I am running two MCMC chains (say chain A and chain B) in parallel, using the Metropolis-Hastings algorithm with acceptance probability:

$P(accept\ x_t) = \min\{1, f(x_t)/f(x_{t-1})\}$.

I would like my proposal distribution to be adaptive, but I know that if I use old values of one chain to choose the proposal (i.e $p(x_t|x_{t-1}) = g(x_1, \dots, x_{t-1})$), I will lose the Markov property (see this post for example), because $p(x_t|x_{t-1}, \dots, x_{1}) \neq p(x_t|x_{t-1})$.

So I'm running two chains in parallel (same target distribution), chain A uses a fixed (non adaptive) proposal while chain B is using an adaptive proposal which is a function of the past values of chain A (i.e. $p(x^B_t|x^B_{t-1}) = g(x^A_1, \dots, x^A_{t-1})$).

To me it looks like the algorithm is correct, because $p(x^B_t|x^B_{t-1}, \dots, x^B_{1}) = p(x^B_t|x^B_{t-1})$, given that the proposal for B is not using past values of the same.

My questions are:

1) Is chain B still Markovian?

2) Even though the two chains have the same target distribution, is using one chain to adapt the other one a good idea? Maybe not because in the initial phase the two chains might be in different area of the parameter space.


  • $\begingroup$ I don't see how a chain could be both Markovian and adaptive. $\endgroup$ – jerad Mar 2 '13 at 22:29
  • $\begingroup$ @jerad The proposal distribution is adaptive, not the chain. It is possible to have an adaptive proposal and a Markov chain, as shown in some of the references here: stats.stackexchange.com/questions/7286/… $\endgroup$ – Matteo Fasiolo Mar 3 '13 at 12:44

Your Answer

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

Browse other questions tagged or ask your own question.