Reconstructing a random scenery observed with random errors along a random walk path
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(2) Reconstructing a random scenery observed with random errors along a random walk path . y. Heinrich Matzinger. Silke W.W. Rolles. January 31, 2002. Abstract. We show that an i.i.d. uniformly colored scenery on Z observed along a random walk path with bounded jumps can still be reconstructed if there are some errors in the observations. We assume the random walk is recurrent and can reach every point with positive probability. At time k, the random walker observes the color at her present location with probability 1 ; and an error Yk with probability . The errors Yk , k 0, are assumed to be stationary and ergodic and independent of scenery and random walk. If the number of colors is strictly larger than the number of possible jumps for the random walk and is suciently small, then almost all sceneries can be almost surely reconstructed up to translations and reections.. 1 Introduction and result. We call a coloring of the integers Z with colors from the set C := f1 2 : : : C g a scenery. Let (Sk k 2 N 0 ) be a recurrent random walk on Z. At time k the random walker observes the color (Sk ) at her current location. Given the color record := ( (Sk ) k 2 N 0 ), can we almost surely reconstruct the scenery without knowing the random walk path? This problem is called scenery reconstruction problem. In general, one can only hope to reconstruct the scenery up to equivalence, where we call two sceneries and 0 equivalent and write 0 if is obtained from 0 by a translation and/or re
(3) ection. Early work on the scenery reconstruction problem was done by Kesten in 14]. He proved that a single defect in a 4-color random scenery can be detected if the scenery is i.i.d. uniformly colored. Reconstruction of typical 2-color sceneries was proved by Matzinger in his Ph.D. thesis 22] (see also 24] and 23]): Almost all i.i.d. uniformly colored sceneries observed along a simple random walk path Eurandom, PO Box 513, 5600 MB Eindhoven, The Netherlands. Matzinger@eurandom.tue.nl y Eurandom, PO Box 513, 5600 MB Eindhoven, The Netherlands. Rolles@eurandom.tue.nl, Fax: 0031-40-247-8190. 1.
(4) (with holding) can be almost surely reconstructed. In 15], Kesten noticed that the proof in 22] heavily relies on the skip-freeness of the random walk. In 21], Lowe, Matzinger and Merkl showed that scenery reconstruction is possible for random walks with bounded jumps if there are suciently many colors. In this article, we prove that scenery reconstruction still works if the observations are seen with certain random errors. We make the same assumptions on scenery and random walk as in 21]: The random walk can reach every integer with positive probability and is recurrent with bounded jumps, and there are strictly more colors than possible single steps for the random walk. To keep the exposition as easy as possible, we assume in addition that for the random walk maximal jump length to the left and maximal jump length to the right are equal we believe that the results of this paper remain true without this assumption. At time k the random walker observes color (Sk ) with probability 1 ; , whereas she observes an error Yk with probability . If the errors are independent of scenery and random walk, the occurences of errors are i.i.d. Bernoulli with parameter and Yk , k 0, is stationary and ergodic, then for all sufciently small, almost all sceneries can be almost surely reconstructed up to translations and re
(5) ections. More precisely, we consider the following setup: Let 2]0 1 . Let be a probability measure over Z with nite support M. With respect to a probability measure P , let S = (Sk k 2 N 0 ) be a random walk starting at the origin with independent -distributed increments. We assume that E S1 ] = 0 and M has greatest common divisor 1 hence S is recurrent and can reach every z 2 Z with positive probability. Let = (k k 2 Z) be a family of i.i.d. random variables, uniformly distributed over C . Let X := (Xk k 2 N 0 ) be a sequence of i.i.d. random variables taking values in f0 1g, Bernoulli distributed with parameter , and let Y := (Yk k 2 N 0 ) be as sequence of random variables taking values in C which is stationary and ergodic under P . We assume that ( S X Y ) are independent. The scenery observed with errors along the random walk path is the process ~ := (~k k 2 N 0 ) dened by ~k := k = (Sk ) if Xk = 0 and ~k := Yk if Xk = 1. Our main theorem reads as follows: Theorem 1.1 If jCj > jMj, then there exists 1 > 0 and a map A : C N0 ;! C Z which is measurable with respect to the canonical sigma algebras, such that P (A(~) ) = 1 for all 2]0 1 . If = 0, there are no errors in the observations. In this case, the assertion of Theorem 1.1 was proved by Lowe, Matzinger, and Merkl in 21]. Closely related coin tossing problems have been investigated by Harris and Keane 7] and Levin, Pemantle, and Peres 17]. The present paper has to a large extend been motivated by their work and a question of Peres who asked for generalizations of the existing random coin tossing results for the case of many biased coins. Let 0 := (0k k 2 N 0 ) be a coin tossing record, obtained in one of the following ways: a) a (two-sided) fair coin is tossed i.i.d., or b) at renewal times of a renewal process a coin with bias is tossed and at all other times a fair coin. Can we almost surely determine from 0 whether we are in case a) or b)? 2.
