Viet Hung Hoang

Kilian Raschel

Pierre Tarrago
email: pierre.tarrago@sorbonne-universite.fr

Pierre Tarrago Harmonic

Harmonic functions for singular quadrant walks

Keywords: 2020 Mathematics Subject Classification. Primary 31C35, 60G50; Secondary 60J45, 60J50, 31C20, 11B39 Discrete harmonic functions, singular random walks in the quarter plane, compensation approach, Green functions, Martin boundary

come   L'archive ouverte pluridisciplinaire

Introduction

Espace probability for random walks in cones. Consider a multidimensional lattice random walk {S(n)} n 1 , i.e., S(n) = X(1) + • • • + X(n) for all n 1. Given a cone C ⊂ R d , d 1, introduce the associated first exit time

τ x = inf{n 0 : x + S(n) / ∈ C} ∞.
If the drift of the increment distribution belongs to the cone, then (ignoring pathological behaviours) for x interior to C, the escape probability (also called survival probability)

(1) P(τ x = ∞) = P(∀n 0, x + S(n) ∈ C) is strictly positive, and defines a discrete harmonic function with Dirichlet boundary conditions.

Figure 1. Various step sets. From left to right: the simplest singular random walk; an arbitrary small step singular random walk; an example of singular walk with bigger jumps.

The question at the origin of the present work is the following: does this natural harmonic function admit an expression in closed form? The harmonicity property is equivalent to a recurrence relation, which in dimension 1 may be easily solved, at least in the bounded jump case. On the other hand, it is known that the behaviour of solutions to multivariate recurrences is much harder and vast [START_REF] Bousquet-Mélou | Linear recurrences with constant coefficients: the multivariate case[END_REF][START_REF] Bousquet-Mélou | Walks confined in a quadrant are not always D-finite[END_REF], a fortiori with boundary conditions depending on a cone; there is no hope, in general, to compute explicitly the escape probability [START_REF] Harmonic | A compensation approach for queueing problems[END_REF].

A glimpse of our results (Part 1). We may now state the contributions of our paper, which consists of two parts. We will introduce a class of singular random walks in dimension 2 (see the next subsection for a precise definition, see also Figure 1), and look at the case of the cone C being the positive quarter plane. In Part 1, we will produce explicit expressions for the escape probability [START_REF] Harmonic | A compensation approach for queueing problems[END_REF]. To that purpose, we will use the compensation approach, as introduced in [2, 1, 3] by Adan, Wessels and Zijm. Before giving more details both on the random walks considered and on the techniques used, let us present an explicit example.

Consider in this paragraph the model p -1,1 = p 1,1 = p 1,-1 = 1 3 , as on Figure 1 (left). To that example, our main result will entail that the escape probability starting at x = (i, j) equals (2) P(τ (i,j) = ∞) = 1 -1 2 i -1 2 j + 1 2 i 5 j + 1 2 j 5 i -1 5 i 13 j -1 5 j 13 i + 1 13 i 34 j + 1 13 j 34 i + • • • .

The integers appearing in the denominators in [START_REF] Adan | Analysis of the symmetric shortest queue problem[END_REF], namely 1, 2, 5, 13, 34, 89, 233, 610, . . . are directly related to Fibonacci sequence (A000045), as the n-th term in the sequence is equal to F 2n-1 . For i = j = 1, this result is derived in [24, Prop. 9], using a functional equation approach.

In this paper, we will prove similar formulas for other singular random walks; in particular, we will see how to define the sequence of denominators in general (providing a nice interplay between probabilistic and arithmetic properties of these singular random walks). As a second step, we will prove that similar expressions hold for infinitely many positive harmonic functions, of the form [START_REF] Adan | A compensation approach for two-dimensional Markov processes[END_REF] h(i, j)

= n 0 c n α i n β j n ,
where the α n , β n , c n are real constants (to be specified).

A glimpse of our results (Part 2). The main objective of our second part is to propose a probabilistic interpretation of the harmonic functions (3) constructed via the compensation approach. Our central result is to prove that the previous harmonic functions actually allow to construct all positive harmonic functions. In other words, the compensation approach yields an exhaustive description of positive harmonic functions for singular random walks. In concrete terms, this means that there is a correspondence between minimal positive harmonic functions and the yellow domain on the left display on Figure 4, exactly the same phenomenon as in the non-singular case [START_REF] Ignatiouk-Robert | Martin boundary of a killed random walk on a quadrant[END_REF]. This result is properly stated in Corollary 20. Among all positive harmonic functions, there is a unique bounded positive harmonic function, which is the escape probability. It corresponds asymptotically by looking at the Green function [START_REF] Bousquet-Mélou | Walks with small steps in the quarter plane[END_REF] G(x, y) =

n 1 P(x + S(n) = y, τ x > n)
along the drift direction.

To prove these results, we use Martin boundary theory, and we believe that several intermediate results are of independent interest. The key idea is to compute the asymptotics of the Green function (4) as x is fixed and y goes to infinity in any direction of the cone. See Theorem 11 for the main statement. While such results have been recently derived in a close context (zero drift random walks [START_REF] Duraj | Martin boundary of random walks in convex cones[END_REF], irreducible non-zero drift random walks [START_REF] Ignatiouk-Robert | Martin boundary of a killed random walk on a quadrant[END_REF]), there was no version in the literature applying to our context. The case of a boundary direction needs a particular attention, due to the interaction with the axes.

From a technical point of view, Part 2 is inspired by the work [START_REF] Ignatiouk-Robert | Martin boundary of a killed random walk on a quadrant[END_REF], [START_REF] Denisov | Random walks in cones[END_REF] and [START_REF] Duraj | Martin boundary of random walks in convex cones[END_REF].

Singular random walks in the quadrant. Throughout the paper, we assume that the distribution of the increments X(i) of the random walk have transition probabilities p i,j in Z 2 such that (see Figure 1):

(i) i,j p i,j = 1 (normalization); (ii) p i,j = 0 for all i -2 or j -2 (small negative jumps); (iii) p -1,-1 = p -1,0 = p 0,-1 = 0 (singular walks); (iv) p -1,1 p 1,-1 = 0; (v) There exists (i, j) with i + j > 0 such that p i,j > 0 (non-degeneracy); (vi) The p i,j admit exponential moments in the following sense: the Laplace transform i,j p i,j e ix+jy is finite in a neighborhood of any point of the curve [START_REF] Denisov | Random walks in cones[END_REF] (moment assumption).

By definition, the kernel of the model is the bivariate polynomial ( 5)

K(α, β) = αβ   i,j -1 p i,j α i β j -1   .
Compensation approach. This technique has been developed in the probabilistic context of stationary distributions for random walks, see [START_REF] Adan | Analysis of the symmetric shortest queue problem[END_REF][START_REF] Harmonic | A compensation approach for queueing problems[END_REF][START_REF] Adan | A compensation approach for two-dimensional Markov processes[END_REF]. It does not aim directly at obtaining a solution for a generating function (as it is usual for quadrant walk problems), but rather tries to find a solution for its coefficients, in our case the escape probability h(i, j) = P(τ (i,j) = ∞) starting from (i, j).

As a discrete harmonic function, h(i, j) satisfies certain recursion relations (coming from harmonicity in our case), which differ depending on whether the state (i, j) lies on the boundary or not:

h(i, j) = k, -1 p k, h(i + k, j + ), ∀i, j 1, (6)
h(i, 0) = 0, ∀i 0, (7) h(0, j) = 0, ∀j 0. ( 8
)
The idea is then to express h(i, j) as a linear combination of products α i β j , for pairs (α, β) such that the recursion relations [START_REF] Bousquet-Mélou | Walks confined in a quadrant are not always D-finite[END_REF] in the interior of the quarter plane hold. This is equivalent to choosing the parameters such that K(α, β) = 0, with K as in [START_REF] Bousquet-Mélou | Linear recurrences with constant coefficients: the multivariate case[END_REF]. The products have to be chosen such that the recursion relations on the boundaries ( 7) and ( 8) are satisfied as well. As it turns out, this can be done by alternatingly compensating for the errors on the two boundaries, which eventually leads to an infinite series of product forms. The typical outcome of the compensation approach is an expression of the form (3). This clearly formally contains the series presented in [START_REF] Adan | Analysis of the symmetric shortest queue problem[END_REF].

As a side note, our work proposes a new example of applicability of the compensation approach, in relation with potential theory and discrete harmonic functions.

Related literature on explicit formulas for escape probabilities. In the quarter plane, the few known formulas concern non-singular walks with certain finite reflection groups (for which an algebraic version of the reflection principle applies); see Figure 2 for three examples. For instance, for the simple random walk (leftmost display on the figure), the escape probability equals [21, Cor. 8]

P(τ (i,j) = ∞) = 1 - p -1,0 p 1,0 i 1 - p 0,-1 p 0,1 j ,
which interestingly corresponds to a finite (four terms) sum in [START_REF] Adan | A compensation approach for two-dimensional Markov processes[END_REF]. Similar expressions hold for the other two models on Figure 2, with a sum involving six and eight terms, respectively.

Figure 2. Three non-singular step sets with finite reflection groups. The rightmost example is assumed to satisfy p -1,1 p 1,-1 = p -1,0 p 1,0 .

Still in the non-singular case, other approaches based on complex analysis techniques (boundary value problems) allow to compute the generating functions i,j 1 h(i, j)x i y j of harmonic functions in terms of certain conformal mappings, see [START_REF] Hoang | Constructing discrete harmonic functions in wedges[END_REF] (zero drift case) and [START_REF] Lecouvey | t-Martin boundary of killed random walks in the quadrant[END_REF] (non-zero drift case).

In the continuous setting (Brownian motion in cones), there is a unified formula [18, Thm C] for the probability of escape from arbitrary cones C ⊂ R d , d 1. More precisely, if the drift d is interior to the cone, then

P(τ x = ∞) = (2π) d/2 exp x -d 2 p C 1, x, d ,
where p C denotes the heat kernel of the cone. The question of finding explicit expressions is then reduced to computing in closed-form the heat kernel, which is more classical. For example, if the cone is a Weyl chamber of type A, the determinantal Karlin-McGregor formula holds, and the survival probability is a finite sum of product forms, as in [START_REF] Adan | A compensation approach for two-dimensional Markov processes[END_REF].

Applications and refinements of our results. We conclude the introduction by mentioning related open questions and potential applications of our results.

• One advantage of the compensation approach is that it is not based on generating functions. However, it would be interesting to understand how the compensation approach is related to Cohen and Boxma techniques [START_REF] Cohen | Boundary value problems in queueing system analysis[END_REF][START_REF] Cohen | Analysis of random walks[END_REF], which we developed in the paper [START_REF] Hoang | Constructing discrete harmonic functions in wedges[END_REF] to study harmonic functions for non-singular random walks.

• In principle, the compensation approach could be applied as well to harmonic functions for reflected random walks in the quarter plane (with Neumann boundary conditions). It should also provide explicit expressions for t-harmonic functions (Dirichlet or Neumann boundary conditions).

• Formulas of the type of (3) hold for singular random walks (our main result) and for a few other (finite group) models, as those represented on Figure 2. This clearly suggests the question of determining the class of models for which harmonic functions may be expressed as in [START_REF] Adan | A compensation approach for two-dimensional Markov processes[END_REF], via the pattern: model → curve → sequence of numbers → harmonic function (via Ansatz (3)).

