Proof. It is sufficient to show that for every \(x\in\{2,\dots,n\}\) and every \(y,z\ge0\), we have \[\begin{align} \mathbb{P}(B(x-1,n)=y\mid B(x,n)=z)=P^{\,z+1}(0,y-z-1), \end{align}\] where \(P^{\,z+1}\) denotes the \((z+1)\)-step transition probability of the Markov chain \(\eta\) defined above. We adapt an idea by Kesten-Kozlov-Spitzer for nearest-neighbor random walks [23] which was later used for bond‑repelling walks on \(\mathbb{Z}\) by Toth in [14]. Fix \(x\in\{2,\dots,n\}\). By definition, \(B(x,n)\) is the number of backward jumps \(x\to x-1\) before the first hit of \(n\). Since the walk must cross the edge \(\{x-1,x\}\) one more time forward than backward in order to get from \(0\) to \(n\) before time \(\tau_n\), the number of forward crossings \(x-1\to x\) up to time \(\tau_n\) is exactly \(B(x,n)+1\).

Conditional on \(\{B(x,n)=z\}\), there are therefore \(z+1\) forward jumps from \(x-1\) to \(x\) by time \(\tau_n\). Each such forward jump from \(x-1\) to \(x\) is called a “failure”, and each jump from \(x-1\) to \(x-2\) is called a “success”. For \(1\le j\le z+1\), let \(\ell_j(x,n)\) be the number of successes that occur after the \((j-1)\)-st failure and before the \(j\)-th failure, where for \(j=1\) this means before the first failure. Then clearly \[\begin{align} B(x-1,n)=\sum_{j=1}^{z+1}\ell_j(x,n). \end{align}\] We now compute the law of the sequence \((\ell_j(x,n))_{1\le j\le z+1}\). Set \[u_0:=0,\quad u_j:=\sum_{i=1}^j\ell_i(x,n)-j\quad\text{for } j\ge1.\] Thus \(u_j=u_{j-1}+\ell_j(x,n)-1\). After the first \(j-1\) failures and the first \(\sum_{i=1}^{j-1}\ell_i(x,n)\) successes have occurred, the edge \(\{x-2,x-1\}\) has been crossed exactly \(2\sum_{i=1}^{j-1}\ell_i(x,n)+1\) times, whereas the edge \(\{x-1,x\}\) has been crossed exactly \(2(j-1)\) times. Therefore, whenever the walk is at \(x-1\) during the \(j\)-th stage, the probability that the next move is a success is \[\frac{e^{-\beta\left(2\sum_{i=1}^{j-1}\ell_i(x,n)+1\right)}}{e^{-\beta\left(2\sum_{i=1}^{j-1}\ell_i(x,n)+1\right)}+e^{-2\beta(j-1)}} = \frac{e^{-\beta(2u_{j-1}+1)}}{1+e^{-\beta(2u_{j-1}+1)}} =p(u_{j-1}),\] and the probability of a failure is \(q(u_{j-1})\). Hence, conditional on the past up to the beginning of the \(j\)-th stage, \[\begin{align} &\mathbb{P}\bigl(\ell_j(x,n)=s_j \,\big|\, \ell_1(x,n)=s_1,\dots,\ell_{j-1}(x,n)=s_{j-1},\, B(x,n)=z\bigr)\\ &\quad= q(u_{j-1}+s_j)\prod_{r=u_{j-1}}^{u_{j-1}+s_j-1}p(r)=\mathbb{P}\bigl(\eta_j=u_{j-1}+s_j-1 \mid \eta_{j-1}=u_{j-1}\bigr). \end{align}\] Therefore, \[\begin{align} &\mathbb{P}\bigl(\ell_1(x,n)=s_1,\dots,\ell_{z+1}(x,n)=s_{z+1}\mid B(x,n)=z\bigr)\\ &\quad= \mathbb{P}\bigl(\eta_1-\eta_0=s_1-1,\;\dots,\;\eta_{z+1}-\eta_z=s_{z+1}-1\mid \eta_0=0\bigr). \end{align}\] Summing over all sequences \((s_1,\dots,s_{z+1})\) with total \(\sum_{j=1}^{z+1}s_j=y\), we obtain \[\mathbb{P}\big(B(x-1,n)=y\mid B(x,n)=z\big) = \sum_{s_1+\dots+s_{z+1}=y} \mathbb{P}\bigl(\eta_1-\eta_0=s_1-1,\dots,\eta_{z+1}-\eta_z=s_{z+1}-1\mid\eta_0=0\bigr).\] The right-hand side is exactly the probability that after \(z+1\) steps the chain \(\eta\), starting at \(0\), is at position \(y-(z+1)\). Hence \[\mathbb{P}(B(x-1,n)=y\mid B(x,n)=z)=P^{\,z+1}(0,\;y-z-1).\] This completes the proof. ◻

Proof. By Lemma 3.b, we have \[\begin{align} P^n(0,m) &= P^n(0,0)\prod_{r=0}^{m-1}\frac{P^n(0,r+1)}{P^n(0,r)} \le P^n(0,0)C_0^m \exp\!\Bigl(-\beta\sum_{r=0}^{m-1}r\Bigr) \le C e^{-c(m+1)^2}. \end{align}\] For \(n=z+1\) and \(m=j-1\), we get for every \(j\ge 1\), \(\mu_z(j)=P^{\,z+1}(0,j-1)\le C e^{-c j^2}.\) After enlarging \(C\) if necessary, the same bound also holds for \(j=0\). Hence ?? is verified.

Recall that by definition, \(\nu(j)=\varrho(j-1)\). Therefore \[\begin{align} \sum_{j\in\mathbb{Z}}|\mu_z(j)-\nu(j)| &= \sum_{j\in\mathbb{Z}}\bigl|P^{\,z+1}(0,j-1)-\varrho(j-1)\bigr|= \sum_{m\in\mathbb{Z}}|P^{\,z+1}(0,m)-\varrho(m)|. \end{align}\] By Lemma 3.a, the last quantity is bounded by \(Ce^{-cz}\), which proves ?? .

Using the fact that \(\nu\) has zero mean, we have \[\begin{align} |m_1(z)| &= \Big|\sum_{j\in\mathbb{Z}}j\bigl(\mu_z(j)-\nu(j)\bigr)\Big|. \end{align}\] Splitting the sum into \(|j|\le z\) and \(|j|>z\), we get \[\begin{align} |m_1(z)| &\le z\sum_{j\in\mathbb{Z}}|\mu_z(j)-\nu(j)| +\sum_{|j|>z}|j|\mu_z(j) +\sum_{|j|>z}|j|\nu(j). \end{align}\] By ?? , the first term is at most \(Cze^{-cz}\). For the \(\mu_z\)-tail, note that \(Y_1\ge 0\), so \(Y_1-Y_0\ge -Y_0\) and therefore \(\mu_z(j)=0\) for \(j<-z\). Using this fact and ?? , we have \[\sum_{|j|>z}|j|\mu_z(j)=\sum_{j>z}|j|\mu_z(j)\le C_0\sum_{j>z}j\,e^{-c_0 j^2}\le Ce^{-cz}.\] Since \(\nu\) has Gaussian tails by definition, \(\sum_{|j|>z}|j|\nu(j)\le Ce^{-cz}\). The two tail sums and the first term are all bounded by \(Ce^{-cz}\) after absorbing the linear factor into the exponential. This proves ?? . The proof for \(m_2(z)\) is analogous. We write \[m_2(z)-\varsigma^2_{\beta} = \sum_{j\in\mathbb{Z}}j^2\bigl(\mu_z(j)-\nu(j)\bigr)\] and split the sum at \(|j|\le z\). The bounded part is controlled by ?? , while the tails are controlled by Gaussian domination. This proves ?? .

By definition, \[\mathbb{P}(|Y_1-Y_0|\ge m\mid Y_0=z) = \sum_{|j|\ge m}\mu_z(j).\] Using ?? and the fact that \(\mu_z(j)=0\) for \(j<-z\), we have \(\sum_{|j|\ge m}\mu_z(j) \le Ce^{-c m}.\) This proves ?? . ◻

Proof. We first show that \(Y\) is irreducible on \(\mathbb{Z}_+\). For every \(z \ge 1\), we have \[Q(0,z)=P(0,z-1)=q(z)\prod_{r=0}^{z-1}p(r)>0,\] Also, for every \(z \ge 0\), \[Q(z,0)=P^{\,z+1}(0,-z-1)\ge \prod_{r=0}^{z}q(-r)>0,\] since the Markov chain \(\eta\) can follow the path \(0\to -1\to -2\to \cdots \to -(z+1)\) with strictly positive probability. Hence, every state communicates with \(0\), and the Markov chain \(Y\) is thus irreducible.

By ?? and ?? , there exist constants \(C,c>0\) such that \[m_1(z)\le C e^{-cz}, \quad \bigl|m_2(z)-\varsigma^2_{\beta}\bigr|\le C e^{-cz}, \quad z \ge 0.\] Hence there exists \(z_0 \in \mathbb{N}\) such that for all \(z \ge z_0\), \(2z\,m_1(z)<m_2(z).\) Moreover, by ?? , the conditional increment law of \(Y\) has a Gaussian upper tail uniformly in the current state. In particular, \(\sup_{z\ge 0}\mathbb{E}[|Y_1-Y_0|^p \mid Y_0=z]<\infty\) for \(p>2\). Applying the recurrence criterion of Markov chains with asymptotically zero drift (see e.g. Theorem 3.2.3 in [24], p. 94), we deduce that the chain \(Y\) is null recurrent. ◻

Proof. Let \[\begin{align} \sigma_0 :=\inf\{n\ge 0:Y_n=0\}\quad\text{and}\quad \widetilde{\sigma}_N:=\inf\{n\ge 0:Y_n\ge N\}\quad \text{for N\ge 1}. \end{align}\] Recall from Lemma 5 that \(Y\) is irreducible and recurrent on \(\mathbb{Z}_+\). Hence for every \(N\ge 1\) and every \(1\le x<N\), we have \[T_N:=\sigma_0\wedge\widetilde{\sigma}_N<\infty \quad\text{a.s. under }\mathbb{P}_x.\]

For \(x\in \mathbb{Z}_+\), define \(\varphi_N(x):=\mathbb{P}_x(\widetilde{\sigma}_N<\sigma_0)\). Note that \(\varphi_N(0):=0\) and \(\varphi_N(x):=1\) for \(x\ge N\). Then, for every \(1\le x\le N-1\), \[\label{eq:hN-harmonic} \varphi_N(x)=\mathbb{E}_x\!\left[\varphi_N(Y_1)\mathbf{1}_{\{Y_1>0\}}\right].\tag{4}\]

We first construct global sub- and super-harmonic barriers.

Step 1: exponential test functions. Set \(\theta_x:=Y_1-x\). By Lemma 4, the jumps of \(Y\) have uniformly exponential tails, so there exists \(\lambda_0>0\) such that \[\label{eq:uniform-exp-moment} \sup_{x\ge 1}\mathbb{E}_x\!\left[e^{\lambda_0|\theta_x|}\right]<\infty.\tag{5}\] In particular, \(\sup_{x\ge 1}\mathbb{E}_x\!\left[|\theta_x|^3e^{\lambda_0|\theta_x|}\right]<\infty.\) Choose \(\gamma\in(0,\lambda_0\wedge c)\) so small that \[\frac{\varsigma^2_{\beta}}{4}\gamma^2-C_1\gamma^3>0,\] where \(C_1<\infty\) is such that the remainder term \(R_x\) in the Taylor expansion \(e^{-\gamma \theta_x}=1-\gamma \theta_x+\frac{\gamma^2}{2}\theta_x^2+R_x\) satisfies \(|R_x| \le \frac{\gamma^3}{6}|\theta_x|^3e^{\gamma|\theta_x|}\) and \(\sup_{x\ge 1}\mathbb{E}_x|R_x|\le C_1\gamma^3.\) Therefore, \[\mathbb{E}_x[e^{-\gamma \theta_x}] = 1-\gamma m_1(x)+\frac{\gamma^2}{2}m_2(x)+O(\gamma^3),\] uniformly in \(x\). Recall that by ?? and ?? , \(m_1(x)=O(e^{-cx})\) and \(m_2(x)=\varsigma^2_{\beta}+O(e^{-cx}).\) Hence there exist constants \(\varepsilon>0\) and \(K_0\in [1,\infty)\) such that \[\label{eq:exp-drift-full} \mathbb{E}_x[e^{-\gamma (Y_1-x)}]\ge 1+2\varepsilon \quad\text{for all }x\ge K_0.\tag{6}\] Since, by Lemma 4, \(Y_1-Y_0\) has uniformly exponential tails, we also have \(\mathbb{P}_x(Y_1=0)\le C e^{-cx}\), for each \(x\ge 1\). After enlarging \(K_0\) if necessary and using \(\gamma<c\), it follows from 6 that \[\label{eq:exp-drift} \mathbb{E}_x[e^{-\gamma (Y_1-x)}\mathbf{1}_{\{Y_1>0\}}]\ge 1+\varepsilon \quad\text{for all }x\ge K_0.\tag{7}\]

Step 2: sub-harmonic and super-harmonic barriers. For each function \(\phi: \mathbb{Z}_+\to\mathbb{R}\), let \[P_+\phi(x):=\mathbb{E}_x[\phi(Y_1)\mathbf{1}_{\{Y_1>0\}}].\] Let \(I(x):=x\). Using ?? and ?? , we notice that \[P_+I(x)-x = \mathbb{E}_x[(Y_1-x)\mathbf{1}_{\{Y_1>0\}}] = m_1(x)+x\,\mathbb{P}_x(Y_1=0) = O(e^{-c'x})\] for some \(c'>0\). Therefore, we can choose sufficiently large \(A>0\) such that \[\label{eq:A-choice} A\varepsilon e^{-\gamma x}\ge 2|P_+I(x)-x| \quad\text{for all }x\ge K_0.\tag{8}\] Define \(\psi_+(0)=0,\psi_-(0)=-A\) and \[\begin{align} \psi_+(x):=x+A e^{-\gamma x}, \quad \psi_-(x):=x-A e^{-\gamma x}, \quad \text{for x\ge K_0} \end{align}\] For \(1\le x\le K_0-1\), define \[\psi_+(x):=\mathbb{E}_x\!\left[\psi_+(Y_{T_{K_0}})\right], \quad \psi_-(x):=\mathbb{E}_x\!\left[\psi_-(Y_{T_{K_0}})\right].\] These expectations are finite because ?? implies that the overshoot at \(T_{K_0}\) has finite mean. Hence, \(\psi_+, \psi_- : \mathbb{Z}_+\to \mathbb{R}\) are well-defined. By 7 and 8 , we notice that for \(x\ge K_0\), \[\begin{align} P_+\psi_+(x)-\psi_+(x) &= \left(P_+I(x)-x\right) +A e^{-\gamma x}\bigl(\mathbb{E}_x[e^{-\gamma(Y_1-x)}\mathbf{1}_{\{Y_1>0\}}]-1\bigr) \ge 0,\\ P_+\psi_-(x)-\psi_-(x) &= \left(P_+I(x)-x\right) -A e^{-\gamma x}\bigl(\mathbb{E}_x[e^{-\gamma(Y_1-x)}\mathbf{1}_{\{Y_1>0\}}]-1\bigr) \le 0. \end{align}\] By the strong Markov property, for every \(1\le x\le K_0-1\), \[P_+\psi_+(x)=\psi_+(x), \quad P_+\psi_-(x)=\psi_-(x).\] Hence, for all \(x\ge 1\), \[\label{eq:global-barriers} P_+\psi_+(x)\ge \psi_+(x), \quad P_+\psi_-(x)\le \psi_-(x).\tag{9}\]

