[R] mgcv, testing gamm vs lme, which degrees of freedom?
Carlo Fezzi
c.fezzi at uea.ac.uk
Fri Jun 18 22:29:22 CEST 2010
Thanks Joris,
I understand your point regarding the need for the two models to be
nested. So, according to your in the example case the LR test is not
appropriate and the two model should be compared with other criteria such
as AIC or BIC for example.
On the other hand, Simon Wood indicated that such a LR test is
(approximately) correct in his previous reply... a am bit confused, which
is the correct approach to test the two models? Is the LR test correct
only if the parametric model is linear in the x variables maybe? In this
case, which is the best appraoch to compare a "gamm" vs a "lme" with
quadratic specification?
Best wishes,
Carlo
> Just realized something: You should take into account that the LR test
> is actually only valid for _nested_ models. Your models are not
> nested. Hence, you shouldn't use the anova function to compare them,
> and you shouldn't compare the df. In fact, if you're interested in the
> contribution of a term, then using anova to compare the model with
> that term and without that term gives you an answer on the hypothesis
> whether that term with spline contributes significantly to the model.
>
>> f2 <- gamm(y ~ s(x), random = list(id=~1), method="ML")
>
>> f3 <- gamm(y ~ x, random = list(id=~1), method="ML" )
>
>> f4 <- gamm(y ~ 1, random = list(id=~1), method="ML" )
>
>> anova(f3$lme,f2$lme)
> Model df AIC BIC logLik Test L.Ratio p-value
> f3$lme 1 4 760 770 -376
> f2$lme 2 5 381 394 -186 1 vs 2 380 <.0001
>
>> anova(f4$lme,f2$lme)
> Model df AIC BIC logLik Test L.Ratio p-value
> f4$lme 1 3 945 953 -470
> f2$lme 2 5 381 394 -186 1 vs 2 568 <.0001
>
>> anova(f3$lme,f4$lme)
> Model df AIC BIC logLik Test L.Ratio p-value
> f3$lme 1 4 760 770 -376
> f4$lme 2 3 945 953 -470 1 vs 2 188 <.0001
>
> This is the correct application of a likelihood ratio test. You see
> that adding the spline increases the df with 1 compared to the linear
> model, as part of the spline gets into the random component. Notice as
> well that the interpretation of a test in case of a random component
> is not the same as in case of a fixed component. If I understood
> correctly, this LR test specifically says something over the effect of
> X, without being interested in the shape of the spline. The
> "significance of a spline" is a difficult concept anyway, as a spline
> can be seen as a form of local regression. It's exactly the use of the
> randomization that allows for a general hypothesis about the added
> value of the spline, without focusing on its actual shape. Hence the
> "freedom" connected to that actual shape should not be used in the df
> used to test the general hypothesis.
>
> Hope this makes sense someway...
>
> Cheers
> Joris
>
>
> On Fri, Jun 18, 2010 at 6:27 PM, Carlo Fezzi <c.fezzi at uea.ac.uk> wrote:
>> Dear Simon,
>>
>> thanks a lot for your prompt reply.
>>
>> Unfortunately I am still confused about which is the correct way to test
>> the two models... as you point out: why in my example the two models
>> have
>> the same degrees of freedom?
>>
>> Intuitively it seems to me the gamm model is more flexible since, as I
>> understand also from you response, it should contain more random effects
>> than the other model because some of the smooth function parameters are
>> represented as such. This should not be taken into account when testing
>> one model vs the other?
>>
>> Continuing with my example, the two models:
>>
>> f2 <- gamm(y ~ s(x), random = list(id=~1), method="ML")
>> f3 <- gamm(y ~ x + I(x^2), random = list(id=~1), method="ML" )
>>
>> Can be tested with:
>>
>> anova(f3$lme,f2$lme)
>>
>> But why are the df the same? Model f2 appears to be more flexible and,
>> as
>> such, should have more (random) parameters. Should not a test of one
>> model
>> vs the other take this into account?
