Biased random walks on subcritical percolation clusters with an infinite open path

Transcription

1 Biased random walks on subcritical percolation clusters with an infinite open path Application for Transfer to DPhil Student Status Dissertation Annika Heckel March 3, 202 Abstract We consider subcritical percolation on the half-plane Z Z + 0 and additionally open the bonds along the line Z {0}. A biased random walk starting at the origin which can move along the open bonds shows anomalous behaviour due to trapping: If the bias is small enough, the speed tends to a positive deterministic limit, but if the bias is large enough, the speed tends to 0. We will also give a generalisation of this result attaching other types of subgraphs of Z d to an infinite line. Setting and main result Consider subcritical bond percolation with p < p c = 2 on the half-plane Z Z+ 0 with nearest-neighbour edges and additionally open all bonds along the line Z {0}. Let G = (V, E) denote the open cluster containing the line Z {0}. Since the percolation is subcritical, G is almost surely an infinite line with finite clusters attached to it. We now start a biased random walk Z n = (X n, Y n ) with bias β > towards the right at the origin (0, 0). More specifically, we assign to each edge e = (x, y)(x, y ) the weight w e = β x x e is open where x x denotes the maximum of x and x. For z Z 2, let π(z) be the sum of the edge weights w e of all edges incident to z. Then the next step of the random walk is distributed proportionally to the edge weights. In other words, if P β ω denotes the law of this random walk for some bond configuration ω, and z, z are vertices joined by an edge e such that π(z) > 0, then P β ω(z n+ = z Z n = z) = w e π(z).

2 This means that if the random walk is at a vertex v = (x, y) V with degree d(v) (in G), if the bond (x, y)(x +, y) is open, it jumps to (x +, y) with β probability β+d(v) and to all other neighbours of v with probability β+d(v). If the bond (x, y)(x +, y) is closed, the next step is distributed uniformly amongst the neighbours of v. Let P β denote the joint law of the percolation configuration and the random walk. We sometimes omit the indices β or ω in P β and P β ω for the sake of simpler notation if the context is clear. We will prove the following theorem: Theorem. There are constants < β l β u < which only depend on p such that a) if < β < β l, then there is a positive constant v = v(p, β) such that lim n X n /n = v P β -almost surely, i.e. the random walk is ballistic. b) if β > β u, then X n /n 0 P β -almost surely. Furthermore, if β l β β u, X n /n v = v(p, β) 0 P β -almost surely. The described anomalous behaviour where a small, but existing bias yields a positive speed while under a large bias the asymptotic speed is zero, is due to the random walk spending a large amount of time in traps or dead ends in the random structure. In our case, the traps are finite percolation clusters extending very far towards the right. Once the random walk moves into one of the traps, a large bias means that it takes a long time to move out of the trap against the direction of the bias. Lyons, Pemantle and Peres 6 considered the case of a random walk on Galton Watson trees with positive extinction probability and showed this behaviour for a random walk with a bias which is directed away from the root. For a biased random walk on an infinite percolation cluster on Z d, which is a model for transport in an inhomogeneous medium under an external field first introduced by Barma and Dhar, this behaviour was proved by Berger, Gantert and Peres 2 and by Snitzman 8. Recent progress has been made on this model in 3 and 4. Later, in Section 3, we will consider the case of an infinite line with more general subgraphs of Z d attached to it which may act as traps, and give a criterion for the ballisticity or non-ballisticity of a biased random walk in these environments. Unlike many other results in this area, our proof does not require regeneration times, and as a consequence the criterion and the derived formula for the asymptotic speed does not depend on such regeneration times and can in many cases be easily calculated. 2

3 This dissertation is organised in the following way. The main result will be proved in Section 2. It will turn out that the key point for the ballisticity or non-ballisticity of the random walk is the expected duration of its first visit to a section of the connected component containing the origin, weighted with the inverse of the length of the section roughly speaking, a section is a finite cluster attached to the line Z {0} and the line segment under it. The term section will be specified in Section 2. and some basic properties of the random walk will be discussed. In Section 2.2, we will prove some lemmas about the distribution of the size of sections. In Section 2.3, the weighted durations of visit lengths to a section will be studied. In Section 2.3., it will be shown that there are values < β l β u < such that for β < β l, the expected weighted duration of the first visit to a section is finite, whereas for β > β u, it is infinite. These will also be the values β l and β u in Theorem. In Section 2.3.2, this will be extended to the weighted overall time spent within a section. Next, in Section 2.3.3, it will be shown that the sequence of weighted overall time the random walk spends in the section containing (x, 0), x Z, is ergodic. This will be needed later on to show that their average tends to their expectation almost surely. Finally, the proof of Theorem will be concluded in Section 2.4. In Section 3, a generalisation of our result will be discussed and our methods will be applied to two other special cases. 2 Proof of the main result 2. Preliminaries and transience of the random walk The idea behind the term section is just a finite open cluster attached to the line Z {0} together with the line segment and smaller clusters below it. More formally, consider the infinite connected open cluster G = (V, E) which contains the vertices in Z {0} together with the open bonds E. We call a bond b between some sites (x, 0) and (x +, 0) on the line Z 0 a bridge if (G, E \ b) is disconnected. Suppose we cut all the bridges, i.e. we look at the graph G = (V, E \ {b b is a bridge}). Then the components of G are called sections. Cutting the bridges separates Z {0} into components, and there cannot be any open paths between different components of Z {0} using only vertices in the upper half-plane because otherwise one the adjacent bonds of the components would not be a bridge. So every section has a corresponding connected line segment. All these line segments are almost surely finite: Otherwise, suppose that w.l.o.g. {(x, 0), (x +, 0),...} is contained in the same section, then for every y x, (y, 0)(y +, 0) is not a bridge, so there is an open path from a vertex (l, 0) to a vertex (r, 0), l x < r, which apart from its end points only uses vertices of the upper half-plane. Since any two paths from (l, 0) to (r, 0) and from (l 2, 0) to (r 2, 0), respectively, intersect if l l 2 r r 2, it follows that 3