(6) Let un denote of a renewal at time n. Harris and Keane in 7] P the probability 2 showed that if 1 surely determine how 0 was n=1 un = 1 then we canPalmost 1 2 produced, whereas this is not possible if n=1 un < 1 and is small enough. Levin, Pemantle, and Peres in 17] showed that to distinguish between a) and b) not only the square-summability of (un ) but also is relevant. They proved that for some renewal sequence (un ) there is a phase transition: There exists a critical parameter c such that for jj > c we can almost surely distinguish between a) and b), whereas for jj < c this is not possible. The problem we address in this paper can be seen as a generalization of the following coin tossing problem: We have C dierent coins 1 2 : : : C each one with C dierent faces 1 2 : : : C . Coin i has distribution i which gives probability 1 ; + =C to face i and probability =C to each remaining face. For all z 2 Z we choose i.i.d. uniformly among 1 2 : : : C a coin
(7) (z ). Let (Sk k 2 N 0 ) be a random walk on Z fullling the conditions described above, independent of
(8) . We generate a coin tossing record 0 := (0k k 2 N0 ) by tossing the coin
(9) (Sk ) at location Sk at time k. Then 0 has the same distribution as ~ dened above, if we choose Yk i.i.d. uniformly distributed over C . Theorem 1.1 implies that we can almost surely determine
(10) up to equivalence from the coin tossing record 0 , as long as is small enough. Research on random sceneries started by work by Keane and den Hollander ( 13] and 5]) who studied ergodic properties of a color record seen along a random walk. Their questions were motivated among others by the work of Kalikow 12] in ergodic theory. More recently, den Hollander, Steif 4], and Heicklen, Homan, Rudolph 8] contributed to this area. A preform of the scenery reconstruction problem is the scenery distinguishing problem (for a description of the problem see 15]) which started with the question whether any two non-equivalent sceneries can be distinguished. This question was asked by Benjamini and independently by den Hollander and Keane. The problem has been investigated by Benjamini and Kesten in 2] and 14]. Howard in 11], 10], 9] also contributed to this area. Recently, Lindenstrauss 18] showed the existence of uncountably many sceneries which cannot be reconstructed. Lowe and Matzinger 20] proved that two-dimensional sceneries can be reconstructed if there are enough colors. In the case of a 2-color scenery and simple random walk with holding, Matzinger 25] showed that the reconstruction can be done in polynomial time. By a result of Lowe and Matzinger 19], reconstruction is possible in many cases even if the scenery is not i.i.d., but has some correlations. In 16], Lenstra and Matzinger showed that scenery reconstruction is still possible if the random walk might jump more than distance 1 with very small probability and the tail of the jump distribution decays suciently fast. The exposition is organized as follows. In Section 2, we introduce some notation and we formally describe our setup. Section 3 describes the structure of the proof of Theorem 1.1: By an ergodicity argument, it suces to nd a partial reconstruction algorithm A0 which reconstructs correctly with probability > 1=2. To construct A0 , we build partial reconstruction algorithms Am , m 1, which reconstruct bigger and bigger pieces of scenery around the origin. 3.
