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Let $\alpha$ denote the acceptance function of the Markov chain generated by the Metropolis-Hastings algorithm with proposal kernel $Q$ and target distribution $\mu$$^1$ and $$\kappa_{\text{aug}}((x,y),A\times B):=1_A(x)(1-\alpha(x,y))Q(x,B)+1_A(y)\alpha(x,y)Q(y,B)$$ for $x,y\in E$ and $A,B\in\mathcal E$.

In general, $$\nu:=\mu\otimes Q$$ is invariant but not reversible$^2$ with respect to $\kappa_{\text{aug}}$. Are we able to show the reversibility under the assumption that $q$ is symmetric?


$^1$ To be precise, let

  • $(E,\mathcal E,\lambda)$ be a measure space;
  • $p$ be a probability density on $(E,\mathcal E,\lambda)$ and $\mu:=p\lambda$;
  • $q:E^2\to[0,\infty)$ be ${\mathcal E}^{\otimes2}$-measurable with $$\int\lambda({\rm d}y)q(x,y)=1\;\;\;\text{for all }x\in E$$ and $$Q(x,\;\cdot\;):=q(x,\;\cdot\;)\lambda\;\;\;\text{for }x\in E;$$
  • $$\alpha(x,y):=\left.\begin{cases}\displaystyle1\wedge\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)\ne0\\1&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x,y\in E$$ and $$\kappa(x,B):=\int_BQ(x,{\rm d}y)\alpha(x,y)+\left(1-\int Q(x,{\rm d}y)\alpha(x,y)\right)1_B(x)\;\;\;\text{for }(x,B)\in E\times\mathcal E.$$

$^2$ i.e. $$\int_{A_1\times B_1}\nu({\rm d}(x,y))\kappa_{\text{aug}}((x,y),A_2\times B_2)=\int_{A_2\times B_2}\nu({\rm d}(x,y))\kappa_{\text{aug}}((x,y),A_1\times B_1)\tag1$$ for all $A_1,B_1,A_2,B_2\in\mathcal E$.

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