Proof. Factors with marks \(0\) or \(1\) are constant. All other factors in \(p_\omega(\gamma)\) are functions of rows in \(\operatorname{Act}(\gamma)\), with repeated departures from one row retaining that same row. The finite-range assumption (1) gives independence from the rows in \(B\), also when \(B\) is infinite. This proves (3); if the active set is empty, \(p_\omega(\gamma)\) itself is constant.

Before the first arrival in \(F\), every transition departs from its complement. Prefix probabilities for the stopped walk therefore use only exterior rows, including when the prefix ends with that arrival. These prefixes determine the stopped path law, so all its measurable event probabilities have the same row dependence. Equivalently, one may kill the walk on arrival in \(F\). An infinite-path event implying avoidance of \(F\) is determined by this killed trajectory: it is false on killed paths and retains its original meaning on surviving paths. This proves the last assertion. ◻

Proof. First test for a scripted cut from time zero. Failure is detected in finite time: either the next \(M\) steps are not the script, or they are the script and the subsequent path first falls below its endpoint. After a detected failure, wait for a strict record above the whole maximum attained by the detection time, and test again. The successive test starts, when finite, are stopping times in the marked path filtration. Equation (5), applied to every finite record prefix, gives conditional success probability \(\xi p_v\) at each test. Hence \[P_0\{\text{the first }j\text{ tests occur and all fail}\} \le (1-\xi p_v)^j.\] On infinite height supremum, every detected failure is followed by another finite test start. Letting \(j\) tend to infinity proves (i). The slab-exit observation implies that the first cut is finite almost surely under \(\widehat P_v\). The tests need not locate the earliest cut; their only role is to establish that some cut exists.

We now identify the law at the actual first cut by finite prefixes. Suppose \(\gamma\) is a regeneration word ending at \(z\). A path beginning with \(\gamma\) whose suffix stays at or above \(h(z)\) has its first cut at that endpoint: all earlier candidates have already been undercut. It also satisfies the original \(D_v\). Conversely, a path in \(D_v\) through its first cut gives such a word. Indeed, every earlier candidate has smaller height than the terminal strict record. If it were not undercut before the terminal record, it could not be undercut afterwards either, since the suffix stays above that record; it would already be a cut.

For any measurable suffix event \(B\), the row separation in (7) and Lemma 2 therefore give \[\begin{split} &P_0\{\text{first-cut prefix }\gamma,\ \text{ translated suffix in }B\}\\ &\hspace{25mm}=W(\gamma)P_0(B\cap D_v). \end{split}\] Dividing by \(p_v\) proves that, under \(\widehat P_v\), the first word has mass \(W(\gamma)\) and its translated suffix has law \(\widehat P_v\), independently of that word. There are only countably many finite marked words, and the first cut is finite almost surely, so these masses sum to one.

The cut is a strict record, hence records of its translated suffix are also records of the full path. A script cannot straddle this boundary: such a script would overlap the script that has just ended. Thus the cuts of the suffix are precisely the subsequent cuts of the full path. Iterating the preceding factorization proves the iid assertion in (ii). Every word is finite and contains at least \(M\) steps, so the successive cut times tend to infinity and the words cover the whole path.

For (iii), the prefix through the raw first cut may have negative heights, but it still ends with the script from a record, and every preceding candidate is undercut within the prefix. Its active departures lie strictly below the start of the terminal script. The same separated-weight calculation gives raw prefix probability \(W(\gamma)p_v\) and, with suffix restriction \(B\), probability \(W(\gamma)P_0(B\cap D_v)\). Their ratio is \(\widehat P_v(B)\). All of these arguments enumerate finite prefixes; none uses a Markov property at a cut selected using the future. ◻

Proof. Assume first that \(p_v>0\). On infinite supremum there is a first cut, and its suffix has the iid word law of Proposition 3. The successive slab bases increase in height by at least \(M\); within each slab the path stays above its base. Since each slab is finite, \(h(X_n)\to+\infty\).

For finite supremum, fix integers \(m,b\ge0\). Consider the event that all heights are at most \(m\), but heights at least \(-b\) are visited infinitely often. From each such visit, a script of \(m+b+1\) consecutive steps \(a_v\), with no mark restriction, crosses \(m\) with conditional probability at least \(\kappa^{m+b+1}\). Take successive visit trials at least \(m+b+1\) times apart. The probability that the first \(j\) trials occur and all fail to cross \(m\) is at most \((1-\kappa^{m+b+1})^j\). The event in question is therefore null. Taking the countable union over \(m,b\) shows that finite supremum forces \(h(X_n)\to-\infty\).