• Is it possible to study more generally the Martin boundary and the Green functions asymptotics of singular random walks in arbitrary cones? What is the exact applicability of the compensation approach? For instance, can it be applied to harmonic functions of higher dimensional models? • The compensation formula (15) that we find in the context of harmonic functions for singular random walks in the quarter plane is reminiscent of the alternating formula for the harmonic function of a space-time Brownian motion (t, B t ) t 0 conditioned to stay in the cone {(t, y), 0 < y < t}, see [START_REF] Defosseux | Affine Lie algebras and conditioned space-time Brownian motions in affine Weyl chambers[END_REF]. This suggests a relation between the singular walks we are considering and stochastic processes conditioned to stay in affine Weyl chambers.

Part 1. Constructing harmonic functions via the compensation approach 2. Properties of the Laplace transform and convergence of the series

This section aims at constructing harmonic functions as series taking the form of (3), by using the compensation approach. We first introduce a curve K ∈ R 2 such that α i β j is a solution of (6) for any (α, β) ∈ K, see Section 2.1. In Section 2.2, we then construct sequences (α n , β n ) n 0 ⊂ K such that their series satisfies the boundary conditions ( 7)- [START_REF] Cohen | Analysis of random walks[END_REF]. We then prove that these series converge and do not depend on the initial starting point of the sequence, see Section 2.3. Finally, in Section 2.4, we study a particular case, where we should renormalize the harmonic function to get a non-zero quantity.

2.1. Level sets of the Laplace transform. We first observe that the product form α i β j is solution to [START_REF] Bousquet-Mélou | Walks confined in a quadrant are not always D-finite[END_REF] if and only if

(9) αβ = k, p k, α k+1 β +1 ,
or equivalently, using the kernel notation [START_REF] Bousquet-Mélou | Linear recurrences with constant coefficients: the multivariate case[END_REF], if and only if K(α, β) = 0. We therefore introduce the algebraic curve (10) K := {(α, β) ∈ R 2 0 : K(α, β) = 0}. See Figure 3 for an example.

To take advantage of some convexity properties, we mainly investigate K through an alternative exponential scaling as follows: [START_REF] Denisov | Random walks in cones[END_REF] G := {(x, y) ∈ R 2 : K(e x , e y ) = 0}.

See Figures 3 and4 for examples of curves G. Let us further denote (12) G + := {(x, y) ∈ R 2 : K(e x , e y ) < 0} and G -:= {(x, y) ∈ R 2 : K(e x , e y ) > 0}.

The following lemma presents some crucial properties of the curve G.

Lemma 1. Under Assumptions (i)-(vi), we have the following assertions:

(i) G + is an unbounded convex domain and includes the ray {t(1, 1) : t < 0};

(ii) G passes through (0, 0) and the two tails of G lie in the third quadrant R 2 -; (iii) G admits a tangent at (0, 0) satisfying the equation

  k, kp k,   x +   k, p k,   y = 0.
Proof. We first prove (i). K(e x , e y ) can be factorized as K(e x , e y ) = e x+y G(x, y), where G(x, y) is defined as [START_REF] Doob | Discrete potential theory and boundaries[END_REF] G(x, y) := k, p k, e kx+ y -1.

Since k, p k, e kx+ y is a moment-generating function of a random variable in Z 2 , it is convex in R 2 . The domain G + is thus convex. By letting x = y < 0, we then have:

G(x, x) = k, p k, e (k+ )x -1 < k, p k, -1 = 0, since p k, = 0 if k + < 0.
Thus, G + is unbounded and includes the ray {t(1, 1) : t < 0}. We now prove (ii). Since G(0, 0) = 0, then (0, 0) ∈ G. By letting x = 0 > y with |y| large enough, we have

G(0, y) = k, p k, e y -1 > p 1,-1 e -y > 0.
Similarly, for all y = 0 > x with |x| large enough, we have

G(x, 0) = k, p k, e kx -1 > p -1,1 e -x > 0.
It implies that both tails of G lie in R 2 -. Finally, (iii) is easily seen from the equation of the tangent at (0, 0), which is ∂ x K(e 0 , e 0 )x + ∂ y K(e 0 , e 0 )y = 0.

Although G + is always convex, the bounded domain delimited by K is not necessarily convex (see Figure 3). This is the main reason for us to use the curve G rather than K.

We now parametrize the curve G by functions. Define f : R 0 → R such that for all x 0, K(e x , e f (x) ) = 0 and f (x)

x. Similarly, we define g : R 0 → R such that K(e g(y) , e y ) = 0 and g(y) y for all y 0. The following lemma describes key properties of f and g. Lemma 2. Under Assumptions (i)-(vi), we have the following assertions:

(i) f and g are well defined, concave, and infinitely differentiable on R -; (ii) max x 0 f (x) > f (0) = 0 and arg max x 0 f (x) includes a unique point, denoted by x 0 . Further, f (x) is strictly increasing on (-∞, x 0 ), strictly decreasing on (x 0 , 0), and lim x→-∞ f (x) = -∞; (iii) max y 0 g(y) > g(0) = 0 and arg max y 0 g(y) includes a unique point, denoted by y 0 . Further, g(y) is strictly increasing on (-∞, y 0 ), strictly decreasing on (y 0 , 0), and lim y→-∞ g(y) = -∞.

Proof. We first prove Item (i). By the convexity of G + , the slope of G at (0, 0), and the tails' position of G (see Lemma 1), it is easily seen that f and g are well defined and concave on R 0 . Since any point (x, y) ∈ G is not the minimiser of the convex function G defined in [START_REF] Doob | Discrete potential theory and boundaries[END_REF], then ∂ x G and ∂ y G cannot vanish simultaneously for any x, y ∈ G. By the construction of f and g, we then have

∂ x G(x, f (x)
) = 0 and ∂ y G(g(y), y) = 0 for all x, y 0. Thus,

f (x) = - ∂ x G(x, f (x)) ∂ y G(x, f (x)) and g (y) = - ∂ y G(g(y), y) ∂ x G(g(y), y)
are well defined for all x, y < 0, and so are all the higher orders derivatives of f and g. We now prove Item (ii). Since

f (0) = - ∂ x K ∂ y K (e 0 , e 0 ) = k, kp k, k, p k, < 0, then max x 0 f (x) > f (0) = 0. If arg max x 0 f (x)
includes at least two distinct points, then it also contains the segment between these points by the concavity of f . This implies that K(exp(x), exp(max x 0 f (x))) is a constant function of x, which does not hold true. Hence, arg max x 0 f (x) includes a unique point x 0 . The strict monotonicity of f on (∞, x 0 ) and (x 0 , 0) then follows. By letting x 1 < x 0 such that f (x 1 ) = 0, we have for all x < x 1 , Thus, f (x) → -∞ as x → -∞. Item (iii) is proven similarly to (ii). The proof is then complete.

f (x) < f (x 0 ) x 0 -x 1 (x -x 1 ) x→-∞ -→ -∞.
2.2. Construction of the product forms. Our main objective here is to prove Lemma 3 below, i.e., to construct sequences

{(a n , b n )} n∈Z ⊂ G such that, setting (14)
(α n , β n ) := e an , e bn ∈ K, the associated series

(15) h(i, j) := n∈Z α i n β j n -α i n+1 β j n
is well defined, positive and satisfies ( 6)- [START_REF] Cohen | Analysis of random walks[END_REF].

To that purpose, we introduce a few useful notations. We first define ( 16)

G 0 := {(x, f (x)) : x ∈ (x 0 , 0]} ∪ {(g(y), y) : y ∈ (y 0 , 0]} ⊂ G,
which corresponds to the yellow part on Figure 4. We first set

f := f |(-∞,x 0 ] -1 and f := f |(0,x 0 ] -1 .
Since f is concave, strictly increasing on (-∞, x 0 ], then f is a well-defined function, convex and strictly increasing on (-∞, f (x 0 )]. Similarly, the function

g := (g |(-∞,y 0 ] ) -1 is convex, strictly increasing on (-∞, g(y 0 )]. We construct a sequence {(a n , b n )} n∈Z contained in G as follows: (a 0 , b 0 ) ∈ G 0 , ( 17
) (a n , b n ) := f • ( g • f ) •(n-1) (b 0 ), ( g • f ) •(n) (b 0 ) , n 1, ( 18
) (a -n , b -n ) := ( f • g) •(n) (a 0 ), g • ( f • g) •(n-1) (a 0 ) , n 1. ( 19
)
Lemma 3. Let {(a n , b n )} n∈Z be any sequence defined by (17), ( 18) and [START_REF] Hoang | Constructing discrete harmonic functions in wedges[END_REF], and set (α n , β n ) = (e an , e bn ) as in [START_REF] Dreyfus | Walks in the quarter plane: genus zero case[END_REF]. Then the series given by (15) converges absolutely for any i, j 1 and satisfies Eq. ( 6)-( 8). Furthermore, h(i, j) 0 for any i, j 1.

Proof. We start by proving the convergence of the series. To that aim, we construct an auxiliary sequence ( a n , b n ) n∈Z as follows:

( a 0 , b 0 ) = (a 0 , b 0 ) ∈ G 0 , ( a n , b n ) = (b 0 -nc 1 -(n -1)c 2 , b 0 -nc 1 -nc 2 ) , n 1, ( a -n , b -n ) = (a 0 -nc 1 -nc 2 , a 0 -(n -1)c 1 -nc 2 ) , n 1,
where we have set c 1 = f (x 0 ) -x 0 > 0 and c 2 = g(y 0 ) -y 0 > 0, with x 0 , y 0 as in Lemma 2.

Let us also put, for any n ∈ Z,

α n , β n = e an , e bn .
Since G + (see [START_REF] Despax | On some lattice random walks conditioned to stay in Weyl chambers[END_REF]) is convex and contains the ray {t(1, 1) : t < 0}, it should also contain the rays {(x 0 , f (x 0 )) + t(1, 1) : t < 0} and {(g(y 0 ), y 0 ) + t(1, 1) : t < 0}. This implies that ( 20)

f (y) y -c 1 and g(x) x -c 2 ,
for all y f (x 0 ) and x g(y 0 ). Thanks to the above inequalities and the monotonicity of g, we first have

a 1 a 1 and b 1 = g(a 1 ) g( a 1 ) b 1 .
By induction argument, we then have

a n = f (b n-1 ) f ( b n-1 ) a n and b n = g(a n-1 ) g( a n-1 ) b n ,
for any n 2. Similarly, a -n a -n and b -n b -n for any n 1. We now have n∈Z

(α i n β j n + α i n+1 β j n ) n∈Z ( α i n β j n + α i n+1 β j n ) = α i 0 β j -1 + α i 0 β j 0 + α i 1 β j 0 + n 1 (e -jc 2 + e jc 1 )α i+j 0 e -(c 1 +c 2 )(i+j) n + n 1 (e ic 2 + e -ic 1 )β i+j 0 e -(c 1 +c 2 )(i+j) n .
The last sums are finite for any i, j 1. Hence h(i, j) converges absolutely for any i, j 1. We now prove that the series h(i, j) satisfies Eq. ( 6)- [START_REF] Cohen | Analysis of random walks[END_REF]. Firstly, all the terms α i n β j n and α i n+1 β j n satisfy Eq. ( 6) (since (α n , β n ) and (α n+1 , β n ) ∈ K), then h(i, j) should also satisfy Eq. ( 6). Secondly, since

h(i, 0) = n∈Z (-α i n + α i n ) = 0 and h(0, j) = n∈Z (β j n -β j n ) = 0,
then h(i, j) satisfies Eq. ( 7) and [START_REF] Cohen | Analysis of random walks[END_REF]. We now prove that h(i, j) 0 for all values of i, j 1, by rewriting

h(i, j) = α i 0 β j 0 1 - β -1 β 0 j - α 1 α 0 i + n 1 α i -n β j -n -β j -(n+1) + n 1 α i n -α i n+1 β j n .
The first term is positive for any i, j 1 such that i+j is large enough, whereas the second and third terms are positive since α -n > β -(n+1) > α -(n+1) and β n > α n+1 > β n+1 for any n 0. Hence, there exists n 0 > 0 such that h(i, j) > 0 for all i, j 1 and i + j n 0 . Thus,

h(i, j) = k, p k, h(i + k, j + ) 0,
for any i, j 1 and i + j = n 0 -1. By induction argument, we have h(i, j) 0 for all i, j 1. The proof is then complete.

