Limit shape formulas for a generalized Seppäläinen–Johansson model

We consider a simpliﬁed model of ﬁrst-passage percolation, involving two families of i.i


Introduction
Let B ij , i, j ≥ 0 be i.i.d Bernoulli(p) random variables, and ξ ij , η ij i, j ≥ 0 be families of independent random variables.We assume that the ξ ij 's are non-negative, integrable and have a common distribution, and likewise for the η ij 's (though the distribution of the ξ ij 's can be different of that of the η ij 's).We place weights on the edges e of Z 2 • If e is the horizontal edge joining (i − 1, j) to (i, j), then e has weight ω e = B ij ξ ij .
• If e is the vertical edge joining (i − 1, j) to (i, j), then e has weight ω e = (1 − B ij )η ij .So for any vertex (i, j), one of the two "incoming" edges (i − 1, j) → (i, j) or (i, j − 1) → (i, j) will have weight 0. Given an up-right path π (i.e a sequence of adjacent vertices which only goes up or right), we define its weight S(π) as

S(π) =
e∈π ω e where the sum is taken over the edges that π traverses; see Figure 1 (4,3).The weight of this path is B 1,0 ξ 1,0 where the minimum is taken over all up-right paths π started at (a, b) and finishing at (m, n).We write F (m, n) for F (0, 0; m, n).A path which achieves the minimum in (1.1) is called a geodesic.
The function F defines a directed metric on Z 2 ≥0 × Z 2 ≥0 , in the sense that for any point (m, n), we have F (m, n; m, n) = 0, and F satisfies the triangle inequality, but the distance can only be measured in one direction; F (a, b; m, n) is only defined when a ≤ m and b ≤ n.
The special case when all the B ij 's are equal to one (i.eonly horizontal edges have non-zero weights) is known as the Seppäläinen-Johansson (SJ) model.It was introduced by Seppäläinen in [11] as a simplified model of directed first-passage percolation, and Johansson showed in [6] that in the special case of Bernoulli weights, the model is completely solvable.Indeed, the law of the last-passage value (that is where we take maximum instead of minimum in (1.1)) is the same as that of the top point of the Krawtchouk ensemble, a discrete orthogonal polynomial ensemble.
By Kingman's subadditive ergodic theorem, there exists a deterministic function f on R 2 ≥0 (which depends on the distribution of the weights) such that almost surely for all x, y ≥ 0. We will refer to f as the limit shape.It follows from the translation invariance of this model that f must be homogeneous (i.e f (cx, cy) = cf (x, y) for all c ≥ 0), and together with the triangle inequality for F , we have that f also satisfies a triangle inequality: The homogeneity plus triangle inequality clearly imply that f must be convex, and so in particular it is continuous.Seppäläinen obtained in [11] the following explicit formula for the limit shape in the SJ model with Bernoulli(p) weights: Similar formulas are known in the SJ model for geometric and exponential weights; see [9] for example.The most general cases for which we can compute explicitly the limit Figure 2: A portion of the random walk web obtained from the B ij 's.If B ij = 0, an edge is placed from (i − 1, j) to (i, j); otherwise, the edge is placed from (i, j − 1) to (i, j).
shape are when ξ ij d = B • X, where B is Bernoulli distributed, and X is a geometric or exponential independent of B; see [8].
The generalized SJ model that we are considering first appears in [1] for ξ ij and η ij exponentially distributed as the zero temperature limit of the beta polymer.The authors show that the distribution function of a point to half-line first-passage percolation version of this model can be written down explicitly as a Fredholm determinant, and as a result, they obtain Tracy-Widom limiting fluctuations.Much more is known about the SJ model for Bernoulli weights.Indeed, in [4], it was shown that the scaled fluctuations of the first-passage value function converges in distribution with respect to uniform convergence on compact sets to the Airy 2 process.Using this, we obtain, in the Bernoulli case, the limiting distribution of the scaled fluctuations of F (nx, ny), for points satisfying x = (1 − p)y/p.
Along the line x = (1−p)y/p, the fluctuations of F behave very differently.If we scale space diffusively, then the fluctuations of F are of order 1, and F has a completely different scaling limit: the Brownian web distance.To see why the Brownian web is involved, consider the subgraph of Z 2 obtained by only keeping the edges e whose corresponding weight ω e is 0. The resulting subgraph is then a system of coalescing random walks moving in the south-west direction; see Figure 2. To each removed edge, we associate a cost of jumping across it given by the weight of that edge.Then F (a, b; m, n) is precisely the smallest total cost one has to pay while traversing through the random walk web to get from (a, b) to (m, n).In the special case where the ξ ′ ij s and η ′ ij s are all equal to 1, this is simply the smallest number of jumps one has to make between the different random walks.The diffusive limit of the random walk web is the Brownian web, and so F converges to a directed metric on the Brownian web.This is proven in [12].See [5] and the references therein for a survey on the Brownian web.

