Using segment function for bernoulli_logit

Modeling
1

Hello,
I am trying to use the segment function to break up my ragged occupancy dataset. However, it appears the data is not being passed through the bernoulli_logit§ portion of the model.
Thank you

data {
  int<lower=0> N; //of observations/ # temporal replication
  int<lower=0> K; // of groups/ # of arrays
  int pos[K]; // position of the first detection of each group
  int y[N]; // observations/ Detection at each site for all temporal replication
  int s[K]; // group size s={45,42,42,41,54,...}
  int x[K]; // Observed occupancy at each site
  
  vector[K] t_burn;
  vector[N] air_temp; 
}


parameters {
  real a_psi;
  real b1_psi;
  
  real a_p; 
  real b1_p;
  
}

transformed parameters {
  vector[K] logit_psi;
  vector[N] logit_p;
  logit_psi = a_psi + b1_psi * t_burn;
  logit_p = a_p + b1_p * air_temp;
 
}
model {
  // Priors
  a_psi ~ normal(0, 1); 
  b1_psi ~ normal(0,1); 

  a_p ~ normal(0, 1);
  b1_p ~ normal(0, 1);

// Likelihood

  for (i in 1:K) {
      if (x[i] == 1) {    // Occurred and observed
      1 ~ bernoulli_logit(logit_psi[i]);
      segment(y, pos[i], s[i]) ~ bernoulli_logit(logit_p[i]); 
    } else {
                            // Occurred and not observed
      target += log_sum_exp(bernoulli_logit_lpmf(1 | logit_psi[i])
                            + bernoulli_logit_lpmf(0 | logit_p[i]),
                            // Not occurred
                            bernoulli_logit_lpmf(0 | logit_psi[i]));
    }
  }
}
2

I guess it’s a bit crude, but the easiest way I can think of testing this is to add print statements:

print(segment(y, pos[i], s[i]));
segment(y, pos[i], s[i]) ~ bernoulli_logit(logit_p[i]);

If this is getting evaluated and the segment is indexing stuff, you’ll see a lot of numbers get printed out, if this either isn’t getting evaluated or the indexing isn’t doing what you expect, you won’t.

3

Thanks bbbales2,

The segment function is indexing correctly. It isn’t getting evaluated how I expect.