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number of levels of grouping factor == number ofobservations

On Tue, 9 Mar 2010, Douglas Bates wrote:

            
It has to be all families.  As someone pointed out earlier, pedigreemm 
won't fit a simple univariate animal LMM unless this is allowed, and if 
one wishes to also estimate biometrical dominance, epistatic terms, 
assortment etc, then one has identified models with 2,3... random effects 
per individual, providing the data is adequate (that is the appropriate 
types of relationships are observed in the pedigrees).  For example, if 
one has pairs of identical twins, and ordinary siblings pairs in the data, 
there are two intraclass correlations which can be used for the 
decomposition into additive and dominance genetic variances rMZ=a+d, 
rSS=0.5*a+0.25*d (obviously the range of allowable r's is bounded).

Cheers, David DUffy.