Some sufficient conditions for infinite collisions of simple random walks on a wedge comb

In this paper, we give some sufficient conditions for the infinite collisions of independent simple random walks on a wedge comb with profile $\{f(n), n\in \ZZ\}$. One interesting result is that if $f(n)$ has a growth order as $n\log n$, then two independent simple random walks on the wedge comb will collide infinitely many times. Another is that if $\{f(n); n\in \ZZ\}$ are given by i.i.d. non-negative random variables with finite mean, then for almost all wedge comb with such profile, three independent simple random walks on it will collide infinitely many times.


Introduction
In this paper, we study the number of collisions of two independent simple random walks on an infinite connected graph with finite degrees. Let X = {X n } and X = {X n } be independent simple random walks starting from the same vertex. As usual, we say that the graph has the infinite collision property if X and X collide infinitely often, i.e., |{n : X n = X n }| = ∞, almost surely. Likewise we say that the graph has the finite collision property if X and X collide finitely many times almost surely. Krishnapur and Peres [5] first finds the example Comb(Z) on which two independent simple random walks will collide finitely many times. Notice that Z ⊂Comb(Z)⊂ Z 2 and both Z and Z 2 have the infinite collision property. This indicates that the infinite collision property is not simply monotone. So we are interested in probing the subgraphs of Comb(Z). Chen, Wei and Zhang [3] shows that Comb(Z, f ) has the infinite collision property when f (n) ≤ n 1 5 . Recently, Barlow, Peres and Sousi [1] gives a sufficient condition (in terms of Green functions) for the infinite collision property and shows that Comb(Z, f ) has the infinite collision property when f (n) ≤ n; while it has the finite collision property when f (n) = n α for each α > 1. Collisions on other graphes, such as random clusters and random trees are studied in [1] and [2]. We will restrict our attention to wedge combs, and give a sufficient condition for a wedge comb to have the infinite collision property.  We like to point out that it is not required that f (n) is monotone in n for n ≥ 0. This would enable us to deal with a large class of wedge combs. As an application of Theorem 1.1, one can improve the results of [3] and [1]. A natural question to ask is what happens if there are more than two independent simple random walks. Suppose that X , X and X are independent simple random walks on a graph. If |{n : X n = X n = X n }| = ∞ almost surely, then we say that the three processes collide together infinitely often and that the graph has the infinitely many triple collisions property. It is shown in [1] that Z has the infinitely many triple collisions property; while Comb(Z, α) has the finitely many triple collisions property for any α > 0. Theorem 1.1 will be proved in the next section. Proofs of Theorems 1.3 and 1.4 will be given in Section 4 and Section 3 respectively. Throughout this paper we use the following notation. For u, v ∈ V, P u is the probability measure of the simple random walk X starting from u, and P u,v is the joint probability measure of the two random walks X and X starting from u and v, respectively. E u and E u,v stand for the corresponding expectations. For vertex x ∈ V, we write x 1 for the first coordinate of x and x 2 for the second coordinate, i.e., x = (x 1 , x 2 ). For a, b ∈ R, a ∨ b = max{a, b} and a ∧ b = min{a, b}. For set A, |A| stands for the number of elements of A.

