Writing log likelihood for WAIC (logistic hierarchal stan model)

Question

I'm creating a new model and I want to compare this with another model using WAIC. I understand that I need to write a generated quantities block. However, I'm struggling to convert the logsumexp of beta. I would greatly appreciate any leads/help. My model block looks like this:

model { 

    //prior for phi,b
    phi ~ cauchy(0,5);
    mu_b ~ normal(0,1);
    sigma_b ~ cauchy(0,1);
    mu ~ normal(0,1);
    sigma ~ cauchy(0,1);

   //model
   log_b_z ~ normal(0, 1);
   theta_raw ~ normal(mu, sigma);

   for (i in 1:n) {
       vector[number_segments] test;    
       for (j in 1:number_segments) {
           test[j] = beta_lpdf(response[i] | p[j][i]*phi, (1-p[j][i])*phi) + log(prob_segment[j]);
       }    
       target += log_sum_exp(test);
   }
}

Have a look at R library [`loo`](https://cran.r-project.org/web/packages/loo/index.html), which provides a method for calculating WAIC based on `stan` models. — Maurits Evers, Mar 07 '18 at 02:25

score 1 · Accepted Answer · answered Mar 07 '18 at 17:58

You need to define a generated quantities block that defines your posterior predictive log likelihood for each data point.

You can do it this way for a mixture with minimal recomputation.

transformed parameters {
  vector[n] log_lik;      
  {
    vector[number_semgnents log_prob_segment = log(prob_segment);
    for (i in 1:n) {
      vector[number_segments] lp = log_prob_segment;
      for (j in 1:number_segments) {
        lp[j] += beta_lpdf(response[i] | p[j, i] * phi, (1 - p[j, i]) * phi);
      log_lik[i] = log_sum_exp(lp);       
   }
}
...
model {
  target += sum(log_lik);
...

You could also define log_lik as a generated quantity---that can be more efficient if you can vectorize the likelihood (which isn't possible yet for mixtures in Stan).

Once you've done that, you can use the loo package to calculate WAIC, etc., as described in the vignette and references.

Writing log likelihood for WAIC (logistic hierarchal stan model)

1 Answers1