General-state MCMC ergodicity and mixing — Examples

The only nonempty $\varphi$-positive set contains state $1$. From $1$ it is already reached, and from $x\ge 2$ the one-step probability of reaching it is $x^{-2}>0$. Thus the chain is $\varphi$-irreducible. The point mass ${\delta }_1$ is invariant because state $1$ is absorbing. Since $\pi$ has only one positive-mass state, no partition into $d\ge 2$ classes all of positive $\pi$-mass exists; the chain is aperiodic under [F2].

Starting from $x\ge 2$, the probability of avoiding absorption through time $n$ is $s_n(x)={\prod }_{j=x}^{x+n-1}(1-\frac{1}{j^2})=\frac{x-1}{x}\frac{x+n}{x+n-1}\mathrm{.}$ The identity follows by writing $1-j^{-2}=((j-1)\mathrm{/}j)((j+1)\mathrm{/}j)$ and canceling adjacent factors. Thus $s_n(x)\to (x-1)\mathrm{/}x>0$, so the probability of ever hitting the $\varphi$-positive set $\{1\}$ is less than one. The chain is not Harris recurrent by [F3].

At time $n$, the mass outside $\{1\}$ is exactly $s_n(x)$. By [F5], $\mathrm{\parallel }{P}^n(x,\cdot )-{\delta }_1{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}=\mathrm{\mid }{P}^n(x,\{1\})-1\mathrm{\mid }=s_n(x),$ which has positive limit. Therefore the stationary-a.e. convergence conclusion cannot be strengthened to all starts without Harris recurrence.

Let $\varphi$ be counting measure. Every nonempty set is reachable from either state at a positive time of at most two steps, so [F2] holds. The uniform law is invariant because each column sum and row sum of the transition matrix equals one.

The partition ${\mathsf{X}}_0=\{0\}$, ${\mathsf{X}}_1=\{1\}$ has positive invariant mass in both classes and transitions alternate deterministically, so it is a cyclic decomposition of period two under [F3].

From $0$, $P^n(0,\cdot )={\delta }_0$ for even $n$ and ${\delta }_1$ for odd $n$. Taking event $\{0\}$ in [F4], the distance to $\pi$ equals $1\mathrm{/}2$ for every $n$. From $1$ the even and odd laws are interchanged, giving the same distance. Hence convergence from either fixed state fails despite irreducibility and invariance; an initial law equal to $\pi$ remains invariant.

The invariant law is $\pi =(1\mathrm{/}2,1\mathrm{/}2)$ by symmetry, and each state has positive holding probability $1-\eta$. Every entry is positive, so it is irreducible by [F4], and each state has positive holding probability, so it is aperiodic. Moreover every row dominates $2\eta \pi$, because its two entries are at least $\eta$. This is a whole-space minorization with $m=1$, $ϵ=2\eta$.

The theorem [F3] gives ${\sup }_x\mathrm{\parallel }{P}_{\eta }^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\le (1-2\eta {)}^n$, so each fixed chain is uniformly geometrically ergodic.

The nonconstant eigenvector $(1,-1)$ has eigenvalue $1-2\eta$. Direct multiplication, or induction, gives $\mathrm{\parallel }{P}_{\eta }^n(0,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}=\frac{1}{2}(1-2\eta {)}^n\mathrm{.}$ For every fixed $n$ this tends to $1\mathrm{/}2$ as $\eta \downarrow 0$. Hence no finite $n$ makes the distance smaller than (for example) $1\mathrm{/}4$ uniformly over the family. Each chain is uniformly ergodic, but the mixing time depends on $\eta$.

Each coordinate marginal under $\pi$ is uniform on $\{0,1\}$. Given $Y=y$, the conditional law of $X$ is the point mass ${\delta }_y$; given $X=x$, the conditional law of $Y$ is ${\delta }_x$. These are valid full conditionals because the target assigns mass only to the two diagonal points.

From $(0,0)$ either coordinate update leaves the state at $(0,0)$; from $(1,1)$ either update leaves it at $(1,1)$. The random-scan mixture in [F3] therefore makes each singleton a closed class. The target is invariant by [F4], but $\{(0,0)\}$ is nonempty and proper, so reducibility follows from [F5].

Starting from $(0,0)$, the time-$n$ law is ${\delta }_{(0,0)}$, whose total-variation distance from $\pi$ is $1\mathrm{/}2$. Thus target invariance alone does not imply convergence from every start, uniqueness of the invariant law, or irreducibility.

For any $A$ with $\lambda (A)>0$, $P(x,A)\ge \sqrt{x}\lambda (A)>0$, so [F1] holds. For $x<1$, the kernel has a positive self-loop of mass $1-\sqrt{x}$; since $\pi (\{1\})=0$, a cyclic decomposition with positive $\pi$-mass classes would contain a $\pi$-positive set of states with positive self-loop probability, contradicting deterministic movement to a disjoint next class. Thus the chain is aperiodic.

