On the range of a two-dimensional conditioned simple random walk

Transcription

1 arxiv: v1 [math.pr] 1 Apr 2018 On the range of a two-dimensional conditioned simple random walk Nina Gantert 1 Serguei Popov 2 Marina Vachkovskaia 2 April 3, Technische Universität München, Fakultät für Mathematik, Boltzmannstr. 3, Garching, Germany gantert@ma.tum.de 2 Department of Statistics, Institute of Mathematics, Statistics and Scientific Computation, University of Campinas UNICAMP, rua Sérgio Buarque de Holanda 651, , Campinas SP, Brazil s: {popov,marinav}@ime.unicamp.br Abstract We consider the two-dimensional simple random walk conditioned on never hitting the origin. This process is a Markov chain, namely it is the Doob h-transform of the simple random walk with respect to the potential kernel. It is known to be transient and we show that it is almost recurrent in the sense that each infinite set is visited infinitely often, almost surely. We prove that, for a large set, the proportion of its sites visited by the conditioned walk is approximately a Uniform[0, 1] random variable. Also, given a set G R 2 that does not surround the origin, we prove that a.s. there is an infinite number of k s such that kg Z 2 is unvisited. These results suggest that the range of the conditioned walk has fractal behavior. Keywords: random interlacements, range, transience, simple random walk, Doob s h-transform AMS 2010 subject classifications: Primary 60J10. Secondary 60G50, 82C41. 1

2 1 Introduction and results We start by introducing some basic notation and defining the conditioned random walk Ŝ, the main object of study in this paper. Besides being interesting on its own, this random walk is the main ingredient in the construction of the two-dimensional random interlacements of [3, 4]. Write x y if x and y are neighbours in Z 2. Let S n, n 0 be two-dimensional simple random walk, i.e., the discrete-time Markov chain with state space Z 2 and transition probabilities defined in the following way: 1, if x y, P xy = 4 1 0, otherwise. We assume that all random variables in this paper are constructed on a common probability space with probability measure P and we denote by E the corresponding expectation. When no confusion can arise, we will write P x and E x for the law and expectation of the 1 random walk started from x. Let τ 0 A = inf{k 0 : S k A}, 2 τ 1 A = inf{k 1 : S k A} 3 be the entrance and the hitting time of the set A by simple random walk S we use the convention inf = +. For a singleton A = {x}, we will write τ i A = τ i x, i = 0, 1, for short. One of the key objects needed to understand the twodimensional simple random walk is the potential kernel a, defined by ax = P0 [S k =0] P x [S k =0]. 4 k=0 It can be shown that the above series indeed converges and we have a0 = 0, ax > 0 for x 0. It it straightforward to check that the function a is harmonic outside the origin, i.e., 1 4 y:y x 1 the simple one, or the conditioned one defined below ay = ax for all x

3 Also, using 4 and the Markov property, one can easily obtain that 1 4 x 0 ax = 1, which implies by symmetry that ax = 1 for all x 0. 6 Observe that 5 immediately implies that as k τ0 0 is a martingale, we will repeatedly use this fact in the sequel. Further, once can show that with γ = the Euler-Mascheroni constant ax = 2 π ln x + 2γ + 3 ln 2 π + O x 2 7 as x, cf. Theorem of [7]. Let us define another random walk Ŝn, n 0 on Z 2 \ {0} in the following way: its transition probability matrix equals compare to 1 ay, if x y, x 0, P xy = 4ax 8 0, otherwise. It is immediate to see from 5 that the random walk Ŝ is indeed well defined. The walk Ŝ is the Doob h-transform of the simple random walk, under the condition of not hitting the origin see Lemma 3.3 of [4] and its proof. Let τ 0, τ 1 be defined as in 2 3, but with Ŝ in the place of S. We summarize the basic properties of the walk Ŝ in the following Proposition 1.1. The following statements hold: i The walk Ŝ is reversible, with the reversible measure µ x := a 2 x. ii In fact, it can be represented as a random walk on the two-dimensional lattice with conductances axay, x, y Z 2, x y. iii Let N be the set of the four neighbours of the origin. 1/aŜk τ 0 N is a martingale. Then the process iv The walk Ŝ is transient. v Moreover, for all x 0 P x [ τ1 x < ] = 1 1 2ax, 9 3

