Reconstructing the environment seen by a RWRE

Consider a walker performing a random walk in an i.i.d. random environment, and assume that the walker tells us at each time the environment it sees at its present location. Given this history of the transition probabilities seen from the walker - but not its trajectory - can we tell if the RWRE is recurrent or transient? Can we reconstruct the law of the environment? We show that in a one-dimensional environment, the law of the environment can be reconstructed, and we know in particular if the RWRE is recurrent or transient.

If we endow the set Ω of all environments with a probability measure P , this process is called a random walk in random environment (RWRE) and P ω is called quenched law. We assume that P = µ ⊗Z is a product measure with marginal µ. We refer to [8] and [9] for results on the RWRE, for instance, a criterion for recurrence and transience; our arguments will not need them.
Of course, the same question may be asked for RWRE on Z d . Let P d denote the set of probability measures on {+e i , −e i , 1 ≤ i ≤ d} where e 1 , . . . , e d are the unit vectors of Z d . For a fixed mapping ω : Z d → P d , the random walk X : N 0 → Z d in the environment ω is the Markov chain starting in 0 and law P ω given by the transition probabilities P ω X(n + 1) = x ± e i | X(n) = x = ω(x, ±e i ).
Suppose that we only observe the sequence (1) ξ := ξ(0), ξ(1), . . . := ω(X(0), ·), ω(X(1), ·), . . . , the history of transition probabilities at the walker's position, but we do not know the trajectory, is it possible to recover the marginal µ? Note that in contrast to the one-dimensional case, recovering µ will not always tell us if the RWRE is recurrent or transient -despite some recent progress, there is still no criterion for recurrence/transience of RWRE in an i.i.d environment on Z d .
These questions are motivated from the classical scenery reconstruction problem, where the walk is a simple random walk on Z d , d ∈ {1, 2}, the "scenery" is a colouring of Z and the walker tells us at each time the colour of its present location. Given this sequence of observations -but not the trajectory of the walker -can we then reconstruct the scenery (up to translations, reflections and rotations)? This problem goes back to Itai Benjamini and Harry Kesten, see [2], and has lead to lot of interesting research, we refer to [4] for some nice (and still open!) problems. One direction of research is to ask about the ergodic properties of the observation sequence given by (1), see [3]. In the one-dimensional case, the original question can be answered in the positive in the sense that an i.i.d. scenery can be reconstructed almost surely, up to translation and reflection, see [6]. In the two-dimensional case, this is possible if the number of colours in the scenery is large enough, see [7]. Typically, scenery reconstruction is easier if the scenery has more colours. Clearly, if d ∈ {1, 2} and each location has a different colour, the trajectory of the walk can be reconstructed from the sequence of observations (up to symmetries, i.e. reflections and rotations). In the same way, in our case, the question will be much easier if µ has a non-atomic part, cf. the argument below.
A related, but different question for RWRE was asked by Omer Adelman and Nathanaël Enriquez in [1]: if we know a single "typical" trajectory of the walk, can we reconstruct the law of the environment? This questions is answered in the positive by [1] for i.i.d. environments on Z d .
If d = 1, the reconstruction of µ is possible, which is made more precise in the following theorem. We denote by M the set of all probability measures on (0, 1) and endow it with the weak topology and the corresponding Borel σ-algebra. for P -almost all ω.
Before we begin the proof, let us consider the simple case, where µ has a non-atomic part. We denote by µ = µ a + µ na the decomposition into the atomic part µ a and the non-atomic part µ na . Note that whether µ na is non-zero can be read off from the observations: the support of µ is almost surely equal to the closure of S = {ξ(0), ξ(1), . . . } and the set of atoms is almost surely given by as only atoms can appear twice in the environment. Now µ na is non-zero if and only if S na = S \ S a = ∅. In this case, observations ξ(n) ∈ S na can be used as perfect markers, as ξ(n) = ξ(m) implies X(n) = X(m). We give an informal description how this allows a reconstruction of µ: • Wait for observations (. . . , ξ(m−2), ξ(m−1), ξ(m), . . . ) in ξ, where ξ(m−2), ξ(m−1) ∈ S na , ξ(m) = ξ(m − 2) and both "markers" ξ(m − 2) and ξ(m − 1) have never before appeared in the sequence of observations.
• If the assumptions are met for the n-th time, denote by η n the value of the corresponding ξ(m).
• When the two markers are seen for the first time, X(m) must be at a point not visited before and ξ(m) is the value of the environment at a point not visited before. Also, the choice of ξ(m) is independent of the earlier entries of ξ, which implies P ω -almost surely The perfect markers in ξ immediately reveal whether X is recurrent or transient. In the recurrent case, the following procedure constructs (a.s.) an environment which is up to translation equal to ω.
• Repeat this step with new end points η 2 , η 3 with a new marker η 3 ∈ S na and concatenate the two obtained blocks of transition probabilities. It may happen that one block has to be contained in the other, which is the case if η 1 appears in the second block or η 3 in the first.