Step 3: a uniform overshoot bound. We claim that \[\label{eq:overshoot-bound} \sup_{N\ge 1}\sup_{1\le x<N} \mathbb{E}_x\!\left[(Y_{T_N}-N)^+;\;\widetilde{\sigma}_N<\sigma_0\right] <\infty.\tag{10}\] Indeed, on \(\{\widetilde{\sigma}_N<\sigma_0\}\), we have \(Y_{T_N-1}<N\) and hence \(Y_{T_N}-N\le (Y_{T_N}-Y_{T_N-1})^+.\) Therefore \[\begin{align} \mathbb{E}_x\!\left[(Y_{T_N}-N)^+;\;\widetilde{\sigma}_N<\sigma_0\right] &\le \sum_{n\ge 1}\mathbb{E}_x\!\left[(Y_n-Y_{n-1})^+;\;T_N=n\right]\\ &= \sum_{n\ge 1}\mathbb{E}_x\!\left[ \mathbf{1}_{\{T_N=n\}} \mathbb{E}\!\left[(Y_n-Y_{n-1})^+\mid Y_{n-1}\right]\right] \le \sup_{z\ge 1}\mathbb{E}_z[(Y_1-z)^+]. \end{align}\] The last quantity is finite by the exponential tail bound in Lemma 4, proving 10 .

Step 4: finite-volume estimates. Fix \(N>K_0\) and \(1\le x<N\). First, since \(\psi_-(Y_{n\wedge T_N})+A\) is a nonnegative supermartingale by 9 , by Fatou’s lemma, \[\mathbb{E}_x[\psi_-(Y_{T_N})]\le \psi_-(x).\] On the event \(\{\widetilde{\sigma}_N<\sigma_0\}\), we have \(Y_{T_N}\ge N\), and therefore \(\psi_-(Y_{T_N})\ge Y_{T_N}-A e^{-\gamma Y_{T_N}}\ge N-A.\) On the event \(\{\sigma_0<\widetilde{\sigma}_N\}\), we have \(Y_{T_N}=0\) and hence \(\psi_-(Y_{T_N})=-A.\) Thus \[\mathbb{E}_x[\psi_-(Y_{T_N})] \ge (N-A)\varphi_N(x)-A(1-\varphi_N(x)) = N \varphi_N(x)-A.\] Hence \[\label{eq:hN-upper} N \varphi_N(x)\le \psi_-(x)+A.\tag{11}\] Similarly, since \(\psi_+(Y_{n\wedge T_N})\) is a nonnegative submartingale, by Fatou’s lemma, we have \[\psi_+(x)\le \mathbb{E}_x[\psi_+(Y_{T_N})].\] On \(\{\sigma_0<\widetilde{\sigma}_N\}\), we have \(Y_{T_N}=0\) and \(\psi_+(Y_{T_N})=0\). On \(\{\widetilde{\sigma}_N<\sigma_0\}\), we have \(\psi_+(Y_{T_N})\le Y_{T_N}+A.\) Therefore, using 10 , \[\mathbb{E}_x[\psi_+(Y_{T_N})] \le N \varphi_N(x)+A \varphi_N(x) +\mathbb{E}_x\!\left[(Y_{T_N}-N)^+;\;\widetilde{\sigma}_N<\sigma_0\right] \le N \varphi_N(x)+C_2,\] for some constant \(C_2<\infty\) independent of \(x,N\). Hence \[\label{eq:hN-lower} N \varphi_N(x)\ge \psi_+(x)-C_2.\tag{12}\]

For \(x\ge K_0\), we have \(\psi_+(x)=x+A e^{-\gamma x}\) and \(\psi_-(x)=x-A e^{-\gamma x}\). Thus 11 –12 imply that \[\label{eq:NhN-large} x-C_3\le N \varphi_N(x)\le x+C_3, \quad K_0\le x<N,\tag{13}\] for some constant \(C_3<\infty\). For \(1\le x<K_0\), 11 implies \[\label{eq:NhN-small} N \varphi_N(x)\le \max_{1\le y<K_0}\psi_-(y)+A=:C_4.\tag{14}\] For \(x\in \mathbb{Z}_+\), define \[h_N(x):=N \varphi_N(x).\] Since also \(h_N(x)\ge 0\), we obtain \(|h_N(x)-x|\le \max\{C_4,K_0-1\}\) for all \(1\le x<K_0\). Combining with 13 , we conclude that there exists \(C<\infty\) such that \[\label{eq:HN-uniform-profile} |h_N(x)-x|\le C \quad\text{for all }N\ge 2,\;1\le x<N.\tag{15}\]

Step 5: compactness and passage to the limit. By 11 and 15 , there exists \(C_5<\infty\) such that \[\label{eq:HN-linear} 0\le h_N(x)\le x+C_5 \quad\text{for all }x\ge 0,\;N\ge 1.\tag{16}\] Fix \(x\ge 0\). Then the sequence \((h_N(x))_{N>x}\) is bounded by 16 . By a diagonal argument, there exists a subsequence \(N_k\to\infty\) and a function \(h:\mathbb{Z}_+\to[0,\infty)\) such that \[h_{N_k}(x)\to h(x) \quad\text{for every }x\ge 0.\] Since \(h_{N_k}(0)=0\), we have \(h(0)=0\).

We now prove harmonicity. Fix \(x\ge 1\). For all \(k\) sufficiently large, \(x<N_k\), and 4 gives \[h_{N_k}(x)=\mathbb{E}_x\!\left[h_{N_k}(Y_1)\mathbf{1}_{\{Y_1>0\}}\right].\] By 16 , \(0\le h_{N_k}(Y_1)\mathbf{1}_{\{Y_1>0\}}\le Y_1+C_5.\) Since \(Y_1\) has finite first moment by Lemma 4, dominated convergence yields \[\lim_{k\to\infty}\mathbb{E}_x\!\left[h_{N_k}(Y_1)\mathbf{1}_{\{Y_1>0\}}\right] = \mathbb{E}_x\!\left[h(Y_1)\mathbf{1}_{\{Y_1>0\}}\right].\] Passing to the limit, we obtain \[h(x)=\mathbb{E}_x\!\left[h(Y_1)\mathbf{1}_{\{Y_1>0\}}\right], \quad x\ge 1.\]

Finally, 13 implies that for every fixed \(x\ge K_0\) and all \(k\) sufficiently large, \(|h_{N_k}(x)-x|\le C_3.\) Letting \(k\to\infty\), we obtain \(|h(x)-x|\le C_3\) for each \(x\ge K_0\). Thus \(h(x)=x+O(1)\) as \(x\to\infty\). This completes the proof. ◻

Proof. Fix \(x,y\in E_K\), and write \(x':=x-K,\;y':=y-K\), which are positive integers. Using the translation invariance, we note that \(p_S^{(K)}(m;x,y)\) is exactly the local probability that the symmetric random walk \((S_n)_{n\ge0}\) starts from \(x'\) and stays positive up to time \(m\) and is at \(y'\) at time \(m\), i.e. \[\begin{align} \label{eq:half-line-identification} p_S^{(K)}(m;x,y) = \mathbb{P}_{x'}^S(S_1>0,\dots,S_{m-1}>0,\;S_m=y'). \end{align}\tag{22}\] Since \((S_n)_{n\ge0}\) is an aperiodic walk on \(\mathbb{Z}\) whose common increment law \(\nu\) has zero mean and finite variance \(\varsigma^2_{\beta}\), the local probability of \(S\) conditioned to stay positive has the exact asymptotic (see, e.g., Theorem 3 in [25]): \[\label{eq:benchmark-local-asymptotic} \begin{align} p_S^{(K)}(m;x,y) &\sim \frac{H(x+1-K)\,H(y-K)}{\varsigma_\beta\sqrt{2\pi}}\, m^{-3/2} \exp\!\Big(-\frac{(x-y)^2}{2\varsigma^2_{\beta}m}\Big)\\ &\sim \frac{H_K(x+1)\,H_K(y)}{\varsigma_\beta\sqrt{2\pi}}\, m^{-3/2}, \quad m\to\infty, \end{align}\tag{23}\] where we recall that \(H\) is the renewal function given by 19 and \(H_K(z):=H(z-K)\) for \(z\in E_K\).

For \(u, v\in \mathbb{Z}\) and \(n \ge 1\), set \[q_n(u,v):=\mathbb{P}_u\bigl(S_1\neq0,\dots,S_{n-1}\neq0,\;S_n=v\bigr).\] By formula (1.5) in [26], there exists a constant \(C\in (0,\infty)\) such that \[\label{eq:uchiyama-origin-bound} q_n(u,v)\le C\,u\,v\,n^{-3/2} \quad \text{for all } u,v\ge1 \text{ and } n\ge1.\tag{24}\] Combining 22 and 24 together with the fact from 20 that \(H_{K}(z)\ge c (z-K)\) for all \(z\in E_K\) with some constant \(c>0\), we have \[\label{eq:half-line-vs-origin} p_S^{(K)}(n;x,y)\le q_n(x-K,y-K) \le C(x-K+1)(y-K) n^{-3/2}\le C H_K(x+1)H_K(y)n^{-3/2}.\tag{25}\] Using this upper bound, we notice that \[\begin{align} &\widehat{p}_S^{(K)}(1;x,y)-\widehat{p}_S^{(K)}(s;x,y) = \sum_{n\ge1}p_S^{(K)}(n;x,y)(1-s^n) \\ &\le C\,H_K(x+1)\,H_K(y)\sum_{n\ge1}n^{-3/2}(1-s^n) \le C\,H_K(x+1)\,H_K(y)\sqrt{1-s}. \end{align}\] Moreover, by the asymptotic 23 , the Abelian theorem yields \[\widehat{p}_S^{(K)}(1;x,y)-\widehat{p}_S^{(K)}(s;x,y) = \frac{\sqrt2}{\varsigma_\beta}\,H_K(x+1)\,H_K(y)\,\sqrt{1-s} + o(H_K(x+1)\,H_K(y)\sqrt{1-s}), \quad s\uparrow1.\] ◻

Proof. Fix \(K\ge0\). For \(x,y\in E_K\), note that \[\begin{align} &Q(x,y)=\mu_x(y-x)\quad \text{and} \quad \bar Q(x,y)=\nu(y-x). \end{align}\] Recall from 20 that there exists \(C_0\ge1\), such that for all \(x\in E_K\), \[C_0^{-1}(x-K+1)\le H_K(x)\le C_0(x-K+1).\] Therefore it suffices to prove that \[\label{eq:weight46sum} \sum_{y\in E_K}|Q_K(x,y)-\bar Q_K(x,y)|\,(y-K+2) \le Ce^{-cx}(x-K+1), \quad x\in E_K.\tag{26}\] Fix \(x\in E_K\). We have \[\begin{align} \sum_{y\in E_K}|Q_K(x,y)-\bar Q_K(x,y)|\,(y-K+2)&= \sum_{y>K}|\mu_x(y-x)-\nu(y-x)|\,(y-K+2)\\ &\le \sum_{j\in\mathbb{Z}}|\mu_x(j)-\nu(j)|\big(x-K+2+|j|\big)\\ &= (x-K+2)\sum_{j\in\mathbb{Z}}|\mu_x(j)-\nu(j)| + \sum_{j\in\mathbb{Z}}|j|\,|\mu_x(j)-\nu(j)|. \end{align}\] The first sum is bounded by \(C(x-K+1)e^{-cx}\) by ?? . For the second sum, splitting it into \(|j|\le x\) and \(|j|>x\), we have \[\begin{align} \sum_{j\in\mathbb{Z}}|j|\,|\mu_x(j)-\nu(j)| &\le x\sum_{j\in\mathbb{Z}}|\mu_x(j)-\nu(j)| + \sum_{|j|>x}|j|\,\mu_x(j) + \sum_{|j|>x}|j|\,\nu(j). \end{align}\] The first term is again bounded by \(Cxe^{-cx}\) using ?? . For the second term, using summation by parts and ?? , we have \[\sum_{|j|>x}|j|\,\mu_x(j) \le \sum_{m>x}\mathbb{P}_x(|Y_1-Y_0|\ge m) \le Ce^{-cx}.\] The third term is bounded by \(Ce^{-cx}\) as \(\nu\) has Gaussian tails. Hence, 26 is verified. This completes the proof. ◻

Proof. Note that \[p_Y^{(K)}(m;x,y)=(Q_K^m)(x,y), \quad p_S^{(K)}(m;x,y)=(\bar Q_K^m)(x,y).\] We notice that for \(m\ge1\), \[\begin{align} Q_K^{m}-\bar Q_K^{m} = Q_K^{m-1}(Q_K-\bar Q_K)+(Q_K^{m-1}-\bar Q_K^{m-1})\bar Q_K. \end{align}\] Using the above identity and induction, we have \[Q_K^m-\bar Q_K^m = \sum_{j=0}^{m-1}Q_K^j(Q_K-\bar Q_K)\bar Q_K^{\,m-1-j}, \quad m\ge1.\] Taking the \((x,y)\)-entry of both sides, we obtain the claimed formula. ◻

