The zero-one law for planar random walks in i.i.d. random environments revisited
aa r X i v : . [ m a t h . P R ] J un THE ZERO-ONE LAW FOR PLANAR RANDOM WALKS IN I.I.D.RANDOM ENVIRONMENTS REVISITED
By Martin P.W. Zerner
Abstract . In this note we present a simplified proof of the zero-one law by Merkl and Zerner (2001) fordirectional transience of random walks in i.i.d. random environments (RWRE) on Z . Also, we indicate how toconstruct a two-dimensional counterexample in a non-uniformly elliptic and stationary environment which hasbetter ergodic properties than the example given by Merkl and Zerner. Introduction
Let us first recall the model of random walks in random environments (RWRE), see also [Zei04]for a survey. For d ≥
1, we denote by P the set of 2 d -dimensional probability vectors, and setΩ = P Z d . Any ω ∈ Ω, written as ω = (( ω ( x, x + e )) | e | =1 ) x ∈ Z d , will be called an environment . It iscalled elliptic if ω ( x, x + e ) > x, e ∈ Z d with | e | = 1 and uniformly elliptic if there existsa so-called ellipticity constant κ >
0, such that ω ( x, x + e ) > κ for all x, e ∈ Z d with | e | = 1.Endowing Ω with the canonical product σ -algebra and a probability measure P turns ω into acollection of random 2 d -vectors, i.e. a random environment . The expectation corresponding to P is denoted by E .Given an environment ω ∈ Ω, the values ω ( x, x + e ) serve as transition probabilities forthe Z d -valued Markov chain ( X n ) n ≥ , called random walk in random environment (RWRE).This process can be defined as the sequence of canonical projections on the sample space ( Z d ) N endowed with the so-called quenched measure P z,ω , which is defined for any starting point z ∈ Z d and any environment ω ∈ Ω and characterized by P z,ω [ X = z ] = 1 and P z,ω [ X n +1 = X n + e | X , X , . . . , X n ] = ω ( X n , X n + e ) P z,ω − a.s. for all e ∈ Z d with | e | = 1 and all n ≥
0. The so-called annealed measures P z , z ∈ Z d , are thendefined as the semi-direct products P z := P × P z,ω on Ω × ( Z d ) N by P z [ · ] := E [ P z,ω [ · ]]. Theexpectations corresponding to P z,ω and P z are denoted by E z,ω and E z , respectively.One of the major open questions in the study of RWRE concerns the so-called , whichwe shall describe now. For ℓ ∈ R d , ℓ = 0, define the event A ℓ := n lim n →∞ X n · ℓ = ∞ o that the walker tends in a rough sense into direction ℓ , which we call to the right . It has beenknown since the work of Kalikow [Ka81], that if the random vectors ω ( x, · ), x ∈ Z d , are i.i.d.under P and P -a.s. uniformly elliptic, then P [ A ℓ ∪ A − ℓ ] ∈ { , } . (1)This was extended in [ZerMe01, Proposition 3] to the elliptic i.i.d. case. We shall call (1)Kalikow’s zero-one law. Primary 60K37, secondary 60F20.
Key words:
Random Environment, Random Walk, Zero-One Law, Transience .