4 and for all x y, x, y 0 P x [ τ0 y < ] = P x [ τ1 y < ] = ax + ay ax y. 10 2ax The statements of Proposition 1.1 are not novel they appear already in [4], but we found it useful to collect them here for the sake of completeness and also for future reference. We will prove Proposition 1.1 in the next section. It is curious to observe that 10 implies that, for any x, P x [ τ 1 y < ] converges to 1 as y. 2 As noted in [4], this is related to the remarkable fact that if one conditions on a very distant site being vacant, then this reduces the intensity near the origin of the two-dimensional random interlacement process by the factor of four. Let be the Euclidean norm. Define the discrete ball Bx, r = {y Z 2 : y x r} note that x and r need not be integer, and abbreviate Br := B0, r. internal boundary of A Z 2 is defined by A = {x A : there exists y Z 2 \ A such that x y}. Now we introduce some more notations and state the main results. For a set T Z + thought of as a set of time moments let Ŝ T = } {Ŝm m T The be the range of the walk Ŝ with respect to that set. For simplicity, we assume in the following that the walk Ŝ starts at a fixed neighbour x 0 of the origin, and we write P for P x0 it is, however, clear that our results hold for any fixed starting position of the walk. For a nonempty and finite set A Z 2, let us consider random variables A Ŝ [0, RA =, A A \ Ŝ [0, VA = = 1 RA; A that is, RA respectively, VA is the proportion of visited respectively, unvisited sites of A by the walk Ŝ. Let us also abbreviate, for M 0 > 0, l A = A 1 max A B y, n. y A ln M 0 n 11 Our main result is the following 4

5 Theorem 1.2. Let M 0 > 0 be a fixed constant, and assume that A Bn \ Bn ln M 0 n. Then, for all s [0, 1], we have, with positive constants c 1,2 depending only on M 0, P[VA s] s ln 1/3 ln 2/3, c 1 + c2 l A 12 and the same result holds with R on the place of V. The above result means that if A Bn \ Bε 0 n is big enough and well distributed, then the proportion of visited sites has approximately Uniform[0, 1] distribution. In particular, one can obtain the following Corollary 1.3. Assume that D R 2 is a bounded open set. Then both sequences RnD Z 2, n 1 and VnD Z 2, n 1 converge in distribution to the Uniform[0, 1] random variable. Indeed, it is straightforward to obtain it from Theorem 1.2 since nd Z 2 is of order n 2 as n note that D contains a disk, and so l nd Z 2 will be of order ln 2M 0 n. We can choose M 0 large enough such that the r.h.s. of 12 goes to 0. Also, we prove that the range of Ŝ contains many big holes. To formulate this result, we need the following Definition 1.4. We say that a set G R 2 does not surround the origin, if there exists c 1 > 0 such that G Bc 1, i.e., G is bounded; there exist c 2,3 > 0 and a function f = f 1, f 2 : [0, 1] R 2 such that f0 = 0, f1 = c 1, f 1,2s c 2 for all s [0, 1], and inf f 1s, f 2 s y c 3, s [0,1],y G i.e., one can escape from the origin to infinity along a path which is uniformly away from G. Then, we have Theorem 1.5. Let G R 2 be a set that does not surround the origin. Then, P [ ng Ŝ[0, = for infinitely many n ] =

