# Is the stationary distribution of the augmented Metropolis-Hastings kernel even reversible in the symmetric proposal case?

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 $$A_1,B_1,A_2,B_2\in\mathcal E$$.