back log transformation
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 ...