6 We also establish some additional properties of the conditioned walk Ŝ, which will be important for the proof of Theorem 1.5 and are of independent interest. Consider an irreducible Markov chain. Recall that a set is called recurrent with respect to the Markov chain, if it is visited infinitely many times almost surely; a set is called transient, if it is visited only finitely many times almost surely. It is clear that any nonempty set is recurrent with respect to a recurrent Markov chain, and every finite set is transient with respect to a transient Markov chain. Note that, in general, a set can be neither recurrent nor transient think e.g. of the simple random walk on a binary tree, fix a neighbour of the root and consider the set of vertices of the tree connected to the root through this fixed neighbour. In many situations it is possible to characterize completely the recurrent and transient sets, as well as to answer the question if any set must be either recurrent or transient. For example, for the simple random walk in Z d, d 3, each set is either recurrent or transient and the characterization is provided by the Wiener s test see e.g. Corollary of [7], formulated in terms of capacities of intersections of the set with exponentially growing annuli. Now, for the conditioned two-dimensional walk Ŝ the characterization of recurrent and transient sets is particularly simple: Theorem 1.6. A set A Z 2 is recurrent with respect to infinite. Ŝ if and only if A is Next, we recall that a Markov chain has the Liouville property, see e.g. Chapter IV of [12], if all bounded harmonic with respect to that Markov chain functions are constants. Since Theorem 1.6 implies that every set must be recurrent or transient, we obtain the following result as its corollary: Theorem 1.7. The conditioned two-dimensional walk Ŝ has the Liouville property. These two results, besides being of interest on their own, will also be operational in the proof of Theorem Some auxiliary facts and proof of Proposition 1.1 For A Z d, recall that A denotes its internal boundary. We abbreviate τ 1 R = τ 1 BR. We will consider, with a slight abuse of notation, the function ar = 2 π ln r + 2γ + 3 ln 2 π 6

7 of a real argument r 1. To explain why this notation is convenient, observe that, due to 7, we may write, as r, x 1 νyay = ar + O 14 r y Bx,r for any probability measure ν on Bx, r. For all x Z 2 and R 1 such that x, y BR and x y, we have P x [τ 1 R < τ 1 y] = ax y ar + O, 15 R 1 y 1 as R. This is an easy consequence of the optional stopping theorem applied to the martingale as n τ0 0, together with 14. Also, an application of the optional stopping theorem to the martingale 1/aŜk τ 0 N yields P x [ τ 1 R < τ 1 r] = ar 1 ax 1 + OR 1 ar 1 ar 1 + Or 1, 16 for 1 < r < x < R <. 1 r x Sending R to infinity in 16 we see that for P x [ τ 1 r = ] = 1 ar + Or ax We need the fact that the walks S and Ŝ are almost indistinguishable on a distant from the origin set. For A Z 2, denote Γ x A to be the set of all finite nearest-neighbour trajectories that start at x A \ {0} and end when entering A for the first time. For V Γ x A write S V if there exists k such that S 0,..., S k V and the same for the conditioned walk Ŝ. We write Γx 0,R for Γ x BR. Lemma 2.1. Assume that V Γ x 0,R ; then we have P x [S V τ 1 R < τ 1 0] = P x [Ŝ V ] 1 + OR ln R Proof. This is Lemma 3.3 i of [4]. If A A are finite subsets of Z 2, then the excursions between A and A are pieces of nearest-neighbour trajectories that begin on A and end on A, see Figure 1, which is, hopefully, self-explanatory. We refer to Section 3.4 of [4] for formal definitions. 7

8 A A Figure 1: Excursions pictured as bold pieces of trajectories of random walks between A and A. Proof of Proposition 1.1. It is straightforward to check i iii directly, we leave this task for the reader. Item iv the transience follows from iii and Theorem of [8]. As for v, we first observe that 9 is a consequence of 10, although it is of course also possible to prove it directly, see Proposition 2.2 of [4]. Indeed, using 8 and then 10, 5 and 6, one can write P x [ τ1 x < ] = 1 4ax = 1 4ax ayp y [ τ1 x < ] y x y x = 1 1 2ax. 1 ay + ax ay x 2 Now, to prove 10, we essentially use the approach of Lemma 3.7 of [4], although here the calculations are simpler. Let us define note that all the probabilities below are for the simple random walk S h 1 = P x [τ 1 0 < τ 1 R], h 2 = P x [τ 1 y < τ 1 R], q 12 = P 0 [τ 1 y < τ 1 R], 8

9 q 21 p 2 y BR 0 q 12 x p 1 1 p 1 + p 2 Figure 2: Trajectories for the probabilities of interest. q 21 = P y [τ 1 0 < τ 1 R], p 1 = P x [τ 1 0 < τ 1 R τ 1 y], p 2 = P x [τ 1 y < τ 1 R τ 1 0], see Figure 2. Using 15 and in addition the Markov property and 5 for 21 we have for x, y 0, x y which implies that ax h 1 = 1 ar + OR 1, 19 ax y h 2 = 1 ar + OR 1 y, 20 ay q 12 = 1 ar + OR 1 y, 21 q 21 = 1 ay ar + OR 1, 22 lim 1 h 1aR = ax, 23 R 9

