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).

• Doesn't the condition on $q$ imply that you will never propose a move to a state where $p(y)=0$?
– πr8
Commented Feb 13, 2020 at 15:40
• @πr8 I guess you're right, but please take a look at the special setting described in my other question: stats.stackexchange.com/q/448447/222528. Consider the proposal scheme described in my edit at the beginning of the question. It seems to be possible that $q_i(y)=0$ and hence $\hat q((i,x),(j,y))=0$. Moreover, I don't see why it shouldn't hold $p(y)=0$. Am I missing something? Commented Feb 13, 2020 at 17:13

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.
• Thank you for your answer. (a) "I see no reason why this should be correct in general." Could you clarify to what precisely this refers? (b) How do you conclude that the target distribution will be a convex combination of $p$ and the "stationary distribution" of $q$? What exactly do you mean by "stationary distribution" of $q$ at all? Say the reference measure for $p$ and $q$ is $\lambda$ so that the target distribution and proposal kernel are $\mu:=p\lambda$ and $Q(x,\;\cdot\;):=q(x,\;\cdot\;)\lambda$. Commented Feb 14, 2020 at 15:15