I've just started to learn to use Stan and rstan. Unless I've always been confused about how JAGS/BUGS worked, I thought you always had to define a prior distribution of some kind for every parameter in the model to be drawn from. It appears that you don't have to do this in Stan based on its documentation though. Here's a sample model that they give here.

data {
  int<lower=0> J; // number of schools 
  real y[J]; // estimated treatment effects
  real<lower=0> sigma[J]; // s.e. of effect estimates 
parameters {
  real theta[J]; 
  real mu; 
  real<lower=0> tau; 
model {
  theta ~ normal(mu, tau); 
  y ~ normal(theta, sigma);

Neither mu nor tau have priors defined. In converting some of my JAGS models to Stan, I've found that they work if I leave many, or most, parameters with undefined priors.

The problem is that I don't understand what Stan is doing when I have parameters without defined priors. Is it defaulting to something like a uniform distribution? Is this one of the special properties of HMC, that it doesn't require a defined prior for every parameter?


2 Answers 2


From (an earlier version of) the Stan reference manual:

Not specifying a prior is equivalent to specifying a uniform prior.

A uniform prior is only proper if the parameter is bounded[...]

Improper priors are also allowed in Stan programs; they arise from unconstrained parameters without sampling statements. In some cases, an improper prior may lead to a proper posterior, but it is up to the user to guarantee that constraints on the parameter(s) or the data ensure the propriety of the posterior.

(See also section C.3 in the 1.0.1 version).

The underlying reason this is okay in Stan but not in BUGS might have to do with the fact that in BUGS, your model "program" is specifying a formal graphical model, while in Stan you're writing a little function to calculate the joint probability density function. Not specifying a proper prior for all variables might screw up the nice formal properties of graphical models.

However, for Hamiltonian MC you just need to (numerically) calculate the joint density function. A flat (even improper) prior only contributes a constant term to the density, and so as long as the posterior is proper (finite total probability mass)—which it will be with any reasonable likelihood function—it can be completely ignored in the HMC scheme.


From the Stan reference v1.0.2 (pg 6, footnote 1)

If no prior were specified in the model block, the constraints on theta ensure it falls between 0 and 1, providing theta an implicit uniform prior. For parameters with no prior specified and unbounded support, the result is an improper prior. Stan accepts improper priors, but posteriors must be proper in order for sampling to succeed.

Both mu and sigma have improper uniform priors.

Under the hood, mu and sigma are treated differently. sigma is defined with a lower bound; Stan samples from log(sigma) (with a Jacobian adjustment for the transformation). For more details on transformations, see Chapter 27 (pg 153).

  • $\begingroup$ Just so I'm clear about this, if STAN samples on the log(sigma) level, the flat prior is still over sigma and not over log(sigma)? $\endgroup$ Nov 25, 2013 at 15:04
  • 1
    $\begingroup$ Yes, the uniform prior is still over sigma, not log(sigma). Stan accomplishes this by transforming the parameter and applying the correct change of variables adjustment by using the Jacobian. $\endgroup$
    – syclik
    Jun 30, 2014 at 21:13

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