The zero-one law for directional transience is the stronger statement that even P [ A ℓ ] ∈ { , } .Except for d = 1, see e.g. [Zei04, Theorem 2.1.2], it is only partially known under which conditionsthis statement holds. For d = 2 and ω ( x, · ), x ∈ Z d , being i.i.d. under P , Kalikow [Ka81]presented it as a question; that case was settled in the affirmative in [ZerMe01], while the case d ≥ Theorem 1. ( see [ZerMe01, Theorem 1]) Let d = 2 , ℓ ∈ R \{ } and let ( ω ( x, · )) , x ∈ Z , bei.i.d. and elliptic under P . Then P [ A ℓ ] ∈ { , } . Let us briefly sketch the proof of Theorem 1 for ℓ = e as given in [ZerMe01], where e isthe first coordinate direction. Assuming that the zero-one law does not hold, one considers twoindependent random walks in the same environment. The first one starts at the origin, thesecond one at a point ( L, z L ) for some L > { ( x , x ) ∈ Z | ≤ x ≤ L } isthen subdivided into three slabs of equal size. By adjusting z L and using d = 2 one can thenforce the paths of the two walkers to intersect at some point x in the middle slab with a positiveprobability, which is bounded away from 0 uniformly in L . In this step some technical result[ZerMe01, Lemma 7] about sums of four independent random variables is used. Now consider athird random walker starting at x . By Kalikow’s zero-one law (1) it eventually needs to go eitherto the left or to the right. Since x has been visited by the first walker which has already traveleda long distance > L to the right and thus most probably will continue to go to the right, thethird walker is also likely to go right. However, by the same argument the third walker shouldalso follow the second walker to the left. This leads to the desired contradiction.The main goal of the present paper is to give a simplified proof of Theorem 1. In Section 2we are going to present a proof in which the slab between 0 and ( L, z L ) is divided into two slabonly. This way the technical lemma [ZerMe01, Lemma 7] is not needed anymore and generaldirections ℓ / ∈ { e , e } can be handled more easily.The same paper [ZerMe01] also provides an example of a stationary elliptic, ergodic environ-ment with P [ A ℓ ] / ∈ { , } . However, the environment in this example has bad mixing properties.In fact, it is not even totally ergodic since it is not ergodic with respect to the subgroup of shiftsin 2 Z . In Section 3 we shall sketch an alternative construction of a counterexample which hasbetter mixing properties than the one given in [ZerMe01].2. A shorter proof of Theorem 1
Without loss of generality we assume k ℓ k = 1. For u ∈ R and ⋄ ∈ { <, ≤ , >, ≥} we considerthe stopping times T ⋄ u := inf { n ≥ | ( X n · ℓ ) ⋄ u } . By Kalikow’s 0-1 law (1), P [ A ℓ ∪ A − ℓ ] ∈ { , } . The case P [ A ℓ ∪ A − ℓ ] = 0 is trivial. So assume P [ A ℓ ∪ A − ℓ ] = 1 . (2)For the proof of the theorem it suffices to show that P [ T < = ∞ ] P [ T > = ∞ ] = P " P " . (3)Indeed, that (3) is sufficient follows from [SzZer99, Proposition 1.2 (1.16)]. For completeness,we repeat the argument here. If (3) holds then either P -a.s. T < < ∞ or P -a.s. T > < ∞ .In the first case, due to translation invariance, T