4 there are infinitely many integers r x such that there is an open path from a vertex (x, 0), x x, to (r, 0) in the upper half-plane. In particular, C((r, 0)) r x, where C((r, 0)) denotes the open cluster containing (r, 0) in the original percolation configuration. Since we are considering subcritical percolation, the cluster size distribution decays exponentially, so there is a constant η > 0 such that for every r, P(C((r, 0)) r x) e η(r x). But then r x P(C((r, 0)) r x) <, so by the Borel-Cantelli lemma, there are almost surely e η only finitely many r x with C((r, 0)) r x. So all line segments are almost surely finite. Let S(x) denote the section containing the vertex (x, 0). For a section S, let l S, r S Z such that (l S, 0) and (r S, 0) are the leftmost and rightmost vertices of S (Z {0}), respectively. Then S contains {l S, l S +,..., r S } {0} completely and no other vertices from Z {0}. By the section definition, (l S, 0)(l S, 0) and (r S, 0)(r S +, 0) are bridges. Let l(s) = r S l S + be the length of the section S on the line Z {0}. Since every section corresponds to a finite connected line segment on Z {0}, the sections appear linearly one after the other along the line Z {0}. Let S 0 := S(0) and number the sections S, S 2,... and S, S 2,... along the line Z 0. Denote by i x the index of the section S(x), i.e. S(x) = S ix. S 0 S S 2 S l 2 r 2 r = l l 0 0 r 0 l r l 2 Figure : Sections (S i ) i Z. We will often consider the corresponding electric network where each edge e is assigned the conductance w e, or equivalently resistance w e, and use the terms weight and conductance interchangeably. For the close relationship between reversible random walks and electric networks, see 7. The random walk is transient if and only if the conductance between the origin and infinity is positive (or equivalently, the resistance is finite). It can easily be seen by bounding the conductance from below by the positive conductance along the open path Z {0} that the random walk is transient. Since G is almost surely a concatenation of finite sections, this implies that for almost all configurations ω, ( ) lim X n = lim X n = P β n n ω-almost surely On the subgraph G {(x, y) x 0}, for all ω such that all sections are finite, the 4

5 resistance between the origin and infinity can be bounded from below by the sum of the resistances of the bridges, which is infinite. Therefore, the random walk is recurrent on the subgraph G {(x, y) x 0}, and hence, for almost all ω, lim X n = P β n ω-a.s. () For technical reasons, we will consider a slightly modified version of our random walk from now on: Instead of starting at the origin (0, 0), we will start at the left border vertex (l S(0), 0) of the section S(0). Both the original and the new version of the random walk almost surely go to infinity and therefore reach the vertex (r S(0), 0) eventually. Hence we can couple them so they coincide from the first time they visit (r S(0), 0). Since this almost surely only takes a finite amount of time and since the distance difference l S(0) is almost surely finite, this does not affect the overall asymptotic speed of the random walk. Therefore we can prove Theorem for the random walk started at (l S(0), 0) instead. Throughout the proof we will need the following lemma about the probability of never returning to a vertex z = (x, 0) on the line Z {0}. Lemma 2. For all n N 0, configurations ω, and x Z, P ω ( k > n : Z k (x, 0) Z n = (x, 0)) β β + 2 =: p esc Proof. The escape probability from a vertex z, which is just the probability that the random walk started at z never returns to z, is given by Ceff z, π(z), where Ceff z, denotes the effective conductance between z and infinity in the corresponding electric network where each edge e is assigned the conductance w e and where π(x) denotes the sum of the weights or conductances of all edges adjacent to x (see for example Theorem 9.25 in 5). β x+ β x+2 β x+3 β x+4... (x, 0) Figure 2: Lower bound for C eff z, along the direct path to infinity. If z = (x, 0) is on the line Z {0}, C eff z, can be bounded from below by the effective conductance of the direct path along the x-axis from z to infinity, which can be calculated with the series rule: ( Cz, eff β x+ + ) ( ) +... = β x+ β i = β x+ ( β ) βx+2 Therefore, the escape probability can be bounded from below as well: i=0 5

6 Cz, eff π(z) βx+ ( β ) β x+ + 2β x = β (0, ) β + 2 Since for almost every environment ω all sections are finite and X n almost surely, the universal lower bound for the escape probability implies the next lemma, which can be seen by considering the points (l Si, 0), i N 0. Lemma 3. For almost every environment ω, it almost surely happens infinitely many times that the random walk visits a section for the first time and never returns to previous sections. More precisely, for almost every ω there are P β ω- almost surely times 0 t 0 < t < t 2 <... such that for every j, there is an i j N such that Z tj S ij and for all t < t j, Z t / S ij and for all t t j, Z t / i<i j S i. 2.2 Sections and the distribution of their size We will need some results about the distribution of the size of the sections later. Lemma 4. For x Z, let diam(s(x)) be the diameter of the section S(x), i.e. diam(s(x)) = max{ x x 2 + y y 2 (x, y ), (x 2, y 2 ) S(x)} Then there is a constant η = η(p) > 0 which does not depend on x such that for all n, P(diam(S(x)) n) e ηn Proof. Note that the cluster C := C((l S(x), 0)) in the original percolation configuration is the same as C((r S(x), 0)). To see this, suppose that (r S(x), 0) / C. Let (r, 0) be the rightmost vertex in C (Z {0}) S. So l S(x) r < r S(x). Since (r, 0)(r +, 0) is not a bridge, there must be a path in the original percolation configuration from a vertex (v, 0) with v r to a vertex (v 2, 0) with v 2 r +. Since (l S(x), 0)(l S(x), 0) is a bridge, v must be at least l S(x). But any open path in the upper half-plane from l S(x), r to r +, ) must intersect C. So (v 2, 0) C S and v 2 > r, a contradiction. Therefore, C spans over the line segment l S(x), r S(x) and the entire section S(x). So for any (x, y ), (x 2, y 2 ) S(x), we have x x 2 + y y 2 C and therefore diam(s(x)) C. Let n 2, and let diam S(x) n, hence C n. Since C = C((l S(x), 0)) also contains (r S(x), 0), we have C((l S(x), 0)) r S(x) l S(x) x l S(x), so C((l S(x), 0)) (x l S(x) ) n. Therefore, P (diam(s(x)) n) P { C((x, 0)) (x y) n} y x 6

