A directional zero–one law under strict ellipticity

Proof. The product in (2) depends only on rows in \operatorname{Dep}(\gamma), which are independent of all the rows defining f. This proves the identity. For two words, conditional independence in the shared environment gives the product of their quenched weights, and (2) factors its environment expectation.

A trajectory stopped on arrival in a set does not use the row at that arrival site. Its finite stopped-prefix probabilities therefore depend only on rows outside the set. The same is true of every measurable event of the stopped trajectory, by the monotone class theorem. This includes events on which the stopping time is infinite.□

Proof. Fix nonnegative integers b,k, a signed coordinate step a\in U with h(a)=1, and an integer m>b+k. It suffices to rule out

Partition the lattice into columns, the cosets of \ZZ a. Every column has only finitely many sites with height in [-k,b]: its successive heights differ by exactly one, regardless of the other coordinates of v.

First suppose that only finitely many columns are visited at heights at least -k. On E_{b,k}, some fixed site x of height in [-k,b] is then visited infinitely often. Fix an environment whose transition entries are all positive. At every visit to this particular site the quenched probability of the script consisting of m steps a is

Choose the first visit as a trial start, and subsequently choose the first visit at least m steps after the preceding start. These are stopping times. By the quenched strong Markov property, the probability that r such trials are finite and all fail is at most (1-q_x(\omega))^r. Infinitely many visits provide infinitely many trials, while a successful script would cross above b. Thus the specified event has quenched probability zero. A countable union over x and integration over the environment handle this first case. There is no need to bound q_x(\omega) uniformly in x or \omega.

For the other case, consider the first visit, for each column, to its portion of height at least -k. These eligibility decisions are determined by the observed path prefix. Select the first eligible time as a trial start, and thereafter the first eligible time at least m steps after the preceding trial start. If infinitely many columns have such visits, infinitely many trial starts remain after these finite exclusions. At an eligible time, every earlier visit in the same column had height below -k. Consequently the m scripted departure sites, all at height at least -k, are distinct and have not appeared earlier. Conditional on any fixed eligible prefix, their rows retain their original independent laws. The annealed conditional probability of the script is therefore exactly

The probability that r selected trials are finite and all fail is at most (1-q_a^m)^r. On E_{b,k} every one fails, so the second case is also null. This argument uses no conditioning on the future event E_{b,k} when evaluating a trial probability.

Taking the countable union over b,k proves the implication: a path from zero with finite height supremum that does not tend to minus infinity belongs to some E_{b,k}. Translation gives the same annealed conclusion from every lattice site. Integrating the quenched probabilities of these null events, and taking a countable intersection over starting sites and integer bounds, gives the last assertion.□

Proof. On A_v, the sequence h(X_n) tends to +\infty and therefore attains a global minimum. Thus A_v is contained in the countable union, over n\ge0 and x\in\ZZ^d, of the events

At least one such event has positive probability. Put q_v(x,\omega)=P_{x,\omega}(h(X_k)\ge h(x)\text{ for all }k\ge0). The quenched Markov property at the deterministic time n gives

Consequently \EE[q_v(x,\omega)]>0, and translation invariance identifies this expectation with p_v. This argument does not use a Markov property at the time of the global minimum.□

Proof. We first establish existence using stopping-time trials. Test the first ordinary strict record. If the path later falls below its height, wait until the first strict record above the entire path seen at that failure time, and test again. The candidates and the times at which failures are detected are stopping times. By (3), the probability that the first k tested candidates are finite and fail is at most (1-p_v)^k. On \{\sup_n h(X_n)=\infty\}, a new candidate is available after every detected failure. Hence a true cut exists there almost surely.

Under \widehat P_v, the height supremum is infinite almost surely by Lemma 2.2: its alternative conclusion of escape to minus infinity is incompatible with D_v. In particular the first true cut is finite under \widehat P_v. This application of Lemma 2.2 is valid before any independence or integrability of slabs has been proved.

We now compute its law. Fix a regeneration word \gamma=(0,\ldots,z) of length m and a measurable event B for the translated infinite suffix. The first true-cut word is \gamma exactly when this prefix occurs and the suffix from z belongs to D_v. For necessity, an earlier record not invalidated by time m could never be invalidated by a future remaining above h(z), so it would be an earlier true cut. For sufficiency, the word’s intermediate records have all failed, while its terminal record is true.

The finite prefix uses only rows below h(z), whereas its suffix constrained by D_v uses only rows at or above h(z). Thus

The event on the left implies the original D_v. Division by p_v therefore gives

Here \Gamma_1 denotes the first relative word. Since \tau_1 is almost surely finite, summing over all such words proves that their weights sum to one. Equation (4) gives an independent suffix with the original conditioned law, and iteration proves the iid assertion. The departure-height claim is part of the word definition.

For the unconditioned first true cut, the same word enumeration allows a prefix that goes below zero. Its endpoint is still a strict global record, and each earlier strict record must have failed before that endpoint. The fresh-suffix factor is again P_0(B\cap D_v), so its conditional translated law is \widehat P_v. No Markov property at a future-dependent cut has been used.□

Proof. For i\ge1, let

Freshness at \rho_i and cancellation of the conditioning probability give

The last inequality follows because D_v forces an infinite supremum and hence reaches every ordinary record index before dropping below zero.

