Recurrence for the frog model with drift on $\mathbb{Z}^d$

In this paper we present a recurrence criterion for the frog model on $\mathbb{Z}^d$ with an i.i.d. initial configuration of sleeping frogs and such that the underlying random walk has a drift to the right.


Introduction
The frog model is a certain model of interacting random walks on a graph. Imagine a graph G = (V, E) with a distinguished vertex x 0 ∈ V , called the origin. At time 0, there is exactly one active frog at x 0 and on each vertex x ∈ V \ {x 0 } there is a number η x ∈ Z + := Z ∩ [0, ∞) of sleeping frogs. The frog at x 0 now starts a nearest-neighbour random walk on the graph G. If it hits a vertex x with η x > 0 sleeping frogs, they all become active at once and start performing nearest-neighbour random walks, independently of each other and of the original frog. More generally, each time an active frogs hits a vertex x ∈ V with η x > 0 sleeping frogs, they all become active at once and start nearest-neighbour random walks, independently of each other and of all other frogs. In this description, the transition function of the underlying random walk is supposed to be the same for all frogs. The frog model is called recurrent, if the probability that the origin x 0 is visited infinitely often equals 1, otherwise the model is called transient. The frog model with V = Z d , E the set of nearest-neighbour edges on Z d , x 0 := 0, η x = 1 for each x ∈ Z d \ {0} and the underlying random walk being simple random walk (SRW) on Z d was studied by Telcs and Wormald [TW99]. They showed in particular that the frog model is recurrent for each dimension d. This result was refined by Popov [Pop01], who considered frogs in a random environment. More precisely, he considered the situation, where there is, for each x ∈ Z d \ {0}, originally one sleeping frog at x with probability p(x) and no frog with probability 1 − p(x), independently of all other vertices, and found the exact rate of decay for the function p(x) to distinguish transience from recurrence. Note that the frog model on Z d with SRW is trivially recurrent for d = 1, 2, due to Pólya's theorem. Thus, in [GS09] Gantert and Schmidt considered the frog model on Z with the underlying random walk having a drift to the right. They considered, both, fixed and i.i.d. random initial configurations (η x ) x∈Z\{0} of sleeping frogs, and derived precise criteria to separate transience from recurrence. In the case of an i.i.d. initial configuration of sleeping frogs they also proved a 0−1 law, which says that the probability of infinitely many returns to 0 equals 1, if E[log + (η 1 )] = ∞, and equals 0, otherwise, independently of the concrete value of the drift. The purpose of the present note is to prove a recurrence criterion in the case of Z d , d ≥ 2, and an i.i.d. initial configuration of sleeping frogs. Thanks to discussion with Serguei Popov, we believe that for the frog model on Z d , d ≥ 2, and an i.i.d. initial configuration of sleeping frogs, in general, transience and recurrence depend on the concrete value of the drift, unless the distribution giving the number of sleeping frogs per site is heavy-tailed enough in which case recurrence holds irrespective of the value of the drift. The purpose of the present note is to prove recurrence if such a criterion on the tails is satisfied, while the full problem of separating transience from recurrence will be addressed in follow-up work. The paper is structured as follows: In Section 2 we give a precise description of the model we consider and state our main theorem, Theorem 2.1. In Section 3 we give the proof of Theorem 2.1 and finally, in Section 4 we give proofs of two auxiliary lemmas, which we need in Section 3 in order to prove Theorem 2.1.

Acknowledgements
We would like to thank Silke Rolles and Nina Gantert for useful discussion und comments.

Setting and main theorem
As mentioned above, we consider recurrence of the frog model on Z d with an i.i.d. initial configuration and such that the underlying random walk has a drift to the right. We denote by S the set of all possible initial configurations of sleeping frogs, i.e.
Further, we denote by p the transition function of the underlying nearest-neighbour random walk. Thus, letting E := {±e j : 1 ≤ j ≤ d}, where e j denotes the j-th standard basis vector in R d , j = 1, . . . , d, we assume that p : and p(x) = 0 for all x ∈ Z d \ E. In order to make the random walk irreducible, we will further assume that 0 < p(e) < 1 holds for all e ∈ E. Additionally, we will abuse notation to write p(x, y) := p(y − x) also for the corresponding transition matrix. Since we assume that the underlying random walk has a drift to the right, we suppose that there is an a ∈ (0, 1) such that (1) m := e∈E p(e)e = ae 1 .
