Integrating a normal RV in Rstan during sampling

Modeling
1

I am trying to do integration in my Stan program.
To this end, I have the following example mode:

data { 
  int<lower=0> N;
  vector[N] y;
  real y_prior;
}
parameters {
  real mu_y;
}
model {
  mu_y ~ normal(y_prior, 1/N);
  y    ~ normal(mu_y, 1);
}

Now, at each sample iteration, I would like to find the CDF/integrate y up to some value (say 0 for simplicity).
Is there any way to do this?
Can one simply add:

generated quantities {
  real y_integ;
  y_integ = normal_cdf(0, mu_y, 1);
}

?
Some experimenting has made me question if this is the correct way of doing it or not.

The model:

\mu_y \sim N(y_{prior}, 1/N)
y \sim N(\mu_y, 1)

Let \Phi_y be the CDF of y include the following term in the sampling:

y_{integ} = \Phi_y(0)
2

The definition of normal_cdf is,

\textrm{normal_cdf}(v \mid \mu, \sigma) = \displaystyle \int_{-\infty}^v \textrm{normal}(z \mid \mu, \sigma) \, \textrm{d}z.

Then \Phi(x) is just \textrm{normal_cdf}(x \mid 0, 1), so that we can also formulate this as

\displaystyle \textrm{normal_cdf}(v \mid \mu, \sigma) = \Phi\left(\frac{v - \mu}{\sigma} \right).

If you define that in generated quantities as you have done, then y_integ will get a posterior distribution of CDF values based on sampled values of mu_y. In this case, it’s a transform of the random variable mu_y.

3

Ok, thank you! I was not sure this was correct but apparently, it was. Thank you for confirming and sorry if this seemed like a trivial question.