[R-sig-ME] estimating AR1 parameters of level one error using lme

Asher Strauss asher.strauss at gmail.com
Tue May 19 10:55:45 CEST 2015


Hi Daniel
and thank you very much for your reply!

I went over the documents you sent me, though couldn't find the sdtalt
package :(

I think using  lme is more suitable for my purposes since it is important
for me include random effects in my model.

I am still stuck with understanding the right code in R to specify a
homogeneous variance structure with correlations structured as AR1 (model
13 in your paper
https://www.researchgate.net/publication/23134911_Multilevel_modelling_Beyond_the_basic_applications
).
As I wrote in my previous message specifying this model in R (if I got the
code right - I assume not...) and in SPSS produces completely different
results.

I will be grateful if you or anybody else can point me towards a solution.

I am repeating here my previous example to make my difficulty easier to
address:

for example using the Glucose data from the nlme package:

data(Glucose)

fGlucose<-filter(Glucose,Meal=="10am")

summary(
lme(
fixed=conc~Time,
random=~1+Time|Subject,
method="REML",
data=fGlucose,
na.action="na.omit",
correlation=corAR1(form=~1+Time|Subject))
)


I get an output of:

Correlation Structure: ARMA(1,0)
 Formula: ~1 + Time | Subject
 Parameter estimate(s):
     Phi1
0.4334469

I am assuming  Phi1 is equivalent to Rho (am I right?). What makes me
suspect I am wrong, and confuses me, is that when estimating AR1
diagonal (homogeneous
level one error variance) and AR1 rho (the auto regression parameter)
 using SPSS I received 1.349 and -0.942 respectively.

here is the SPSS syntax I am using  on the same data set (Glucose):

COMPUTE filter_$=(Meal="10am").
VARIABLE LABELS filter_$ 'Meal="10am" (FILTER)'.
VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'.
FORMATS filter_$ (f1.0).
FILTER BY filter_$.
EXECUTE.

MIXED conc WITH Time
 /CRITERIA=CIN(95) MXITER(100) MXSTEP(10) SCORING(1)
SINGULAR(0.000000000001) HCONVERGE(0, ABSOLUTE) LCONVERGE(0, ABSOLUTE)
PCONVERGE(0.000001, ABSOLUTE)
  /FIXED=INTERCEPT Time | SSTYPE(3)
  /METHOD=REML
  /PRINT=SOLUTION TESTCOV
  /RANDOM=INTERCEPT Time | SUBJECT(Subject)
  /REPEATED=Time | SUBJECT(Subject)COVTYPE(AR1).

It would be really great if you or anybody else can assist me here!
Thank you!

Best
Asher Strauss

On Mon, May 18, 2015 at 9:01 PM, Daniel Wright <Daniel.Wright at act.org>
wrote:

> It may also be convenient to use the gls function in nlme.
>
> This is used in
> https://www.researchgate.net/publication/23134911_Multilevel_modelling_Beyond_the_basic_applications
> and in http://www.ats.ucla.edu/stat/r/examples/alda/ch7.htm which is the
> UCLA page for Singer and Willet's wonderful book.
>
>
> -----Original Message-----
> From: R-sig-mixed-models [mailto:r-sig-mixed-models-bounces at r-project.org]
> On Behalf Of Asher Strauss
> Sent: Saturday, May 16, 2015 1:19 AM
> To: r-sig-mixed-models at r-project.org
> Subject: [R-sig-ME] estimating AR1 parameters of level one error using lme
>
> Hi all,
> I am new to using R for mixed effects (have been using SPSS until now),
> please for give me if this is a trivial question.
>
> I am trying to understand how to estimate AR1 parameters of level one
> error using lme (I have understood that specifying level one
> variance-covariance matrix is not easily possible in lmer, is this true?).
>
> When using SPSS one estimates two parameters: AR1 diagonal and AR1 rho. I
> am searching for an equivalent command in R.
>
> for example using the Glucose data from the nlme package:
>
> data(Glucose)
>
> fGlucose<-filter(Glucose,Meal=="10am")
>
> summary(
> lme(
> fixed=conc~Time,
> random=~1+Time|Subject,
> method="REML",
> data=fGlucose,
> na.action="na.omit",
> correlation=corAR1(form=~1+Time|Subject))
> )
>
>
> I get an out put of:
>
> Correlation Structure: ARMA(1,0)
>  Formula: ~1 + Time | Subject
>  Parameter estimate(s):
>      Phi1
> 0.4334469
>
> is Phi1 equivalent to Rho?  I do not believe so, since when estimating AR1
> diagonal and AR1 rho using SPSS I received 1.349 and -0.942 respectively.
>
> here is the SPSS syntax I am using:
> COMPUTE filter_$=(Meal="10am").
> VARIABLE LABELS filter_$ 'Meal="10am" (FILTER)'.
> VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'.
> FORMATS filter_$ (f1.0).
> FILTER BY filter_$.
> EXECUTE.
>
> MIXED conc WITH Time
>  /CRITERIA=CIN(95) MXITER(100) MXSTEP(10) SCORING(1)
> SINGULAR(0.000000000001) HCONVERGE(0, ABSOLUTE) LCONVERGE(0, ABSOLUTE)
> PCONVERGE(0.000001, ABSOLUTE)
>   /FIXED=INTERCEPT Time | SSTYPE(3)
>   /METHOD=REML
>   /PRINT=SOLUTION TESTCOV
>   /RANDOM=INTERCEPT Time | SUBJECT(Subject)
>   /REPEATED=Time | SUBJECT(Subject)COVTYPE(AR1).
>
> Thank you!
> Asher
>
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>
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