Since the transition function p will be kept fixed throughout, we omit it from the notation. For a fixed η ∈ S we denote by P η a probability measure on a suitable measurable space (Ω, F ), which describes the evolution of the frog model with initial configuration η and underlying random walk given by the transition function p as described in the introduction. We refrain from giving a mathematical construction of the frog model with respect to η but refer the interested reader to [Pop03]. Now, let µ be a probability distribution on (Z + , P(Z + )) and let P µ be the corresponding product measure on (Z The corresponding expectation operator will be denoted by E µ . Finally, we denote by P the P µ -mixture of the measures P η , i.e. (2) Thus, the measure P describes the evolution of the frog model with respect to a random i.i.d. initial configuration η. From (2) we can make the following easy but important observation: An event A ∈ F holds P -a.s. if and only if it holds P η -a.s. for P µ -a.a. η ∈ S.
With this notation at hand, we are ready to state the main result of this note: Theorem 2.1. If, additionally to the above assumptions, the distribution µ is such that
Furthermore, for x, y ∈ Z d we denote by f (x, y) the probability that the underlying random walk ever hits y, if it starts at x. Thus, if we denote this random walk by (X n ) n≥0 , then f (x, y) = P (∃ n ≥ 0 : X n = y|X 0 = x). If we choose, according to our assumptions, ε > 0 such that ε ≤ p(±e) ≤ 1 − ε holds for each e ∈ E, then we have the following lower bound for the probabilities f (x, y): where we denote by |x| := max 1≤j≤d |x j | the maximum norm of a vector x = (x 1 , . . . , x d ) ∈ R d . This follows from the fact that one can get from x to y in at most d|y − x| steps. If y lies to the right of x, then one can do better. More precisely, we have the following bound.
A proof of Lemma 3.1 is given in Section 4. The following lemma about the behaviour of maxima of nonnegative i.i.d. random variables is one of the cornerstones of the proof of Theorem 2.1. Throughout, we denote by |A| the cardinality of the set A.
where c 2 , c 3 > 0 and β > 1 are constants (Here, β needs not necessarily be an integer). For i ∈ N define Then, for each finite constant c > 0 it holds that The proof of Lemma 3.2 is given in Section 4. Now we can proceed to the proof of Theorem 2.1, which uses a technique from [Pop01]. Choose the positive integer α such that where c 1 is the constant from Lemma 3.1. Further, we define Let us repeat the following important observation from [Pop01]: For recurrence of the frog model, everything that matters is the trajectories of the activated frogs. The actual moment that a certain frog gets activated is unimportant. Thus, if we know that a certain frog starting from vertex x ∈ Z d will sooner or later be at vertex y, we will say that the frogs at vertex y are activated by a frog from x, even if it is not the first frog to visit vertex y. We will call a vertex x ∈ Z d active if at least one active frog ever visits x. Fix k ∈ N with k ≥ 2 and define the event A k := at a certain moment and at some vertex x k ∈ V k \ V k−1 at least α (d+1)(k−1) frogs get activated by the initial frog starting from the origin .
In the following, we will implicitly be conditioning on the event A k . Note that the event A k only depends on the randomness coming from the path of the initial frog and from the values of the η x , where x ∈ V k . Define We will try to construct inductively sets and all the sites in D i are visited by frogs starting from B i−1 , i ∈ N. Furthermore, denoting for each i ∈ N and y ∈ F k+i−1 by ζ y the indicator of the following event {at least one active frog starting from B i−1 eventually visits y}, we require that holds for each i ∈ N. Note that by the definition of the sets F n in (3) we have and hence, since α 2 ≥ 4, we get Note that by (13) for all i ∈ N Thus, in principle, there are enough vertices in F k+i−1 to form disjoint sets B i and D i as required. The next thing to do is prove that, in fact, with high enough probability enough vertices in F k+i−1 are visited by frogs starting from B i−1 and also that the number of activated frogs is large enough for (11) to occur. Suppose that for 0 ≤ j ≤ i the sets B j and D j have already been succesfully constructed. We will soon be more precise about what this exactly means. For i ∈ Z + we define events G i,1 , G i,2 and G i as follows: Let If G i,1 happens than we can construct the set D i+1 by choosing exactly α (d+1)(i+k) vertices from F k+i that are visited by frogs starting from B i according to (15) and let B i+1 := F k+i \ D i+1 . Then, we define We will call the ith inductive step succesful if G i happens (given that A k , G 0 , . . . , G i−1 happen). As just explained, in this case it is possible to form subsets B i+1 , D i+1 of F k+i with all the desired properties. In what follows we will implicitly be conditioning on the event A k ∩G 0 ∩. . .∩G i−1 but will suppress this from the formulas for ease of notation. Also, for the computations which follow the following remark from [Pop01] will be crucial: Suppose that there are disjoint subsets A, B ⊆ Z d and we know that for each x ∈ A there is a frog starting from a vertex y ∈ B which activates the frogs at vertex x. Then, all the frogs starting from A are independent, since we only allow for interaction when an active frog is waking up a sleeping frog.
