I am currently reviewing some work and have come across the following, which seems wrong to me. Two mixed models are fitted (in R) using lmer. The models are non-nested and are compared by likelihood-ratio tests. In short, here is a reproducible example of what I have:

Resp = rnorm(100)
A = factor(rep(1:5,each=20))
B = factor(rep(1:2,times=50))
C = rep(1:4, times=25)
m1 = lmer(Resp ~ A + (1|C), REML = TRUE)
m2 = lmer(Resp ~ B + (1|C), REML = TRUE)

As far as I can see, lmer is used to compute the log-likelihood and the anova statement tests the difference between the models using a chi-square with the usual degrees of freedom. This does not seem correct to me. If it is correct, does anyone know of any reference justifying this? I am aware of methods relying on simulations (Paper by Lewis et al., 2011) and the approach developed by Vuong (1989) but I do not think that this is what is produced here. I do not think that the use of the anova statement is correct.


1 Answer 1


This is not correct in two ways:

  1. (Ordinary) likelihood ratio test can only be used to compare nested models;
  2. We cannot compare mean models under REML. (This is not the case here, see @KarlOveHufthammer's comments below.)

In the case of using ML, I am aware of using AIC or BIC to compare the non-nested models.

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    $\begingroup$ Regarding point 2, the anova() function in R does not compare the two models fitted under REML; it refits them using ML and then perform the test. See lme4:::anova.merMod, which contains the line mods <- lapply(mods, refitML). (But you are still right that anova() can’t be used to compare the two models, as they are not nested.) $\endgroup$ Commented Jan 29, 2014 at 19:32
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    $\begingroup$ also note that there is some disagreement on nesting: Brian Ripley says nesting is essential for AIC comparison (see p. 20 of linked document for discussion), while Anderson and Burnham (see p. 2) disagree .. $\endgroup$
    – Ben Bolker
    Commented Jan 29, 2014 at 23:43
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    $\begingroup$ @BenBolker Another reference (see also this and this) for the use of AIC with non nested models, as long as you consider all the normalising constants as well as non-pathological models. In the context of LMM, however, you have to use some modifications of the AIC. $\endgroup$ Commented Jan 29, 2014 at 23:52
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    $\begingroup$ Link mangled: I think stats.ox.ac.uk/~ripley/ModelChoice.pdf should work. $\endgroup$
    – Ben Bolker
    Commented Jan 30, 2014 at 20:38
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    $\begingroup$ @BenBolker Well, Brian Ripley is quite opinionated. However, he hasn't provided a devastating argument against the use of AIC for non nested models :). Sorry for repeating your link. $\endgroup$ Commented Jan 30, 2014 at 23:46

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