Can a Metropolis-Hastings estimator converge if the proposal density vanishes whenver the target density vanishes? I'm running the Metropolis-Hastings algorithm for a target and proposal density $p$ and $q$, respectively, for which it holds $$p(y)=0\Rightarrow q(x,y)=0\;\;\;\text{for all }y.\tag1$$
By definition of the Metropolis-Hastings acceptance function $\alpha$, $$\alpha(x,y)=\left.\begin{cases}\displaystyle1\wedge\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)>0\\1&\text{, otherwise}\end{cases}\right\},$$ this means that every proposal $y$ with $p(y)=0$ will be accepted. Won't this prevent convergence to the equilibrium distribution?
I would like to use the importance sampling estimator described in this paper, but $(1)$ is obviously a contradiction to the crucial Assumption 1 in the paper. Can we fix this by modifying the estimator in a suitable sense? I'd need that the estimator remains asymptotic unbiased (as it is the case under Assumption 1 as shown in Theorem 14 in the paper).
 A: I see no reason why this should be correct in general. The target distribution will then be a convex combination of $p$ and of the stationary distribution of $q(\cdot,\cdot)$, if any. If $q(\cdot,\cdot)$ is a transient or null recurrent Markov kernel, the chain does not even converge to a distribution.
For instance, take $p$ as a Uniform $(-1,1)$ density and $q(\cdot,\cdot)$ as the standardized Gaussian random walk kernel. The transition is then
$$
x_{t+1}=\begin{cases}
y_t\sim\mathcal N(x_t,1) &\text{if }|y_t|>1 \text{ or }u_t\le \frac{\mathbb{I}_{(-1,,1)}(y_t)q(y_t,x_t}{\mathbb{I}_{(-1,,1)}(x_t)q(x_t,y_t)}=1\\
x_t &\text{otherwise}
\end{cases}$$
which means that $$x_{t+1}=y_t\sim\mathcal N(x_t,1)$$
with probability one and the Markov chain is the (null recurrent) Gaussian random walk.
Obviously if one imposes
$$p(y)=0\Rightarrow q(x,y)=0\;\;\;\text{for all }y.\tag1$$
then the probability that $Y\sim q(x,y)$ belongs to $$\{y;\,p(y)=0\}$$
is zero, hence this specification of $\alpha(x,y)$ when $p(y)=0$ is irrelevant.