10 lim 2aR = ax y, R 24 lim 12aR = ay, R 25 lim 21aR = ay. R 26 Observe that, due to the Markov property, it holds that h 1 = p 1 + p 2 q 21, h 2 = p 2 + p 1 q 12. Solving these equations with respect to p 1, p 2, we obtain p 1 = h 1 h 2 q 21 1 q 12 q 21, 27 p 2 = h 2 h 1 q 12 1 q 12 q Let us denote h 1 = 1 h 1, h 2 = 1 h 2, q 12 = 1 q 12, q 21 = 1 q 21. Next, using Lemma 2.1, we have that P x [ τ 1 y < τ 1 R] = P x [τ 1 y < τ 1 R τ 1 R < τ 1 0] 1 + or 1 = P x[τ 1 y < τ 1 R < τ 1 0] 1 + or 1 P x [τ 1 R < τ 1 0] = p 21 q or 1 1 h 1 = h 2 h 1 q 12 1 q 21 1 q 12 q 21 1 h or 1 = h 1 + q 12 h 2 h 1 q 12 q 21 q 12 + q 21 q 12 q 21 h Since P x [ τ 1 y < ] = lim R P x [ τ 1 y < τ 1 R], using we obtain 10 observe that the product terms in 29 are of smaller order and will disappear in the limit. We now use the ideas contained in the last proof to obtain some refined bounds on the hitting probabilities for excursions of the conditioned walk. Let us assume that x n ln M 0 n and y A, where the set A is as in Theorem 1.2. Also, abbreviate R = n ln 2 n. 10

11 Ex 0 Ex1 A Bn \ B 0 n ln M 0 n Ex 2 Bn Bn ln 2 n Figure 3: Excursions and their visits to A Lemma 2.2. In the above situation, we have P x [ τ 1 y < τ 1 R] = 1+Oln 3 n axar + ayar ax yar axay. ax2ar ay 30 Proof. Analogously to 29, using together with Lemma 2.1, we obtain that P x [ τ 1 y < τ 1 R] = P x [τ 1 y < τ 1 R τ 1 R < τ 1 0] 1 + On 1 = B 1 B On 1, where ay ax B 1 = ar + OR 1 ar + OR 1 + ay ar + OR 1 y ax y ar + OR 1 y axay ar + OR 1 ar + OR 1 y = 1 + OR ln R 1 ay axar + ayar ax yar axay ar 1 + Oln 3 na 2 R 11

12 = 1 + Oln 3 n ay ar axar + ayar ax yar axay, a 2 R and B 2 = ax ay ar + OR 1 ar + OR 1 y + ay ar + OR 1 a 2 y ar + OR 1 ar + OR 1 y = 1 + OR ln R 1 ax ar 2ayaR a2 y + Oln 1 n 1 + Oln 3 na 2 R = 1 + Oln 3 n ax ar 2ayaR a2 y. a 2 R Gathering the pieces, we obtain Proofs of the main results We start with Proof of Theorem 1.2. First, we describe informally the idea of the proof. We consider the visits to the set A during excursions of the walk from Bn to Bn ln 2 n, see Figure 3. The crucial argument is the following: the randomness of VA comes from the number of excursions and not from the excursions themselves. If the number of excursions is around c, then it is possible to show ln using a standard weak-lln argument that the proportion of uncovered sites in A is concentrated around e c. On the other hand, that number of excursions can be modeled roughly as Y, where Y is an Exponential1 random variable. ln Then, P[VA s] P[Y ln s 1 ] = s, as required. We now give a rigorous argument. Let Ĥ be the conditional entrance measure for the conditioned walk Ŝ, i.e., Ĥ A x, y = P x [Ŝ τ1 A = y τ 1 A < ]. 31 Let us first denote the initial piece of the trajectory by Ex 0 = Ŝ[0, τn ]. Then, we consider a Markov chain Ex k, k 1 of excursions between Bn and Bn ln 2 n, defined in the following way: for k 2 the initial site of Ex k is chosen according to the measure ĤBn z k 1,, where z k 1 Bn ln 2 n is the last 12