At a cut, its height is the maximum of the entire past. The subsequent ordinary records are therefore exactly the records of the translated suffix. By Lemma 2.4, the successive true-record indices are sums of iid positive integers S_1,S_2,\ldots with the law of S. Let

Summing (5) yields \widehat E_v[N(n)/n]\ge p_v. If \mathbb E_{\nu_v}S=\infty, the strong law applied to every bounded truncation S_j\wedge a gives (S_1+\cdots+S_k)/k\to\infty. Inverting these sums gives N(n)/n\to0 almost surely. Since 0\le N(n)/n\le1, dominated convergence contradicts the preceding lower bound. Thus the mean of S is finite. The ordinary strong law and the same inversion now give N(n)/n\to1/\mathbb E_{\nu_v}S; dominated convergence proves \mathbb E_{\nu_v}S\le1/p_v.

Each ordinary record height gain is at most one: its final step has height increment at most \|v\|_\infty=1, and starts no higher than the preceding maximum. Adding the gains within one slab gives L\le S. Finally, L>0 almost surely, so its mean is strictly positive.□

Proof. Normalize v as above. Lemma 2.3 gives p_v>0. On infinite height supremum, Lemma 2.4 provides a first true cut followed by iid slabs. Their positive widths have finite, strictly positive mean by Lemma 2.5, so the successive base heights tend to infinity. Every later slab stays at or above its base, and consequently h(X_n)\to+\infty.

The finite-supremum implication is exactly Lemma 2.2. These two alternatives exhaust all paths and prove the sign union.□

Proof. Call an integer s a bad start if an interval beginning at s, of length at most n, has sum greater than a times its length. Among starts 1,\ldots,M, take the leftmost bad start, choose a witness interval, skip all its indices, and repeat. The chosen intervals are disjoint, cover every bad start, and lie in \{1,\ldots,M+n\}. Their total length is at least the number of bad starts. Therefore

Taking expectations, using stationarity, and letting M\to\infty proves the assertion.□

Proof of Proposition 3.1. The construction has two parts. A displacement allowance a and a number N of slabs will be chosen so that most initial portions have total radius at most a times their length, but some slab has a much larger excursion with probability bounded below. That excursion supplies a record for a small tilt of the opposite height direction. The continuation then avoids the old prefix in two successive regions: it stays beyond the tilted record for a fixed multiple of N slabs, and after their final true cut it stays at negative height. The estimates below arrange these two separations with a continuation probability independent of a and N.

Rescale v so that \|v\|_\infty=1. For either sign, under \widehat P_{\sigma v} let (L_i,R_i)_{i\ge1} denote the widths and radii of its iid slabs, and put H_k=\sum_{i=1}^kL_i, with H_0=0. Write E_\sigma for expectation in this slab sequence. Lemma 2.5 gives 0<E_\sigma L<\infty. A point in slab k+1 has signed height at least H_k, and the displacement of any point through slab k has norm at most \sum_{j=1}^kR_j.

Scales with controlled initial portions. Suppose, to obtain a contradiction, that

The function I is increasing and concave, I(0)=0, and I(t)=o(t). The last assertion follows by bounded convergence from R<\infty almost surely.

By the two width strong laws we may fix c>0 and finite C,C_0\ge0 such that, in each sign,

Indeed choose c below both means and C above both means, then choose C_0 to bound the almost surely finite maximal deviations with the displayed probability. Fix, in this order, an integer C_1, numbers b,B, and \eta>0, so that

The multiplier C_1 gives the opposite continuation enough slabs to descend below height zero. The coefficient b makes its height descent dominate its lateral displacement in the tilted functional; B ensures that a large excursion creates a new tilted record. Finally, \eta controls the probability that an initial portion exceeds its displacement allowance. These constants remain fixed throughout the proof.

For large a, let N=N(a) be the smallest positive integer such that I(aN)\ge\eta a. Such an integer exists. Since I(am)/a\to0 for every fixed m, we have N(a)\to\infty. For sufficiently large a, minimality and concavity give

In either sign and for m\ge0, define \mathcal R_m to be the event that, for every k\le m,

We claim that

To see this, put n=C_1N, A=an, and truncate each radius to \widehat R_j=\min(R_j,A). Concavity and (8) give

Lemma 3.2 bounds by 2C_1\eta the probability that any of the first n averages of the truncated radii exceeds a. Also

Intersecting these estimates with (6) proves (9). In particular, \widehat P_{\sigma v}(\mathcal R_{i-1})>.8 for i\le N, and \mathcal R_{i-1} depends only on the first i-1 slabs.

A large slab after a controlled initial portion. Choose a lower index i_0 so large that

This is possible by integrability of the widths. Increase i_0 if necessary so that, for every i\ge i_0,

After fixing i_0, take a sufficiently large that N\ge i_0 and I(Bai_0)<\eta a/4.

For i_0\le i\le N, call

the large-slab event, and put \lambda_\sigma=\sum_{i=i_0}^N\widehat P_{\sigma v}(G_i). Integrating the decreasing radius tails gives

For the last bound, use I(BaN)\le B I(aN). For the lower bound, use I(Ba(N+1))\ge I(aN)\ge\eta a, then subtract (10). The result is

Thus one sign, which may depend on the scale, satisfies \lambda_\sigma\ge\eta/(4B).