Proof. Fix \(K\ge0\), \(u\in W_K\) and \(z\in E_K\). By Lemma 7, \(p_S^{(K)}(n;x,z)\le C\,H_K(x+1)\,H_K(z)\,n^{-3/2}\) for each \(x\in E_K, n\ge1\). Hence \[\begin{align} f_S^{(K)}(n;u,z)=\sum_{x\in E_K}Q(u,x)\,p_S^{(K)}(n;x,z)& \le C\,H_K(z)\,n^{-3/2}\sum_{x\in E_K}Q(u,x)\,H_K(x+1)\\ &= C\,A_{K,u}\,H_K(z)\,n^{-3/2}, \end{align}\] which proves ?? . By the definition of \(\widehat f_S^{(K)}\), we have \[\begin{align} \widehat f_S^{(K)}(1;u,z)-\widehat f_S^{(K)}(s;u,z) &= \sum_{x\in E_K}Q(u,x)\sum_{n\ge0}p_S^{(K)}(n;x,z)(1-s^n)\\ &= \sum_{x\in E_K}Q(u,x)\Bigl(\widehat p_S^{(K)}(1;x,z)-\widehat p_S^{(K)}(s;x,z)\Bigr). \end{align}\] By Lemma 7, for each fixed \(x, z\in E_K\), \[\begin{align} &\widehat p_S^{(K)}(1;x,z)-\widehat p_S^{(K)}(s;x,z) = \frac{\sqrt2}{\varsigma_\beta}\,H_K(x+1)\,H_K(z)\,\sqrt{1-s} + o\!\bigl(H_K(x+1)H_K(z)\sqrt{1-s}\bigr),\quad s\uparrow1,\\ &0\le \widehat p_S^{(K)}(1;x,z)-\widehat p_S^{(K)}(s;x,z) \le C\,H_K(x+1)\,H_K(z)\,\sqrt{1-s}, \quad 0\le s<1. \end{align}\] Since \(\sum_{x\in E_K}Q(u,x)\,H_K(x+1)=A_{K,u}<\infty,\) the dominated convergence theorem implies \[\begin{align} \widehat f_S^{(K)}(1;u,z)-\widehat f_S^{(K)}(s;u,z) &= \frac{\sqrt2}{\varsigma_\beta}\,H_K(z)\,\sqrt{1-s} \sum_{x\in E_K}Q(u,x)\,H_K(x+1) + o\!\bigl(H_K(z)\sqrt{1-s}\bigr)\\ &= \frac{\sqrt2\,A_{K,u}}{\varsigma_\beta}\,H_K(z)\,\sqrt{1-s} + o\!\bigl(H_K(z)\sqrt{1-s}\bigr), \end{align}\] which is exactly ?? . Finally, using ?? , we have \[\begin{align} 0\le \widehat f_S^{(K)}(1;u,z)-\widehat f_S^{(K)}(s;u,z) &= \sum_{n\ge1}f_S^{(K)}(n;u,z)(1-s^n)\\ &\le C\,A_{K,u}\,H_K(z)\sum_{n\ge1}n^{-3/2}(1-s^n)\le C\,A_{K,u}\,H_K(z)\,\sqrt{1-s}. \end{align}\] This verifies ?? . ◻

Proof. Using the Duhamel’s formula in Lemma 9, \[\begin{align} p_Y^{(K)}(m;x,z)-p_S^{(K)}(m;x,z) &= \sum_{j=0}^{m-1}\sum_{a,b\in E_K} p_Y^{(K)}(j;x,a)\,\Delta_K(a,b)\,p_S^{(K)}(m-1-j;b,z). \end{align}\] Multiply by \(Q(u,x)\) and sum over \(x\in E_K\), we get \[\begin{align} f_Y^{(K)}(m;u,z) &= f_S^{(K)}(m;u,z) +\sum_{j=0}^{m-1}\sum_{a,b\in E_K} f_Y^{(K)}(j;u,a)\,\Delta_K(a,b)\,p_S^{(K)}(m-1-j;b,z). \label{eq:A-recursion} \end{align}\tag{28}\] By Lemma 7, \(p_S^{(K)}(m-1-j;b,z)\le C\,H_K(b+1)H_K(z)\,(m-j)^{-3/2}.\) Therefore, using Lemma 8, \[\begin{align} \sum_{b\in E_K}|\Delta_K(a,b)|\,p_S^{(K)}(m-1-j;b,z) &\le C H_K(z)(m-j)^{-3/2} \sum_{b\in E_K}|\Delta_K(a,b)|\,H_K(b+1)\\ &\le Ce^{-ca}H_K(a)H_K(z)(m-j)^{-3/2}. \end{align}\] Also, by Lemma 10, \(f_S^{(K)}(m;u,z)\le C {A}_{K,u}\,H_K(z)\,m^{-3/2}.\) Taking absolute values in 28 and using the previous bound, we obtain \[\begin{align} f_Y^{(K)}(m;u,z) &\le C A_{K,u}H_K(z)m^{-3/2} + C H_K(z)\sum_{j=0}^{m-1}(m-j)^{-3/2} \sum_{a\in E_K}f_Y^{(K)}(j;u,a)e^{-ca}H_K(a). \label{eq:A-ineq} \end{align}\tag{29}\] Define \[M_m:=\sup_{1\le r\le m}\sup_{z\in E_K}\frac{f_Y^{(K)}(r;u,z)}{A_{K,u}H_K(z)\,r^{-3/2}}, \quad m\ge1.\] We have \[\begin{align} \sum_{a\in E_K}f_Y^{(K)}(j;u,a)e^{-ca}H_K(a) &\le A_{K,u}M_{m-1}j^{-3/2}\sum_{a\in E_K}e^{-ca}H_K(a)^2. \end{align}\] Since \(a\in E_K\) implies \(a\ge K+1\) and \(H_K(a)\le C(a-K+1)\), the sum on the right-hand side is bounded by \(Ce^{-cK}\). Thus for \(j\ge1\), \[\sum_{a\in E_K}f_Y^{(K)}(j;u,a)e^{-ca}H_K(a)\le Ce^{-cK}A_{K,u}M_{m-1}j^{-3/2}.\] The \(j=0\) term is treated directly. Since \(f_Y^{(K)}(0;u,a)= Q(u,a)\) and \(a\ge K+1\), we have \[\sum_{a\in E_K}f_Y^{(K)}(0;u,a)e^{-ca}H_K(a)\le Ce^{-cK}A_{K,u}.\] Substituting these bounds into 29 , we obtain \[\begin{align} f_Y^{(K)}(m;u,z) &\le C {A}_{K,u}H_K(z)m^{-3/2} + C e^{-cK}A_{K,u}H_K(z)\Biggl[ m^{-3/2} + M_{m-1}\sum_{j=1}^{m-1}j^{-3/2}(m-j)^{-3/2} \Biggr]. \end{align}\] Since \(\sum_{j=1}^{m-1}j^{-3/2}(m-j)^{-3/2}\le C m^{-3/2},\) we conclude that \[f_Y^{(K)}(m;u,z)\le C\Bigl( 1+e^{-cK}(1+M_{m-1})\Bigr)A_{K,u}H_K(z)m^{-3/2}.\] Thus \(M_m\le C\big(1+ e^{-cK}(1+M_{m-1})\big).\) Choosing \(K_0\) sufficiently large such that \(C e^{-cK}\le \frac{1}{4}\) for all \(K\ge K_0\), we obtain \[M_m\le C+\frac{1}{4}(1+ M_{m-1}).\] A simple induction yields \(\sup_m M_m<\infty\), and thus \[f_Y^{(K)}(n;u,z)\le C\,A_{K,u}\,H_K(z)\,n^{-3/2}, \quad z\in E_K, u\in W_K, \;n\ge1.\] Moreover, we have \[\begin{align} 0\le \widehat f_Y^{(K)}(1;u,z)-\widehat f_Y^{(K)}(s;u,z)= \sum_{n\ge0}f_Y^{(K)}(n;u,z)(1-s^n)&\le C\,A_{K,u}\,H_K(z)\sum_{n\ge1}n^{-3/2}(1-s^n)\\ &\le C \,A_{K,u}\,H_K(z)\sqrt{1-s}. \end{align}\] This completes the proof. ◻

Proof. We divide the proof into several steps.

Step 1: Expansion of \(\widehat f_Y^{(K)}\) via Duhamel formula. Fix \(K\ge K_0\), \(u\in W_K\), and \(z\in E_K\). summing 28 over \(m\ge0\) yields \[\label{eq:gf-main} \widehat f_Y^{(K)}(s;u,z) = \widehat f_S^{(K)}(s;u,z) + \sum_{a\in E_K}\widehat f_Y^{(K)}(s;u,a)\,\kappa(s;a,z),\tag{30}\] where for \(a,z\in E_K\), we define \[\kappa(s;a,z):= s\sum_{b\in E_K}\Delta_K(a,b)\,\widehat{p}_S^{(K)}(s;b,z).\] Subtracting 30 at \(s\) from the corresponding identity at \(1\), we obtain \[\begin{align} \label{eq:U-eq} \widehat f_Y^{(K)}(1;u,z)-\widehat f_Y^{(K)}(s;u,z) &= \widehat f_S^{(K)}(1;u,z)-\widehat f_S^{(K)}(s;u,z) + \sum_{a\in E_K}\big(\widehat f_Y^{(K)}(1;u,a)-\widehat f_Y^{(K)}(s;u,a) \big)\kappa(1;a,z)\\ \nonumber &+\sum_{a\in E_K}\widehat f_Y^{(K)}(s;u,a)\bigl(\kappa(1;a,z)-\kappa(s;a,z)\bigr). \end{align}\tag{31}\]

We first record the basic bounds. By Lemma 11, \[\label{eq:fYhat-bound} \widehat f_Y^{(K)}(s;u,a)\le \widehat f_Y^{(K)}(1;u,a)\le C\,A_{K,u}\,H_K(a), \quad a\in E_K,\;0\le s<1.\tag{32}\] By Lemma 7, \(\widehat{p}_S^{(K)}(1;b,z)\le C\,H_K(b+1)\,H_K(z)\), for each \(b,z\in E_K.\) Using this bound together with Lemma 8, we get \[\begin{align} \label{eq:beta1-bound} |\kappa(1;a,z)| \le \sum_{b\in E_K}|\Delta_K(a,b)|\,\widehat{p}_S^{(K)}(1;b,z) \le C e^{-ca}H_K(a)\,H_K(z), \quad a,z\in E_K. \end{align}\tag{33}\]

Next, by Lemma 7, \[\widehat{p}_S^{(K)}(1;b,z)-\widehat{p}_S^{(K)}(s;b,z) = \frac{\sqrt2}{\varsigma_\beta}\,H_K(b+1)\,H_K(z)\,\sqrt{1-s} + o\bigl(H_K(b+1)\,H_K(z)\,\sqrt{1-s}\bigr)\] as \(s\uparrow1\), and \(|\widehat{p}_S^{(K)}(1;b,z)-\widehat{p}_S^{(K)}(s;b,z)| \le C\,H_K(b+1)\,H_K(z)\,\sqrt{1-s}.\) Also, \[\begin{align} \kappa(1;a,z)-\kappa(s;a,z) &= \sum_{b\in E_K}\Delta_K(a,b)\,\bigl(\widehat{p}_S^{(K)}(1;b,z)-\widehat{p}_S^{(K)}(s;b,z)\bigr)\\ &+ (1-s)\sum_{b\in E_K}\Delta_K(a,b)\,\widehat{p}_S^{(K)}(s;b,z). \end{align}\] Since \(\widehat{p}_S^{(K)}(s;b,z)\le \widehat{p}_S^{(K)}(1;b,z)\le C\,H_K(b+1)\,H_K(z)\), the second term is bounded by \[C(1-s)e^{-ca}H_K(a)H_K(z) = o\bigl(e^{-ca}H_K(a)H_K(z)\sqrt{1-s}\bigr)\] as \(s\uparrow1\). Set \[\delta_K(a):= \frac{\sqrt2}{\varsigma_\beta}\sum_{b\in E_K}\Delta_K(a,b)\,H_K(b+1), \quad a\in E_K,\] and note that \(|\delta_K(a)|\le C e^{-ca}H_K(a)\), for each \(a\in E_K.\) Hence \[\label{eq:beta-expansion} \kappa(1;a,z)-\kappa(s;a,z) = \delta_K(a)\,H_K(z)\,\sqrt{1-s} + o\bigl(e^{-ca}H_K(a)\,H_K(z)\,\sqrt{1-s}\bigr),\tag{34}\] as \(s\uparrow1\), with the uniform bound \[\label{eq:beta-bound} |\kappa(1;a,z)-\kappa(s;a,z)| \le C e^{-ca}H_K(a)\,H_K(z)\,\sqrt{1-s}.\tag{35}\]

Step 2: Functional equation for normalized generating functions. Now fix \(K\ge 0\) and \(u\in W_K\). For \(y\in E_K\) and \(0\le s<1\), define \[\begin{align} F_y^Y(s)&:=\frac{\widehat f_Y^{(K)}(1;u,y)-\widehat f_Y^{(K)}(s;u,y)}{H_K(y)\sqrt{1-s}}, \quad F^S_y(s):=\frac{\widehat f_S^{(K)}(1;u,y)-\widehat f_S^{(K)}(s;u,y)}{H_K(y)\sqrt{1-s}},\\ W_y(s)&:=\frac{\sum_{a\in E_K}\widehat f_Y^{(K)}(s;u,a)\,\bigl(\kappa(1;a,y)-\kappa(s;a,y)\bigr)}{H_K(y)\sqrt{1-s}}. \end{align}\] We first notice that by Lemma 11, \[\label{eq:F-bound} |F^Y_z(s)|\le C\,A_{K,u}, \quad z\in E_K,\;0\le s<1.\tag{36}\] Also, by Lemma 10, for every \(z\in E_K\), \[\label{eq:R-bound} F^S_z(s)\to \frac{\sqrt2\,A_{K,u}}{\varsigma_\beta} \quad\text{as }s\uparrow1\quad\text{and}\quad |F^S_z(s)|\le C\,A_{K,u}, \quad 0\le s<1.\tag{37}\] For \(W_z(s)\), we notice that by 32 and 35 , \[\begin{align} \label{W46bound} |W_z(s)|\le C A_{K,u}. \end{align}\tag{38}\] Using 34 and applying dominated convergence in the sum over \(a\), we obtain that for every fixed \(z\in E_K\), \[W_z(s)\to L_{K,u}:=\sum_{a\in E_K}\widehat f_Y^{(K)}(1;u,a)\,\delta_K(a)\quad \text{as s\uparrow1}.\] This sum is finite since \(|\delta_K(a)|\le C e^{-ca}H_K(a)\) and \(\widehat f_Y^{(K)}(1;u,a)\le C A_{K,u}H_K(a)\) by Lemma 11.