It remains to obtain \(p_v>0\) from a positive probability of \(A_v\). On \(A_v\), the height sequence tends to infinity, so its global minimum is attained at some finite time, even if its possible values are not discrete. Thus for some deterministic \(n\ge0\) and \(x\in\mathbb Z^d\), the event \[\{X_n=x,\ h(X_{n+j})\ge h(x)\text{ for all }j\ge0\}\] has positive annealed probability. If \(q_x(\omega)\) denotes the quenched probability from \(x\) of staying above \(h(x)\), the quenched Markov property at time \(n\) expresses this probability as \(E_Q[P_{0,\omega}(X_n=x)q_x(\omega)]\). It is bounded above by \(E_Qq_x=p_v\), by stationarity. Therefore \(p_v>0\), and the two alternatives just proved give (9). ◻

Proof. Let \(S(\gamma)\) be the number of ordinary records in a regeneration word \(\gamma\), excluding its initial boundary and including its terminal record. Each increase of the running maximum is at most one, because every step has height increment at most one. Therefore \(L(\gamma)\le S(\gamma)\). The record indices of successive cuts, with index zero at the initial boundary, are sums of iid positive integer increments with law \(S\): the terminal height of each word is the maximum so far, so record counts concatenate across boundaries.

For \(i\ge0\), let \(\rho_i\) be the time of ordinary record index \(i\) in a raw walk, with \(\rho_0=0\) and \(\rho_i=\infty\) if that record is never reached. Put \[D'_{v,i}=\{\rho_i<\infty,\ h(X_j)\ge0 \text{ for }0\le j\le\rho_i\}.\] A script from record \(i\) creates exactly \(M\) consecutive strict records. Hence a cut of record index \(i+M\), on a path in \(D_v\), occurs exactly when the prefix realizes \(D'_{v,i}\), the script follows, and its suffix stays above its endpoint. Summing (6) over the possible record prefixes and dividing by \(p_v\) gives \[ \widehat P_v\{i+M\text{ is a cut index}\} =\xi P_0(D'_{v,i})\ge\xi p_v. \tag{11}\] The inequality holds because on \(D_v\) the supremum is infinite almost surely, so every ordinary record is reached.

Let \(N_{\mathrm{rec}}(n)\) count the cut indices in \(\{1,\ldots,n\}\). Equation (11) implies \[\liminf_{n\to\infty} \frac{E_{\widehat P_v}N_{\mathrm{rec}}(n)}{n}\ge\xi p_v>0.\] If \(E_{\nu_v}S=\infty\), however, the partial sums of iid copies of \(S\), divided by their number, tend to infinity almost surely. To see this, apply the strong law to \(\min(S,K)\) for each positive integer \(K\), then let \(K\to\infty\). Inverting these partial sums shows \(N_{\mathrm{rec}}(n)/n\to0\) almost surely. Since this ratio lies in \([0,1]\), bounded convergence gives the same limit in expectation, a contradiction. Thus \(E_{\nu_v}S<\infty\), and \(L\le S\) proves (10). ◻

Proof. Call an integer \(j\ge1\) bad if some interval starting at \(j\), of length at most \(n\), has average greater than \(a\). Among the bad starts in \(\{1,\ldots,m\}\), select the leftmost one and a witnessing interval. Continue by selecting the leftmost bad start strictly to the right of the selected interval, and repeat. These disjoint intervals cover all bad starts in \(\{1,\ldots,m\}\) and lie in \(\{1,\ldots,m+n\}\). Their total length is at least the number of those bad starts, while the sum of the \(Z_i\) over them is at least \(a\) times their total length. Consequently \[a\,\#\{j\le m:j\text{ is bad}\}\le\sum_{i=1}^{m+n}Z_i.\] Taking expectations, using stationarity, dividing by \(m\), and letting \(m\to\infty\) proves the assertion. ◻

Proof. Both nonbacktracking probabilities \(p_v,p_{-v}\) are positive by Proposition 4. For \(\sigma\in\{+,-\}\) write \[P_\sigma^*=\widehat P_{\sigma v},\qquad E_\sigma^*=E_{P_\sigma^*}.\] Here \(\sigma v\) means \(v\) or \(-v\), respectively. Under either law, let \(L_i,\mathcal R_i\) be the width and radius of its \(i\)th regeneration word, and put \[H_0=0,\qquad H_k=\sum_{i=1}^k L_i.\] An unsubscripted \(L\) or \(\mathcal R\) denotes the corresponding variable for one word. All constants chosen below are independent of the scale parameters \(i_0\) and \(a\).