In this sign consider

Independence of the i-th slab from its preceding slabs gives E_\sigma Z\ge.8\lambda_\sigma. The summands of Z need not be independent, but Z\le\sum_i\1_{G_i}, whose summands are independent. Hence

The constant \delta is independent of i_0 and a.

We have therefore found, with a probability independent of the scale, a slab whose spatial excursion is much larger than the accumulated displacement before it, while its height width is at most its index. The next step converts that excursion into a stopping time from which an opposite-going path can use fresh environment rows.

A tilted record. Write h(x)=\sigma v\cdot x for the sign just selected. On \mathcal R_{i-1}\cap G_i, the initial site of slab i has norm at most a(i-1). Some point of the slab consequently has norm greater than (B-1)ai. At that point some signed coordinate u\in\{\pm e_1,\ldots,\pm e_d\} satisfies

Throughout this slab its running h-maximum is at least H_{i-1}\ge ci/2, and at most H_i\le(C+2)i, by (11).

Define the linear functional

Before slab i, nonnegative height and \mathcal R_{i-1} give Y(X)\le a(i-1). At the selected point,

There is thus a strict Y-record while the running h-maximum lies in the deterministic window

For each fixed choice of \sigma,u,i_0,a, let T be the first positive integer time such that the following three conditions hold: the entire prefix has nonnegative h-height; Y(X_T) is a strict record; and the running h-maximum belongs to W. These are conditions on the observed prefix, so T is a stopping time for (\cF_n). It need not be a true regeneration time. The preceding construction and a union over the 2d signed coordinates show that some u satisfies

Set p_*:=\min(p_v,p_{-v})>0. Removing the conditioning yields

The site X_T has not appeared in its earlier prefix, since it is a strict Y-record.

Preserving the maximum with a fresh continuation. Since \|v\|_\infty=1, there is a nearest-neighbor step e with h(e)=-1. From X_T force M repetitions of this step, where the fixed integer M will be chosen shortly. Each step increases Y by at least ba-1. For all sufficiently large a, its successive sites have strictly increasing Y-coordinates, starting at the strict record X_T. Consequently every scripted departure site is distinct and unused by the prefix. Conditional on any fixed prefix realizing T, the script has the exact raw probability

Here independence is across departure rows, not across the entries in a single row. The constant \alpha is positive because U is finite and every transition entry is positive almost surely. It is not a lower bound on the transition entries in individual environments.

At the script endpoint consider a translated opposite-going template with law \widehat P_{-\sigma v}, and require \mathcal R_{C_1N} for its first C_1N true-cut slabs. This requirement includes existence of those cuts and concerns the entire template; they are not declared to be stopping times. It has conditional-template probability greater than .8. At any point of its slab k+1, where 0\le k<C_1N, the coordinate displacement in direction u is at least -a(k+1), while the displacement in the original height h is at most -ck+C_0. Its total Y-increment from the launching record is consequently at least

Choose M once and for all so that Mb>1+bC_0+1. Since bc>1, the expression in (14) is strictly positive for all k in question and all sufficiently large a. Thus these template sites avoid the entire original prefix, whose Y-values are at most the launching record.

There are two additional separations to check. First, the whole opposite template stays at or below its own initial h-height, whereas every script departure lies strictly above that height. It therefore avoids all script departures, including in its infinite tail. Second, after C_1N template slabs its absolute h-height is at most

for large N. This endpoint is a true opposite cut, so its remaining tail stays below zero. That tail avoids the original prefix, all of whose heights are nonnegative. No assumption that the script itself stays above zero is needed.

These inequalities make the use of fresh rows exact. Condition on a fixed finite prefix realizing T, and then on the forced script. The conditional environment differs from the original iid law only at departure sites of these two finite words. Under an independent raw law, the probability of D_{-\sigma v}\cap\mathcal R_{C_1N} is at least .8p_{-\sigma v}. Every path in that event avoids all the old departure sites after translation to the script endpoint, by the preceding separations. Lemma 2.1 therefore identifies its probability with the corresponding fresh-row probability after the prefix and script. More explicitly, kill both the continuation and an independent raw chain on arrival at the same set of old departure sites. The finite-word identity gives equality of their cylinder probabilities, hence of their killed path laws. Now restrict to D_{-\sigma v}\cap\mathcal R_{C_1N}, whose paths never hit that set. Repeated sites within the template are retained in its raw path weights.

It follows that, from every qualifying prefix, the script followed by the required continuation has conditional probability at least .8p_*\alpha^M. This argument uses finite-prefix conditioning and departure-row factorization, not a Markov property at a future-dependent true cut. The script decreases h, and the template never exceeds its own starting height. Hence the final supremum of h on this continuation is its running supremum at T, which belongs to W.