Main results
Let F H (m, n) be the first-passage value at (m, n) in the SJ model with weights B ij ξ ij on horizontal edges (i − 1, j) → (i, j) and F V (m, n) the first-passage value at (m, n) for the SJ model on vertical edges with weights (1 − B ij )η ij (that is, we change all the weights on vertical edges to be 0 for F H and all the weights on horizontal edges to be 0 for F V ).Let f H and f V be the corresponding limit shapes.Clearly, we have F H (m, n) ≤ F (m, n), since we can simply ignore any weights picked up along vertical edges when following a geodesic for F and this gives a valid path for F H . Similarly, we also have F V (m, n) ≤ F (m, n).Passing to the limit, this implies the corresponding bound on limit shapes: for all x, y ≥ 0. Our first result is that equality holds in (2.1).
As we will see later, f H (x, y) = 0 for all x ≤ (1 − p)y/p and f V (x, y) = 0 for all x ≥ (1 − p)y/p, so Theorem 2.1 can be rewritten as Since the scaled first-passage functions converge to their limit shape, it follows that for any x > (1 − p)y/p, and a similar asymptotic holds with F V when x < (1 − p)y/p.Thus the plane is divided into two regions, one where essentially only the horizontal weights matter, and one where only the vertical weights matter.Numerical simulations seem to show that the error term is actually of much smaller order than o(n), and that the equality max(f H (x, y), f V (x, y)) = f (x, y) almost holds even in the prelimit.
Next we consider the special case where at least one of the ξ ij 's or η ij 's are Bernoulli distributed (without loss of generality, we can assume it is the ξ ij 's).In this situation, we can show that the error in (2.2) is in fact o(n 1/3 ).As a consequence, we obtain the following limiting fluctuations result.Theorem 2.2.Suppose that B ij ξ ij follow a Bernoulli(p) distribution.Assume 0 < p < 1, and let x, y be positive and satisfying x > (1 − p)y/p.Define Here T W GUE is the Tracy-Widom GUE distribution.
Theorem 2.2 is known for the SJ model; this was proven by Dauvergne, Nica and Virág in [4].In fact, they prove a functional version of this result; we state it here in our language as it will be of use in the proof of Theorem 2.2. Then with respect to uniform convergence on compact sets.Here A 1 (t) is the top line of the Airy line ensemble.
The Airy line ensemble is a collection of continuous stochastic processes A 1 , A 2 , . . .
for all t.Präfer and Spohn first constructed a process with the same finite distributions in [10] as the scaling limit of the polynuclear growth model, however it was only shown in [3] that there is a version of this process with continuous sample paths that satisfies (2.5).That is, the processes A 1 , A 2 , . . .can be coupled together in such a way that they are almost surely non-intersecting.The top line A 1 is usually called the Airy 2 process, and is the scaling limit of the largest eigenvalue in the Dyson Brownian motion; see [7].In particular, A 1 (0) follows the Tracy-Widom GUE distribution.For our purposes, this last fact and the continuity of the paths of A 1 are the only properties of the Airy 2 process that we will need.
Setting t = 0 in (2.4) shows that the fluctuations of the SJ first-passage value converge in distribution to the Tracy-Widom GUE distribution.Our contribution is to extend this result to the generalized SJ model.The scaling exponents and coefficients in (2.4) are exactly the same in both models, as well as the limiting distribution.This further reinforces this behaviour of F being almost completely determined by only one set of weights on either side of the critical line.
It is believed that (2.4) should be universal, in the sense that the scaling exponents and limiting distribution should be the same for any choice of distribution of the weights, and thus the SJ model should be in the Kardar-Parisi-Zhang (KPZ) universality class.As the proof of Theorem 2.2 will show, for any set of weights for which (2.4) holds for the SJ model, the convergence of the marginal at 0 also holds in the generalized case.