The sign of harmonic funtions h(i, j) can be asserted more specifically.

Remark 4. If the walk has jumps inside the positive quadrant, i.e., there exist k, 0 such that k + > 0 and p k, > 0, then h(i, j) > 0 for all i, j 1. On the other hand, if the walk does not have any jumps inside the positive quadrant, i.e., p k, = 0 for all k, 0, then h(i, j) > 0 for i, j 1 and i + j large enough, but h(i, j) can vanish for some small i, j.

2.3. Dependency on the starting point. In Lemma 3, we assumed that the starting point (a 0 , b 0 ) belongs to G 0 . In this subsection, we prove that we do not obtain more harmonic functions with starting points in G \ G 0 .

We construct a sequence {(a n , b n )} n∈Z as follows:

(a 0 , b 0 ) ∈ G, (21)
a n = a n-1 s.t. (a n , b n-1 ) ∈ G, b n = b n-1 s.t. (a n , b n ) ∈ G, (22) b
-n = b -(n-1) s.t. (a -(n-1) , b n ) ∈ G, a -n = a -(n-1) s.t. (a -n , b -n ) ∈ G, n 1. (23) Let us define O(G 0 ) := {{(a n , b n )} n∈Z satisfying (21)-(23) s.t. (a 0 , b 0 ) ∈ G 0 }.
In other words, O(G 0 ) is the set of orbits formed by the sequences {(a n , b n )} n∈Z satisfying ( 21)-( 23) and initiated from a point in G 0 .

We have the following lemma.

Lemma 5. Any sequence {(a n , b n )} n∈Z satisfying (21)-( 23) belongs to the set O(G 0 ). As a result, the series h(i, j) defined in [START_REF] Duraj | Martin boundary of random walks in convex cones[END_REF] does not depend on the initial value (α 0 , β 0 ), but only depends on the orbits in O(G 0 ).

Proof. Assume that {(a n , b n )} n∈Z satisfies (21)-(23) with (a 0 , b 0 ) / ∈ G 0 . It is sufficient to show that there always exists n 0 ∈ Z such that (a n 0 , b n 0 ) ∈ G 0 or (a n 0 , b n 0 -1 ) ∈ G 0 . If 0 a 0 < g(y 0 ), then (a 0 , b -1 ) = a 0 , (g| [y 0 ,0] ) -1 (a 0 ) ∈ G 0 . Similarly, if 0 b 0 < f (x 0 ), then (a 1 , b 0 ) = (f | [x 0 ,0] ) -1 (b 0 ) ∈ G 0 . If x 0 < a 0 < 0, then (a 0 , b -1 ) = (a 0 , f (a 0 )) ∈ G 0 . If y 0 < b 0 < 0, (a 1 , b 0 ) = (g(b 0 ), b 0 ) ∈ G 0 .
If a 0 < x 0 and b 0 < y 0 , without loss of generality, we will further assume that b 0 = f (a 0 ). Since f (x) x + c 1 and g(y) y + c 2 , for all x x 0 and y y 0 , then there exists n 0 1 such that

a n 0 = (g • f ) •(n) (a 0 ) > x 0 or b n 0 = f • (g • f ) •(n) (a 0 ) > y 0 .
Hence, a n 0 or b n 0 falls into the preceding cases. The rest of the proof then follows trivially.

2.4. Behavior on the boundary. We will show that the harmonic functions defined in [START_REF] Duraj | Martin boundary of random walks in convex cones[END_REF], considered as a function of (a 0 , b 0 ) ∈ G 0 , will converge (after normalization) to a non-trivial positive harmonic function as (a 0 , b 0 ) tends to the boundary of G 0 , which consists of the points (x 0 , f (x 0 )) and (g(y 0 ), y 0 ), with x 0 and y 0 introduced in Lemma 2.

For that matter, let us re-introduce the sequence {(a n , b n )} n∈Z in ( 17)-( 19) as functions of one variable y:

b 0 (y) = y ∈ (y 0 , 0), a 0 (y) = g(y), b -1 (y) = g • g(y), (a n (y), b n (y)) = f • ( g • f ) •(n-1) (y), ( g • f ) •(n) (y) , n 1, (a -n (y), b -(n+1) (y)) = f • ( g • f ) •(n-1) • b -1 (y), ( g • f ) •(n) • b -1 (y) , n 1.
We further denote (α n (y), β n (y)) = e an(y) , e bn(y) , n ∈ Z,

h y (i, j) = n 0 α n (y) i β n (y) j -α n+1 (y) i β n (y) j ,
for all i, j 0, y ∈ (y 0 , 0). Following the above construction, we remark that

a -n (y), b -(n+1) (y) = a n • b -1 (y), b n • b -1 (y)
for any n 1. Then, putting for any i, j 0 and y ∈ (y 0 , 0)

t i,j (y) = n 0 α n (y) i β n (y) j -α n+1 (y) i β n (y) j , one has h y (i, j) = t i,j (y) -t i,j (b -1 (y)).
The following proposition shows the convergence of harmonic functions h y (i, j) as y → y + 0 . Proposition 6. For any i, j 0, lim

y→y + 0 h y (i, j) = 0.
Moreover, for any i, j 0, t i,j (y) is differentiable on [y 0 , 0) and

lim y→y + 0 h y (i, j) y -y 0 = 2t i,j (y 0 ).
Furthermore, t i,j (y 0 ) 0 for all i, j 1.

The remainder of this section is devoted to proving Proposition 6. We first study the derivatives of some important functions.

Lemma 7. We have:

(i) lim y→y + 0 b -1 (y) = y 0 and b -1 (y 0 ) = -1;
(ii) f is strictly positive and strictly increasing on (-∞, f (x 0 )). Furthermore,

lim y→-∞ f (y) = 1;
(iii) g is strictly positive and strictly increasing on (-∞, g(y 0 )). Furthermore,

lim x→-∞ g (x) = 1.
Proof. We first prove Item (i). Since g(y) is strictly concave on (-∞, 0) and admits its maximum value at y 0 , then g (y 0 ) = 0 and

g(y) = g(y 0 ) + g (y 0 ) 2 (y -y 0 ) 2 + o((y -y 0 ) 2 )
as y → y 0 , where g (y 0 ) < 0. Hence, lim

y→y + 0 b -1 (y) -y 0 b 0 (y) -y 0 2 = lim y→y + 0 g(b -1 (y)) -g(y 0 ) g(b 0 (y)) -g(y 0 ) = 1. This implies b -1 (y) → y 0 as b 0 (y) = y → y + 0 . Since b -1 (y) < y 0 < b 0 (y) for all y ∈ (y 0 , 0), then b -1 (y 0 ) = -1.
We move to the proof of Item (ii) (Item (iii) would be proven similarly). Since f is strictly convex and strictly increasing on (-∞, f (x 0 )), then f is strictly increasing and positive on (-∞, f (x 0 )). Putting α(β) = e f (log β) , β 0, we first study the behavior of α (β) as β → 0. Since α(β) satisfies K(α(β), β) = 0, one may differentiate once and twice the equation and evaluate them at β = 0 as follows:

∂ α K(0, 0)α (0) + ∂ β K(0, 0) = 0, ∂ αα K(0, 0)α (0) 2 + 2∂ αβ K(0, 0)α (0) + ∂ ββ K(0, 0) + ∂ α K(0, 0)α (0) = 0.
While the value of α (0) cannot be deduced from the first equation, α (0) can be solved explicitly from the second:

α (0) = 1 -p 0,0 ± (1 -p 0,0 ) 2 -4p 1,-1 p -1,1 2p 1,-1 = 0, ∞.
These solutions correspond to the behavior of the two branches (e f (y) , e y ) and (e g(y) , e y ) as y → -∞. This implies that

f (y) = α (e y ) e y e f (y) y→-∞ -→ α (0) 1 α (0) = 1.
The following lemma shows the convergence of t i,j (y).

Lemma 8. For any i, j 1 and y ∈ [y 0 , 0), the series

t i,j (y) = n 0 α n (y) i β n (y) j -α n+1 (y) i β n (y) j converges absolutely.
Proof. By Lemma 7, the derivatives of α n (y) and β n (y) are well defined on [y 0 , 0) for all n 0. Since f (y), g(x) ↓ 1 as x, y → -∞, then

a n (y) = f • b n-1 (y) n-1 k=1 g • a k (y) • f • b k-1 (y) = o(r n ), b n (y) = n k=1 g • a k (y) • f • b k-1 (y) = o(r n ),
as n → ∞, for any r > 1. Let us now fix i, j 1. Recall from the proof of Lemma 3 that

α n (y) i β n (y) j y i+j e ic 2 e -(i+j)(c 1 +c 2 ) n = O e -(i+j)(c 1 +c 2 ) n ,
as n → ∞, where c 1 and c 2 are positive constants. Hence,

α n (y) i β n (y) j = ia n (y) + jb n (y) α n (y) i β n (y) j o(r n ),
as n → ∞, for any y ∈ [y 0 , 0) and r > 0. Similarly, we also have

α n+1 (y) i β n (y) j o(r n ),
as n → ∞, for any y ∈ [y 0 , 0) and r > 0. Thus, the series

t i,j (y) = n 0 ia n (y) + jb n (y) α n (y) i β n (y) j -ia n+1 (y) + jb n (y) α n+1 (y) i β n (y) j
converges absolutely for any i, j 1.

We now prove Proposition 6.

