Proof. Since \(\tau_{k,m} + 1 = \sum_{j \in \mathbb{Z}} L(\tau_{k,m},i)\), we see that \[\begin{align} \frac{\tau^n_{x_i,h_i}}{2\sigma n^{3/2}} + \frac{1}{2\sigma n^{3/2}} &= \frac{1}{2\sigma n^{3/2}}\sum_{k\in \mathbb{Z}} L\left(\tau^n_{x_i,h_i}, k \right) \nonumber \\ & \overset{\eqref{dRKpath}}{=} \frac{1}{n} \sum_{k \in \mathbb{Z}} \Lambda^n_{x_i,h_i}\left(k/n\right) = \int_\mathbb{R}\Lambda^n_{x_i,h_i}(y) \, dy =\int_{\mu^{n,-}_{x_i,h_i}}^{\mu^{n,+}_{x_i,h_i}}\Lambda^n_{x_i,h_i}(y) \, dy. \label{txn-integral} \end{align}\tag{12}\] The statement now follows from 8 and Theorem 4, from which we need the joint convergence of \((\Lambda^n_{x_i,h_i}(\cdot))_{1\le i\le N}\) and the tightness of distributions of \(\mu_{x_i,h_i}^{n,\pm}\), \(1\le i\le N\). We omit the details as they are standard. ◻

Proof. Let \(H_n=L(n,X_n)\) and recall 13 . We shall first prove that \[\label{XgHg-local} \begin{align} \lim_{n\to\infty} &n^{3k/2} P\left( X_{\lfloor \gamma_i n^{3/2} \rfloor} = \lfloor (2\sigma)^{-2/3} a_i n \rfloor, \, H_{\lfloor \gamma_i n^{3/2} \rfloor} = \lfloor (2\sigma)^{2/3} h_i \sqrt{n} \rfloor + 1, \, \forall i\leq k \right) \\ &= E\left[ \prod_{i=1}^k \lambda_i e^{-\lambda_i \mathfrak{t}_{a_i,h_i}} \right], \qquad \forall a_1,a_2,\ldots,a_k \in \mathbb{R}, \, \forall h_1,h_2,\ldots,h_k\geq 0, \end{align}\tag{14}\] and then show that the right-hand side is the joint density of \(\left( \mathfrak{X}(\gamma_i), \mathfrak{H}(\gamma_i) \right)\), \(1\leq i \leq k\). Indeed, we can interpret the expression under the limit as a density of a \(2k\)-dimensional random vector with a piecewise constant density, so that the above convergence of densities by Scheffe’s lemma gives us the weak convergence of this sequence of \(2k\)-dimensional random vectors to \(\left( \mathfrak{X}(\gamma_i), \mathfrak{H}(\gamma_i) \right)\), \(1\leq i \leq k\). The statement of the theorem follows by noticing that the random variables \(\left( \mathcal{X}_n(\gamma_i), \mathcal{H}_n(\gamma_i) \right)\), \(i\leq k\) can be coupled with joint random variables whose density is given by the left side of 14 in such a way that the maximum difference of any coordinate is \(\mathcal{O}(n^{-1/2})\).

To prove 14 , note that \(\lfloor \gamma_i n^{3/2} \rfloor \sim \text{Geo}(1-e^{-\lambda_i n^{-3/2}})\) so that by conditioning on the values of these independent geometric random variables we get \[\begin{align} &P\left( X_{\lfloor \gamma_i n^{3/2} \rfloor} = \left\lfloor \tfrac{a_i n}{(2\sigma)^{2/3}} \right\rfloor, H_{\lfloor \gamma_i n^{3/2} \rfloor} = \lfloor (2\sigma)^{2/3} h_i \sqrt{n} \rfloor + 1, \, \forall i\leq k \right) \\ &= \sum_{j_1,j_2,\ldots,j_k \geq 0} \left( \prod_{i=1}^k (1-e^{-\lambda_i n^{-3/2}})\,e^{-\lambda_i n^{-3/2}j_i } \right)\\ &\makebox[4.7cm]{\;} \times P\left( X_{j_i} = \left\lfloor \tfrac{a_i n}{(2\sigma)^{2/3}} \right\rfloor, H_{j_i} = \lfloor (2\sigma)^{2/3} h_i \sqrt{n} \rfloor+1, \, \forall i\leq k \right)\\ &= \sum_{j_1,\ldots,j_k \geq 0} \left( \prod_{i=1}^k (1-e^{-\lambda_i n^{-3/2}})\,e^{-\lambda_i n^{-3/2}j_i} \right) P\left( \tau_{\lfloor (2\sigma)^{-2/3} a_i n \rfloor, \lfloor (2\sigma)^{2/3} h_i \sqrt{n} \rfloor} = j_i , \, \forall i\leq k \right) \\ &= \prod_{i=1}^k (1-e^{-\lambda_i n^{-3/2}}) E\left[ \prod_{i=1}^k \exp\left\{ -\lambda_i n^{-3/2}\tau_{\lfloor (2\sigma)^{-2/3} a_i n \rfloor, \lfloor (2\sigma)^{2/3} h_i \sqrt{n} \rfloor} \right\} \right]. \end{align}\] Therefore, it follows from Corollary 1 that \[\begin{align} &\lim_{n\to\infty} n^{3k/2} P\left( X_{\lfloor \gamma_i n^{3/2} \rfloor} = \lfloor (2\sigma)^{-2/3} a_i n \rfloor, \, H_{\lfloor \gamma_i n^{3/2} \rfloor} = \lfloor (2\sigma)^{2/3} h_i \sqrt{n} \rfloor+1, \, \forall i\leq k \right) \\ &= \lim_{n\to\infty} \left( \prod_{i=1}^k n^{3/2}(1-e^{-\lambda_i n^{-3/2}}) \right) E\left[ \prod_{i=1}^k \exp\left\{ -\lambda_i n^{-3/2} \tau_{\lfloor (2\sigma)^{-2/3} a_i n \rfloor, \lfloor (2\sigma)^{2/3} h_i \sqrt{n} \rfloor} \right\} \right] \\ &= \left( \prod_{i=1}^k \lambda_i \right) E\left[ \prod_{i=1}^k \exp\left\{ -2\sigma\lambda_i \mathfrak{t}_{(2\sigma)^{-2/3}a_i, (2\sigma)^{-1/3}h_i} \right\} \right]= \left( \prod_{i=1}^k \lambda_i \right) E\left[ \prod_{i=1}^k e^{ - \lambda_i \mathfrak{t}_{a_i, h_i}} \right], \end{align}\] where in the last equality we used that \(( \mathfrak{t}_{c a_i, \sqrt{c} h_i} )_{i\leq k} \overset{\text{Law}}{=} ( c^{3/2}\, \mathfrak{t}_{a_i,h_i} )_{i\leq k}\) for any \(c>0\), which follows from 8 , the fact that the curves \((\Lambda_{a_i,h_i}(\cdot))_{i\leq k}\) are independent coalescing reflected/absorbed Brownian motions, and then the scaling property of Brownian motion.

We have completed the proof of 14 , and now it remains to show that
\(((a_1,h_1),\ldots,(a_k,h_k)) \mapsto E\left[ \prod_{i=1}^k \lambda_i e^{-\lambda_i \mathfrak{t}_{a_i,h_i}} \right]\) is the joint density for \(\left( \mathfrak{X}(\gamma_i), \mathfrak{H}(\gamma_i) \right)\), \(1\leq i \leq k\). To this end, fix a bounded continuous function \(\phi: \mathbb{R}^k \times \mathbb{R}_+^k \to \mathbb{R}\) and note that (using the notation \(\mathbf{a}=(a_1,\ldots,a_k)\), \(\mathbf{h} = (h_1,\ldots,h_k)\), \(\mathbf{da} = da_1\cdots da_k\) and \(\mathbf{dh} = dh_1 \cdots dh_k\)) \[\begin{align} & \int_{\mathbb{R}^k} \int_{\mathbb{R}_+^k} \phi(\mathbf{a},\mathbf{h}) E\left[ \prod_{i=1}^k \lambda_i e^{-\lambda_i \mathfrak{t}_{a_i,h_i}} \right] \mathbf{dh} \, \mathbf{da}\\ &= \left( \prod_{i=1}^k \lambda_i \right) \int_{\mathbb{R}^k} \int_{\mathbb{R}_+^k} \phi(\mathbf{a},\mathbf{h}) P\left( \gamma_i > \mathfrak{t}_{a_i,h_i}, \, \forall i\leq k \right) \mathbf{dh} \, \mathbf{da} \\ &= \left( \prod_{i=1}^k \lambda_i \right) \int_{\mathbb{R}^k} \int_{\mathbb{R}_+^k} \phi(\mathbf{a},\mathbf{h}) P\left( \mathfrak{L}(\gamma_i,a_i) > h_i, \, \forall i\leq k \right) \mathbf{dh} \, \mathbf{da} \\ &= \left( \prod_{i=1}^k \lambda_i \right) \int_{\mathbb{R}^k} \int_{\mathbb{R}_+^k} \int_{\mathbb{R}_+^k} \left( \prod_{i=1}^k \lambda_i e^{-\lambda_i s_i} \right) \phi(\mathbf{a},\mathbf{h}) P\left( \mathfrak{L}(s_i,a_i) > h_i, \, \forall i\leq k \right) \mathbf{ds} \, \mathbf{dh} \, \mathbf{da} \\ &= \int_{\mathbb{R}_+^k} \left( \prod_{i=1}^k \lambda_i^2 e^{-\lambda_i s_i} \right) E\left[ \int_{\mathbb{R}^k} \int_{\mathbb{R}_+^k} \phi(\mathbf{a}, \mathbf{h}) \mathbb{1}_{\{ h_i < \mathfrak{L}(s_i,a_i), \, \forall i \leq k \}} \, \mathbf{dh} \, \mathbf{da} \right] \mathbf{ds} \\ &= \int_{\mathbb{R}_+^k} \left( \prod_{i=1}^k \lambda_i^2 e^{-\lambda_i s_i} \right) E\left[ \int_{\mathbb{R}^k} \int_{\prod_{i=1}^k [0,\mathfrak{L}(s_i,a_i))} \phi(\mathbf{a}, \mathbf{h}) \, \mathbf{dh} \, \mathbf{da} \right] \mathbf{ds}. \end{align}\] By the generalization of the occupation time formula, see 5 , \[\label{intLtHt} \int_{\mathbb{R}^k} \int_{\prod_{i=1}^k [0,\mathfrak{L}(s_i,a_i))} \phi(\mathbf{a}, \mathbf{h}) \, \mathbf{dh} \, \mathbf{da} = \int_{\prod_{i=1}^k [0,s_i] } \phi\left( \mathfrak{X}(t_1),\ldots, \mathfrak{X}(t_k), \mathfrak{H}(t_1),\ldots,\mathfrak{H}(t_k) \right) \, \mathbf{dt}.\tag{15}\] Hence, \[\begin{align} & \int_{\mathbb{R}^k} \int_{\mathbb{R}_+^k} \phi(\mathbf{a},\mathbf{h}) E\left[ \prod_{i=1}^k \lambda_i e^{-\lambda_i \mathfrak{t}_{a_i,h_i}} \right] \mathbf{dh} \, \mathbf{da}\\ &=\int_{\mathbb{R}_+^k} \left( \prod_{i=1}^k \lambda_i^2 e^{-\lambda_i s_i} \right) E\left[ \int_{ \prod_{i=1}^k [0,s_i] } \phi\left( \mathfrak{X}(t_1),\ldots, \mathfrak{X}(t_k), \mathfrak{H}(t_1),\ldots,\mathfrak{H}(t_k) \right) \, \mathbf{dt} \right] \, \mathbf{ds} \\ &= E\left[ \int_{\mathbb{R}_+^k} \phi\left( \mathfrak{X}(t_1),\ldots, \mathfrak{X}(t_k), \mathfrak{H}(t_1),\ldots,\mathfrak{H}(t_k) \right) \left( \prod_{i=1}^k \int_{t_i}^\infty \lambda_i^2 e^{-\lambda_i s_i} \, ds_i \right) \mathbf{dt} \right] \\ &= E\left[ \int_{\mathbb{R}_+^k} \phi\left( \mathfrak{X}(t_1),\ldots, \mathfrak{X}(t_k), \mathfrak{H}(t_1),\ldots,\mathfrak{H}(t_k) \right) \left( \prod_{i=1}^k \lambda_i e^{-\lambda_i t_i} \right) \mathbf{dt} \right]\\ &= E\left[ \phi\left( \mathfrak{X}(\gamma_1),\ldots, \mathfrak{X}(\gamma_k), \mathfrak{H}(\gamma_1),\ldots,\mathfrak{H}(\gamma_k) \right) \right]. \end{align}\] This completes the proof of the theorem. ◻

