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A coalescing coupling of probability distributions of two Markov chains keeps them identical after a meeting time $T$. The coupling inequality for total variation gives $\Vert P^t(x,\cdot )-P^t(y,\cdot ){\Vert }_{\mathrm{TV}}\le \mathbb{P}(T>t)$. A uniform bound on $\mathbb{E}T$, combined with Markov inequality, bounds the mixing time of a Markov chain.

We prove uniqueness in the sense of at most one Gibbs measure satisfying the printed conditional distributions. No shift-invariance assumption is needed in advance. Write a history as $\eta =({ϵ}_{s-1},{ϵ}_{s-2},\dots )$. The variation hypothesis makes $U$ continuous and bounded on the finite-alphabet sequence space; let $u_{-}\le U\le u_{+}$ and $V=u_{+}-u_{-}$. The conditional distribution of the first spin, obtained by marginalizing the prescribed two-spin law, must beIn particular $q(a\mid \eta )\ge \delta :=e^{-2V}/k>0$ for every symbol and history. Any two candidate measures have this same time-homogeneous one-step kernel, by conditional marginalization and the tower property.

If $\eta ,{\eta }^{\prime }$ agree in their most recent $r$ symbols, each two-spin energy in this formula changes by at mostBoth numerator and denominator therefore change by factors between $e^{-D{\lambda }^{-r}}$ and $e^{D{\lambda }^{-r}}$. Consequently the total variation distance of the two kernels is at most $e^{2D{\lambda }^{-r}}-1\le B{\lambda }^{-r}$, where $B=e^{2D}-1$. Uniform positivity also bounds it by $1-k\delta$. PutThese bounds are decreasing, strictly below one, and summable. We now give the summable-memory coupling argument rather than assuming a small one-step contraction coefficient.

Starting from any two histories at a remote time, use a maximal coupling of the next-spin laws. If the actual histories have agreed for $r$ consecutive symbols, their probability of disagreement is at most $b_r$. Use shared independent uniform random numbers to construct a dominating match-length process $R$: it increases from $r$ to $r+1$ with probability $1-b_r$ and resets to zero with probability $b_r$. Since $b_r$ decreases, the actual consecutive-match length can be kept at least $R$ at every step. When the dominating process does not reset, the actual spins can be coupled to agree.

A new run from zero has probabilityof never resetting. The product is positive because $b_r<1$ and $\sum b_r<\infty$: its tail logarithm converges absolutely, and its finite initial factors are positive. After each reset a fresh run has the same success probability, so the probability of $m$ failed runs is $(1-p_{\ast }{)}^m$. There are almost surely only finitely many resets. Every failed run has finite duration, and thus the time of the final reset is finite almost surely. Its distribution depends only on $(b_r)$, not on the two starting histories. It follows that the probability of any disagreement in a fixed future block tends uniformly to zero as the block's distance from the initial time tends to infinity.

Now take two candidate Gibbs measures, sample their respective histories before time $-N$, and couple their forward kernels as above. Their finite-block laws really are obtained by these kernels: successive conditional expectations use the common conditional distribution $q$. For any fixed coordinate block $[s,t]$, the coupling inequality for total variation bounds the distance between the two candidate block laws by the dominating probability of a reset at or after step $N+s$. That bound tends to zero as $N\to \infty$. Hence all probabilities of finite cylinder sets agree; cylinder sets generate the product Borel sigma-algebra, so the measures agree.

Thus there is at most one measure satisfying these conditional specifications. There is a separate compatibility issue: for an arbitrary unnormalized symbolic potential the finite-horizon conditional distributions printed in the question need not be consistent as the right endpoint changes. The argument proves the requested uniqueness whenever such a law exists; it does not silently infer existence from the variation estimate. A normalized one-sided kernel or the usual compatible Gibbs specification supplies the standard existence formulation. For a concrete incompatibility, take two symbols and $U(a,\eta )=\log W_{{\eta }_1,a}$ with $W=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$. This has zero variation once two symbols agree. From a history ending in symbol one, the first-symbol probability obtained from a two-spin block is $2/5$, but from a three-spin block it is $5/13$. Thus the printed finite-horizon specification can indeed have no measure, even though its uniqueness implication remains valid.

