Heavy-tailed random walks on complexes of half-lines

We study a random walk on a complex of finitely many half-lines joined at a common origin; jumps are heavy-tailed and of two types, either one-sided (towards the origin) or two-sided (symmetric). Transmission between half-lines via the origin is governed by an irreducible Markov transition matrix, with associated stationary distribution $\mu_k$. If $\chi_k$ is $1$ for one-sided half-lines $k$ and $1/2$ for two-sided half-lines, and $\alpha_k$ is the tail exponent of the jumps on half-line $k$, we show that the recurrence classification for the case where all $\alpha_k \chi_k \in (0,1)$ is determined by the sign of $\sum_k \mu_k \cot ( \chi_k \pi \alpha_k )$. In the case of two half-lines, the model fits naturally on $\mathbb{R}$ and is a version of the oscillating random walk of Kemperman. In that case, the cotangent criterion for recurrence becomes linear in $\alpha_1$ and $\alpha_2$; our general setting exhibits the essential non-linearity in the cotangent criterion. For the general model, we also show existence and non-existence of polynomial moments of return times. Our moments results are sharp (and new) for several cases of the oscillating random walk; they are apparently even new for the case of a homogeneous random walk on $\mathbb{R}$ with symmetric increments of tail exponent $\alpha \in (1,2)$.


Introduction
We study Markov processes on a complex of half-lines R + × S, where S is finite, and all half-lines are connected at a common origin. On a given half-line, a particle performs a random walk with a heavy-tailed increment distribution, until it would exit the half-line, when it switches (in general, at random) to another half-line to complete its jump.
To motivate the development of the general model, we first discuss informally some examples; we give formal statements later.
The one-sided oscillating random walk takes place on two half-lines, which we may map onto R. From the positive half-line, the increments are negative with density proportional to y −1−α , and from the negative half-line the increments are positive with density proportional to y −1−β , where α, β ∈ (0, 1). The walk is transient if and only if α + β < 1; this is essentially a result of Kemperman [15].
The oscillating random walk has several variations and has been well studied over the years (see e.g. [16][17][18][19]). This previous work, as we describe in more detail below, is restricted to the case of two half-lines. We generalize this model to an arbitrary number of half-lines, labelled by a finite set S, by assigning a rule for travelling from half-line to half-line.
First we describe a deterministic rule. Let T be a positive integer. Define a routing schedule of length T to be a sequence σ = (i 1 , . . . , i T ) of T elements of S, dictating the sequence in which half-lines are visited, as follows. The walk starts from line i 1 , and, on departure from line i 1 jumps over the origin to i 2 , and so on, until departing i T it returns to i 1 ; on line k ∈ S, the walk jumps towards the origin with density proportional to y −1−α k where α k ∈ (0, 1). One simple example takes a cyclic schedule in which σ is a permutation of the elements of S. In any case, a consequence of our results is that now the walk is transient if and only if k∈S µ k cot(πα k ) > 0, (1.1) where µ k is the number of times k appears in the sequence σ. In particular, in the cyclic case the transience criterion is k∈S cot(πα k ) > 0. It is easy to see that, if S contains two elements, the cotangent criterion (1.1) is equivalent to the previous one for the one-sided oscillating walk (α 1 + α 2 < 1). For more than two half-lines the criterion is non-linear, and it was necessary to extend the model to more than two lines in order to see the essence of the behaviour.
More generally, we may choose a random routing rule between lines: on departure from half-line i ∈ S, the walk jumps to half-line j ∈ S with probability p(i, j). The deterministic cyclic routing schedule is a special case in which p(i, i ) = 1 for i the successor to i in the cycle. In fact, this set-up generalizes the arbitrary deterministic routing schedule described above, as follows. Given the schedule sequence σ of length T , we may convert this to a cyclic schedule on an extended state-space consisting of µ k copies of line k, and then reading σ as a permutation. So the deterministic routing model is a special case of the model with Markov routing, which will be the focus of the rest of the paper.
Our result again will say that (1.1) is the criterion for transience, where µ k is now the stationary distribution associated to the stochastic matrix p(i, j). Our general model also permits two-sided increments for the walk from some of the lines, which contribute terms involving cot(πα k /2) to the cotangent criterion (1.1). These twosided models also generalize previously studied classical models (see e.g. [15,16,22]). Again, it is only in our general setting that the essential nature of the cotangent criterion (1.1) becomes apparent.
Rather than R + × S, one could work on Z + × S instead, with mass functions replacing probability densities; the results would be unchanged.
The paper is organized as follows. In Section 2 we formally define our model and describe our main results, which as well as a recurrence classification include results on existence of moments of return times in the recurrent cases. In Section 3 we explain how our general model relates to the special case of the oscillating random walk when S has two elements, and state our results for that model; in this context the recurrence classification results are already known, but the existence of moments results are new even here, and are in several important cases sharp. The present work was also motivated by some problems concerning many-dimensional, partially homogeneous random walks similar to models studied in [5,6]: we describe this connection in Section 4. The main proofs are presented in Sections 5, 6, and 7, the latter dealing with the critical boundary case which is more delicate and requires additional work. We collect various technical results in Appendix A.

