Geometric tail for hitting in a finite irreducible chain

Fix one $a\in A$. For $x\mathrm{\notin }A$, irreducibility gives a nonempty set $\{n\ge 1:p^{(n)}(x,a)>0\}$; let $n_x$ be its least element. Set $n_x=0$ for $x\in A$. By [F2] and [F6], ${\mathbb{P}}_x(T_A\le n_x)>0$ for every $x$: it is $1$ on $A$, and off $A$ the event $\{X_{n_x}=a\}$ has positive probability. Since $E$ is finite, $m:=\max (1,{\max }_{x\in E}{n}_x)$ is finite and $q_x:={\mathbb{P}}_x(T_A\le m)>0$ for every $x$. Thus $\epsilon :=\min (1\mathrm{/}2,{\min }_{x\in E}{q}_x)$ lies in $(0,1)$ and $q_x\ge \epsilon$ uniformly. The witness lengths are least natural numbers, so this finite construction makes no choice-function assumption.

Define the bounded path functional $H_m(\omega )={\mathbf{1}}_{\{{\omega }_j\mathrm{\notin }A\text{ for }0\le j\le m\}}$ and $r_m(x):={\mathbb{E}}_x{H}_m(X_0,X_1,\dots )={\mathbb{P}}_x(T_A>m)$. Step 1.1 gives $r_m(x)=1-q_x\le 1-\epsilon$ for every $x$. For $k\ge 0$ let $B_k=\{T_A>km\}\in {\mathcal{F}}_{km}$ by [F3]. The event $B_{k+1}$ is $B_k$ intersected with avoidance of $A$ during the next $m$ steps. Applying [F5] at time $km$ and integrating over $B_k$ yields ${\mathbb{P}}_x(B_{k+1})={\mathbb{E}}_x[{\mathbf{1}}_{B_k}{r}_m(X_{km})]\le (1-\epsilon ){\mathbb{P}}_x(B_k)$. This unconditional recursion also holds when a survival event has probability zero.

Since ${\mathbb{P}}_x(B_0)\le 1$, induction in step 2.1 gives ${\mathbb{P}}_x(T_A>km)\le (1-\epsilon {)}^k$ for every $k\ge 0$. Since $T_A$ is integer-valued (with $+\mathrm{\infty }$ allowed), its strict tail is constant on each interval $[j,j+1)$; integrating that tail in [F7] gives ${\mathbb{E}}_x{T}_A={\sum }_{j\ge 0}{\mathbb{P}}_x(T_A>j)$. Writing $j=km+s$ with $0\le s<m$ and using monotonicity of the tail, ${\mathbb{E}}_x{T}_A\le m{\sum }_{k\ge 0}(1-\epsilon {)}^k=\frac{m}{\epsilon }<\mathrm{\infty }\mathrm{.}$ This bound is uniform in $x$.

The same estimate gives the scaffold's finite-class return consequence. For this clause, let $E$ be countable and let $C\subseteq E$ be finite and nonempty, satisfy ${\sum }_{z\in C}p(y,z)=1$ for each $y\in C$, and have an irreducible restricted matrix. For each $y\in C$, closure and the finite-dimensional iterated law [F6] imply ${\mathbb{P}}_y(X_0,\dots ,X_n\in C)=1$ for every $n$ and identify the joint law of $(X_0,\dots ,X_n)$ with that of the restricted matrix on $C$. In particular, each event $\{T_{\{a\}}>n\}$ has the same probability under the ambient and restricted chains; summing these integer tails by [F7] shows their hitting-time expectations agree. Fix $a\in C$ and put $M={\max }_{y\in C}{\mathbb{E}}_y{T}_{\{a\}}<\mathrm{\infty }$ by applying the argument of steps 1.1–3.1 to this finite restricted chain. For every $n\ge 0$, the bounded future-path Markov identity at time $1$, applied to avoidance of $a$ in the next $n+1$ coordinates, gives ${\mathbb{P}}_a(T_a^{+}>n+1)={\sum }_{y\in C}p(a,y){\mathbb{P}}_y(T_{\{a\}}>n)\mathrm{.}$ Summing this identity over $n\ge 0$ and using the same integer-valued tail identity from [F7] as in step 3.1 gives ${\mathbb{E}}_a{T}_a^{+}=1+{\sum }_{y\in C}p(a,y){\mathbb{E}}_y{T}_{\{a\}}\le 1+M<\mathrm{\infty }$. Hence ${\mathbb{P}}_a(T_a^{+}<\mathrm{\infty })=1$, so [F8] makes $a$ recurrent. Since $a$ was arbitrary, every state of $C$ is recurrent.

The nonempty-target hypothesis is necessary: for $A=\mathrm{\varnothing }$, $T_A=\mathrm{\infty }$ and the finite-mean conclusion fails. If $A=E$ or the starting state lies in $A$, then $T_A=0$ and the tail bound holds immediately; for a one-state chain these are the only target cases. Deterministic and other degenerate rows are covered by the same positive accessibility witnesses and the uniform block estimate. At $k=0$ the asserted bound is just ${\mathbb{P}}_x(T_A>0)\le 1$. AC is used only through [F5] and [F6]; all path-length witnesses above are least natural numbers. The theorem is not an iff statement.