A coupling of probability distributions is a joint probability distribution of random variables $(X,Y)$ with the specified marginal distributions: $\mathbb{P}(X\in A)=\mu (A)$ and $\mathbb{P}(Y\in A)=\nu (A)$. No independence is required.

For every set $A$, the difference of the two indicator functions vanishes when $X=Y$. ConsequentlyMaximizing over $A$ gives the coupling inequality for total variation:

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

Use the monotone grand coupling in every coordinate: at each step use the same attempted move for chains started from the bottom and top states, censoring moves that leave the interval. The two copies coalesce after the lower copy has accumulated enough upward drift and boundary censoring has removed their initial separation. Away from the boundary, one step has mean displacementStandard exponential concentration for sums of bounded independent increments shows that for every fixed $C>6$ the one-coordinate coupling time $\tau$ satisfiesAll other initial states lie between the extremal copies. Coupling the $d$ coordinates independently and using the union bound givesbecause $\log d=o(n)$. The coupling inequality for total variation therefore gives

Couple two noisy voter model chains by choosing the same update vertex, the same refresh coin and refreshed spin, and, for a voter update, the same chosen neighbor. Let $D_t$ be their Hamming distance. If $G$ is $r$-regular, conditioning on the current disagreement set givesbecause every disagreeing vertex is counted in exactly $r$ neighbor sets. ThereforeThe coupling inequality for total variation and $1_{\{{\sigma }_t\ne {\sigma }_t^{\prime }\}}\le D_t$ give

Reflection in $L$ interchanges $x$ and $y$. Relabel the two points if necessary so that $x$ lies on the side of $L$ containing the center $a$; the absolute difference is symmetric in $x$ and $y$. Couple a Brownian motion $B$ started at $x$ to one $B^{\prime }$ started at $y$ by setting $B_t^{\prime }=\varphi (B_t)$ before $T_L$ and $B_t^{\prime }=B_t$ afterwards. The Brownian reflection coupling has the correct marginal laws because reflection is an isometry and the Strong Markov property applies at $T_L$. Once the paths meet on $L$, they agree forever.

If the path from $x$ does not hit the target circle before $T_L$, the coupled paths meet before the relevant uncoupled hitting outcomes can differ. The coupling inequality for total variation therefore gives

In the diamond coordinates, the horizontal line through $(u,v)$ has $2(n-|u|)+1$ points and the vertical line has $2(n-|v|)+1$ points. There is a numerical constant $a>0$, independent of $n$ and of the starting point, such that the walk enters the middle diamond $D_{n/2}$ within two steps with probability at least $a$. To see this, suppose $|u|\le |v|$, so $|u|\le n/2$. With probability at least $1/2$ the first move uses the longer horizontal line, and a fixed positive fraction of its choices have $|v^{\prime }|\le n/4$. From $(u,v^{\prime })$, the vertical line has length at least $3n/2$, is chosen with probability bounded below, and a fixed positive fraction of its points satisfy $|u^{\prime }|+|v^{\prime }|\le n/2$. The case $|v|\le |u|$ is symmetric.

Now couple two copies. Independently use the preceding construction until both lie in $D_{n/2}$, an event having probability at least $a^2$ in two steps. From $(u,v)$ and $(u^{\prime },v^{\prime })$ in that diamond, their horizontal line ranges overlap in at least $n+1$ values. A maximal coupling of the next moves can therefore, with probability bounded below, send them to $(u,w)$ and $(u^{\prime },w)$ for the same $|w|\le n/2$. Their vertical lines are then identical, and another maximal coupling makes the two states equal with probability bounded below. Thus there are constants $k$ and $b>0$, independent of $n$, for which the two copies coalesce during every block of $k$ steps with conditional probability at least $b$.

The coupling time consequently has a geometric tail. By the coupling inequality for total variation, for each fixed $0<\epsilon <1$,Since a nontrivial chain has mixing time bounded below by a positive constant, the mixing time has constant order.