7 Since we consider subcritical percolation, the cluster size distribution decays exponentially, so there is a positive constant η such that for any z Z Z + 0 and any n 2, P( C(z) n) e η n. Therefore, the probability above can be bounded by P ( C((y, 0)) n) + P ( C((y, 0)) x y) x n<y x ne η n + y Z:y x n y Z:y x n e η n e η (x y) = ne η n + e η Since P(diam(S(x)) n) < for all n, this implies that there is a constant η = η(p) > 0 such that for all n, P(diam(S(x)) n) e ηn Since the number S(x) of vertices in S(x) is bounded by (diam(s(x)) + ) 2, it follows immediately that P( S(x) n) e η( n ). In particular, E S(x) = P( S(x) n) < (2) n= Recall that i x denotes the index of the section S(x), i.e. S(x) = S ix. Then the index behaves roughly linear in x, which is proved in the following lemma. Lemma 5. i x lim x x = lim x x x y=0 l(s(y)) = E < almost surely. Proof. It is a known fact and can easily be checked that if (X i ) i Z is an ergodic sequence of random variables with values in (E, E) and φ : (E N, E N ) (E, E ) is measurable, then the sequence (φ((x i+j ) i Z )) j Z is also ergodic. Therefore, the sequence ( l(s(x)) ) x Z is ergodic: Note that is a measurable function in the independent bond states, and these can be seen as an i.i.d. (and hence ergodic) sequence of bond state vectors. Shifting the bond configuration to the left x times and applying the same measurable function gives l(s(x)). The second equality now follows from Birkhoff s ergodic theorem, and since, the expectation is finite. For the first equality, note that the terms l(s(y)) for all y such that S(y) = S(x) add up to. So the sum x y=0 l(s(y)) counts the number of sections until (x, 0), except possibly a part of S 0 and of S ix, so i x x y=0 l(s(y)) i x + 7

8 Dividing by x and taking the limit gives the first equality. Lemma 6. n lim n n i=0 x l(s i ) = lim = x i x E Proof. Note that Hence, almost surely, n r Sn l(s i ) r Sn + l(s 0 ) i=0 n r Sn lim l(s i ) = lim n n n n i=0 = lim r Sn n i rsn i x = lim x x since r Sn as n. The second equality follows from the previous lemma. 2.3 Durations of visits to a section and their expectations We will see later that the ballisticity of the random walk will depend on the expected amount of time the random walk spends within a section during its first visit to this section, divided by the length of the section. We will first investigate when this expectation is finite or infinite for different values of β Expected duration of the first visit to a section and definition of β l and β u Let T i (x) be the time the random walk spends in S(x) during its i th visit to S(x). If there is no i th visit to S(x), let T i (x) = 0. More specifically, for x Z, let s x 0 = 0, and for i, define inductively: t x i = inf{t s x i Z t S(x)} s x i = inf{t t x i Z t / S(x)} Now for all x Z, i N, let T i (x) = s x i tx i if t x i < and let T i (x) = 0 if t x n i =. Since by (), X n almost surely, the random walk almost surely visits S(x) at least once for all x 0, so t x < and T (x) > 0. Let T (x) = i=0 T i(x) denote the overall amount of time spent in S(x). Recall that we are considering a modified version of the random walk that starts at the vertex (l S(0), 0). It will turn out that the key to the ballisticity of this random walk (and therefore of the original random walk) will be the expectation 8

9 of T (0) divided by the length of the section S(0), i.e. the expectation E T (0). We will later show that the speed of the random walk is positive if and only if this expectation is finite. To simplify notation later, let W i (x) = weighted visit lengths. T i(x) l(s(x)) and W (x) = T (x) l(s(x)) be the The aim of this section is to show that there are values < β l β u < such that for < β < β l, EW (0) is finite, whereas it is infinite for β > β u. These will then also be the constants β l and β u in Theorem. Whenever it needs to be made clear that an expectation corresponds to P β for a specific β, it is denoted by E β. Usually the index β is omitted. We want to study EW (0) more closely using expected return times to get bounds for the critical values in a similar way as it was done for β u in 2. However, since the possible exits from our traps (the sections) do not have the same x-coordinate, they may have very different weights, and this complicates things considerably. We must consider random walks started at the left border vertex and at the right border vertex of a section separately. So consider the time T L the random walk, started at time 0 at (l S(0), 0), takes to exit S(0) either via (l S(0), 0) or via (r S(0) +, 0), i.e. the smallest t such that Z t = (l S(0), 0) or Z t = (r S(0) +, 0). Let W L = T L. Then, since the random walk starts at (l S0, 0), EW (0) = EW L (3) Also consider the time T R it takes the random walk, started at (r S(0), 0), to exit section S(0) either via (l S(0), 0) or via (r S(0) +, 0), and let W R = T R. Let A be a possible configuration for S(0), i.e. finite percolation clusters and a corresponding line segment that form a possible shape for the section S(0), so that the leftmost point of A (Z {0}) is the origin. Let p A = P(S(0) A), where S(0) A means that S(0) has the same shape as A but is shifted along the x-axis (see Figure 3). Let l(a) be the length of A on the x-axis. Then (l(a), 0) is the rightmost vertex of A (Z {0}). Let A denote the union of A and the adjacent vertices (0, ) and (l(a), 0) together with the edges (, 0)(0, 0) and (l(a), 0)(l(A), 0). For x A, let π (x) be the sum of all edge weights of edges incident to x within A. Let π (A ) denote the sum of all weights π (x) of all vertices x in A. For a configuration A, consider a weighted random walk moving on A started at (0, 0). Let EW L A = E T L l(a) A denote the expected weighted time it takes this random walk to reach the exit vertices (, 0) or (l(a), 0). Now consider the same for a random walk started at (l(a), 0), and denote the corresponding 9

