anova() and the difference between (x | y) and (1 | y:x) in lme4
Dear Hans, I'm not sure if one can consider (A|B) and (1|B) + (1|A:B) to be nested. (1|B) + (A|B) and (1|B) + (1|A:B) are nested. (1|A:B) is the same as (A|B) with constrains on the variance-covariance matrix. Best regards, ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 Thierry.Onkelinx at inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -----Oorspronkelijk bericht----- Van: r-sig-mixed-models-bounces at r-project.org [mailto:r-sig-mixed-models-bounces at r-project.org] Namens Hans Ekbrand Verzonden: woensdag 11 juni 2014 23:17 Aan: r-sig-mixed-models at r-project.org Onderwerp: Re: [R-sig-ME] anova() and the difference between (x | y) and (1 | y:x) in lme4
On Wed, Jun 11, 2014 at 10:38:38AM -0400, Ben Bolker wrote:
On 14-06-11 10:21 AM, ONKELINX, Thierry wrote:
Dear Hans, I assume that var1 is a factor variable. The difference is in the distribution of the random effects. (1|var1:var2) : all random intercept come from the same univariate normal distribution rnorm(mean = 0, sd = sigma) (0 + var1|var2): the random intercepts come from a multivariate normal distribution: rmvnorm(mean = 0, sigma = Sigma). Sigma is a positive definite matrix (0 + var1|var2) is a bit easier to understand because the BLUP's have the same interpretation of those of (1|var1:var2) The bottom-line is that (var1|var2) and (1|var1:var2) allow the same model fit but (var1|var2) makes less assumptions at the cost of estimation more parameters. (var1|var2) requires n * (n + 1) / 2 parameters, with n = number of levels of var1. (1|var1:var2) requires just 1 parameter.
[...]
(2) the 'unstructured' (var1|var2) and 'grouped/positive compound symmetry' models (1|var1:var2) are in principle nested (all off-diagonals equal to zero, all diagonals identical -> simpler model), so you should be able to use a likelihood ratio test/ANOVA to test.
Would that hold even if I include a random intercept term for var2 (=country) in the 'grouped/positive compound symmetry' model? below.poverty.line ~ 1 + employment.type + (1 | country:employment.type) + (1 | country) + gender + age + age.2 + n.adults.minus.n.children + n.children + education + household.type
anova(fit.hierarchical, fit.flat.plus.random.intercept)
Data: my.df
Models:
fit.flat.plus.random.intercept: below.poverty.line ~ 1 + employment.type + (1 | country:employment.type) +
fit.flat.plus.random.intercept: (1 | country) + gender + age + age.2 + n.adults.minus.n.children +
fit.flat.plus.random.intercept: n.children + education + household.type
fit.hierarchical: below.poverty.line ~ 1 + employment.type + (employment.type |
fit.hierarchical: country) + gender + age + age.2 + n.adults.minus.n.children +
fit.hierarchical: n.children + education + household.type
Df AIC BIC logLik deviance Chisq Chi Df Pr(>Chisq)
fit.flat.plus.random.intercept 19 38808 38988 -19385 38770
fit.hierarchical 38 38804 39163 -19364 38728 42.161 19 0.001686 **
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