The next lemma gives an upper bound on the probability that the event G i,1 does not happen (conditionally on the event A k ∩ G 0 ∩ . . . ∩ G i−1 ).
Proof of Lemma 3.4. By inequalities (13) and (18) we have Thus, using the simple inequality P (X ≤ a) ≤ P (X ≤ b) if a < b we obtain Now note that we have for all d ≥ 1. Note that c does not depend on k. Hence, by (29), Chebyshev's inequality, inequalities (14), (19) and the second inequality in (20) we have for each i ≥ 1.
where c 4 = c −2 α 2d is also independent of k. For i = 0 we obtain the desired upper bound (26) by using the first inequality in (20) instead of the second one.
Next, we aim at bounding below the conditional probability of G i,2 given that G i,1 happens. Note that if G i,1 happens, the set B i+1 is well-defined and also we have where Y 1 , Y 2 , . . . are i.i.d. with the same distribution µ as the η x and we write a i := α (d+1)(k+i) , i ∈ N, for short. This follows directly from independence and (10).
Since the Y j are nonnegative and have infinite mean, we know from Cramér's theorem (see Theorem 2.2.3 and the following Remark (c)in [DZ10]) that with the notation S n := n j=1 Y j , n ∈ N, we have (32) P (S n ≤ n) ≤ 2 exp −nb , n ∈ N , where b = I(1) > 0 is the value at 1 of the Legendre-Fenchel transform I(x) of the cumulant generating function of Y 1 . That I(1) > 0 also follows from the fact that Y 1 is nonnegative and has infinite mean. From (31) and (32) we conclude that for each i ≥ 0 where we let b := I(1) > 0. Now, using ) and a i ≥ iα k , from Lemma 3.4 and (33) we immediately infer the following lemma.
Lemma 3.5. With the constant c 4 = c 4 (α, d) from Lemma 3.4 we have and note that From Lemma 3.4 and the multiplication rule for conditional probabilites, we obtain that under our initial assumption that the event A k happens we have where we have used the simple inequality Proof of Proposition 3.6. First note that, if k ≥ 1 is fixed, the sets D i , i ∈ N, satisfy D i ⊆ F k+i−1 and, hence, we have D i ∩ V k = ∅ and also D i ∩ j∈Z + B j = ∅ for each i ∈ N. The event A k does not depend on the values of the random variables η x for x / ∈ V k . Furthermore, the event j∈Z + G j only depends on the η x such that x ∈ A k ∪ j∈Z + B j . Thus, after conditioning on A k and on j∈Z + G j , by independence, we still have the i.i.d. property for the η x , where x ∈ i∈Z + D i . This will allow us to apply Lemma 3.2 below. Note that for each fixed configuration η x , x ∈ i∈Z + D i , by (4) we have where δ := ε dα 2k ∈ (0, 1) and . Then, by using Lemma 3.2 with L i := D i , c := − log δ, c 2 = α (k−1)(d+1) , c 3 = d + 1, r = d+1 2 and β = α we obtain that P µ -a.s.
Hence, P µ -a.s., there is a strictly increasing sequence (i m ) m∈N of positive integers such that for all m ∈ N (42) M im ≥ exp cα 2im .
By construction, for each i ∈ N, the frogs in D i get activated by frogs starting from B i−1 . Hence, by the remark before Lemma 3.3, all frogs in ∞ i=1 D i are independent. Hence, from (43) and the second Borel-Cantelli lemma we conclude that P µ -a.s.
Thus, also P 0 is visited infinitely often ∞ i=0 G i = 1 , as claimed. Now, note that from (39) and Proposition 3.6 we have P 0 is visited infinitely often ≥ P 0 is visited infinitely often Since lim k→∞ g(k) = 0 by (44) the proof of Theorem 2.1 will be completed, if we can show that P -a.s. the event A k happens for arbitrarily large k ∈ N. This is guaranteed by the following lemma.
Lemma 3.7. We have Proof of Lemma 3.7. Denote by π the path of the initial frog starting from the origin. By the properties of the underlying random walk, clearly, π contains infinitely many different vertices. We are going to use Lemma 3.2 with J = π, Y x = η x , x ∈ π, and r = (d + 1)/2. The pairwise disjoint sets L i , i ∈ N, are constructed inductively as follows: Let L 1 contain the first α 2 − 1 pairwise different vertices in π \ {0}. Clearly, L 1 ⊆ V 1 . If L i−1 for i ≥ 2 has already been constructed, let L i contain exactly the next α 2i − α 2i−2 vertices in π, which are not contained in V i−1 . Then, Note that the sets L i satisfy l i := |L i | ≥ c 2 α 2i , where c 2 = 1 − α −2 . Hence, from Lemma 3.2 (with c 3 = 2, c = 1, β = α and r = (d + 1)/2) we conclude that P µ -a.s.