Next, define \[G_z(a):=\frac{H_K(a)\,\kappa(1;a,z)}{H_K(z)}, \quad a,z\in E_K.\] Then 31 becomes \[\label{eq:F-eq} F_z^Y(s) = F^S_z(s) + \sum_{a\in E_K}G_z(a)\,F_a^Y(s) + W_z(s).\tag{39}\] Moreover, by 33 , \(|G_z(a)|\le C e^{-ca}H_K(a)^2\), for each \(a,z\in E_K\). Since \(a\ge K+1\) on \(E_K\) and \(H_K(a)\le C(a-K+1)\), we have \(\sup_{z\in E_K}\sum_{a\in E_K}|G_z(a)|\le Ce^{-cK}.\) Therefore, for sufficiently large \(K_0\), we have \[\label{G46bound} \sup_{z\in E_K}\sum_{a\in E_K}|G_z(a)|\le \frac{1}{2} \quad\text{for all }K\ge K_0.\tag{40}\]

Define function \(\psi_s: E_K\to \mathbb{R}\) by \[\psi_s(z):=F^S_z(s)+W_z(s), \quad z\in E_K.\] Then 39 becomes \[F_z^Y(s)=\psi_s(z)+\sum_{a\in E_K}G_z(a)\,F_a^Y(s).\] By 37 and 38 , we have \(\sup_{z\in E_K}|\psi_s(z)|\le C A_{K,u}\), for each \(0\le s<1.\) Also, for each fixed \(z\in E_K\), \[\begin{align} \label{def46g} \psi_s(z)\to \widetilde{\psi}(z)\equiv \frac{\sqrt2\,A_{K,u}}{\varsigma_\beta}+L_{K,u} \quad\text{as }s\uparrow1. \end{align}\tag{41}\]

Step 3: Solution to the functional equation. Let \(\mathcal{G}\) be the bounded linear operator on \(\ell^\infty(E_K)\) defined by \[(\mathcal{G}\varphi)(z):=\sum_{a\in E_K}G_z(a)\,\varphi(a).\] By 40 , we have \(\|\mathcal{G}\|\le \frac{1}{2}\), and thus \((I-\mathcal{G})^{-1}=\sum_{m\ge0}\mathcal{G}^m\) on \(\ell^\infty(E_K)\). Therefore, \[F_z^Y(s)=\big((I-\mathcal{G})^{-1}\psi_s\big)(z)=\sum_{m\ge0}(\mathcal{G}^m \psi_s)(z).\] For each fixed \(m\) and fixed \(z\), the series defining \((\mathcal{G}^m \psi_s)(z)\) is absolutely summable, and dominated convergence yields \((\mathcal{G}^m \psi_s)(z)\to (\mathcal{G}^m \widetilde{\psi})(z)\) as \(s\uparrow1,\) where \(\widetilde{\psi}\) is the constant function defined in 41 . Moreover, \(|(\mathcal{G}^m \psi_s)(z)|\le \|\mathcal{G}\|^m\sup_{y\in E_K}|\psi_s(y)| \le C\,2^{-m},\) uniformly in \(s\). Therefore, by dominated convergence in \(m\), \[F_z^Y(s)\to C_{K,u}(z):=\sum_{m\ge0}(\mathcal{G}^m \widetilde{\psi})(z) \quad\text{as }s\uparrow1\] for every fixed \(z\in E_K\). Recalling the definition of \(F_z^Y(s)\), we conclude that \[\widehat f_Y^{(K)}(1;u,z)-\widehat f_Y^{(K)}(s;u,z)=C_{K,u}(z)\,H_K(z)\,\sqrt{1-s} + o\bigl(H_K(z)\,\sqrt{1-s}\bigr), \quad s\uparrow1.\]

Moreover, by 36 , \(|F_z^Y(s)|\le C\,A_{K,u}\) for all \(z\in E_K\) and \(0\le s<1\). Passing to the limit \(s\uparrow1\), we obtain that \(C_{K,u}(z)\le C\,A_{K,u}\) for all \(z\in E_K\). This completes the proof. ◻

Proof. By Lemma 12, \(C_{K,u}(y)\le C\,A_{K,u}\) for all \(y\in E_K, u\in W_K\). Hence \[\Lambda_K(u,v) = \sum_{y\in E_K}C_{K,u}(y)\,H_K(y)\,Q(y,v) \le C\,A_{K,u}\sum_{y\in E_K}H_K(y)\,Q(y,v) = C\,A_{K,u}\,B_{K,v}<\infty.\]

Fix \(K\ge K_0\) and \(u,v\in W_K\). For \(n\ge2\), a first return of \(Y\) to \(W_K\) at time \(n\) and location \(v\), started from \(u\in W_K\), must proceed as follows:

• the first step moves from \(u\in W_K\) into some \(x\in E_K\),

• the process has an excursion inside \(E_K\) of length \(n-2\) from \(x\) to some \(y\in E_K\),

• the final step moves from \(y\in E_K\) into \(v\in W_K\).

Hence, for all \(u,v\in W_K\) and all \(n\ge2\), we have \[\label{lem:first-return-decomposition} \mathcal{K}_Y^{(K)}(n;u,v) = \sum_{x\in E_K}\sum_{y\in E_K} Q(u,x)\,p_Y^{(K)}(n-2;x,y)\,Q(y,v)=\sum_{y\in E_K} f_Y^{(K)}(n-2;u,y)\,Q(y,v).\tag{42}\] Note also that \(\mathcal{K}_Y^{(K)}(1;u,v)=Q(u,v)\). Therefore, \[\begin{align} \sum_{n=1}^{\infty}\mathcal{K}_Y^{(K)}(n;u,v)(1-s^n) &=Q(u,v)(1-s) + \sum_{n\ge2}\sum_{y\in E_K} f_Y^{(K)}(n-2,u,y)\,Q(y,v)\,(1-s^n) \\ &= Q(u,v)(1-s)+ \sum_{y\in E_K}Q(y,v)\sum_{m\ge0}f_Y^{(K)}(m;u,y)\,(1-s^{m+2}). \end{align}\] Now write \(1-s^{m+2}=(1-s^m)+s^m(1-s^2)\). Therefore \[\begin{align} \sum_{n=1}^{\infty}\mathcal{K}_Y^{(K)}(n;u,v)(1-s^n) &= Q(u,v)(1-s)+ \sum_{y\in E_K}Q(y,v)\sum_{m\ge0}f_Y^{(K)}(m;u,y)(1-s^m) \notag\\ &\quad +(1-s^2)\sum_{y\in E_K}Q(y,v)\sum_{m\ge0}f_Y^{(K)}(m;u,y)s^m. \label{eq:K-split} \end{align}\tag{43}\]

We treat the second term and the last term on the right-hand side separately. For the second term, Lemma 12 gives, for each fixed \(y\in E_K\), \[\sum_{m\ge0}f_Y^{(K)}(m;u,y)(1-s^m) = C_{K,u}(y)H_K(y)\sqrt{1-s} + o(H_K(y)\sqrt{1-s}), \quad s\uparrow1.\] Moreover, by Lemma 11, \(\sum_{m\ge0}f_Y^{(K)}(m;u,y)(1-s^m) \le C\,A_{K,u}\,H_K(y)\,\sqrt{1-s}\) for each \(y\in E_K\) and \(m\ge1\). Since \(\sum_{y\in E_K}H_K(y)\,Q(y,v)=B_{K,v}<\infty,\) the dominated convergence theorem yields \[\begin{align} \sum_{y\in E_K}Q(y,v)\sum_{m\ge0}f_Y^{(K)}(m;u,y)(1-s^m) &= \sqrt{1-s}\sum_{y\in E_K}C_{K,u}(y)H_K(y)Q(y,v) + o(\sqrt{1-s}) \notag\\ &= \Lambda_K(u,v)\,\sqrt{1-s} + o(\sqrt{1-s}). \label{eq:first-main-term} \end{align}\tag{44}\]

For the last term in 43 , since \(\sum_{m\ge0}f_Y^{(K)}(m;u,y)s^m\le \sum_{m\ge0}f_Y^{(K)}(m;u,y)\le C\,A_{K,u}\,H_K(y)\) by Lemma 11, we obtain \[\begin{align} \nonumber 0\le (1-s^2)\sum_{y\in E_K}Q(y,v)\sum_{m\ge0}f_Y^{(K)}(m;u,y)s^m &\le C\,(1-s^2)\,A_{K,u}\sum_{y\in E_K}H_K(y)\,Q(y,v) \\ &\le C\,(1-s)\,A_{K,u}\,B_{K,v}=o(\sqrt{1-s}). \label{eq:second-term-small} \end{align}\tag{45}\] Finally, combining 43 , 44 , and 45 , we conclude that \[\sum_{n=1}^{\infty}\mathcal{K}_Y^{(K)}(n;u,v)\,(1-s^n) = \Lambda_K(u,v)\,\sqrt{1-s} + o(\sqrt{1-s}), \quad s\uparrow1.\] This completes the proof. ◻

Proof. Recall that the stopping times \((T_n)_{n\ge0}\) are defined by 46 . Fix \(u,v \in W_K\) and \(n \ge 0\). If \(Y_0 = u\) and \(Y_n = v\), then either \(n = 0\) and \(u = v\), or else there exists a unique \(r\ge1\) such that \(T_r=n.\) By the strong Markov property at the return times \(T_1,T_2,\dots,T_{r-1}\), the corresponding excursion increments are independent and each has law given by the first-return kernel \(\mathcal{K}_Y^{(K)}\). Therefore, \[\begin{align} &\mathbb{P}_u^Y\bigl(T_r=n,\;Y_n=v,\;T_1=n_1,\dots,T_r-T_{r-1}=n_r,Y_{T_1}=u_1,\dots,Y_{T_{r-1}}=u_{r-1}\bigr)\\ &\quad= \mathcal{K}_Y^{(K)}(n_1;u,u_1)\, \mathcal{K}_Y^{(K)}(n_2;u_1,u_2)\cdots \mathcal{K}_Y^{(K)}(n_r;u_{r-1},v). \end{align}\] Summing over all \(r\ge1\), all compositions \(n_1+\cdots+n_r=n\) with \(n_i\ge1\), and all intermediate states \(u_1,\dots,u_{r-1}\in W_K\), we obtain \[\mathbf{1}_{\{n=0,\;u=v\}} + \sum_{r=1}^{n} \;\sum_{\substack{n_1,\dots,n_r\ge 1\\ n_1+\cdots+n_r=n}} \;\sum_{u_1,\dots,u_{r-1}\in W_K} \mathcal{K}_Y^{(K)}(n_1;u,u_1)\, \mathcal{K}_Y^{(K)}(n_2;u_1,u_2)\cdots \mathcal{K}_Y^{(K)}(n_r;u_{r-1},v),\] where term \(\mathbf{1}_{\{n=0,\;u=v\}}\) accounts for the trivial case \(n=0\). Taking the matrix generating function, we obtain the result of the lemma. ◻

Proof. Recall from the matrix renewal decomposition in Lemma 13, \[\widehat{\mathcal{U}}(s)=\bigl(I-\widehat{\mathcal{K}}(s)\bigr)^{-1}.\] We first identify the singular behavior of \(\widehat{\mathcal{K}}(s)\) near \(s=1\). Note that \(P_K=\widehat{\mathcal{K}}(1)\). By Proposition 2, \[P_K(u,v)-\widehat{\mathcal{K}}(s)(u,v) = \Lambda_K(u,v)\,\sqrt{1-s} + o(\sqrt{1-s}), \quad s\uparrow1.\] Equivalently, \[\widehat{\mathcal{K}}(s) = P_K-\Lambda_K\,\sqrt{1-s} + o(\sqrt{1-s}), \quad s\uparrow1.\] Let \(\Pi_K:={\boldsymbol{1}}\,{\boldsymbol{\pi}}_K\) be the rank-one spectral projection of the stochastic matrix \(P_K\) corresponding to the eigenvalue \(1\). Let \(\lambda(s)\) denote the eigenvalue of \(\widehat{\mathcal{K}}(s)\) converging to \(1\) as \(s\uparrow1\). Then \[\begin{align} \label{lambda46idn} \lambda(s) = 1-{\boldsymbol{\pi}}_K\Lambda_K{\boldsymbol{1}}\,\sqrt{1-s} + o(\sqrt{1-s}) = 1-\gamma_K\,\sqrt{1-s} + o(\sqrt{1-s}), \quad s\uparrow1. \end{align}\tag{47}\] By continuity of the eigenvalues of a finite matrix, there exists \(\delta>0\) such that all eigenvalues of \(\widehat{\mathcal{K}}(s)\) other than \(\lambda(s)\) have modulus at most \(1-\delta\) for all \(s\) sufficiently close to \(1\). Let \(\Pi(s)\) be the spectral projection corresponding to \(\lambda(s)\), and write \[\widehat{\mathcal{K}}(s)=\lambda(s)\Pi(s)+N(s),\] where \(\Pi(s)N(s)=N(s)\Pi(s)=0\) and \(\rho(N(s))\le 1-\delta.\) Then \[I-\widehat{\mathcal{K}}(s) = (1-\lambda(s))\Pi(s)+(I-N(s))(I-\Pi(s)),\] and therefore \[\bigl(I-\widehat{\mathcal{K}}(s)\bigr)^{-1} = \frac{1}{1-\lambda(s)}\,\Pi(s) + (I-N(s))^{-1}(I-\Pi(s)).\] The second term is bounded as \(s\uparrow1\) because \(\rho(N(s))\le 1-\delta\). Moreover, since \(\widehat{\mathcal{K}}(s)\to P_K\), we have \(\Pi(s)\to \Pi_K\) as \(s\uparrow1\). Hence \[\bigl(I-\widehat{\mathcal{K}}(s)\bigr)^{-1} = \frac{1}{1-\lambda(s)}\,\Pi_K+O(1), \quad s\uparrow1.\] Using the expansion 47 , we obtain \[\widehat{\mathcal{U}}(s) = \frac{1}{\gamma_K}\,(1-s)^{-1/2}\,\Pi_K+O(1), \quad s\uparrow1.\] Notice that the coefficients of \((1-s)^{-1/2}=\sum_{n=0}^{\infty} a_n s^n\) satisfy \[a_n=\frac{\binom{2n}{n}}{4^n}\sim \frac{1}{\sqrt{\pi}}\,n^{-1/2}, \quad n\to\infty.\]