Proof of Theorem 7. Suppose that \((\mathcal{X}_{n_j}(\cdot),\mathcal{H}_{n_j}(\cdot)) \underset{j\to \infty}{\Longrightarrow} (Y(\cdot),Z(\cdot))\) along a subsequence \((n_j)_{j\geq 1}\). By Proposition 6 the process \((Y(\cdot),Z(\cdot))\) must be continuous, and it will be enough to show that \((Y(\cdot),Z(\cdot))\) has the same finite dimensional distributions as \((\mathfrak{X}(\cdot),\mathfrak{H}(\cdot))\). To this end, note that it follows from Theorem 8 and the convergence assumption that for any bounded continuous function \(\phi:\mathbb{R}^k\times\mathbb{R}_+^k \to \mathbb{R}\), any \(\lambda_1,\lambda_2,\ldots, \lambda_k > 0\), and independent random variables \(\gamma_1, \gamma_2,\ldots,\gamma_k\) with \(\gamma_i\sim \text{Exp}(\lambda_i)\), \(1\le i\le k\), which are also independent from \((Y(\cdot),Z(\cdot))\) and \((\mathfrak{X}(\cdot),\mathfrak{H}(\cdot))\), \[E\left[ \phi\left( Y(\gamma_1), \ldots, Y(\gamma_k), Z(\gamma_1), \ldots, Z(\gamma_k)\right) \right]=E\left[ \phi\left( \mathfrak{X}(\gamma_1), \ldots, \mathfrak{X}(\gamma_k) , \mathfrak{H}(\gamma_1), \ldots, \mathfrak{H}(\gamma_k) \right) \right],\] or, equivalently, \(\forall \lambda_1,\lambda_2,\ldots,\lambda_k > 0\) \[\begin{align} &\int_{[0,\infty)^k} \left( \prod_{i=1}^k \lambda_i e^{-\lambda_i t_i} \right) E\left[ \phi\left( \mathfrak{X}(t_1), \ldots, \mathfrak{X}(t_k), \mathfrak{H}(t_1), \ldots, \mathfrak{H}(t_k)\right) \right] \, dt_1 \ldots dt_k \\ &\qquad = \int_{[0,\infty)^k} \left( \prod_{i=1}^k \lambda_i e^{-\lambda_i t_i} \right) E\left[ \phi\left( Y(t_1), \ldots, Y(t_k),Z(t_1), \ldots, Z(t_k) \right) \right] \, dt_1 \ldots dt_k. \end{align}\] That is, for any bounded continuous function \(\phi\), the functions \[\begin{align} (t_1,\ldots,t_k)&\mapsto E\left[ \phi\left( \mathfrak{X}(t_1),\ldots,\mathfrak{X}(t_k),\mathfrak{H}(t_1),\ldots,\mathfrak{H}(t_k) \right) \right]\quad \text{and} \\ (t_1,\ldots,t_k)&\mapsto E\left[ \phi\left(Y(t_1),\ldots,Y(t_k),Z(t_1),\ldots,Z(t_k) \right) \right] \end{align}\] (which are also bounded and continuous) have the same \(k\)-dimensional Laplace transforms and are thus equal for all \((t_1,\ldots,t_k) \in [0,\infty)^k\). This is enough to conclude that \((Y(\cdot),Z(\cdot))\) has the same finite dimensional distributions as \((\mathfrak{X}(\cdot),\mathfrak{H}(\cdot))\), and since this is true for any subsequential limit \((Y(\cdot),Z(\cdot))\) of \((\mathcal{X}_n(\cdot),\mathcal{H}_n(\cdot))\) we can conclude that indeed \((\mathcal{X}_n(\cdot),\mathcal{H}_n(\cdot)) \underset{n\to\infty}{\Longrightarrow} (\mathfrak{X}(\cdot),\mathfrak{H}(\cdot))\). ◻

Proof. Due to the relationship between local times of directed edges and local times of sites in 16 , it suffices to prove a corresponding joint diffusion limit for directed edge local times. That is, we will prove the joint convergence with the site local time profiles \(\Lambda_{x_i,h_i}^{n,+}(\cdot)\) replaced by the directed edge local time processes \(\left(\mathcal{E}^+_{\tau_{x_i,h_i}^n}(\lfloor y n \rfloor)/(\sigma\sqrt{n}) \right)_{y\geq x_i}\). Moreover, by the Markov property for the directed edge local times there is no loss of generality in assuming that \(x_i=x\) for all \(i\). Indeed, by induction on the number of points, if \(x_1\leq x_2 \leq \cdots \leq x_{k-1} \leq x_k\), then by conditioning on the values of \(h_{i,k}^n = \mathcal{E}^+_{\tau_{x_i,h_i}^n}(\lfloor x_k n \rfloor)/(\sigma\sqrt{n})\) it’s enough to prove the joint convergence of the curves started from the pairs \((x_k,h_{1,k}^n), (x_k,h_{2,k}^n),\ldots,(x_k,h_{k-1,k}^n), (x_k,h_k)\).

Fix an \(x\in\mathbb{R}\). Details of the proof depend on the sign of \(x\) but the difference between the three cases (\(x<0,\;x=0,\;x>0\)) is very minimal. For that reason, we shall only consider the case \(x<0\), in which we have reflection on \([x,0)\) and absorption on \([0,\infty)\). The other two cases have no reflection interval and are simpler. The proof proceeds by induction on \(N\). The case \(N=1\) is contained in Theorem 11. See also 17 and discussion below it.

We shall first treat the case \(N=2\) in detail. Fix \(x<0\), \(\varepsilon\in(0,1/100)\), and \(0\le h_1<h_2\). We start by lining up several facts about two sequences of rescaled reflected coalescing random walks with independent integer-valued increments \(\zeta_i(j),\;j\in\mathbb{N},\;i=1,2\), with mean \(0\), variance \(\sigma^2\in(0,\infty)\), and a finite third moment. For \(i=1,2\) let \[\lim_{n\to\infty}Y^n_i(0)=h_i\;\text{a.s},\quad Y^n_i\left(\frac{j}{n}\right)=\left|Y^n_i\left(\frac{j-1}{n}\right)+\frac{\zeta_i(j)}{\sigma\sqrt{n}}\right|,\quad j\ge 1.\] At each step we first connect \(Y^n_i\left(\frac{j-1}{n}\right)\) and \(Y^n_i\left(\frac{j-1}{n}\right)+\frac{\zeta_i(j)}{\sigma\sqrt{n}}\) and then reflect at \(0\) if the second point is negative. We get a continuous piecewise linear curve \(Y^n_i(y),\;y\ge 0\), which hits \(0\) in the interval \(((j-1)/n, j/n)\) if the expression inside the absolute value above is negative. We also define \(\widetilde{Y}^n_2\), to be the walk which starts from \(Y^n_2(0)-\lfloor n^\varepsilon \rfloor/(\sigma\sqrt{n})\) and uses the same increments as \(Y^n_2\). Note that \(\widetilde{Y}^n_2\left(\frac{j}{n}\right)=Y^n_2\left(\frac{j}{n}\right)-\lfloor n^\varepsilon \rfloor/(\sigma\sqrt{n})\) up to (not including) the first time the right hand side falls below \(0\). We produce \(\widetilde{Y}^n_2(y),\;y\ge 0\), by the same kind of linear interpolation as above and define the hitting times of \(0\) after time \(|x|\) and the crossing time of \(Y^n_1\) and \(\widetilde{Y}^n_2\) respectively as follows: \[\begin{align} &\nu^n_1=\inf\{y\ge |x|: Y^n_1(y)=0\},\;\;\widetilde{\nu}^n_2=\inf\{y\ge |x|: \widetilde{Y}^n_2(y)=0\},\\ & \widetilde{c}^n_{1,2}=\inf\{y\ge 0: \widetilde{Y}^n_2(y)=Y^n_1(y)\}\;\left(=\inf\{y\ge 0: Y^n_2(y)-Y^n_1(y)=\lfloor n^\varepsilon \rfloor/(\sigma\sqrt{n})\}\right). \end{align}\]

• By independence of \(Y^n_1(\cdot)\) and \(\widetilde{Y}^n_2(\cdot)\), convergence of one-dimensional processes (see [32]¹), and the continuous mapping theorem, \[(Y^n_1(\cdot),\widetilde{Y}^n_2(\cdot),\nu^n_1,\widetilde{\nu}^n_2,\widetilde{c}^n_{1,2})\Longrightarrow (|W_1(\cdot)|,|W_2(\cdot)|,\nu_1,\nu_2,c_{1,2})\quad\text{ as n\to\infty},\] where the independent Brownian motions \(W_1(\cdot)\) and \(W_2(\cdot)\) start at \(h_1\) and \(h_2\) respectively and the last three stopping times are defined in the same way as for \(Y^n_1(\cdot)\) and \(\widetilde{Y}^n_2(\cdot)\). The continuous mapping theorem is applicable, since the set of discontinuities of the last three functions on the path space of a standard two dimensional Brownian motion has measure \(0\).

• We define a mapping \(\phi\) which takes a pair of paths \((f_1,f_2)\in C([0,\infty))^2,\;f_1,f_2\ge 0\), \(f_1(0)\le f_2(0)\), and returns a pair of coalescing paths \(\phi(f_1,f_2)\) as follows. Let \(c(f_1,f_2):=\inf\{y\ge 0:f_1(y)=f_2(y)\}\in[0,\infty]\) and \(\nu_0(f_i)=\inf\{y\ge|x|:f_i(y)=0\}\in[|x|,\infty]\). We shall refer to \(\nu_0\) as a “marked time”. Set \[\phi(f_1,f_2)(y):= \begin{cases} (f_1(y),f_2(y)),&\text{if }y\le c(f_1,f_2);\\ (f_1(y),f_1(y)),&\text{if }y>c(f_1,f_2)\;\text{and }c(f_1,f_2)\le \nu_0(f_1);\\ (f_2(y),f_2(y)),&\text{if }y>c(f_1,f_2)\;\text{and }c(f_1,f_2)>\nu_0(f_1). \end{cases}\] In words, at the time when the two paths meet we keep only the lower path if they meet before or at the marked time of the lower path and we keep only the upper path if they meet after the marked time of the lower path. This mapping is a.s.continuous for the pair of independent reflected Brownian motions. Hence, just as above, we can conclude the convergence of a pair of rescaled coalescing reflected random walks together with their marked and coalescence times to a pair of coalescing reflected Brownian motions together with their marked and coalescence times.

From now on we assume that \(\zeta_i(j)\sim\pi\), \(i=1,2\), \(j\in\mathbb{N}\). Set \[\label{S} S^n_i(j)=\mathcal{E}^+_{\tau^n_{x,h_i}}(j+\lfloor xn \rfloor),\;j\ge 0; \quad W^n_i(y) = \frac{S^n_i(yn)}{\sigma\sqrt{n}},\quad yn\in \mathbb{N}_0, \quad i=1,2,\tag{19}\] and linearly interpolated in between so that \(W^n_i\in C([0,\infty))\). For convenience we set \(S^n_0(j)=0\) for all \(j\ge 0\). Note that \(W^n_i(0)\to h_i\), \(i=1,2\), a.s.as \(n\to\infty\), see 18 .

Recall that \(\varepsilon\in(0,1/100)\) and define for \(i=1,2\) \[\omega^{n,4\varepsilon}_i=\min\{j\ge \lfloor |x|n \rfloor: S^n_i(j)\le n^{4\varepsilon}\}, \quad \omega^{n,\varepsilon}_{1,2}=\min\{j\ge 0: S^n_2(j)-S^n_1(j)\le n^\varepsilon\}.\] We consider the pair \((S^n_1,S^n_2)\) and enlarge the probability space as necessary to construct a pair of crossing marked and reflected Markov chains \((R^n_1,\widetilde{R}^n_2)\) as follows. Let \(R^n_1(0)=S^n_1(0)\) and \(\widetilde{R}^n_2(0)=S^n_2(0)-\lfloor n^\varepsilon \rfloor\).

Up until \(\omega^{n,\varepsilon}_{1,2}\wedge\omega^{n,4\varepsilon}_1\) we set \(R^n_1(j)=S^n_1(j)\), couple the increments of \(S^n_2\) with independent increments \(\zeta_2(j)\) at each step \(j\), and set \(\widetilde{R}^n_2(j)=\widetilde{R}^n_2(0)+\sum_{k=1}^j \zeta_2(j)\). Given the initial data, the coordinate processes of \((R^n_1,\widetilde{R}^n_2)\) are independent, and \((R^n_1(j),\widetilde{R}^n_2(j))=(S^n_1(j),S^n_2(j)-\lfloor n^\varepsilon \rfloor)\) for \(j<\omega^{n,\varepsilon}_{1,2}\wedge\omega^{n,4\varepsilon}_1\), provided that the coupling does not break down.