• Continuing in this way, we obtain in the limit an environmentω, which is up to translation either ω orω, whereω(z) = ω(−z) is the reflected environment.
• To decide on the orientation, consider all movements from a point z withω z = 1 2 . A proportion ofω z of those movements needs to be made to the right.

The main idea: the environment as a random walk
We now assume that µ is a purely atomic measure. We follow [5]. There can only be countably many support points η 1 , η 2 , . . . which can be found in S. We denote by N ∈ [1, ∞] the cardinality of S and exclude the deterministic environments where N = 1. Let T be the rooted tree with root o where each vertex has exactly N neighbours. We label the vertices by a mapping ϕ : T → S which satisfies • ϕ restricted to the neighbours of any vertex v is a bijection. That is, each vertex has exactly one neighbour which is labeled by a specific η i .
Given an environment ω, we define R : Z → T to be the bi-infinite path on T with R(0) = o and ϕ(R(z)) = ω(z) for all z ∈ Z. Due to the second property in the definition of ϕ, this determines R uniquely. As the environment is chosen under P in an i.i.d. way, R performs under P a random walk on T , starting at the root. In each step, both on the positive and negative time axis, R moves from a vertex v to a neighbour w with probability µ(ϕ(w)). Roughly speaking, this provides us with an embedding of the environment into the tree. Note that since we do not observe ω, we do not know the path of R.
In a second step, the random walk X on Z can be represented as a random walk T on the trajectory of R. Given X, there is exactly one T : for all n ∈ N 0 . The crucial point is that although we observe neither X nor R, we know the path of T , as the labels of vertices visited by T must coincide with the observation ξ, we have which we observe. In other words, as X performs a random walk on Z and yields ξ, the process T moves along a path on the tree giving the same observation ξ when reading the labels of the vertices provided by ϕ. Due to the structure of the labeling, there is only one such path.
The first observation at time 0 will be given by η 0 , so we choose our labeling ϕ such that ϕ(0) = ϕ(1) = η 0 . This determines ϕ uniquely on the whole integer line. The steps of R representing this part of the environment are given by (R(0), . . . , R(10)) = (0, 1, 2, 1, 2, 3, 4, 5, 4, 3, 4), see Figure 2 for an illustration. To keep this example simple, we do not consider R in negative time, which corresponds to ω on the negative integers. Next, say the first steps of X are as follows: (X(0), . . . , X(10)) = (0, 1, 2, 3, 4, 3, 4, 5, 6, 7, 6) This path gives us the observations (ξ(0), . . . , ξ(10)) = (η 0 , η 0 , η 1 , η 0 , η 1 , η 0 , η 1 , η 1 , η 0 , η 0 , η 0 ), which, given our choice of ϕ, implies the following movement of T : In order to make statements about the movement of X when we only observe T , we look for specific crossings of finite paths by T . For a generic process S : I → W with I ⊂ Z and W a tree, we call (i 1 , i 2 ) a crossing of (w 1 , w 2 ) by S, when S(i 1 ) = w 1 , S(i 2 ) = w 2 and S(i) / ∈ {w 1 , w 2 } for min{i 1 , i 2 } < i < max{i 1 , i 2 }. We call this crossing positive, if i 1 < i 2 and negative otherwise. The crossing is said to be straight, if |i 2 − i 1 | is equal to the path distance between w 1 and w 2 . Consider again the example above, where (0, 5) is a crossing of (0, 3) by R. Since R steps back during the time interval (0, 5), this is not a straight crossing. On the other hand, (4, 7) is a straight crossing of (2, 5) by R. Of central importance to us are straight crossings of a path (v 1 , v 2 ) in the tree by T , as T can only move in a straight way on the trajectory of R if R moves in a straight way on the tree T .
If (i 1 , i 2 ) is a straight crossing of (v 1 , v 2 ) by T then (i 1 , i 2 ) is a straight crossing by X of a straight crossing by R, that is, there are (z 1 , z 2 ) such that (i 1 , i 2 ) is a straight crossing of (z 1 , z 2 ) by X and (z 1 , z 2 ) is a straight crossing of (v 1 , v 2 ) by R.
For a stochastic process Z on a tree, we denote by the hitting time of vertex v. For m ≥ 0, define I m = (4m+1, 4m+4) and let W m be the indicator random variable which is 1 if the first crossing of I m by T is straight and 0 otherwise. By ( * ), is an indicator variable equal to 1 if and only if the first crossing of I m by R is straight and W X m is 1 if and only if the first crossing by X of the first crossing of I m by R is straight. We will show that W 0 , W 1 , . . . are independent and identically distributed, this time following [6].