Proof of Proposition 6. Since b -1 (y) → y 0 as y → y + 0 by Lemma 7(i), then h y (i, j) = t i,j (y) -t i,j (b -1 (y)) → 0, as y → y + 0 . By L'Hôpital's rule, we now deduce h y (i, j) y -y 0

y→y + 0 -→ t i,j (y 0 ) -t i,j (b -1 (y 0 ))b -1 (y 0 ) = 2t i,j (y 0 ).
Since h y (i, j) is harmonic and non-negative for all y ∈ (y 0 , 0), this guarantees that t i,j (y 0 ) is harmonic and non-negative for all i, j 1. We now prove that t i,j (y 0 ) > 0 for all i, j 1 and i + j large enough. Since for any n 0 one has α n (y 0 ) > α n+1 (y 0 ), then [START_REF] Mishna | Two non-holonomic lattice walks in the quarter plane[END_REF] jb n (y 0 )α n (y 0 ) i β n (y 0 ) j -jb n (y 0 )α n+1 (y 0 ) i β n (y 0 ) j > 0, for any n 0 and i, j 1. We then show that

(25) ia n (y 0 )α n (y 0 ) i β n (y 0 ) j -ia n+1 (y 0 )α n+1 (y 0 ) i β n (y 0 ) j > 0,
or equivalently,

α n (y 0 ) i > α n+1 (y 0 ) i f • b n (y 0 ) • g • a n (y 0 ) ,
for any n 0, j 1 and i large enough. Indeed, recall that α n (y 0 ) α n+1 (y 0 ) = e an(y 0 )-a n+1 (y 0 ) e f (x 0 )-x 0 +g(y 0 )-y 0 > 0, for any n 0, and

( f • b n (y 0 )) • ( g • a n (y 0 )) ↓ 1 as n → ∞.
Hence, for any n 0 and i large enough,

α n (y 0 ) α n+1 (y 0 ) i > f • b n (y 0 ) • g • a n (y 0 ) ,
which is equivalent to Inequality [START_REF] Ney | The Martin boundary for random walk[END_REF]. Both ( 24) and ( 25) imply that there exists n 0 > 1 such that t i,j (y 0 ) > 0 for any j 1 and i n 0 . The harmonicity then implies that t i,j (y 0 ) > 0 for any i, j 1 and i + j n 0 + 1.

We want to point out that Remark 4 also holds true for the harmonic function t i,j (y 0 ), that is, t i,j (y 0 ) > 0 for all i, j 1 if there exist k, 0 such that k + 1 and p k, > 0.

Explicit expression for walks with small steps

This section aims at giving an explicit expression for all terms appearing in the harmonic functions [START_REF] Duraj | Martin boundary of random walks in convex cones[END_REF], in the case where the random walks only have small jumps, that is, the positive transition probabilities can only be in the set {p -1,1 , p 1,-1 , p 1,0 , p 0,1 , p 1,1 }, see Figure 1.

The following lemma presents a uniformization of the zero set of K(α, β).

:= p 0,1 + 2p -1,1 p 1,0 , c := p 2 0,1 -4p -1,1 p 1,1 , b := p 1,0 + 2p 1,-1 p 0,1 , c := p 2 1,0 -4p 1,-1 p 1,1 .
We further have the involutions α(s) = α(1/s) and β(s) = β(1/(ρ 2 s)).

Proof. We will find a rational uniformization of the algebraic curve

(26) α 2 β 2 K 1 α , 1 β = 0.
We first rewrite

α 2 β 2 K 1 α , 1 β = p 1,1 + p 1,0 β + p 0,1 α + p 1,-1 β 2 + p -1,1 α 2 -αβ = (p 1,-1 ) β 2 + (-α + p 1,0 ) β + p -1,1 α 2 + p 0,1 α + p 1,1 .
The above polynomial admits the discriminant

δ(α) := (-α + p 1,0 ) 2 -4p 1,-1 p -1,1 α 2 + p 0,1 α + p 1,1 = aα 2 -2bα + c = b 2 -ac 4a 2 aα -b √ b 2 -ac 2 -4 ,
where a, b, c are defined in Lemma 9, and

b 2 -ac = 4p 1,-1 (1 -4p -1,1 p 1,-1 )p 1,1 + p 2 1,0 p -1,1 + p 2 0,1 p 1,-1 + p 0,1 p 1,0 > 0.
By setting

2 aα -b √ b 2 -ac = s + 1 s , that is, α = √ b 2 -ac 2a s + 1 s + b a , s ∈ C,
one can easily find β satisfying Eq. ( 26) as follows:

β = α(s) -p 0,1 + δ • α(s) 2p 1,-1 = b 2 -a c 2a ρs + 1 ρs + b a ,
where b, c, ρ are defined in Lemma 9. The proof is then complete.

By the above lemma, one can describe the curve K in [START_REF] Defosseux | Affine Lie algebras and conditioned space-time Brownian motions in affine Weyl chambers[END_REF] as

K = {(α(s), β(s)) : s > 0}.
We remark that at s = 1, the discriminant δ • α(1) = 0, then β(1) is the double root of Eq. ( 26) as an equation of β. Similarly, at s = 1/ρ, α(1/ρ) is a double root of Eq. ( 26) as an equation of α. Thus, defining K 0 as the analogue of G 0 (see [START_REF] Einmahl | Extensions of results of Komlos, Major, and Tusnady to the multivariate case[END_REF]) before the exponential change of variable, one easily deduces the description 23) and [START_REF] Dreyfus | Walks in the quarter plane: genus zero case[END_REF], with (α 0 , β 0 ) ∈ K 0 , can be described explicitly as

K 0 = {(α(s), β(s)) : s ∈ (1/ρ, 1)}. Proposition 10. Any sequence {(α n , β n )} n∈Z satisfying (21)-(
{(α n , β n )} n∈Z = α(ρ 2n s), β(ρ 2n s) n∈Z ,
with s ∈ (1/ρ, 1). As a consequence, the harmonic function formed by (α n , β n ) n∈Z in (15) becomes

h(i, j) = k∈Z α(ρ 2n s) i β(ρ 2n s) j -α(ρ 2n+2 s) i β(ρ 2n s) j , i, j 1.
Proof. We will prove the proposition by induction argument. Since (α 0 , β 0 ) ∈ K 0 , then (α 0 , β 0 ) = (α(s), β(s)), with s ∈ (1/ρ, 1). Now assume that (α n , β n ) = (α(ρ 2n s), β(ρ 2n s)) with n ∈ Z. Since β n = β(ρ 2n s) = β 1/(ρ 2n s) by the involution and (α n+1 , β n ) is a root of K(α, β), then

α n+1 = α 1 ρ 2n+2 s = α(ρ 2n+2 s).
Since (α n+1 , β n+1 ) is also a root of K(α, β), then β n+1 = β(ρ 2n+2 s). Similarly, the involutions of Lemma 9 also imply that α n-1 = α(ρ 2n-2 s) and β n-1 = β(ρ 2n-2 s). The proof is then complete.

A concrete example. We now return to our simplest walk with the transition probabilities p 1,1 = p 1,-1 = p -1,1 = 1/3 and provide a full proof of the formula (2). The zero set of the associated kernel

K(α, β) = 1 3 (x 2 y 2 + x 2 + y 2 ) -xy
admits the uniformization presented in Lemma 9, with

α(s) = √ 5 s + 1/s , β(s) = √ 5 ρs + 1/(ρs) and ρ = 3 + √ 5 2
.

The sequence {(α n , β n )} n∈Z in Proposition 10 can be expressed as functions of s ∈ (1/ρ, 1):

(α n (s), β n (s)) = √ 5 ρ 2n s + 1/(ρ 2n s) , √ 5 ρ 2n+1 s + 1/(ρ 2n+1 s) , n ∈ Z.
Consequently, this leads to harmonic functions depending on s, denoted by

h s (i, j) := n∈Z α n (s) i β n (s) j -α n+1 (s) i β n (s) j .
Let us look at two specific examples. Firstly, the survival probability (2) is the harmonic function associated to the initial point

α 0 √ 5 -1 2 , β 0 √ 5 -1 2 = (1, 1) ∈ K 0 ,
and can be expressed explicitly as We now give the expression for the normalized positive harmonic function constructed in Proposition 6. Recall that such a function is obtained by differentiating the harmonic function h s (i, j) with respect to the variable log β 0 (s) and evaluating as b 0 → log β 0 (1). (The logarithms appear in our computation because Proposition 6 was stated under (log α, log β)-coordinates instead of (α, β)-coordinates.) We have:

P(τ (i,j) = ∞) = h √ 5-1 2 (i, j) = • • • - 1 5 i 13 j + 1 5 i 2 j - 1 1 i 2 j + 1 1 i 1 j - 1 2 i 1 j + 1 2 i 5 j - 1 13 i 5 j + • • • In this formula, the sequence ρ n s + 1 ρ n s √ 5 n 0 = {1, 1, 2,
h s (i, j) log β 0 (s) -log β 0 (1) s→1 - -→ dh s (i, j) ds ds d (log β 0 (s)) s=1 ,
where

ds d (log β 0 (s)) s=1 = β 0 (1) β 0 (1) = - 3 √ 5 , dh s (i, j) ds s=1 = n∈Z i α n (1)α n (1) i-1 -α n+1 (1)α n+1 (1) i-1 β n (1) j + j α n (1) i -α n+1 (1) i β n (1)β n (1) j-1 , (α n (1), β n (1)) = √ 5 ρ 2n + 1 ρ , √ 5 ρ 2n+1 + 1 ρ 2n+1 , (α n (1), β n (1)) =   - √ 5 ρ 2n -1 ρ 2n ρ 2n + 1 ρ 2n 2 , - √ 5 ρ 2n+1 -1 ρ 2n+1 ρ 2n+1 + 1 ρ 2n+1 2    , n ∈ Z.
In these formulas, we want to point out that the sequence

ρ n + 1 ρ n n 0 = {2,
L 0 = 2, L 1 = 1, L n = L n-1 + L n-2 , n 2,
see A005248. Moreover, the sequence

{u n } n 0 = √ 5 ρ n - 1 ρ n n 0 = {0,
u 0 = 0, u 1 = 5, u n = 3u n-1 -u n-2 , n 2.

Part 2. Green functions and Martin boundary for singular random walks in the quadrant

In this part, we describe the Martin boundary inside the cone and prove that it coincides with the set of harmonic functions constructed in Part 1.

Presentation of the results

A brief account on Green functions and Martin boundary theory. Let us first recall the definition of the Green function of the random walk inside the cone Z 2

>0 , which plays an important role in the construction of the Martin boundary. For x, y ∈ Z 2 >0 , set

G(x, y) = n 1 P(x + S(n) = y, τ x > n) and G(x, y) = n 1 P(x + S(n) = y),
where τ x = inf{n 0 : x + S(n) ∈ Z 2 >0 } ∞, and set x 0 = (1, 1). The Martin kernel associated to the reference point x 0 is then defined as [START_REF] Spitzer | Principles of random walks, 2nd Edition[END_REF] K M (x, y) = G(x, y) G(x 0 , y) .

The Martin boundary ∂

Z 2 >0
M S of the random walk S killed outside of Z 2 >0 is the boundary in the topological space {f : Z 2 >0 → R} (with the topology of point-wise convergence) of the set of maps

K M ( • , y) : y ∈ Z 2 >0 . The space ∂ Z 2 >0
M S is a compact measurable subspace of the set of real functions on Z 2 >0 , see [START_REF] Doob | Discrete potential theory and boundaries[END_REF], and for any non-negative function h on Z 2

>0

which is harmonic with respect to S killed outside Z 2 >0 , there exists a positive measure µ h on ∂

Z 2 >0 M S such that h( • ) = ∂ Z 2 >0 M S K M ( • , ω)dµ h (ω).
One sees from ( 27) that a convenient way to compute the asymptotics of K M ( • , y) as y goes to infinity is to first study the behavior of G(x, y) when y goes to infinity in the cone.

