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Under a stationary initial law, reversal leaves the law of a finite reversible-chain path unchanged. The first time an ordered occurrence of state $a$ then state $b$ appears therefore has the same distribution as the reverse ordered occurrence. Decomposition at the first entrance relates this symmetry to differences of mean hitting times. If positive diagonal return times are used instead of zero diagonal entrance times, their stationary-weighted terms cancel by Kac's lemma.

The chain is a reversible Markov chain, so detailed balance makes the stationary path probabilityA path contains an ordered occurrence of $a$ followed later by $b$ exactly when its reversal contains $b$ followed later by $a$. These occurrences need not be adjacent. Therefore $\mathrm{Pr}(\tau (a,b)\le m)=\mathrm{Pr}(\tau (b,a)\le m)$ for every $m$, and in particular the two finite expectations are equal. This is ordered-state visit times in a reversible chain.

To decompose the expectations, distinguish the entrance time $h(i,j)$, which allows time zero and has $h(j,j)=0$, from the positive return convention $k(j,j)$ in part (i). The Strong Markov property at the first visit to $a$ or $b$ givesEquating them yields the difference formula with $h$. Replacing $h$ by $k$ in the sum adds ${\pi }_{a}k(a,a)-{\pi }_{b}k(b,b)=1-1=0$, again by Kac's lemma. HenceFor a numerical consistency check, the other entrance times are $k(b,a)=15/11$ and $k(c,a)=16/11$, while $k(b,b)=3$. Both sides of the displayed identity equal $-6/11$.

First realize the chain as a random walk on a multigraph: put three parallel edges between $a,b$, two between $a,c$, and one between $b,c$. There are six edges and vertex degrees $5,4,3$. Choosing an incident edge uniformly gives exactly the stated transition probabilities. The stationary distribution of a graph random walk is the degree divided by twelve, $\pi =(5,4,3)/12$, and detailed balance follows because each edge multiplicity contributes the same flux in both directions.

For the first positive return time, Kac's lemma givesFor passage to $b$, put $h_a=k(a,b)$ and $h_c=k(c,b)$. The first-step equations are $h_a=1+(2/5)h_c$ and $h_c=1+(2/3)h_a$, since reaching $b$ stops the clock. ThusA multigraph is essential: a simple graph on three vertices cannot have six edges.

Conditional on $X_0=x$, the random variable $T$ is the first return time $T_x^{+}$. The Kac's lemma mean-recurrence formula for a finite irreducible Markov chain givesAveraging over $X_0\sim \pi$ therefore yields

Let the common initial state be $x$ and the robber's final state be $y$. Part (a) gives their joint distribution asLet ${\tau }_y=\min \{n\ge 0:X_n=y\}$ be the cop's hitting time of $y$ when starting from $x$. For a chain started from $y$, condition on its first step to obtainHence the expected catch time, averaged over the displayed joint law, isUsing Kac's lemma once more givesThe convention ${\tau }_y=0$ correctly counts an immediate catch when the robber's reverse step leaves it at the common starting state.

A complete excursion from $B$ along any one arm and back to $B$ has mean length $L=2n$. This follows from the same path recurrence, or from Kac's lemma: the graph has $3n$ edges, the stationary distribution of simple random walk assigns mass $\mathrm{deg}(B)/(2|E|)=3/(6n)=1/(2n)$ to $B$, so its mean return time is $L=2n$.

Let $x$ be the mean time to hit $E$ starting at $B$. On each visit to $B$, the first step reaches $E$ with probability $1/3$; with probability $2/3$ it begins a wrong-arm excursion of mean duration $L$, after which the problem restarts. ThusThe walk first takes mean time $n^2$ to reach $B$, so the requested time from $A$ is

The renewal-reward form of Kac's lemma says that the expected number of visits to state $i$ during one return cycle from $k$ to $k$ isHere time spent means the number of discrete time instants at which the chain occupies $i$ between consecutive visits to $k$.

A distribution $\pi$ is an invariant distribution whenfor every state $j$. The pair $(\pi ,P)$ satisfies detailed balance whenfor every $i,j$. Summing this identity over $i$ givesso detailed balance implies invariance.

For an irreducible positive recurrent Markov chain, Kac's lemma relates the invariant mass to the mean recurrence time:

Let $h_i={\mathbb{E}}_i{T}_0$ be the expected hitting time of state $0$. For $i\ge 1$, first-step analysis givesWith $d_i=h_i-h_{i-1}$ this becomesThe minimal nonnegative solution hasso

Equivalently, Kac's lemma gives ${\mathbb{E}}_0{T}_0^{+}=1/{\pi }_0$. A first step from $0$ either returns immediately with probability $1-p$ or moves to $1$ with probability $p$, soSubstituting ${\pi }_0=(q-1)/(q-1+q^{2}p)$ gives the same result.