Disjoint Crossings, Positive Speed and Deviation Estimates for First Passage Percolation

Consider bond percolation on the square lattice Z 2 where each edge is independently open with probability p. For some positive constants p0 ∈ (0, 1), 1 and 2, the following holds: if p > p0, then with probability at least 1− 1 n 4 there are at least 2 n log n disjoint open left-right crossings in Bn := [0, n] 2 each having length at most 2n, for all n ≥ 2. Using the proof of the above, we obtain positive speed for first passage percolation with independent and identically distributed edge passage times {t(ei)}i satisfying E (log t(e1)) + < ∞; namely, lim sup n T pl (0,n) n ≤ Q a.s. for some constant Q < ∞, where T pl (0, n) denotes the minimum passage time from the point (0, 0) to the line x = n taken over all paths contained in Bn. Finally, we also obtain corresponding deviation estimates for nonidentical passage times satisfying infi P(t(ei) = 0) > 2 3 .


Introduction
Consider bond percolation in Z 2 where each bond is independently open with probability p.A bond that is not open is said to be closed.For integer M ≥ 1 and > 0, let R n, (M ) denote the rectangle [0, n]×[0, (M log n) 1+ ].Here x represents the smallest integer greater than x.All constants mentioned henceforth are independent of n.A path Π = (e 1 , ..., e t ) of edges contained in R n, (M ) is said to be an open left-right crossing if every e i is open and e 1 intersects the left side of R n, (M ) and e t touches the right side of R n, (M ).For proof of (i), we use a contour argument and we use oriented paths to control the length of the crossings in the proof of (ii).For further analysis regarding critical values for oriented percolation, we refer to Durrett (1984) and references therein.
(ii) If p > p 0 , there are positive constants γ 2 = γ 2 (p, δ) and C 2 = C 2 (p, δ) so that with probability at least 1 − C2 n δ , there are at least γ2n log n open left-right crossings of B n , each containing at most 2n edges, for all n ≥ 2.
As an application of Proposition 1.1, we obtain positive speed and deviation estimates for first passage percolation.

First passage percolation
Consider the square lattice Z 2 with edges {e i } i≥1 .The passage times {t(e i )} i≥1 are independent and identically distributed (i.i.d.) having the same distribution as a random variable X.For a path π containing k edges g 1 , ..., g k , let T (π) := k i=1 t(g i ) denote its passage time.Let T ll (0, n) = min π T (π) be the minimum passage time from the line x = 0 to the line x = n where the minimum is taken over all paths π contained in B n .Similarly, T pl (0, n) and T pp (0, n) = min π T (π) be, respectively, the minimum passage time from (0, 0) to the line x = n and from (0, 0) to (n, 0), again over all paths contained in B n .Clearly, T ll (0, n) ≤ T pl (0, n) ≤ T pp (0, n) a.s.
We have the following result.
≤ Q 2 a.s.(iii) For every > 0, we have If we think of n Tpp(0,n) , n T pl (0,n) or n T ll (0,n) as the corresponding speed of first passage percolation, then (i)-(iv) of the above result implies positive speed even if expected passage time is not finite.
Suppose now the passage time is zero with large probability.It is then intuitive to expect infinite speed and we have the following result.Proposition 1.4.If P(X = 0) > 2 3 and E(log X) + < ∞, then Finally, we obtain deviation estimates for first passage percolation with independent passage times but not necessarily identically distributed.Theorem 1.5.Suppose inf i P(t(e i ) = 0) > 2 3 and inf i P(t(e i ) < ∞) = 1.
for all n ≥ 1 and for some positive constants β 1 and C 1 .In particular, we have for all n ≥ 1 and for some positive constants In particular, we have for all n ≥ 1 and for some positive constants β 4 , β 5 > 1 and C 3 .
The proof of the above theorem uses Proposition 1.1(ii).
The paper is organized as follows: In Section 2, we prove Theorems 1.3 and 1.5 assuming Proposition 1.1.In Section 3, we prove Proposition 1.1.

Proof of Theorems 1.3 and 1.5
Proof of Theorem 1.3: We prove (iv) first.If EX < ∞, then the result follows from strong law of large numbers since This proves (iv).We now prove (i)-(iii).
(i) Since P(X < ∞) = 1, we choose a finite N large enough so that P(X ≤ N ) > p 0 .For M ≥ 1 let R n,0 (M ) = [0, n] × [1, M log n + 1] be the shifted rectangle.Also for i ≥ 1, let f i denote the edge from (i − 1, 0) to (i, 0).We set an edge e in R n,0 (M ) to be open if its passage time t(e) ≤ N. Set δ = 2 in Proposition 1.1 and let A n denote the event that the rectangle R n,0 (M ) contains an open left-right crossing containing less than 2n edges, where M = M 3 is as in Proposition 1.1(ii).Since we only need to hit the line x = n from the line x = 0, we have that where Thus by Borel-Cantelli Lemma, we have that P(lim inf n A n ) = 1 and this implies that as n → ∞.This proves (i) with Q 1 = 2N.
To prove (ii)-(iii), we assume henceforth that X < ∞ a.s. and EX = ∞.Choose N finite, sufficiently large so that P(X ≤ N ) > p 0 .Set the edge e to be open if t(e) ≤ N.
Clearly each edge is independently open with probability at least p 0 .
We prove (iii) first and obtain (ii) as a Corollary.
(iii) Let {h 1,i } 1≤i≤M log n+1 be the set of edges forming the left vertical side of R n,0 (M ) and including the edge from (0, 0) to (0, 1).Similarly let {h n,i } 1≤i≤M log n+1 be the set of ECP 19 (2014), paper 52.edges forming the right vertical side of R n,0 (M ) and including the edge from (n, 0) to (n, 1).We then have that where and J n is as in (2.1).From (2.2), the third term goes to zero a.s. as n → ∞.For the first two terms, we apply Feller's theorem (Theorem 8.9, Durrett ( 2001)) with a n = exp n−1