We set Σ = R 2 0 ∩ S 1 , and for u ∈ Σ, denote by φ(u) the unique solution of (28) E exp φ(u), S(1) = 1 and E S(1) exp φ(u), S(1) = r u u := µ u for some r u > 0. Remark that φ is bijective from Σ to G 0 (whose definition may be found in ( 16)), with

φ -1 (a, b) = ∇K(a, b) ∇K(a, b)
for (a, b) ∈ G 0 . For u ∈ Σ, we then write P u for the probability measure on {S(n)} n 1 given by the transition probabilities

P u (S(1) ∈ A) = E exp φ(u), S(1) 1 S(1)∈A
for A ⊂ Z 2 , and we denote by E u the corresponding expectation. We then write Σ u for the covariance matrix of S(1) under P u (recall that we denote by µ u the drift of S(1) under P u , see (28)).

Statement of the main result.

The main result of this section is the following theorem, which gives an asymptotics of G(x, y) when y goes to infinity along any direction in the quarter plane. Denote by e 1 = (1, 0) and e 2 = (0, 1) the two standard basis vectors.

Theorem 11. For any x ∈ Z 2 >0 and {y(n

)} n 1 with |y(n)| n→∞ ---→ ∞,
• for all t 1,

P u |S(n) -nµ u | > tn 4 exp(-η tn).
Proof. Using Markov inequality, we get (denoting S(n) = (S 1 (n), S 2 (n)))

P u |S 1 (n) -nµ u 1 | > tn 2 exp(-nΛ * u,1 (t))
, where Λ * u,1 is the Legendre transform of the function

α ∈ R → log E u e α(S 1 (1)-µ u 1 )
= log E e (α+φ(u) 1 )S 1 (1) -αµ u 1 . An easy computation yields that

Λ * u,1 (t) = Λ 1 (t + µ u 1 ) -φ(u) 1 (t + µ u 1 ),
where Λ 1 is the Legendre transform of α → log E e αS 1 (1) . Since u → µ u 1 and u → φ(u)

1 are C 2 , u → Λ * u,1 (t) is C 2 .
Since for all u ∈ Σ, we have Λ * u,1 (0) = (Λ * u,1 ) (0) = 0, and by convexity (Λ * u,1 ) (t) > 0 for all (u, t) ∈ Σ × R, we deduce the existence of η > 0 such that for all u ∈ Σ and all t 1, Λ * u,1 (t) > ηt 2 . Hence, uniformly on u ∈ Σ, for t 1,

P u |S 1 (n) -nµ u 1 | > tn 2 exp(-nηt 2 ).
Doing the same for S 2 (n) yields the result. Likewise, by strict convexity there exists η > 0 such that Λ * u,1 (t) > η t for t 1, independently of u. The second result is then deduced as the first one.

Finally, we will use a uniform bound on the Green function G on Z 2 , namely, there exists C > 0 such that for all x, y ∈ Z 2 , (30) G(x, y) C.

The existence of such a bound is just a consequence of the transience of the walk {S(n)} n 1 .

Asymptotics of the Green function on the half-space

The most delicate part of the description of the Martin boundary is related to the behavior of the Green function along the half-axes R >0 e 1 and R >0 e 2 . This situation is dealt by first giving an asymptotics of the Green function on the half-planes H 1 and H 2 .

For x, y ∈ H i , denote by (31) τ i x = inf{n 0 : x + S(n) ∈ H i } and introduce the Green kernels on the half-plane

G i (x, y) = n 1 P(x + S(n) = y, τ i x > n) and G i u (x, y) = n 1 P u (x + S(n) = y, τ i x > n).
The goal of this section is to prove the following result. For z ∈ R 2 , denote by z 1 (resp. z 2 ) its first (resp. second) coordinate.

Proposition 14. Let ε, η > 0 be small enough and i ∈ {1, 2}. As y goes to ∞ with

|y i | = o(|y| 1/2+ε ), then uniformly on x ∈ H i with |x| = o(|y| 1/2-η ), G i (x, y) ∼ B(u(y))x ī |y| 3/2 V (y ī)e u(y),x-y ,
where u(y) = y |y| , u → B(u) is a continuous, positive function on Σ and V will be introduced in Proposition 15.

We only prove it for i = 1, the other case being similar. The proof of Proposition 14 is decomposed in three cases, depending on the distance of the endpoint to the horizontal axis. To prove this result, we need two local limit theorems for the walk x + S(n) conditioned to stay in H 1 .

The first local limit theorem concerns the case where the distance of the endpoint x + S(n) to the horizontal axis is similar to the fluctuation scale √ n. The equivalence part of this result is given by [START_REF] Duraj | Martin boundary of random walks in convex cones[END_REF]Lem. 3.2]

P e 1 (x + S(n) = y, τ x > n) ∼ κx 2 V (y 2 ) 2π det(Σ e 1 )n 2 exp - C e 1 (y -nµ e 1 ) 2 2n ,
where V is a positive harmonic function with respect to {-S 2 (n)} n 1 killed at the boundary. Moreover, there exists C > 0 such that for all x, y ∈ H 1 with |x| = o( √ n),

P e 1 (x + S n = y, τ x > n) C x 2 V (y 2 ) n 2 exp - C u (y -nµ e 1 ) 2 2n + o n -2 .
Remark that in the latter proposition, C only depends on the behavior of |x| √ n . By [START_REF] Denisov | Random walks in cones[END_REF]Lem. 13 (a)], we also have as z goes to ∞ (32)

V (z) ∼ z.

The second local limit theorem deals with the case where the distance of the endpoint x + S(n) to the horizontal axis is large compared to the fluctuation scale √ n.

Proposition 16. For η small enough, as n goes to infinity, uniformly on x, y ∈ H

1 with |x| = o( √ n) and u ∈ Σ with n -1/2 = o(u 2 ) and u 2 n -1/2+η , P u (x + S(n) = y, τ x > n) = c(u)x 2 µ u 2 Σ u 11 2πn det(Σ u ) exp - C u (y -nµ u ) 2 2n + o x 2 µ u 2 n ,
with o(•) uniform on u and x, y ∈ H 1 . Moreover, there exists C > 0 such that for all x, y ∈ H 1 with |x| = o( √ n),

P u (x + S n = y, τ x > n) Cx 2 µ u 2 t exp - C u (y -nµ u ) 2 2n + o n -2 .
The proof of the latter proposition is very similar to the one of Proposition 15 and is given in Appendix A. 5.1. Asymptotics at mesoscopic distance. Since the support of the random walk S is {(i, j) ∈ Z 2 : i, j -1} by our assumptions (ii), we have almost surely S 2 (1) -1. Such a random variable is called skip-free in the literature [START_REF] Brown | Some results for skip-free random walk[END_REF] and has interesting properties allowing to explicitly compute the survival probability.

For z > 0, denote by τ z = inf{n 0 : z + S 2 (n) = 0}, and remark that τ 1 x = τ x 2 for x ∈ H 1 , see (31). Then, the skip-free assumption yields that

(SF1) if µ u 2 > 0, P u (τ x 2 < ∞) = P u (τ 1 < ∞) x 2 = c x 2 u
, where c u ∈ (0, 1) is the unique real in (0, 1) such that E u (c S 2 (1) ) = 1; (SF2) as µ u 2 goes to 0, one has 1

-c u ∼ 2 µ u 2 Σ u 22 .
We have the following first asymptotic result for the Green function on H 1 , when the endpoint remains far from the horizontal axis.

Lemma 17. Let ε > 0. For any x ∈ H 1 , as t → ∞, uniformly on u ∈ Σ with µ u 2 t -1/2+ε , √ t G u (x, tµ u ) ∼ A(u)P u (τ 1 = ∞) ∼ B(u)x 2 u 2 .
Proof. For x ∈ H 1 , we have τ 1 x = inf{n 0 :

x 2 + S 2 (n) 0}, see (31). Set t u = tµ u . Then, for n t = t 1-ε , G u (x, t u ) = nt n=1 P u x + S(n) = t u , τ 1 x > n + E u G u (x + S(n t ), t u ), τ x 2 > n t = nt n=0 P u (x + S(n) = t u , τ x 2 > n) + E u G u (x + S(n t ), t u ), τ x 2 > n t -E u G u (x + S(τ x ), t u ), n t < τ x 2 < ∞ =: S 1 + S 2 + S 3 .
By Lemma 13,

S 1 nt n=1 P u (|S(n) -nµ u | | tµ u -x -nµ u |) nt n=1 P u |S(n) -nµ u | > |tµ u | 2n n 4 nt n=1 exp - η|tµ u | 2 (2n) 2 n t exp - η|tµ u | 2 (2n t ) 2 .
Hence, since

n t t 1-ε , S 1 = O exp(-ct 2ε )
for some c > 0. Then, set α = t 1/2-ε/4 and write

S 2 = E u G u (x + S(n t ), t u ), |S(n t ) -n t µ u | α, τ 1 x > n t + E u G u (x + S(n t ), t u ), |S(n t ) -n t µ u | > α, τ 1 x > n t =: S 21 + S 22 .
First, by (30) and Lemma 13, (33)

S 22 CP u (|S(n t ) -n t µ u | > α) 4C exp(-η(α/n t ) 2 n t ) = O(exp(-ηt ε/2 )).
Then, on the event {|S(n t ) -n t µ u | α}, as t goes to infinity |x+S(nt)-ntµ| t = o(t 1/2-ε ). Hence, by Theorem 12, uniformly on x + S(n t ) on the event {|S(n t )| α},

√ t G u (x + S(n t ), t u ) t→∞ ---→ A(u).
Hence,

S 21 ∼ t→∞ A(u) √ t P u |S(n t ) -n t µ u | α, τ 1
x > n t as t goes to ∞. Split the latter probability as

P u |S(n t ) -n t µ u | α, τ 1 x > n t = P u |S(n t ) -n t µ u | α, n t < τ 1 x < ∞ + P u τ 1 x = ∞ -P u |S(n t ) -n t µ u | > α, τ 1 x = ∞ .
First,

P u |S(n t ) -n t µ u | > α, τ 1 x = ∞ P u (|S(n t ) -n t µ u | > α) = O(exp(-ηt ε/2 ))
as in (33). Then, on the event {|S(n t ) -n t µ u | α, τ x > n t }, v := x + S(n t ) satisfies

v 2 x 2 + n t µ u 2 -α ct 1-ε µ u 2 -t 1/2-ε/4 ct 1/2
for some constant c > 0, thanks to the condition µ u 2 t -1/2+ε . Hence, by (SF1) and (SF2), on the event {|S(n t ) -n t µ u | α, τ 1

x > n t }, we have

(34) P τ 1 x+S(nt) < ∞| |S(n t ) -n t µ u | α, τ 1 x > n t 1 -c 1 µ u 2 Σ u 22 c 2 t 1/2 exp(-ct ε )
for some constant c > 0, where we used again that µ u 2 t -1/2+ε in the last inequality. This implies that

P u |S(n t )-n t µ u | α, n t < τ 1 x < ∞ = O exp(-ct ε ) . Finally, by (SF1) and (SF2) with µ u 2 = o(1), µ u 2 t -1/2+ε , P u (τ x 2 = ∞) = 1 -c x 2 u 1 -1 -ct -1/2+ε x 2 c x 2 t -1/2+ε
for some constants c, c depending on x, and we finally have

S 21 ∼ t→∞ A(u) √ t P u (τ 1 x = ∞) and then S 2 ∼ t→∞ A(u) √ t P u (τ 1 x = ∞).
Note that the latter term will be the main contribution to the asymptotics of G u (x, t u ). Finally, by (30),

S 3 = E u G u (x + S(τ x ), t u ), n t < τ 1 x < ∞ CP u n t < τ 1 x < ∞ C P u |S(n t ) -n t µ u | α, n t < τ 1 x < ∞ + P u |S(n t ) -n t µ| > α .
By (34), the first term is bounded by exp(-ct ε ) and by Lemma 13 the second term is bounded by exp(-ηt ε/2 ). Hence,

S 3 = o exp(-ηt ε/2 ) .
Putting together the results on S 1 , S 2 and S 3 yields then

G u (x, t u ) ∼ t→∞ A(u) √ t P u (τ 1 x = ∞)
uniformly on u such that u 2 t -1/2+ε . The second equivalence in Lemma 17 is a consequence of (SF1) and (SF2). 5.2. Asymptotics of the Green function at microscopic distance. The proof essentially follows the original proof of Ney and Spitzer [START_REF] Ney | The Martin boundary for random walk[END_REF].