>>
>> Sorry if this may sound dull, many thanks for your help,
>>
>> Carlo
>>
>>
>>
>>> On Wednesday 16 June 2010 20:33, Carlo Fezzi wrote:
>>>> Dear all,
>>>>
>>>> I am using the "mgcv" package by Simon Wood to estimate an additive
>>>> mixed
>>>> model in which I assume normal distribution for the residuals. I would
>>>> like to test this model vs a standard parametric mixed model, such as
>>>> the
>>>> ones which are possible to estimate with "lme".
>>>>
>>>> Since the smoothing splines can be written as random effects, is it
>>>> correct to use an (approximate) likelihood ratio test for this?
>>> -- yes this is ok (subject to the usual caveats about testing on the
>>> boundary
>>> of the parameter space) but your 2 example models below will have the
>>> same
>>> number of degrees of freedom!
>>>
>>>> If so,
>>>> which is the correct number of degrees of freedom?
>>> --- The edf from the lme object, if you are testing using the log
>>> likelihood
>>> returned by the lme representation of the model.
>>>
>>>> Sometime the function
>>>> LogLik() seems to provide strange results regarding the number of
>>>> degrees
>>>> of freedom (df) for the gam, for instance in the example I copied
>>>> below
>>>> the df for the "gamm" are equal to the ones for the "lme", but the
>>>> summary(model.gam) seems to indicate a much higher edf for the gamm.
>>> --- the edf for the lme representation of the model counts only the
>>> fixed
>>> effects + the variance parameters (which includes smoothing
>>> parameters).
>>> Each
>>> smooth typically contributes only one or two fixed effect parameters,
>>> with
>>> the rest of the coefficients for the smooth treated as random effects.
>>>
>>> --- the edf for the gam representation of the same model differs in
>>> that
>>> it
>>> also counts the *effective* number of parameters used to represent each
>>> smooth: this includes contributions from all those coefficients that
>>> the
>>> lme
>>> representation treated as strictly random.
>>>
>>> best,
>>> Simon
>>>
>>>
>>>> I would be very grateful to anybody who could point out a solution,
>>>>
>>>> Best wishes,
>>>>
>>>> Carlo
>>>>
>>>> Example below:
>>>>
>>>> ----
>>>>
>>>> rm(list = ls())
>>>> library(mgcv)
>>>> library(nlme)
>>>>
>>>> set.seed(123)
>>>>
>>>> x <- runif(100,1,10) # regressor
>>>> b0 <- rep(rnorm(10,mean=1,sd=2),each=10) # random intercept
>>>> id <- rep(1:10, each=10) # identifier
>>>>
>>>> y <- b0 + x - 0.1 * x^3 + rnorm(100,0,1) # dependent variable
>>>>
>>>> f1 <- lme(y ~ x + I(x^2), random = list(id=~1) , method="ML" ) # lme
>>>> model
>>>>
>>>> f2 <- gamm(y ~ s(x), random = list(id=~1), method="ML" ) # gamm
>>>>
>>>> ## same number of "df" according to logLik:
>>>> logLik(f1)
>>>> logLik(f2$lme)
>>>>
>>>> ## much higher edf according to summary:
>>>> summary(f2$gam)
>>>>
>>>> -----------
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
>>>> http://www.R-project.org/posting-guide.html and provide commented,
>>>> minimal,
>>>> self-contained, reproducible code.
>>>
>>> --
>>>> Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY
>>>> UK
>>>> +44 1225 386603 www.maths.bath.ac.uk/~sw283
>>>
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Joris Meys
> Statistical consultant
>
> Ghent University
> Faculty of Bioscience Engineering
> Department of Applied mathematics, biometrics and process control
>
> tel : +32 9 264 59 87
> Joris.Meys at Ugent.be
> -------------------------------
> Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
>
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