The effect of weights on the resulting estimates

Modeling
1

I am hoping to understand how weights will effect the fit that STAN develops. I know that weights aren’t full Bayesian, but I’m willing to sacrifice that.

Suppose I have two Bernoulli observations that are a function of a latent variable, and weight them differently in the model. Will this mean that the model will optimize to ‘care’ about the higher-weighted observation more, and/or by multiplying the log-liklihood by a weight am I implicitly changing the value of the probability (in the log liklihood).

For example, in the following model:

data{
  int result[2];
  vector[2] weights;
  vector[2] breaks;


}

parameters{
  real mu;

}

model {
  real p;
  for(i in 1:2){
    p = normal_cdf(breaks[i],mu,3);
    target += bernoulli_lpmf(result[i]|p) * weights[i];
}
/// the log density of a bernoulli is  x *log(p) + (1-x)*log(1-p) where x is the result
/// if I multiply this by a weight, am I implicitly telling STAN that the probability is actually p ^ weights[i] ?



}

Additionally, suppose weights[1] = .5 and weights[2] = 1 ; then will this Stan model care more about the 2nd observation than it will about the 1st observation?

Thanks, appreciate any help!

2

if I multiply this by a weight, am I implicitly telling STAN that the probability is actually p ^ weights[i] ?

Stan samples from a distribution defined by the unnormalized log density target.

So if before your model was:

target += log(q(theta));

And now your model is:

target += log(q(theta)) * w;

Then before you were generating samples from a distribution proportional to q(theta) and now you are generating samples from a distribution proportional to q(theta)^w (assuming these are proper distributions and can be normalized and Stan is mixing and whatnot).

Well Stan just samples whatever the target defines. In this case the model weights the first likelihood half as much as the second, yes.