13 site of the excursion Ex k 1 ; also, the initial site of Ex 1 is the last site of Ex 0 ; the weights of trajectories are chosen according to 8 i.e., each excursion is an Ŝ-walk trajectory. It is important to observe that one may couple Ex k, k 1 with the true excursions of the walk Ŝ in an obvious way: one just picks the excursions subsequently, each time tossing a coin to decide if the walk returns to Bn. Let ψ n = min P x [ τn = ] x Bn ln 2 n be the minimal probability to avoid Bn, starting at sites of Bn ln 2 n. Using 17 it is straightforward to obtain that P x [ τn = ] = for any x Bn ln 2 n, and so it also holds that ψ n = ln 1 + On ln ln 1 + On ln Let us consider a sequence of i.i.d. random variables η k, k 0 such that P[η k = 1] = 1 P[η k = 0] = ψ n. Let N = min{k : η k = 1}, so that N is a Geometric random variable with mean ψn 1. Now, 32 implies that P x [ τn = ] ψ n O ln n for any x Bn ln 2 n, so it is clear 2 that N can be coupled with the actual number of excursions N in such a way that N N a.s. and P[N N] On Note that this construction preserves the independence of N from the excursion sequence Ex k, k 1 itself. Define A R k Ex0 Ex 1... Ex k =, A and V k = A \ Ex0 Ex 1... Ex k A = 1 R k 2 Let Z n, n 1 be a sequence of {0, 1}-valued random variables adapted to a filtration F n, n 1 and such that P[Z n+1 = 1 F n ] [p, p + ε] a.s.. Then it is elementary to obtain that the total variation distance between the random variable min{k : Z k = 1} and the Geometric random variable with mean p 1 is bounded above by Oε/p. 13

14 to be the proportions of visited and unvisited sites in A with respect to the first k excursions together with the initial piece Ex 0. Now, it is straightforward to check that 30 implies that, for any x Bn and y A P x [ τ1 y < τ 1 n ln 2 n ] ln ln = 1 + O 34 and, for y, z Bn \ B n such that y z = n/b with b 2 ln M 0 n 2 ln M 0 n P z [ τ1 y < τ 1 n ln 2 n ] = 2 ln + ln b 1 + O For y A and a fixed k 1 consider the random variable ξ k y = 1{y / Ex 0 Ex 1... Ex k }, ln so that V k = A 1 y A ξk y. Now 34 implies that, for all j 1, and 35 implies that for any y A. Let µ k y µ k P[y / Ex j ] = 1 ln 1 + O ln ln P[y / Ex 0 Ex 1 ] = 1 O = Eξ y k. Then we have y = P[y / Ex 0 Ex 1... Ex k ] ln = 1 O 1 = exp k 1 + O k 1 + ln Next, we need to estimate the covariance of ξ y k First note that, for any x Bn ln 1 + O ln,. 35 ln k and ξ k z in case y z n ln M 0 n. [ P x {y, z} Ex1 = ] [ ] = 1 P x [y Ex 1 ] P x [z Ex 1 ] + P x {y, z} Ex1 ln ln [ ] = O + P x {y, z} Ex1 14

15 by 34, and P x [ {y, z} Ex1 ] = Px [ max{ τ1 y, τ 1 z} < τ 1 n ln 2 n ] = P x [ τ1 y < τ 1 z < τ 1 n ln 2 n ] + P x [ τ1 z < τ 1 y < τ 1 n ln 2 n ] P x [ τ1 y < τ 1 n ln 2 n ] P y [ τ1 z < τ 1 n ln 2 n ] + P x [ τ1 z < τ 1 n ln 2 n ] P z [ τ1 y < τ 1 n ln 2 n ] ln M 0 ln ln 2 = O. Therefore, similarly to 36 we obtain Eξ k y ξ z k = exp 2k 1 + O ln 1 + O k 1 + ln ln, which, together with 36, implies after some elementary calculations that, for all y, z A such that y z n ln M 0 n uniformly in k, since ln covξ k y ln, ξ z k = O ln 2 + k exp 2k ln = O ln uniformly in k. Recall the notation l A from 11. Now, using Chebyshev s inequality, we write [ A 1 P ] µ k > ε y A ξ k y ε A 2 Var ξ y k = ε A 2 = ε A 2 y,z A y A y covξ y k, ξ z k y,z A, y z < n ln M 0 n covξ y k, ξ z k + 15 y,z A, y z n ln M 0 n covξ y k, ξ z k 37