On the other hand, using the fact that \(Q(z,0)\) is decreasing in \(z\), one can show by stochastic domination that \(\mathcal{U}_{n}(0,0)=\mathbb{P}_0(Y_{n}=0)\) is non-increasing in \(n\). Therefore, by the Karamata’s Tauberian theorem for nonnegative power series (see Corollary 1.7.3 in [27]), we obtain \[\mathcal{U}_n(0,0) \sim \frac{{\boldsymbol{\pi}}_K(0)}{\gamma_K}\cdot\frac{1}{\sqrt{\pi}}\,n^{-1/2}.\] Since, by Lemma 2, \(r_n=\mathbb{P}_0(Y_{n-1}=0)=\mathcal{U}_{n-1}(0,0),\) we conclude that \[r_n \sim \frac{{\boldsymbol{\pi}}_K(0)}{\sqrt{\pi}\,\gamma_K}\,n^{-1/2}.\] ◻

Proof. For \(1\le a<L\), define \[G_L(y,a):= \mathbb{E}_y\!\left[\sum_{k=0}^{T_L-1}\mathbf{1}_{\{Y_k=a\}}\right].\] Then \[\mathbb{E}_y\!\left[\sum_{k=0}^{T_L-1}\mathbf{1}_{\{Y_k\in A_j\}}\right] = \sum_{a=1}^{j-1}G_L(y,a).\] By the strong Markov property at time \(\sigma_a\), we note that \(G_L(y,a)=\mathbb{P}_y(\sigma_a<T_L)\,G_L(a,a)\le G_L(a,a).\) Hence, to complete the proof, it is sufficient to show that \[G_L(a,a)\le C a \quad\text{for all }1\le a<L/2.\]

Let \(p_{a,L}:=\mathbb{P}_a(T_L<\sigma_a^+).\) The number of visits to \(a\) before \(T_L\) is geometric with success parameter \(p_{a,L}\), and therefore \(G_L(a,a)=1/{p_{a,L}}.\) Hence, it is enough to prove that \[p_{a,L}\ge \frac{c}{a} \quad\text{for all }1\le a<L/2,\] with a constant \(c>0\) independent of \(L\). On the other hand, since \(p_{a,L}\ge \mathbb{P}_a(\sigma_0<\sigma_a^+)\), it suffices to show that \[p_a:=\mathbb{P}_a(\sigma_0<\sigma_a^+)\ge \frac{c}{a} \quad\text{for all }a\ge 1.\]

By Lemma 4, the law of the increment \(Y_1-x\) converges exponentially fast to the centered law \(\nu\). Since \(\nu\) is non-degenerate, there exist an integer \(r\ge 1\), a constant \(\eta>0\), and \(a_0<\infty\) such that \[\mathbb{P}_x(Y_1\le x-r)\ge \eta \quad\text{for all }x\ge a_0.\] Fix \(a\ge a_0+r\), and let \(\widetilde{\sigma}_a:=\inf\{n\ge 0:Y_n\ge a\}\). For any \(z\le a-r\), consider the stopping time \(\tau:=\sigma_0\wedge \widetilde{\sigma}_a\). Since \(h\) is non-negative and harmonic for the chain killed at \(0\), the process \(h(Y_{n\wedge\tau})\) is a non-negative martingale. Hence \[h(z)=\mathbb{E}_z[h(Y_\tau)] \ge \mathbb{E}_z[h(Y_{\widetilde{\sigma}_a});\widetilde{\sigma}_a<\sigma_0].\] By Lemma 6, \(h(x)=x+O(1)\) as \(x\to\infty\). Therefore there exists \(C_1<\infty\) such that \(|h(x)-x|\le C_1\) for all \(x\ge 1\), and hence \(h(Y_{\widetilde{\sigma}_a})\ge a-C_1\) on \(\{\widetilde{\sigma}_a<\sigma_0\}.\) It follows that \(h(z)\ge (a-C_1)\,\mathbb{P}_z(\widetilde{\sigma}_a<\sigma_0),\) and thus \[\mathbb{P}_z(\sigma_0<\widetilde{\sigma}_a)\ge 1-\frac{h(z)}{a-C_1}.\] Since \(z\le a-r\) and \(h(z)\le z+C_1\le a-r+C_1\), we obtain \[\mathbb{P}_z(\sigma_0<\widetilde{\sigma}_a)\ge \frac{r-2C_1}{a-C_1}.\] Choose \(r>2C_1\). Then there exists \(c_1>0\) such that \(\inf_{z\le a-r}\mathbb{P}_z(\sigma_0<\widetilde{\sigma}_a)\ge c_1/{a}\) for \(a\ge a_0+r\). Therefore \[p_a \ge \mathbb{P}_a(Y_1\le a-r)\, \inf_{z\le a-r}\mathbb{P}_z(\sigma_0<\widetilde{\sigma}_a) \ge \frac{\eta c_1}{a}, \quad \text{for }a\ge a_0+r.\] For the finitely many values \(1\le a<a_0+r\), irreducibility and accessibility of \(0\) from \(a\) before returning to \(a\) imply that \(p_a>0\). Hence \(c_0:=\min_{1\le a<a_0+r} a\,p_a>0.\) Therefore, \[p_a\ge \frac{c}{a} \quad\text{for all }a\ge 1 \quad\text{with } c:=\min\{c_0,\eta c_1\}>0.\] It follows that for \(1\le a<L/2\), \[G_L(a,a)\le \frac{a}{c},\] and hence \(G_L(y,a)\le a/c\). Summing over \(a\in A_j\), we have \[\sup_{1\le y<L} \mathbb{E}_y\!\left[\sum_{k=0}^{T_L-1}\mathbf{1}_{\{Y_k\in A_j\}}\right] \le \frac{1}{c}\sum_{a=1}^{j-1}a \le C j^2.\] This proves the lemma. ◻

Proof. Recall that for \(x\ge 0\), \[h_N(x):=N\,\mathbb{P}_x(\widetilde{\sigma}_N<\sigma_0).\] From the proof of Lemma 6, there exists \(C_0<\infty\) such that \[\label{eq:HN-profile-proof} |h_N(x)-x|\le C_0 \quad\text{for all }N\ge 2,\;1\le x<N.\tag{48}\]

We first prove the mean exit-time bound \[\label{eq:mean-exit-xN} \mathbb{E}_x[T_N]\le C_1\,xN \quad\text{for all }N\ge 2,\;1\le x\le N/2.\tag{49}\] Notice that \[\mathbb{E}_x[T_N]=\sum_{a=1}^{N-1}G_N(x,a) \quad\text{with}\quad G_N(x,a):= \mathbb{E}_x\!\left[\sum_{k=0}^{T_N-1}\mathbf{1}_{\{Y_k=a\}}\right]\quad\text{for } 1\le a<N.\] We claim that there exists \(C_2<\infty\) such that \[\begin{align} \tag{50} G_N(x,a)&\le C_2\min\{x,a\} \quad\text{for all }1\le a\le N/2,\;1\le x<N,\quad\text{and}\\ \tag{51} G_N(x,a)&\le C_2\,\frac{x}{N}\,(N-a+1) \quad\text{for all }N/2<a<N,\;1\le x\le N/2. \end{align}\]

We start with the case \(a\le N/2\). By the strong Markov property at time \(\sigma_a\), \[G_N(x,a)=\mathbb{P}_x(\sigma_a<T_N)\,G_N(a,a).\] Let \(p_{a,N}:=\mathbb{P}_a(T_N<\sigma_a^+).\) Then the number of visits to \(a\) before \(T_N\) is geometric with success parameter \(p_{a,N}\), so \[G_N(a,a)=\frac{1}{p_{a,N}}.\] As in the proof of Lemma 14, we have \(p_{a,N}\ge \mathbb{P}_a(\sigma_0<\sigma_a^+)\ge c/a\) for all \(a\ge 1\), with a constant \(c>0\) independent of \(N\). Hence \[G_N(a,a)\le \frac{a}{c}.\] If \(x\ge a\), this gives \(G_N(x,a)\le G_N(a,a)\le C a=C\min\{x,a\}\). If \(x<a\), then on the event \(\{\sigma_a<T_N\}\) the chain must reach level \(a\) before hitting \(0\), and therefore \[\mathbb{P}_x(\sigma_a<T_N)\le \mathbb{P}_x(\widetilde{\sigma}_a<\sigma_0)=\frac{h_a(x)}{a}.\] By 48 , \(h_a(x)/{a}\le (x+C_0)/a\le C{x}/{a}\) for each \(1\le x<a.\) Thus \[G_N(x,a)\le \mathbb{P}_x(\sigma_a<T_N)\,G_N(a,a)\le C\,\frac{x}{a}\cdot a=Cx.\] This proves 50 .

We now consider the case \(N/2<a<N\) and \(1\le x\le N/2\). Again, \[\begin{align} \label{GN46inq} G_N(x,a)\le \mathbb{P}_x(\sigma_a<T_N)\,G_N(a,a). \end{align}\tag{52}\] As before, \[\mathbb{P}_x(\sigma_a<T_N)\le \mathbb{P}_x(\widetilde{\sigma}_a<\sigma_0)=\frac{h_a(x)}{a}\le C\,\frac{x}{a}\le C\,\frac{x}{N}.\] It remains to bound \(G_N(a,a)\).

We show that \[\begin{align} \label{p46aN95bound} p_{a,N}\ge \frac{c'}{N-a+1} \quad\text{for all }N/2<a<N. \end{align}\tag{53}\] By Lemma 4, the increment law of \(Y_1-Y_0\) converges exponentially fast to the centered non-degenerate law \(\nu\). Hence there exist an integer \(r\ge 1\), a constant \(\eta>0\), and \(z_0<\infty\) such that \[\mathbb{P}_z(Y_1\ge z+r)\ge \eta \quad\text{for all }z\ge z_0.\] We choose \(r>2C_0+1\). First, consider the finitely many values of \(N\) with \(N\le 2z_0+2r\). Since the state space \(\{1,\dots,N-1\}\) is finite and \(0\) and \([N,\infty)\) are both accessible from every interior state, there exists a constant \(C_3<\infty\) such that \[G_N(a,a)\le C_3(N-a+1) \quad\text{for all }N\le 2z_0+2r,\;N/2<a<N.\] Thus it remains to treat the case \(N>2z_0+2r\). Then \(a>N/2\) implies \(a\ge z_0+r\). If \(a\in[N-r,N)\), then \[p_{a,N}\ge \mathbb{P}_a(Y_1\ge N)\ge \eta\ge \eta\,\frac{1}{N-a+1}.\] Now assume \(N/2<a<N-r\). Starting from \(a\), on the event \(\{Y_1\ge a+r\}\) the chain jumps to some state \(z\ge a+r\). If \(z\ge N\), then \(T_N<\widetilde{\sigma}_a\) already. If \(a+r\le z<N\), let \[\tau:=\widetilde{\sigma}_N\wedge \inf\{k\ge 0:Y_k\le a\}.\] Since \(0\le h_N\le N\), the process \(h_N(Y_{m\wedge\tau})\) is a bounded martingale, and therefore \[h_N(z)=\mathbb{E}_z[h_N(Y_\tau)].\] On the event \(\{\widetilde{\sigma}_N<\inf\{k\ge 0:Y_k\le a\}\}\), we have \(Y_\tau\ge N\) and hence \(h_N(Y_\tau)=N\). Otherwise \(Y_\tau\le a\), and by 48 , \(h_N(Y_\tau)\le a+C_0\). Thus \[h_N(z)\le (a+C_0)+(N-a-C_0)\, \mathbb{P}_z\!\left(\widetilde{\sigma}_N<\inf\{k\ge 0:Y_k\le a\}\right).\] Since \(z\ge a+r\) and 48 gives \(h_N(z)\ge z-C_0\ge a+r-C_0\), it follows that \[\mathbb{P}_z\!\left(\widetilde{\sigma}_N<\inf\{k\ge 0:Y_k\le a\}\right) \ge \frac{r-2C_0}{N-a-C_0}.\] Because \(a<N-r\) and \(r>2C_0+1\), the denominator is positive, and therefore \[\mathbb{P}_z\!\left(\widetilde{\sigma}_N<\inf\{k\ge 0:Y_k\le a\}\right)\ge \frac{c_3}{N-a+1}\] for some \(c_3>0\) independent of \(N\) and \(a\). Combining with \(\mathbb{P}_a(Y_1\ge a+r)\ge \eta\), we obtain 53 . Thus \[G_N(a,a)=\frac{1}{p_{a,N}}\le C\,(N-a+1).\] Together with 52 , this implies 51 .

Combining 50 and 51 , we obtain that for \(1\le x\le N/2\), \[\begin{align} \mathbb{E}_x[T_N] &= \sum_{a=1}^{N-1}G_N(x,a)= \sum_{a=1}^{\lfloor N/2\rfloor}G_N(x,a) + \sum_{a=\lfloor N/2\rfloor+1}^{N-1}G_N(x,a)\\ &\le C_2\sum_{a=1}^{\lfloor N/2\rfloor}\min\{x,a\} + C_2\frac{x}{N}\sum_{a=\lfloor N/2\rfloor+1}^{N-1}(N-a+1)\le Cx N. \end{align}\] This proves 49 .

We now prove the claim of the lemma. Fix \(x\ge 1\) and \(n\ge 1\). If \(x>\sqrt{n+1}\), then the claimed bound is immediate after increasing the constant. Thus we may assume \(x\le \sqrt{n+1}\). Set \(N:=2\lceil\sqrt{n+1}\rceil.\) Then \(x\le N/2\). Since \(\{\sigma_0>n\}\subseteq \{\widetilde{\sigma}_N<\sigma_0\}\cup \{T_N>n\},\) we have \[\mathbb{P}_x(\sigma_0>n)\le \mathbb{P}_x(\widetilde{\sigma}_N<\sigma_0)+\mathbb{P}_x(T_N>n).\] By definition of \(h_N\) and 48 , \[\mathbb{P}_x(\widetilde{\sigma}_N<\sigma_0)=\frac{h_N(x)}{N}\le \frac{x+C_0}{N}\le C_4\,\frac{x}{N}.\] Also, by Markov’s inequality and 49 , \[\mathbb{P}_x(T_N>n)\le \frac{\mathbb{E}_x[T_N]}{n}\le C_1\,\frac{xN}{n}.\] Since \(N:=2\lceil\sqrt{n+1}\rceil\), the last two bounds imply \[\mathbb{P}_x(\sigma_0>n)\le C\,\frac{x}{\sqrt{n+1}}.\] This proves the lemma. ◻

Proof. We first prove that there exists a positive constant \(C_1<\infty\) such that \[\label{eq:half-line-green-Aj} \sup_{y\ge1} \mathbb{E}_y\!\left[ \sum_{k=0}^{\sigma_0-1}\mathbf{1}_{\{Y_k\in A_j\}} \right] \le C_1j^2 \quad\text{for all } j\ge1.\tag{54}\] The case \(j=1\) is trivial. Let \(j\ge2\) and fix \(y\ge1\). Choose \(L\) sufficiently large so that \(L>y\) and \(j\le L/2\). By Lemma 14, \[\mathbb{E}_y\!\left[ \sum_{k=0}^{T_L-1}\mathbf{1}_{\{Y_k\in A_j\}} \right] \le C_1j^2.\] Since the Markov chain \((Y_k)_{k\ge 0}\) is recurrent by Lemma 5, we have \(\sigma_0<\infty\) a.s. under \(\mathbb{P}_y\). Hence \(T_L\uparrow\sigma_0\) a.s. as \(L\to\infty\). Letting \(L\to\infty\) and using monotone convergence, we get \[\mathbb{E}_y\!\left[ \sum_{k=0}^{\sigma_0-1}\mathbf{1}_{\{Y_k\in A_j\}} \right] \le C_1 j^2.\] Taking the supremum over \(y\ge1\), we obtain 54 .

