[R-meta] AICc or variance components, which one matters more?
Luke Martinez
m@rt|nez|ukerm @end|ng |rom gm@||@com
Sat Nov 13 15:23:36 CET 2021
Dear Wolfgang,
Thank you for your response.
>> I don't know why you think a model is better if the estimates of its
variance components are non-zero.
Because of two things. 1) It maybe more desirable/efficient for a random
factor to explain some heterogeneity (as in g1) rather than not (as in g2).
2) The CI for the variance component (using the full data) is narrower for
lab in g1 relative to that in g2.
>> Personally, I would rather think about the meaning of these models and
use the one that makes more sense in the given context.
Sure, but that's not always as straightforward. For instance, imagine in
your current data, the dominant structure is the hierarchy of A over B with
a few exceptions where this hierarchy can break (meaning A and B can also
be considered crossed).
In such a situation, one may think that if my data was much larger, then
the few exceptions in the current data may actually turn out to be the
dominant structure (meaning A and B being crossed may be a more
representative model of the population structure of A and B) or may be
that's just overthinking.
So when meaning alone is not that conclusive, you may a bit rely on the
comparison across a set of candidate models. Here, AICc and how much
heterogeneity is explained by each model may at least help you pick the one
that more fully describes your current data.
Curious to know your thoughts?
All the best,
Luke
Sat, Nov 13, 2021, 6:56 AM Viechtbauer, Wolfgang (SP) <
wolfgang.viechtbauer using maastrichtuniversity.nl> wrote:
> Please see my comments below.
>
> Best,
> Wolfgang
>
> >-----Original Message-----
> >From: R-sig-meta-analysis [mailto:
> r-sig-meta-analysis-bounces using r-project.org] On
> >Behalf Of Philippe Tadger
> >Sent: Saturday, 13 November, 2021 11:03
> >To: Luke Martinez; R meta
> >Subject: Re: [R-meta] AICc or variance components, which one matters more?
> >
> >Dear Luke
> >
> >I will suggest to choose the model that can provide a reliable
> >estimation of the variances. Usually a variance estimation equal to zero
> >or a correlation estimation equal to 1 or -1 are potential warning
> >signal, your model is in the limit of the parameter space: so you can
> >having an underestimation and/or a non-identifiable model. How can you
> >be sure? Using the profile() function in each model. In your case model
> >g1 is identifiable, but g2 is not identifiable. I'll choose g1.
>
> This is not correct. Both variance componments are identifiable in both
> models, as the profile plots indicate. The estimate of the lab variance
> component in g2 just happens to be 0 (the peak of the likelihood profile is
> at 0).
>
> >On 13/11/2021 05:24, Luke Martinez wrote:
> >> Hello Colleagues,
> >>
> >> I've fit two candidate models (g1 and g2).
> >>
> >> With g1, I get two non-zero variance components for 'lab' and 'study'.
> >>
> >> With g2, I get a zero variance component for 'lab' and a non-zero one
> >> for 'study'.
> >>
> >> So, from the perspective of variance components, g1 seems like a better
> model.
>
> I don't know why you think a model is better if the estimates of its
> variance components are non-zero.
>
> >> But when I compare the models using AICc, g2 seems like a better model.
>
> Personally, I would rather think about the meaning of these models and use
> the one that makes more sense in the given context.
>
> >> I wonder, then, which criterion should I use to choose the model (AICc
> >> or the variance components)?
> >>
> >> Thanks,
> >> Luke
> >> For reproducibility, I'm showing a somewhat similar situation with a
> >> small section of my data below.
> >>
> >> m="
> >> lab study yi vi es_id
> >> 1 1 1.04 0.48503768 1
> >> 1 1 0.96 0.51076604 2
> >> 1 2 1.71 0.05767389 3
> >> 2 2 1.52 0.07539841 4
> >> 1 3 1.91 0.31349510 5
> >> 2 4 3.01 0.67910095 6
> >> 2 4 3.62 0.50670360 7
> >> 9 5 0.99 0.18297170 8
> >> 9 5 0.43 0.37225851 9
> >> 9 6 1.68 0.39072390 10
> >> 9 6 1.25 0.02879550 11"
> >> dd <- read.table(text=m,h=T)
> >>
> >> (g1=rma.mv(yi, vi, random = ~1|lab/study, data = dd))
> >> estim sqrt nlvls fixed factor
> >> sigma^2.1 0.1245 0.3529 3 no lab
> >> sigma^2.2 0.2535 0.5035 7 no lab/study
> >>
> >> (g2=rma.mv(yi, vi, random = list(~1|lab, ~1|study), data = dd))
> >> estim sqrt nlvls fixed factor
> >> sigma^2.1 0.0000 0.0000 3 no lab
> >> sigma^2.2 0.5127 0.7160 6 no study
> >>
> >> fitstats(g1,g2)[5,]
> >> g1 g2
> >> AICc: 30.85992 29.73897
> >> (for full data, AICc of g2 is more noticeably smaller than that of g1)
>
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