16 Let ε A 2 A By, n ln ln M 0 n + A 2 O y A ln ε 2 l A + ε 2 O. 38 Φ s = min { k : V k s } be the number of excursions necessary to make the unvisited proportion of A at most s. We have P[VA s] = P[Φ s N] = P[Φ s N, N = N] + P[Φ s N, N N] = P[Φ s N] + P[Φ s N, N N] P[Φ s N, N N], so, recalling 33, P[VA s] P[Φ s N] P[N N] On Next, we write P[Φ s N] = E P[ N Φ s Φ s ] = E1 ψ n Φs, 40 and concentrate on obtaining lower and upper bounds on the expectation in the right-hand side of 40. For this, assume that s 0, 1 is fixed and abbreviate 1/3 ln δ n = kn 1 = 1 δ n ln s, ln k n + 1 = 1 + δ n ln s ; ln we also assume that n is sufficiently large so that δ n 0, 1 2 and 1 < k n < k + n. Now, according to 36, µ k± n y = exp 1 ± δ n ln s O = s exp ln s 1 ± δ n + O k n ± k n ± 1 ln + ln

17 = s 1 + O δ n ln s 1 ln ln s 1, so in both cases it holds that observe that s ln s 1 1/e for all s [0, 1] µ k± n y = s + O δ n + With a similar calculation, one can also observe that We then write, using 41 P[Φ s > k n + ] = P[V k+ n > s] [ = P A 1 y A ln = s + Oδ n ψ n k± n = s + Oδ n. 42 ξ k+ n y ] > s [ = P A 1 ξ k+ n y µ k+ n y > s A ] 1 µ k+ n y y A y A [ = P A ] 1 ξ k+ n y µ k+ n y > Oδ n. 43 y A Then, 38 implies that P[Φ s > k n + ] O l A ln Quite analogously, one can also obtain that P[Φ s < kn ] O Using 42 and 44, we then write l A ln E1 ψ n Φs E 1 ψ n Φs 1{Φ s k + n } 1 ψ n k+ n P[Φ s k n + ] ln 1/3 s O 1 O 2/3 ln 1/ /3 ln 1/ l A ln 2/3 ln 1/3 +, 46 17

18 and, using 42 and 45, E1 ψ n Φs = E 1 ψ n Φs 1{Φ s k n } + E 1 ψ n Φs 1{Φ s < k n } 1 ψ n k n + P[Φ s < kn ] ln 1/3 s + O ln 2/3 ln 1/3 + 1 O l A Therefore, using also 39 40, we obtain 12, thus concluding the proof of Theorem 1.2. Next, we will prove Theorems 1.6 and 1.7, since the latter will be needed in the course of the proof of Theorem 1.5. Proof of Theorem 1.6. Clearly, we only need to prove that every infinite subset of Z d is recurrent for Ŝ. Basically, this is a consequence of the fact that, due to 10, lim y P x 0 [ τ1 y < ] = 1 2 for any x 0 Z 2. Indeed, let Ŝ0 = x 0 ; since A is infinite, by 48 one can find y 0 A and R 0 such that {x 0, y 0 } BR 0 and P x0 [ τ1 y 0 < τ 1 R 0 ] 1 3. Then, for any x 1 BR 0, we can find y 1 A and R 1 > R 0 such that y 1 BR 1 \ BR 0 and P x1 [ τ1 y 1 < τ 1 R 1 ] 1 3. Continuing in this way, we can construct a sequence R 0 < R 1 < R 2 <... depending on the set A such that, for each k 0, the walk Ŝ hits A on its way from BR k to BR k+1 with probability at least 1, regardless of the past. This 3 clearly implies that A is a recurrent set. Proof of Theorem 1.7. Indeed, Theorem 1.6 implies that every subset of Z 2 must be either recurrent or transient, and then Proposition 3.8 in Chapter 2 of [10] implies the Liouville property. Still, for the reader s convenience, we include the 48 18