The result of the lemma is trivial when \(M=1\). We assume that \(M\ge2\) and set \(h:=\lfloor M/2\rfloor\). For \(m\in\{M,\dots,2M\}\), by the Markov property at time \(h\), \[\begin{align} \mathbb{P}_0(Y_m\in A_j,\sigma_0^+>m) &= \mathbb{E}_0\!\left[ \mathbf{1}_{\{\sigma_0^+>h\}} \mathbb{P}_{Y_h}(Y_{m-h}\in A_j,\sigma_0>m-h) \right]. \end{align}\] On the event \(\{\sigma_0^+>h\}\), we have \(Y_h\ge1\). Therefore, summing over \(m=M,\dots,2M\) and using 54 , we get \[\begin{align} \label{est1} \sum_{m=M}^{2M} \mathbb{P}_0(Y_m\in A_j,\sigma_0^+>m) &\le \mathbb{P}_0(\sigma_0^+>h) \sup_{y\ge1} \sum_{s\ge0} \mathbb{P}_y(Y_s\in A_j,\sigma_0>s)\le C_1 j^2\,\mathbb{P}_0(\sigma_0^+>h). \end{align}\tag{55}\]

It remains to bound \(\mathbb{P}_0(\sigma_0^+>h)\). If \(h\le1\), this probability is at most \(1\). If \(h\ge2\), then by the Markov property at time \(1\), \[\mathbb{P}_0(\sigma_0^+>h) = \sum_{x\ge1}\mathbb{P}_0(Y_1=x)\mathbb{P}_x(\sigma_0>h-1).\] By Lemma 15, \[\mathbb{P}_x(\sigma_0>h-1)\le C_2\frac{x}{\sqrt h}.\] Since \(Y_1\) has finite first moment under \(\mathbb{P}_0\) by Lemma 4, we obtain \[\begin{align} \label{est2} \mathbb{P}_0(\sigma_0^+>h)\le C_3 (h+1)^{-1/2}. \end{align}\tag{56}\] Combining 55 and 56 , we get \[\sum_{m=M}^{2M} \mathbb{P}_0(Y_m\in A_j,\sigma_0^+>m) \le Cj^2(M+1)^{-1/2}.\] This completes the proof. ◻

Proof. When \(j^2>n/2\), the result of the lemma is trivial. Thus, in the rest of the proof, we assume that \[\label{eq:j2-small} 1\le j^2\le n/2.\tag{57}\] Recall that \(B(1,n)\) is the number of backward jumps of the TSAW \((\widetilde{X}_k)_{k\ge 0}\) from \(1\) to \(0\) before the first hit of \(n\). Since each jump from \(1\) to \(0\) creates exactly one further visit to \(0\) after time \(0\), the event \(\{\tau_n<\tau_0^{(j)}\}\) is equivalent to the event that there are at most \(j-1\) such backward jumps before \(\tau_n\). On the other hand, by Lemma 2, \((B(n,n),B(n-1,n),\dots,B(1,n)) \stackrel d= (Y_0,Y_1,\dots,Y_{n-1}),\) and in particular, \(B(1,n)\stackrel d=Y_{n-1}\) under \(\mathbb{P}_0\). Therefore, \[\label{eq:gen-ruin-Y} r_n^{(j)}=\mathbb{P}\bigl(B(1,n)\le j-1\bigr)=\mathbb{P}_0(Y_{n-1}\le j-1)=\sum_{a=0}^{j-1}\mathbb{P}_0(Y_{n-1}=a).\tag{58}\]

By Proposition 3, there exists a constant \(C_0<\infty\) such that \[\label{eq:r95n-upper} r_n=\mathbb{P}_0(Y_{n-1}=0)\le C_0 n^{-1/2} \quad\text{for all }n\ge 1.\tag{59}\]

We next prove that there exists a constant \(C_1<\infty\) such that, for all \(m\ge1\), \[\label{eq:sigma0-plus-upper-direct} \mathbb{P}_0(\sigma_0^+>m)\le C_1(m+1)^{-1/2}.\tag{60}\] The case \(m=1\) is trivial. For \(m\ge2\), by the Markov property at time \(1\), \[\mathbb{P}_0(\sigma_0^+>m) = \sum_{x\ge1}\mathbb{P}_0(Y_1=x)\mathbb{P}_x(\sigma_0>m-1).\] Note that, by Lemma 15, we have \(\mathbb{P}_x(\sigma_0>m-1)\le C{x}/{\sqrt m}\) for some positive constant \(C<\infty\). Since \(Y_1\) has finite first moment under \(\mathbb{P}_0\) by Lemma 4, we thus obtain \[\mathbb{P}_0(\sigma_0^+>m) \le C m^{-1/2}\sum_{x\ge1}x\,\mathbb{P}_0(Y_1=x) \le C_1 m^{-1/2}.\] This verifies 60 .

By the strong Markov property, for \(a\ge 1\), \[\begin{align} \mathbb{P}_0(Y_{n-1}=a) &= \sum_{\ell=0}^{n-2}\mathbb{P}_0(Y_\ell=0)\, \mathbb{P}_0(Y_{n-\ell-1}=a,\sigma_0^+>n-\ell-1). \label{eq:last-zero-decomp-correct} \end{align}\tag{61}\] Since \(\mathbb{P}_0(Y_\ell=0)=r_{\ell+1}\), summing 61 over \(1\le a\le j-1\), we have \[\begin{align} r_n^{(j)}-r_n &= \sum_{a=1}^{j-1}\mathbb{P}_0(Y_{n-1}=a) \notag\\ &= \sum_{\ell=0}^{n-2}r_{\ell+1} \mathbb{P}_0(Y_{n-\ell-1}\in A_j,\sigma_0^+>n-\ell-1) \notag\\ &= \sum_{m=1}^{n-1} r_{n-m}\, \mathbb{P}_0(Y_m\in A_j,\sigma_0^+>m). \label{eq:sum-over-m} \end{align}\tag{62}\]

For \(m\ge1\), let \[a_m:=\mathbb{P}_0(Y_m\in A_j,\sigma_0^+>m).\] We split the sum in 62 into the ranges \(m<j^2\) and \(m\ge j^2\).

Range 1: \(m<j^2\). By 60 , \(a_m\le \mathbb{P}_0(\sigma_0^+>m)\le C_1(m+1)^{-1/2}.\) Using also 59 and 57 , we have \[\begin{align} I_1 &:= \sum_{m=1}^{j^2-1} r_{n-m}a_m \le C_2\sum_{m=1}^{j^2-1}(n-m)^{-1/2}(m+1)^{-1/2} \le C_3 n^{-1/2}\sum_{m=1}^{j^2-1}(m+1)^{-1/2} \le C_4\frac{j}{\sqrt n}. \end{align}\]

Range 2: \(j^2\le m\le n-1\). By 59 , \[I_2:= \sum_{m=j^2}^{n-1} r_{n-m}a_m \le C_0\sum_{m=j^2}^{n-1}\frac{a_m}{\sqrt{n-m}}.\] We split this sum further. First consider \(j^2\le m\le n/2\). Since \((n-m)^{-1/2}\le C_5n^{-1/2}\) on this range, by Lemma 16, we have \[\begin{align} \label{I246bound1} \sum_{j^2\le m\le n/2}\frac{a_m}{\sqrt{n-m}} &\le C_5 n^{-1/2} \sum_{j^2\le m\le n/2}a_m \le C_6 n^{-1/2} \sum_{\substack{k\ge0\, :\, 2^k j^2\le n/2}} j^2(2^k j^2+1)^{-1/2} \le C_7\frac{j}{\sqrt n}. \end{align}\tag{63}\] It remains to consider \(n/2<m\le n-1\). Note that \[\sum_{j^2\le m\le n/2}\frac{a_m}{\sqrt{n-m}}=\sum_{1\le k<n/2}\frac{a_{n-k}}{\sqrt k}.\] We split the latter sum into \(k<j^2\) and \(k\ge j^2\). If \(k<j^2\), then by 60 and 57 , we notice that \[a_{n-k}\le C_1(n-k+1)^{-1/2}\le C_8 n^{-1/2}.\] Therefore \[\sum_{1\le k<j^2}\frac{a_{n-k}}{\sqrt k} \le C_8 n^{-1/2}\sum_{1\le k<j^2}k^{-1/2} \le C_9\frac{j}{\sqrt n}.\] Finally, consider \(j^2\le k<n/2\). Notice that \[\begin{align} \sum_{j^2\le k<n/2}\frac{a_{n-k}}{\sqrt k} &\le \sum_{\substack{l\ge0\, :\, 2^l j^2<n/2}} (2^l j^2)^{-1/2} \sum_{2^l j^2\le k<2^{l+1}j^2}a_{n-k}. \end{align}\] For each \(l\) in the last sum, the indices \(n-k\) lie in the interval \([\lfloor n/2\rfloor,n-1]\). Hence, using Lemma 16 with \(M=\lfloor n/2\rfloor\), we have \[\sum_{2^l j^2\le k<2^{l+1}j^2}a_{n-k} \le \sum_{m=\lfloor n/2\rfloor}^{n-1}a_m \le C_{10} j^2(n+1)^{-1/2}.\] Consequently, \[\begin{align} \label{I246bound2} \sum_{j^2\le m\le n/2}\frac{a_m}{\sqrt{n-m}}=\sum_{j^2\le k<n/2}\frac{a_{n-k}}{\sqrt k} &\le C_{10} j^2(n+1)^{-1/2} \sum_{\substack{l\ge0\, :\, 2^l j^2<n/2}} (2^l j^2)^{-1/2} \le C_{11}\frac{j}{\sqrt n}. \end{align}\tag{64}\] Combining 63 and 64 , we get \[I_2\le C_{12}\frac{j}{\sqrt n}.\] Combining 62 , the bounds on \(I_1\) and \(I_2\), and 59 , we obtain \[r_n^{(j)} \le r_n+I_1+I_2 \le C_0 n^{-1/2}+C_{13}\frac{j}{\sqrt n} \le C_{14}\frac{j}{\sqrt n}.\] This completes the proof. ◻

Proof. It is sufficient to show that for every \(x\in\{1,\dots,n-1\}\) and every \(y,z\ge0\), \[\label{eq:forward-kernel} \mathbb{P}(F(x+1,n)=y\mid F(x,n)=z)=P^{\,z}(-1,y-z-1).\tag{66}\]

Fix \(x\in\{1,\dots,n-1\}\) and condition on the event \(\{F(x,n)=z\}\). Then the edge \(\{x-1,x\}\) is crossed exactly \(z\) times from left to right before \(\tau_0^+\). These \(z\) forward crossings split the trajectory into \(z\) successive stages. During the \(j\)-th stage, the walk starts at \(x\) immediately after the \(j\)-th jump from \(x-1\) to \(x\), makes some number of jumps from \(x\) to \(x+1\), and eventually leaves the stage by a jump from \(x\) to \(x-1\). For \(1\le j\le z\), let \(\ell_j(x,n)\) be the number of jumps from \(x\) to \(x+1\) during the \(j\)-th stage. Then \[F(x+1,n)=\sum_{j=1}^{z}\ell_j(x,n).\] Fix integers \(s_1,\dots,s_z\ge 0\) such that \(s_1+\cdots+s_z=y,\) and define \[u_0:=-1, \quad u_j:=\sum_{i=1}^j s_i-j-1, \quad 1\le j\le z.\] Thus \(u_j=u_{j-1}+s_j-1\). At the beginning of the \(j\)-th stage, the edge \(\{x-1,x\}\) has been crossed \(2j-1\) times, while the edge \(\{x,x+1\}\) has been crossed \(2\sum_{i=1}^{j-1}s_i\) times. Thus the probability of exactly \(s_j\) jumps from \(x\) to \(x+1\) during this stage and then one jump from \(x\) to \(x-1\) is \(P(u_{j-1},u_j)\). Hence, \[\begin{align} &\mathbb{P}\bigl(\ell_j(x,n)=s_j \,\big|\, \ell_1(x,n)=s_1,\dots,\ell_{j-1}(x,n)=s_{j-1},\,F(x,n)=z\bigr) = P(u_{j-1},u_j). \end{align}\] Therefore, \[\begin{align} &\mathbb{P}\bigl(\ell_1(x,n)=s_1,\dots,\ell_z(x,n)=s_z\mid F(x,n)=z\bigr)= P(u_0,u_1)\,P(u_1,u_2)\cdots P(u_{z-1},u_z). \end{align}\] Similarly as in the proof of Lemma 2, summing over all sequences \((s_1,\dots,s_z)\) such that \(s_1+\cdots+s_z=y\), we obtain \[\mathbb{P}(F(x+1,n)=y\mid F(x,n)=z)=P^{\,z}(-1,y-z-1),\] which is exactly 66 . The identity \(\mathbb{P}(Z_n\ge1)=r_n\) follows from 65 . This proves the lemma. ◻

Proof. Let \(\upsilon_0:=\inf\{k\ge 1:Z_k=0\}\) be the absorbing time of \(Z\). Then for \(n\ge 1\), we have \(\{\upsilon_0>n\}=\{Z_n\ge 1\}\). By Lemma 17, \(\mathbb{P}(\upsilon_0>n)=\mathbb{P}(Z_n\ge 1)=r_n.\) Hence, by Proposition 3, \[\label{eq:forward-survival} \mathbb{P}(\upsilon_0>n)\le C_0\,n^{-1/2} \quad\text{for } n\ge 1.\tag{67}\]

