit by REML
> Formula: H.I.Index ~ (1 | Sample.Name) + (1 | Operator) + (1 |
> Operator:Run)
> Data: x
> AIC BIC logLik MLdeviance REMLdeviance
> 96.73522 109.0108 -44.36761 90.80064 88.73522
> Random effects:
> Groups Name Variance Std.Dev.
> Operator:Run (Intercept) 4.2718e-11 6.5359e-06
> Operator (Intercept) 5.4821e-04 2.3414e-02
> Sample (Intercept) 5.0352e+00 2.2439e+00
> Residual 8.5436e-02 2.9229e-01
> number of obs: 159, groups: Operator:Run, 20; Operator, 4; Sample.Name,
> 4
> Fixed effects:
> Estimate Std. Error t value
> (Intercept) 0.0020818 1.1222683 0.001855
> There is a difference between lmer and lme is for the factor
> Operator:Run.
It's just a matter of round-off. It is possible for the ML or REML
estimates of a variance component to be zero, as is the case here, but
the current computational methods do not allow the value zero because
this will cause some of the matrix decompositions to fail. In lmer we
use a constrained optimization with the relative variance (variance of
a random effect divided by the residual variance) constrained to be
greater than or equal to 5e-10, which is exactly the value you have
here.
I'll add code to the model fitting routine to warn the user when
convergence to the boundary value occurs. I haven't done that in the
past because it is not always easy to explain what is occurring.
For a model with variance components only, like yours, convergence on
the boundary means that an estimated variance component is zero. In
the case of bivariate or multivariate random effects the
variance-covariance matrix can be singular without either of the
variances being zero.
The bottom line for you is that the estimated variance for
Operator:Run is zero and you should reduce the model to y ~ (1|Sample)
+ (1|Operator)
I cannot find where the problem is. Could anyone point me
> out if my model specification is correct for the problem I am dealing
> with. I am pretty new user to lme and lmer. Thanks for your help!
>
>
>
>
>
> Wilson Wang
>
>
>
>
>
>
> [[alternative HTML version deleted]]
>
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