1. If \(\omega^{n,\varepsilon}_{1,2}<\omega^{n,4\varepsilon}_1\) then

   1. at time \(\omega^{n,\varepsilon}_{1,2}\) we decouple \(\widetilde{R}^n_2\) from \(S^n_2\) and simply let \(\widetilde{R}^n_2\) use increments \(\zeta_2(j)\), i.e., \(\widetilde{R}^n_2(j)=|\widetilde{R}^n_2(j-1)+\zeta_2(j)|\) for all \(j>\omega^{n,\varepsilon}_{1,2}\);

   2. \(R^n_1\) continues to follow \(S^n_1\) up to time \(\omega^{n,4\varepsilon}_1\), after which it uses independent increments \(\zeta_1(\cdot)\) and follows a reflected marked random walk.

2. If \(\omega^{n,\varepsilon}_{1,2}\ge \omega^{n,4\varepsilon}_1\) then

   1. at time \(\omega^{n,4\varepsilon}_1\) we decouple \(R^n_1\) from \(S^n_1\), \(R^n_1\) continues to use independent increments \(\zeta_1(\cdot)\) and follows a reflected marked random walk;

   2. we continue to couple the increments of \(\widetilde{R}^n_2\) and \(S^n_2\) up to time \(\omega^{n,4\varepsilon}_2\), starting from which we let \(\widetilde{R}^n_2\) use increments \(\zeta_2(\cdot)\) and continue as a reflected marked random walk.

We agree that if any of the couplings above breaks down before the specified decoupling time, we shall just use the corresponding independent \(\pi\)-distributed increments for the rest of the time. Since a single curve gets absorbed within \(n^{1+\varepsilon}\) units of time with probability tending to 1, Lemma 7 implies that the above couplings will not break down before the specified decoupling time with probability tending to 1 as \(n\to\infty\).

Diffusively rescaling the process \((R^n_1,\widetilde{R}^n_2)\) and “connecting the dots” in the same way as before we get a sequence of pairs of independent reflected marked processes which converge in distribution together with their rescaled marked and crossing times to a pair of independent marked reflected Brownian motions and their marked and crossing times. Indeed, by the result for \(N=1\) and fact (I), each sequence of the rescaled coordinate processes together with their rescaled marked times converges to a reflected Brownian motion and its marked time. Since the increments of the coordinate processes are independent, the claimed convergence follows. By (II), we get the convergence of rescaled marked coalescing reflected processes \((R^n_1,\widetilde{R}^n_2)\) and sequences of their rescaled marked and coalescence times to a pair of coalescing reflected Brownian motions together with their marked and coalescence times.

Our next task is to sort out the marked and coalescence times and conclude the convergence of \((W^n_1(\cdot),W^n_2(\cdot))\) together with their coalescence and absorption times to a pair of coalescing independent reflected/absorbed Brownian motions together with their coalescence and absorption times.

We shall use results from Appendix 8 to show that the coalescence and absorption times we are interested in are close (in the sense that after scaling, with probability tending to 1, the difference is negligible) to the crossing and marked times for which the convergence has been shown above.

Let us start with scenario (i) (coalescence before the marked time for \(R^n_1\)). There are two cases: (i1) \(S^n_2(\omega^{n,\varepsilon}_{1,2})-\lfloor n^\varepsilon \rfloor\le n^{4\varepsilon}\) and (i2) \(S^n_2(\omega^{n,\varepsilon}_{1,2})-\lfloor n^\varepsilon \rfloor> n^{4\varepsilon}\). In case (i1), with probability tending to 1, by the result for a single curve, [3], \(S^n_2\) will hit zero (but not necessarily gets absorbed) in at most \(\lceil n^{28\varepsilon}\rceil\) additional steps. This means that \(S^n_1\) and \(S^n_2\) must have coalesced within the same number of steps, which is negligible after division by \(n\) as \(n\to\infty\). This shows the convergence of the coalescence times in this case.

In case (i2), we have that \(\widetilde{R}^n_2(\omega^{n,\varepsilon}_{1,2}),\,R^n_1(\omega^{n,\varepsilon}_{1,2})> n^{4\varepsilon}\). It follows from Corollary 4 (with \(b= \lfloor n^\varepsilon \rfloor\)) that with probability tending to 1, the crossing time of \(R^n_1\) and \(\widetilde{R}^n_2\) and the actual coalescence time of \(S^n_1\) and \(S^n_2\) will differ by less than \(n^{7\varepsilon}\), which is negligible after division by \(n\).

In both cases, again by the result for a single curve, [3], with probability tending to 1, the common absorption time of the coalesced processes differs from the marked time of \(R^n_1\) by no more than \(n^{28\varepsilon}\), which is negligible after division by \(n\).

Under scenario (ii) (\(S^n_1\) gets within \(n^{4\varepsilon}\) from \(0\) after time \(\lfloor |x|n \rfloor\) and before coalescence), again by [3], with probability tending to 1 the marked time of \(R^n_1\) will be within \(n^{28\varepsilon}\) from the absorption time of \(S^n_1\). This difference is negligible after division by \(n\). The crossing time of \(R^n_1\) and \(\widetilde{R}^n_2\) will be irrelevant for our purposes, and the absorption time of \(S^n_2\) will be within \(n^{28\varepsilon}\) of the marked time of \(\widetilde{R}^n_2\) with probability tending to 1.

From the above we conclude the desired convergence for the case \(N=2\). Suppose that we have the claimed result for all \(1\le k\le N-1\) and any choice of \(x<0\) and \(0\le h_1<h_2<\ldots<h_{N-1}\). Let \(h_N>h_{N-1}\) and consider the walks \((S^n_i(\cdot) )_{i\leq N}\) and their rescaled versions \((W^n_i(\cdot) )_{i\leq N}\). Let \(\omega^{n,\varepsilon}_{N-1,N}=\min\{j\ge 0: S^n_N(j)-S^n_{N-1}(j)\le n^\varepsilon\}\). We shall take the pair \((S^n_{N-1}, S^n_N)\) and construct \((R^n_{N-1}, \widetilde{R}^n_N)\) using the same algorithm as for \((R^n_1,\widetilde{R}^n_2)\) above. By construction, given the initial data, the reflected marked random walk \(\widetilde{R}^n_N\) is independent from all \((S^n_i(\cdot) )_{i\leq N-1}\). By the same argument as for \(N=2\), we can apply (II) to rescaled coalescing marked reflected processes \((R^n_{N-1}, \widetilde{R}^n_N)\) and in the same way as for \(N=2\) deduce the convergence of \((W^n_{N-1},W^n_N)\) together with their coalescence and absorption times to a pair of coalescing independent reflected absorbed Brownian motions starting from \((h_{N-1},h_N)\) together with their coalescence and absorption times. The induction hypothesis, the convergence of the initial data to deterministic values, and the fact that the increments of \(\widetilde{R}^n_N\) are independent from all \((S^n_i(\cdot) )_{i\leq N-1}\) yield the statement of the theorem for \(x<0\). ◻

Proof. Given any finite set of points \((x_1,h_1),\ldots,(x_N,h_N)\), it follows from Theorem 11 that the joint distribution of the curves \((\Lambda_{x_i,h_i}^n(\cdot))_{i\leq N}\) and the endpoints \(\{\mu_{x_i,h_i}^{n,*} \}_{i\leq N,\, * \in \{+,-\} }\) is tight. Moreover, any subsequential limit must have the property that the limiting curves \(\Lambda_{x_i,h_i}(\cdot)\) are continuous and the subsequential limits of the endpoints \(\mu_{x_i,h_i}^{n,*}\) are equal to the endpoints of the curves \(\Lambda_{x_i,h_i}(\cdot)\). Furthermore, it follows from the forward (resp.backward) joint GRKT that the subsequential limits of the merge points \(\mu_{(x_i,h_i),(x_j,h_j)}^{n,+}\) (resp.\(\mu_{(x_i,h_i),(x_j,h_j)}^{n,-}\)) must be the merge points of the curves \(\Lambda_{x_i,h_i}(\cdot)\) and \(\Lambda_{x_j,h_j}(\cdot)\).

Therefore, it remains only to show that the subsequential limit of the joint distribution of the curves is unique. To this end, for each \(i\leq N\), \(m_i\in\mathbb{N}\), let \((y_{i,j})_{j\leq m_i}\) be a finite collection of spatial points. We need to show that the limiting distribution of the collection of random variables \(\Lambda_{x_i,h_i}^n(y_{i,j})\), \(i\leq N\), \(j\leq m_i\), is determined by the forward (or backward) joint GRKT. If \(y_{i,j} \geq x_i\) then it is part of the forward curve \(\Lambda_{x_i,h_i}^{n,+}(\cdot)\) and the joint convergence of these finite dimensional distributions is covered by Theorem 12. However, the following lemma shows that the finite dimensional distributions of the discrete backward curves \(\Lambda_{x_i,h_i}^{n,-}(\cdot)\) are determined by the finite dimensional distributions of the discrete forward curves.

Before giving the proof of Lemma 1 we shall use it to justify our last claim. If \(y< x\), then the event on the left side of ?? involves the backward curve from \((x,h)\) while the event on the right side of ?? only involves the forward curve from \((y,h')\).

Now, suppose the spatial points \((y_{i,j})_{i\leq N, j\leq m_i}\) are ordered so that \(y_{i,j} < x_i\) for \(j\leq m_i'\) and \(y_{i,j} \geq x_i\) for \(m_i' < j \leq m_i\), and for each \(i\leq N\) and \(j\leq m_i\) fix \(b_{i,j}>0\). Then it follows from Lemma 1 that for \(n\) sufficiently large we have \[\begin{align} &P\left( \Lambda_{x_i,h_i}^n(y_{i,j}) \leq b_{i,j}, \quad \forall i\leq N, j\leq m_i \right)\\ &= P\left( \bigcap_{i\leq N} \left( \bigcap_{j\leq m_i'} \left\{ \Lambda_{x_i,h_i}^n(x_i) \leq \Lambda_{y_{i,j},b_{i,j}}^n(x_i) \right\} \cap \bigcap_{m_i' < j \leq m_i} \left\{ \Lambda_{x_i,h_i}^n(y_{i,j}) \leq b_{i,j} \right\} \right) \right). \end{align}\] Since the events in the probability on the right depend only on the finite dimensional distributions of the forward curves it then follows from Theorem 12 that \[\begin{align} \lim_{n\to\infty} &P\left( \Lambda_{x_i,h_i}^n(y_{i,j}) \leq b_{i,j}, \quad \forall i\leq N, j\leq m_i \right) \nonumber \\ &= P\left( \bigcap_{i\leq N} \left( \bigcap_{j\leq m_i'} \left\{ h_i < \Lambda_{y_{i,j},b_{i,j}}(x_i) \right\} \cap \bigcap_{m_i' < j \leq m_i} \left\{ \Lambda_{x_i,h_i}(y_{i,j}) \leq b_{i,j} \right\} \right) \right). \label{Lnxhy-lim} \end{align}\tag{20}\] A few points of further explanation are due regarding the application of Theorem 12 in this last equality.

Let \(m_i' < j \leq m_i\). Since \(\Lambda_{x_i,h_i}(\cdot)\) is a reflected/absorbed Brownian motion and \(b_{i,j} > 0\), it follows that \(\Lambda_{x_i,h_i}(y_{i,j})\) doesn’t have an atom at \(b_{i,j}\). Thus Theorem 12 justifies replacing the events \(\{\Lambda_{x_i,h_i}^n(y_{i,j}) \leq b_{i,j} \}\) by the corresponding events \(\{\Lambda_{x_i,h_i}(y_{i,j}) \leq b_{i,j} \}\) in the limit.

Let \(j\leq m_i'\). To justify replacing the events \(\{ \Lambda_{x_i,h_i}^n(x_i) \leq \Lambda_{y_{i,j},b_{i,j}}^n(x_i) \}\) with the corresponding events \(\{ h_i < \Lambda_{y_{i,j},b_{i,j}}(x_i)\}\) we need to consider two cases.

Case \(h_i>0\) or \(x_i \leq 0\)

this case is justified by noting that \(\Lambda_{x_i,h_i}^n(x_i) = \frac{\lfloor 2\sigma h_i \sqrt{n} \rfloor+1}{2\sigma h_i \sqrt{n}} \to h_i\) as \(n\to \infty\) and that since \(\Lambda_{y_{i,j},b_{i,j}}(\cdot)\) is a reflected/absorbed Brownian motion then the distribution of \(\Lambda_{y_{i,j},b_{i,j}}(x_i)\) doesn’t have an atom at \(h_i\) if either \(h_i>0\) or if \(x_i \leq 0\).

