Lower and upper limits of deterministic node

[Luis Baroja]

Hi everyone,
I'm trying to truncate deterministic node theta.t to ensure its values lie between 0 and 1.
I've tried to solve the issue by employing T(0,1) function at the end of line 6, which doesn't work, and I'm not sure how to use ~dinterval() to address this particular problem.
Here the code:

model{
  # PRIORS
  alpha~dunif(0,1)
  beta~dunif(0,1)
  # Deterministic Node
  for(j in 1:n.intervals){
    theta.t[j]<-exp(-alpha*retention.intervals[j])+beta
  }
  # DATA
  for(i in 1:n.participants){
    for(j in 1:n.intervals){
      k[i,j]~dbin(theta.t[j],18)
    }
  }
  }

Thanks in advance!

[Martyn Plummer]

You can either add a data step (as below) or supply these values with the data

data {
   breaks[1] <- 0
   breaks[2] <- 1
   for (j in 1:n.intervals) {
       trunc[j] <- 1
   }
}
model {
   ...
}

Then in the model you add

    for(j in 1:n.intervals) {
        trunc[j] ~ dinterval(theta.t[j], breaks)
    }

This adds the information that theta.t[j] is observed to lie between breaks[1] (0) and breaks[2] (1).

Note that formally this is censoring ([i]a posteriori[/i] constraint) not truncation ([i]a priori[/i] constraint). Happily in your example neither alpha nor beta have unobserved parents, so censoring and truncation are equivalent. You should not do this in a model where you are trying to learn about hyper-parameters of alpha and beta.