For the independence, note that conditioned on R (or on ω, i.e. under the quenched law P ω ), the random variables W X 0 , W X 1 , . . . depend only on the path of X between ladder times of Xthe times when X reaches a point z ∈ Z where a crossing of a new I m by R begins. That is, if z 1,m = τ R (4m + 1), z 2,m = τ R (4m + 4), then W X m depends only on and these collections of increments of X are independent for different m, since [z 1,m , z 2,m ] are disjoint intervals. This implies that W X 0 , W X 1 , . . . are independent conditioned on R. Moreover, the conditional probability of the event {W X m = 1} depends only on the path segment of R in the time interval [z 1,m , z 2,m ] given by the random variable which again are independent for different m. In particular, W R 0 , W R 1 , . . . are independent. Note that although X may leave the corresponding path segment of R during [τ X (z 1,m ), τ X (z 2,m )], this does not influence the distribution of W R m . Consequently, we have which proves the independence.
We now evaluate the probability P (W m = 1) = P (W X m = 1| W R m = 1)P (W R m = 1).
By our definition, I m is labeled as (η 0 , η 1 , η 1 , η 0 ) and R moves to a neighbour labeled by η i with probability λ i . Let P m denote the law of R when starting at the left end point 4m + 1 and let E m be the event that R reaches 4m + 4 before returning to 4m + 1. In order to reach 4m + 4 before returning to 4m + 1, R needs to make two steps to the right (with probability λ 2 1 ), then make any number k of steps between 4m + 3 and 4m + 2 and back before moving to 4m + 4. This gives Given that the first crossing of I m by R is straight, the probability of {W X m = 1} depends on whether X moves on the positive or on the negative integers. If the first crossing of I m by R happens during an interval (t m,1 , t m,2 ) in positive time (which corresponds to the environment on the positive integers), the process X needs to move to the right for T to cross I m . In this case the first crossing of I m by T is a crossing by X of a positive crossing by R. If on the other hand the first crossing of I m by R is by the trajectory in negative time, X performs a crossing of the crossing by R by moving to the left and the corresponding crossing of R is negative. Let D m be the event that t m,1 > 0, then as X moves from t m,1 (or t m,1 + 1) to t m,1 + 1 (or t m,1 + 2) with probability η 0 (η 1 , respectively) and in the other direction with probability 1 −η 0 (and 1 −η 1 ). Given D c m , we need to interchange η i and 1 − η i , which by our choice of I m leads to Using that {W R m = 1} is independent of D m , we get P (W X m = 1| W R m = 1) = 1 − η 1 (1 − η 1 ) and therefore, This proves that W X 0 , W X 1 , . . . are independent identically Bernoulli-distributed random variables. By the law of large numbers, we have P ω -almost surely Since the W m are functions of ξ, this convergence provides us with a measurable mapping which, given ξ yields λ 1 . In the case N = 2, this already determines the measure µ.
In the general case N ≥ 2 we reduce this to the procedure above. Fix two values η 0 , η 1 ∈ S to which µ assigns weights λ 0 and λ 1 . The intervals I m are now replaced by disjoint vertex-sets (I m (η 0 , η 1 )) m≥0 in the tree T , such that each set I m (η 0 , η 1 ) contains exactly four neighbouring vertices v 1,m , . . . , v 4,m with strictly increasing distance from the root and labels ϕ(v 1,m ) = ϕ(v 4,m ) = η 0 and ϕ(v 2,m ) = ϕ(v 3,m ) = η 1 . When T crosses the m-th of such a set for the first time without leaving this set of vertices, let W m (η 0 , η 1 ) be equal to 1 if this crossing is straight and 0 otherwise. The same arguments as in the case N = 2, this time conditioning on a movement of R within I m (η 0 , η 1 ), show that W 0 (η 0 , η 1 ), W 1 (η 0 , η 1 ), . . . are again independent and The law of large numbers allows us to recover λ 1 and repeating this with different choices of values η 1 shows that we can recover any λ i . The (countable) combination of all these operations yields a weight vector (λ 0 , λ 1 , . . . ) as a measurable function of ξ by which we can define A(ξ) = N k=0 λ k δ η k .
For N = 3, transience of R is proven in Lemma 5 in [5]. If N > 3, T contains a subtree on which R is transient, so R is transient on T as well.
Finally, we give a statement and an open question for the case d ≥ 2. In order to make sure that the RWRE visits infinitely many sites, assume that µ is concentrated on the subset P d = {γ ∈ P d : γ(e i ) > 0, γ(−e i ) > 0, 1 ≤ i ≤ d}, and let M d be the set of probability measures on P d .
Theorem 3. Assume d ≥ 2 and assume that µ has a non-atomic part. Then, there exists a measurable mapping A : ( P d ) N 0 → M d , such that for any measure µ ∈ M d P ω A(ξ) = µ = 1 for P -almost all ω.
The proof of Theorem 3 goes along the same lines as the informal description after Theorem 1, which showed how to reconstruct µ in the case where µ has a non-atomic part.
Question 4. Assume d ≥ 2 and assume that µ is purely atomic. Is there a measurable mapping A : ( P d ) N 0 → M d , such that for any measure µ ∈ M d P ω A(ξ) = µ = 1 for P -almost all ω?