Hi

I’v fitted a neg binom model using Stan’s function neg_binomial_2_lpmf(*y* | *mu*, *phi*). *phi* is the precision. Now I would like to create posterior predictive samples in R using rnbinom(*n*, *size*, *prob*, *mu*). It seems clear that the *mu* from neg_binomial_2_lpmf is the same as for rnbinom. Then, I guess I provide *size*, the “dispersion parameter” according to the rnbinom help file. But how is the link between the precision *phi* and the dispersion parameter *size*? The help file says “The variance is mu + mu^2/size”. I can solve variance = *mu*+*mu*^2/*size* = 1/precision = 1/*phi* for *size*, but then I get negative values for *size*, hence something is wrong here. Is *phi* simply 1/*size*?

many thanks, Pius

Nooo! You would like to create posterior predictive samples in a generated quantities block in Stan! Switching between probability distributions like this is error prone anyway. You can avoid that with generated quantities!

I got \phi = n which agrees with you. \phi is required to be positive in Stan and n is required to be positive in R, so where does the negative size come from?