MCMC central limit theorems, MCSE, and long-run variance — Examples

Let $\mathsf{X}=\{0,1\}$, $P(0,\{0\})=P(1,\{1\})=1$, and let $\pi (\{0\})=\pi (\{1\})=1\mathrm{/}2$. The identity kernel leaves $\pi$ invariant, so the chain initialized with $X_0\sim \pi$ is stationary by [F2]. Set $f(0)=-1$ and $f(1)=1$. Then $\pi (f)=0$ and $\pi (f^2)=1$.

The transition is deterministic, hence $X_k=X_0$ for every $k$ and $S_n:={\sum }_{k=0}^{n-1}f(X_k)=nf(X_0)$. Thus $S_n\mathrm{/}\sqrt{n}=\sqrt{n}f(X_0)$ takes values $\pm \sqrt{n}$ with equal probabilities. For each fixed $t>0$, once $n>t^2$ its distribution function at $t$ is $1\mathrm{/}2$, while any $N(0,{\sigma }^2)$ has distribution function strictly greater than $1\mathrm{/}2$ at $t$ when ${\sigma }^2>0$ and equal to $1$ when ${\sigma }^2=0$. Since $t$ is a continuity point of every such normal law, [F4] rules out convergence to any normal distribution.

Choose $\varphi =\pi$ and $A=\{1\}$. Then $\varphi (A)=1\mathrm{/}2>0$, but $P^n(0,A)=0$ for every $n\ge 1$ because the chain stays at $0$. By [F5] the chain is not $\varphi$-irreducible and is not Harris recurrent relative to $\varphi$. This counterexample shows only that finite variance without an ergodicity assumption does not guarantee a root-$n$ CLT.

For $n\ge 2$, $n^{1\mathrm{/}2}-1<b_n\le n^{1\mathrm{/}2}$, so $b_n\mathrm{/}{n}^{1\mathrm{/}2}\to 1$. Thus $n^{2\mathrm{/}3}(\log n{)}^3\mathrm{/}{b}_n\sim n^{1\mathrm{/}6}(\log n{)}^3\to \mathrm{\infty }$, which violates the required limit in [F1] at $\delta =1$.

Since $1\le b_n\le n^{1\mathrm{/}2}<n$ for $n\ge 2$, the finite OBM procedure in [F2] is defined. Step 1.1 establishes failure of one source sufficient condition only, not inconsistency of the estimator.

Put $v=v_{ϵ}\mathrm{/}(1-a^2)$. If $X_k\sim N(0,v)$ independently of ${ϵ}_{k+1}$, then $a{X}_k+{ϵ}_{k+1}$ is normal with mean zero and variance $a^{2}v+v_{ϵ}=v$. Thus $\pi =N(0,v)$ is invariant, and initializing with $\pi$ gives a stationary chain by [F1].

Iterating the recursion gives $X_n=a^{n}x+{\sum }_{j=1}^n{a}^{n-j}{ϵ}_j$ when $X_0=x$, so $P^n(x,\cdot )=N(a^{n}x,v_n)$ with $v_n=v(1-a^{2n})$. For each fixed $x$, $a^{n}x\to 0$ and $v_n\to v$, so the Gaussian densities $p_n^x(y)$ converge pointwise to the density $p(y)$ of $N(0,v)$. For sufficiently large $n$, $v\mathrm{/}2\le v_n\le v$ and $\mathrm{\mid }{a}^{n}x\mathrm{\mid }\le 1$. On $\mathrm{\mid }y\mathrm{\mid }\le 2$, $p_n^x(y)\le (\pi v{)}^{-1\mathrm{/}2}$; on $\mathrm{\mid }y\mathrm{\mid }>2$, $\mathrm{\mid }y-a^{n}x\mathrm{\mid }\ge \mathrm{\mid }y\mathrm{\mid }\mathrm{/}2$, hence $p_n^x(y)\le (\pi v{)}^{-1\mathrm{/}2}{e}^{-y^2\mathrm{/}(8v)}$. This gives one integrable majorant for $p_n^x$; with $p$ added it also dominates $\mathrm{\mid }{p}_n^x-p\mathrm{\mid }$. By [F2], $\int \mathrm{\mid }{p}_n^x-p\mathrm{\mid }dy\to 0$. Therefore $\mathrm{\parallel }{P}^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\le \int \mathrm{\mid }{p}_n^x-p\mathrm{\mid }dy\to 0$ for every $x$.

The Borel sigma-algebra of $\mathbb{R}$ has the countable generating algebra of finite unions of rational half-open intervals and rational rays. Applying [F3] to step 1.2 gives ${\alpha }_X(n)\to 0$.

Let $E$ be an event invariant under every power of the canonical left shift and put $p=\mathbb{P}(E)$. Finite-coordinate cylinders form a generating pi-system. The events approximable in probability by such cylinders form a sigma-algebra: complements preserve symmetric-difference probability, and for a countable union one first truncates by continuity from below and then approximates the finitely many sets. Hence for every $\eta >0$ there are $j$ and $A\in \sigma (X_0,\dots ,X_j)$ with $\mathbb{P}(E\mathrm{△}A)<\eta$. For $m>j$, $B={\theta }^{-m}E$ is measurable with respect to $\sigma (X_m,X_{m+1},\dots )$, stationarity gives $\mathbb{P}(B)=p$, and shift invariance gives $E=B$. Thus $\mathrm{\mid }p-p^2\mathrm{\mid }=\mathrm{\mid }\mathbb{P}(E\cap B)-p\mathbb{P}(B)\mathrm{\mid }\le 2\eta +{\alpha }_X(m-j)$. Let $m\to \mathrm{\infty }$ using step 2.1 and then $\eta \downarrow 0$; we obtain $p(1-p)=0$. By [F7], every invariant event having probability zero or one is precisely ergodicity of the canonical shift.

Set $f(x)=x$ and $g(x)=x\mathrm{/}(1-a)$. Both are in $L^2(\pi )$ and $\pi (f)=0$. The kernel action is $Pg(x)=\mathbb{E}[(ax+{ϵ}_1)\mathrm{/}(1-a)]=ax\mathrm{/}(1-a)$, so $g-Pg=f$. Also $D_{k+1}=(a{X}_k+{ϵ}_{k+1})\mathrm{/}(1-a)-a{X}_k\mathrm{/}(1-a)={ϵ}_{k+1}\mathrm{/}(1-a)$, whence $\mathbb{E}[D_{k+1}^2]=v_{ϵ}\mathrm{/}(1-a{)}^2$. Steps 1.1 and 3.1 verify the stationary and ergodic hypotheses of [F5], so this is the Poisson CLT variance.

Iterating the recursion gives $X_k=a^k{X}_0+{\sum }_{j=1}^k{a}^{k-j}{ϵ}_j$. Independence implies ${\gamma }_k=\mathrm{Cov}(X_0,X_k)=v{a}^k$. Since $\mathrm{\mid }a\mathrm{\mid }<1$, the covariance series is absolutely summable, and [F6] gives ${\sigma }_f^2=v+2v{\sum }_{k\ge 1}{a}^k=v+2va\mathrm{/}(1-a)=v_{ϵ}\mathrm{/}(1-a{)}^2$. This agrees with step 4.1, including $a=0$ and negative $a$.