1
$\begingroup$

In the case of the Bayesian estimation of GARCH(1, 1) model with Student–t or a Skewed distributions for innovations, is it more correct to assume a uniform distribution for the parameters or to assume the truncated Normal prior?

In other word, what is the difference between the following two settings that are usually found in the literature? Why do we assume a normal as a prior distribution?

    model{ 
  for (t in 1:N){
    y[t] ~dt(mu, tau[t], nu)
    a[t] <- y[t] - mu
    tau[t] <-1/h[t]
  }
  for (t in 2:N){
    h[t] <- alpha0 +alpha1 * pow(a[t-1],2)+ beta1 * h[t-1]
  }

  #prior

  mu ~ dnorm(0, 0.001)
  h[1] ~ dunif(0, 0.0012)
  alpha0 ~ dunif(0, 0.2)
  alpha1 ~ dunif(0.00001, 0.8)
  beta1 ~ dunif(0.00001, 0.8)
  nu ~ dunif(1,30)

or

#prior 2
       mu ~ dnorm(0, 0.001)
  h[1] ~ dunif(0, 0.0012)
  alpha0 ~ dnorm(0, 0.001)I(0,)
  alpha1 ~ dnorm(0.00001, 0.001)I(0,)
  beta1 ~ dnorm(0.00001, 0.001)I(0,)
  nu ~ dnorm(0,0.001)I(0,)
$\endgroup$
1
$\begingroup$

From a Bayesian perspective, the question does not make sense as there is no "better" or "worse" (or "correct" versus "incorrect") prior. The entire statistical analysis is based on and hence conditional on the choice of the prior as a reference measure. Provided the posterior is proper, both analyses are coherent. If one wants to check which prior is most compatible with the data, one can use a Bayes factor to this effect, but the resulting "prior" is data-dependent and therefore not a "prior" any longer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.