Case \(h_i=0\) and \(x_i>0\)

in this case, first of all note that \[\left\{ \Lambda_{x_i,0}^n(x_i) \leq \Lambda_{y_{i,j},b_{i,j}}^n(x_i) \right\} = \{ \tau^n_{y_{i,j},b_{i,j}} > \tau^n_{x_i,0} \} = \left\{ n\mu^{n,+}_{y_{i,j},b_{i,j}} \geq \lfloor x_i n \rfloor \right\}.\] Since Theorem 12 implies that \(\mu^{n,+}_{y_{i,j},b_{i,j}} \Rightarrow \mathfrak{m}^+_{y_{i,j},b_{i,j}}\), and since \(\mathfrak{m}^+_{y_{i,j},b_{i,j}}\) has a continuous distribution on \((0,\infty)\) it follows that we can replace the event \(\{ \Lambda_{x_i,0}^n(x_i) \leq \Lambda_{y_{i,j},b_{i,j}}^n(x_i)\}\) with the event \(\{ \mathfrak{m}^+_{y_{i,j},b_{i,j}} > x \}\) in the limit. Finally, note that \(\{ \mathfrak{m}^+_{y_{i,j},b_{i,j}} > x_i \} = \{\Lambda_{y_{i,j},b_{i,j}}(x_i) > 0 \}\).

Finally, since it follows from Corollary 3 that \[\left\{ \Lambda_{x_i,h_i}(y_{i,j}) < b_{i,j} \right\} \subset \{h_i < \Lambda_{y_{i,j},b_{i,j}}(x_i)\} \subset \left\{ \Lambda_{x_i,h_i}(y_{i,j}) \leq b_{i,j} \right\},\] and since \(P(\Lambda_{x_i,h_i}(y_{i,j}) = b_{i,j}) = 0\) when \(b_{i,j}>0\) we can conclude from 20 that \[\begin{align} &\lim_{n\to\infty} P\left( \Lambda_{x_i,h_i}^n(y_{i,j}) \leq b_{i,j}, \quad \forall i\leq N, j\leq m_i \right) = P\left( \Lambda_{x_i,h_i}(y_{i,j}) \leq b_{i,j}, \quad \forall i\leq N, j\leq m_i \right). \end{align}\] This completes the proof of the corollary, pending the proof of Lemma 1. ◻

Proof of Lemma 1. For any integers \(j\neq k\) and \(m\geq 0\) we claim that \[\label{backpath} L(\tau_{k,m},j) = \min\{ \ell \geq 0: L(\tau_{j,\ell},k) \geq m+1 \}.\tag{21}\] Indeed, this can be justified by noting that \[L(\tau_{k,m},j) = 0 \iff \tau_{k,m} < \tau_{j,0} \iff L(\tau_{j,0},k) \geq m+1,\] and that for \(\ell \geq 1\), \[L(\tau_{k,m},j) = \ell \iff \tau_{j,\ell-1} < \tau_{k,m} < \tau_{j,\ell} \iff L(\tau_{j,\ell-1},k) < m+1 \leq L(\tau_{j,\ell},k).\] Translating the equation 21 to the rescaled path \(\Lambda_{x,h}^n(\cdot)\) we have \[\begin{align} \Lambda^n_{x,h}(y) &= \frac{L(\tau_{\lfloor xn \rfloor,\lfloor 2\sigma h \sqrt{n} \rfloor}, \lfloor yn \rfloor)}{2 \sigma\sqrt{n}} \\ &= \frac{1}{2\sigma\sqrt{n}} \min\left\{ \ell \geq 0: L(\tau_{\lfloor yn \rfloor,\ell}, \lfloor xn \rfloor) \geq \lfloor 2\sigma h\sqrt{n} \rfloor + 1 \right\} \\ &= \inf\left\{ b\geq 0: L(\tau_{\lfloor yn \rfloor,\lfloor 2\sigma b \sqrt{n} \rfloor}, \lfloor xn \rfloor) \geq \lfloor 2\sigma h\sqrt{n} \rfloor + 1 \right\} \\ &= \inf\left\{ b\geq 0: \Lambda_{y,b}^n(x) \geq \frac{\lfloor 2\sigma h \sqrt{n} \rfloor+1}{2\sigma\sqrt{n}} \right\} = \inf\left\{ b\geq 0: \Lambda_{y,b}^n(x) \geq \Lambda_{x,h}^n(x) \right\}. \end{align}\] From this we see that \[\Lambda_{x,h}^n(y) \leq h' \iff \inf\left\{ b\geq 0: \Lambda^n_{y,b}(x) \geq \Lambda_{x,h}^n(x) \right\} \leq h' \iff \Lambda_{x,h}^n(x) \leq \Lambda_{y,h'}^n(x).\] In the last equivalence we used the fact that \(b\mapsto\Lambda_{y,b}^n(x)\) is non-decreasing and right continuous. ◻

Proof. First of all, note that \(R_{K,n} \supset R_{K,n,-} \cap R_{K,n,+}\), where \[\begin{align} R_{K,n,\pm} &= \left\{ \tau_{\pm Kn,0} > n^{3/2} \text{ and } \max_{k\in \mathbb{Z}} L(\tau_{\pm Kn,0},k) \leq 2\sigma K \sqrt{n} \right\} \\ &= \left\{ \int_\mathbb{R}\Lambda_{\pm K,0}^n(x) \, dx > (1 + n^{-3/2})/(2\sigma) \text{ and } \sup_{x \in \mathbb{R}} \Lambda_{\pm K,0}^n(x) \leq K \right\}, \end{align}\] where we used 12 in the last equality. It follows from Theorem 4 and Corollary 1, along with symmetry considerations and scaling properties of Brownian motion, that \[\begin{align} \limsup_{n\to\infty} P\left( R_{K,n}^c \right) &\leq 2 P\left( \int_\mathbb{R}\Lambda_{-K,0}(x) \, dx \leq \frac{1}{2\sigma} \text{ or } \sup_{x \in \mathbb{R}} \Lambda_{-K,0}(x) > K \right) \\ &= 2 P\left( \int_\mathbb{R}\Lambda_{-1,0}(x) \, dx \leq \frac{1}{2\sigma K^{3/2}} \text{ or } \sup_{x \in \mathbb{R}} \Lambda_{-1,0}(x) > \sqrt{K} \right). \end{align}\] Since the last probability above can be made arbitrarily small by taking \(K\) sufficiently large, the conclusion of the lemma follows easily. ◻

Proof. Let \(\mathcal{T}_{\delta}^{(K)} = \left\{ \mathfrak{t}_{x,h}: (x,h)\in (\delta \mathbb{Z}\cap [-K,K]) \times (\sqrt{\delta} \mathbb{Z}\cap [0,K]) \right \}\). It follows from Corollary 1 that \(\mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)})\) converges in distribution to \(2\sigma\mathop{\mathrm{gap}}(\mathcal{T}_{\delta}^{(K)})\). Since the map \((x,h) \mapsto \mathfrak{t}_{x,h}\) is almost surely one-to-one, then \(P\left(\mathop{\mathrm{gap}}(\mathcal{T}_{\delta}^{(K)}) > 0 \right) = 1\) and thus there exists a \(\delta'>0\) such that \(P\left(\mathop{\mathrm{gap}}(\mathcal{T}_{\delta}^{(K)}) > \frac{\delta'}{2\sigma} \right) \geq 1-\frac{\eta}{3}\), and the conclusion of the lemma then follows easily. ◻

Proof. It is enough to show that for any \(\varepsilon,\eta>0\) there is a \(\delta'>0\) such that \[\label{tightcond} \limsup_{n\to\infty} P\Bigg( \max_{\substack{k,\ell\leq n^{3/2} \\ |k-\ell|\leq \delta' n^{3/2}}} |X_k - X_\ell| > \varepsilon n \Bigg) < \eta.\tag{22}\] To this end, we will later show that we can choose parameters \(K,\delta,\) and \(\delta'\) such that the conclusions to Lemmas 2 and 3 hold. Thus, we will only need to control the fluctuations of the path of the walk on the event \(R_{K,n} \cap \{ \mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)}) > \delta' \}\). For any times \(k<\ell\leq n^{3/2}\) with \(|k-\ell|\leq \delta' n^{3/2}\), it follows from Remark 14 that on the event \(R_{K,n} \cap \{ \mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)}) > \delta' \}\) there are times \(\tau,\tau' \in \mathcal{T}_{\delta,n}^{(K)}\) and \(\tau'' \in \mathcal{T}_{\delta,n}\) that are “consecutive” in the set \(\mathcal{T}_{\delta,n}\) in that \(\mathcal{T}_{\delta,n} \cap \left((\tau,\tau')\cup(\tau',\tau'')\right) = \varnothing\) and such that either (i) \(\tau n^{3/2}\leq k < \ell < \tau' n^{3/2}\) or (ii) \(\tau n^{3/2} \leq k < \tau' n^{3/2}\le \ell < \tau''n^{3/2}\). If one also has that \(|X_k-X_\ell| > \varepsilon n\) then this implies that either there is a \(j \in (\tau n^{3/2},\tau' n^{3/2})\) with \(|X_j-X_{\tau n^{3/2}}| > \varepsilon n/4\) or there is a \(j \in (\tau'n^{3/2},\tau''n^{3/2})\) with \(|X_j-X_{\tau'n^{3/2}}|>\varepsilon n/4\). Indeed, if \(|X_k-X_\ell|>\varepsilon n\) and (i) is true then by the triangle inequality \(\max\{|X_\ell-X_{\tau n^{3/2}}|,|X_k-X_{\tau n^{3/2}}|\}>\varepsilon n/2>\varepsilon n/4\). If \(|X_k-X_\ell|>\varepsilon n\) and (ii) is true then \[\varepsilon n<|X_k-X_\ell|\le |X_k-X_{\tau n^{3/2}}|+|X_{\tau n^{3/2}}-X_{\tau' n^{3/2}-1}|+1+|X_{\tau' n^{3/2}}-X_\ell|,\] so at least one of the terms on the right should be larger that \(\varepsilon n/4\).

Thus, for all sufficiently large \(n\) we have that \[\label{Hunion} \Bigg\{ \max_{\substack{k,\ell\leq n^{3/2} \\ |k-\ell|\leq \delta' n^{3/2}}} |X_k - X_\ell| > \varepsilon n \Bigg\} \cap R_{K,n} \cap \left\{ \mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)}) > \delta' \right\} \subset \bigcup_{\substack{x \in \delta\mathbb{Z}\cap [-K,K]\\h \in \sqrt{\delta}\mathbb{Z}\cap [0,K]}} A^{(K)}_{x,h,\delta,\varepsilon,n},\tag{23}\] where \[\begin{align} A^{(K)}_{x,h,\delta,\varepsilon,n} &= \left\{ \exists j> \tau^n_{x,h} : \, \left|X_j - X_{\tau^n_{x,h}} \right| > \frac{\varepsilon}{4} n \text{ and } \mathcal{T}_{\delta,n} \cap \left( \frac{\tau^n_{x,h}}{n^{3/2}}, \frac{j}{n^{3/2}} \right) = \varnothing \right\} \\ &\makebox[5cm]{\;}\cap \left\{ \max_{i \in \mathbb{Z}} L(\tau^n_{x,h}, i ) \leq 2\sigma K \sqrt{n} \right\}. \end{align}\] We claim that there exists a constant \(c_0>0\) which does not depend on \(x,h,\delta\) or \(\varepsilon\) such that for all \(K>0\), \(x \in \delta\mathbb{Z}\cap [-K,K]\), \(h \in \sqrt{\delta}\mathbb{Z}\cap [0,K]\), and \(n\) sufficiently large we have \[\label{PHbound} P(A^{(K)}_{x,h,\delta,\varepsilon,n}) \leq 2 (1-c_0)^{\lfloor \frac{\varepsilon}{5\delta} \rfloor}.\tag{24}\] Postponing the proof of 24 for now, we will show how to use this to finish the proof of 22 . Since there are at most \((2K/\delta+1)(K/\sqrt{\delta}+1) \leq \frac{6K^2}{\delta^{3/2}}\) indices \((x,h)\) in the union on the right side of 23 , it follows from 23 and 24 that for \(n\) sufficiently large we have \[\begin{align} P\Bigg( \max_{\substack{k,\ell\leq n^{3/2} \\ |k-\ell|\leq \delta' n^{3/2}}} |X_k - X_\ell| > \varepsilon n \Bigg) \leq P(R_{K,n}^c) + P\left( \mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)} ) \leq \delta' \right) + \frac{12 K^2}{\delta^{3/2}}(1-c_0)^{\lfloor \frac{\varepsilon}{5\delta} \rfloor}. \end{align}\] Given \(\varepsilon,\eta>0\) we can first choose \(K\) large enough (depending only on \(\eta\)) so that \(P(R_{K,n}^c) \leq \frac{\eta}{3}\). Next, we choose \(\delta>0\) small enough (depending on \(\varepsilon\) and \(K\)) so that \(\frac{12 K^2}{\delta^{3/2}}(1-c_0)^{\lfloor \frac{\varepsilon}{5\delta} \rfloor} < \frac{\eta}{3}\), and finally we choose \(\delta'>0\) depending on \(K, \eta\) and \(\delta\) as in Lemma 3 so that \(P( \mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)} \leq \delta' ) \leq \frac{\eta}{3}\). This completes the proof of 22 , pending the proof of 24 .