Model and results
Consider (X n , ξ n ; n ∈ Z + ), a discrete-time, time-homogeneous Markov process with state space R + × S, where S is a finite non-empty set. The state space is equipped with the appropriate Borel sets, namely, sets of the form B × A where B ∈ B(R + ) is a Borel set in R + , and A ⊆ S. The process will be described by: • an irreducible stochastic matrix labelled by S, P = (p(i, j); i, j ∈ S); and • a collection (w i ; i ∈ S) of probability density functions, so w i : R → R + is a Borel function with R w i (y)dy = 1. We view R + × S as a complex of half-lines R + × {k}, or branches, connected at a central origin O := {0} × S; at time n, the coordinate ξ n describes which branch the process is on, and X n describes the distance along that branch at which the process sits. We will call X n a random walk on this complex of branches.
To simplify notation, throughout we write P x,i [ · ] for P[ · | (X 0 , ξ 0 ) = (x, i)], the conditional probability starting from (x, i) ∈ R + × S; similarly we use E x,i for the corresponding expectation. The transition kernel of the process is given for (x, i) ∈ R + × S, for all Borel sets B ⊆ R + and all j ∈ S, by The dynamics of the process represented by (2.1) can be described algorithmically as follows. Given (X n , ξ n ) = (x, i) ∈ R + × S, generate (independently) a spatial increment ϕ n+1 from the distribution given by w i and a random index η n+1 ∈ S according to the distribution p(i, · ). Then, In words, the walk takes a w ξn -distributed step. If this step would bring the walk beyond the origin, it passes through the origin and switches onto branch η n+1 (or, if η n+1 happens to be equal to ξ n , it reflects back along the same branch).
The finite irreducible stochastic matrix P is associated with a (unique) positive invariant probability distribution (µ k ; k ∈ S) satisfying j∈S µ j p(j, k) − µ k = 0, for all k ∈ S. (2.2) For future reference, we state the following.
(A0) Let P = (p(i, j); i, j ∈ S) be an irreducible stochastic matrix, and let (µ k ; k ∈ S) denote the corresponding invariant distribution.
Our interest here is when the w i are heavy tailed. We allow two classes of distribution for the w i : one-sided or symmetric. It is convenient, then, to partition S as S = S one ∪ S sym where S one and S sym are disjoint sets, representing those branches on which the walk takes, respectively, one-sided and symmetric jumps. The w k are then described by a collection of positive parameters (α k ; k ∈ S).
For a probability density function v : R → R + , an exponent α ∈ (0, ∞), and a constant c ∈ (0, ∞), we write v ∈ D α,c to mean that there exists c : If v ∈ D α,c is such that (2.3) holds and c(y) satisfies the stronger condition c(y) = c + O(y −δ ) for some δ > 0, then we write v ∈ D + α,c . Our assumption on the increment distributions w i is as follows.
(A1) Suppose that, for each k ∈ S, we have an exponent α k ∈ (0, ∞), a constant c k ∈ (0, ∞), and a density function v k ∈ D α k ,c k . Then suppose that, for all y ∈ R, w k is given by We say that X n is recurrent if lim inf n→∞ X n = 0, a.s., and transient if lim n→∞ X n = ∞, a.s. An irreducibility argument shows that our Markov chain (X n , ξ n ) displays the usual recurrence/transience dichotomy and exactly one of these two situations holds; however, our proofs establish this behaviour directly using semimartingale arguments, and so we may avoid discussion of irreducibility here.
Throughout we define, for k ∈ S, Our first main result gives a recurrence classification for the process. (a) Suppose that max k∈S χ k α k ≥ 1. Then X n is recurrent.
(iii) Suppose in addition that the densities v k of (A1) belong to D + α k ,c k for each k. Then k∈S µ k cot(χ k πα k ) = 0 implies that X n is recurrent.
In the recurrent cases, it is of interest to quantify recurrence via existence or non-existence of passage-time moments. For a > 0, let τ a := min{n ≥ 0 : X n ≤ a}, where throughout the paper we adopt the usual convention that min ∅ := +∞. The next result shows that in all the recurrent cases, excluding the boundary case in Theorem 2.1(b)(iii), the tails of τ a are polynomial.
Theorem 2.2. Suppose that (A0) and (A1) hold. In cases (a) and (b)(i) of Theorem 2.1, there exists 0 < q < ∞ such that for all x > a and all k ∈ S, Remark 2.3. One would like to precisely locate q ; here we only locate q within an interval, i.e., we show that there exist q 0 and q 1 with 0 < q 0 < q 1 < ∞ such that E x,k [τ q a ] < ∞ for q < q 0 and E x,k [τ q a ] = ∞ for q > q 1 . The existence of a critical q follows from the monotonicity of the map q → E x,k [τ q c ], but obtaining a sharp estimate of its value remains an open problem in the general case.
We do have sharp results in several particular cases for two half-lines, in which case our model reduces to the oscillating random walk considered by Kemperman [15] and others. We present these sharp moments results (Theorems 3.2 and 3.4) in the next section, which discusses in detail the case of the oscillating random walk, and also describes how our recurrence results relate to the known results for this classical model.