10 S(0) (l S(0), 0) (r S(0), 0) A (, 0) (l(a), 0) (0, 0) (l(a), 0) Figure 3: S(0) is of shape A. A is A with the exit vertices (, 0) and (l(a), 0). expectation by EW R A = E T R l(a) A. Then we have EW L = A EW R = A p A EW L A p A EW R A Lemma 7. For all β >, E β W L < E β W R < Proof. For every bond configuration ω, the probability that a random walk started at (l S(0), 0) never returns to (l S(0), 0) is at least p esc by Lemma 2. Since X n almost surely for almost every bond configuration ω, and since the random walk needs to pass through (r S(0), 0) if X n, it follows that the probability of its first step being up or towards the right and of then visiting (r S(0), 0) before returning to (l S(0), 0) (if ever) is at least p esc. Since this is true for almost every bond configuration (all configurations where all sections are well-defined and finite), it is true for every possible finite configuration A. So for every finite possible configuration A, the probability that the random walk started at (0, 0) takes it first step up or towards the right and visits (l(a), 0) before it returns to (0, 0) is at least p esc. For it to reach the vertices (, 0) or (l(a), 0), it needs to pass through (0, 0) or (l(a), 0) first. Hence, with the strong Markov property, the expected time for it to reach (, 0) or (l(a), 0) is at least p esc times the expected time it takes for the random walk 0

11 to reach (, 0) or (l(a), 0) if started from (l(a), 0). Therefore, ET L A p esc ET R A and hence, EW R = A T R p A E l(a) A T L p A E p esc l(a) A A = p esc EW L So if EW L <, it follows that EW R < as well since p esc > 0. Note that the converse statement (E β W R < E β W L < ), if it is true at all, can not be deduced in the same manner since there is no positive lower bound for the probability to reach (0, 0) before returning to (l(a), 0) which holds for all finite possible configurations A. Finally, consider a weighted random walk on A where both exit vertices (, 0) and (l(a), 0) are merged into one new vertex s which inherits the edges (, 0)(0, 0) and (l(a), 0)(l(A), 0) and their weights from (, 0) and (l(a), 0), and start the random walk at the new vertex s. The weight of each vertex x s in this new graph is still π (x), and let π (s) = π ((, 0))+π ((l(a), 0)) = + β l(a) be the weight of the new vertex s. The sum of all vertex weights is still π (A ). Let T be the return time of the random walk to s if it is started at s at time 0, and denote the corresponding weighted time by W, then EW A = E T l(a) A = l(a) ET A. Note that π defines an invariant measure for the Markov chain given by the transition probabilities of the random walk on A. So π divided by the overall weight π(a ) is the unique invariant distribution of the Markov chain. The expected return time ET A is then given by the inverse of this invariant distribution at s (see for example Theorem 7.5 in 5), so ET A = π (A ) π (s) = π (A ). Therefore, we have +β l(a) EW A = π (A ) l(a)( + β l(a) ) (4) Starting at s, the random walk moves to (0, 0) with probability and to +β l(a) (l(a), 0) with probability βl(a), and then moves from there according to +β l(a) the edge weights until it returns to s. Hence, And therefore, ET A = + + β l(a) ET L A + βl(a) + β l(a) ET R A EW A = l(a) + + β l(a) EW L A + βl(a) + β l(a) EW R A (5)

12 It follows with (4) that π (A ) l(a) = + βl(a) l(a) + EW L A + β l(a) EW R A EW L A (6) Let EW = A π (A ) p A l A ( + β l(a) ) = A p A EW A (7) Then we have the following lemma: Lemma 8. For any β >, E β W = E β W L = Proof. Suppose EW L <, then by Lemma 7, EW R <. For any configuration A, by (5), EW A + EW L A + EW R A, and therefore EW + EW L + EW R <. Hence, EW = implies EW L =. Now we can finally prove the two lemmas which are the aim of this section: Lemma 9. There is a constant β l = β l (p) > such that for all < β < β l, T (0) E β W (0) = E < Proof. By (3), E β W (0) = E β W L. By (6), for every A, E β W L A π (A ) l(a). Hence, it suffices to show that there is a β l such that for all < β < β l, A p A π (A ) l(a) <. If A is a configuration for S(0), then A (diam(a) + ) 2 (with diam(a) defined as in Lemma 4). Also, the weight π (x) of any vertex within A is bounded by 4β diam(a)+. So, since l(a), p A π (A ) p A ((diam(a) + ) 2 + 2) 4β diam(a)+ l(a) A A = p A ((n + ) 2 + 2) 4β n+ n A:diam(A)=n = (n 2 + 2n + 3) 4β n+ p A n n A:diam(A)=n 24n 2 β n+ P(diam(S(0)) = n) 24β n n 2 ( βe η) n, using Lemma 4 in the last step. This sum is convergent for < β < e η. Lemma 0. There is a constant β u = β u (p) such that for all β > β u, T (0) E β W (0) = E β = 2

13 Proof. Since by (3), E β W (0) = E β W L, and by Lemma 8, E β W = implies E β W L =, it suffices to show that there is a β u such that for all β > β u, E β W =. Let A n be the section in the form of a hook consisting of the open edges ((0, 0), (0, )) and ((0, ), (, )),..., ((n, ), (n, )), with l(a n ) = (see Figure 4). β β 2... β n β... (0, 0) (n, 0) Figure 4: The hook A n and the box B n. Then π (A n) l(a n )( + β l(an) ) = 2 e E(A ) w e + β = βn+ β 2 > 2 βn+ β 2 = 2 ( + β + + β + β β n ) + β = (8) We want to give a lower bound for p An = P(S(0) A). For this, consider the box B n = {0,..., n} {0, } as in Figure 4. Note that S(0) A n is equivalent to the bonds ((0, 0), (0, )) and ((0, ), (, )),..., ((n, ), (n, )) being open and all other bonds in B n being closed (so that S(0) has the shape A n ) and that the boundary bonds are closed and there is no open path from a vertex (i, 0), i < 0, to a vertex (j, 0), j > n, which contains none of the vertices in B n. This ensures that the bonds (, 0)(0, 0) and (0, 0)(, 0) are bridges, so S(0) is indeed hook-shaped and the hook is not just part of a larger section. For n N, let D n be the event that there is an open path from a vertex (i, 0), i < 0, to a vertex (j, 0), j > n, which contains none of the vertices in B n. Then D n is independent from the bond states inside and on the boundary of B n and therefore, p An = p n+ ( p) 2n+3 P(D C n ) D n implies that there is a j n such that C((j, 0)) j, where C((j, 0)) is the open cluster containing (j, 0) in the original percolation configuration. Since we are considering subcritical percolation, this probability decays exponentially in j, so P( C((j, 0)) j) e η j for some η > 0, and hence P(D n ) P { C((j, 0)) j} e η j = j n j n 3 e ηn e η n 0