Proof of Theorem 1.1
Let X = {X n , n ≥ 0} be a simple random walk on Comb(Z, f ). Write X n = (U n , V n ) for n ≥ 0. Then {U n } is a random process on Z. Define a sequence of stopping times: T 0 = 0 and is a simple random walk on Z (by the strong Markov property).
Let X be another simple random walk on Comb(Z, f ), independent of X , and U n , V n , W k , T k , θ n be the corresponding counterparts. Define another sequence of stopping times by setting σ 0 = 0 and σ m+1 := inf{n > σ m : U n = U n and (U n = U n−1 or U n = U n−1 ) }.
for all N ∈ N and for all u, v ∈ V N with u 1 + u 2 + v 1 + v 2 being even.
By the strong Markov property, {Z τ m , m ≥ 0} is a simple random walk on Z. Suppose that u, v ∈ V N . Write u = (u 1 , u 2 ) and v = (v 1 , v 2 ). Then |u 1 |, |v 1 | ≤ N and Notice that U n + V n + U n + V n is always even under the assumption that U 0 the local time of 0 by {Z τ m , m ≥ 0}. As a result of the previous argument, For each w ∈ Z, let P w be the probability measure of a simple random walk W on Z starting from w. Use ξ(·, ·) to denote the local time by W . By (9.11) of [7], there exists x ∈ N such that, for any w ∈ Z, | w| ≤ 2N , Furthermore, by Theorem 2.13 of [7] and the argument that W is a simple random walk on Z, we can find d ∈ N such that for any N ∈ N and u ∈ V N , Combining (2.3), (2.5) with (2.6), we arrive at the desired conclusion.
The above lemma shows that, with a small exception, the number of collisions is bounded by departure times θ and θ of U and U linearly. In the following applications of Lemma 2.1, = 1/2, we can find d ∈ N, such that for all N ∈ N and for all u, v ∈ V N with u 1 + u 2 + v 1 + v 2 being even, Obviously, d ≥ 2. We fix such d throughout this section.
Next, we want to estimate the probability that there is at least one collision of X and X once U and U collide. To this end, we define a sequence of events. For m ≥ 0, let and collides with X at a vertex with height greater than or equal to v/2 after time σ m but before X or X exits the segment. Here, by height (of a vertex) we mean the distance from the vertex to the x-axis. The next lemma shows that these events have good bounds.
We now turn to the proof of the second inequality. Define the number of collisions of X and X before one of them exits Then =2E (u,0) ( number of visits to (u, v) by X before returning to (u, 0) ).
In the previous arguments, the inequality follows from the property of reversible Markov chain and the fact that the reversible measure for x ∈ [1, f (u)] ∩ Z is just the degree of the vertex. Therefore P (u,v) (X n = (u, x), n < τ 0 ) ≤ 2P (u,x) (X n = (u, v), n < τ 0 ).
By Theorem 9.7 of [7], The second inequality of the lemma will follow once we show that there exists c 2 > 0 independent of u, v such that (2.14) When the event Ψ 0 occurs, there is a collision at vertex (u, w) for some w with w ≥ v/2. Conditioned on this event, the total number of collisions in the set {u} × [v/3, f (u)] will be greater than the number of collisions that take place before the first time that one of the random walks exits this interval. A lower bound could be obtained by considering two independent simple random walks in an interval, starting at v/2. Before hitting either v/3 or 2v/3, the average number of collisions is the average number of returning to the starting point before exiting the interval, which is exactly the Green function of a simple random walk starting at v/2, before exiting the interval (v/3, 2v/3). This is of order v.
By (2.13) and (2.14), we conclude that P (u,0),(u,v) (Ψ 0 ) ≤ 2/(c 2 v). This completes the proof of the case that f (u) ≥ 2v > 0 and v is even. The proof can be modified to treat other cases and is omitted here.
Now we are ready to state a key lemma. To be concise, we set As a result,f is a strictly positive and increasing function on Z + .

Lemma 2.3.
There exists c > 0 such that, for all N ∈ N and for all u, v ∈ V N with u 1 + u 2 + v 1 + v 2 being even, If H > 0 then X and X collide before time θ dN ∧ θ dN . We shall use the second moment method to estimate the probability of {H > 0}. Calculate directly as follows.
Here the equation (2.15) is by the strong Markov property. Applying Lemma 2.2 and the strong Markov property again, we have the following estimates.
By the Hölder inequality, We have reached the conclusion of the lemma.
Proof of Theorem 1.1. It suffices to consider the case that two independent simple random walks on Comb(Z, f ) starting from the same vertex (0, 0). Notice that as X , X start from (0, 0), We write P = P (0,0),(0,0) and t = θ d m ∧ θ d m for short. By the strong Markov property and Lemma 2.3, there exists c > 0 such that for all m ∈ N We will show below that (1.1) implies that Hence (2.16) holds in either case and the proof of the theorem is completed.

Proof of Theorem 1.4
In this section we will show a slightly more general result from which Theorem 1.4 will follow. To prove the theorem we introduce yet another independent simple random walk X on Comb(Z, f ). For any u, v, w ∈ V, we write P u,v,w for the joint probability measure of the three independent simple random walks X , X and X starting from u, v and w, respectively. Let E u,v,w be the corresponding expectation. By the assumption of Theorem 3.1, there exists c > 4, such that for all n ∈ N, Hence we fix f and c which satisfy (3.1) throughout this section. Since Recall that θ n is defined in (2.1).

Lemma 3.2.
For any > 0, there exists d ∈ N \ {1} such that P u (θ dn ≤ n 2 ) ≤ for all n ∈ N and for all u ∈ V n .
Proof. For each w ∈ Z, let P w be the probability measure of a simple random walk W on Z starting from w. By Theorem 2.13 of [7] and the argument that W is a simple random walk on Z, there exists d ∈ N \ {1} which depends only on , such that for all n ∈ N and for all u ∈ V n P u (θ d n ≤ n 2 ) ≤P u max For x ∈ V, let τ x = inf{m ≥ 0 : X m = x} be the hitting time of x by X .

