Chapter 11 Exercises
Leukaemia: semiparametric survival models
Solutions
For the leukaemia data in Example 11.7.2, compare the piecewise constant hazard models to a simple exponential survival model with a constant hazard. Is there any benefit in the piecewise constant models?
model{
for (i in 1:N) {
for (j in 1:T) {
Y[i,j] <- step(obs.t[i] - t[j] + eps)
dN[i,j] <- Y[i,j]*step(t[j+1] - obs.t[i] - eps)*ind[i]
}
}
dt[1] <- t[1]
for (j in 2:(T+1)) {
dt[j] <- t[j] - t[j-1]
}
for (j in 1:T) {
for (i in 1:N) {
dN[i,j] ~ dpois(Idt[i,j])
Idt[i,j] <- Y[i,j] * exp(beta*Z[i]) * lam * dt[j]
}
}
cumhaz.treat[1] <- 0
cumhaz.placebo[1] <- 0
for (j in 2:(T+1)) {
cumhaz.treat[j] <- cumhaz.treat[j-1] + lam * dt[j]*exp(beta*-0.5)
cumhaz.placebo[j] <- cumhaz.placebo[j-1] + lam * dt[j]*exp(beta*0.5)
S.treat[j] <- exp(-cumhaz.treat[j])
S.placebo[j] <- exp(-cumhaz.placebo[j])
}
lam ~ dgamma(0.001, 0.001)
beta ~ dnorm(0.0, 0.000001)
dummy <- period4[1]
}
list(N=42, T=23, eps=1.0E-10,
obs.t=c(1,1,2,2,3,4,4,5,5,8,8,8,8,11,11,12,12,15,17,22,23,
6,6,6,6,7,9,10,10,11,13,16,17,19,20,22,23,25,32,32,34,35),
ind=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
1,1,1,1,1,0,1,0,1,0,0,1,1,0,0,0,1,1,0,0,0,0,0),
Z=c(0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,
0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,
-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,
-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5,-0.5),
t=c(1,2,3,4,5,6,7,8,9,10,11,12,13,15,16,17,19,20,22,23,25,32,34,35),
period4=c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4))
list( beta = 0.0, lam = 0.1 )
node mean sd MC error 2.5% median 97.5% start sample
S.placebo[2] 0.8912 0.02237 1.998E-4 0.8437 0.8923 0.9315 100 5901
S.placebo[3] 0.7948 0.03972 3.552E-4 0.7118 0.7963 0.8676 100 5901
S.placebo[4] 0.7093 0.05295 4.742E-4 0.6005 0.7106 0.8081 100 5901
S.placebo[5] 0.6333 0.0628 5.635E-4 0.5067 0.6341 0.7527 100 5901
S.placebo[6] 0.5659 0.06989 6.287E-4 0.4275 0.5658 0.7011 100 5901
S.placebo[7] 0.5059 0.07474 6.744E-4 0.3606 0.5049 0.6531 100 5901
S.placebo[8] 0.4525 0.07777 7.041E-4 0.3043 0.4505 0.6083 100 5901
S.placebo[9] 0.405 0.07935 7.212E-4 0.2567 0.402 0.5666 100 5901
S.placebo[10] 0.3628 0.07976 7.281E-4 0.2166 0.3588 0.5278 100 5901
S.placebo[11] 0.3251 0.07925 7.268E-4 0.1827 0.3201 0.4916 100 5901
S.placebo[12] 0.2915 0.07802 7.193E-4 0.1542 0.2857 0.4579 100 5901
S.placebo[13] 0.2615 0.07624 7.067E-4 0.1301 0.2549 0.4265 100 5901
S.placebo[14] 0.2108 0.07155 6.712E-4 0.09258 0.203 0.3701 100 5901
S.placebo[15] 0.1895 0.06883 6.499E-4 0.07811 0.1811 0.3447 100 5901
S.placebo[16] 0.1704 0.06598 6.272E-4 0.0659 0.1616 0.3211 100 5901
S.placebo[17] 0.138 0.06008 5.793E-4 0.04691 0.1287 0.2786 100 5901
