[R-sig-ME] back log transformation

espesser robert.espesser at lpl-aix.fr
Thu Mar 24 11:37:39 CET 2011

Dear all,

This subject has been  previously discussed, but I am not sure I proceed
the right way with the use of the variances.

Here is the  summary of my lmer model :

Linear mixed model fit by REML

Formula: log(TIME) ~ LONG + ACCO  + (1 | SUJET)

Data: dssPUISS
AIC   BIC logLik deviance REMLdev
899.6 934.1 -442.8    856.7   885.6
Random effects:
Groups   Name        Variance Std.Dev.
SUJET    (Intercept) 0.019090 0.13817
Residual             0.130297 0.36097
Number of obs: 1018, groups: SUJET, 24

Fixed effects:
Estimate Std. Error t value
(Intercept)   5.77423    0.04462  129.42
LONG          0.02883    0.01129    2.55
ACCO1        -0.05722    0.02272   -2.52

LONG is continuous .
ACCO is a 2 levels factor .

I would proceed so:

1) To compute TIME at this specific point :

sujet== "s3"
long == 6
acco == "1"

TIME = exp( intercept + 6*LONG + ACCO1
+  estimate_of_s3_intercept +  0.5*var(Residual)  )

with var( Residual)  ==  0.130297

Is it correct ?

2) I am  mainly interested to back-transform the fixed effects, at the
same point.

2.1) I would use:

TIME = exp( intercept + 6*LONG + ACCO1
+ 0.5*var(SUJET) +0.5*var(Residual) )

with var(SUJET) == 0.019090

2.2) Suppose  there was a second  random intercept (say b)  in my model,
I would use:

TIME = exp( intercept + 6*LONG + ACCO1
+ 0.5*var(SUJET) + 0.5*var(b) +  0.5*var(Residual)  )

Are these 2 expressions correct ?

2.3)
Suppose there was a random slope in the model, something like:

log(TIME) ~ LONG + ACCO  + (LONG | SUJET)

How can I get TIME  on the original scale ?

3) Related question :

To  extract the stddev of the SUJET  random intercept ,  I use:

attr(VarCorr(MyModel.lmer)\$SUJET,"stddev")

Is it the right  way ?

Thank you very much for your help .
Robert

--
Robert Espesser
CNRS UMR 6057 - Université de Provence
5 Avenue Pasteur
13100 AIX-EN-PROVENCE

Tel: +33 (0)442 95 36 26