Proposition 18. Let (θ t ) t 0 a positive function going to zero at ∞ and ε > 0. For η > 0 small enough, as t goes to ∞, uniformly on u ∈ Σ with

t -1/2 θt u 2 t -1/2+η and x ∈ H 1 with |x| = o(t 1/2-ε ), √ t G u (x, tµ u ) ∼ B(u)x 2 u 2 .
Proof. Write y = tµ u and split G u (x, tµ u ) as

G u (x, tµ u ) = t-A √ t -1 n=1 P u x + S(n) = y, τ 1 x > n + t+A √ t n= t-A √ t P u x + S(n) = y, τ 1 x > n + ∞ n= t+A √ t +1 P u x + S(n) = y, τ 1 x > n := S 1 + S 2 + S 3 .
First, by the assumption on

u 2 , n -1/2 = o(u 2 ) and u 2 Cn -1/2+η for t -A √ t n t + A √ t ,
= o( √ n) and t -A √ t n t + A √ t and u ∈ Σ with t -1/2 θt u 2 t -1/2+ε , P u x + S(n) = y, τ x > n = c(u)x 2 µ u 2 Σ u 11 2πn det(Σ u ) exp - C u (y -nµ u ) 2 2n + o x 2 µ u 2 n .
Applying the latter to S 2 yields

√ tS 2 = √ t   t+A √ t n= t-A √ t c(u)x 2 µ u 2 Σ u 11 2πn det(Σ u ) exp - C u (y -nµ u ) 2 2n + o x 2 µ u 2 n   = √ t c u x 2 µ u 2 Σ u 11   t+A √ t n= t-A √ t 1 2πn det(Σ u ) exp - C u (y -nµ u ) 2 2n   + x 2 µ u 2 o (1)
.

By [25, Eq. (2.20)], √ t t+A √ t n= t-A √ t 1 2πn det(Σ u ) exp - C u (y -nµ u ) 2 2n t→∞ ---→ 1 2π det(Σ u ) C u µ u + ε(A),
uniformly in u ∈ Σ, with ε(A) A→∞ ----→ 0 and |x| = t 1/2-ε . Hence, we also have

(35) √ tS 2 ∼ t→∞ c u x 2 µ u 2 Σ u 11 2π det(Σ u ) C u µ u (1 + ε(A)).
Then, for 0 < δ < 1, split S 1 as

S 1 = δt n=1 P u x + S(n) = y, τ 1 x > n + t-A √ t -1 n= δt +1 P u x + S(n) = y, τ 1 x > n =: S 11 + S 12 .
By Lemma 13, the first term is bounded above as

√ t x 2 µ u 2 S 11 √ t δt n=1 P u x + S(n) = y Ct 3/2 exp -c(1 -δ)t t→∞ ---→ 0
uniformly in |x| < t 1/2-ε , A > 0 and u ∈ Σ (see [25, Eq. (2.22)]). By Proposition 16 with r = 3, the second term is bounded as

√ t x 2 µ u 2 S 12 √ t t-A √ t -1 n= δt +1 Cx 2 µ 2 n exp -c u y -nµ u 2 + o n -2 √ t t-A √ t -1 n= δt +1 Cx 2 µ 2 n exp -c u y -nµ u 2 + o 1 t . By [25, Eq. (2.26)], lim sup t→∞ √ t t-A √ t -1 n= δt +1 1 n exp -c u y -nµ u 2 = ε (A)
uniformly on u ∈ Σ, with ε (A) going to 0 when A going to ∞. Hence, lim sup

t→∞ √ t x 2 µ u 2 t-A √ t -1 n= δt +1 P u (x + S(n) = y, τ 1 x > n) = ε (A).
Finally, (36) lim sup

t→∞ √ t x 2 µ u 2 S 1 = ε (A),
uniformly on u ∈ Σ and x ∈ H 1 with |x| < t 1/2-ε . One proves similarly that (37) lim sup

t→∞ √ t x 2 µ u 2 S 3 = ε (A),
uniformly on u ∈ Σ and x ∈ H 1 with |x| < t 1/2-ε . Putting (35), ( 36) and (37) together, we get

1 -ε(A) -2ε (A) < lim inf √ t G u (x, tµ u )Σ u 11 2π det(Σ u ) C u µ u c u x 2 µ u 2 lim sup √ t G u (x, tµ u )Σ u 11 2π det(Σ u ) C u µ u c u x 2 µ u 2 < 1 + ε(A) + 2ε (A)
for all A > 0. Letting A go to ∞, we thus get

lim t→∞ √ t G u (x, tµ u ) x 2 µ u 2 = c u Σ u 11 2π det(Σ u ) C u µ u uniformly on u ∈ Σ and x ∈ H 1 with |x| = o(t 1/2-ε ).
We give a similar asymptotic result along the horizontal axis.

Proposition 19. Let B > 0. Uniformly on x ∈ H 1 with |x| = o(t 1/2-ε ), y = tµ e 1 + t e 2 with 0 t B √ t,
G e 1 (x, y) ∼ 1 |y| 3/2 B(e 1 )V (y 2 )x 2 e -φ(u(y))-φ(e 1 ),y , with V introduced in (32) and B(e 1 ) > 0.

Proof. The proof follows the lines of the one of Proposition The result is then deduced.

Identification of the Martin Boundary

We can now describe the Martin boundary ∂ M >0 is homeomorphic to G 0 through the map

(a 0 , b 0 ) ∈ G 0 → h (a 0 ,b 0 ) h (a 0 ,b 0 ) (1, 1) ∈ ∂ Z 2 >0 M S.
More precisely, for (a 0 , b 0 ) ∈ G 0 , writing u = ∇K(a 0 ,b 0 ) ∇K(a 0 ,b 0 ) ,

h (a 0 ,b 0 ) (x) = e (a 0 ,b 0 ),x P u (τ x = ∞) = e (a 0 ,b 0 ),x -E e (a 0 ,b 0 ),x+S(τx) , τ x < ∞ and if (a 0 , b 0 ) ∈ ∂G 0 , with e i = ∇K(a 0 ,b 0 ) ∇K(a 0 ,b 0 ) , i ∈ {1, 2}, h (a 0 ,b 0 ) (x) = x īe (a 0 ,b 0 ),x -E e (a 0 ,b 0 ),x+S(τx) (x ī + S ī(τ x )) .
Proof of Theorem 11. We only prove the first statement and sketch the proof of the second one, which is similar to the first proof. Let V (u 0 ) be a neighborhood of

u 0 ∈ Σ such that d(V (u 0 ), ∂Σ) > 0. Set t n = |y(n)| |µ u(y(n)) |
and u n = y(n) |y(n)| . Then, for m n 1 to adjust later,

G un (x, y(n)) = mn m=1 P un (x + S(m) = y(n), τ x > m) + E un (G un (x + S(m n ), y(n)), τ x > m n ) = mn m=0 P un (x + S(m) = y(n), τ x > m) + E un G u (x + S(m n ), y(n)), τ x > m n -E un G u (x + S(τ x ), y(n)), m n < τ x < ∞ (38) =: S 1 + S 2 + S 3 .
By the second part of Lemma 13, as long as m n < t n /3 and for |y(n)| large enough

S 1 mn n=1 P un |S(m) -mµ u(n) | t n µ u(n) -x -mµ u(n) mn n=1 P un |S(m) -mµ u(n) | > |t n µ u(n) | 2m m mn n=1 exp - η |tµ u(n) | 2m m n exp - η |t n µ u(n) | 2m n .
Hence, as m n = o(t

1/4 n ), ( 39
) S 1 = o(exp(- √ t n )).
Then, write

S 2 =E un G un (x + S(m n ), y(n)), |S(m n ) -m n µ u(n) | 2m n , τ x > m n + E un G un (x + S(m n ), y(n)), |S(m n ) -m n µ u(n) | > 2m n , τ x > m n =S 21 + S 22 .
First, by (30) and Lemma 13,

S 22 CP un (|S(m n ) -m n µ u(n) | > 2m n ) C exp(-2η m n ).
Then, set u = y(n)-(x+S(mn)) |y(n)-(x+S(mn))| ∈ S 1 . There exists t 0 such that on the event {|S(m n )nµ u | 2m n }, for t n t 0 , u ∈ V (u 0 ). Then, by Theorem 12, as t n goes to infinity with

m n = o(t 1/4 n ), uniformly on x + S(m n ) on the event {|S(m n )| 2m n } and using the fact that u n → u 0 , √ t G un (x + S(m n ), y(n)) t→∞ ---→ A(u 0 ).
Hence,

S 21 ∼ n→∞ E un A(u 0 ) √ t n 1 |S(mn)-mnµ u(n) | 2mn,τx>mn = A(u 0 ) √ t n P un |S(m n ) -m n µ u(n) | 2m n , τ x > m n .
Since

P un |S(m n ) -m n µ u(n) | 2m n , τ x > m n = P un (τ x > m n ) -P un |S(m n ) -m n µ u(n) | > 2m n , τ x > m n , with P un (τ x > m n ) ∼ n→∞ P un (τ x = ∞) ∼ n→∞ P u 0 (τ x = ∞) and P un (|S(m n ) -m n µ u(n) | > 2m n , τ x > m n ) exp(-2η m n ) by Lemma 13, S 21 ∼ n→∞ A(u)P u 0 (τ x = ∞) √ t n ,
and as long as m n = o(t

1/4 n ), and log(t n ) = o(m n ), ( 40
) S 2 ∼ t→∞ A(u 0 )P u 0 (τ x = ∞) √ t .
As in the proof of Lemma 17, the latter term will be the main contribution to the asymptotics of G un (x, y(n)). Finally, by (30),

S 3 = E un G un (x + S τx , y(n)), m n < τ x < ∞ CP un [m n < τ x < ∞] C P un |S(m n ) -m n µ u(n) | cm n , m n < τ x < ∞ + P un (|S(m n ) -m n µ| > cm n ) ,
for c small enough so that inf u∈V (u 0 ) d(µ u(n) , ∂R 2 >0 ) := d > c. On the one hand, it follows from Lemma 13 that there exists η such that