Suppose that (12) fails. The truncated mean \[I(t)=\sum_{\sigma\in\{+,-\}}E_\sigma^*\min(\mathcal R,t), \qquad t\ge0,\] is then nondecreasing and unbounded. It is concave, \(I(0)=0\), and \(I(t)=o(t)\): the last assertion follows by dominated convergence from the almost-sure finiteness of each regeneration word.

We will obtain a fixed positive probability that the final supremum of one of the two signed heights lies in a bounded window, and then place these windows arbitrarily far apart. The unbounded truncated mean will produce a large spatial excursion after a controlled initial segment, and hence a record for a tilted linear functional. A marked connector will let us append a separated continuation in the opposite direction while preserving the height maximum already attained.

Proof. The terminal displacement of a word of law \(\nu_{\sigma v}\) is integrable by Proposition 7. Write its mean as \(b_\sigma'\in\mathbb R^d\). Since \[b_\sigma'\cdot(\sigma v)=E_{\nu_{\sigma v}}L>0,\] this vector is nonzero. Under \(\widehat P_{\sigma v}\), the \(k\)th cut location divided by \(k\) tends almost surely to \(b_\sigma'\). Integrability also gives \(\mathcal R_k/k\to0\) almost surely: for each \(\varepsilon>0\), the sum of \(\widehat P_{\sigma v}(\mathcal R_k>\varepsilon k)\) is finite, and Borel–Cantelli applies. Between consecutive cuts the displacement from the earlier cut is bounded by the next radius. Since the number of completed words tends to infinity along the trajectory, it follows that \(|X_n|\to\infty\) and \[\frac{X_n}{|X_n|}\longrightarrow r_\sigma:=\frac{b_\sigma'}{|b_\sigma'|}.\] Under the raw law on \(A_{\sigma v}\), the first cut is finite and its translated suffix has law \(\widehat P_{\sigma v}\) by Proposition 3; hence the same conclusion holds there.

The signs of the projections on \(v\) make \(r_+\) and \(r_-\) distinct. Suppose they are not opposite. Then \(w=r_++r_-\) is nonzero and \[w\cdot r_+=w\cdot r_-=1+r_+\cdot r_->0.\] Proposition 4 gives \(P_0(A_v\cup A_{-v})=1\), so the limiting rays imply \(P_0(A_w)=1\). Rescale \(w\) to have \(\ell^\infty\) norm one and apply the cut construction in that direction.

Under \(\widehat P_w\), the limiting ray exists almost surely because this law is a conditioning of the raw law just considered. Its value is unchanged on deleting any finite number of regeneration words and translating the remaining path: such operations do not affect a limiting direction when positions tend to infinity in norm. Consequently it is a tail random variable of the iid words under \(\widehat P_w\): for every \(k\), its value can be computed from the words after \(k\). The independent tail zero–one law makes it deterministic. Since \(P_0(A_w)=1\), the raw walk has a first cut in direction \(w\) almost surely, with translated suffix law \(\widehat P_w\). Its limiting ray must therefore have that same deterministic value almost surely. This contradicts the two distinct rays \(r_+,r_-\), each occurring with positive raw probability. Hence \(r_-=-r_+\). ◻

Proof. After normalizing \(v\), Proposition 4 shows that both signs of escape in direction \(v\) have positive probability. Choose a coordinate with nonzero projection on the opposite rays from Lemma 8. Along one ray this coordinate tends to \(+\infty\), and along the other it tends to \(-\infty\). Each ray occurs with positive probability, giving the assertion. ◻

Proof. The raw probability of \(\mathcal D_H^\sigma\) is \(\xi u_{H-M}^\sigma\). Its terminal script begins at a record and ends at a candidate of height \(H\). By the script gap and Lemma 2, imposing a suffix that stays at signed height at least \(H\) multiplies this probability by \(p_\sigma\). Conversely, a cut at \(H\) must be the end of the script beginning at the first hit of \(H-M\): coordinate records are first hits, and each scripted step increases the signed height by one. Dividing by \(p_\sigma\) to condition on \(D_\sigma\) proves (27). No cut can have width less than \(M\).