L,s ( L ∈ N , s ∈ {− , , +1 } ) the set of all finite nearest-neighbor paths that startat z L and leave the strip { x | ≤ x · ℓ ≤ x L } on the opposite side through a vertex x withsign ( x · ℓ ⊥ ) = s , we rewrite N L as N L = X s = ± X π ∈ Π L,s P ,z L h T ≥ L < T < , { X n | n ≤ T ≥ L } ∩ π = ∅ , ( X n ) n follows π,s = sign (cid:16) y L − X T ≥ L · ℓ ⊥ (cid:17) i . Using the disjointness of the paths we get by independence in the environment, N L = X s = ± X π ∈ Π L,s P (cid:2) T ≥ L < T < , { X n | n ≤ T ≥ L } ∩ π = ∅ , s = sign (cid:0) y L − X T ≥ L · ℓ ⊥ (cid:1)(cid:3) P z L [( X n ) n follows π ] ≤ X s = ± X π ∈ Π L,s P (cid:2) T ≥ L < T < , s = sign (cid:0) y L − X T ≥ L · ℓ ⊥ (cid:1)(cid:3) P z L [( X n ) n follows π ]= X s = ± P (cid:2) T ≥ L < T < , s = sign (cid:0) y L − X T ≥ L · ℓ ⊥ (cid:1)(cid:3) P z L (cid:2) T ≤ < T >x L , s = sign (cid:0) X T ≤ · ℓ ⊥ (cid:1)(cid:3) -1 LAW FOR PLANAR RWRE 5 = P " { z L P " (cid:9) z L + P " n z L P " (cid:9) z L ( ) ≤ P [ T ≥ L < T < ] X s = ± P z L (cid:2) T ≤ < T >x L , s = sign (cid:0) X T ≤ · ℓ ⊥ (cid:1)(cid:3) = 12 P " P " (cid:9) z L + P " (cid:9) z L ≤ P [ T ≥ L < T < ] P z L [ T ≤ < T >x L ]= 12 P " P " z L ≤ P [ T ≥ L < T < ] P [ T ≤− L < T > ]= 12 P " P " ( ) ≤ P [2 L < T < ] P [2 L < T > ] −→ L →∞ P [ T < = ∞ ] P [ T > = ∞ ] = 12 P " P " . Hence, due to (11), 12 P [ T < = ∞ ] P [ T > = ∞ ] ≤ lim sup L →∞ C L . For the proof of (3) it therefore suffices to showlim L →∞ C L = 0 . (12)By considering the possible locations of the intersections of the two paths, we estimate C L by C L ≤ C L + C x L L = P " x () z L + P " x () z L , where C ba := P ,z L [ ∃ x : a ≤ x · ℓ ≤ b, H ( x ) ≤ T ≥ L < T < , H ( x ) ≤ T ≤ < T >x L ] . Due to symmetry and translation invariance it suffices to show for the proof of (12) that C L → ∞ .To this end let ε > r ( x, ω ) := P x,ω [ A ℓ ]. Then C L ≤ C L , + C L , , where (13) C L , := P [ ∃ x : r ( x, ω ) ≤ ε, H ( x ) ≤ T ≥ L < ∞ ] = P " x , r ( x, ω ) ≤ ε and C L , := P z L [ ∃ x : x · ℓ ≤ L, r ( x, ω ) ≥ ε, H ( x ) < ∞ ] = P " x z L , r ( x, ω ) ≥ ε . In order to bound C L , , consider σ := inf { n ≥ | r ( X n , ω ) ≤ ε } . Note that σ is a stopping timew.r.t. the filtration ( F n ) n ≥ , where F n is the σ -field generated by X , . . . , X n and ω . Therefore,by the strong Markov property, C L , = P [ σ ≤ T ≥ L < ∞ ] = E [ P X σ ,ω [ T ≥ L < ∞ ] , σ ≤ T ≥ L , σ < ∞ ] . Since for all x ∈ Z and almost all ω , P x,ω -a.s. { T ≥ L < ∞} ց A ℓ as L → ∞ due to (2), we getby dominated convergencelim L →∞ C L , = E (cid:2) P X σ,ω [ A ℓ ] , σ < ∞ (cid:3) = E [ r ( X σ , ω ) , σ < ∞ ] ≤ ε (14)by definition of σ . Now consider C L , . By translation invariance C L , ≤ P [ ∃ x : x · ℓ ≤ − L, r ( x, ω ) ≥ ε, H ( x ) < ∞ ] ( ) , ( ) ≤ P [ T ≤− L < ∞ , A ℓ ] + P [ ∃ n ≥ L : r ( X n , ω ) ≥ ε, A − ℓ ] . (15)Obviously, the first term in (15) goes to 0 as L → ∞ . The same holds for the second term sincedue to the martingale convergence theorem, P -a.s. r ( X n , ω ) = P [ A ℓ | F n ] → A ℓ as n → ∞ , cf.[ZerMe01, Lemma 5]. Together with (13) and (14) this yields lim sup L C L ≤ ε . Letting ε ց L C L = 0. This finishes the proof of (12). (cid:3) A stationary and totally ergodic counterexample
The stationary and ergodic environment constructed in [ZerMe01] is based on two disjointtrees which together span Z . The branches of these trees are paths of coalescing random walkswhich for one tree go either up or right and for the other tree go either left or down. In orderto allocate enough space for both trees some periodicity was introduced which destroyed totalergodicity of the environment.In this section we shall sketch an alternative construction which gives a totally ergodic en-vironment. It has been inspired by the Poisson tree considered in [FeLaTh04, Section 3]. Themain difference to the tree in [FeLaTh04] is that the nodes and leaves of our tree do not form aPoisson process but a point process which has been obtained from a Poisson process by a localthinning procedure as follows: We first color the vertices of Z independently black with somefixed probability 0 < p < p = 1 /
