[R] random effects with lmer() and lme(), three random factors

Xianqun (Wilson) Wang xwang at aviaradx.com
Mon Jul 31 21:27:39 CEST 2006

Dr. Bates,

Thanks for the notes! It helps. So now I see consistent resluts from
both lme and lmer. Since I have several response variables to look at, I
will reduce the model separately. 

Speaking of the model reduction, it is clear in this example that the
trivial variance of Operator:Run could be ignored. For a
non-trivial-variance multivariate model, I wonder if there is any
function (like step function for lm and glm) would work with lme or lmer
and allow me to do the AIC-based model selection.   


-----Original Message-----
From: dmbates at gmail.com [mailto:dmbates at gmail.com] On Behalf Of Douglas
Sent: Saturday, July 29, 2006 7:51 AM
To: Xianqun (Wilson) Wang
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] random effects with lmer() and lme(), three random

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

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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