We now return to the proof of 24 . The event \(A^{(K)}_{x,h,\delta,\varepsilon,n}\) involves a long excursion either to the right or left of \(\lfloor xn \rfloor\). To simplify things we will consider an event which only has long excursions to the right; the corresponding event with long excursions to the left is handled similarly. More precisely, letting \[\begin{align} A^{(K),+}_{x,h,\delta,\varepsilon,n} &= \left\{ \exists j> \tau_{x,h}^n: \, X_j > \lfloor xn \rfloor + \frac{\varepsilon}{4} n \text{ and } \mathcal{T}_{\delta,n} \cap \left( \frac{\tau^n_{x,h}}{n^{3/2}}, \frac{j}{n^{3/2}} \right) = \varnothing \right\} \\ &\makebox[5.3cm]{\;} \cap \left\{ \max_{i \in \mathbb{Z}} L( \tau_{x,h}^n, i ) \leq 2\sigma K \sqrt{n} \right\}, \end{align}\] we will show that there is a \(c_0>0\) such that for all \(K>0\), \(x \in \delta\mathbb{Z}\cap [-K,K]\), \(h \in \sqrt{\delta}\mathbb{Z}\cap [0,K]\), and \(n\) large \[\label{PH43bound} P(A^{(K),+}_{x,h,\delta,\varepsilon,n}) \leq (1-c_0)^{\lfloor \frac{\varepsilon}{5\delta} \rfloor}.\tag{25}\]

To prove 25 , the key observation is that, on the event \(A_{x,h,\delta,\varepsilon,n}^{(K),+}\), for any \((x',h') \in (\delta\mathbb{Z})\times (\sqrt{\delta}\mathbb{Z}_+)\) with \(\tau_{x',h'}^n > \tau_{x,h}^n\) we must have that \(\mu_{(x,h),(x',h')}^{n,+} > \frac{\lfloor xn \rfloor}{n} + \frac{\varepsilon}{4}\) (since \(n \mu_{(x,h),(x',h')}^{n,+}\) is the maximum of the walk between times \(\tau_{x,h}^n\) and \(\tau_{x',h'}^n\)) and thus \(\Lambda_{x,h}^n(y) < \Lambda_{x',h'}^n(y)\) for all \(\lfloor xn \rfloor\leq \lfloor yn \rfloor \leq \lfloor xn \rfloor + \frac{\varepsilon n}{4}\). In particular, for any \(k \in \mathbb{Z}\) with \(\lfloor xn \rfloor \leq \lfloor k\delta n \rfloor < \lfloor (k+1)\delta n \rfloor \leq \lfloor xn \rfloor+\frac{\varepsilon n}{4}\), let \(\ell_k\) be the non-negative integer such that \(\tau_{k\delta,\ell_k\sqrt{\delta}}^n \leq \tau_{x,h}^n < \tau_{k\delta, (\ell_k+1)\sqrt{\delta}}^n\), or equivalently \[\Lambda_{k\delta,\ell_k\sqrt{\delta}}^n(k\delta) \leq \Lambda_{x,h}^n(k \delta) < \Lambda_{k\delta,(\ell_k+1)\sqrt{\delta}}^n(k\delta).\] (Note also that the definition of \(A_{x,h,\delta,\varepsilon,n}^{(K),+}\) implies that \(\ell_k \sqrt{\delta} \leq K\).) Then on the event \(A_{x,h,\delta,\varepsilon,n}^{(K),+}\) we have that \[\mu_{(k\delta,\ell_k\sqrt{\delta}),(k\delta,(\ell_k+1)\sqrt{\delta})}^{n,+} \geq \mu_{(x,h),(k\delta,(\ell_k+1)\sqrt{\delta})}^{n,+} \geq \frac{\lfloor xn \rfloor}{n} + \frac{\varepsilon}{4} \geq \frac{\lfloor (k+1)\delta n \rfloor}{n}.\] (See Figure 3 for a visual depiction of this.)

Therefore, if we define \[S_{x,h,k} = \bigcup_{0\leq \ell \leq \frac{K}{\sqrt{\delta}} } \left\{ \ell \sqrt{\delta} \leq \Lambda_{x,h}(k \delta) < (\ell+1)\sqrt{\delta}, \, \mathfrak{m}_{(k\delta,\ell\sqrt{\delta}),(k\delta,(\ell+1)\sqrt{\delta})}^{+} \geq (k+1)\delta\right\},\] then it follows from the Joint GRKT (Theorem 4) that \[\label{PA43bound} \limsup_{n\to\infty} P\left( A_{x,h,\delta,\varepsilon,n}^{(K),+} \right) \leq P\left( \bigcap_{k=\frac{x}{\delta}}^{\frac{x}{\delta} + \lfloor \frac{\varepsilon}{5\delta} \rfloor} S_{x,h,k} \right).\tag{26}\] Note that if we fix some \(k_0 > x/\delta\) then the event \(\bigcap_{k=\frac{x}{\delta}}^{k_0-1} S_{x,h,k}\) contains information about \(\Lambda_{x',h'}(y)\) for finitely many pairs \((x',h') \in (\delta\mathbb{Z}) \times (\sqrt{\delta}\mathbb{Z}_+)\) and \(y \in [x', k_0\delta]\). On the other hand, the only part of the event \(S_{x,h,k_0}\) that depends on this information concerns the value of \(\Lambda_{x,h}(k_0\delta)\). Therefore, it follows from the Markov property that \[\begin{align} P\left( S_{x,h,k_0} \, \biggl| \, \bigcap_{k=\frac{x}{\delta}}^{k_0-1} S_{x,h,k} \right) &= \sum_{0\leq \ell \leq \frac{K}{\sqrt{\delta}} } P\left( \ell \sqrt{\delta} \leq \Lambda_{x,h}(k_0\delta) < (\ell+1)\sqrt{\delta} \, \biggl| \, \bigcap_{k=\frac{x}{\delta}}^{k_0-1} S_{x,h,k} \right) \\ &\qquad\qquad\qquad\qquad \times P\left( \mathfrak{m}_{(k\delta,\ell\sqrt{\delta}),(k\delta,(\ell+1)\sqrt{\delta})}^{+} \geq (k+1)\delta\right) \\ &\leq \max_{\ell \leq \frac{K}{\sqrt{\delta}} } P\left( \mathfrak{m}_{(k\delta,\ell\sqrt{\delta}),(k\delta,(\ell+1)\sqrt{\delta})}^{+} \geq (k+1)\delta\right) \\ &= \max_{\ell \leq \frac{K}{\sqrt{\delta}} } P\left( \mathfrak{m}_{(k,\ell),(k,\ell+1)}^{+} \geq k+1 \right), \end{align}\] where the last equality follows from the scaling property of Brownian motion. Repeatedly applying this we get that \[\label{PcapSbound} P\left( \bigcap_{k=\frac{x}{\delta}}^{\frac{x}{\delta} + \lfloor \frac{\varepsilon}{5\delta} \rfloor} S_{x,h,k} \right) \leq \left\{ \sup_{k\in\mathbb{Z}} \sup_{\ell \in \mathbb{Z}_+} P\left( \mathfrak{m}_{(k,\ell),(k,\ell+1)}^{+} \geq k+1 \right) \right\}^{\left\lfloor \frac{\varepsilon}{5\delta} \right\rfloor }\tag{27}\] By the properties of the construction of the reflected/absorbed Brownian motions we have that \[P\left( \mathfrak{m}_{(k,\ell),(k,\ell+1)}^{+} \geq k+1 \right) = \begin{cases} P\left( \mathfrak{m}_{(-1,\ell),(-1,\ell+1)}^{+} \geq 0 \right) & \text{if } k\leq -1 \\ P\left( \mathfrak{m}_{(0,\ell),(0,\ell+1)}^{+} \geq 1 \right) & \text{if } k\geq 0. \end{cases}\] All the probabilities on the right are less than 1 and as \(\ell \to \infty\) they converge to the probability that two independent Brownian motions started a distance 1 apart do not meet in 1 unit of time, which is also strictly less than 1. Thus, it follows that there is a \(c_0>0\) such that the supremum inside the braces in 27 is less than \(1-c_0\). Finally, it follows from 26 and 27 that 25 holds for \(n\) sufficiently large. ◻

Proof. Recall 13 and the notation \(H_n=L(n,X_n)\). The idea of the proof is rather simple. Since Proposition 15 implies that the walk doesn’t move a large distance in a short time, in order for the local time process \((H_n)_{n\in\mathbb{Z}_+}\) to change by much in a short time either (1) the local time profile has to have a large fluctuation in a short distance or (2) the local time at a fixed location should increase very rapidly in a short amount of time. We will use the joint GRKT, Theorem 4, to show that both of these are unlikely to occur.

It is enough to show that for any \(\varepsilon, \eta>0\) there is a \(\delta'>0\) such that \[\label{tightcond-H} \limsup_{n\to\infty} P\Bigg( \max_{\substack{k,\ell\leq n^{3/2} \\ |k-\ell|\leq \delta' n^{3/2}}} |H_k - H_\ell| > \varepsilon\sqrt{n} \Bigg) < \eta.\tag{28}\] We note that similarly to 23 \[\label{Gunion} \Bigg\{ \max_{\substack{k,\ell\leq n^{3/2} \\ |k-\ell|\leq \delta' n^{3/2}}} |H_k - H_\ell| \geq \varepsilon\sqrt{n} \Bigg\} \cap R_{K,n} \cap \left\{ \mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)}) > \delta' \right\} \subset \bigcup_{\substack{x \in \delta\mathbb{Z}\cap [-K,K]\\h \in \sqrt{\delta}\mathbb{Z}\cap [0,K]}} G^{(K)}_{x,h,\delta,\varepsilon,n},\tag{29}\] where \[\begin{align} G^{(K)}_{x,h,\delta,\varepsilon,n} &= \left\{ \exists j > \tau_{x,h}^n: \, \left| H_j - H_{\tau_{x,h}^n} \right| > \frac{\varepsilon}{4} \sqrt{n} \text{ and } \mathcal{T}_{\delta,n} \cap \left( \frac{\tau_{x,h}^n}{n^{3/2}}, \frac{j}{n^{3/2}} \right) = \varnothing\right\} \\ &\makebox[5cm]{\;}\cap \left\{ \max_{i \in \mathbb{Z}} L(\tau^n_{x,h} , i ) \leq 2\sigma K \sqrt{n} \right\}. \end{align}\] To bound \(P(G^{(K)}_{x,h,\delta,\varepsilon,n})\) we will fix an \(M\geq 1\) (to be chosen later) and then use 24 to conclude that for \(n\) sufficiently large we have \[\label{PG-PGH} P(G^{(K)}_{x,h,\delta,\varepsilon,n}) \leq 2(1-c_0)^{\lfloor \frac{4\varepsilon^2}{5\delta M} \rfloor} + P\left( G^{(K)}_{x,h,\delta,\varepsilon,n} \setminus A^{(K)}_{x,h,\delta,\frac{4\varepsilon^2}{M},n} \right).\tag{30}\] To bound the last probability, note that on the event \(G^{(K)}_{x,h,\delta,\varepsilon,n}\setminus A^{(K)}_{x,h,\delta,\frac{4\varepsilon^2}{M},n}\) not only does the local time at the present location change by at least \(\frac{\varepsilon}{4}\sqrt{n}\) before the next stopping time of the form \(\tau_{x',h'}^n\) with \((x',h') \in \delta\mathbb{Z}\times \sqrt{\delta} \mathbb{Z}_+\), but it does so without moving a distance of more than \(\frac{\varepsilon^2}{M} n\). Now, since \[\begin{align} \left| H_j - H_{\tau_{x,h}^n} \right| &= \left| L(j,X_j) - L\left(\tau_{x,h}^n, \lfloor xn \rfloor \right) \right| \\ &\leq \left| L\left(j,X_j \right) - L(\tau_{x,h}^n,X_j) \right| + \left| L(\tau_{x,h}^n,X_j) - L\left(\tau_{x,h}^n, \lfloor xn \rfloor \right) \right|, \end{align}\] we see that if the event \(G^{(K)}_{x,h,\delta,\varepsilon,n} \setminus A^{(K)}_{x,h,\delta,\frac{4\varepsilon^2}{M},n}\) occurs, then either

1. there is an integer \(k\) with \(|k-\lfloor xn \rfloor| \leq n(\varepsilon^2/M)\) such that \(|L(\tau_{x,h}^n, k) - L(\tau_{x,h}^n, \lfloor xn \rfloor) | \geq \frac{\varepsilon}{8}\sqrt{n}\), or equivalently, \(\sup\limits_{y: |y-x| \leq \varepsilon^2/M} |\Lambda_{x,h}^n(y) - \Lambda_{x,h}^n(x) | \geq \frac{\varepsilon}{16\sigma}\),

2. or there is a time \(j > \tau_{x,h}^n\) such that \(|L(j,X_j) - L(\tau_{x,h}^n,X_j)| \geq \frac{\varepsilon}{8}\sqrt{n}\) and such that there are no stopping times \(\tau_{x',h'}^n \in (\tau_{x,h}^n, j)\) with \((x',h') \in \delta\mathbb{Z}\times \sqrt{\delta}\mathbb{Z}_+\).