Proof of Theorem 2.1
Throughout the next two sections, we will usually simplify notation by omitting the floor function when evaluating F or F H at non-integer points.It is then understood that if a and b are not integers, then F (a, b) is defined to be F (⌊a⌋, ⌊b⌋).The main ingredient which goes in the proof of Theorem 2.1 is the following curious identity.Lemma 3.1.Let B ij ∈ {0, 1}, and let ξ ij , η ij be collections of real numbers.Let F (m, n) be the first passage value from (0, 0) to (m, n) with weights and let F H (m, n) be the first passage value from (0, 0) to (m, n) where the weights on the horizontal edges are ω ij and the weights on the vertical edges are 0 except for the weights on the y-axis which are ω 0j .Suppose that ξ ij ≥ 0 for all i, j ≥ 0, that is the ξ ij 's and η ij 's are all non-negative except for the η ij 's lying on the y-axis which are all non-positive.Then for all m, n.
Thus if one replaces the weights on the y-axis with non-positive weights, then the first-passage value corresponds exactly to the first-passage value on horizontal edges.
That is, we can completely ignore any vertical edges not on the y-axis; given a geodesic from (0, 0) to (m, n), it cannot pass through a vertical edge of non-zero weight except for edges on the y-axis.Lemma 3.1 is deterministic and holds for arbitrary collections of numbers ξ ij , η ij and B ij satisfying the conditions in the lemma.A similar result holds with weights on the x-axis changed to being non-positive and first-passage on vertical edges.
Proof.Let X ij , Y ij be the horizontal and vertical increments for F : all i, j.Since any path from (0, 0) to (i, j) must pass through exactly one of the vertices (i − 1, j) or (i, j − 1), it is easy to see that F and F H satisfy the recursions ).
(3.1) Using (3.1), we obtain recursions for X ij and Y ij : and Now, suppose that X i,j−1 and Y i−1,j are both non-negative.By definition of the model, at least one of ω ij or ω ij must be 0. If Likewise, if both X i,j−1 and Y i−1,j are non-negative and ω ij = 0, then and if instead ω ij = 0, then Note also from these recursions that X ij and Y ij are then both non-negative.Using (3.1) for F H and the fact that vertical edges have weight 0 in this model, we find and So the increments for F and F H satisfy the exact same recursion provided the increments for F are non-negative.The increments are indeed non-negative; this is clear when i = 0 or j = 0 since ξ ij ≥ 0 and η i0 ≤ 0 for all i and j, and the general case follows by double induction on (i, j).Finally the boundary conditions are the same since the weights on the edges of both axes are the same.
We apply Lemma 3.1 to the special case where the weights on the y-axis are zero.Since the first-passage value is the same as F H (m, n) when we change all the weights on the y-axis to be zero, there must be a geodesic π for F H (m, n) which does not pass through any vertical edge of non-zero weight except possibly on the y-axis.We obviously have S(π) ≥ F (m, n), and the only difference between these two are the extra weights picked up by π along the y-axis: where Z = min{k ≥ 0 : (1, k) ∈ π} is the position where π exits the y-axis (when Z = 0, we interpret the sum as being 0).Since π is a geodesic for F H (m, n), we have Z ≤ D(m, n), where D(m, n) is the top-most departure point from the y-axis that a geodesic for F H can take: Note that D(m, n) is independent of the η ij 's since it is defined completely in terms of first-passage percolation with the weights B ij ξ ij on horizontal edges.Thus, along with the inequality (1 − B 0j ) ≤ 1, we have the upper bound Together with the obvious lower bound F (m, n) ≥ F H (m, n), Theorem 2.1 will be proved for y ≤ (1 − p)x/p provided that we can show in probability (this will imply F (nx, ny)/n → f H (x, y) in probability, but since we know that F (nx, ny)/n → f (x, y) almost surely by the subadditive ergodic theorem and almost sure convergence implies convergence in probability, we thus have F (nx, ny)/n → f H (x, y) almost surely).As mentioned above, D(m, n) is independent of the η ij 's, so by the law of large numbers, (3. We can see that E(m, n) has the same law as D(m, n), since it corresponds exactly to the bottom-most departure point for first-passage percolation with down-left paths from (m, n) to (0, 0).So (3.4) is equivalent to showing that E(nx, ny) n → 0 in probability.To get Theorem 2.1 for x < (1 − p)y/p, one can use the same argument as above but with first-passage percolation on vertical edges instead and by considering the right-most departure from the x-axis and left-most entry to the line y = n.The proof for this case is exactly the same so we will not write it down.
Lemma 3.2.Let ω ij be i.i.d. and non-negative, and let f H be the limit shape for the SJ model on horizontal edges with weights ω ij .Let p = P(ω ij = 0).Then f H (x, y) > 0 if and only if x > p 1−p y.
Proof.First assume that 0 < x < py/(1 − p) (the case where x = 0 is trivial, since F H (0, n) = 0 for all n).Note in particular that this implies p > 0 in that case.In order for F H (m, n) to be 0, there has to be a path from the origin to (m, n) which only visits edges of weight 0. Every time the path sees a horizontal edge of weight 0, it will take it, otherwise it will keep moving up until it sees an edge of weight 0. The number of up steps it needs to take before it sees such an edge has the geometric distribution on {0, 1, . . .} with probability of success p, and it needs to take m right steps.So where Z 1 , . . ., Z m are i.i.d.Geo(p) random variables.Take ⌊nx⌋ and ⌊ny⌋ instead of m and n.Then for θ > 0, we have, by Markov's inequality, There is some 0 < ǫ < .
By this condition on x and y, we have θ = − log(1 − ǫp) > 0. Substituting θ in the above, we then find Let g(ǫ) be the expression in square brackets above.Then < 0 so g is strictly decreasing.Since g(0) = 0, it follows that g(ǫ) < 0 for every 0 < ǫ < 1, and therefore the above probabilities are summable in n.By the Borel-Cantelli lemma, it follows that F H (⌊nx⌋, ⌊ny⌋) is 0 for all but finitely many n almost surely, and thus the limit shape must satisfy f H (x, y) = 0.This proves that f H (x, y) = 0 for all x < py/(1 − p), and by continuity, we obtain this for x = py/(1 − p) as well.
Define new weights ω ij as follows: Then we have ω ij ≥ ω ij for all i, j, and ω ij is c times a Bernoulli(1 − s) for some 0 ≤ s < q.
With f H the limit shape of the ω ij 's (which we can compute explicitly using (1.2)), we then have With the hypotheses of the previous lemma and the assumption that p > 0, we have for any fixed x that the function y → f H (x, y) is strictly decreasing on [0, (1 − p)x/p].
We are now ready to show (3.5) and conclude the proof of Theorem 2.1.Let q = P(B ij ξ ij = 0).Then for a point (x, y) satisfying x > qy/(1 − q), we have that the function z → f H (x, z) is strictly decreasing in a neighbourhood of y by Lemma 3.3.So for all ǫ > 0 small enough, f H (x, y − ǫ) > f H (x, y), and because almost surely, it follows that By the remarks leading up to (3.5), this shows that f (x, y) = f H (x, y) for x > qy/(1 − q).As explained previously, a symmetric argument using first-passage percolation on vertical edges shows that we also have f (x, y) = f V (x, y) for x < (1 − r)y/r where r = P((1 − B ij )η ij = 0).By continuity of the limit shape and Lemma 3.2, we also have f (x, y) = f H (x, y) = 0 for x = qy/(1 − q) and f (x, y) = f V (x, y) = 0 for x = (1 − r)y/r.the case where both ξ ij and η ij are positive almost surely, we are done (since then we have q = 1 − p and r = p).If not, then we still have to deal with points in between the lines x = qy/(1 − q) and x = (1 − r)y/r.However on those lines, we have f (x, y) = 0, and f is non-negative and convex.So f must also be zero in between those lines, and therefore equals both f H and f V there.This shows that f (x, y) = max(f H (x, y), f V (x, y)) in all cases and concludes the proof of Theorem 2.1!