The finite final maximum. Together with (13), this proves that for every sufficiently large i_0, after choosing a large enough, some sign \sigma satisfies

The constant \rho does not depend on the scale. For each sign let M_\sigma=\sup_{n\ge0}\sigma v\cdot X_n, allowing the value +\infty. Equation (15) implies

for every sufficiently large i_0. Both terms on the left tend to zero as i_0\to\infty, a contradiction. Thus I(t) is bounded, and monotone convergence proves Proposition 3.1.□

Proof. By Lemma 2.6, both signs of v have positive probability and P_0(A_v\cup A_{-v})=1. Normalize \|v\|_\infty=1. For \sigma\in\{+,-\}, let D_j^\sigma be the displacement of the jth slab of the \sigma v sequence, and let R_j^\sigma be its radius. Proposition 3.1 and the bound |D_j^\sigma|\le R_j^\sigma give the mean vector

In particular b_\sigma\ne0. The strong law gives convergence of the kth slab endpoint divided by k to b_\sigma. Moreover, for each \varepsilon>0,

so R_k^\sigma/k\to0 almost surely. Between the kth and (k+1)st endpoints, the displacement from the former is at most R_{k+1}^\sigma. The number of completed slabs tends to infinity. Thus the entire trajectory, after its first true cut, has limiting unit direction

Lemma 2.4 transfers this conclusion to the raw law on A_{\sigma v}. This argument uses slab count rather than elapsed time and requires no moment of a slab’s duration.

The two rays are distinct, since their projections on v have opposite signs. The following argument proves the special case of Simenhaus’s antipodality principle needed here [13, Proposition 1]. We claim that r_-=-r_+. Otherwise w=r_++r_- satisfies

The preceding endpoint and radius estimates then imply A_w on both original directional events, so P_0(A_w)=1. Under \widehat P_w, the limiting ray therefore exists almost surely. By Lemma 2.4, this law is represented by an iid sequence of finite relative slab words. Deleting finitely many words removes a finite time prefix and translates all remaining positions by a fixed vector. Since A_w holds, positions tend to infinity in norm, and these operations preserve the limiting ray. The ray is consequently a tail variable of the iid word sequence. The tail zero–one law, applied to a countable determining family of Borel subsets of the unit sphere, makes it deterministic. No displacement moment in direction w is needed here.

The raw walk has a first true w cut almost surely, and its conditional future there has law \widehat P_w by Lemma 2.4. Its raw limiting ray must therefore be that same deterministic ray. This contradicts the positive probabilities of the two distinct rays r_+ and r_-. The claim follows. Choose a coordinate with nonzero projection on r_+; on A_v and A_{-v} the corresponding coordinate tends to opposite infinities, proving Proposition 4.1.□

Proof. On \{T_H<T_{-1}\}, all departures before T_H have height less than H. The future non-backtracking event from the fresh endpoint has probability p_\sigma. Its factor cancels the denominator in conditioning on D_\sigma, giving the cut probability u_H. The hitting events are nested in H, and a non-backtracking walk reaches every height, proving the inequalities.

A finite path first hitting H before -1 has a unique slab decomposition. Its cuts are the strict record times whose heights are never undercut before the terminal time, together with the initial boundary. Between two successive cuts all intermediate strict records are invalidated before the second cut. These are exactly the qualifying words of Lemma 2.4. Conversely, concatenating any such list of total width H gives one of these bridges. Distinct words have disjoint departure-height intervals, so their raw weights multiply. Their total mass is u_H, which proves (17) and the bridge identification. Reversing the list preserves its total width and the product of its word weights. It is a bijection on admissible lists, proving the last assertion. At H=0 the list is empty.□

Proof. The slab cut heights form a renewal process with increments L. Partitioning according to its last cut at or before n gives

Subtracting this identity at n+1 from the identity at n yields

Every term on the left is nonnegative, and F(0)=1. Thus u_n-u_{n+1}\le F(n+1)\le F(n); telescoping proves (18). Finally \sum_{k\ge0}F(k)=EL<\infty, and monotonicity of F gives (19) by grouping this sum into dyadic intervals.□

Proof. Initially allow the first positive common width K to be infinite. If it is reached after i positive and j negative words, that event is determined by those two finite prefixes: their widths agree, and no earlier positive partial sums agree. The unused tails are therefore independent tapes with their original laws, independent of the completed block. To see this without any stopping-time convention, condition separately on each pair (i,j), factor the law of the unused tails, and sum over (i,j). Repeating the argument gives the renewal property at every common cut that exists.

Let c_n=u_n^+u_n^- be the probability that n is a common cut, including c_0=1. Partitioning according to the last common cut at or before n gives the elementary identity

Indeed, after a common cut at n-i, the unused pair of tapes has its original law, and K>i says precisely that there is no further common cut through n. The identity remains valid if the next gap is infinite. Since c_j\ge p_+p_->0, it follows that

Letting n tend to infinity proves (21). In particular K is finite almost surely. Restarting at each successive common cut now justifies the infinite iid common-block sequence.

Let N_+ be the number of positive words in the first common block. The event \{N_+\ge i\} says that none of H_1^+,\ldots,H_{i-1}^+ belongs to the negative renewal set. It depends only on the preceding positive words and the negative tape, and is independent of the whole next word A_i. Tonelli’s Theorem therefore gives

The last inequality uses the integer bound L\ge1. The same argument applies to the negative tape. Since |a(\gamma)|\le R(\gamma), this proves (22) and the moment bound for S.

Finally, across the block the positive lower anchor is its old anchor minus V_+ in lateral coordinates, whereas the negative anchor moves by V_-. Their negative-minus-positive difference increases by V_++V_-, as asserted.□

Proof. There are two costs to control: imposing the random chunk heights, and then retaining the aligned starting displacement Z. We first condition on H alone. For a specified pair of slab lists of common total width h, let r be the number of their common positive cumulative widths, including h. The pair is uniquely parsed into its first r common blocks. If r belongs to the count window for chunk i, its actual unconditioned mass is

It has zero mass otherwise. The independent exact-cut bridge pair at height h has mass given by the same product divided by u_h^+u_h^-. Thus the conditional density of the actual pair, on its support, is

Common cuts here are counted in the original downward list order, before any list is reversed for chronological placement.

