Birth–death recurrence through scale products

For $x<y$, the path that takes consecutive upward steps from $x$ to $y$ has probability ${\prod }_{k=x}^{y-1}{p}_k>0$; for $x>y$, consecutive downward steps have probability ${\prod }_{k=y+1}^x{q}_k>0$. By [F8], each such path gives positive $n$-step probability, so [F5] shows that every pair of states communicates. Thus the chain is irreducible. Every $q_j\mathrm{/}{p}_j$ is finite and strictly positive, so every $s_m$ is well-defined and positive, and $1\le S\le +\mathrm{\infty }$.

Fix integers $M>i\ge 1$ and set $E_M=\{0,1,\dots ,M\}$ and ${\tau }_M=T_{\{0,M\}}$. On $E_M$ define the matrix ${\widehat{p}}_M$ by making $0$ and $M$ absorbing and retaining the original row $(q_j,r_j,p_j)$ at each $1\le j<M$. For $x\in E_M$ and $B\subseteq E_M$, set ${\widehat{K}}_M(x,B)={\sum }_{y\in E_M}{\widehat{p}}_M(x,y){\delta }_y(B)\mathrm{.}$ Each row is a finite nonnegative weighted sum of Dirac probability measures; its weights sum to one, so [F10]–[F12] show that ${\widehat{K}}_M$ is a probability kernel. Under each original ${\mathbb{P}}_x$, $x\in E_M$, define $Y_n=X_{n\wedge {\tau }_M}$. This process stays in $E_M$. When ${\tau }_M\le n$ it is already at an absorbing endpoint; when ${\tau }_M>n$, it is at an interior state and its next original transition remains in $E_M$. Applying [F9] to the bounded functional $G(\omega )={\mathbf{1}}_B({\omega }_1)$ gives the conditional law of the next original state. Splitting on the ${\mathcal{F}}_n$-measurable events $\{{\tau }_M\le n\}$ and $\{{\tau }_M>n\}$ gives ${\mathbb{E}}_x[{\mathbf{1}}_{\{Y_{n+1}\in B\}}\mid {\mathcal{F}}_n]={\mathbf{1}}_{\{{\tau }_M\le n\}}{\mathbf{1}}_B(Y_n)+{\mathbf{1}}_{\{{\tau }_M>n\}}K(X_n,B);$ on the second event $X_n=Y_n$ is interior and its birth–death row, given by [F4], is exactly the corresponding ${\widehat{K}}_M$ row; on the first event $Y_n$ is an absorbing endpoint and the first term is its row probability. Thus this conditional expectation is ${\widehat{K}}_M(Y_n,B)$, so $Y$ is a finite-state Markov chain with kernel ${\widehat{K}}_M$ and deterministic start $x$.

For $1\le j<M$, the event that $Y$ takes $j$ consecutive downward steps from $j$ to $0$ has probability $q_j{q}_{j-1}\cdots q_1>0$ by [F8]. Let ${\epsilon }_M={\min }_{1\le j<M}{\prod }_{k=1}^j{q}_k>0.$ At any block time $kM$, if $Y_{kM}=j$ is interior, the conditional probability of following those $j$ downward steps is at least ${\epsilon }_M$ by [F9]; the boundary is then hit within the next $M$ steps. If the chain is already at a boundary, it has already hit one. Therefore, writing ${\sigma }_M=T_{\{0,M\}}$ for $Y$, ${\mathbb{P}}_x({\sigma }_M>(k+1)M)\le (1-{\epsilon }_M){\mathbb{P}}_x({\sigma }_M>kM),{\mathbb{P}}_x({\sigma }_M>kM)\le (1-{\epsilon }_M{)}^k\mathrm{.}$ This holds for every $x\in E_M$; hence $Y$ hits $\{0,M\}$ almost surely from every state.

Define $H(0)=0$ and $H(i)={\sum }_{m=0}^{i-1}{s}_m$ for $i\ge 1$. For $i\ge 1$, $H(i+1)-H(i)=s_i,H(i)-H(i-1)=s_{i-1},p_i{s}_i=q_i{s}_{i-1},$ where the last equality follows from the product definition of $s_i$. Consequently $p_i(H(i+1)-H(i))=q_i(H(i)-H(i-1)),$ which, together with $p_i+q_i+r_i=1$, gives $H(i)=p_{i}H(i+1)+r_{i}H(i)+q_{i}H(i-1)\mathrm{.}$ Thus $H\mathrm{/}H(M)$ is harmonic at every interior state of the finite chain, equals zero at $0$, and equals one at $M$.

