Let $\alpha$ denote the acceptance function of the Markov chain $(X_n)_{n\in\mathbb N_0}$ generated by the Metropolis-Hastings algorithm with proposal kernel $Q$ target distribution $\mu$$^1$, $(Y_n)_{n\in\mathbb N}$ denote the corresponding proposal sequence, $f\in\mathcal L^2(\mu)$ and $$W_nf:=\frac1n\sum_{i=1}^n((1-\alpha(X_{i-1},Y_i))f(X_{i-1})+\alpha(X_{i-1},Y_i)f(Y_i))\;\;\;\text{for }n\in\mathbb N\tag1.$$
I'm searching for a central limit theorem yielding that $$\sqrt nW_n f\xrightarrow{n\to\infty}\mathcal N(0,\sigma^2(f))\tag2$$ for some $\sigma^2(f)\ge0$.
This is my approach so far: $$Z_n:=(X_{n-1},Y_n)\;\;\;\text{for }n\in\mathbb N$$ is a time-homogeneous Markov chain with transition kernel \begin{equation}\begin{split}&\kappa_{\text{aug}}((x,y),A\times B)\\&\;\;\;\;:=(1-\alpha(x,y))\delta_x(A)Q(x,B)+\alpha(x,y)\delta_y(A)Q(y,B)\;\;\;\text{for }x,y\in E\text{ and }A,B\in\mathcal E\end{split}\end{equation} and stationary distribution $\nu:=\mu\otimes Q$. Let $$A_ng:=\frac1n\sum_{i=1}^ng(Z_i)\;\;\;\text{for }g\in L^2(\nu)\text{ and }n\in\mathbb N$$ and $$H((x,y),\;\cdot\;):=(1-\alpha(x,y))\delta_{(x,\:y)}+\alpha(x,y)\delta_{(y,\:x)}\;\;\;\text{for }x,y\in E.$$
Note that $Q$ may be treated as a linear operator from $L^2(\mu)$ to $L^2(\mu)$ with adjoint $$(Q^\ast\varphi)(x,y):=\varphi(x)\;\;\;\text{for }\varphi\in L^2(\mu)$$ and $$\kappa_{\text{aug}}^n=HQ^\ast\kappa^{n-1}Q\;\;\;\text{for all }n\in\mathbb N\tag3.$$ My idea is now to let $f_0:=f-\mu f$ and $g:=HQ^\ast f_0$ so that $$A_ng=W_nf\;\;\;\text{for all }n\in\mathbb N.\tag4$$ From general theory we now know that a central limit holds under the Maxwell-Woodroofe condition $$\sum_{N=1}^\infty N^{-\frac32}\left\|\sum_{n=0}^{N-1}\kappa_{\text{aug}}^ng\right\|_{L^2(\nu)}<\infty\tag5$$ yielding that $$\sqrt nA_ng\xrightarrow{n\to\infty}\mathcal N(0,\sigma_{\text{aug}}^2(g))\tag6,$$ where $$\sigma_{\text{aug}}^2(g):=\lim_{n\to\infty}n\operatorname{Var}[A_ng]=\left\|g\right\|_{L^2(\nu)}^2+2\sum_{n=1}^\infty\left\langle\kappa_{\text{aug}}^ng,g\right\rangle_{L^2(\nu)}\tag7.$$ Are we able to simplify the expression for the asymptotic variance $\sigma_{\text{aug}}^2(g)$ in the present situation?
We may note that, by $(3)$, $$\left\langle\kappa_{\text{aug}}^ng,g\right\rangle_{L^2(\nu)}=\left\langle HQ^\ast\kappa^nf_0,HQ^\ast f_0\right\rangle_{L^2(\nu)}\;\;\;\text{for all }n\in\mathbb N_0.\tag8$$ Moreover, $\nu$ is reversible with respect to $H$ and hence $H$ is $L^2(\nu)$-self-adjoint.
$^1$ To be precise, let
- $(E,\mathcal E,\lambda)$ be a measure space;
- $p$ be a probability density on $(E,\mathcal E,\lambda)$ and $\mu:=p\lambda$;
- $q:E^2\to[0,\infty)$ be ${\mathcal E}^{\otimes2}$-measurable with $$\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)\delta_x(B)\;\;\;\text{for }(x,B)\in E\times\mathcal E.$$