Let D_0 be the averaged conditional relative entropy of (\mathcal A,\mathcal B) given H with respect to these independent bridges. The chunks are independent, so (27) and u_h^\pm\le1 give

For the last bound, \mathbf E H_i\le8n\mathbf E K, and a positive integer variable with mean a has entropy at most 1+\log(a+1), by comparison with a geometric distribution. Since the reference conditional on H factors between the tapes,

Introducing Z while retaining its actual marginal costs precisely its conditional mutual information with the lists. The chain rule and the independence of the buffer give

Conditional on \mathcal A,H, the variable Z is obtained by adding the independent d_0. Data processing in (28) therefore yields I(Z;\mathcal B\mid H)\le D_0.

The direct bound \mathcal H(Z\mid H)=O(\log n) would be too large after summing over scales. Instead, add the negative tape’s displacement to Z. Their sum is the total common-block gap, whose entropy can be compared with that of the buffer. Conditioning on the negative tape costs at most D_0, as just proved:

For the last equality, the two lists in each chunk determine its common count. Knowing these lists and Z therefore fixes both T-U and d_0=Z-X_A. The buffer pair is independent of the chunks.

The vector Z+X_B is the sum of the first T common gap increments. The count T, like U, is independent of the underlying iid tape; we are not conditioning on the observed chunks in the following identities. Consequently

Here n\le U<2n and 10n\le T\le14n-4, so T takes at most 4n values. Monotonicity of h_k gives

Subtract \mathcal H(d_0) in (30) and use \mathcal H(U,d_0)=\mathcal H(d_0)+\mathcal H(U\mid d_0). Together with (29) and (31), this gives explicitly

The bound for D_0 is uniform in n, so its doubled contribution is absorbed into C. This proves Lemma 5.2.□

Proof. Fix (H,Z) in its retained support; in particular H_1,H_3\ge n. Write N=N_*. By list-reversal invariance in Lemma 4.2, the placed chronological chunks have their corresponding raw bridge laws.

Retain the complete positive lower and middle chunks, concatenated into a path \alpha from height 0 to its first hit of N-H_1. Retain the negative upper and middle chunks only up to their first arrival at a departure site of \alpha. A middle contact ensures that this arrival occurs before the negative middle endpoint at height H_3-1. The contact site y must therefore satisfy

Thus each path has made at least n progress in its own signed height before reaching the contact, and neither has crossed the global barriers -1,N.

We explain the comparison with two paths in one environment before using any quenched exit probabilities. The unused positive upper chunk and negative lower chunk integrate to one. The remaining four bridge normalizations are bounded by p_+^{-2}p_-^{-2}. Consecutive bridges within a sign have disjoint departure layers, so their raw weights multiply to the raw weight of their concatenation. Summing over the negative completion after its retained prefix, and dropping its remaining bridge requirements, costs at most that prefix’s raw weight: in each fixed environment the total conditional completion probability is at most one, and this inequality can then be averaged. This argument allows the completion to revisit its prefix’s departure rows and does not factor their weights independently. The first-hit chunk boundaries determine the splits, so no extra multiplicity is introduced.

The retained negative prefix has no departure site in \operatorname{Dep}(\alpha): its terminal point is the first arrival there. Repeated sites within either path cause no difficulty. Lemma 2.1 therefore identifies the product of their raw weights with the weight of the two paths sampled independently conditional on a common iid environment. For each \alpha the first contact fixes the negative prefix uniquely, and \alpha itself is stopped at a fixed first-hit height. We may thus sum these joint prefix weights without overcounting, and extend the paths as raw trajectories in the common environment.

Only now, in that shared environment, define for interior sites

where the barriers are absolute first-coordinate heights. If Q_y^\omega\ge1/2 at the contact, the positive raw path has visited such a site after hitting n and before exiting the global slab. Conditional on the environment, stop it at its first such visit. The probability of a subsequent exit at -1 is at least 1/2. Consequently the annealed probability of this visit event is at most

If Q_y^\omega<1/2, the negative marginal gives the analogous bound 2(u_n^--u_N^-). Indeed the almost-sure finite-height-interval exit property established in Section 4 holds quenched for all these starting sites and barriers outside one null set. The probability of the other exit is therefore greater than 1/2, with no common lower bound on the transition rows. Measured from the negative start N-1, the global barriers N and -1 have signed relative heights -1 and N, respectively. Lateral translation by Z does not change its annealed law.

Combining these alternatives with the bounded normalizations gives, uniformly in (H,Z),

where the last step is Lemma 4.3. Finally \mathbf E N\le12n\mathbf E K. Average over the retained joint marginal of (H,Z) and take the minimum with 1.□

Proof of Proposition 5.1. For l\ge2, take n=2^{l-2} and abbreviate \delta_l=\delta_n. The event E_{\mathrm{mid}} contains the original contact event in common blocks (2^l,2^{l+1}]. If \delta_l=0, absolute continuity from Lemma 5.2 makes its actual probability zero. Otherwise the binary relative-entropy inequality gives

For completeness, applying data processing to an event with actual and reference probabilities a,b gives binary divergence a\log(a/b)+(1-a)\log((1-a)/(1-b)). It is at least a\log(1/b)-\log2. Use b\le\delta_l and absorb \log2 into C.