Two half-lines become one line
In the case of our general model in which S consists of two elements, S = {−1, +1}, say, it is natural and convenient to represent our random walk on the whole real line R. Namely, if ω(x, k) := kx for x ∈ R + and k = ±1, we let Z n = ω(X n , ξ n ).
The simplest case has no reflection at the origin, only transmission, i.e. p(i, j) = 1{i = j}, so that µ = ( 1 2 , 1 2 ). Then, for B ⊆ R a Borel set, and, similarly, writing w − (y) for w −1 (−y), for x ∈ R + , and hence we have for x ∈ R \ {0} and Borel B ⊆ R, We may make an arbitrary non-trivial choice for the transition law at Z n = 0 without affecting the behaviour of the process, and then (3.1) shows that Z n is a time-homogeneous Markov process on R. Now Z n is recurrent if lim inf n→∞ |Z n | = 0, a.s., or transient if lim n→∞ |Z n | = ∞, a.s. The one-dimensional case described at (3.1) has received significant attention over the years. We describe several of the classical models that have been considered.

Examples and further results
Homogeneous symmetric random walk The most classical case is the following.
In this case, Z n describes a random walk with i.i.d. symmetric increments.  Since it deals with a sum of i.i.d. random variables, Theorem 3.1 may be deduced from the classical theorem of Chung and Fuchs [7], via e.g. the formulation of Shepp [22]. The method of the present paper provides an alternative to the classical (Fourier analytic) approach that generalizes beyond the i.i.d. setting. (Note that Theorem 3.1 is not, formally, a consequence of Shepp's most accessible result, Theorem 5 of [22], since v does not necessarily correspond to a unimodal distribution in Shepp's sense.) With τ a as defined previously, in the setting of the present section we have τ a = min{n ≥ 0 : |Z n | ≤ a}. Use E x [ · ] as shorthand for E[ · | Z 0 = x]. We have the following result on existence of passage-time moments, whose proof is in Section 6; while part (i) is well known, we could find no reference for part (ii).
Our main interest concerns spatially inhomogeneous models, i.e., in which w x depends on x, typically only through sgn x, the sign of x. Such models are known as oscillating random walks, and were studied by Kemperman [15], to whom the model was suggested in 1960 by Anatole Joffe and Peter Ney (see [15, p. 29]).

One-sided oscillating random walk
The next example, following [15], is a one-sided oscillating random walk : In other words, the walk always jumps in the direction of (and possibly over) the origin, with tail exponent α from the positive half-line and exponent β from the negative half-line. The following recurrence classification applies. and v − ∈ D + β,c − , then the case α + β = 1 is recurrent. Theorem 3.3 was obtained in the discrete-space case by Kemperman [15, p. 21]; it follows from our Theorem 2.1, since in this case k∈S µ k cot(χ k πα k ) = 1 2 cot(πα) + 1 2 cot(πβ) = sin(π(α + β)) 2 sin(πα) sin(πβ) .
The special case of (Osc1) in which α = β was called antisymmetric by Kemperman; here Theorem 3.3 shows that the walk is transient for α < 1/2 and recurrent for α > 1/2. We have the following moments result, proved in Section 6.

Mixed oscillating random walk
A final model is another oscillating walk that mixes the one-and two-sided models: In the discrete-space case, Theorem 2 of Rogozin and Foss [16, p. 159] gives the recurrence classification.

Additional remarks
It is possible to generalize the model further by permitting the local transition density to vary within each half-line. Then we have the transition kernel for all Borel sets B ⊆ R. Here the local transition densities w x : R → R + are Borel functions. Variations of the oscillating random walk, within the general setting of (3.2), have also been studied in the literature. Sandrić [18,19] supposes that the w x satisfy, for each x ∈ R, w x (y) ∼ c(x)|y| −1−α(x) as |y| → ∞ for some measurable functions c and α; he refers to this as a stable-like Markov chain. Under a uniformity condition on the w x , and other mild technical conditions, Sandrić [18] obtained, via Foster-Lyapunov methods similar in spirit to those of the present paper, sufficient conditions for recurrence and transience: essentially lim inf x→∞ α(x) > 1 is sufficient for recurrence and lim sup x→∞ α(x) < 1 is sufficient for transience. These results can be seen as a generalization of Theorem 3.1. Some related results for models in continuous-time (Lévy processes) are given in [20,21,23]. Further results and an overview of the literature are provided in Sandrić's PhD thesis [17].