We first prove the second-moment bound. For \(z\ge 1\), using the transition kernel of \(Z\), we have \[\begin{align} \mathbb{E}[(Z_{k+1}-Z_k)^q\mid Z_k=z] &= \sum_{j\in\mathbb{Z}} j^q\,P^{\,z}(-1,j-1) \quad\text{for } q\in\{1,2\}. \end{align}\] By the same argument as in Lemma 4, there exist constants \(C_1<\infty\) and \(c_1>0\) such that \[\label{eq:forward-step-estimates} \left|\mathbb{E}[Z_{k+1}-Z_k\mid Z_k=z]\right|\le C_1e^{-c_1 z}, \quad \left|\mathbb{E}[(Z_{k+1}-Z_k)^2\mid Z_k=z]-\varsigma_\beta^2\right|\le C_1e^{-c_1 z} \quad \text{for }z\ge 1.\tag{68}\] Therefore, \[\begin{align} \mathbb{E}[Z_{k+1}^2-Z_k^2\mid Z_k=z] &= 2z\,\mathbb{E}[Z_{k+1}-Z_k\mid Z_k=z] +\mathbb{E}[(Z_{k+1}-Z_k)^2\mid Z_k=z] \end{align}\] is uniformly bounded above in \(z\ge 1\). Hence there exists \(C_2<\infty\) such that \[\mathbb{E}[Z_{k+1}^2\mid Z_k=z]\le z^2+C_2 \quad\text{for all }z\ge 1.\] Since \(0\) is absorbing, the same inequality also holds for \(z=0\). Therefore \(Z_{k\wedge\upsilon_0}^2-C_2(k\wedge\upsilon_0)\) with \(k\ge 1,\) is a supermartingale. Thus, for every \(n\ge 1\), \(\mathbb{E}[Z_{n\wedge\upsilon_0}^2-C_2(n\wedge\upsilon_0)] \le \mathbb{E}[Z_1^2-C_2].\) Since \(Z_{n\wedge\upsilon_0}=Z_n\) and \(Z_1=1\), this yields \[\mathbb{E}[Z_n^2]\le 1-C_2+C_2\,\mathbb{E}[n\wedge\upsilon_0].\] Using 67 , we obtain \[\mathbb{E}[n\wedge\upsilon_0] = \sum_{k=0}^{n-1}\mathbb{P}(\upsilon_0>k) \le 1+\sum_{k=1}^{n-1}C_0\,k^{-1/2} \le C_3\sqrt n.\] Therefore, \(\mathbb{E}[Z_n^2]\le C_4\sqrt n.\) Finally, using the Cauchy–Schwarz inequqality and 67 , we obtain \[\begin{align} \mathbb{E}[Z_n] &= \mathbb{E}[Z_n;\upsilon_0>n]\le \bigl(\mathbb{E}[Z_n^2]\bigr)^{1/2}\,\mathbb{P}(\upsilon_0>n)^{1/2}\le (C_4\sqrt n)^{1/2}(C_0n^{-1/2})^{1/2} \le C_5. \end{align}\] This completes the proof. ◻

Proof. By definition, \(e=\{e^-, e^+\}\in \mathcal{C}_{\rm CP}(\rho)\) if and only if \(\tau_{e^+}{(e^+)}<\tau_\rho^+{(e^+)}\), that is, the extension process \(\mathbf{X }^{(e^+)}\) hits \(e^+\) before returning to \(\rho\). As the subtree \(\mathcal{P}_{e_1}\cup \mathcal{P}_{e_2}\) is finite, the extension process \(\widehat \mathbf{X }\) visits \(\mathcal{P}_e\) infinitely many times. By the restriction principle in Lemma 1, the extension process \(\mathbf{X }^{(e^+)}\) coincides with the restriction of \(\widehat \mathbf{X }\) to \(\mathcal{P}_e\). Hence the event \(e\in \mathcal{C}_{\rm CP}(\rho)\) is exactly the event that the first excursion of \(\widehat \mathbf{X }\) from \(\rho\) crosses the edge \(\{e^-,e^+\}\) downward at least once, that is, \(N_e\ge 1\). ◻

Proof. Let \(n:=|e|\). We first estimate \(\mathbb{P}(N_e\ge 1).\) Consider the restriction of \(\widehat{\mathbf{X }}\) to the path \(\mathcal{P}_e\). By the restriction principle, this restriction process has the same law as the one-dimensional TSAW on \(\{0,1,\dots,n\}\) up to its first return to \(0\), after identifying \(\rho\) with \(0\) and \(e^+\) with \(n\). In particular, the crossings from \(e^-\) to \(e^+\) by \(\widehat{\mathbf{X }}\) before the first return to \(\rho\) are exactly the forward crossings of the last edge \(\{n-1,n\}\) by the restriction process before the first return to \(0\). Therefore \[\mathbb{P}(N_e\ge 1)=r_n.\] By Proposition 3, there exists a constant \(C_1>0\) such that \[\label{eq:mu-denom} \mathbb{P}(N_e\ge 1)\ge C_1\,n^{-1/2}.\tag{70}\]

We now estimate the moments of \(N_e\). Let \[F(n,n):= \sum_{k=0}^{\tau_0^+-1} \mathbf{1}_{\{\widetilde{X}_k=n-1,\;\widetilde{X}_{k+1}=n\}}\] be the number of forward crossings of the last edge by the one-dimensional TSAW on \(\{0,1,\dots,n\}\) before the first return to \(0\). By the restriction principle, we note that \[\label{eq:Ne-F-law} N_e\stackrel d=F(n,n).\tag{71}\] By Lemma 17, \(F(n,n)\) has the same distribution as \(Z_n\). Consequently, for every integer \(q\ge 1\), \[\label{eq:mu-num-moment} \sum_{j\ge 1} j^q\,\mathbb{P}(N_e=j) = \mathbb{E}[Z_n^q].\tag{72}\] By Lemma 18, there exists a constant \(C_2<\infty\) such that \[\label{eq:fY-moment-used} \mathbb{E}[ Z_n]\le C_2,\quad \mathbb{E}[ Z_n^2]\le C_2\sqrt n.\tag{73}\] Combining 70 , 72 , and 73 , we obtain \[\begin{align} \sum_{j\ge1} j\,\vartheta_e(j) &= \frac{\sum_{j\ge1} j\,\mathbb{P}(N_e=j)}{\mathbb{P}(N_e\ge1)} \le C\,\sqrt n, \\ \sum_{j\ge1} j^2\,\vartheta_e(j) &= \frac{\sum_{j\ge1} j^2\,\mathbb{P}(N_e=j)}{\mathbb{P}(N_e\ge1)} \le C\,n. \end{align}\] This completes the proof. ◻

Proof. For \(i\in \{1,2\}\), set \(m_i:=|e_i|-|e|+1\) and write the unique path from \(e^+\) to \(e_i^+\) as \((u_{i,1},u_{i,2},\dots,u_{i,m_i})\) with \(u_{i,1}=e^+\) and \(u_{i,m_i}=e_i^+\). Let \(L_i\) be the number of crossings from \(e^+\) to \(u_{i,2}\) by \(\widehat{\mathbf{X }}\) before time \(\widehat\tau_\rho^+\).

Fix \(j\ge 1\). We now work on the event \(\{N_e=j\}\), i.e. the number of crossings from \(e^+\) to \(e^-\) by \(\widehat{\mathbf{X }}\) before \(\widehat\tau_\rho^+\) is exactly \(j\). We first estimate the conditional moments of \(L_i\) on this event. By the restriction principle, the restriction of \(\widehat{\mathbf{X }}\) to \(\mathcal{P}_e\) coincides with the extension process \(\mathbf{X }^{(e^+)}\) up to the first return to \(\rho\). In particular, \(N_e\) is also equal to the number of crossings by \(\mathbf{X }^{(e^+)}\) from \(e^+\) to \(e^-\) before its first return to \(\rho\). Since \(e^+\) is the endpoint of the path \(\mathcal{P}_e\), right after each visit to \(e^+\), the extension process \(\mathbf{X }^{(e^+)}\) jumps from \(e^+\) to \(e^-\) deterministically. Therefore, by the strong construction of \(\mathbf{X }^{(e^+)}\), \(N_e\) is independent of the exponential variables \(\bigl\{\xi(e^+,y,\ell): y\sim e^+,\;\ell\ge0\bigr\}\) and of the exponential variables with first coordinate in \(\{u_{1,2},\dots,u_{1,m_1}\}\cup\{u_{2,2},\dots,u_{2,m_2}\}.\) Fix \(i\in\{1,2\}\). For \(\ell\ge0\), define \[T_0(\ell):= \sum_{q=0}^{\ell} \frac{\xi(e^+,e^-,q)}{w(2q+1)}\quad\text{and}\quad T_i(\ell):= \sum_{q=0}^{\ell} \frac{\xi(e^+,u_{i,2},q)}{w(2q)}.\] Let \[\overline{L}_i:= \sum_{\ell\ge0}\mathbf{1}_{\{T_i(\ell)<T_0(j-1)\}},\] which is equal to the number of crossings from \(e^+\) to \(u_{i,2}\) before the \(j\)-th crossing from \(e^+\) to \(e^-\) when only the two oriented edges \((e^+,e^-)\) and \((e^+,u_{i,2})\) are kept, with the same exponential clocks on these two oriented edges. Since the crossing order from \(e^+\) is obtained by ordering the clock times corresponding to all oriented edges with tail \(e^+\), the number of crossings from \(e^+\) to \(u_{i,2}\) before the \(j\)-th crossing by \(\widehat{\mathbf{X }}\) from \(e^+\) to \(e^-\) is at most \(\overline{L}_i\). Hence, on \(\{N_e=j\}\), we have \[L_i\le \overline{L}_i.\] Moreover, \(\overline{L}_i\) is independent of \(\{N_e=j\}\).

By the same argument as in Lemma 17, we have \[\mathbb{P}(\overline{L}_i=y)=P^j(-1,y-j-1) \quad\text{for each } y\ge0.\] Equivalently, if \((\eta_n)\) is the Markov chain with transition kernel \(P\) which starts from \(-1\), then \(\overline{L}_i\) has the same distribution as \(j+1+\eta_j\). By Lemma 3 and the remark following it, applied with initial state \(-1\), there exists a constant \(C_1<\infty\) such that, for all \(j\ge1\), \[\mathbb{E}[\overline{L}_i]\le C_1 j\quad\text{and} \quad\mathbb{E}[\overline{L}_i^2]\le C_1 j^2.\] Consequently, \[\label{eq:Li-moments} \mathbb{E}[L_i\mid N_e=j]\le C_1j,\quad \mathbb{E}[L_i^2\mid N_e=j]\le C_1j^2, \quad i=1,2.\tag{74}\]

By definition of \(\mathcal{C}_{\rm CP}(\rho)\) and by the restriction principle, for \(i\in\{1,2\}\), the event \(\{e_i\in\mathcal{C}_{\rm CP}(\rho)\}\) is the event that the extension process \(\mathbf{X }^{(e_i^+)}\) reaches \(e_i^+\) before its first return to \(\rho\), which is the same as the event that the restriction of \(\widehat{\mathbf{X }}\) to \(\mathcal{P}_{e_i^+}\) reaches \(e_i^+\) before \(\widehat\tau_\rho^+\).

We now condition on the event \(\{N_e=j\}\) and on the exponential variables \(\{\xi(e^+,y,\ell):y\sim e^+,\ell\ge0\}\). Under this conditioning, the values of \(L_1\) and \(L_2\) are determined. Moreover, the exponential variables with first coordinate in \(\{u_{1,2},\dots,u_{1,m_1}\}\) and the exponential variables with first coordinate in \(\{u_{2,2},\dots,u_{2,m_2}\}\) are independent. For fixed \(i\in\{1,2\}\), under the conditioning above, the restriction of \(\widehat{\mathbf{X }}\) to the path \((e^+,u_{i,2},\dots,u_{i,m_i})\) has exactly \(L_i\) crossings from \(e^+\) to \(u_{i,2}\) before time \(\widehat\tau_\rho^+\). Hence the event that the restriction of \(\widehat{\mathbf{X }}\) to \(\mathcal{P}_{e_i^+}\) reaches \(e_i^+\) before \(\widehat\tau_\rho^+\) is the same as the event that the restriction of \(\widehat{\mathbf{X }}\) to \((e^+,u_{i,2},\dots,u_{i,m_i})\) reaches \(e_i^+\) before its \(L_i\)-th return to \(e^+\). By the restriction principle, after identifying \(e^+\) with \(0\) and \(e_i^+\) with \(m_i-1\), the conditional probability of this event is equal to \(r_{m_i-1}^{(L_i)}\), where we use the convention that \(r_m^{(0)}:=0\). Therefore \[\begin{align} &\mathbb{P}\bigl(e_1,e_2\in\mathcal{C}_{\rm CP}(\rho)\mid N_e=j\bigr) = \mathbb{E}\left[ r_{m_1-1}^{(L_1)}r_{m_2-1}^{(L_2)} \,\middle|\, N_e=j \right]. \label{eq:qj-ruin-bound} \end{align}\tag{75}\] By Proposition 4, we have, for all \(\ell\ge0\), \[r_{m_i-1}^{(\ell)} \le C_2\left(1\wedge \frac{\ell}{\sqrt{m_i}}\right), \quad i\in \{1,2\}.\] It thus follows from 75 that \[\begin{align} &\mathbb{P}\bigl(e_1,e_2\in\mathcal{C}_{\rm CP}(\rho)\mid N_e=j\bigr) \le C_2\, \mathbb{E}\left[ \left(1\wedge \frac{L_1}{\sqrt{m_1}}\right) \left(1\wedge \frac{L_2}{\sqrt{m_2}}\right) \,\middle|\, N_e=j \right]. \label{eq:qj-L-bound} \end{align}\tag{76}\]

It remains to estimate the last expectation. Without loss of generality assume \(m_1\le m_2\).