(11) Section 4 contains the proofs of the theorems from Section 3. The core of the reconstruction is an algorithm Algn which reconstructs a nite piece of scenery around the origin given as input nitely many observations, stopping times and a small piece of scenery which has been reconstructed earlier. Section 5 contains the denition of Algn . In Section 6, we show that Algn fullls its task with high probability.. 2 Notation and setup In this section, we collect frequently used notation. Sets and functions: The cardinality of a set D is denoted by jDj. We write f jD for the restriction of a function f to a set D. For a sequence S = (si i 2 I ) we write jSj := jI j for the number of components of S . If si is an entry of S , we write si 2 S sometimes we write s(i) instead of si . For events Bk , k 1, we 1 write lim inf k!1 Bk := 1 n=1 \k=n Bk for the event that all but nitely many Bk 's occur. Integers and integer intervals: N denotes the set of natural numbers by denition, 0 2= N . We set N 0 := N f0g. If x 2 R, we denote by bxc the largest integer x. Unless explicitly stated otherwise, intervals are taken over the integers, e.g. a b] = fn 2 Z : a n bg, a b = fn 2 Z : a n < bg. Sceneries: We x C 2, and denote by C := f1 ::: C g the set of colors. A scenery is an element of C Z. A piece of scenery is an element of C I for a subset I of Z here I need not be an integer interval. The cardinality of the set I is called the length of the piece of scenery. We denote by (1)I the piece of scenery in C I which is identically equal to 1. For I = fi1 i2 : : : ik g Z with i1 < i2 < : : : < ik and a piece of scenery 2 C I we dene ! to be the piece of scenery read from left to right and to be read from right to left: ! := ( (ij ) j 2 1 k]) and := ( (ik;j+1 ) j 2 1 k]). Equivalence of sceneries: Let 2 C I and 0 2 C I be two pieces of sceneries. We say that and 0 are equivalent and write 0 i I and I 0 have the same length and there exists a 2 Z and b 2 f;1 1g such that for all k 2 I we have that a + bk 2 I 0 and k = a0 +bk . We call and 0 strongly equivalent and write 0 if I 0 = a + I for some a 2 Z and k = a0 +k for all k 2 I . We say occurs in 0 and write v 0 if 0 jJ for some J I 0 . We write
(12) 0 if 0 jJ for some J I 0 . If the subset J is unique, we write
(13) 1 0 . Random walks, random sceneries, and random errors: Let be a probability measure on Z with nite support M. We assume that jMj < jCj, i.e. the number of colors is strictly larger than the number of possible jumps of the random walk. We assume max M = j min Mj, and we write L := max M for the maximal jump length of the random walk. Let 2 ZN0 denote the set of all paths with jump sizes Sk+1 ; Sk 2 M for all k 2 N 0 . We denote by Qx the distribution on ( 2 )N0 of a random walk (Sk k 2 N 0 ) starting at x with i.i.d. inP crements distributed according to . We assume that k2M (k) = 0 and M 0. 4.