P un (|S(m n ) -m n µ u(n) | > cm n ) exp(-η m n )
for all m n 1 and u ∈ V (u 0 ). On the other hand,

P un (|S(m n ) -m n µ u(n) | cm n , m n < τ x < ∞) sup y∈B(x+S(mn),cm n ) P un (τ y < ∞) sup y∈B(x+S(mn),cm n ) P un (τ i y < ∞),
for i ∈ {1, 2}, where τ i y = inf{n 0 :

x i + S i (n) 0}.
Since S is skip-free and there exists µ i > 0 such µ u i µ i for all u ∈ V (u 0 ), by (SF1) and (SF2) there exists c u 0 > 0 independent of u ∈ V (u 0 ) such that P un (τ i y < ∞) c y u 0 . Hence, by the choice of c we have

(41) sup y∈B(x+S(mn),cm n ) P un (τ y < ∞) exp(-c u 0 (d -c)m n ).
Finally, there exists c > 0 independent of u ∈ V (u 0 ) such that S 3 exp(-cm n ), and as long as log(

t n ) = o(m n ), (42) S 3 = o( √ t n ).
Putting (39), ( 40) and (42) together yields x ), y), τ ī x < τ i x ) in (38). The main term S 21 is then asymptotically equivalent to

√ t n G un (x, y(n)) ---→ t→∞ A(u 0 )P u 0 (τ x = ∞),
A(e i )V (y 2 (n))f (y 2 / |y(n)|) |y(n)| 3/2 E e i (x i + S i (m n ), τ x > m n ).
Then, using that x ī is harmonic for S killed at τ i under P e i (because S is skip-free),

E e i (x ī + S ī(m n ), τ x > m n ) = E e i (x ī + S ī(m n ), τ i x > m n ) -E e i (x ī + S ī(m n ), τ i x > m n τ ī) = x ī -E e i E e i x ī + S ī(τ ī) + S ī(m n -τ ī), τ i x+S(τ ī x ) > m n -τ ī x , τ ī x m n , τ ī < τ i x ) ,
where S has the same law of S and is independent of τ ī. By harmonicity of x ī for S killed at τ i , we get

E e i x ī + S ī(τ ī) x + S ī(m n -τ ī x ), τ i xī+Sī(τ ī x ) > m n -τ ī x = x ī + S ī(τ ī x ),
so that

E e i (x ī + S ī(m n ), τ x > m n ) = x ī -E e i x ī + S ī(τ ī), τ ī m n , τ ī x < τ i x . Since P(m n < τ ī x < ∞) = o(exp(-t ε n ))
for some ε > 0 as in (41), we have

E e i (x ī + S ī(m n ), τ x > m n ) ∼ x ī -E e i x ī + S ī(τ ī x ), τ ī x < τ i x = x ī -E e i (x ī + S ī(τ x )) = -E e i (S ī(τ x )) ,
where we used that x ī + S ī(τ i x ) = 0 because S is skip-free.

Proof of Corollary 20. By Theorem 11, if {y

(n)} n 0 is such that |y(n)| → ∞ with y(n) |y(n)| → u ∈ Σ, then if u ∈ {e 1 , e 2 }, for x ∈ Z 2 >0 we have K(x, y(n)) → e φ(u),x-x 0 P un (τ x = ∞) P un (τ x 0 = ∞) = C(u)e φ(u),x P un (τ x = ∞) = C(u) e φ(u),x -E e φ(u),x+S(τ ) , τ < ∞ (43) := hφ(u) (x), for some C(u) > 0 and if u = e i , i ∈ {1, 2}, K(x, y(n)) → e φ(e i ),x-x 0 x i -E e i (x i + S i (τ x )) x 0i -E e i (x 0i + S i (τ x 0 )) = C(e i ) x i e φ(e i ),x -E e φ(e i ),x+S(τ ) (x i + S i (τ x )) := hφ(e i ) (x), (44)
for some C(e i ) > 0. Remark that as d(x, ∂Z >0 ) → ∞, P un (τ x = ∞) → 1 and

E e i (x ī + S ī(τ i x ), τ ī x > τ i x ) → E e i (x ī + S ī(τ i x )) = x i , so that if u ∈ {e 1 , e 2 }, hφ(u) (x) ∼ d(x,∂Z >0 )→∞ e φ(u),x and if u = e i , i ∈ {1, 2}, hφ(e i ) (x) ∼ d(x,∂Z >0 )→∞ x īe φ(e i ),x . Since u → φ(u) is injective, we deduce that u → hφ(u) is injective. Hence, Φ : u → hφ(u) is a bijection from Σ to ∂ Z 2 >0 M S.
Remark that Φ is continuous: this is clear from (43) on Σ \ {e 1 , e 2 }, and (44) for e 1 is obtained as a limit of (43) as u goes to e 1 (the derivative appearing because of the term C(u)). Since Σ is compact and Φ is a continuous bijection,

Φ is a homeomorphism from Σ to ∂ Z 2 >0 M S. It remains to prove that h φ(u) h φ(u) (1,1) = hφ(u) . Let us first suppose that u ∈ Σ \ {e 1 , e 2 }. The function h φ(u) is harmonic in Z 2
>0 , so that h φ(u) = Σ hφ(w) dm u (w) for some positive measure m u on Σ. We have h φ(u) (x) ∼ e φ(u),x as d(x, ∂Z >0 ) → ∞. Hence, since hφ(w) (x) ∼ e φ(w),x (resp. hφ(e i ) (x) ∼ x īe φ(e i ),x as d(x, ∂Z >0 ) → ∞ for w ∈ Σ \ {e 1 , e 2 } (resp. w = e i , i ∈ {1, 2}) we must have m u = cδ u for come constant c. Since

h φ(u) (1,1) h φ(u) (1,1) = hφ(u) (1, 1) = 1, c = 1 and thus h φ(u) h φ(u) (1, 1)
= hφ(u) .

Then for i ∈ {1, 2}, by continuity of Φ and Proposition 6,

h φ(e i ) h φ(e i ) (1, 1) = lim u→e i h φ(u) h φ(u) (1, 1) = lim u→e i hφ(u) = hφ(e i ) .
Since φ is a homeomorphism from Σ to G 0 with φ -1 (a 0 , b 0 ) = ∇K(a 0 ,b 0 ) ∇K(a 0 ,b 0 ) , we get the statement of the corollary.

Appendix A. Proof of Proposition 16

The proof of Proposition 16 is very similar to the one of Theorem 5 in [START_REF] Denisov | Random walks in cones[END_REF] which provides a local limit theorem for zero-drift random walks in cones (the cone being a half-space for us) : the two differences are the presence of a small drift in our case and the fact that we are dealing with walks with small negative jumps in the vertical direction, which greatly simplifies the proof. The first step is to get good estimates on the heat kernel of the Brownian motion with drift inside R × R >0 .

Lemma 21. Let (B t ) t 0 be a Brownian motion with drift µ and covariance Σ. Denote by τ B

x the exit time from R × R 0 for x + B t . Then, writing C = Σ -1/2 and K B (x, y)dy = dP x + B t = y, τ B x > t , we have for x, y ∈ R × R 0

K B t (x, y) = 1 -e - 2x 2 y 2 tΣ 11 exp -C(y-x-tµ) 2 2t
2πt det(Σ) .

In particular, for all x, y ∈ R × R 0 Proof. This lemma is a consequence of Girsanov theorem and the reflection formula. For v ∈ R 2 , denote by s v the orthogonal symmetry with respect to a one-dimensional vector space generated by v, so that s v (x) = x -2 x,v ⊥ v ⊥ ,v ⊥ v ⊥ for any v ⊥ = 0 with v ⊥ , v = 0.

K B t (x,
Recall that if W is a standard two-dimensional Brownian motion with zero drift and τ v x = inf{t 0 : x + W t ∈ Rv}, the reflection formula yields dP(x + W t = y, τ v x > t) The rest of the lemma is deduced by using the formula 1 -e -u u for u ∈ R and 1 -e -u ∼ u as u goes to 0.

= 1 2πt exp - y -x 2 2t -exp - s v (y) -x 2 2t = 1 2πt   exp - y -x 2 2t -exp   - y -x -2 y,v ⊥ v ⊥ ,v ⊥ v ⊥ 2 2t     = 1 2πt   exp - y -x 2 2t -exp   - y -x 2 -4 y-x,v ⊥ • y,v ⊥ v ⊥ ,v ⊥ + 4 y,v ⊥ 2 v ⊥ ,v ⊥ 2t     = 1 -e - 2 y,v ⊥ x,v ⊥ t v ⊥ ,v
We next turn to the proof of Proposition 16.

Proof of Proposition 16. The proof of this proposition is exactly the same as the one of [START_REF] Denisov | Random walks in cones[END_REF]Thm 5] and [START_REF] Duraj | Martin boundary of random walks in convex cones[END_REF]Prop. 2.4]. We introduce the main steps without proving them in details, since there are exactly the same as in the aforementioned references.

Conditioned central limit theorem starting away from the boundary. Since S(n) has all moments uniformly bounded with respect to u, by [START_REF] Einmahl | Extensions of results of Komlos, Major, and Tusnady to the multivariate case[END_REF], for all r 3, γ > 0 there exists γ > 0 small enough and a coupling between S n -nµ and a Brownian motion B u with variance Σ u such that P( sup

0 t n |S t -tµ -B u t | > n 1/2-γ ) Cn -r ,
with C independent of u ∈ Σ. As in [START_REF] Denisov | Random walks in cones[END_REF]Lem. 20] we deduce from this by choosing γ = 1/4 + 2ε and using Lemma 21 that for any A ⊂ R 2 compact with non-empty interior, uniformly on x ∈ H 1 with x 2 n 1/4-ε with ε small (only depending on γ, r) , for some C, c > 0 and r large enough.

P u (x + S(n) ∈ nµ + √ nA, τ 1 x > n) ∼ nµ+ √ nA∩R×R 0 K B u n (x,
Conditioned central limit theorem starting close the boundary. Introduce the stopping times ν x n = inf(m 1, x 2 + S 2 (m) n 1/4-ε ). Then, as in [11, Lemma 14] P(ν n n 1/2-ε ) = O(e -Cn ε ) and by Lemma 13, P(ν n n 1/2-ε , |S(ν n ) -ν n µ u | n 1/4-ε/4 ) 1 -exp(-Cn -ε ) for some ε > 0. Hence, we have by (45)

P u (x + S(n) ∈ nµ + √ nA, τ 1 x > n) =E P x + S(ν x n ) ∈ x + S(ν x n ) + (n -ν x n )µ + √ n(A + n -1/2 (ν x n µ -S(ν x n )), τ 1 x+S(ν x n ) > n -ν x n , ν x n < τ 1 x =E P x + S(ν x n ) ∈ x + S(ν x n ) + (n -ν x n )µ + √ n(A + o(n -1/2-ε 4 )), τ 1 x+S(ν x n ) > n -ν x n , ν n < n 1/2-ε , ν x n < τ 1 x + o exp -Cn -ε ∼ c(u)E x 2 + S 2 (ν x n ), ν n < n 1-ε , ν x n < τ 1 x µ 2 2πn Σ u 11 det(Σ u ) A exp - C u y 2 2 dy,
uniformly on u ∈ Σ and x ∈ H 1 , x = o( √ n). Then, using that P(|x 2 + S 2 (n)|, ν n n 1/2-ε ) = O(e -Cn ε ) as in [START_REF] Denisov | Random walks in cones[END_REF]Lem. 16] and the fact that S 2 (k) -kµ u 2 is a martingale, ν x n ∧ τ 1

x is a stopping time and x 2 + S 2 (τ 1 x ) = 0 (because S is skip-free), we get

E x 2 + S 2 (ν x n ), ν n < n 1/2-ε , ν x n < τ 1 x = E x 2 + S 2 (ν x n ∧ n 1/2-ε ), ν x n < τ 1 x + O(e -Cn ε ) = E x 2 + S 2 (ν x n ∧ τ 1 x ∧ n 1/2-ε ) + O(e -Cn ε ) = x 2 + µ u 2 E(ν x n ∧ τ 1 x ∧ n 1/2-ε ) + O(e -Cn ε ).
Since µ u 2 n -1/2+η , for η small enough we have Similarly, we get for x ∈ H 1 and t 0,

µ u 2 E(ν x n ∧ τ 1 x ∧ n 1/2-ε ) n -1/2+η
P u (|S(n) -nµ u | √ nt, τ 1 x > n) Cx

Figure 3 .