If \(m_1\le j^2\) and \(m_2\le j^2\), then the expectation in 76 is at most \(1\), while \[\left(1\wedge \frac{j}{\sqrt{m_1}}\right) \left(1\wedge \frac{j}{\sqrt{m_2}}\right)=1.\]

If \(m_1\le j^2<m_2\), then \[\left(1\wedge \frac{L_1}{\sqrt{m_1}}\right) \left(1\wedge \frac{L_2}{\sqrt{m_2}}\right) \le \frac{L_2}{\sqrt{m_2}}.\] Using 74 , we get \[\mathbb{E}\left[ \left(1\wedge \frac{L_1}{\sqrt{m_1}}\right) \left(1\wedge \frac{L_2}{\sqrt{m_2}}\right) \,\middle|\, N_e=j \right] \le C_1\frac{j}{\sqrt{m_2}}= C_1\left(1\wedge \frac{j}{\sqrt{m_1}}\right) \left(1\wedge \frac{j}{\sqrt{m_2}}\right).\]

Finally, suppose that \(j^2<m_1\le m_2\). Then \[\left(1\wedge \frac{L_1}{\sqrt{m_1}}\right) \left(1\wedge \frac{L_2}{\sqrt{m_2}}\right) \le \frac{L_1L_2}{\sqrt{m_1m_2}}.\] By Hölder’s inequality and 74 , \[\mathbb{E}[L_1L_2\mid N_e=j] \le \Bigl(\mathbb{E}[L_1^2\mid N_e=j]\mathbb{E}[L_2^2\mid N_e=j]\Bigr)^{1/2} \le C_1 j^2.\] Thus \[\mathbb{E}\left[ \left(1\wedge \frac{L_1}{\sqrt{m_1}}\right) \left(1\wedge \frac{L_2}{\sqrt{m_2}}\right) \,\middle|\, N_e=j \right] \le C_1\frac{j^2}{\sqrt{m_1m_2}}=C_1\left(1\wedge \frac{j}{\sqrt{m_1}}\right) \left(1\wedge \frac{j}{\sqrt{m_2}}\right)\]

Combining the three cases with 76 , we obtain the desired estimate. ◻

Proof of Proposition 6. Fix two edges \(e_1,e_2\in E\), and let \(e:=e_1\wedge e_2\) be the last common edge of \(\mathcal{P}_{e_1}\) and \(\mathcal{P}_{e_2}\). If \(e=e_1\) or \(e=e_2\), then ?? is immediate, since for example when \(e=e_1\), \[\mathbb{P}\bigl(e_1,e_2\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr) = \mathbb{P}\bigl(e_2\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr),\] while \(\mathbb{P}\bigl(e_1\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr)=1.\) Thus it remains to consider the case where both \(e_1\) and \(e_2\) lie strictly below \(e\) in two distinct descendant subtrees of \(e^+\). Set \[n:=|e|, \quad m_i:=|e_i|-|e|+1\in\mathbb{N} \quad\text{for } i\in \{1,2\}.\] By Lemma 20, \(\{e\in\mathcal{C}_{\rm CP}(\rho)\}=\{N_e\ge1\}.\) Therefore, using the law of total probability, \[\begin{align} &\mathbb{P}\bigl(e_1,e_2\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr) = \sum_{j\ge1}\vartheta_e(j)\, \mathbb{P}\bigl(e_1,e_2\in \mathcal{C}_{\rm CP}(\rho)\mid N_e=j\bigr). \label{eq:branch-mixture} \end{align}\tag{77}\]

By Lemma 21, there exists a constant \(C_1<\infty\) such that \[\label{eq:qj-upper-per} \mathbb{P}\bigl(e_1,e_2\in \mathcal{C}_{\rm CP}(\rho)\mid N_e=j\bigr) \le C_1 \Bigl(1\wedge \frac{j}{\sqrt{m_1}}\Bigr) \Bigl(1\wedge \frac{j}{\sqrt{m_2}}\Bigr) \quad\text{for all } j\ge1.\tag{78}\] Also, by Proposition 7, there exists a constant \(C_2<\infty\) such that \[\label{eq:mu-upper-per} \sum_{j\ge1} j\,\vartheta_e(j)\le C_2\sqrt{n} \quad\text{and}\quad \sum_{j\ge1} j^2\,\vartheta_e(j)\le C_2n.\tag{79}\] Substituting 78 into 77 , we obtain \[\begin{align} &\mathbb{P}\bigl(e_1,e_2\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr)\le C_1 \sum_{j\ge 1} \vartheta_e(j) \Bigl(1\wedge \frac{j}{\sqrt{m_1}}\Bigr) \Bigl(1\wedge \frac{j}{\sqrt{m_2}}\Bigr). \label{eq:sum-to-bound} \end{align}\tag{80}\]

We now estimate the sum on the right-hand side. Without loss of generality, assume that \(m_1\le m_2\).

Case 1: \(m_1\le n\) and \(m_2\le n\). Since \(\Bigl(1\wedge \frac{j}{\sqrt{m_1}}\Bigr) \Bigl(1\wedge \frac{j}{\sqrt{m_2}}\Bigr)\le 1\), the sum is bounded by \(C_1\sum_{j\ge 1} \vartheta_e(j)=C_1.\) Therefore, \[\label{eq:sum-case1} \mathbb{P}(e_1,e_2\in\mathcal{C}_{\rm CP}(\rho)\mid e\in\mathcal{C}_{\rm CP}(\rho)) \le C_1 \prod_{i=1}^2\Bigl(1\wedge \sqrt{\frac{n}{m_i}}\Bigr).\tag{81}\]

Case 2: \(m_1\le n< m_2\). Using the fact that \(\Bigl(1\wedge \frac{j}{\sqrt{m_1}}\Bigr) \Bigl(1\wedge \frac{j}{\sqrt{m_2}}\Bigr)\le \frac{j}{\sqrt{m_2}}\) and 79 , we obtain \[\label{eq:sum-case2} \mathbb{P}(e_1,e_2\in\mathcal{C}_{\rm CP}(\rho)\mid e\in\mathcal{C}_{\rm CP}(\rho)) \le \frac{C_1}{\sqrt{m_2}} \sum_{j\ge1} j\,\vartheta_e(j)\le C_3 \sqrt{\frac{n}{m_2}}= C_3 \prod_{i=1}^2\Bigl(1\wedge \sqrt{\frac{n}{m_i}}\Bigr).\tag{82}\]

Case 3: \(n< m_1\le m_2\). Using the fact that \(\Bigl(1\wedge \frac{j}{\sqrt{m_1}}\Bigr) \Bigl(1\wedge \frac{j}{\sqrt{m_2}}\Bigr)\le \frac{j^2}{\sqrt{m_1m_2}}\) and 79 , we obtain \[\label{eq:sum-case3} \mathbb{P}(e_1,e_2\in\mathcal{C}_{\rm CP}(\rho)\mid e\in\mathcal{C}_{\rm CP}(\rho)) \le \frac{C_1}{\sqrt{m_1m_2}} \sum_{j\ge1} j^2\,\vartheta_e(j) \le C_4\,\frac{n}{\sqrt{m_1m_2}} = C_4 \prod_{i=1}^2\Bigl(1\wedge \sqrt{\frac{n}{m_i}}\Bigr).\tag{83}\]

Combining 81 , 82 , and 83 , we conclude from 80 that \[\label{eq:joint-upper-product-form} \mathbb{P}\bigl(e_1,e_2\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr) \le C_5 \prod_{i=1}^2\Bigl(1\wedge \sqrt{\frac{n}{m_i}}\Bigr).\tag{84}\]

It remains to compare the right-hand side with the product of \(\mathbb{P}(e_i\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho))\) for \(i\in \{1,2\}\). Since \(e_i\) is below \(e\), we have \(\{e_i\in \mathcal{C}_{\rm CP}(\rho)\}\subseteq \{e\in \mathcal{C}_{\rm CP}(\rho)\}\) for \(i\in \{1,2\}\), and hence \[\label{eq:cond-marginal-ratio} \mathbb{P}\bigl(e_i\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr) = \frac{\mathbb{P}(e_i\in \mathcal{C}_{\rm CP}(\rho))}{\mathbb{P}(e\in \mathcal{C}_{\rm CP}(\rho))}.\tag{85}\] By Proposition 3, there exist constants \(c_1>0\) and \(C_6<\infty\) such that for all \(f\in E\), \[c_1\,|f|^{-1/2}\le \mathbb{P}(f\in \mathcal{C}_{\rm CP}(\rho))=r_{|f|}\le C_6\,|f|^{-1/2}.\] Since \(|e_i|=n+m_i-1\), applying this to 85 , we obtain \[\begin{align} \mathbb{P}\bigl(e_i\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr) &\ge c_2\,\sqrt{\frac{n}{n+m_i-1}} \quad \text{for }i\in \{1,2\}. \label{eq:marginal-lower} \end{align}\tag{86}\] Finally, for every \(m\ge 1\) and \(n\ge 1\), we note that \(1\wedge \sqrt{\frac{n}{m}} \le \sqrt{2}\,\sqrt{\frac{n}{n+m-1}}\). Combining this inequality with 84 and 86 , we get \[\begin{align} \mathbb{P}\bigl(e_1,e_2\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr) &\le C_7 \prod_{i=1}^2 \sqrt{\frac{n}{n+m_i-1}}\\ &\le M\, \mathbb{P}\bigl(e_1\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr) \mathbb{P}\bigl(e_2\in \mathcal{C}_{\rm CP}(\rho)\mid e\in \mathcal{C}_{\rm CP}(\rho)\bigr), \end{align}\] for some constant \(M<\infty\) independent of \(e_1, e_2\). This proves ?? . ◻

Proof of Theorem 1. By Lemma 19, \[\mathbb{P}(\tau_\rho^+=\infty) = \mathbb{P}\bigl(|\mathcal{C}_{\rm CP}(\rho)|=\infty\bigr),\] where \(\mathcal{C}_{\rm CP}(\rho)\) is the cluster of the root in the ruin percolation. Consequently, the TSAW is transient if and only if the cluster \(\mathcal{C}_{\rm CP}(\rho)\) is infinite with positive probability, and recurrent if and only if it is almost surely finite.

By Proposition 6, the ruin percolation is quasi‑independent, so the criteria of Proposition 5 apply. Fix an edge \(e=\{e^{-},e^{+}\}\in E\) with \(|e|\ge 1\). By definition, \[\mathbf{Q}(\rho\leftrightarrow e^{+}) = \mathbb{P}\bigl(\tau_{e^{+}}(e^{+})<\tau_\rho^+(e^{+})\bigr).\] Since \(\mathbf{X }^{(e^{+})}\) has the same law as TSAW on \(\{0,1,2,\cdots,|e|\}\), it follows that \[\mathbf{Q}(\rho\leftrightarrow e^{+})=r_{|e|},\] where we recall that \(r_n\) is the ruin probability that the TSAW on \(\{0, 1,2,\cdots, n\}\) hits \(n\) before returning to \(0\). By Proposition 3, there exists a constant \(c_*\in (0,\infty)\) such that \(r_n\sim c_*\,n^{-1/2}\) as \(n\to\infty\). Hence, \[\label{eq:Q-asymptotic} \mathbf{Q}(\rho\leftrightarrow e^{+})=\mathbb{P}\bigl(\tau_{e^{+}}<\tau_\rho^+\bigr) \sim c_* |e|^{-1/2}.\tag{87}\]

Recall the definition of the adapted conductances: \[c(e) = \begin{cases} 1, & |e|=1,\\[1ex] \dfrac{\mathbf{Q}(\rho\leftrightarrow e^{+})} {\mathbf{Q}(e\;\text{is closed}\mid \rho\leftrightarrow e^{-})}, & |e|>1. \end{cases}\] Since \[\mathbf{Q}(e\;\text{is closed}\mid \rho\leftrightarrow e^{-}) = 1- \frac{\mathbf{Q}(\rho\leftrightarrow e^{+})}{\mathbf{Q}(\rho\leftrightarrow e^{-})},\] and by 87 , \[\frac{\mathbf{Q}(\rho\leftrightarrow e^{+})}{\mathbf{Q}(\rho\leftrightarrow e^{-})} = 1 - \frac{1}{2|e|}+o(|e|^{-1}),\] we obtain \[\mathbf{Q}(e\text{ is closed}\mid \rho\leftrightarrow e^-) = \frac{1}{2|e|}+o(|e|^{-1}).\] Combining this with 87 , we get \[\label{eq:c-asymptotic} c(e)\sim 2c_*\,|e|^{1/2}.\tag{88}\]

We distinguish the following two cases:

Case 1: when \({\rm br}_r(\mathcal{T})<\tfrac12\). Assume \({\rm br}_r(\mathcal{T})<\frac{1}{2}\). Choose \(\gamma\) such that \({\rm br}_r(\mathcal{T})<\gamma<\tfrac12.\) By the definition of the branching‑ruin number, \[\inf_{\pi\in\Pi}\sum_{e\in\pi}|e|^{-\gamma}=0.\] Since \(|e|^{-1/2}\le |e|^{-\gamma}\) for \(\gamma<\tfrac12\), and using 87 , we have \[\inf_{\pi\in\Pi}\sum_{e\in\pi}\mathbf{Q}(\rho\leftrightarrow e^{+}) = 0.\] By Proposition 5(i), \(\mathbf{Q}(\rho\leftrightarrow \infty)=0\) and the cluster \(\mathcal{C}_{\rm CP}(\rho)\) is thus almost surely finite. Lemma 19 therefore implies that the TSAW is a.s. recurrent.

Case 2: when \({\rm br}_r(\mathcal{T})>\tfrac12\). Assume \({\rm br}_r(\mathcal{T})>\frac{1}{2}\), and fix \(\gamma\) such that \(\tfrac12<\gamma<{\rm br}_r(\mathcal{T})\). By the max-flow min-cut Theorem, there exists a non-zero flow \((\theta(e))_{e\in E}\) such that \(\theta(e)\le |e|^{-\gamma}.\) By 88 , we also have \(c(e)\sim 2c_*\,|e|^{1/2}\). Hence \[\sup_{v\in V}\sum_{e\in \mathcal{P}_v}\frac{\theta(e)}{c(e)}\le C \sum_{n=1}^{\infty}\frac{1}{n^{\gamma+1/2}}<\infty.\] Hence \[\sum_{e\in E}\frac{\theta(e)^2}{c(e)}\le \int_{\partial \mathcal{T}} V_{\theta}(\xi)\,\mathrm{d}\mathfrak{m}_{\theta}(\xi) < \infty.\] where \(V_{\theta}(\xi):=\sum_{e\in \xi} \frac{\theta(e)}{c(e)}\) for each \(\xi\in \partial \mathcal{T}\) and \(\mathfrak{m}_\theta\) is the harmonic measure induced by flow \(\theta\) on \(\partial \mathcal{T}\) (see e.g., Proposition 16.1 in [28]). Therefore, by Proposition 5(ii), we get \(\mathbf{Q}(\rho\leftrightarrow \infty)>0\), and the cluster \(\mathcal{C}_{\rm CP}(\rho)\) is thus infinite with positive probability. Lemma 19 therefore implies that the TSAW is a.s. transient. ◻