In MCMC, can I accept proposals from another MCMC process without trying to approximate the proposal distribution? I'm trying to sample a Markov chain which takes proposal from another Markov chain. Normally one would have a proposal distribution one could sample from. However, say in this case the proposals come from a complicated Bayesian Network for which it is infeasible to infer the proposal distribution. Sampling the parameters of the Bayesian Network (using Gibb's say) could approximate the proposal distribution. Could we instead of approximating the proposal distribution use the samples from the Gibb's process as candidates in the outer Markov chain? Would the overall process still converge to the true distribution?
 A: If I understand correctly the problem, taking $\pi$ as the target density and $Q(\theta,\theta^\prime)$ as the proposal Markov kernel, an MCMC algorithm that would be associated with the transition at time $t$:
$$\theta^{(t+1)}=\begin{cases}
\theta^\prime\sim Q(\theta_{(t)},\theta^\prime) &\text{with probability } \dfrac{\pi(\theta^\prime)}{\pi(\theta_{(t)}}\,\dfrac{Q(\theta^\prime,\theta_{(t)})}{Q(\theta_{(t)},\theta^\prime)}\wedge 1\\
\theta_{(t)} &\text{otherwise}
\end{cases}$$
would be the standard Metropolis-Hastings algorithm. Moving to the transition

*

*Select a component index $i$ with uniform probabilities

*Take
$$\theta^{(t+1)}=\begin{cases}
\theta^\prime\sim Q_i(\theta_{(t)},\theta^\prime)&\text{with probability } \dfrac{\pi(\theta^\prime)}{\pi(\theta_{(t)}}\,\dfrac{Q_i(\theta^\prime,\theta_{(t)})}{Q_i(\theta_{(t)},\theta^\prime)}\wedge 1\\
\theta_{(t)} &\text{otherwise}
\end{cases}$$
where $Q_i(\cdot,\cdot)$ denotes the conditional kernel on the $i$-th component of $\theta$ associated with $Q(\cdot,\cdot)$, would work. Were the conditional kernel symmetric, the ratio would then cancel. If in the opposite, it is assymmetric and the densities $Q_i(\cdot,\cdot)$ and $Q(\cdot,\cdot)$ are intractable, I see no generic algorithm taking advantage of this conditional proposal.

If the stationary distribution of $Q_i(\theta_{(t)},\cdot)$, say $\tilde\pi_i(\theta|\theta_{(t)})$, is known, then the step
$$\theta^{(t+1)}=\begin{cases}
\theta^\prime\sim Q_i(\theta_{(t)},\theta^\prime)&\text{with probability } \dfrac{\pi(\theta^\prime)}{\pi(\theta_{(t)}}\,\dfrac{\tilde\pi_i(\theta_{(t)}|\theta^\prime)}{\tilde\pi_i(\theta^\prime|\theta_{(t)})}\wedge 1\\
\theta_{(t)} &\text{otherwise}
\end{cases}$$
would work as well.