Many-dimensional random walks
The next two examples show how versions of the oscillating random walk of Section 3 arise as embedded Markov chains in certain two-dimensional random walks.
n ), n ∈ Z + , a nearest-neighbour random walk on Z 2 with transition probabilities Suppose that the probabilities are given for x 2 = 0 by, (the rest being zero) and for Figure 1 for an illustration. Set τ 0 := 0 and define recursively τ k+1 = min{n > τ k : ξ τn . We show that X n is a discrete version of the oscillating random walk described in Section 3. Indeed, |ξ (2) n | is a reflecting random walk on Z + with increments taking values −1, 0, +1 each with probability 1/3. We then (see e.g. [9, p. 415]) have that for some constant c ∈ (0, ∞), Suppose that ξ (1) 0 = x > 0. Since between times τ 0 and τ 1 , ξ (1) n is monotone, we have Here, by the Azuma-Hoeffding inequality, for some ε > 0 and all r ≥ 1, Similarly, Combining these bounds, and using the symmetric argument for {ξ (1) where u(r) = (c + o(1))r −1/2 . Thus X n satisfies a discrete-space analogue of (Osc1) with α = β = 1/2. This is the critical case identified in Theorem 3.3, but that result does not cover this case due to the rate of convergence estimate for u; a finer analysis is required. We conjecture that the walk is recurrent.
We present two variations on the previous example, which are superficially similar but turn out to be less delicate. First, modify the random walk of the previous example by supposing that (4.1) holds but replacing the behaviour at See the left-hand part of Figure 2 for an illustration. The embedded process X n now has, for all x ∈ Z and for r ≥ 0, where u(r) = (c/2)(1 + o(1))r −1/2 . Thus X n is a random walk with symmetric increments, and the discrete version of our Theorem 3.1 (and also a result of [22]) implies that the walk is transient. This walk was studied by Campanino and Petritis [5,6], who proved transience via different methods. Next, modify the random walk of Example 4.1 by supposing that (4.1) holds but replacing the behaviour at x 2 = 0 by p(x 1 , 0; x 1 , 1) = p(x 1 , 0; x 1 , −1) = 1/2 if x 1 ≥ 0, and p(x 1 , 0; x 1 , −1) = 1 for x 1 < 0. See the right-hand part of Figure 2 for an illustration. This time the walk takes a symmetric increment as at (4.3) when x ≥ 0 but a one-sided increment as at (4.2) when x < 0. In this case the discrete version of our Theorem 3.6 (and also a result of [16]) shows that the walk is transient.
One may obtain the general model on Z + × S as an embedded process for a random walk on complexes of half-spaces, generalizing the examples considered here; we leave this to the interested reader.

Recurrence classification in the non-critical cases 5.1 Lyapunov functions
Our proofs are based on demonstrating appropriate Lyapunov functions; that is, for suitable ϕ : R + × S → R + we study Y n = ϕ(X n , ξ n ) such that Y n has appropriate local supermartingale or submartingale properties for the one-step mean increments First we note some consequences of the transition law (2.1). Let ϕ : R + ×S → R + be measurable. Then, we have from (2.1) that, for (x, i) ∈ R + × S, Our primary Lyapunov function is roughly of the form x → |x| ν , ν ∈ R, but weighted according to an S-dependent component (realised by a collection of multiplicative weights λ k ); these weights provide a crucial technical tool.
For ν ∈ R and x ∈ R we write Then, for parameters λ k > 0 for each k ∈ S, define for x ∈ R + and k ∈ S, Now for this Lyapunov function, (5.1) gives Depending on whether i ∈ S sym or i ∈ S one , the above integrals can be expressed in terms of v i as follows. For i ∈ S sym ,