(14) has greatest common divisor 1, consequently the random walk is recurrent and can reach every integer with positive probability. The scenery := (k k 2 Z) is i.i.d. with k uniformly distributed on C . Let X := (Xk k 2 N 0 ) be a sequence of i.i.d. Bernoulli random variables with values in f0 1g. If Xk = 0, then at time k the random walk observes color (Sk ), whereas if Xk = 1 an error occurs in the observations at time k: the random walker observes Yk , where Y := (Yk k 2 N 0 ) is a sequence of random variables taking values in C . We assume that; ( S X Y ) are independent and realized as canonical projections on := C Z 2 f0 1gN0 C N0 with the product algebra generated by the canonical projections and probability measures Px := Z Qx BN0 , 2 0 1], x 2 Z here denotes the uniform distribution on C , B the Bernoulli distribution with parameter on f0 1g and a probability measure on C N0 such that the left-shift is measure-preserving and ergodic with respect to . We abbreviate P := P0 and P := P0 . . We call := (k := (Sk ) k 2 N 0 ) the scenery observed along the random walk path sometimes we write S instead of . We dene ~ := (~k k 2 N 0 ), the scenery observed with errors along the random walk path, by. ~k :=. . k if Xk = 0 Yk if Xk = 1:. For a xed scenery 2 C Z we set P := Q0 BN0 , where denotes the Dirac measure at . Thus P is the canonical version of the conditional probability P (j ). We use P and P (j ) as synonyms i.e. we never work with a dierent version of the conditional probability P (j ). Admissible paths: Let I = i1 i2 ] be an integer interval. We call a path R 2 ZI admissible if Ri+1 ; Ri 2 M for all i 2 i1 i2 ; 1]. We call R(i1) the starting point, R(i2 ) the endpoint, and jI j the length of R. Words: We call the elements of C := n2N0 C n words. If w 2 C n, we say that w has length n and write jwj = n. Ladder intervals, ladder paths, and ladder words: A ladder interval is a set of the form I \ (a + LZ) with a bounded interval I and a modulo class a + LZ 2 Z=LZ. Let I be a ladder interval. We call a path R of length jI j which traverses I from left to right or from right to left a ladder path or a straight crossing of I . The ladder words of a scenery over I are ( jI )! and ( jI ) . Filtration and shift: We dene a ltration over : G := (Gn n 2 N 0 ) with Gn := (~k k 2 0 n]) is the natural ltration of the observations with errors. We dene the shift : C N0 ! C N0 , 7! ( + 1).. 2.1 Conventions about constants. All constants keep their meaning throughout the whole article. Unless otherwise stated, they depend only on C and . Constants , , ", ", c1 , c2 and n1 play a special role in the constructions below we state here how they are chosen. All other constants are denoted by ci , i 3, i , "i , i 1. 5.
(15) We choose > 0. We choose c2 2]1 CC;1 and " 2]0 "max with "max := min f1=30 "1=90 ln C ; ln c2 ; ln(C ; 1)]=(90 ln C )g . where "1 is as in Lemma 6.7. We choose c1 2 N to be a multiple of 36 with c1 27= ln C ; ln c2 ; ln(C ; 1) ; 90" ln C ]. We set " := c1 ". We choose > max f 1 + ; 3c1 ln min ]=ln 2g, where we abbreviate min := minf(i) : i 2 Mg. Finallyp we choose n1 2 N , n1 minf25 c3g, large 2n P1enough;cthat b n c 1 = 2 n 5 m c1 L2 for all n n1 and "2 (n1 )+(2"3 (n1 )) + m=2 c4 e < 1=2 holds, where c3 is dened in Theorem 3.5, "2 (n1 ) in Lemma 4.3, "3 (n1 ) in Theorem 3.3 and c4 and c5 in Lemma 4.4.. 3 The structure of the reconstruction In order to prove Theorem 1.1, we reduce the problem of reconstructing the scenery successively to simpler problems. Theorems 3.1 and 3.2 below show that it suces to nd algorithms which do only partial reconstructions. Proofs are postponed to later sections: Sections 5 and 6 are dedicated to the proof of Theorem 3.5, all other statements of this section are proved in Section 4. Our rst theorem states that it suces to nd a reconstruction algorithm A0 which reconstructs correctly with probability > 1=2:. Theorem 3.1 If there exist 1 > 0 and a measurable map A0 : C N0 ;! C Z such that P (A0 (~) ) > 1=2 for all 2]0 1 , then there exists a measurable map A : C N0 ;! C Z such that P (A(~) ) = 1 for all 2]0 1 . The idea is to apply the reconstruction algorithm A0 to all the shifted observations i (~), i 0. By the hypothesis and an ergodicity argument, as k tends to innity the proportion of sceneries A0 (i (~)) for i 2 0 k which are equivalent to is strictly bigger than the proportion of sceneries which are not equivalent to . Therefore we are able to reconstruct the scenery. We build the algorithm A0 required by Theorem 3.1 by putting together a hierarchy of partial reconstruction algorithms Am , m 1. The algorithm Am tries to reconstruct a piece of scenery around the origin of length of order 2nm with (nm m 2 N ) recursively dened as follows: We choose n1 as in Section 2.1, and we set for m 1 p nm+1 := 2b nm c :. 6. (3.1).
