[R-sig-ME] back log transformation

[espesser]

Thank you very much  for yours answers , Ben.

First, here is  the reference of the thread  about back log- transformation,
  initiated by Christina Bogner:

http://finzi.psych.upenn.edu/R-sig-mixed-models/2009q1/002066.html

If I have understand well , computing  just  the exp()  without the 
addition of variances, as in:

TIME = exp( intercept + 6*LONG + ACCO1)

  gives  (approximatively  ?)  the estimated  median of TIME .
   I believed that I got the geometric mean of TIME for the conditions ( 
long==6, acco == "1"),
at least for a non-mixed linear model.
I failed to  find clear information/explanation about this, so I 
appreciate reference on the topic.

Thank you again for your help

R


Le 24/03/2011 20:54, Ben Bolker a écrit :
> On 03/24/2011 06:37 AM, espesser wrote:
>>   Dear all,
>>
>> This subject has been  previously discussed, but I am not sure I proceed
>> the right way with the use of the variances.
>    Can you give a reference to the previous discussion please?
>
>
>> 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 ?
>
>     Is the 0.5*var(Residual) to get the mean (rather than the median) of
> TIME on the original scale ?  It seems reasonable but I wonder if you
> could simplify your life a little bit by predicting the median rather
> than the median ...
>
>> 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
>    Don't quite know what you mean here.  It seems you're thinking about
> estimating a marginal mean (unknown subject) rather than a conditional
> mean.  Your approach seems reasonable but I wouldn't want to swear it
> was right ...
>
>>
>> 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 ?
>>
>    This gets stickier.  The second 'random intercept' is from a second
> random effect grouping factor?  If the random effects are independent,
> this seems plausible -- otherwise the variance of the sum will not be
> equal to the sum of the variances ...
>
>
>> 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 ?
>    If you want the marginal mean (i.e., something analogous to what you
> are doing above), then you need to calculate the variance -- e.g. if the
> value  is  (a+b*x + e_a + e_b*x + e_i) where e_a, e_b are random
> intercept and slope and e_i is residual error, then **if** they were
> all independent the variance would be var_a + var_b*x^2 + var_e.
> However, a and b are generally correlated so I believe it would be
> var_a + var_b*x^2 + 2*cov(a,b)*x + var_e.
>>
>> 3) Related question :
>>
>> To  extract the stddev of the SUJET  random intercept ,  I use:
>>
>> attr(VarCorr(MyModel.lmer)$SUJET,"stddev")
>>
>    Yes.
>
>    As mentioned above, I think your life would be a bit easier if you
> just decided that you wanted the median (which is invariant under
> transformation) rather than the mean on the back-transformed scale ...


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

Tel: +33 (0)442 95 36 26