Estimates of functional increments
In the course of our proofs, we need various integral estimates that can be expressed in terms of classical transcendental functions. For the convenience of the reader, we gather all necessary integrals in Lemmas 5.1 and 5.2; the proofs of these results are deferred until Section A.2. Recall that the Euler gamma function Γ satisfies the functional equation zΓ(z) = Γ(z + 1), and the hypergeometric function m F n is defined via a power series (see [1]).
The next result collects estimates for our integrals in the expected functional increments (5.4) and (5.5) in terms of the integrals in Lemma 5.1.
For α ∈ (0, 2) and ν > −1 we have For α ∈ (0, 1) and ν > −1 we have Moreover, if v ∈ D + α,c then stronger versions of all of the above estimates hold with Proof. These estimates are mostly quite straightforward, so we do not give all the details. We spell out the estimate in (5.6); the others are similar. We have With the substitution u = 1+y x , this last expression becomes Let ε ∈ (0, c). Then there exists y 0 ∈ R + such that |c(y) − c| < ε for all y ≥ y 0 , so that |c(ux − 1) − c| < ε for all u in the range of integration, provided x ≥ y 0 . Writing for the duration of the proof, we have that For u ≥ 1+x x and x ≥ y 0 , we have as x → ∞, it follows that for any ε > 0 we may choose x sufficiently large so that and since ε > 0 was arbitrary, we obtain (5.6) from (5.11).
We also need the following simple estimates for ranges of α when the asymptotics for the final two integrals in Lemma 5.4 are not valid.
there exists a solution (θ k ; k ∈ S) with θ k ∈ R for all k to the system of equations if and only if k∈S µ k b k = 0. Moreover, if a solution to (5.13) exists, we may take θ k > 0 for all k ∈ S.
Proof. As column vectors, we write µ = (µ k ; k ∈ S) for the stationary probabilities as given in (A0), b = (b k ; k ∈ S), and θ = (θ k ; k ∈ S). Then in matrix-vector form, (5.13) reads (P − I)θ = b, while µ satisfies (2.2), which reads (P − I) µ = 0, the homogeneous system adjoint to (5.13). (Here I is the identity matrix and 0 is the vector of all 0s.) A standard result from linear algebra (a version of the Fredholm alternative) says that (P − I)θ = b admits a solution θ if and only if the vector b is orthogonal to any solution x to (P − I) x = 0; but, by (A0), any such x is a scalar multiple of µ. In other words, a solution θ to (5.13) exists if and only if µ b = 0, as claimed.
Also, all lower triangular elements of L are positive, i.e. L k, > 0 for 0 ≤ < k ≤ M .
Proof. We construct L inductively. Item (i) demands (5.14) In the case m = M , with the usual convention that an empty sum is 0, the demand (   We start with the case max k∈S χ k α k < 1. We will obtain a local supermartingale by choosing the λ k carefully. Lemma 5.7, which shows how the stationary probabilities µ k enter, is crucial; a similar idea was used for random walks on strips in Section 3.1 of [10]. Next is our key local supermartingale result in this case.
Since i ν,α 2,1 > 0 for all α > 0 and all ν ∈ (0, α), Lemma 5.6 shows that Df ν (x, i) will be negative for all i and all x sufficiently large provided that we can find ν such that (P λ) i λ i + R (α i , ν) < 0 for all i. By (5.12), writing θ = (θ k ; k ∈ S), by (5.15). Therefore, by Lemma 5.6, we can always find sufficiently small ν > 0 and a vector λ = λ(ν) with strictly positive elements for which Df ν (x, i) ≤ −εx ν−α i for some ε > 0, all i, and all x sufficiently large. Maximizing over i gives part (i). The argument for part (ii) is similar. Suppose ν ∈ (−1, 0). This time, k∈S µ k a k = δ for some δ > 0, and we set b k = −a k + δ, so that k∈S µ k b k = 0 once more. Lemma 5.7 now shows that we can find θ k so that j∈S p(i, j)θ j − θ i + a i = δ, for all i ∈ S. (5.16) With this choice of θ k we again set λ k = 1 + θ k ν; note we may assume λ k > 0 for all k for ν sufficiently small. Again, Df ν (x, i) will be non-negative for all i and all x sufficiently large provided that we can find λ and ν such that (P λ) i λ i + R (α i , ν) < 0 for all i. Following a similar argument to before, we obtain with (5.16) that Thus we can find for ν < 0 close enough to 0 a vector λ = λ(ν) with strictly positive elements for which Df ν (x, i) ≤ −εx ν−α i for all i and all x sufficiently large.
Now we examine the case max k∈S χ k α k ≥ 1.
Proposition 5.11. Suppose that (A0) and (A1) hold, and max k∈S χ k α k ≥ 1. Then there exist ν ∈ (0, α ), λ k > 0 (k ∈ S), ε > 0, and x 0 ∈ R + such that for all i ∈ S, Before starting the proof of this proposition, we introduce the following notation. For k ∈ S denote by a k = π cot(πχ k α k ) and define the vector a = (a k ; k ∈ S). For i ∈ S and A ⊆ S, write P (i, A) = j∈A p(i, j). Define S 0 = {i ∈ S : χ i α i ≥ 1} and recursively, for m ≥ 1, The rest of the proof is devoted into establishing this fact.
In the case i ∈ S M , the condition in (5.20) reads, by (5.21), .

