Poisson-equation martingale decomposition and MCMC CLT

By [F1], each $D_{k+1}$ is a square-integrable martingale difference, so $\mathbb{E}[D_{k+1}\mid {\mathcal{F}}_k]=0$. Since $f=g-Pg$ $\pi$-almost everywhere and each $X_k$ has law $\pi$, the identity holds at every $X_k$ simultaneously outside a countable union of null events. Therefore $f(X_k)=g(X_k)-g(X_{k+1})+D_{k+1}$ almost surely.

Set $q(x)=P(g^2)(x)-(Pg(x){)}^2$ on the $\pi$-full set where these terms are finite, and set $q=0$ elsewhere. By [F1], $q(X_k)$ is the conditional variance of $D_{k+1}$, is nonnegative, and $\pi (q)=\mathbb{E}[D_1^2]={\tau }_f^2$. Thus $q\in L^1(\pi )$. The process $(q(X_k){)}_{k\ge 0}$ is a measurable factor of the stationary ergodic path process $(X_k)$: the inverse image of any shift-invariant event for $q(X)$ is shift-invariant for $X$. Hence it too is stationary and ergodic by [F10], and [F2] gives $n^{-1}{\sum }_{k=0}^{n-1}q(X_k)\to {\tau }_f^2$ almost surely.

Stationarity and $g\in L^2(\pi )$ give $\mathbb{E}[(g(X_0)-g(X_n){)}^2\mathrm{/}n]\le 4\pi (g^2)\mathrm{/}n\to 0$. The $L^2$-norm of the endpoint divided by $\sqrt{n}$ therefore tends to zero, so [F9] gives convergence to zero in probability.

Summing step 1.1 for $k=0,\dots ,n-1$ cancels the intermediate $g(X_k)$ terms and yields $S_n:={\sum }_{k=0}^{n-1}f(X_k)=g(X_0)-g(X_n)+M_n$, where $M_n:={\sum }_{k=0}^{n-1}{D}_{k+1}$.

Suppose ${\tau }_f^2>0$. For row $n$, set $Z_{n,j}=D_j\mathrm{/}\sqrt{n{\tau }_f^2}$ for $1\le j\le n$, set $Z_{n,j}=0$ for $j>n$, and use ${\mathcal{F}}_{n,j}={\mathcal{F}}_{\min (j,n)}$ for every $j\ge 0$. These are square-integrable martingale-difference rows by step 1.1. Their terminal variance clock is ${\mathrm{\Gamma }}_{n,\mathrm{\infty }}=(n{\tau }_f^2{)}^{-1}{\sum }_{k=0}^{n-1}q(X_k)\to 1$ almost surely by step 1.2. For every $\epsilon >0$, stationarity of the transition pairs $(X_{j-1},X_j)$ gives ${\sum }_{j=1}^n\mathbb{E}[Z_{n,j}^2{\mathbf{1}}_{\{\mathrm{\mid }{Z}_{n,j}\mathrm{\mid }>\epsilon \}}]={\tau }_f^{-2}\mathbb{E}[D_1^2{\mathbf{1}}_{\{\mathrm{\mid }{D}_1\mathrm{\mid }>\epsilon \sqrt{n{\tau }_f^2}\}}]\to 0$ by [F8], since $D_1^2$ is integrable. Extending each row by zeros makes it an infinite array with the same terminal sum and clock. Therefore [F3] gives $M_n\mathrm{/}\sqrt{n{\tau }_f^2}\Rightarrow N(0,1)$; scaling by ${\tau }_f$ using [F11] yields $M_n\mathrm{/}\sqrt{n}\Rightarrow N(0,{\tau }_f^2)$.

If ${\tau }_f^2=0$, stationarity and [F1] imply $\mathbb{E}[D_{k+1}^2]=0$, so $D_{k+1}=0$ almost surely and hence $M_n=0$. By step 2.1, $S_n=g(X_0)-g(X_n)$. Step 1.3 shows $S_n\mathrm{/}\sqrt{n}\to 0$ in probability; [F7] then gives convergence to the point mass $N(0,0)$.

If ${\tau }_f^2>0$, step 2.2 gives $M_n\mathrm{/}\sqrt{n}\Rightarrow N(0,{\tau }_f^2)$. Combining this with step 1.3, the decomposition in step 2.1, and [F7] gives $S_n\mathrm{/}\sqrt{n}\Rightarrow N(0,{\tau }_f^2)$.

Assume additionally that the covariance series is absolutely convergent. Stationarity and expansion of the finite square give $n^{-1}\mathrm{Var}(S_n)={\gamma }_0+2{\sum }_{k=1}^{n-1}(1-k\mathrm{/}n){\gamma }_k\to {\gamma }_0+2{\sum }_{k=1}^{\mathrm{\infty }}{\gamma }_k$, where ${\gamma }_k={\mathrm{Cov}}_{\pi }(f(X_0),f(X_k))$; [F8] applies to the series because $\mathrm{\mid }{\gamma }_k\mathrm{\mid }$ is summable. On the other hand, step 2.1 writes $S_n=R_n+M_n$ with $R_n=g(X_0)-g(X_n)$. For $i<j$, $D_{i+1}$ is ${\mathcal{F}}_j$-measurable and $\mathbb{E}[D_{j+1}\mid {\mathcal{F}}_j]=0$, so [F1, F6] give $\mathbb{E}[D_{i+1}{D}_{j+1}]=0$. Stationarity gives $\mathrm{Var}(M_n)={\sum }_{j=0}^{n-1}\mathbb{E}[D_{j+1}^2]=n{\tau }_f^2$. Also $\mathrm{Var}(R_n)\le 4\pi (g^2)$ and [F6] gives $\mathrm{\mid }\mathrm{Cov}(R_n,M_n)\mathrm{\mid }\le \sqrt{\mathrm{Var}(R_n)\mathrm{Var}(M_n)}=O(\sqrt{n})$. Dividing $\mathrm{Var}(S_n)=\mathrm{Var}(R_n)+\mathrm{Var}(M_n)+2\mathrm{Cov}(R_n,M_n)$ by $n$ therefore shows that its limit is ${\tau }_f^2$. Steps 3.1 and 3.2 give the CLT, so [F5] identifies its variance with the displayed covariance sum.