Width integrability gives

Among 2\le l\le J, at most C\sqrt J indices have \delta_l>J^{-1/2}. They contribute at most C\sqrt J to m(J). For every other index with positive \delta_l, the logarithm in (33) is at least \tfrac12\log J. Since 14n\le16n, monotonicity and bounded overlap give

using (26). Summing (33), and adding the initial scale, proves

□

Proof. Write B=\log(m+1). We first show that, at every almost surely finite stopping time \rho,

For a fixed coordinate put a=A_\rho(e) and p_e=w_\rho(e), so that 0\le p_e\le a\le1. Conditional maximal inequality and integration over levels give, when a>0,

If a=0, the martingale has conditional expectation zero at every later time and contributes zero. The conditional maximal inequality follows first for bounded stopping times and finite horizons; localization and monotone convergence give the displayed version. Summing over e and adding the missing mass 1-\sum_e p_e bounds the resulting sum by the entropy of a probability vector on m+1 elements, hence by B. This proves (36).

At each time the increase of a coordinate maximum is at most its current value. Consequently

Set b=2B+1 and consider the successive thresholds 1+kb, k\ge1. At their first strict crossings, the overshoot is at most one. After a crossing, reaching the next threshold requires a further increase greater than b-1=2B. Conditional Markov inequality and (36) therefore bound the conditional probability of this next crossing by 1/2. The first crossing has the same bound since A_0\le1. Applying this argument at finite horizons and then letting the horizon increase yields

Because B\ge\log2, this is an exponential tail on scale B, and its integral proves (34).

Finally, for p>0, conditional Jensen inequality gives

Multiplication by p and an adjustment of the absolute constant give (35). This last argument uses only a marginal exponential moment, so it applies to events in a larger joint space without any assertion that the martingales remain martingales in an enlarged filtration.□

Proof. Let \mathcal E_N=\{e\in\ZZ^{d-1}:\|e\|\le N^2\}. Its cardinality satisfies

We first express the event in (37) by raw path weights, then transfer those weights to a shared environment, and only afterward introduce quenched exit probabilities.

The lower prefix and the endpoint likelihood. Enumerate a word a_0 from 0 to its first arrival at height k, before height -1. Its departure sites all have height less than k; write z for its terminal site. For each e\in\mathcal E_N, enumerate a word a from z to the first contact arrival. The required words stay in heights k,\ldots,N-1 and reach height r before or at their terminal time. Let \operatorname{Dep}(a) denote the departure sites of a, excluding its terminal site unless that site occurred earlier as a departure.

For any such positive prefix define

This is a deterministic function of the two words and e. The two departure sets of a_0 and a are disjoint, so the positive prefix and completion have raw probability

For a raw negative walk started at (N-1,e), let \mathcal B(a) be the event that it never rises above N-1, avoids \operatorname{Dep}(a) at every time, and visits the terminal site of a. The first-contact condition is exactly this avoidance and visit condition. In particular, if the terminal site of a occurred earlier as a departure, the event \mathcal B(a) is empty. Summing (39) times the raw probability of \mathcal B(a), over these words and endpoints, and dividing by u_N^+p_- gives the probability on the left of (37).

Transfer and unique prefix counting. For a fixed word a, the quenched probability of \mathcal B(a) uses no rows at \operatorname{Dep}(a). Indeed the walk can be killed on first arrival at this set, before any row there is used. This remains valid for the infinite avoidance event by passage from finite paths. The departure-row identity of Lemma 2.1 therefore gives

where p_\omega(a) is the quenched product of transition probabilities along the word. This identity has not been restricted by any environment-dependent exit-probability condition.

Keep W(a_0) as a scalar outside the new experiment. In a fresh i.i.d. environment sample a raw walk Z from z and a raw negative walk from (N-1,e), independently conditional on the environment. We call Z the comparison path. Formula (40) expresses the sum over a as an integral over these two full paths. For each realized pair, at most one prefix a of Z satisfies the avoidance and visit conditions: its terminal is the first point of Z in the range of the entire negative path. This proves uniqueness pathwise, without using an optional-stopping assertion at contact. The restrictions that a stays above k-1 and reaches r are retained at this stage.

Stopped posteriors on the comparison path. Fix a_0 and consider the original raw positive law conditional on this lower prefix. In its natural continuation filtration, with no environment revealed, the coordinates

are nonnegative martingales whose sum is at most one. Stop them on first arrival at height k-1 or N. All departures up to this stop have height at least k. These rows are fresh under conditioning on a_0, so the stopped continuation has precisely the same path law as Z stopped at

Arrival at the lower boundary uses a row at height k, not a departure row at k-1. Thus the equality includes the terminal arrival, as illustrated in Figure 2. The stopping time is almost surely finite by Lemma 2.2: a path that never leaves this finite height interval would have finite height supremum without tending to negative height infinity. The corresponding quenched assertion holds for almost every environment, simultaneously for all lattice starting sites and integer height intervals.