19 proof here. Assume that h : Z 2 \ {0} R is a bounded harmonic function for Ŝ. Let us prove that lim inf hy = lim sup hy, 49 y y that is, h must have a limit at infinity. Indeed, assume that 49 does not hold, which means that there exist two constants b 1 < b 2 and two infinite sets B 1, B 2 Z 2 such that hy b 1 for all y B 1 and hy b 2 for all y B 2. Now, on one hand hŝn is a bounded martingale, so it must a.s. converge to some limit; on the other hand, Theorem 1.6 implies that both B 1 and B 2 will be visited infinitely often by Ŝ, and so hŝn cannot converge to any limit, thus yielding a contradiction. This proves 49. Now, if lim y hy = c, then it is easy to obtain from the Maximum Principle that hx = c for any x. This concludes the proof of Theorem 1.7. Finally, we are able to prove that there are big holes in the range of Ŝ: Proof of Theorem 1.5. Clearly, if G does not surround the origin in the sense of Definition 1.4, then G Bc 1 \ Bc 3. For the sake of simplicity, let us assume that G B1 \ B1/2; the general case can be treated in a completely analogous way. Consider the two sequences of events E n = { τ 1 2 3n 1 G < τ 1 2 3n, Ŝj > 2 3n 1 for all j τ 1 2 3n }, E n = { Ŝj > 2 3n 1 for all j τ 1 2 3n } and note that E n E n and 2 3n 1 G Ŝ[0, = on E n. Our goal is to show that a.s. an infinite number of events E n, n 1 occurs. Observe, however, that the events in each of the above two sequences are not independent, so the basic second Borel-Cantelli lemma will not work. In the following, we use a generalization of the second Borel-Cantelli lemma, known as the Kochen-Stone theorem [6]: for any sequence of events A 1, A 2, A 3,... it holds that [ P k=1 ] 1{A k } = lim sup k k i=1 P[A i] 2 k i,j=1 P[A i A j ]. 50 Let us prove that there exists a positive constant c 4 such that P[E n ] c 4 n for all n

20 Indeed, since G B1 \ B1/2 does not surround the origin, by comparison with Brownian motion it is elementary to obtain that, for some c 5 > 0, P x [ τ1 2 3n 1 G < τ 1 2 3n ] > c 5 for all x B2 3n 1. Lemma 2.1 then implies that, for some c 6 > 0, P x [ τ1 2 3n 1 G < τ 1 2 3n ] > c 6 52 for all x B2 3n 1. Let us denote, recalling 7, γ = π 1 2 2γ+3 ln 2. Using 17, we then obtain 2 ln 2 ln 2 2γ+3 ln 2 π = [ P z Ŝ j > 2 3n 1 for all j 0 ] = 1 a23n 1 + O2 3n a2 3n + O2 3n 1 = 1 + o2 3n. 53 3n + γ for any z B2 3n. The inequality 51 follows from 52 and 53. Now, we need an upper bound for P[E m E n ], m n. Clearly, E m E n E m E n, and note that the event E m E n means that the particle hits B2 3n before B2 3m 1 starting from a site on B2 3m, and then never hits B2 3n 1 starting from a site on B2 3n. So, again using 17 and Lemma 2.1, we write analogously to 53 and also omitting a couple of lines of elementary calculations P[E m E n ] P[E m E n] = a23m 1 1 a2 3m 1 + O2 3m a2 3m 1 1 a2 3n 1 + O2 3m 1 a23n 1 + O2 3n a2 3n + O2 3n 1 = 1 + o2 3m. 54 3n m + 13m + γ Now, 51 implies that k i=1 P[E i] c 9 ln k, and 54 implies after some elementary calculations that k i,j=1 P[E i E j ] c 10 ln 2 k. So, applying 50 to the sequence of events E n, n 1, we obtain that [ ] P 1{E k } = c 11 > 0. k=1 Now, note that, again due to Proposition 3.8 in Chapter 2 of [10], the Liouville property implies that every tail event must have probability 0 or 1, and so the probability in the above display must be equal to 1. This concludes the proof of Theorem

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