# [R] mgcv, testing gamm vs lme, which degrees of freedom?

Carlo Fezzi c.fezzi at uea.ac.uk
Fri Jun 18 18:27:37 CEST 2010

```Dear Simon,

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)
>>
>> -----------
>>
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