In the second case, let \(k \in \mathbb{Z}\) be such that if \(X_j \geq 0\) then \(\lfloor k\delta n \rfloor \leq X_j < \lfloor (k+1)\delta n \rfloor\) or if \(X_j < 0\) then \(\lfloor (k-1)\delta n \rfloor < X_j \leq \lfloor k\delta n \rfloor\), and then let \(\ell \in \mathbb{Z}_+\) be such that \(\tau_{k\delta,\ell\sqrt{\delta}}^n \leq \tau_{x,h}^n < j < \tau_{k\delta,(\ell+1)\sqrt{\delta}}^n\). Moreover, note that the conditions in the event \(G^{(K)}_{x,h,\delta,\varepsilon,n} \setminus A^{(K)}_{x,h,\delta,\frac{4\varepsilon^2}{M},n}\) ensure that \(|k| \leq (K+\frac{\varepsilon^2}{M})/\delta\leq 2K/\delta\) and \(\ell \leq K/\sqrt{\delta}\). Then, it follows that \[L\left(\tau_{k\delta,(\ell+1)\sqrt{\delta}}^n, X_j \right) - L\left(\tau_{k\delta,\ell\sqrt{\delta}}^n, X_j \right) \geq L(j,X_j) - L(\tau_{x,h}^n,X_j) \geq \frac{\varepsilon}{8}\sqrt{n}.\] If \(X_j\geq 0\) then this implies that \(\sup_{y\in [k\delta,(k+1)\delta)} \Lambda_{k\delta,(\ell+1)\sqrt{\delta}}^n(y) - \Lambda_{k\delta,\ell\sqrt{\delta}}^n(y) \geq \frac{\varepsilon}{16\sigma}\) while if \(X_j < 0\) then it implies that \(\sup_{y\in ((k-1)\delta,k\delta]} \Lambda_{k\delta,(\ell+1)\sqrt{\delta}}^n(y) - \Lambda_{k\delta,\ell\sqrt{\delta}}^n(y) \geq \frac{\varepsilon}{16\sigma}\). Therefore, it follows from the above consideration together with Theorem 4 that \[\begin{align} &\limsup_{n\to\infty} P\left( G^{(K)}_{x,h,\delta,\varepsilon,n} \setminus A^{(K)}_{x,h,\delta,\frac{4\varepsilon^2}{M},n} \right) \leq P\left( \sup_{y: |y-x| \leq \frac{\varepsilon^2}{M}} |\Lambda_{x,h}(y) - h | \geq \frac{\varepsilon}{16\sigma} \right) \tag{31} \\ &\quad + \frac{12 K^2}{\delta^{3/2}} \max_{\substack{0\leq k \delta\leq K+\frac{\varepsilon^2}{M} \\ 0\leq \ell\sqrt{\delta} \leq K }} P\left( \sup_{y\in [k\delta, (k+1)\delta)} \Lambda_{k\delta, (\ell+1)\sqrt{\delta}}(y) - \Lambda_{k\delta, \ell\sqrt{\delta}}(y) \geq \frac{\varepsilon}{16\sigma} \right). \tag{32} \end{align}\] Recalling that the curves \(\Lambda_{x,h}(\cdot)\) are reflected/absorbed Brownian motions, it follows that the second probability in 31 can be bounded in terms of a Brownian motion \(B(\cdot)\): \[P\left( \sup_{y: |y-x| \leq \frac{\varepsilon^2}{M}} |\Lambda_{x,h}(y) - h | \geq \frac{\varepsilon}{16\sigma} \right) \leq 4P\left( \sup_{0\leq t\leq \frac{\varepsilon^2}{M}} B(t) \geq \frac{\varepsilon}{16\sigma} \right) = 8 \overline{\Phi}\left( \frac{\sqrt{M}}{16\sigma} \right).\] Similarly, for the probabilities in 32 , since \(\Lambda_{k\delta,\ell\sqrt{\delta}}(\cdot)\) and \(\Lambda_{k\delta,(\ell+1)\sqrt{\delta}}(\cdot)\) are independent coalescing, reflected/absorbed Brownian motions then it follows that (as long as \(\sqrt{\delta} < \frac{\varepsilon}{16\sigma}\)) \[\begin{gather} P\left( \sup_{y\in [k\delta, (k+1)\delta)} \Lambda_{k\delta, (\ell+1)\sqrt{\delta}}(y) - \Lambda_{k\delta, \ell\sqrt{\delta}}(y) \geq \frac{\varepsilon}{16\sigma} \right) \\ \leq P\left( \sup_{0\leq t < \delta} \left(\sqrt{\delta} + \sqrt{2} B(t)\right) \geq \frac{\varepsilon}{16\sigma} \right) = 2 \overline{\Phi}\left(\frac{1}{\sqrt{2}} \left( \frac{\varepsilon}{16\sigma\sqrt{\delta}} - 1 \right) \right). \end{gather}\] Combining these estimates with 29 , 30 , 31 and 32 we get that \[\begin{align} \limsup_{n\to\infty} &P\left( \max_{\substack{k,\ell\leq n^{3/2} \\ |k-\ell|\leq \delta' n^{3/2}}} |H_k - H_\ell| > \varepsilon\sqrt{n} \right) \leq \limsup_{n\to\infty} \left( P(R_{K,n}^c) + P\left( \mathop{\mathrm{gap}}(\mathcal{T}_{\delta,n}^{(K)} ) \leq \delta' \right) \right) \nonumber\\ &\qquad + \frac{6 K^2}{\delta^{3/2}} \left\{ 2(1-c_0)^{\lfloor \frac{4\varepsilon^2}{5\delta M} \rfloor} + 8 \overline{\Phi}\left( \frac{\sqrt{M}}{16\sigma} \right) + \frac{24 K^2}{\delta^{3/2}} \overline{\Phi}\left( \frac{1}{\sqrt{2}} \left( \frac{\varepsilon}{16\sigma\sqrt{\delta}} - 1 \right) \right) \right\} . \label{KMd} \end{align}\tag{33}\] Finally, we show how the parameters \(K\), \(M\), \(\delta\), and \(\delta'\) can be carefully chosen to make the right side smaller than \(\eta\).

1. First, choose \(K = K(\eta)\) as in Lemma 2.

2. Next, choose \(M\geq 1\) and \(\delta>0\) (depending on \(K,\varepsilon,\) and \(\eta\)) so that 33 is less than \(\eta/3\). For instance, this can be done by letting \(M=\delta^{-1/2}\) and letting \(\delta>0\) be sufficiently small.

3. Finally, choose \(\delta'>0\) (depending on \(K\) and \(\delta\)) as in Lemma 3.

 ◻

Proof. For the distribution of the initial condition in ?? , note that at time \(\tau_{k,m}\) the walk has taken \(m\) previous steps away from site \(k\). Thus, \(\mathcal{E}^+_{\tau_{k,m}}(k)=j\) if and only if \(j\) of these steps have been to the right and \(m-j\) of these steps have been to the left. Translating this to the corresponding urn process this means that the change in discrepancy must be \(\mathfrak{D}_m-\mathfrak{D}_0=2j-m\) (or \(\widetilde{\mathfrak{D}}_m-\widetilde{\mathfrak{D}}_0=2j-m\) in the case \(k=0\)).

For the transition probabilities in ?? , we will explain only the case when \(\ell \leq -1\) as the other cases are similar. In this case since the walk starts at \(X_0=0\) and ends at \(X_{\tau_{k,m}} = k < \ell < 0\) then the walk will have taken one more step from \(\ell\) to \(\ell-1\) than it took from \(\ell-1\) to \(\ell\) by time \(\tau_{k,m}\), and so the information \(\mathcal{E}_{\tau_{k,m}}^+(\ell-1)=i\) means that the distribution of \(\mathcal{E}_{\tau_{k,m}}^+(\ell)\) is determined by the urn process \((\mathfrak{R}_j, \mathfrak{B}_j), j\geq 0\), for \(j\leq \beta_{i+1}\). More precisely, \(\mathcal{E}_{\tau_{k,m}}^+(\ell) = i+x\) if and only if there are \(i+x\) red balls drawn before the \((i+1)\)-st blue ball drawn, or equivalently if \(\mathfrak{D}_{\beta_{i+1}}-\mathfrak{D}_0 = x-1\). Since at sites \(\ell \leq -1\) the discrepancy urn process starts at \(\mathfrak{D}_0 = 1\) then we have that \(P\left( \mathcal{E}_{\tau_{k,m}}^+(\ell) = i+x \mid \mathcal{E}_{\tau_{k,m}}^+(\ell-1) = i \right) = P_1\left( \mathfrak{D}_{\beta_{i+1}} = x \right)\). ◻

Proof. The one dimensional case \(N=1\) was proved in [3]. In fact, in the proof of that Lemma it was shown that the renewal times of the Markov chain \((\mathfrak{D}_{\beta_n})_{n\ge 0}\) have exponential moments (see (3.13) in [3]). Thus, it follows from [33] that for every \(j \in \mathbb{Z}\) there is a constant \(M(j)<\infty\) such that \[\label{TV-M} \sum_{k \in \mathbb{Z}} \left| P_j\left( \mathfrak{D}_{\beta_n} = k \right) - \pi(k) \right| \leq M(j) e^{-C_2 n}, \qquad \forall n\geq 1,\tag{35}\] and, moreover, that the constants \(M(j)\) are such that \[\label{Mbar} \overline{M} := \sum_{k \in \mathbb{Z}} \pi(k) M(k) < \infty.\tag{36}\] The proof of 34 with \(C_3 = \max\{M(0), M(1), \overline{M} \}\) then follows rather easily from 35 and 36 using induction on \(N\geq 1\). ◻

Proof. It follows from the definition of \(\mathfrak{t}_{x,h}\) and the relation \(h = \Lambda_{x,h}(x) = \mathfrak{L}(\mathfrak{t}_{x,h},x)\) that \[\label{txhprop} s > \mathfrak{t}_{x,h} \iff \mathfrak{L}(s,x) > h.\tag{37}\] Next, note that \[\begin{align} \Lambda_{x,h}(y) = 0 &\iff \mathfrak{L}(\mathfrak{t}_{x,h},y) = 0 = \mathfrak{L}(\mathfrak{t}_{y,0},y) \\ &\iff \mathfrak{t}_{x,h} < \mathfrak{t}_{y,0} \qquad \text{(recalling that \mathfrak{t}_{x,h} \neq \mathfrak{t}_{y,0}, a.s.)}\\ &\iff \mathfrak{t}_{x,h} < \mathfrak{t}_{y,h'} \quad \forall h'>0 \qquad\text{(using \mathfrak{t}_{y,0} = \lim_{h'\to 0^+} \mathfrak{t}_{y,h'} and \mathfrak{t}_{x,h} \neq \mathfrak{t}_{y,0} a.s.)} \\ &\overset{\eqref{txhprop}}{\iff} \Lambda_{y,h'}(x) = \mathfrak{L}(\mathfrak{t}_{y,h'},x) > h, \quad \forall h'>0\\ &\iff \inf\left\{ h'\geq 0: \Lambda_{y,h'}(x) > h \right\} = 0. \end{align}\] Similarly, if \(\Lambda_{x,h}(y) = \mathfrak{L}(\mathfrak{t}_{x,h},y) = b\) then the definition of \(\mathfrak{t}_{y,b}\) implies that \(\mathfrak{t}_{x,h} \leq \mathfrak{t}_{y,b}\), but since these times are almost surely not equal then we have \(\mathfrak{t}_{x,h} < \mathfrak{t}_{y,b}\). Moreover, for any \(h' \in [0,b)\) we have that \(b > h' = \mathfrak{L}(\mathfrak{t}_{y,h'},y)\) so that \(\mathfrak{t}_{y,h'} < \mathfrak{t}_{x,h}\) for all \(h'\in [0,b)\). Conversely, suppose that \(\mathfrak{t}_{y,h'} < \mathfrak{t}_{x,h} < \mathfrak{t}_{y,b}\) for all \(h' \in [0,b)\). The second inequality implies that \(\Lambda_{x,h}(y) = \mathfrak{L}(\mathfrak{t}_{x,h},y) \leq \mathfrak{L}(\mathfrak{t}_{y,b},y) = b\), while the first inequality implies (by 37 ) that \(\Lambda_{x,h}(y) = \mathfrak{L}(\mathfrak{t}_{x,h},y) > h'\). Since this holds for all \(h' \in [0,b)\) these together imply that \(\Lambda_{x,h}(y) = b\). Using the above argument to get the first \(\iff\) in the display below, we see that \[\begin{align} \Lambda_{x,h}(y) = b &\iff \mathfrak{t}_{y,h'} < \mathfrak{t}_{x,h} < \mathfrak{t}_{y,b}, \quad \forall h' \in [0,b) \\ &\iff \mathfrak{t}_{y,h'} < \mathfrak{t}_{x,h} < \mathfrak{t}_{y,h''}, \quad \forall h' \in [0,b) \text{ and } \forall h'' > b \\ &\overset{\eqref{txhprop}}{\iff} \Lambda_{y,h'}(x) \leq h < \Lambda_{y,h''}(x), \quad \forall h' \in [0,b) \text{ and } \forall h'' > b \\ &\iff \inf\left\{ h'\geq 0: \Lambda_{y,h'}(x) > h \right\} = b. \end{align}\] Note that the second line holds since \(h''\mapsto \mathfrak{t}_{y,h''}\) is right continuous by the definition in 6 and the fact that \(t\mapsto \mathfrak{L}(t,y)\) is almost surely continuous as shown in [1]. This completes the proof of the lemma. ◻