On introducing the constant
Using the fact that for < m we have ρ ,m = exp(−νL m, ) and for > m we have ρ ,m = exp(νL ,m ), the above expression becomes, up to o(ν) terms .
Introducing the upper triangular matrix U = (U m,n ; 0 ≤ m < n ≤ M ) defined by U m,n = max i∈Sm P (i,Sn) Hence by Corollary 5.9, there exist non-trivial solutions for the lower triangular matrix L within an algorithmically determined region L. The positivity of the lower triangular part of L implies that the components of λ are ordered: λ (m) < λ (m+1) for 0 ≤ m < M − 1. By choosing λ (M ) sufficiently large, we can guarantee that the first term (and consequently the entire sequence) satisfies λ (0) > 1.
We are almost ready to complete the proof of Theorem 2.1, excluding part (b)(iii); first we need one more technical result concerning non-confinement.
Proof. We claim that for each x ∈ R + , there exists ε x > 0 such that Indeed, given (x, i) ∈ R + × S, we may choose j ∈ S so that p(i, j) > 0 and we may choose z 0 ≥ x sufficiently large so that, for some which gives (5.23). The local escape property (5.23) implies the lim sup result by a standard argument.
Proof of Theorem 2.1. We are not yet ready to prove part (b)(iii): we defer that part of the proof until Section 7.
The other parts of the theorem follow from the supermartingale estimates in this section together with the technical results from Section A.1. Indeed, under the conditions of part (a) or (b)(i) of the theorem, we have from Proposition 5.11 or Proposition 5.10(i) respectively that for suitable ν > 0 and λ k , Thus we may apply Lemma A.1, which together with Lemma 5.12 shows that lim inf n→∞ X n ≤ x 0 , a.s. Thus there exists an interval I ⊆ [0, x 0 + 1] such that (X n , η n ) ∈ I × {i} i.o., where i is some fixed element of S. Let τ 0 := 0 and for k ∈ N define τ k = min{n > τ k−1 + 1 : (X n , η n ) ∈ I × {i}}. Given i ∈ S, we may choose j, k ∈ S such that p(i, j) > δ 1 and p(j, k) > δ 1 for some δ 1 > 0; let γ = α i ∨ α j . Then we may choose δ 2 ∈ (0, 1) and z 0 ∈ R + such that v i (z) > δ 2 z −1−γ and v j (z) > δ 2 z −1−γ for all z ≥ z 0 . Then for any ε ∈ (0, 1), uniformly in k. Thus Lévy's extension of the Borel-Cantelli lemma shows X n < ε infinitely often. Thus, since ε ∈ (0, 1) was arbitrary, lim inf n→∞ X n = 0, a.s. On the other hand, under the conditions of part (b)(ii) of the theorem, we have from Proposition 5.10(ii) that for suitable ν < 0 and λ k , for any x 1 sufficiently large. Thus we may apply Lemma A.2, which shows that for any ε > 0 there exists x ∈ (x 1 , ∞) for which, for all n ≥ 0, by Lemma 5.12. Since ε > 0 was arbitrary, we get lim inf m→∞ X m ≥ x 1 , a.s., and since x 1 was arbitrary we get lim m→∞ X m = ∞, a.s.
6 Existence or non-existence of moments 6

.1 Technical tools
The following result is a straightforward reformulation of Theorem 1 of [2]. Lemma 6.1. Let Y n be an integrable F n -adapted stochastic process, taking values in an unbounded subset of R + , with Y 0 = x 0 fixed. For x > 0, let σ x := inf{n ≥ 0 : Y n ≤ x}. Suppose that there exist δ > 0, x > 0, and γ < 1 such that for any n ≥ 0, Then, for any p The following companion result on non-existence of moments is a reformulation of Corollary 1 of [2]. Lemma 6.2. Let Y n be an integrable F n -adapted stochastic process, taking values in an unbounded subset of R + , with Y 0 = x 0 fixed. For x > 0, let σ x := inf{n ≥ 0 : Y n ≤ x}. Suppose that there exist C 1 , C 2 > 0, x > 0, p > 0 and r > 1 such that for any n ≥ 0, on {n < σ x } the following hold: Then for any q > p, E[σ q x ] = ∞ for x 0 > x.

Proof of Theorem 2.2
Proof of Theorem 2.2. Under conditions (a) or (b)(i) of Theorem 2.1, we have from Propositions 5.11 or 5.10 respectively that there exist positive λ k and constants ε > 0, β > 0 and ν ∈ (0, β) such that, Let Y n = f ν (X n , ξ n ). Then Y n is bounded above and below by positive constants times (1+X n ) ν , so we have that (6.1) holds for x sufficiently large with γ = 1−(β/ν). It follows from Lemma 6.1 that E[σ p x ] < ∞ for p ∈ (0, ν/β), which gives the claimed existence of moments result.
It is not hard to see that some moments of the return time fail to exist, due to the heavy-tailed nature of the model, and an argument is easily constructed using the 'one big jump' idea: a similar idea is used in [12]. We sketch the argument. For any x, i, for all y sufficiently large we have P x,i [X 1 ≥ y − x] ≥ εy −ᾱ . Given such a first jump, with uniformly positive probability the process takes time at least of order y β to return to a neighbourhood of zero (where β can be bounded in terms of α); this can be proved using a suitable maximal inequality as in the proof of Theorem 2.10 of [12]. Combining these two facts shows that with probability of order y −ᾱ the return time to a neighbourhood of the origin exceeds order y β . This polynomial tail bound yields non-existence of sufficiently high moments.
We can now complete the proofs of Theorems 3.2 and 3.4.
We sketch the argument for E[σ p ] = ∞ when p > 1/2. For ν ∈ (1, α), it is not hard to show that Df ν (x, i) ≥ 0 for all x sufficiently large, and we may verify the other conditions of Lemma 6.2 to show that E[σ p ] = ∞ for p > 1/2.
Proof of Theorem 3.4. Most of the proof is similar to that of Theorem 3.2, so we omit the details. The case where a different argument is required is the non-existence part of the case α ≥ 1. We have that for some ε > 0 and all y sufficiently large, P x,i [X 1 ≥ y] ≥ ε(x + y) −α . A similar argument to Lemma 5.5 shows that for any ν ∈ (0, 1), for some C ∈ R + , E[X ν n+1 − X ν n | X n = x] ≥ −C. Then a suitable maximal inequality implies that with probability at least 1/2, started from X 1 ≥ y it takes at least cy ν steps for X n to return to a neighbourhood of 0, for some c > 0. Combining the two estimates gives which implies E[σ p ] = ∞ for p ≥ α/ν, and since ν ∈ (0, 1) was arbitrary we can achieve any p > α.
7 Recurrence classification in the critical cases