Proof of Theorem 2.2
We will abbreviate things in the statement of Theorem 2.3 by writing τ = τ (x, y), χ = χ(x, y) and ρ = ρ(x, y).Write F n (t) for the left-hand side of (2.4).Thus we have F n (t) ⇒ A 1 (t) with respect to uniform convergence on compact sets.
By the Skorokhod representation theorem [2, Theorem 6.7], the F n and A 1 can be coupled together on the same probability space such that F n → A 1 uniformly on compact sets, almost surely.We henceforth work with this particular coupling.Fix ǫ > 0 and define 0) (we again use the convention that F k = F ⌊k⌋ when k is not an integer).Indeed, let K be a compact subset of R which contains all the t n 's.Then The first term on the right-hand side above converges to 0 since F n → A 1 uniformly on K, and the second term converges to 0 because A 1 is continuous.We have   Now subtract (4.2) from (4.1) and divide by n 1/3 .After some rearranging, this yields Since we are in the Bernoulli case, f (x, y) is given by (1.2), and so we find by our assumption on x, y and p.This together with (4.3) implies that for all sufficiently large n, we must have F H (nx, ny) = F H (nx, ny − ǫn 1/3 ).
Consequently, the bottom-most entry point E(nx, ny) is at most ǫn We are now ready to conclude.Let G be a bounded, uniformly continuous function on R, and let ǫ > 0. Then there is a δ > 0 such that |G(x)−G(y)| < ǫ whenever |x−y| < δ.

0 ( 3 . 4 )
3) is equivalent to D(nx, ny) n → in probability.It will be more convenient to work with the bottom-most entry point E(m, n) to the line x = m:
and define X H ij and Y H ij similarly for F H . we're given all the increments of a model, we can deduce what the first passage values are by just adding/subtracting the increments: 1/3, and this implies almost surely, and because D(nx, ny) has the same distribution as E(nx, ny), we deduce Let us note that it is fairly straightforward to generalize the above argument to obtain thatE(nx + τ n 2/3 t, ny) n 1/3 → 0 (4.5)uniformly for t in a compact set almost surely.However, while it is true that E(m, n) and D(m, n) have the same distribution for a fixed endpoint (m, n), it is not the case that the joint laws of {E(m, k) : k ∈ S} and {D(m, k) : k ∈ S} are the same for k varying in some set of integers S. We therefore cannot conclude that (4.5) holds for D; this is the only obstacle in obtaining uniform convergence as in(2.4)for the generalized SJ model.