M
. Indeed for a m ≤ n < a m+1 , we have Since Since E(log X) + < ∞, we have from Feller's theorem that the right hand side of (2.5) is zero a.s.This implies that I 1,n −→ 0 a.s. (2.6) as n → ∞.Since I 1,n has the same distribution as I 2,n we have that From (2.4), the result (iii) then follows with Q 2 = 2N.
(ii) The analysis is the same as above except that we obtain instead of (2.5), since we only need to hit the line x = n.From (2.6) and (2.2), the result then follows.
Proof of Proposition 1.4: Fix δ = 2 in Proposition 1.1(i) and let A n denote the event that the rectangle R n,0 (M ) defined in proof of Theorem 1.3 contains a left-right crossing containing only edges with zero passage time.Here M = M 2 is as in Proposition 1.1(i).
where I 1,n , I 2,n and J n are as in (2.4).The result follows by an analogous analysis following (2.4).
Proof of Theorem 1.5: We consider the shifted rectangle R n, . Let A n denote the event that R n,1 (M ) has a left-right crossing consisting of edges with zero passage times.Following an analogous analysis as preceding (2.3) we obtain (2.8) (i) follows by an analogous analysis as Proposition 1.4.

Proof of Proposition 1.1
Proof of Proposition 1.1(i) : We use a counting argument and again the fact that either an open left-right crossing occurs or a closed dual top-bottom crossing occurs but not both.As above, we fix a midpoint x of an edge at the bottom side of R n,0 (M ) and suppose that there is a dual top-bottom crossing of R n,0 (M ) with length k ≥ M log n, intersecting x.There are at most 4.3 k−1 dual paths intersecting x and each is closed with probability (1 − p) k .Since there are at most n choices for x, we have that a closed dual top bottom crossing occurs with probability at most provided M is large.The sum above is convergent since 1 − p < 1 3 .Fixing such an M proves the result.
Proof of Proposition 1.1(ii) : Fix p 0 > 0 and p > p 0 .We use comparison with oriented percolation.We draw oriented arrows from (i, j) to (i + 1, j − 1) and from (i, j) to (i + 1, j + 1).To draw arrows from (i, j) to (i + 1, j − 1), we let S 1 and S 2 be the bonds from (i, j) to (i, j − 1) and from (i, j − 1) to (i + 1, j − 1), respectively.Let E i denote the event that S i is open.We draw arrow from (i, j) to (i + 1, j − 1) if E 1 ∩ E 2 holds.An analogous procedure is used for drawing oriented arrows from (i, j) to (i + 1, j + 1).
We have that P p (E 1 ∩ E 2 ) = p 2 ≥ p 2 0 .We start from the left side of R n,0 (M ) and continue this oriented percolation process iteratively.Let P or denote the corresponding probability measure and let LR n denote the event that there exists an oriented leftright crossing of R n,0 (M ).If LR n occurs, there is an open left-right crossing of R n,0 (M ) containing at most 2n edges in the original bond percolation.Using a contour argument as in Durrett (1984), we have that provided p 0 and M are large.Fixing such a p 0 and M establishes the result.
To obtain the estimate on P or (LR n ), we let C denote the collection of all oriented paths starting from the left side E lef t of R n,0 (M ).We grow the cluster from vertices with even y coordinate in E lef t and for every vertex x ∈ C, there is an oriented path from E lef t to x.As in Durrett (1984), we place a square S x on each vertex x ∈ Z 2 of C. If x = (i, j), then S x has endvertices (i, j − 1), (i + 1, j), (i, j + 1) and (i − 1, j).The ECP 19 (2014), paper 52.edges of S x are oriented in such a way that the square S x forms a clockwise oriented contour around x.If two oriented edges in opposite directions coincide, they "cancel" each other and we draw nothing.There is an outermost contour Π of ∪ x∈C S x that is oriented clockwise and encloses E lef t .Oriented arrows with at least one end-vertex in C and crossing Π are called boundary arrows and we say such arrows were terminated in the cluster growing process.
Suppose that there is no oriented left-right crossing of R n,0 (M ).Let z u and z f denote the rightmost points of intersection of Π and the top and bottom edge of R n,0 (M ), respectively.Let Π 1 denote the part of Π from z u to z f .The path Π 1 is contained in R n,0 (M ).We write LR c n = 1≤j≤n−1 A j ∩ LR c n , where A j denotes the event that Π 1 cuts the horizontal segment [j − 1, j] × {M K n } of the top edge of R int T .Suppose that A j ∩ LR c n occurs and suppose that Π 1 contains k ≥ M log n oriented edges.We count up and down arrows as in Durrett (1984) and obtain a subset Π 2 of Π 1 consisting of at least k 32 edges, each cutting a boundary arrow that was independently terminated.Since the number of choices of Π 1 is at most 4. for all n sufficiently large and for some constant α > 0, provided p 0 > (1 − 3 −32 ) 1/2 .Fixing such an p 0 , we get that