R Under development (unstable) (2024-03-03 r86036 ucrt) -- "Unsuffered Consequences"
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> library(coxme)
Loading required package: survival
Loading required package: bdsmatrix

Attaching package: 'bdsmatrix'

The following object is masked from 'package:base':

    backsolve

> aeq <- function(x, y, ...) all.equal(as.vector(x), as.vector(y), ...)
> #
> # A set of tests using user-defined matrices
> #  The lung data is handy, though models of time that ignore status aren't
> #  very sensible.
> #
> fit1 <- lmekin(time ~ age + (1|ph.ecog), lung, vfixed=.5)
> 
> # Solve using the extended normal equations
> #  The loglik is sum(y - hat y)^2 + b'P b/2 where the penalty matrix
> #  is all zeros for the fixed effects, and A-inverse for the random
> #  where b is the coefficient vector and A is the variance of the random effects
> #  In our case A-inverse/2 = identity matrix.  
> # If the variance is known this equation is easy to solve.
> efit <- function(p, pmat) {
+     tlung <- na.omit(lung[,c("time", "age", "ph.ecog")])
+     cmat <- cbind(0,0, rbind(0,0, pmat))
+     x <- with(tlung, cbind(1, age, ph.ecog==0, ph.ecog==1,
+                           ph.ecog==2, ph.ecog==3))
+     as.vector(solve(t(x) %*% x + p*cmat, t(x) %*% tlung$time))
+ }
> efit1 <- efit(2, diag(4))
> 
> aeq(fixef(fit1), efit1[1:2])
[1] TRUE
> aeq(unlist(ranef(fit1)), efit1[3:6])
[1] TRUE
> 
> # Now put in a more complex pmat (linear trend constraint)
> # Since coxme is using the inverse variance as the penalty, we
> #  need to give the inverse penalty as the "variance".
> # (This is an lmekin test with non-zero off diagonal elements 
> #  in the variance for the random effect, which is key to working
> #  with kinship matrices.)
> pmat <- matrix(c(2,-1,0,0, -1,2,-1,0, 0, -1, 2, -1, 0,0,-1,2),4)
> dimnames(pmat) <- list(0:3, 0:3)
> 
> fit2 <- lmekin(time ~ age + (1|ph.ecog), lung, vfixed=.5,
+                varlist= coxmeMlist(solve(pmat), rescale=F))
> efit2 <- efit(2, pmat)
> aeq(fixef(fit2), efit2[1:2])
[1] TRUE
> aeq(unlist(ranef(fit2)), efit2[3:6])
[1] TRUE
> 
> 
> proc.time()
   user  system elapsed 
   1.46    0.12    1.59