7) and white otherwise. Then those black points x ∈ Z for which the set x + {± e , ± e , − e ± e } contains another black point are removedsimultaneously, i.e. painted white again. The remaining set of black points is called B ⊆ Z .Obviously, the random variables x ∈ B , x ∈ Z , are only finite range dependent.Now each black point grows a gray line to the right until the line’s tip reaches a neighbor ofanother black point, see also Figure 1. The set of the gray points obtained this way is called G ⊆ Z . More formally, for x ∈ Z let g ( x ) := inf { n ≥ | x + ne ∈ ( B + { e , − e , − e } ) } and set G := { x + ke | x ∈ B, ≤ k ≤ g ( x ) } . Note that almost surely all g ( x ) , x ∈ Z , are finite. Now consider the set T := B ∪ G . Lemma 2. If x ∈ T then x + e ∈ T or { x + e , x + e + e } ⊆ T or { x − e , x + e − e } ⊆ T .Similarly, if x ∈ T c then x − e ∈ T c or { x + e , x − e + e } ⊆ T c or { x − e , x − e − e } ⊆ T c .Moreover, T c = ∅ .Proof. First note that x ∈ B ⇒ x + e ∈ T (16)since either x + e is black or it will be painted gray as the right neighbor of a black point. -1 LAW FOR PLANAR RWRE 7 Figure 1.
The black points constitute some in a certain way thinned discretePoisson process on a torus. From each black point a gray line grew to the rightuntil its tip became a ℓ -neighbor of a black point.For the first statement of the lemma let x ∈ T and assume x + e ∈ T c . Then by (16), x ∈ T \ B = G . Hence it suffices to show that x ∈ G, x + e ∈ T c ⇒ (( x + e ∈ B ∧ x + e + e ∈ T ) ∨ ( x − e ∈ B ∧ x + e − e ∈ T )) , (17)where ∧ and ∨ denote “and” and “or”. Since x is gray but x + e is white, the gray line to which x belongs must have stopped growing in x . This means that one of the neighbors x + e , x + e or x − e must be black. Hence, since x + e is white x + e or x − e must be black. By constructionof B , only one of them can be black. By symmetry we may assume x + e ∈ B . Then, due to(16), x + e + e ∈ T . Thus (17) has been shown and the first statement follows.For the second statement of the lemma let x ∈ T c and assume x − e ∈ T . Then applying(16) to x − e instead of x yields x − e ∈ T \ B = G . Consequently, an application of (17) to x − e instead of x yields without loss of generality (due to symmetry) that x − e + e ∈ B and x + e ∈ T . Now it suffices to show that x − e , x − e − e ∈ T c . Neither of these points can beblack by construction of B since x − e + e is already black. So it suffices to show that neitherof them is gray. This is done by contradiction. Assume that one of them is gray. If x − e weregray then x − e − e would have to be black or gray. By construction of B , x − e − e cannotbe black since x − e + e is already black. Hence x − e − e would have to be gray, too. Sowe may assume that x − e − e is gray. By construction of the gray lines, there is some k ≥ x − ke − e is black and all the points x − ie − e (1 ≤ i < k ) in between are gray.Now recall that x − e is gray, too. Hence by the same argument, there is some m ≥ x − me is black and all the points x − ie (1 ≤ i < m ) in between are gray. By construction of B , k and m cannot be equal, since this would give two black points, x − ke − e and x − ke ,on top of each other. So assume 2 ≤ k < m . The case 2 ≤ m < k is treated similarly. Then thegray line starting at the black point x − me passes through the neighbor x − ke of the blackvertex x − ke − e . Hence, it has to stop there, i.e. the next point x − ( k − e − e cannot begray, which it