Proof. First of all, since \(\Lambda_{y,h'}(x) = \lim_{\varepsilon\to 0^+} \Lambda_{y,h'+\varepsilon}(x)\), almost surely (see Remark 2 on p. 387 and Theorem 4.3 in [1]), it follows that \[\Lambda_{y,h'}(x) < h \; \Longrightarrow \;h' \notin \overline{\{ b\geq 0: \Lambda_{y,b}(x) > h \}}\; \overset{\text{Lem.\,} \ref{lem:TSRMweave}}{\iff} \;h' < \Lambda_{x,h}(y)\] This proves the first inclusion in ?? . The second inclusion follows easily from Lemma 6 because \(\Lambda_{x,h}(y) > h'\) implies that \(h' \notin \{b\geq 0: \Lambda_{y,b}(x) > h \}\). Finally, the third inclusion in ?? follows from the first inclusion by taking complements of both sets and reversing the roles of \((x,h)\) and \((y,h')\). ◻

Proof. Note that ?? follows from ?? and the invariance principle for random walks. We now prove ?? . \[\begin{align} P(\gamma>b^{2-\delta})&\ge P\left(\gamma>b^{2-\delta},S_*(\lfloor b^{2-\delta} \rfloor)>b/2\right)=P\left(\gamma>b^{2-\delta},Y_*(\lfloor b^{2-\delta} \rfloor)>b/2\right)\\ &=P\left(\gamma>b^{2-\delta}\mid Y_*(\lfloor b^{2-\delta} \rfloor)>b/2\right)P\left(Y_*(\lfloor b^{2-\delta} \rfloor)>b/2\right)\\ &\ge \left(1-C_3e^{-C_2b/2}\right)^{\lfloor b^{2-\delta} \rfloor}P\left(Y_*(\lfloor b^{2-\delta} \rfloor)-Y(0)>-b/2\right)\to 1\;\;\text{as}\;b\to\infty. \end{align}\] In the last line we used Proposition 10 with \(N=1\) (proved already in [3]) and the invariance principle for a mean \(0\) random walk. ◻

Proof. By 39 and 41 , \[\begin{gather} E_{n_1,m}\left[\pm \Delta(1)+C_6e^{-C_5 m}(e^{-C_5 \Delta(1)}-1)\right]\ge \\-C_4e^{-C_5(n_1\wedge m)} +C_6e^{-C_5 m}\left(\sum_{k\in\mathbb{Z}}p_\Delta(k|n_1,n_1+m)e^{-C_5 k}-1\right)\ge 0. \end{gather}\] ◻

Proof. Note that if \(n_1<b\) then \(\check{S}_\Delta\) follows a simple symmetric random walk for which both statements hold by the standard Gambler’s ruin estimates. Thus, we shall assume that \(n_1\ge b\). Without loss of generality we also suppose that \(b>a_1+1+C_7\), where \(a_1\) is defined just above 41 and \(C_7\) is from ?? .

(a) Let \(a_1<m<b\) and \(f(s)=-s+C_6e^{-C_5s}\). Note that the process \(f(\check{S}_\Delta(j\wedge \check{\theta}_{(a_1,b)})),\;j\ge 0\), is a submartingale relative to filtration \({\mathcal{G}}_j:=\sigma(\widetilde{S}_1(\ell),\check{S}_\Delta(\ell),\chi(\ell),\; 0\le\ell\le j)\), \(j\ge 0\). Indeed, either \(\widetilde{S}_1\) stays above \(b\) and, thus, above \(\check{S}_\Delta\) up to time \(\check{\theta}_{(a_1,b)}\), and the conditions of Lemma 8 are satisfied, or, if prior to \(\check{\theta}_{(a_1,b)}\) the process \(\widetilde{S}_1\) goes below \(b\), \(\check{S}_\Delta\) switches to a simple symmetric random walk. By the optional stopping theorem and monotonicity of the function \(f\), \[\begin{align} b-m&<f(m)-f(b)\le E_{n_1,m}(f(\check{S}_\Delta(\check{\theta}_{(a_1,b)}))-f(b))\\ &\le E_{n_1,m}(f(\check{S}_\Delta(\check{\theta}_{(a_1,b)}))-f(b)\mid \check{S}_\Delta(\check{\theta}_{(a_1,b)})\le a_1)P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(a_1,b)})\le a_1)\\ &< \left(b+C_6-E_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(a_1,b)})\mid \check{S}_\Delta(\check{\theta}_{(a_1,b)})\le a_1)\right)P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(a_1,b)})\le a_1) \end{align}\] Since \(\check{S}_\Delta(\check{\theta}_{(a_1,b)})\ge 0\), we get that for \(a_1<m<b\) \[\label{lb1} P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(a_1,b)})\le a_1)>\frac{b-m}{b+C_6}.\tag{42}\] Note that, by the remark made at the start of the proof, the above inequality holds for all \((n_1,m)\in\{(n_1,m)\in \mathbb{Z}_+^2\mid n_1<b\;\text{or}\;(n_1\ge b\;\text{and}\;a_1<m<b)\}\)

Let us consider the case \(0<m\le a_1\). Using 42 we get \[\begin{gather} P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(0,b)}) = 0)\\ \ge P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(0,a_1]}) = 0)+\sum_{m\ge 0}P_{n_1,m}\left(\check{S}_\Delta(\check{\theta}_{(0,a_1]})> a_1, \widetilde{S}_1(\check{\theta}_{(0,a_1]})=m\right)\\ \times\sum_{a_1<j<b}P_{n_1,m}\left(\check{S}_\Delta(\check{\theta}_{(0,a_1]})=j\mid \check{S}_\Delta(\check{\theta}_{(0,a_1]})> a_1, \widetilde{S}_1(\check{\theta}_{(0,a_1]})=m\right)\frac{b-j}{b+C_6}\\ \times\inf_{0<m'\le a_1,n_1'\in\mathbb{Z}_+} P_{n_1',m'}(\check{S}_\Delta(\check{\theta}_{(0,b)}) = 0) \end{gather}\] Rearranging the right hand side and using ?? we see that \[\begin{gather} P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(0,b)}) = 0)\ge P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(0,a_1]}) = 0)\\+ \frac{b-a_1-1-C_7}{b+C_6}P_{n_1,m}\left(\check{S}_\Delta(\check{\theta}_{(0,a_1]})> a_1\right)\inf_{0<m'\le a_1,n_1'\in\mathbb{Z}_+} P_{n_1',m'}(\check{S}_\Delta(\check{\theta}_{(0,b)}) = 0)\\=\left(1-\frac{b-a_1-1-C_7}{b+C_6}\inf_{0<m'\le a_1,n_1'\in\mathbb{Z}_+} P_{n_1',m'}(\check{S}_\Delta(\check{\theta}_{(0,b)}) = 0)\right)P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(0,a_1]}) = 0)\\+\frac{b-a_1-1-C_7}{b+C_6}\inf_{0<m'\le a_1,n_1'\in\mathbb{Z}_+}P_{n_1',m'}(\check{S}_\Delta(\check{\theta}_{(0,b)}) = 0). \end{gather}\] We note that for all \(m\le a_1\) and \(n_1\in\mathbb{Z}_+\) \[\begin{align} P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(0,a_1]}) = 0)&\ge P_{n_1,m}(\check{S}_\Delta(1) = 0)\\ & \ge P_{n_1,a_1}(\check{S}_\Delta(1)=0\mid \check{S}_1(1)\ge n_1)P_{n_1}(\check{S_1}(1)\ge n_1)\\ &\ge \left(1+\lambda^{-2a_1-1}\right)^{-a_1}P_{n_1}(\check{S_1}(1)\ge n_1). \end{align}\] Since for all large \(n_1\) the distribution of \(\check{S}_1(1) - \check{S}_1(0)\) is close to \(\pi\), we get that \[\inf_{0<m'\le a_1,n_1'\in\mathbb{Z}_+} P_{n_1',m'}(\check{S}_\Delta(\check{\theta}_{(0,a_1]}) = 0)=:C_{10}>0,\] and \[\label{lb2} \inf_{0<m'\le a_1,n_1'\in\mathbb{Z}_+} P_{n_1',m'}(\check{S}_\Delta(\check{\theta}_{(0,b)}) = 0)\ge \frac{C_{10}(b+C_6)}{C_{10}b+C_6+(1-C_{10})(a_1+1+C_7)}.\tag{43}\] Inequalities 42 and 43 imply part (a).

(b) We shall first show that there is a constant \(C_{11}>0\) such that for all \(n_1\in\mathbb{Z}_+\) and integers \(a\) and \(m\) satisfying \(0\le a<m<b\), \[\label{lb3} P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(a,b)})\ge b)>C_{11}\frac{m-a}{b-a}.\tag{44}\] Let \(a_2=a_1\vee C_6\), \(g(s)=s+C_6e^{-C_5s}\), and assume that \(a_2\le a<b\) and \(\check{S}_\Delta(0)=m\in(a,b)\). Consider the submartingale \(g(\check{S}_\Delta(j\wedge\check{\theta}_{(a,b)})),\;j\ge 0\). Note that, by convexity of \(g\) and the fact that \(g(0)=C_6<a<g(a)\), if \(0\le\check{S}_\Delta(j)<a\) then \(g(\check{S}_\Delta(j))<g(a)\). Using this observation and ?? we get \[g(m)-g(a)\le E_{n_1,m}\left(g(\check{S}_\Delta(\check{\theta}_{(a,b)})-g(a)\right)\le (b+C_7+C_6-a) P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(a,b)})\ge b)\] and conclude that 44 holds with some constant \(C_{12}>0\) and \(a_2\le a<m<b\). Replacing the constant \(C_{12}\) with a smaller one, \[C_{11}:=\min_{0\le a'<m'\le a_2}\; \inf_{n_1'\in\mathbb{Z}_+}P_{n_1',m'}(\check{S}_\Delta(\check{\theta}_{(a',a_2]})>a_2)\,C_{12}>0,\] we obtain 44 for all \(0\le a<m<b\) and \(n_1\in\mathbb{Z}_+\). A lower bound for the infimum in the previous line is obtained similarly to the bound just above 43 . Since in this case the exit is through the upper boundary point, we shall restrict to the event \(\check{S}_1(1)\le n_1'\) instead of \(\check{S}_1(1)\ge n_1'\).