 3 Figure 3. The model p 1,1 = 5/6, p 1,-1 = p -1,1 = 1/12: the interior of K (left) is not convex, while the interior of G (right) is convex.

Figure 4 .

 4 Figure 4. Left: an example of curve G (blue color) and the subpart (yellow) G 0 corresponding to the Martin boundary. Right: an example of the compensation approach procedure.

Lemma 9 .

 9 One has{(α, β) ∈ C 2 : K(α, β) = 0} = {(α(s), β(s)) : s ∈ C} , a := 1 -4p -1,1 p 1,-1 , ρ := 1+

) 2 2n ∼|y| e 2 + O y 2 |y| 2 .

 2n2 A(u(y)) exp(-φ(u(y)) -φ(e 1 ), y ) + ε(A), with ε(A) going to 0 as A goes to infinity. Hence, lim A→∞ lim t→∞ √ tS A = A(u(y)) exp(-φ(u(y)) -φ(e 1 ), y ). Concluding the proof as the one of Proposition 18, we get G e 2 (x, y) ∼ t→∞ B(e 2 ) |y| 3/2 V (y 2 )x 2 e -φ(u(y))-φ(e 1 ),y , with B(e 2 ) = κ A(e 2 ). Proof of Proposition 14. When |y| = o(y 2 ), the statement of Proposition 14 is obtained by putting together Lemma 17 and Proposition 18 with u = u(y) and t = |y| |µ u(y) | , using that µ u(y) 2 ∼ y 2 |y| and V (z) ∼ z as z goes to infinity. When y 2 B|y 1 | 1/2 for some B > 0, use Proposition 19 with t = y 1 |µ e 1 | ∼ |y| |µ e 1 | to get G(x, y) ∼ B(e 1 )x 2 V (y 2 ) |y| 3/2 e φ(e 1 ),y -φ(u(y)),y with B(e 1 ) > 0. The result is deduced provided we prove that φ(e 1 ), y = φ(u(y)), x + o(1) uniformly on y = o(|x| 1/2-ε ). As y goes to infinity, u(y) = e 1 + y 2 |y| e 2 + o y 2 |y| . Since φ is C 2 , K(φ(u)) = 1 and ∇K • φ(u) ∈ Ru. For δ > 0 small enough with u + δ ∈ Σ, we have φ(u + δ) = φ(u) + u ⊥ h u (δ) + O(|δ| 2 ) for some linear functional h u . Hence, φ(u(y)) = φ(e 1 ) + α y 2 Therefore, for x = o(|y| 1/2-ε ), φ(u(y)), x = φ(e 1 ), y + o y 2 |y| 1/2+ε = φ(e 1 ), y + o(1).

Z 2 >0S. 2 >0M

 22 We first prove Theorem 11, and then deduce from the latter the following description: Corollary 20. The Martin boundary ∂ Z S of {S(n)} n 0 killed outside of the quarter plane Z 2

  and setting B(u 0 ) = A(u 0 )|µ u 0 | yields the desired limit for G un , since t n = |y(n)| |µ un | ∼ n→∞ |y(n)| |µ u 0 | . Using (29) yields then the final result. The proof of the second statement is similar. The unique change is to use Proposition 14 instead of Theorem 12, and the renewal relation G(x, y) = G i (x, y) -E( G(x + S(τ ī

  C)e Cµ,C(x-y) -t Cµ 2 2 dP(Cx + B t = Cy, τ Ce 1 x > t).Then, since Σ 1/2 e 2 , Ce 1 = 0, we can apply the previous computation to getK B t (x, y) =(det C)e Cµ•(x-y)-Cx 2 -2t C(y -x), Cµ + t 2 Cµ, Cµ 2t -x -tµ) 2 2t.

2 µ 2 o µ u 2 x 2 n

 22 exp -c u t 2 dy + o(n -r ).Local limit theorem starting close the boundary. Using the same proof as the one of[START_REF] Denisov | Random walks in cones[END_REF] Thm 5] with the local limit theorem from Spitzer [27, Ch. 7 P10] which is valid for any random walk whose steps set generates Z 2 , we deduceP u (x + S(n) = y, τ 1 x > n) = c(u)x 2 µ 2 Σ u 11 2πn det(Σ u ) exp -C u (y -nµ u ) 2 2n +with o(•) uniform on u and x, y ∈ H 1 with |x| √ n, andP u (x + S(n) = y, τ 1 x > n) Cx 2 µ 2 nexp -c u y -nµ u 2 dy + o n -(1+r) .

  with o(•) and C only depending on A and θ t . Hence, by Proposition 16, uniformly for x, y ∈ H 1 with |x|

  18, using Proposition 15 instead of Proposition 16. The only different step is the computation of ( tµ e 1 + t e 2 -nµ e 1

	By [25], uniformly on t	B	√	t,
	lim t→∞	t+A n= t-A √ √ t	t	1 2πn det(Σ e 1 )	exp -	C e 1
					√ t	 	t+A n= t-A √ √ t	t	1 2πn det(Σ u )	exp -	C u ( tµ u -nµ u ) 2 2n	 
	which has to be replaced by
	√	t	  n= t-A t+A √ √ t	t	1 2πn 2 det(Σ e 1 )	exp -	C e 1 ( tµ e 1 + t e 2 -nµ e 1 ) 2 2n	  := S A .
	As t goes to infinity and for t -A	√	t	n	t + A √	t , n ∼ t, and thus
	S A ∼	1 √ t	  n= t-A t+A √ √ t	t	1 2πn det(Σ e 1 )	exp -	C e 1 ( tµ e 1 + t e 2 -nµ e 1 ) 2 2n	  .

  Write B t = CB t and B t = B t -tCµ, so that B is a Brownian motion with covariance identity and drift Cµ and B is a standard Brownian motion. Using the change of variable u → Cu and Girsanov theorem applied to B, we getK B t (x, y) = dP(x + B t = y, τ B x > t) = (det C)dP(Cx + B t = Cy, τ Ce 2

	2πt	⊥	exp -	y -x 2 2t	.

  (remark that we use the fact that nµ + √ nA ∈ R × R 0 for n large enough, due to the condition n -1/2 = o(u 2 )) and (46)P u (|S(n)| t, τ 1 x > n) Cx 2 µ 2 exp -ct 2 dy + O(n -r ),

				y)dy
	(45)	∼	c(u)x 2 µ 2 2πΣ u 11 det(Σ u ) A	exp -	C u y 2 2	dy,

  n 1/2-ε = o(1),uniformly on u ∈ Σ in the range of the Lemma. Finally, uniformly on such u and x ∈ H 1 ,

	|x| = o( √	n),
	P u (x + S(n) ∈ nµ +	√	nA, τ 1 x > n) ∼ n→∞	c(u)x 2 µ 2 2πnΣ u 11 det(Σ u ) A	exp -	C u y 2 2	dy.

Acknowledgments

Part of this work has been written during a stay at the University of Münster of the first and second authors, who would like to thank the Institute for Mathematical Stochastics for the very good working conditions. We thank the editors for their kind invitation to submit a paper to this special issue in honour of J. W. Cohen. We further thank Ivo Adan, Onno Boxma, Johan van Leeuwaarden and Michel Mandjes for continuous and fruitful interactions since 2010. We thank Ivo Adan for useful discussions at an early stage of the project. KR thanks Alin Bostan and Lucia Di Vizio for interesting discussions on singular random walks.

This project has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme under the Grant Agreement No. 759702, from Centre Henri Lebesgue (programme ANR-11-LABX-0020-0), from the ANR DeRerumNatura (ANR-19-CE40-0018) and ANR CORTIPOM (ANR-21-CE40-0019).

• if y (n) |y(n)| = u n with lim u n := u 0 ∈ Σ \ {e 1 , e 2 }, then lim n→∞ |y(n)|e -φ(un(y(n))),x-y(n) G(x, y(n)) = A(u 0 )P u 0 (τ x = ∞);

• if y(n) |y(n)| = u n with lim u n = e i , i ∈ {1, 2}, then, with ī = 3 -i,

with u → A(u) a continuous, positive function on Σ and V introduced in Proposition 15.

The proof of this theorem is postponed to Section 6. The main difficulty concerns the case of an asymptotic direction along the boundary axes, where the survival probability vanishes. The latter case is solved by first studying the Green function of the random walk in the half-planes H 1 := Z × Z >0 and H 2 := Z >0 × Z. We achieve the latter in Section 5.

To conclude this section, we collect a few useful estimates on the classical random walk on Z 2 . 4.3. Preliminary estimates, and the Ney and Spitzer theorem. For all u ∈ Σ, we introduce the modified Green kernels

For all x, y ∈ Z 2 >0 , we have (29) G u (x, y) = e φ(u),y-x G(x, y) and G u (x, y) = e φ(u),y-x G(x, y).

We recall the following result from Ney and Spitzer [25, Thm 2], using our notation µ u = E u (S(1)).

Theorem 12. There exists a continuous function A : Σ → R >0 such that, as t → ∞, uniformly on u ∈ Σ and x ∈ Z 2 with |x| = o(t 1/2 ), √ t G u (x, tµ u ) → A(u).

We should emphasize that the initial proof of Ney and Spitzer requires the random walk {S(n)} n 1 to be irreducible, and is valid only for fixed x ∈ Z 2 . However, the only reason for the first requirement is the use of the local large deviation limit theorem [25, Thm 2.1], which has since been proven for any random walk with finite generating function and whose support generates Z 2 , see [START_REF] Spitzer | Principles of random walks, 2nd Edition[END_REF]Ch. 7,P10]. Likewise, the hypothesis of a fixed x can be relaxed to the condition |x| = o(t 1/2 ) (see the proof of Proposition 19 for a similar computation).

We will apply several times the following lemma, which gives a large deviation bound which is uniform on P u , u ∈ Σ. Lemma 13. There exist constants η, η > 0 such that for all u ∈ Σ, n 1,

• for all t ∈ (0, 1), P u |S(n) -nµ u | > tn 4 exp(-ηt 2 n);