is. This gives the desired contradiction and proves the second statement.For the last statement T c = ∅ we show that x ∈ B implies x + e + e ∈ T c or x + e − e ∈ T c .Firstly, by construction of B , not both x + e + e and x + e − e can be black. Secondly, noneof them can be gray. Indeed, assume that for example x + e + e were gray. As above, thiswould imply that its left neighbor x + e would be black or gray, too. However, by constructionof B , x + e cannot be black since x is already black. Hence x + e and x + e + e would belongto the same gray line starting at some black point x − ke + e with k ≥
1. However, this linewould have to stop in x + e and not extend to x + e + e since x + e is a neighbor of the blackpoint x , which would give a contradiction. (cid:3) Now we define for any vertex x ∈ Z its ancestor a ( x ) as follows: For x ∈ T we set a ( x ) := x + e if x + e ∈ T , x + e else if x + e , x + e + e ∈ T , x − e else if x − e , x + e − e ∈ T and for x ∈ T c we define a ( x ) := x − e if x − e ∈ T c , x + e else if x + e , x − e + e ∈ T c , x − e else if x − e , x − e − e ∈ T c .Due to Lemma 2 the function a : Z → Z is well defined and determines two disjoint infinitetrees with sets of vertices T and T c , respectively. (The thinning of the Poisson process at thebeginning was necessary to prevent the black and gray tree to disconnect the white complementinto finite pieces, possibly leaving it without an infinite component.) Moreover,( a n ( x ) · e ) n ≥ is monotone increasing for x ∈ T and decreasing for x ∈ T c with a ( x ) · e ≥ x · e + 1 for x ∈ T and a ( x ) · e ≤ x · e − x ∈ T c . (18)It can be shown, cf. [FeLaTh04, Theorem 3.1(d)], that all the branches in the tree on T arealmost surely finite, i.e. the length h ( x ) := sup { n ≥ | ∃ y a n ( y ) = x } of the longest line ofdescendants of x in T is almost surely finite for all x ∈ T . Moreover, it can be shown, cf.[FeLaTh04, Theorem 3.1(b)], that the tree on T is connected. This implies that all the branchesof the white tree on T c have finite height h ( x ) as well. The rest of the construction is the sameas in [ZerMe01, pp. 1730, 1732]: We define the environment in terms of the ancestor function a for x, y ∈ Z with | x − y | = 1 by ω ( x, y ) = (cid:26) − / ( h ( x ) + 4) if y = a ( x )1 / ( h ( x ) + 4) else.By Borel Cantelli there is a positive constant c such that P x,ω [ ∀ n X n = a n ( x )] > c for all x ∈ Z and almost all ω . Consequently, due to (18), and since neither tree is empty, P (cid:20) lim inf n →∞ X n · e n ≥ (cid:21) > P (cid:20) lim inf n →∞ X n · ( − e ) n ≥ (cid:21) > P [ A e ] / ∈ { , } . Since ( ω ( x, · )) x ∈ Z has been obtained from B by the applica-tion of a deterministic function which commutes with all spatial shifts in Z and since B itselfis stationary and totally ergodic with respect to all shifts, ( ω ( x, · )) x ∈ Z is stationary and totally -1 LAW FOR PLANAR RWRE 9 ergodic as well. We refrained from investigating the mixing properties of this environment, whichhas been done for a similar counterexample for d ≥ Open problems:
The gap between positive and negative results concerning the directionalzero-one law in d = 2 could be narrowed by answering one of the following questions: (1) Isthere a stationary and ergodic counterexample to the directional zero-one law for d = 2, whichis uniformly elliptic? For d ≥ d = 2 be extended to sta-tionary, ergodic and uniformly elliptic environments which have weaker independence propertiesthan finite range dependence? References [BrZeiZer06]
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