The iid slab law from Proposition 3 now gives (28). Its numerator and the constraint on total width are unchanged when the list is reversed. Alternatively, for each fixed prefix through \(H\) realizing \(\mathcal D_H^\sigma\), the cut suffix supplies the same factor \(p_\sigma\). Conditional on the cut at \(H\) under \(\widehat P_\sigma\), the prefix therefore has mass \(W(\text{prefix})/\bar u_H^\sigma\). This is exactly its raw conditional mass given \(\mathcal D_H^\sigma\).

The parsing can be recovered from the finite prefix alone. Select the candidate ends whose heights are not undercut later in the prefix, and split at those ends. A suffix staying at or above \(H\) cannot change this selection, because \(H\) is a strict record height. Between successive selected candidates, every earlier candidate is undercut, which is precisely the defining condition for a slab. A script cannot straddle a selected boundary: it would overlap the script ending there. Thus these are exactly the slab boundaries, and the parsing is unique.

Finally, active departures of a slab ending at signed height \(b\) lie strictly below \(b-M\). All departures of the following slabs lie at or above \(b\). These row sets are more than \(R\) apart in infinity distance. Applying Lemma 2 successively, and using translation invariance, factors the weight of any prescribed concatenation. The same argument applies after grouping the slabs into bridges of prescribed widths. Each group boundary is reached on a first hit, so the grouping introduces no extra path multiplicity. ◻

Proof. Fix a sign and suppress it. Partition according to the last renewal at or before \(m\). The next width exceeds the distance from that renewal to \(m\), so \[\sum_{k=0}^m F(k)\bar u_{m-k}=1.\] Subtracting the identity at \(m+1\), and using \(\bar u_0=1\), gives \[\sum_{k=0}^m F(k) (\bar u_{m-k}-\bar u_{m+1-k})=F(m+1).\] By (27), every adjacent difference in this sum is nonnegative except the one with \(m-k=M-1\), which equals \(-\xi\). Take \(m=n+M\). The exceptional index is \(k=n+1\), while the term \(k=0\) equals \(\xi(u_n-u_{n+1})\). Dropping the other nonnegative terms yields \[\xi(u_n-u_{n+1}) \le F(n+M+1)+\xi F(n+1) \le(1+\xi)F(n).\] Summation from \(n\) to \(n'-1\), using that \(F\) decreases, proves the first bound. The second follows by grouping the ordinary tail sum \(\sum_{k\ge0}F(k)=E_{\nu_\sigma}L<\infty\) into dyadic intervals. ◻

Proof. Let \(K\) be the first positive common width, allowing \(K=\infty\) for now. For each fixed pair of finite prefixes that realize \(K=k\), the fact that \(k\) is their first common width is decided by those prefixes. Their unused tails are independent tapes with the initial laws. Enumerating these prefix pairs therefore gives the renewal recursion \[c_0=1,\qquad c_n=\sum_{k=1}^n\mathbf P(K=k)c_{n-k}\quad(n\ge1), \qquad c_n=\bar u_n^+\bar u_n^-.\] Set \(f(t)=\sum_{k\ge1}\mathbf P(K=k)t^k\) for \(0<t<1\). Nonnegative power-series summation gives \[\sum_{n\ge0}c_nt^n=\frac{1}{1-f(t)}.\] Equation (27) implies \(c_n\ge\xi^2p_+p_->0\) for every \(n\ge M\). Hence \(\sum c_nt^n\ge\xi^2p_+p_-t^M/(1-t)\), and consequently \((1-f(t))/(1-t)\) stays bounded as \(t\uparrow1\). First this forces \(f(t)\to1\), proving \(K<\infty\) almost surely. We can then write \[\frac{1-f(t)}{1-t} =\mathbf E\frac{1-t^K}{1-t}\ \uparrow\ \mathbf E K,\] which proves finite mean. The same prefix factorization at each common renewal proves that the successive pairs of lists are iid. The lower bound \(K\ge M\) follows from the minimum slab width. The strong law gives the assertion about \(W'_i\).