In particular, P µ -a.s. for each k 0 ∈ N there exists a k ≥ k 0 such that implying that P -a.s. the event A k happens for arbitrarily large values of k.

Proofs of auxiliary lemmas
This section is devoted to the proofs of Lemmas 3.2 and 3.1. In order to prove Lemma 3.2 we need some facts about the behaviour of the maxima of nonnegative i.i.d. random variables, some of which rely on the following simple lemma on real sequences: (i) m n ≥ u −1 (n) for infinitely many n ∈ N (ii) y n ≥ u −1 (n) for infinitely many n ∈ N Proof of Lemma 4.1. Of course, (ii) trivially implies (i). So let us prove the converse. Let n 0 := inf{n ∈ N : m n ≥ u −1 (n)} . By (i) n 0 is finite and m n 0 = y n 0 . Hence, there is an n ∈ N such that y n ≥ u −1 (n). It thus suffices to show that for each n 1 ∈ N with y n 1 ≥ u −1 (n 1 ) there is a further n 2 > n 1 such that y n 2 ≥ u −1 (n 2 ). Since u −1 is unbounded, there is a k ∈ N such that u −1 (k) > y n 1 . By (i) there is an n > k such that m n ≥ u −1 (n) > u −1 (k) > y n 1 , since u −1 is also increasing. Now, choose n 2 ∈ {k + 1, . . . , n} minimal such that m n 2 ≥ u −1 (n). Then, m n 2 −1 < u −1 (n) and u −1 (n 2 ) ≤ u −1 (n) ≤ m n 2 = max(m n 1 −1 , y n 2 ) = y n 2 , since m n 1 −1 < u −1 (n).
For a sequence (Y j ) j∈N of nonnegative random variables and n ∈ N we define Lemma 4.2. Let (Y i ) i∈N be an i.i.d. sequence of nonnegative random variables and let u : [0, ∞) → [0, ∞) be an increasing and invertible function.
We first prove (a). Since the Y n are identically distributed and also u −1 is increasing, we have From the first Borel-Cantelli lemma we conclude that P (Y n ≥ u −1 (n) infinitely often ) = 0 and from Lemma 4.1 we obtain P (M ′ n ≥ u −1 (n) infinitely often ) = 0, which is equivalent to the assertion. Now, we turn to the proof of (b). By assumption we have By independence, the second Borel-Cantelli lemma implies that P (M ′ n ≥ u −1 (n) infinitely often ) ≥ P (Y n ≥ u −1 (n) infinitely often ) = 1 . Proof. (a) follows from Lemma 4.2 (a) by choosing u(x) = (log + (x)/c) r and noting that M ′ n < exp cn 1/r eventually implies max 1≤i≤⌊Ln r ⌋ Y i < exp cL 1/r n eventually. Similarly, (b) follows from Lemma 4.2 (b). To prove (c) choose a set G with P (G) = 1 according to (b) such that for all ω ∈ G there exists a strictly increasing sequence (n k ) k∈N (depending on ω) with M ′ n k (ω) ≥ exp cn 1/r k for all k ∈ N , wherec := c(inf i≥2 s i−1 /s i ) −1/r < ∞ by the assumptions on the sequence (s i ) i∈N . Then, for each ω ∈ G and for infinitely many values of i ∈ N there is a k = k i such that s i−1 < n k ≤ s i . The claim now follows from the chain of inequalities Note that by disjointness of the sets L i we have for the cardinality of k≤i L i : wherec > 0 is a constant depending only on c 2 , c 3 and β. Hence, for each i ∈ N, M ⋆ i is stochastically larger than M ′ s i from Corollary 4.3 (c) and the integer sequence (s i ) i∈N satisfies the above assumptions. In particular, we have Sketch of the proof of Lemma 3.1. First note that the probability f (x, y) is also the probability that the continuous time random walk (CTRW) (X t ) t>0 = (X (1) t , . . . , X (d) t ) t>0 corresponding to p ever visits y if it is starting at x. The benefit of working in continuous time here is that for CTRW the coordinates are independent, which is not true for discrete time random walks. Because of (1), letting τ := inf{t > 0 : X (1) t = y 1 }, we know that P x (τ < ∞) = 1. Furthermore, f (x, y) = P x ∃ t > 0 : X t = y ≥ P x X τ = y) = ∞ 0 P x X τ = y τ = t P x τ ∈ dt = ∞ 0 P x (X (2) t , . . . , X (d) t ) = (y 2 , . . . , y d ) P x τ ∈ dt ≥ γ 2 (y 1 −x 1 )