Use these same posterior functions on the observed prefixes of Z, and freeze them at \sigma. At arrival at k-1 their values need not be zero: they still describe success before the original lower barrier -1, and that completion may use rows affected by a_0. We neither recompute them in the fresh environment nor continue them through a departure below k. Put

Lemma 6.1 and (38), applied in the path filtration of the stopped comparison walk, imply uniformly in a_0 that

for every event V on the joint space of the fresh environment and Z. In particular, \EE S_*\le C\log(2N). The event-weight inequality, rather than a martingale assertion after revealing the environment, permits the next step.

Quenched exit-probability bins. For k\le y_1<N define

Almost surely every transition at every site is strictly positive. For each interior site y, a finite sequence of downward axial steps reaches k-1 before N with positive quenched probability; a finite sequence of upward axial steps reaches N before k-1 with positive quenched probability. Thus 0<Q_y^\omega<1, without a common bound away from either endpoint. Partition these values into dyadic bins [a,2a), where a=2^{-j} and j\ge1, and write \mathcal D_a^\omega=\{y:k\le y_1<N,\ a\le Q_y^\omega<2a\} for the sites in one bin. Let V_a be the event that, at or after its first arrival at height r and before \sigma, the comparison path visits an interior site with Q_y^\omega\ge a. With the environment fixed, stop at the first such visit. The quenched Markov property gives

The equality follows from nested first-hit events and almost sure exit from the height interval, again by Lemma 2.2. These events use the fresh comparison law, not the law conditioned on the lower prefix.

For almost every environment, the raw negative walk has probability at most 2a of both satisfying D^- and hitting \mathcal D_a^\omega. To see this, take its first visit to that set. On D^- the negative walk has finite first-coordinate supremum, so Lemma 2.2 implies that its first coordinate tends to -\infty. The quenched form of that lemma holds outside a single environment-null set for all lattice starting sites. Consequently, after that first bin visit it must hit k-1 before N, even if it had previously visited k-1. Its conditional probability is at most Q_y^\omega<2a by the quenched Markov property.

We now use the unique prefix established above. In the transferred integral, classify its contact site by its Q^\omega bin and bound its endpoint factor w(a_0a;e) by M(e). Dropping the exact-contact condition leaves V_a for the comparison path and D^-\cap\{\hbox{the negative path hits }\mathcal D_a^\omega\} for the negative path. There is no sum over competing prefixes. After integrating the negative path, every endpoint e leaves the same fresh joint marginal \PP(d\omega)P_{z,\omega}(dZ). Thus the bin’s total contribution, summed over endpoints, is at most

If \Delta=0, every V_a is null by (42), proving the required zero bound. Suppose \Delta>0. For a<\Delta, sum (43) using \EE S_*\le C\log(2N) and \sum_{a<\Delta}a\le2\Delta. For a\ge\Delta, use (41) and P(V_a)\le\Delta/a to bound the contribution by

There are O(\log(e/\Delta)) such dyadic bins, and their logarithms sum to O(\log^2(e/\Delta)). Thus, for each fixed a_0, all endpoint and upper-prefix contributions together are at most

Finally \sum_{a_0}W(a_0)=u_k^+\le1, and 1/(u_N^+p_-)\le1/(p_+p_-). Restoring these factors proves (37). The constants lose only these fixed non-backtracking probabilities and the endpoint-cardinality factor. In particular, the argument has not required a lower bound for Q_y^\omega or an inverse-transition moment.□

Proof. All constants below may depend on the two slab laws. Choose a large integer J and put

For sufficiently large J, we have 3\le G<J. Depth means distance below the initial top of the arrangement. The choice 2^{-G}=O((\log J)^{-4}) compensates for the logarithmic losses below, while G=O(\log\log J) leaves enough contact scales.

A bridge ending at a prescribed depth. Let \mathcal C_N be the event that the positive tape has a cut at depth N, and write

The renewal identity gives \mathbf P(\mathcal C_N)=u_N^+\ge p_+>0. Under \mathbf Q_N, reverse the list of positive slabs through this cut and translate its bottom to the origin. Call the resulting chronological path Y, and let T_a denote its first hit of height a. Lemma 4.2 says that Y has law P_0(\,\cdot\mid T_N<T_{-1}), stopped at T_N. The steps within each slab retain their original order. Denote the lateral endpoint by E, so that the endpoint is (N,E). The negative trajectory starts at (N-1,E) and, relative to that starting point, has the independent law \widehat P_{-e_1}.

The endpoint has a useful elementary tail bound. On \mathcal C_N there are at most N positive slabs before depth N, since their widths are positive integers. The norm of their total lateral displacement is at most the sum of their radii, hence at most the sum of the first N positive tape radii. Proposition 3.1 and Markov’s inequality give

This uses a spatial radius moment only.

An uncovered scale forces a late first contact. Call l\ge0 a contact depth label if a contact occurs at some integer depth in [2^l,2^{l+1}). For G<l\le J, let U_l be the event that none of the labels l-G,l-G+1,\ldots,l is a contact depth label. Equivalently, there is no contact at depths in [2^{l-G},2^{l+1}).

Fix a positive cut depth

After the list reversal, this cut occurs at the bridge’s first hit T_k, and the remaining bridge stays at or above height k. Indeed, every earlier slab has departure heights below k, while every later slab has departure heights at least k. This is a property of the words, without a stopping-time assertion about the cut.