Logarithmic Lyapunov functions
In this section we prove Theorem 2.1(b)(iii). Throughout this section we write a k := cot(χ k πα k ) and suppose that max k∈S χ k α k < 1, that k∈S µ k a k = 0, and that v i ∈ D + α i ,c i for all i ∈ S, that is, for y > 0, v i (y) = c i (y)y −1−α i , with α i ∈ (0, ∞) and c i (y) = c i + O(y −δ ), where δ > 0 may be chosen so as not to depend upon i.
To prove recurrence in the critical cases, we need a function that grows more slowly than any power; now the weights λ k are additive rather than multiplicative. For x ∈ R write g(x) := log(1 + |x|). Then, for x ∈ R + and k ∈ S, define where λ k > 0 for all k ∈ S. Also write h(x, k) := (g(x, k)) 1/2 = (log(1 + |x|) + λ k ) 1/2 .
Lemma 5.7 shows that there exist λ k > 0 (k ∈ S) such that we fix such a choice of the λ k from now on. We prove recurrence by establishing the following result.
Lemma 7.1. Suppose that the conditions of Theorem 2.1(b)(iii) hold, and that (λ k ; k ∈ S) are such that (7.2) holds. Then there exists x 0 ∈ R + such that It is not easy to perform the integrals required to estimate Dh(x, i) (and hence establish Lemma 7.1) directly. However, the integrals for Dg(x, i) are computable (they appear in Lemma 5.2), and we can use some analysis to relate Dh(x, i) to Dg(x, i). Thus the first step in our proof of Lemma 7.1 is to estimate Dg(x, i).
For the Lyapunov function g, (5.1) gives for x ∈ R + and i ∈ S, where we have used the fact that g(x) is defined for all x ∈ R and symmetric about 0, and we have introduced the notation The next lemma, proved in the next subsection, estimates the integrals in (7.4).
Lemma 7.2. Suppose that v i ∈ D + α i ,c i . Then, for ∈ {one, sym}, for α ∈ (0, 1/χ i ), for some η > 0, as x → ∞, Note that Lemma 7.2 together with (7.3) shows that (7.5) by (7.2). This is not enough by itself to establish recurrence, since the sign of the O(x −α i −η ) term is unknown. This is why we need the function h(x, i).

Proof of recurrence in the critical case
Proof of Lemma 7.2. To ease notation, we drop the subscripts i everywhere for the duration of this proof. From (7.4) we obtain, for x > 0, Let α ∈ (0, 2). The claim in the lemma for = sym will follow from the estimates since then we obtain from (7.6) with (7.7) and Lemma 5.2 that which yields the stated result via the digamma reflection formula (equation 6.3.7 from [1, p. 259]). We present here in detail the proof of only the final estimate in (7.7); the others are similar. Some algebra followed by the substitution u = y/(1 + x) shows that the third integral in (7.7) is There is a constant C ∈ R + such that for all x sufficiently large, using Taylor's theorem for log and the fact that c(y) is uniformly bounded. On the other hand, for u > Here we have that Combining these estimates and using the fact that α ∈ (0, 2), we obtain the final estimate in (7.7). The claim in the lemma for = one follows after some analogous computations, which we omit.
Finally, there exists ε > 0 such that, for all i, j ∈ S and all x sufficiently large, Proof. The proof is based on the observation that, since (h(x, i)) 2 = g(x, i), Thus, for y ≥ 0, 2h(x, i) , (7.8) since g(x + y) − g(x) ≥ 0 and h(x + y, i) ≥ h(x, i). This gives the first inequality in the lemma. Similarly, for y ∈ [0, x], 2h(x, i) , since g(x − y) − g(x) ≤ 0 and h(x − y, i) ≤ h(x, i). This gives the second inequality, and also yields the third inequality once combined with the y ∈ [0, x] case of (7.8).
Finally, for y ≥ x note that Also note that, for y ≥ x > 0, g(x, i) = g(x) + λ i and g(y − x, j) = g(y − x) + λ j , so So the sign of the expression in (7.9) is non-positive for y ≤ ψ(x) := x − 1 + (1 + x)e λ i −λ j and non-negative for y ≥ ψ(x), and g(ψ(x) − x, j) − g(x, i) = 0. (7.10) By the monotonicity in y of the denominator, the expression in (7.9) satisfies both for y ∈ [0, ψ(x)] and for y ∈ [ψ(x), ∞). Here h(ψ(x) − x, j) = h(x, i), by (7.10). Hence we obtain the bound To improve on this estimate, suppose that y ≥ Kx where K ∈ N is such that Kx > ψ(x). Then, using the fact that the numerator in (7.9) is positive, we may choose K ∈ N such that for all j and all y ≥ Kx, for K sufficiently large (depending on max j λ j ) and and all x sufficiently large. Here It follows that, for some ε > 0, for all x sufficiently large, for y ≥ Kx, The final inequality in the lemma now follows since, for all x sufficiently large, where we used the substitution u = 2+y 1+x . For K sufficiently large the term inside the logarithm is uniformly positive, and the claimed bound follows. Now we may complete the proofs of Lemma 7.1 and then Theorem 2.1(b)(iii).
Proof of Lemma 7.1. Lemma 7.3 together with (7.5) shows that, for all x sufficiently large.
Proof of Theorem 2.1(b)(iii). Lemma 7.1 with Lemma A.1 shows that lim inf n→∞ X n ≤ x 0 , a.s., and then a similar argument to that in the proof of parts (a) and (b)(i) of Theorem 2.1 shows that lim inf n→∞ X n = 0, a.s.

