## Not understood [R for BE/BA]

Hello!

» Sounds neat, will check if I can somehow do such a thing.

I found Lindstrom&Bates, equation 3.4 or J. GURKA, 2006, equation 5.

Some experiment:

m = matrix(c(5, 1, 0, 0, 1, 2, 0, 0, 0, 0, 5, 3, 0, 0, 3, 4), nrow = 4, ncol = 4) m      [,1] [,2] [,3] [,4] [1,]    5    1    0    0 [2,]    1    2    0    0 [3,]    0    0    5    3 [4,]    0    0    3    4 log(det(m))  4.59512 log(det(m[1:2, 1:2])) + log(det(m[3:4, 3:4]))  4.59512 solve(m)            [,1]       [,2]       [,3]       [,4] [1,]  0.2222222 -0.1111111  0.0000000  0.0000000 [2,] -0.1111111  0.5555556  0.0000000  0.0000000 [3,]  0.0000000  0.0000000  0.3636364 -0.2727273 [4,]  0.0000000  0.0000000 -0.2727273  0.4545455 solve(m[1:2, 1:2])            [,1]       [,2] [1,]  0.2222222 -0.1111111 [2,] -0.1111111  0.5555556 solve(m[3:4, 3:4])            [,1]       [,2] [1,]  0.3636364 -0.2727273 [2,] -0.2727273  0.4545455

So Vi can be result of function ƒ(Zi, θ); Zi is known for all subjects, θ - vector of optimizables parameters.

» My approach is very simple: I make a case for saying that when T is not replicated while R is
» replicated, then it is not possible to make a meaningful model when V=ZGZt+R (as long as we count on R having the withins and G having the betweens).

Mmm... I can't say that ZGZ'+R is meaningful. If you construct V manualy it can still be represented as ZGZ'+R if ρ not equal 0. If used random model with covariation (ρ≠0) nondiagonal elements of G depends on diagonal and when you construct V some nondiagonal elements depends on both variance parameters. Then we add diagonal R matrix and have combination (parts of V): σ2BT2WT, σ2BR2WR, σ2BR, σBTBR*ρ (we have no σ2BT) so we can estimate σ2WT from σ2BT2WT and σ2BT is stable because it is part of σBTBR*ρ (Yes estimate of σ2WT is indirect). If ρ = 0 we have σBTBR*ρ = 0, and we can estimate only σ2BT2WT.

And I think that V can be decomposed to ZGZ'+R if you know Z, ρ (ρ≠0) and σ2BR in model with covariance and diagonal R. Ing. Helmut Schütz 