Let \(N_+\) be the number of positive words in the first common block. Since every width is at least one, \(N_+\le K\). The event \(\{N_+\ge i\}\) is determined by the first \(i-1\) positive widths and the entire negative tape: none of those positive width sums may be a negative width sum. It is therefore independent of the \(i\)th positive word and its radius \(\mathcal R_i^+\). Tonelli’s theorem gives \[\mathbf E\sum_{i=1}^{N_+}\mathcal R_i^+ =\sum_{i\ge1}\mathbf P(N_+\ge i)E_{\nu_+}\mathcal R =\mathbf E N_+\,E_{\nu_+}\mathcal R<\infty.\] The identical argument for the negative tape proves the claimed integrability, using Proposition 7. ◻

Proof. All entropies and expectations in this proof refer to the original tape-and-count experiment. First omit \(Z\), and let \(D_0\) denote the relative entropy between the resulting marginals of \(P^{(n)}\) and \(Q^{(n)}\). A pair of lists of common width \(h\) determines the number of its common renewals, counting the terminal width \(h\) and excluding zero, and hence determines its parsing into common blocks. If that number is in the count window for chunk \(i\), the actual unconditioned probability of the list pair is \[\frac1n \prod_{\gamma\in A_i}W(\gamma) \prod_{\gamma\in B_i}W(\gamma).\] If the number is outside the window, the probability is zero. This follows from the common-block construction and the slab weights: the specified pair of lists determines exactly one allowed count. In contrast, the product of the two bridge laws at width \(h\) has the same product of weights divided by \(\bar u_h^+\bar u_h^-\). Thus, conditional on \(H_i=h\), the density relative to the independent bridges is, on the actual support, exactly \[ \frac{\bar u_h^+\bar u_h^-} {nP^{(n)}(H_i=h)}. \tag{38}\] Independence of the three actual chunks now gives \[\begin{align*} I(\mathcal A;\mathcal B\mid H) &\le D_0\\ &=\sum_{i=1}^3 \left(\mathscr H(H_i)-\log n +\mathbb E\log(\bar u_{H_i}^+\bar u_{H_i}^-)\right) \le\sum_{i=1}^3(\mathscr H(H_i)-\log n) \le C. \tag{39}\end{align*}\] The information inequality follows from (34). For the last bound, use \(\mathbb E H_i=\mathbf E K_1\,\mathbb E I_i=O(n)\) and compare the law of \(H_i\) with weights proportional to \(e^{-h/n}\) on the nonnegative integers. The normalizer is \(O(n)\), so \(\mathscr H(H_i)\le\log n+C\).

Reintroducing \(Z\) costs its conditional information with the lists: \[ \begin{split} D(P^{(n)}\Vert Q^{(n)}) &=D_0+I(Z;\mathcal A,\mathcal B\mid H)\\ &=D_0+\mathscr H(Z\mid H)-\mathscr H(d_0). \end{split} \tag{40}\] Here the lists determine \(X_A\), and the buffer is independent of the lists and widths; conditional on those data, \(Z\) is a translate of \(d_0\). Likewise, given \((\mathcal A,H)\), the variable \(Z\) is conditionally independent of \(\mathcal B\). Conditional data processing and (39) imply \[I(Z;\mathcal B\mid H) \le I(\mathcal A;\mathcal B\mid H)\le D_0.\] Every variable whose entropy is evaluated in the next calculation has finite entropy; the conditioning lists need not: \[\begin{align*} \mathscr H(T,Z+X_B) &\ge\mathscr H(T,Z+X_B\mid\mathcal B,H) =\mathscr H(T,Z\mid\mathcal B,H)\\ &\ge\mathscr H(Z\mid H)-D_0 +\mathscr H(T\mid Z,\mathcal A,\mathcal B,H)\\ &=\mathscr H(Z\mid H)-D_0+\mathscr H(I_0\mid d_0). \tag{41}\end{align*}\] For the final identity, each pair of chunk lists determines its common block count, hence \(I_1,I_2,I_3\). The lists and \(Z\) also determine \(d_0=Z-X_A\). Independence from the buffer then identifies the remaining conditional entropy of \(T\) with \(\mathscr H(I_0\mid d_0)\).