14 Let N = N(p) N be large enough so that for all n N, the last term is at most 2. So for all n N, p An = p n+ ( p) 2n+3 P(D C n ) 2 pn+ ( p) 2n+3. (9) Then E β W (7) = (8),(9) = π (A ) p A l(a)( + β l(a) ) π (A p n) An l(a A n N n )( + β l(an) ) ( ) 2 pn+ ( p) 2n+3 2 βn+ β 2 n N p( p) 3 β 2 p n ( p) 2n (β n+ ) n N The last sum is infinite for any β p ( p) Expected total visit length We will now prove that the expected total weighted visit length W (0) to S(0) is finite if and only if the expectation of the weighted first visit length EW (0) is finite. Since W (0) W (0), it suffices to show the following lemma. Lemma. If E β W (0) = E β T (0) <, then E β W (0) <. Proof. Let β > be such that E β W (0) <. This also means that the expected weighted visit length EW R if the random walk is started at (r S(0), 0), averaged over all possible configurations for S(0) as defined in the previous section, is finite (see Lemma 7). Let C = C(p, β) < be an upper bound for both of these expectations. Every time the random walk enters the section S(0), it goes to (r S(0), 0) with probability at least p esc (if it is not already there), and then with probability at least p esc exits towards the right from (r S(0), 0) and never returns to S(0). So the probability of never returning to section S(0) after the current visit is ( 2 at least p 2 esc = β β+2) for any environment where S(0) is finite. So if J denotes the total number of visits to S(0), for all finite possible configurations A and j, Hence, since J, P(J j + J j, S(0) A) p 2 esc P(J j S(0) A) ( p 2 esc) j (0) 4

15 If we only condition on J j, the j th visit is not distributed as the first visit starting from one of the border vertices (without conditions) because the condition J j gives information on the environment and therefore on the distribution of the visit length. However, this is not a problem when also conditioning on a fixed configuration A for S(0) and we can use the upper bound EW j (0) J j, S(0) A EW L A + EW R A, where W L and W R are the weighted visit lengths when starting at the (l S(0), 0) and (r S(0), 0), respectively. As mentioned above, we know that A p A(EW L A+EW R A) 2C. Hence, E W (0) = A = A (0) j p A E W (0) S(0) A = p A E W j (0) J j S(0) A A j p A E W j (0) J j, S(0) A P(J j S(0) A) j ( p 2 esc ) j A p A (E W L A + E W R A ) 2C p 2 esc < Ergodicity Lemma We want to use the convergence or divergence of the expectation of W (x) later by taking averages which tend to EW (x). To make this possible, we need the ergodicity of the sequence (W (x)) x N0. ( ) Lemma 2. The sequence (W (x)) x N0 = T (x) l(s(x)) is ergodic if the random x N 0 walk is started at the left border vertex (l S(0), 0) of S(0). Proof. The sequence is stationary: We started the random walk at (l S(0), 0) and the first visit to any section S(x), x 0 starts at (l S(x), 0). Now consider the sequence (W (x)) x. Since X n almost surely, the random walk visits S() eventually and starts its first visit at (l S(), 0). The distribution of (W (x)) x now only depends on (S(x + )) x Z in the same way that the distribution of (W (x)) x 0 depends on (S(x)) x Z, and as (S(x + )) x Z and (S(x)) x Z have the same distribution, so do (W (x)) x and (W (x)) x 0. So (W (x)) x N0 is stationary. So we only need to prove that the σ-algebra I of shift-invariant events (in the sequence (W (x)) x N0 ) is almost surely trivial. More specifically, consider the space ( Q N 0, F = σ ( )) (W (x)) x N0 with the shift operator τ : (xn ) n N0 (x n+ ) n N0. Let P be the probability measure on this space generated by the sequence (W (x)) x N0, that is for every A F, P(A) := P β ( (W (x)) x N0 A ). Then the σ-algebra of shift-invariant events is I = {A F τ (A) = A} 5

16 Let A I, then we need to prove P(A) = P β ( (W (x)) x N0 A ) {0, }. It is possible to generate the bond environment ω and the random walk on it in the following way: Assign to each bond i.i.d. Bernoulli random variables which describe the bond states as usual. Independently from this, assign to each vertex 2 4 independent infinite sequences of i.i.d. random variables: Each sequence corresponds to a possible configuration of open bonds incident to this vertex (except when the vertex is isolated), and each entry is a random variable valued {left, right, up, down} which is distributed according to the corresponding bond configuration. We can then piece together the values of all these random variables to generate the random walk: First work out the environment, then for every vertex take the infinite sequence which corresponds to its local bond configuration and use the n th entry for the decision if the random walk visits this vertex for the n th time, which way does it jump?. Let d denote a possible outcome of all jump decision variables and ω a bond state configuration as usual. It is almost surely the case that all sections are finite and that the random walk tends to infinity, and by Lemma 3 it almost surely happens infinitely many times that the random walk visits a section for the first time and never returns to previous sections afterwards. Let E be the event that all these events occur (so E has probability ), and let E be the event that E holds and we cannot reach a configuration (ω, d) / E by changing only finitely many of the bond states and jump decisions. Then E also holds almost surely otherwise, since we only have countably many bond states and jump decisions and each of them can only take finitely many values, there would be a finite set of bond state and jump decision random variables so that conditional on them taking certain values (which has positive probability), E has probability strictly less than. But this contradicts P(E ) =. So we have P(E) =. Now let A = { (ω, d) (W (x)) x N0 A } E Then, since P β (E) =, we have P β (A ) = P β ( (W (x)) x N0 A ) = P(A). We claim that A is a tail event in the independent bond state and jump decision random variables. This suffices to prove the claim since then by Kolmogorov s 0- law, P(A ) {0, }. To see this, suppose (ω, d) A, and suppose in (ω, d ) finitely many of the outcomes of the bond states and jump decision variables are changed. Then (ω, d ) E, so all sections are still finite, the random walk tends to positive infinity and it happens infinitely many times that it visits a section for the first time and never goes back. Denote by (Z t ) t N0 the original and by (Z t) t N0 the new random walk, and by S (x), T (x) the new sections and visit lengths, and T (x) l(s (x)) let W (x) =. Changing finitely many bond states affects at most finitely many sections (which may be altered, split, merged). Let x 0 N 0 such that (Z t ) t N0 never returns to previous sections after its first visit to S(x 0 ), and such that S(x 0 ) is the first such section after the 6

