Dear List,
I am trying to use a random effects logistic regression model to model
hospital mortality. I want to assume that the hospital-specific intercepts
follow a t-distribution, rather than the more standard normal
distribution. The hospital-specific intercepts are contained in a variable
called 'intercept'. I am running the model in Classic BUGS. When, I
compile the model, I am given the following error message:
Error in file: t.bug
for node: intercept[1]
-- error --
Value of parameter outside range 1.0E-10 .. 1.0E+10.
I am attaching the code used below.
The t-distribution is allowed to take on values over the entire real line.
Thus, I don't understand the error message. Can anyone point out my
error. Any help would be appreciated.
Thanks very much,
BUGS Code:
model mixture;
const N = 12824, #number of patients
M = 109; #number of hospitals
var totdeath[N], denom[N], age[N], female[N], shock[N], diabcomp[N], chf[N],
cvd[N], arf[N], crf[N], malig[N], carddys[N], pulmoned[N],
intercept[M], #hospital-specific random effects.
beta[11], #regression coefficients for risk factors
inst[N], #hospital id number
p[N], #patient's probability of death
d, tau; #required for t-distribution
data inst, totdeath, denom, age, female, shock, diabcomp, chf, cvd, arf,
crf,
malig, carddys, pulmoned in "omid2000.dat";
inits beta in "beta.in", intercept in "intercept.in";
{
for (i in 1:N){
totdeath[i] ~ dbin(p[i],denom[i]);
logit(p[i]) <- intercept[inst[i]] + beta[1]*age[i] +
beta[2]*female[i] + beta[3]*shock[i] + beta[4]*diabcomp[i] +
beta[5]*chf[i] + beta[6]*cvd[i] + beta[7]*arf[i] + beta[8]*crf[i] +
beta[9]*malig[i] + beta[10]*carddys[i] + beta[11]*pulmoned[i];
}
for (j in 1:M){
intercept[j] ~ dt(d,tau,4);
}
d ~ dnorm(0,1.0E-6);
tau ~ dgamma(1.0E-3,1.0E-3);
beta[1] ~ dnorm(0.07,1);
beta[2] ~ dnorm(0.13,1);
beta[3] ~ dnorm(2.73,1);
beta[4] ~ dnorm(0.28,1);
beta[5] ~ dnorm(0.27,1);
beta[6] ~ dnorm(0.53,1);
beta[7] ~ dnorm(1.22,1);
beta[8] ~ dnorm(0.30,1);
beta[9] ~ dnorm(0.86,1);
beta[10] ~ dnorm(0.18,1);
beta[11] ~ dnorm(0.34,1);
}
Thanks very much,
Peter Austin
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