Lemma 3.3.
For > 0, there exists c 1 ∈ N such that P u τ v > c 1 n 2 ≤ for all n ∈ N and for all u, v ∈ V n .
Proof. Fix 0 < < 1, n ∈ N and u ∈ V n . Let h, c 1 ∈ N such that Suppose that the random walk X starts from vertex u ∈ V n . Since θ hn = τ (hn,0) ∧ τ (−hn,0) for u ∈ V n , then Hence we are going to estimate probabilities of the three events above.
Because Comb(Z, f ) is a tree, there exists a unique simple path from v to (hn, 0). If u is a vertex on the path, then by the gambler's ruin problem and the fact that X is a recurrent process, where dist(·, ·) is the graph distance. By the assumption (3.3), If u is not on the path, then with probability one X will reach the path first at v, (v 1 , 0) or (u 1 , 0). A slight change in the proof actually shows that (3.5) holds for these cases. So in all cases, (3.5) gives an upper bound of the probability of the first event. Similarly, the probability of the second event is also small.
We now study the third event. LetG be the spanning tree of V hn , which is a connected subgraph of Comb(Z, f ). The process X before time θ hn is also a simple random walk onG. Regard G as an electric network that each edge is assigned a unit conductance. Then by Proposition 10.6 of [6], where c G is twice of the total conductance and ℜ(u ↔ {−hn, hn}) is the effective resistance between u and {−hn, hn}. Since G is a tree, we can easily get that Consequently, E u (θ hn ) ≤ 2|V hn | 2 ≤ 2c 2 h 2 n 2 . By the Markov inequality and (3.3), we get the probability of the third event Combining (3.4)-(3.7) together we complete the proof of the lemma.
By Lemmas 3.2 and 3.3, we can find c 1 ∈ N and d ∈ N \ {1} such that for all n ∈ N and for all u, v ∈ V n . Fix such c 1 and d throughout this section. We are now ready to proceed to a similar key argument as Lemma 2.3.

Lemma 3.4.
There exists c 2 > 0 such that for all N ∈ N\{1} and for all u, v, w ∈ V N with (−1) We will prove that Then by the Hölder inequality, Now we begin to prove (3.9). For x ∈ V N , let q n (u, x) = P u (X n = x, θ dN > n). Then q 2n (x, x) is decreasing in n by the spectral theory (or referring to [1]). By the strong Markov property, for any c 1 N 2 ≤ n ≤ 2c 1 N 2 with u 1 + u 2 + x 1 + x 2 + n and k + n being even, As a result, The third inequality is by the Hölder inequality. Applying the Hölder inequality and (3.8) again, we have the following estimates.
Combining the two estimates together, we obtain that The first part of (3.9) is proved by choosing c * 1 = c 1 c −5 d −3 /1024. We now turn to the second moment, i.e., the second part of (3.9). Since that Comb(Z, f ) is a graph with uniformly bounded degree, by Corollary 14.6 of [8], there exists c * 0 > 0 such that P x (X k = y) ≤ c * 0 / k for all x, y ∈ V. Hence, By the inequality above and the strong Markov property, (c * 0 ) 2 k    ≤ 1 + 2(c * 0 ) 2 log(2c 1 ) + 4(c * 0 ) 2 log N E u,v,w (H).
We have thus verified the second part of (3.9) and finished the proof.
Proof of Theorem 3.1. It suffices to consider the case that three independent simple random walks on Comb(Z, f ) starting from the same vertex (0, 0). For each m ≥ 1, denote by Υ m the event Then by the strong Markov property and Lemma 3.4 there exists c 2 > 0 such that =P X t ,X t ,X t X n = X n = X n for some n The second inequality is proved similarly as Lemma 5.2 of [1].
Proof of Theorem 1.3. By Lemma 4.2, P( Z k,h > 0) ≤ c c k log β k for each k ≥ 2 and 1 ≤ h ≤ k log β k.
Summing over all k ≥ 2 and over all h ranging over powers of 2 and satisfying 1 ≤ h ≤ k log β k, we get that k h power of 2 Hence the total number of sets Q k,h with at least one collision is almost surely finite. Since E( Z k,h ) ≤ E(Z k,h ) < ∞, the number of collisions in each set Q k,h is a.s. finite. So, the total number of collisions in {(k, w) ∈ V : k ≥ 2, w ≥ 1} ⊆ k,h Q k,h is a.s. finite. We omit the details of the rest of the proof here.