S.placebo[18] 0.1242 0.05712 5.549E-4 0.03957 0.1148 0.2595 100 5901
S.placebo[19] 0.1009 0.05134 5.064E-4 0.02817 0.09145 0.2251 100 5901
S.placebo[20] 0.09105 0.04855 4.827E-4 0.02376 0.0816 0.2097 100 5901
S.placebo[21] 0.07419 0.04326 4.37E-4 0.01692 0.06498 0.1819 100 5901
S.placebo[22] 0.03682 0.02812 3.007E-4 0.005147 0.02928 0.1107 100 5901
S.placebo[23] 0.03027 0.02475 2.689E-4 0.003664 0.02331 0.09601 100 5901
S.placebo[24] 0.02746 0.02321 2.542E-4 0.003091 0.0208 0.08943 100 5901
S.treat[2] 0.975 0.008222 1.184E-4 0.9562 0.976 0.9886 100 5901
S.treat[3] 0.9508 0.01599 2.305E-4 0.9143 0.9526 0.9773 100 5901
S.treat[4] 0.9272 0.02333 3.366E-4 0.8742 0.9297 0.9662 100 5901
S.treat[5] 0.9042 0.03026 4.37E-4 0.8359 0.9074 0.9552 100 5901
S.treat[6] 0.8819 0.0368 5.32E-4 0.7993 0.8856 0.9443 100 5901
S.treat[7] 0.8602 0.04297 6.218E-4 0.7643 0.8644 0.9335 100 5901
S.treat[8] 0.8391 0.04878 7.065E-4 0.7308 0.8437 0.9228 100 5901
S.treat[9] 0.8185 0.05426 7.866E-4 0.6988 0.8234 0.9123 100 5901
S.treat[10] 0.7985 0.05941 8.62E-4 0.6682 0.8037 0.9019 100 5901
S.treat[11] 0.7791 0.06425 9.332E-4 0.6389 0.7844 0.8916 100 5901
S.treat[12] 0.7602 0.0688 0.001 0.6109 0.7655 0.8815 100 5901
S.treat[13] 0.7417 0.07308 0.001063 0.5841 0.7472 0.8714 100 5901
S.treat[14] 0.7064 0.08083 0.001178 0.5341 0.7118 0.8516 100 5901
S.treat[15] 0.6894 0.08435 0.00123 0.5107 0.6947 0.8419 100 5901
S.treat[16] 0.6729 0.08763 0.001279 0.4883 0.678 0.8323 100 5901
S.treat[17] 0.6412 0.09354 0.001368 0.4465 0.6459 0.8134 100 5901
S.treat[18] 0.6259 0.0962 0.001408 0.4269 0.6304 0.8042 100 5901
S.treat[19] 0.5966 0.1009 0.00148 0.3903 0.6005 0.7859 100 5901
S.treat[20] 0.5826 0.1031 0.001512 0.3732 0.5861 0.777 100 5901
S.treat[21] 0.5555 0.1068 0.001569 0.3412 0.5583 0.7593 100 5901
S.treat[22] 0.4713 0.1156 0.001707 0.2494 0.471 0.7007 100 5901
S.treat[23] 0.4499 0.1171 0.001732 0.228 0.4487 0.6848 100 5901
S.treat[24] 0.4396 0.1177 0.001742 0.218 0.4379 0.677 100 5901
beta 1.55 0.4107 0.005211 0.791 1.542 2.391 100 5901
lam 0.05299 0.01065 1.478E-4 0.03431 0.05226 0.07563 100 5901
Dbar Dhat pD DIC
dN 214.881 212.856 2.025 216.905
total 214.881 212.856 2.025 216.905
Judging from a DIC of around 217, compared to 221 for the piecewise constant model, there does not seem to be a significant improvement in fit from allowing the hazard to vary over time in this way. Note that even under the model with four hazard pieces, the posterior distributions of the four different hazards lam[1],...,lam[4] are very similar, except perhaps that there are higher hazards in the last of the four periods. Since this dataset is small, it is not surprising that there is insufficient power to identify a much more complex model.