17 changes to the bond states and jump decision variables. This means that (S (x 0 ), S (x 0 + ),...) = (S(x 0 ), S(x 0 + )...) and the jump decision variables for all z x x 0 S(x) remain unchanged. Then, as (Z t) t N0 also reaches the section S(x 0 ) eventually, from this point onwards Z and Z agree. Hence for all x x 0, T (x) = T (x) and l(s (x)) = l(s(x)), so (W (x)) x x0 = ( W (x) ) x x 0 Since A is shift-invariant, τ x 0 (A) = A, so (W (x)) x 0 A (W (x)) x x0 A Hence (W (x)) x 0 A, and since we also have (ω, d ) E, it follows that (ω, d ) A. So A is a tail event in the independent bond state and jump decision random variables, and as described before, the claim follows with Kolmogorov s 0- law. With Birkhoff s ergodic theorem, we have the following corollary: Corollary 3. x T (y) T (0) lim x x l(s(y)) = E y=0 almost surely 2.4 Finishing the proof Let T i be the total time spent in section S i, so (T i ) i N0 is a (random) subsequence of (T (x)) x N0 where repetitions have been eliminated. Then T (x) = T ix. For all x, the terms W (y) = T (y) l(s(y)) with y such that S(y) = S(x) add up to T (x). Let (r i, 0) denote the right border vertex of a section S i. Then Hence, r n + n r n y=0 y=0 n W (y) i=0 r n W (y) n r n + n i=0 r n T i T 0 + y=0 W (y) T i T 0 n + r n + n r n r n + y=0 W (y) r By Lemma 6, we have lim nn x n = lim x i x = E almost surely, so with Corollary 3, since T 0 < almost surely, n E T (0) lim T i = a.s. () n n E i=0 7

18 Looking at () and Lemma 6, the following proposition seems immediate (but a formal proof requires some more work): Proposition 4. X n lim n n = E T (0) almost surely where := 0. Of course this proposition, together with Lemmas 9 and 0, suffices to prove Theorem. For the proof of Proposition 4, we need two more technical lemmas. The first one ensures that sections do not extend too far over their border vertices. Lemma 5. Let u > 0, and for n N 0, let G n be the event that S n > un or S n > un. Then ( ) P lim sup G n = P n n=0 k n G k = 0, i.e. there are almost surely only finitely many n such that G n holds. Proof. First note that for every u > 0, x= P( S(x) > u x ) = x= P( S(0) > u x ) ( ) u + P( S(0) x ) ( ) u u + P( S(0) x) x= = + u + 2 ( u + ) E S(0) < x= by (2). So by the Borel-Cantelli Lemma there almost surely only finitely many x such that S(x) > u x. Note that we cannot simply use the same argument for the sequence (S i ) i Z because the S i are not identically distributed. By r Lemma 6, Sn =: C > 0 almost surely. So there almost surely n E is a (random) n 0 that for all n n 0, r Sn < 2Cn. If S n > un infinitely often, this implies S(r Sn ) > un > u 2C r n for infinitely many n > n 0, and hence S(x) > u 2C x for infinitely many x. By the previous argument for u = u 2C, this has probability 0. So almost surely S n > un only finitely many times. The corresponding statement for S n follows because (S n ) n N has the same distribution as (S n ) n N. 8

19 The second lemma ensures that the random walk does not go back and forth too much between distant sections. Lemma 6. Let q >. For n N 0, let H n be the event that the random walk visits the section S n again after its first visit to S qn. Then ( ) P lim sup H n = P H k = 0, n n=0 k n i.e. there are almost surely only finitely many n such that H n occurs. Proof. Condition on a bond configuration ω such that all sections are finite (which has probability ). For any k Z, if the random walk is started at (l k, 0) or (r k, 0), with probability at least p esc it never returns to its starting point (see Lemma 2). Since by (), X n almost surely, this implies that the random walk almost surely does not enter S k again (since it would then have to pass through (l k, 0) and (r k, 0) again in order to go to + later). Hence, the probability that the random walks ever enters S k when started at (l k, 0) or (r k, 0) is at most p esc. Whenever the random walk enters a section S k from the right, it does so via (r k, 0). Therefore, we can conclude via induction with the strong Markov property of the random walk (conditioned on the bond configuration ω) that for any k, l N, l < k, the probability of returning to S l after the first visit to S k is at most ( p esc ) k l. In particular, P ω (H n ) ( p esc ) qn n ( p esc ) qn n, and therefore, P ω (H n ) ( p esc ) (q )n = ( p esc )( ( p esc ) q ) < n=0 n=0 Hence by the Borel-Cantelli lemma, P ω (lim sup n H n ) = 0 for almost every ω, and therefore, P (lim sup n H n ) = 0. Now we have all the ingredients to prove Proposition 4. Proof of Proposition 4. For n N 0, denote by F n the first time and by L n the last time the random walk is in section S n. Condition on the probability event that all clusters are finite and that X n. Then for all n 0, F n and L n are almost surely well-defined and finite, and F 0 < F <..., L 0 < L <.... If T denotes the total amount of time spent in sections S k with k < 0, then F n L n n T i + T + (2) i=0 n T i (3) i=0 9