Subtract \(\mathscr H(I_0,d_0)=\mathscr H(d_0)+\mathscr H(I_0\mid d_0)\) from (41) and use (40). The result is \[ D(P^{(n)}\Vert Q^{(n)}) \le 2D_0+\mathscr H(T,Z+X_B)-\mathscr H(I_0,d_0). \tag{42}\] Now \(Z+X_B\) is the sum of the lateral increments through the first \(T\) common blocks. Both \(T\) and \(I_0\) are independent of the underlying block sequence. It follows that \[\begin{align*} \mathscr H(T,Z+X_B)-\mathscr H(I_0,d_0) &=\mathscr H(T)+\mathbb E h_T'-\log n-\mathbb E h_{I_0}'\\ &\le C+h_{14n}'-h_n'. \end{align*}\] Indeed, \(T<14n\), \(I_0\ge n\), and \(T\) has at most \(4n\) possible values. Monotonicity of \(h_j'\) proves the displayed bound. Finally, the term \(2D_0\) is bounded uniformly in \(n\) by (39), proving (37). ◻

Proof. Fix \((H,Z)\) in the support of its retained marginal. Conditional on these data, all six reference lists are independent. Reversing the positive list order preserves its bridge law by (28). Retain the positive path formed by chunks \(3\) and \(2\), and call it \(\alpha\). It runs from the origin through its first hit of height \(N-H_1\). For the negative path formed by chunks \(1\) and \(2\), retain only the prefix through its first arrival within distance \(R\) of \(\mathop{\mathrm{Dep}}(\alpha)\); call the retained prefix \(\beta\) and its terminal site \(y'\). On \(E_{\rm mid}\) such an arrival exists no later than a negative departure in chunk \(2\), and hence before the two negative chunks terminate. This first proximity need not itself join the two middle chunks; the outer chunk widths will nevertheless put both visits at progress at least \(n\).

The geometry puts this encounter away from both starting sides. The two negative chunks stay at or below \(N-1\) and end at their first hit of \(H_3-1\). Therefore \[H_3\le y_1'\le N-1.\] Choose a departure \(y\) of \(\alpha\) within distance \(R\) of \(y'\). Since \(\alpha\) ends on its first hit of \(N-H_1\), \[ H_3-R\le y_1\le N-H_1-1, \qquad N-1-y_1'\ge H_1-R. \tag{45}\] The inequalities \(H_1,H_3\ge Mn\) and \(M>R+1\) show that both visits occur at or after the respective path has reached signed progress \(n\) from its start. Up to these visits neither path has hit either absolute height \(-1\) or \(N\).

We next express the probability of these prefixes using two raw walks in one environment. The unused positive chunk \(1\) and negative chunk \(3\) integrate to one. The four remaining bridge normalizations, \[\bar u_{H_3}^+\bar u_{H_2}^+ \bar u_{H_1}^-\bar u_{H_2}^-,\] have a fixed positive lower bound by (27). Within each sign, concatenating the two bridges gives the product of their raw weights, by the separation of active departures established in Section 4.

For fixed \(\alpha\) and a retained negative prefix \(\beta\), sum over all allowed negative completions to the terminal height \(H_3-1\). Their total raw weight is at most \(W(\beta)\): full words to that height end on its first hit, so the corresponding continuation events are disjoint. Nor is there multiplicity from the lists. The intermediate chunk boundary is a first hit of its prescribed height, and the bridge parsing into slabs is unique. The same uniqueness holds for the positive concatenation. Thus, after dropping any further completion restrictions, the conditional contact probability is at most a fixed constant times a sum of products \[ W(\alpha)W(\beta) =\mathbb E_Q\bigl[p_\omega(\alpha)p_\omega(\beta)\bigr]. \tag{46}\] The equality holds for each retained pair: every departure of \(\beta\) precedes its first proximity arrival and is therefore at distance strictly greater than \(R\) from every departure of \(\alpha\). The terminal arrival \(y'\) uses no departure row. Lemma 2 then applies, with repeated departures within either path retained in their respective weights.

The right side of (46) is the probability of the specified prefixes for two raw marked walks, started at \(0\) and \((N-1,Z)\), independently conditional on a shared environment of law \(Q\). Summing these probabilities introduces no overcounting. A full positive trajectory has at most one prefix through its first hit of the fixed height \(N-H_1\); that prefix determines \(\mathop{\mathrm{Dep}}(\alpha)\), and a full negative trajectory has at most one prefix through its first arrival within distance \(R\) of that set.

