If we run the Metropolis-Hastings algorithm for a target distribution $\mu$ with proposals from a quasi-Monte Carlo sequence $(y_n)_{n\in\mathbb N}$ (such as a Sobol sequence) and the generated chain is denoted by $(X_n)_{n\in\mathbb N_0}$, will $$\frac1n\sum_{i=0}^{n-1}f(X_i)\xrightarrow{n\to\infty}\int f\:{\rm d}\mu\tag1$$ still hold?
The obvious question is: How do we need to compute the acceptance probability $\alpha$? Since the proposals are non-random, they don't have a density with respect to the reference measure with respect to which $\mu$ has a density $p$. So, do we simply use $$\alpha(x,y)=\min\left(1,\frac{p(y)}{p(x)}\right)\tag2$$ as in the "random walk Metropolis-Hastings algorithm" (where the proposal density is assumed to be symmetric)?
I was able to find two works on this subject:
- In his thesis MARKOV CHAIN MONTE CARLO ALGORITHMS USING COMPLETELY UNIFORMLY DISTRIBUTED DRIVING SEQUENCES Tribble proves consistency of the Metropolis-Hastings estimator when the input sequence of uniformly distributed random numbers is replaced by a "completely uniformly distributed" sequence. However, his result needs is established only under the assumption of a finite state space.
- Tribble also published a paper A quasi-Monte Carlo Metropolis algorithm together with Owen. I think they described what they intend to do in equations $(4)$ and $(5)$, but I honestly don't get what they do and I don't find an answer to the question I've raised above about the definition of $\alpha$. What would their $\Psi_i$ look like, when the quasi-Monte Carlo sequence is, say, a Sobol sequence?