20 Now fix u > 0 and q > and let N be a (random) integer large enough so that for all n N, G n and H n from Lemmas 5 and 6 do not hold. N < exists almost surely. Let t 0 = max{f Nq, L N } <, and let t t 0. Then there are (random) indices n, n 2 such that F n t < F n + and L n2 t < L n2 +. By the definition of t 0, we have n Nq and n 2 N. Note that n, n 2 as t. Since n Nq, H n q does not hold because n q Nq q = N. So the random walk does not return to any sections S n with n < n q after time F n because it would have to pass through S n q to get to any of the earlier sections. Also, for any section S n with n q n n, G n does not hold, so S n un < un. So for all vertices (x, y) in sections S n with n q n n, x l n q un. Hence, X t l n q un n q i= and so with (2), since t < F n +, X t t n q i=0 l(s i ) l(s 0 ) un n i=0 T i + T + l(s i ) un = n q i=0 l(s i ) l(s 0 ) un n q n n q n q i=0 l(s i ) l(s 0) n u = n n i=0 T i + T + n (4) With Lemma 6 and (), since n as t and l(s 0 ) and T are almost surely finite, the right-hand side tends to ( qe ) u T (0) E E = q ue E T (0) almost surely as t. Since (4) holds for all t t 0, we have almost surely: lim inf t X t t q ue Since u > 0 and q > were arbitrary and E almost surely, lim inf t X t t E E T (0) T (0) <, it follows that (5) For the other direction, since L n2 t < L n2 +, the walk never returns to S n2 after time t. Since n 2 N, H n2 + does not hold, so at time t the walk must be in a section S n with n 2 n < q(n 2 + ) because it will return to S n2 + at 20

21 time L n2 + > t. For any such n, G n does not hold, so S n un uq(n 2 + ). Let a = q(n 2 + ), then So with (3), since t L n2, a X t r a + uq(n 2 + ) l(s i ) + uq(n 2 + ) i=0 X t t a n2 i=0 l(s i) + uq(n 2 + ) i=0 T i = a n 2 + a a n 2 + i=0 l(s i) + uq n2 i=0 T i n 2 + (6) With Lemma 6 and (), since n 2 as t, and the right-hand side tends to q E T (0) E E lim sup t + uq X t t = q + uqe E T (0) a n 2 + = q(n 2+) n 2 + q, almost surely as t. Since (6) holds for all t t 0, we have almost surely: q + uqe Since u > 0 and q > were arbitrary and E almost surely, lim sup t X t t E E Together with (5), this implies the claim. T (0) T (0) <, it follows that 3 Generalisations The same proof, except for some of the results in Section 2.2 which were specific to subcritical percolation clusters and the definition of β u and β l, can be used to show a more general result. Only the ergodicity lemma in Section needs to be proved slightly differently for this more general case. Take the line Z {0} and attach subgraphs of Z d to the vertices they may overlap each other, or we can take them to be disjoint (and not embed the whole graph into Z d ). Suppose this is done in an ergodic way (the sequence of environments of (x, 0), x Z needs to be ergodic). Define sections as before, and suppose that almost surely every section is finite and the expectation of the size of the section containing the origin is finite. 2

22 Then the proof we used for the case of subcritical percolation on the half-plane carries over to this more general case we only need to slightly adjust the proof of the ergodicity lemma in Section by proving it in two steps, first conditioning on a fixed environment ω and proceeding as before, then showing that the set of all environments ω such that P ω ((W (x)) x Z A) = is in turn shift-invariant and therefore has probability 0 or. It follows that a random walk with bias β in the direction of the line Z {0} started at the leftmost vertex of S(0) (Z {0}) (and hence a walk started at the origin) is ballistic if and only if E <, where S(0), T (0) and so T (0) on are defined as before. The speed almost surely tends to T (0) E We will now apply our results and methods to two other special cases to determine the asymptotic behaviour of the speed of the biased random walk for different β in these contexts. In these cases, the critical value for the bias can be calculated explicitly. Example 7. We attach to each vertex (x, 0) the most basic kind of trap consisting of a hook of length L(x), i.e. a graph with the edges (0, 0)(0, ), (0, )(, ), (, )(2, ),... (L(x) 2, L(x) ), as in Figure 5. We attach them so they do not overlap. Let the L(x) be i.i.d. (or alternatively, they could also be chosen in an ergodic way), and let EL(0) <. Then the critical value for a biased random walk on the resulting graph is the radius of convergence r of the probability generating function G L(0) (z) = Ez L(0) of L(0). In other words, if < β < r, the speed of the random walk tends to a positive deterministic limit, while for β > r, it tends to 0. At the critical point, the speed tends to 0 if and only if G L(0) (r) =.. L(0) (0, 0) Figure 5: Line with i.i.d. hook-shaped traps. Proof. In this example the section S(0) consists exactly of (0, 0) and the hook attached to it. In particular, =, so the weighted visit lengths to S(x) are just the normal visit lengths. Every time the random walk is at (0, 0), it exits S(0) with probability β+ β+2 > 0. So the number of times it is at (0, 0) during one visit is geometrically distributed and has finite expectation. In between visits to (0, 0), the random walk makes independent excursions within 22