A.1 Semimartingale results
Lemma A.1. Let (X n , ξ n ) be an F n -adapted process taking values in R + × S. Let f : R + ×S → R + be such that lim x→∞ f (x, i) = ∞ for all i ∈ S, and E f (X 0 , ξ 0 ) < ∞.
Suppose that there exist x 0 ∈ R + and C < ∞ for which, for all n ≥ 0, Proof. First note that, by hypothesis, E f (X 1 , ξ 1 ) ≤ E f (X 0 , ξ 0 ) + C < ∞, and, iterating this argument, it follows that E f (X n , ξ n ) < ∞ for all n ≥ 0. Fix n ∈ Z + . For x 0 ∈ R + in the hypothesis of the lemma, write λ = min{m ≥ n : is an (F m , m ≥ n)-adapted non-negative supermartingale. Hence, by the supermartingale convergence theorem, there exists Y ∞ ∈ R + such that lim m→∞ Y m = Y ∞ , a.s. In particular, Since n ∈ Z + was arbitrary, the result follows: Lemma A.2. Let (X n , ξ n ) be an F n -adapted process taking values in R + × S. Let f : R + × S → R + be such that sup x,i f (x, i) < ∞ and lim x→∞ f (x, i) = 0 for all i ∈ S. Suppose that there exists x 1 ∈ R + for which inf y≤x 1 f (y, i) > 0 for all i and E[f (X n+1 , ξ n+1 ) − f (X n , ξ n ) | F n ] ≤ 0, on {X n ≥ x 1 }, for all n ≥ 0.
Proof. Fix n ∈ Z + . For x 1 ∈ R + in the hypothesis of the lemma, write λ = min{m ≥ n : X m ≤ x 1 } and set Y m = f (X m∧λ , ξ m∧λ ). Then (Y m , m ≥ n) is an (F m , m ≥ n)adapted non-negative supermartingale, and so converges a.s. as m → ∞ to some Y ∞ ∈ R + . Moreover, by the optional stopping theorem for supermartingales, Here we have that, a.s., Combining these inequalities we obtain min i inf y≤x 1 f (y, i)P[λ < ∞ | F n ] ≤ Y n , a.s.
In particular, on {X n ≥ x > x 1 }, we have Y n = f (X n , ξ n ) and so Since lim y→∞ f (y, i) = 0 and inf y≤x 1 f (y, i) > 0, given ε > 0 we can choose x > x 1 large enough so that max i sup y≥x f (y, i) min i inf y≤x 1 f (y, i) < ε; the choice of x depends only on f , x 1 , and ε, and, in particular, does not depend on n. Then, on {X n ≥ x}, P[λ < ∞ | F n ] < ε, as claimed.

A.2 Proofs of integral computations
Proof of Lemma 5.1. Consider i ν,α 0 . With the change of variable v = 1/u, we get Hence we obtain the given expression for i ν,α 2,1 . Next consider i ν,α 1 . By considering separately the asymptotics of the integrand as u ↓ 0 and u ↑ 1, we see that i ν,α 1 is finite provided α ∈ (0, 2) and ν > −1. .
Lemma A.3. Let A ⊆ R + be a Borel set, and let f and g be measurable functions from A to R. Suppose that there exist constants g − and g + with 0 < g − < g + < ∞ such that g − ≤ g(u) ≤ g + for all u ∈ A. Then, Proof. It suffices to suppose that A |f (u)|du < ∞. Then A |f (u)g(u)|du ≤ g + A |f (u)|du < ∞, and