There is a contact after T_k and before T_N: the last positive departure is (N-1,E), the starting departure site of the negative trajectory. Let \tau be the first contact since T_k, defined pathwise against the entire negative trajectory. Its depth is at most s<2^{l+1}. On U_l, it must therefore be strictly less than 2^{l-G}. With

the nearest-neighbor path must have reached height r before this contact. Thus

No restriction on drawdowns from the running maximum is needed.

For each deterministic s\in I_l, let \mathcal A_{l,s} be the event in (46), together with |E|\le N^2, whether or not s is a cut. Proposition  6.2, with upper stopping height N, applies to this event. Its width-probability difference satisfies

because s-2^{l-G}\ge2^{l-1} and Lemma 4.3 applies. The same Lemma gives

Writing \phi(x)=x\log^2(e/x) for 0<x\le1 and \phi(0)=0, the first-contact estimate is

Averaging over the cuts. Let \mu_+=E_+L. The renewal-count strong law yields, almost surely under the positive tape law,

Fix c_0=1/(4\mu_+), and let \mathcal B_J be the event that each interval I_l, G<l\le J, contains at least c_0 2^l positive cuts. Since G=G(J)\to\infty, the eventual almost-sure estimate implies \mathbf P(\mathcal B_J^c)=o(1). Moreover,

In particular, no independence between the cut at N and the density event is required.

On \mathcal B_J\cap\{|E|\le N^2\}, every uncovered label l produces \mathcal A_{l,s} for each actual cut s\in I_l. Consequently,

The right side has dropped the cut requirement on s. This permits the deterministic-barrier estimate (49); the number of summands is O(2^l) and cancels the averaging factor.

We next check that the logarithms in this estimate do not require a quantitative tail rate. For J\ge3, put

where a zero summand is interpreted as zero. Terms with a_l\ge J^{-2} contribute at most 4A\log^2 J. On (0,J^{-2}), the function x\log^2(1/x) is increasing, so the remaining at most J terms contribute at most 4J^{-1}\log^2 J. Therefore

For sufficiently large J, 2^{-G}a_l\le e^{-1} uniformly in l, since a_l\le A. The function \phi is increasing on [0,e^{-1}]. Thus (47) gives, uniformly over s\in I_l,

Taking expectations in (51), then using (45), (49), and (50), gives

Indeed, \log(2N)=O(J), while the choice  (44) gives 2^{-G}=O(\log^{-4}(2+J)). By (52), the last term is O(J/\log^2(2+J)).

Removing the conditioning. Let H_J count contact depth labels in \{1,\ldots,J\}. Equation (53) and Markov’s inequality imply that with \mathbf Q_N-probability tending to one, at most J/4 of the tested labels are uncovered. Since G=o(J), at least J/2 are then covered. A single contact depth label h can cover only the labels h,h+1,\ldots,h+G. Hence

For large J, all depths counted here are less than 2^{J+1}<N. On \mathcal C_N they are therefore fully represented in the positive list ending at depth N. Undoing the translation and list reversal leaves exactly the contacts in the original opposed arrangement. Thus (54) implies

Only a probability bounded away from zero is asserted after removing the conditioning. This is enough for an expectation lower bound.

From depths to common-block indices. Write K_1,K_2,\ldots for the common-block widths and W_i=K_1+\cdots+K_i, with W_0=0. By Lemma 4.4, \mu:=\mathbf E K_1<\infty and W_i/i\to\mu almost surely. In particular, eventually

An integer depth z\ge1 belongs to the unique block

The departure-height convention makes this true even at a common cut depth. For every sufficiently large z, the strong-law bounds give

The finite initial blocks, however large their widths, disappear once z exceeds their cumulative width.

Define the dyadic index label of a block i>1 by j(i)=\lceil\log_2 i\rceil-1, so that 2^{j(i)}<i\le2^{j(i)+1}. It follows from (56) that there is a deterministic integer c_1, depending only on \mu, such that eventually

Almost surely this holds above a finite random depth. Therefore the probability that it holds for every contact with \lceil\sqrt J\rceil\le l\le J tends to one.

Intersect this event with the event in (55). The intersection still has probability at least p_+-o(1). Discarding the first \lceil\sqrt J\rceil depth labels loses o(J/(G+1)) labels. Each remaining contact depth label supplies a contact block whose index label lies within c_1 of it. Conversely, one index label can account for at most 2c_1+1 depth labels. Thus, on the intersection, at least c_2J/(G+1) of the index labels 1,\ldots,J+c_1 contain a contact, for a fixed c_2>0 and large J. Taking expectations yields

Since G+1=O(\log\log(3+J)), replacing J+c_1 by the argument of m proves the proposition.□

Proof of Theorem 1.1. Fix any nonzero real direction \ell. Suppose that 0<P_0(A_\ell)<1. Lemma 2.6 implies that P_0(A_{-\ell})>0. The spatial-radius estimate and Proposition 4.1 then supply a coordinate direction in which both signs have positive probability. Reflecting and relabeling that coordinate gives the setting of Propositions 5.1 and 7.1. They imply simultaneously

for all sufficiently large J, which is impossible. Thus P_0(A_\ell)\in\{0,1\}.

The reduction applies to every fixed real \ell without rational approximation. Strict ellipticity supplied the fresh-row probabilities; the spatial-radius moment was derived under coexistence. Neither a moment of a slab’s duration nor a speed assumption entered the argument.□