To realize refreshes, use the marked state space $\mathsf{X}\times \{0,1\}$ and the probability kernel $K((x,b),A\times \{1\})=\sqrt{x}\lambda (A)$, $K((x,b),A\times \{0\})=(1-\sqrt{x})1_A(x)$. Every measurable set is a disjoint union of these two slice forms, so [F7] verifies countable additivity and measurable evaluations. Its first-coordinate transition is $P$. Apply [F8] with initial state $(x,0)$. Direct multiplication of the transition probabilities gives probability $(1-\sqrt{x}{)}^{t-1}\sqrt{x}\lambda (A)$ that the first marked refresh occurs at time $t\ge 1$ with new state in $A$. Summing over $t$ shows that the waiting time is finite almost surely and the new state has law $\lambda$. The same finite-history calculation, iterated for any finite number of marked refreshes, gives independent $\lambda$ refresh states and almost-surely finite waiting times. A countable intersection then gives infinitely many refreshes almost surely. For $\lambda (A)>0$, the probability that the first $N$ refresh draws all miss $A$ is $(1-\lambda (A){)}^N\to 0$. Hence every positive-$\lambda$ set is hit almost surely from every start, proving [F3].

The one-step law under $\pi$ has a holding part with density $(1-\sqrt{x})\mathrm{/}(2\sqrt{x})$ and a refresh part of total mass $\int \sqrt{x}\pi (dx)=1\mathrm{/}2$, distributed as $\lambda$. For every measurable $A$, $\int P(x,A)\pi (dx)={\int }_A\frac{1-\sqrt{x}}{2\sqrt{x}}dx+\frac{1}{2}\lambda (A)={\int }_A\frac{dx}{2\sqrt{x}}=\pi (A)\mathrm{.}$ Thus $\pi$ is invariant. By [F6], convergence holds from every state.

Fix any $r\in (0,1)$ and set $B_r=(0,(1-r{)}^2)$. Since $\pi (B_r)=1-r>0$, this set meets every full-$\pi$-measure set on which a proposed geometric bound with rate $r$ holds and its multiplier is finite. For each $x\in B_r$, a refresh draw has no atom at $x$, so $P^n(x,\{x\})=(1-\sqrt{x}{)}^n$, while $\pi (\{x\})=0$. Hence $\mathrm{\parallel }{P}^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\ge (1-\sqrt{x}{)}^n$. Choose $x$ in the intersection just described. Then $1-\sqrt{x}>r$, so no finite $M(x)$ can satisfy $(1-\sqrt{x}{)}^n\le M(x)r^n$ for every $n\ge 1$. As this holds for every $r\in (0,1)$, the chain is not geometrically ergodic, despite convergence from every start.

The transition rows sum to one. Every adjacent move has positive probability, so every state reaches every other state, proving counting-measure irreducibility. Each state has a positive holding probability, so no positive-mass cyclic family can move deterministically to a disjoint successor. The proposed masses sum to one. At state $0$, the incoming mass is ${\pi }_{0}P(0,0)+{\pi }_{1}P(1,0)=3\mathrm{/}8+1\mathrm{/}8=1\mathrm{/}2={\pi }_0$. For $n\ge 1$, it is ${\sum }_j{\pi }_{j}P(j,n)={\pi }_{n-1}\mathrm{/}4+{\pi }_n\mathrm{/}4+{\pi }_{n+1}\mathrm{/}2=2^{-n-2}+2^{-n-3}+2^{-n-3}=2^{-n-1}={\pi }_n\mathrm{.}$ Thus $\pi P=\pi$. Also ${\pi }_{n}P(n,n+1)=2^{-n-1}\mathrm{/}4=2^{-n-3}={\pi }_{n+1}P(n+1,n)$; these detailed-balance equalities and ${\sum }_{n\ge 0}{2}^{-n-1}=1$ provide a second check.

Let $C=\{0\}$, $\nu =P(0,\cdot )$, $m=1$, and $ϵ=1$. Then $P(0,\cdot )=\nu$, so $C$ is small. Set $V(n)=(\sqrt{2}{)}^n$. For $n\ge 1$, $\frac{PV(n)}{V(n)}=\frac{1}{4}\sqrt{2}+\frac{1}{2}\frac{1}{\sqrt{2}}+\frac{1}{4}=\frac{1}{4}+\frac{\sqrt{2}}{2}<\frac{24}{25}\mathrm{.}$ Since $\sqrt{2}\mathrm{/}2<71\mathrm{/}100$ (squaring gives $1\mathrm{/}2<5041\mathrm{/}10000$), the preceding ratio is less than $96\mathrm{/}100=24\mathrm{/}25$. At $0$, $PV(0)=3\mathrm{/}4+\sqrt{2}\mathrm{/}4<5\mathrm{/}4<49\mathrm{/}25=24\mathrm{/}25+b$ with $b=1$. Hence $PV\le (24\mathrm{/}25)V+1_C$.

The hypotheses of [F5] now hold, so there are $M<\mathrm{\infty }$ and $r<1$ with $\mathrm{\parallel }{P}^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\le MV(x)r^n$ for all $x,n$. The example supplies a concrete certificate: the target is invariant and the drift/minorization conditions are checked directly.