By step 2.1, the finite-state chain $Y$ satisfies the all-start almost-sure boundary-hitting hypothesis of [F13]. Apply [F13] with boundary $\{0,M\}$ and payoff zero at $0$, one at $M$. Its hitting payoff is ${\mathbb{P}}_i(T_M<T_0)$, and [F14] identifies the unique bounded harmonic extension as $H(i)\mathrm{/}H(M)$ by step 2.2. Therefore ${\mathbb{P}}_i(T_M<T_0)=\frac{H(i)}{H(M)}\mathrm{.}$

The stopped path $Y$ equals the original path $X$ through its first hit of $\{0,M\}$. Hence from the interior start $i$ the event that $Y$ first hits $M$ rather than $0$ is exactly the original event $\{T_M<T_0\}$. Thus the probability in step 3.1 is the original-chain probability, not a new boundary convention.

As $M>i$ increases, the events $A_M=\{T_M<T_0\}$ decrease: reaching $M+1$ before $0$ requires first reaching $M$ before $0$. If $T_0<\mathrm{\infty }$, the path has a finite maximum before its first hit of $0$, so it fails to belong to $A_M$ for every $M$ above that maximum. Conversely, for each fixed $M$, step 2.1 shows that the first hit of $\{0,M\}$ is almost surely finite; on $\{T_0=\mathrm{\infty }\}$ this forces $T_M<T_0$ almost surely. Taking the countable intersection over $M>i$ proves ${\mathbb{P}}_i\text{\hspace{0.17em}⁣}(\{T_0=\mathrm{\infty }\}\mathrm{△}{\bigcap }_{M>i}{A}_M)=0.$ By [F15] and steps 3.1 and 4.1, ${\mathbb{P}}_i(T_0=\mathrm{\infty })={\lim }_{M\to \mathrm{\infty }}\frac{H(i)}{H(M)}\mathrm{.}$ Since $H(M)={\sum }_{m=0}^{M-1}{s}_m\uparrow S$, this limit is $H(i)\mathrm{/}S$ when $S<+\mathrm{\infty }$ and zero when $S=+\mathrm{\infty }$.

From state $0$, the first step is a self-loop with probability $r_0$, which is an immediate positive-time return, or a move to $1$ with probability $p_0$. In the latter case, the future path returns to $0$ exactly when it hits $0$ from start $1$; applying [F9] to the bounded event functional ${\mathbf{1}}_{\{T_0<\mathrm{\infty }\}}$ gives ${\mathbb{P}}_0(T_0^{+}<\mathrm{\infty })=r_0+p_0{\mathbb{P}}_1(T_0<\mathrm{\infty })=1-p_0{\mathbb{P}}_1(T_0=\mathrm{\infty })\mathrm{.}$ If $S=+\mathrm{\infty }$, step 5.1 makes this return probability one, so $0$ is recurrent by [F7]. If $S<+\mathrm{\infty }$, step 5.1 gives ${\mathbb{P}}_1(T_0=\mathrm{\infty })=H(1)\mathrm{/}S=1\mathrm{/}S>0$; since $p_0>0$, the return probability is strictly less than one and $0$ is not recurrent. This proves both directions of the stated equivalence.

The state space is the fixed infinite set ${\mathbb{N}}_0$, so empty and one-state spaces do not instantiate the claim. Zero transition weights are exactly the off-neighbor entries and $q_0=0$; zero holding probabilities are allowed. The finite auxiliary chain has absorbing endpoint rows, while the original interior rows have $p_i,q_i>0$. The hitting time $T_0$ includes time zero by [F6], but the escape formula starts at $i\ge 1$; the return time $T_0^{+}$ from $0$ is strictly positive. The scale terms are all positive, $H(1)=1$, and $S\ge 1$, so the finite-denominator branch is defined. AC [A1] is used for canonical chain laws [F1], finite-dimensional laws [F8], bounded future-path conditioning [F9], and the Dirichlet theorem [F13], [F14]; once these Markov facts are available, the finite path, product, and difference calculations are choice-free. Both iff directions are proved in step 6.1.