23 {0,..., L(0) } {}. So the overall first visit length to S(x) is finite if and only if the expected length of each excursion is. This is just the expected return time to (0, 0), which as before is given by the overall weight of S(0) divided by the weight of (0, 0) within S(0): + + β + β β L(0) E = E and this is convergent if and only if G L(0) (β) is. + βl(0) β = G L(0)(β) β + β 2 β The reason the calculations worked out so easily in this example is that there was just one attachment vertex for the traps and sections. In this case, the model is equivalent to a random walk on Z with waiting times whose distribution at each vertex is determined randomly first and which change in a certain way with β. Once there are several attachment vertices in S(0) (Z {0}), calculations quickly get more complicated. In some cases the critical point can still be calculated with a little more effort, as in the next example. Example 8. Consider bond percolation on the ladder Z {0, } where all the bonds (x, 0)(x +, 0) are always open, the bonds (x, 0)(x, ) are open with probability p and the bonds (x, )(x+, ) are open with probability q, 0 < p, q <, independently. Then β c := ( p)q is the critical value for the biased random walk starting at the origin with bias β in positive x-direction: If < β < β c, the speed tends to a positive deterministic limit, and if β β c, the speed tends to 0 almost surely. Sketch proof. It can easily be checked that indeed the sequence of environments of (x, 0), x Z, is ergodic and almost surely every section is finite and the expectation of the size of the section containing (0, 0) is finite, so our result applies. So if we start the random walk at the leftmost vertex of S(0) (Z {0}) (as before, this can be done without loss of generality and the result carries over to starting the walk at (0, 0)), the walk is ballistic if and only if E T (0) <. As in Figure 6, let (l, 0) and (r, 0) be the border vertices of S(0) (Z {0}), = r l+ the length of the section, (x l, ) and (x r, ) the border vertices of S(0) Z ({}). Note that the definition of sections implies x l l r x r and that {l, l+,..., r} {0} and {x l, x l +,..., x r } {} are completely contained in S(0) and (l, 0), (l, ) and (r, 0), (r, ) are open. Otherwise there would be bridges between (l, 0) and (r, 0) or S(0) would not be connected. Now T (0) is bounded by the sum of + T r, where T r is the first time the random walk either exits S(0) or visits one of the vertices in {(r, 0), (r, )}, and the expected time it takes the random walk to exit S(0) when started on the vertices (r, 0) and (r, ), respectively. If we only look at the first coordinate of the random walk on ({x l,..., r} {0, }) S(0), the random walk may stay where it is for an amount of time which 23

24 (x l, ) (x r, ) (l, 0) (0, 0) (r, 0) Figure 6: The section S(0) on a ladder. is geometrically distributed with finite mean m (if it has the option to move up and down in the second coordinate), and then it moves left with probability +β and right with probability β +β, or it moves right with probability if it cannot go left. So we can couple the walk with a walk on Z which jumps left with probability β +β and right with probability +β by making the same left/right jumps after the waiting times except when our original walk can only go right, in that case the new walk on Z may still go left. Then the random walk on Z has speed β β+ = v, so the time T r this walk takes to reach r (or (l, 0)), divided by r l, has expectation at most β+ β. Since T r r l, the +T r expectation of = +T r r l+ T r r l is also at most β β+. Our original random walk moves to the right whenever the new one does, and taking the possible waiting times with mean at most m into account, which are independent of the left/right movements, this expectation changes by a factor of at most + m. So E +Tr β+ ( + m) β <. Therefore, if the expected visit length when started at one of the vertices (r, 0) or (r, ) divided by is finite, so is E T (0). The converse statement is also true due to the positive escape probability and is proved in the same way as the corresponding Lemma 7 for subcritical percolation clusters. So start a random walk at s {(r, 0), (r, )}. Every time the random walk is at s, it can exit the section S(0) in at most two steps. Hence, the probability that it is there for the last time before exiting the section is at least β+2 β β+2 a lower bound which does not depend on the shape of the section. So if the expected weighted return time to s within S(0) is finite, so is the expected weighted total visit length when starting the random walk at s. If the random walk takes a step to the left and stays within {x l, x l +,..., r} {0, }, the expected weighted return time is the overall weight of this subgraph of S(0) divided by the weight of s and by, which is bounded from above by a constant. So if the expected weighted return time to s when the random walk stays in S(0) ({r, r +,..., x r } {0, }) is finite, so is the overall weighted visit length when starting the random walk at s. For the converse statement, note that whenever x r > r, when starting at s the probability of stepping into r +, x r {} and staying within S(0) (r, x r {0, }) during the entire visit is bounded by a constant from below. So if the expected overall weighted visit length is finite, so is the weighted return time 24

25 to s in the subgraph r, x r {}. This is just 2(β r + β r+ + β r β xr ) E π(s) 2β r (β xr r+ ) = E π(s)(β ) where the weight π(s) within S(0) (r, x r {0, }) is either β r or β r + β r+. Recapitulating, this means that E T (0) <, and therefore the random walk is ballistic, if and only if E 2β r (β xr r+ ) π(s)(β ) < for s {(r, 0), (r, )}. Since π(s) {β r, β r + β r+ } this is equivalent to E β xr r <, and this is a quantity we can calculate explicitly: β x r r E = β j P(l = m, x l = (m + n), r = i, x r = i + j) i + m + i,j,m,n 0 = i,j,m,n 0 q i+j+n+m ( q) 2 p 2 ( p) j+n β j (i + m + ) = p 2 ( q) 2 (q( p)β) j j 0 p 2 ( q) = q( p) (q( p)β) j j 0 i,m,n 0 This is convergent if and only if β < ( p)q. q i+m+n ( p) n (i + m + ) It seems reasonable to conjecture that also for subcritical percolation clusters, there is just one critical point β c. The proof in the last example does not directly carry over to this case though since the expected weighted time to reach the right border vertex if started at the left border vertex of a section may not be finite. References M. Barma and D. Dhar. Directed diffusion in a percolation network. Journal of Physics C: Solid State Physics, 6:45 458, N. Berger, N. Gantert, and Y. Peres. The speed of biased random walk on percolation clusters. Probability Theory and Related Fields, 26:22 242, A. Fribergh. The speed of a biased random walk on a percolation cluster at high density. The Annals of Probability, 38(5):77 782, A. Fribergh and A. Hammond. Phase transition for the speed of the biased random walk on the supercritical percolation cluster. Preprint, 20. Available at 25

26 5 A. Klenke. Probability Theory. Springer, London, R. Lyons, R. Pemantle, and Y. Peres. Biased random walks on Galton- Watson trees. Probability Theory and Related Fields, 06: , R. Lyons with Y. Peres. Probability on Trees and Networks. Cambridge University Press, 202. In preparation. Current version available at 8 A.-S. Sznitman. On the anisotropic walk on the supercritical percolation cluster. Communications in Mathematical Physics, 240:23 48,