Only in this shared-environment experiment do we classify an encounter by quenched exit probabilities. For \(0\le x_1\le N-1\), define \[q_x^\omega=P_{x,\omega}(T_{-1}<T_N),\] where \(T_a\) here is the first hit of absolute first-coordinate height \(a\). Slab exit is almost surely finite, so \(1-q_x^\omega=P_{x,\omega}(T_N<T_{-1})\). At the sites in (45), if \(q_y^\omega\ge1/2\), the positive walk has visited, after reaching progress \(n\), a site with probability at least \(1/2\) of exiting through its starting side’s barrier \(-1\). If \(q_y^\omega<1/2\), join \(y'\) to \(y\) by a coordinate path of at most \(dR\) steps inside the slab. Uniform ellipticity gives \[1-q_{y'}^\omega \ge\kappa^{dR}(1-q_y^\omega) \ge\kappa^{dR}/2.\] In that case the negative walk has visited, after reaching its signed progress \(n\), a site with probability at least \(\kappa^{dR}/2\) of exiting through its own starting side’s barrier \(N\).

For either sign \(\sigma\), consider the raw marginal event that, after its first hit of signed progress \(n\) and before slab exit, the walk visits a site whose quenched probability of exit through the starting side’s barrier is at least \(c=\kappa^{dR}/2\). With the environment fixed, stop at the first such visit. The probability of the indicated exit after this visit is at least \(c\). Integrating the resulting inequality over environments bounds the visit probability by \[ c^{-1}(u_n^\sigma-u_N^\sigma). \tag{47}\] Indeed, from each of the two starting sites, the barriers have signed relative heights \(-1,N\). The difference \(u_n^\sigma-u_N^\sigma\) is exactly the raw probability of hitting signed height \(n\) and then exiting at \(-1\) before \(N\): a path reaching \(N\) must first reach \(n\), and exit from the finite slab is almost surely finite. Stationarity identifies the negative walk’s translated raw probabilities with \(u_n^-,u_N^-\).

Every retained pair belongs to at least one of these two raw visit events. Combining (46)–(47) therefore gives \[Q^{(n)}(E_{\rm mid}\mid H,Z) \le C\sum_{\sigma\in\{+,-\}}(u_n^\sigma-u_N^\sigma) \le CN(F_+(n)+F_-(n)),\] where the last inequality is Lemma 11. This estimate uses the raw marginal visit probabilities after the prefix transfer; it does not condition an exit probability on the contact event. Finally, the retained marginal has \[\mathbb E N=\mathbf E K_1\,\mathbb E(I_1+I_2+I_3)=O(n).\] Averaging over \((H,Z)\) and also using the trivial probability bound one proves (44). ◻

Proof. Apply data processing in Proposition 13 to the indicator of \(E_{\rm mid}\). If \(p=P^{(n)}(E_{\rm mid})\) and \(q_n=Q^{(n)}(E_{\rm mid})\), binary relative entropy satisfies \[D(\operatorname{Bern}(p)\Vert\operatorname{Bern}(q_n)) \ge p\log(1/q_n)-\log2.\] Together with Proposition 14, this yields, when \(\delta_n>0\), \[ P^{(n)}(E_{\rm mid})\log(1/\delta_n) \le C+h_{14n}'-h_n'. \tag{49}\] If \(\delta_n=0\), then \(q_n=0\), and the finite relative entropy in (37) forces \(P^{(n)}(E_{\rm mid})=0\). The same observation handles \(q_n=0\) when \(\delta_n>0\).

For \(l\ge2\), take \(n=2^{l-2}\). The width-tail summability in Lemma 11 implies \[\sum_{l\ge2}\delta_{2^{l-2}}<\infty.\] Among \(2\le l\le J\), at most \(O(\sqrt J)\) indices can have \(\delta_{2^{l-2}}>J^{-1/2}\). At every other index with positive \(\delta_{2^{l-2}}\), the logarithm in (49) is at least \(\tfrac12\log J\). The corresponding entropy differences have a telescoping bound: \[\begin{align*} \sum_{l=2}^J \bigl(h_{14\cdot2^{l-2}}'-h_{2^{l-2}}'\bigr) &\le\sum_{l=2}^J \bigl(h_{2^{l+2}}'-h_{2^{l-2}}'\bigr)\\ &\le4h_{2^{J+2}}'=O(J), \end{align*}\] by monotonicity and (35). Since \((4n,8n]=(2^l,2^{l+1}]\), use (43), sum (49) over the remaining indices, and bound each exceptional probability by one. Adding the initial scale gives \[m(J)\le 1+O(\sqrt J)+O(J/\log J)=O(J/\log J),\] as claimed. ◻