Non-trapping, 38 , and the properties of \(\pi_2\) imply that there is a constant \(\check{\sigma}>0\) such that \[\label{A44618} 2\check{\sigma}^2\le \inf_{n_1,m\in\mathbb{Z}_+}E_{n_1,m}\left[(\check{S}_\Delta(1)-m)^2\right].\tag{45}\] Now choose \(a_3\in\mathbb{Z}_+\) such that for all \(m\ge a_3\), \[\label{A44619} 2mC_4e^{-C_5m}<\check{\sigma}^2.\tag{46}\] Then 39 ensures that for all \(m>a_3\) and \(n_1\in\mathbb{Z}_+\), \[E_{n_1,m}\left[\check{S}_\Delta^2(1)-\check{\sigma}^2\right]\ge E_{n_1,m}\left[(\check{S}_\Delta(1)-m)^2-2m|\check{S}_\Delta(1)-m|+m^2-\check{\sigma}^2\right]\ge m^2\] and, hence, \(\check{S}_\Delta^2(j)-\check{\sigma}^2j\), with \(\check{S}_\Delta(0)>a_3\) and \(j=0,1,\ldots,\check{\theta}_{(a_3,b)}\) is a submartingale. Therefore, \[E_{n_1,m}\left[\check{S}_\Delta^2(\check{\theta}_{(a_3,b)}\wedge T)-\check{\sigma}^2(\check{\theta}_{(a_3,b)}\wedge T)\right]\ge m^2> 0,\quad\forall T\in\mathbb{N}.\] Using ?? we get \[\label{A44623} E_{n_1,m}\left[\check{\theta}_{(a_3,b)}\right]=\lim_{T\to\infty}E_{n_1,m}\left[\check{\theta}_{(a_3,b)}\wedge T\right]\le\frac{1}{\check{\sigma}^2}(b+C_7)^2.\tag{47}\] Representing the exit time from \([0,b)\) recursively as exits from \([0,a_3]\) and from \((a_3,b)\) and using \(C_{11}/b\) as a lower bound on the right hand side of 44 we get that \[\begin{align} E_{n_1,m}\left[\check{\theta}_{[0,b)}\right]&\le C_{11}^{-1}b\left(\sup_{0\le m\le a_3, n_1\in\mathbb{Z}_+}E_{n_1,m}\left[\check{\theta}_{[0,a_3]}\right]+\sup_{a_3< m<b, n_1\in\mathbb{Z}_+}E_{n_1,m}\left[\check{\theta}_{(a_3,b)}\right]\right)\\ &\le C_{11}^{-1}b\left(\sup_{0\le m\le a_3, n_1\in\mathbb{Z}_+}E_{n_1,m}\left[\check{\theta}_{[0,a_3]}\right]+\frac{1}{\check{\sigma}^2}(b+C_7)^2\right). \end{align}\] To complete the proof of (b) we only need to show that the supremum in the last expression is finite. The finiteness of the supremum follows by comparison with a geometric random variable from the inequality \[\label{lbe} \inf_{0\le m\le a_3, n_1\in\mathbb{Z}_+}P_{n_1,m}\left(\check{\theta}_{[0,a_3]}\le (a_3+1)\right)>0,\tag{48}\] Markov property of \((\widetilde{S}_1,\widetilde{S}_2)\), and the definition of \(\check{S}_\Delta\). To show 48 we note that, when \(\widetilde{S}_1\) is large, its next step distribution is close to \(\pi\). Thus, given any fixed \(\ell\in\mathbb{N}\), there is an \(n_2\in\mathbb{N}\) such that the probability that after the next step \(S_1\) will be within \(\ell\) units from its current position is bounded away from zero whenever the current position is larger than \(n_2\). For a given \(\ell\), on this event, the probability that \(\check{S}_\Delta\) enters \((a_3,\infty)\) in the same step is bounded away from \(0\). If \(\widetilde{S}_1\) is below \(b\) then \(\check{S}_\Delta\) follows a simple symmetric random walk, so the probability that it exits \([0,a_3]\) within the next \(a_3+1\) units of time is bounded away from \(0\). There are finitely many integer values between \(b\) and \(n_2\), hence, due to non-trapping, if \(\widetilde{S}_1\) is in \([b,n_2]\) then the probability that in the next step \(\check{S}_\Delta\) enters \((a_3,\infty)\) is again bounded away from 0. We conclude that 48 holds. ◻

Proof. Let \(n_1\ge b^4\) and \(0< m\le b\). Note that \(\{S_{1*}(b^7)\ge b^4/2\}\subset\{\widetilde{S}_{1*}(b^7)\ge b^4/2\}\) and \[\begin{align} P_{n_1,m} &(\theta_{(0,\infty)}\ge b^7)\\ &= P_{n_1,m}(\theta_{(0,\infty)}\ge b^7,\theta_{(0,\infty)}<\theta_{[0,b^2)})+P_{n_1,m}(\theta_{(0,\infty)}\ge b^7,\theta_{(0,\infty)}>\theta_{[0,b^2)})\\ &\le P_{n_1,m}(\theta_{[0,b^2)}\ge b^7)+P_{n_1,m}(S_\Delta(\theta_{(0,b^2)})\ge b^2)\\ & \le 1-P_{n_1,m}(\theta_{[0,b^2)}< b^7, S_{1*}(b^7)\ge b^4/2, \widetilde{S}_{1*}(b^7)\ge b^4/2)\\ &\qquad\qquad +1-P_{n_1,m}(S_\Delta(\theta_{(0,b^2)}) \leq 0,S_{1*}(b^7)\ge b^4/2,\theta_{[0,b^2)}<b^7, \widetilde{S}_{1*}(b^7)\ge b^4/2) \\ & = 1-P_{n_1,m}(\check{\theta}_{[0,b^2)}< b^7, \widetilde{S}_{1*}(b^7)\ge b^4/2,S_{1*}(b^7)\ge b^4/2)\\ &\qquad\qquad +1-P_{n_1,m}(\check{S}_\Delta(\theta_{(0,b^2)})= 0,\widetilde{S}_{1*}(b^7)\ge b^4/2,\check{\theta}_{[0,b^2)}<b^7,S_{1*}(b^7)\ge b^4/2) \\ & \le 1-P_{n_1,m}(\check{\theta}_{[0,b^2)}< b^7)+P_{n_1,m}(S_{1*}(b^7)< b^4/2)\\ &\qquad\;\; +1-P_{n_1,m}(\check{S}_\Delta(\theta_{(0,b^2)})= 0)+P_{n_1,m}(\check{\theta}_{[0,b^2)}\ge b^7)+P_{n_1,m}(S_{1*}(b^7)< b^4/2)\\ & \le 2P_{n_1,m}(\check{\theta}_{[0,b^2)}\ge b^7)+1-P_{n_1,m}(\check{S}_\Delta(\theta_{(0,b^2)})= 0)+2P_{n_1,m}(S_{1*}(b^7)< b^4/2)\\ & \le \frac{2E_{n_1,m}[\check{\theta}_{[0,b^2)}]}{b^7}+1-P_{n_1,m}(\check{S}_\Delta(\check{\theta}_{(0,b^2)}= 0)+2P_{n_1,m}(S_{1*}(b^7)< b^4/2)\\ &\le \frac{2C_9}{b}+1-\frac{b^2-b}{b^2+C_8}+2P_{n_1,m}(S_{1*}(b^7)< b^4/2), \end{align}\] where in the last inequality we used Lemma 10 with \(b^2\) instead of \(b\). The stated result follows by applying ?? with \(b\) replaced by \(b^4\) and \(\delta=1/4\) and letting \(b\) go to infinity. ◻

Proof. We first prove ?? . First of all, note that a simple computation yields that \[\label{d-ud} \frac{P_x\left( \mathfrak{D}_1=x-1 \right)}{P_x\left( \mathfrak{D}_1=x+1, \, \mathfrak{D}_2=x \right)} = \frac{1+\lambda^{2x+1}}{\lambda^{2x-1}} = \lambda^{-2x+1} + \lambda^2, \quad \forall x \in \mathbb{Z}.\tag{51}\] Using this, we get that for any \(x\in \mathbb{Z}\), \(y\le 0\), and \(n\geq 1\), \[\begin{align} P_x\left( \mathfrak{D}_{\beta_n} = y-1 \right) &=P_x\left(\mathfrak{D}_{\beta_n}=y-1,\,\mathfrak{D}_{\beta_n-1} = y\right) \nonumber \\ & =\sum_i P_x\left(\mathfrak{D}_{i+1}=y-1,\,\mathfrak{D}_i = y,\,\beta_{n-1}\leq i < \beta_n\right) \nonumber \\ &= \sum_i P_{y}\left( \mathfrak{D}_1 = y-1 \right)P_x\left( \mathfrak{D}_i=y,\,\beta_{n-1}\leq i < \beta_n \right) \nonumber \\ &= \left( \lambda^{-2y+1} + \lambda^2 \right) \sum_i P_{y}\left( \mathfrak{D}_1=y+1, \mathfrak{D}_2 = y \right) P_x\left(\mathfrak{D}_i=y,\, \beta_{n-1}\leq i < \beta_n \right) \nonumber \\ &\leq \left( \lambda+ \lambda^2 \right) P_x\left( \mathfrak{D}_{\beta_n} = y \right). \label{lt-1} \end{align}\tag{52}\] Since the multiplicative factor \(\lambda+\lambda^2\) is strictly less than 1 for all \(y\) sufficiently negative, the inequality ?? follows easily by iterating the bound in 52 .

For the proof of ?? , we let \[r_n(y) = \sup_{x\leq y} \frac{P_x\left( \mathfrak{D}_{\beta_n} = y+1 \right)}{P_x\left( \mathfrak{D}_{\beta_n} = y \right)}, \qquad y \in \mathbb{Z},\;n\geq 1.\] To bound \(r_1(y)\), note that 51 implies \[\begin{align} P_x(\mathfrak{D}_{\beta_1} = y+1) &= P_x\left( \mathfrak{D}_{y-x+1} = y+1 \right) P_{y+1}\left( \mathfrak{D}_1 = y+2, \, \mathfrak{D}_2 = y+1 \right)\nonumber \\ &= P_x\left( \mathfrak{D}_{y-x+1} = y+1 \right) P_{y+1}\left( \mathfrak{D}_1 = y \right) \frac{\lambda^{2y+1}}{1+\lambda^{2y+3}} \nonumber \\ &= P_x(\mathfrak{D}_{\beta_1} = y) \frac{\lambda^{2y+1}}{1+\lambda^{2y+3}}\;\;\Rightarrow\;\;r_1(y) = \frac{\lambda^{2y+1}}{1+\lambda^{2y+3}} \label{r1}. \end{align}\tag{53}\] Next, we will use 53 and an induction argument to get bounds on \(r_n(y)\) for \(n\geq 2\). For \(x\leq y\) and \(n\geq 2\) note that \[\begin{align} P_x\left( \mathfrak{D}_{\beta_n} = y+1 \right) &= P_x\left( \mathfrak{D}_{\beta_n} = y+1,\, \mathfrak{D}_{\beta_{n-1}} \leq y+1 \right) + P_x\left( \mathfrak{D}_{\beta_n} = y+1,\, \mathfrak{D}_{\beta_{n-1}} = y+2 \right) \\ &= \sum_i P_x\left( \beta_{n-1}\leq i < \beta_n, \, \mathfrak{D}_i = y+1 \right) P_{y+1}\left(\mathfrak{D}_1 = y+2, \, \mathfrak{D}_2 = y+1 \right) \\ &\qquad + P_x\left( \mathfrak{D}_{\beta_{n-1}} = y+2 \right)P_{y+2}\left( \mathfrak{D}_1 = y+1 \right) \\ &\leq \sum_i P_x\left( \beta_{n-1}\leq i < \beta_n, \, \mathfrak{D}_i = y+1 \right) P_{y+1}\left(\mathfrak{D}_1 = y \right) \frac{\lambda^{2y+1}}{1+\lambda^{2y+3}} \\ &\qquad + r_{n-1}(y+1)P_x\left( \mathfrak{D}_{\beta_{n-1}} = y+1 \right)P_{y+1}\left( \mathfrak{D}_1 = y \right) \frac{1+\lambda^{2y+1}}{1+\lambda^{2y+3}} \\ &\leq \left( \frac{\lambda^{2y+1}}{1+\lambda^{2y+3}} + \frac{1+\lambda^{2y+1}}{1+\lambda^{2y+3}} r_{n-1}(y+1) \right) P_x\left( \mathfrak{D}_{\beta_n} = y \right). \end{align}\] where the second to last inequality follows from the definition of \(r_{n-1}(y+1)\) and the fact that \(\frac{P_{y+2}\left( \mathfrak{D}_1 = y+1 \right)}{P_{y+1}\left( \mathfrak{D}_1 = y \right)} = \frac{1+\lambda^{2y+1}}{1+\lambda^{2y+3}}\). Thus, we have shown that \[r_n(y) \leq \frac{\lambda^{2y+1}}{1+\lambda^{2y+3}} + \frac{1+\lambda^{2y+1}}{1+\lambda^{2y+3}} r_{n-1}(y+1), \quad\forall n\geq 1, \, y\in \mathbb{Z}.\] From this and 53 we get the estimate \[r_n(y) \leq \sum_{k=1}^n \frac{(1+\lambda^{2y+1})\lambda^{2(y+k)-1}}{(1+\lambda^{2(y+k)-1})(1+\lambda^{2(y+k)+1})} \leq \frac{(1+\lambda^{2y+1})\lambda^{2y+1}}{1-\lambda^2}.\] Therefore, \(r_n(y)\le \frac{\lambda^{2y+1}}{1-\lambda}\) for all \(y\geq 0\) and \(n\in \mathbb{N}\), and the claim in ?? follows. ◻