(16) Denition 3.1 For m 1 and a measurable map f : C N0 ! C ;3 2nm 3 2nm ] we dene. m n n Ereconst )
(17) j ;4 2nm 4 2nm ]g : (3.2) f := f j ;2 m 2 m ]
(18) f (~ m Ereconst f is the event that the reconstruction procedure f reconstructs cor-. rectly a piece of scenery of length of order 2nm around the origin. Note that any nite piece of scenery occurs somewhere with probability 1 because the scenery is i.i.d. uniformly colored. Therefore it is crucial to reconstruct a piece of scenery around the origin. Theorem 3.2 Suppose there exist 1 > 0 and a sequence of measurable maps Am : C N0 ! C ;3 2nm 3 2nm ] , m 1, such that for all 2]0 1 ; m m+1 m lim inf Ereconst inf Ereconst (3.3) Am = lim Am \ Ecenter P ; a.s. m!1 m!1 . . m+1 := Am+1 (~ where Ecenter )j ;3 2nm 3 2nm ] = Am (~) . Suppose further that. P. . 1 ; . m=1. m Ereconst Am. c. !. < 1=2 for all 2]0 1 :. (3.4). Then there exists a measurable map A0 : C N0 ;! C Z such that P (A0 (~) ) > 1=2 for all 2]0 1 . In the following, we explain how we construct maps Am satisfying the assumptions of Theorem 3.2. The task of A1 is to reconstruct a piece of scenery of length of order 2n1 around the origin with high probability. It is shown by Lowe, Matzinger, and Merkl in 21] that the whole scenery can be reconstructed with probability one in case there are no errors in the observations. They only prove existence of a reconstruction procedure, but do not explicitly construct an algorithm. In 26] we construct an algorithm which even works in polynomial time: A nite piece of scenery around the origin can be reconstructed with high probability from nitely many error-free observations the number of observations needed is polynomial in the length of the piece of scenery which is reconstructed. We prove: Theorem 3.312For innitely many n 2 N there exists a measurable map n 0 2. 2 n. Ainitial : C ! C ;3 2n 3 2n] such that. "3 (n) := P. . ; . j ;2n 2n ]
(19) Aninitial j 0 2 212n
(20) j ;4 2n 4 2n ] c. satises limn!1 "3 (n) = 0. As an immediate consequence of Theorem 3.3 a piece of scenery around the origin can be reconstructed with high probability even if there are errors in the observations. As long as the probability to see an error at a particular time is suciently small, the probability to see no errors in the rst 2 212n observations is close to 1. The following corollary makes this precise:. 7.
(21) Corollary 3.1 Let Aninitial and "3 (n) be as in Theorem 3.3. There exist 2 (n) > 0 such that for all 2]0 2 (n) ; . P j ;2n 2n ]
(22) Aninitial ~j 0 2 212n
(23) j ;4 2n 4 2n] c 2"3(n): 1 . The maps Am , m 2, will be dened We will choose A1 := Aninitial inductively. Given a partial reconstruction algorithm Am we dene stopping times which tell us when the random walker is in some sense \close" to the origin: We compare Am (~) with Am (t (~)), i.e. we compare the output of Am. if the input consists of the observations collected by the random walker starting at the origin and the observations starting at time t. If both outputs agree up to equivalence on a suciently large subpiece, then with a high chance, the random walker is - on an appropriate scale - close to the origin. The stopping times constructed from Am are used to reconstruct a piece of scenery around the origin of length of order 2nm+1 which is much larger than the piece of scenery reconstructed by Am recall our choice of nm (3.1). Whenever the stopping times indicate that the random walk is \close" to the origin, we collect signicant parts of the observations of length c1 nm . If we have suciently many stopping times, the random walk will walk over the same piece of scenery over and over again. This allows us to lter out the errors in the observations. Once this is done, the obtained words are put together like in a puzzle game. The words are used to extend the piece of scenery of length of order 2nm which has been reconstructed by Am . Formally we dene stopping times in the following way: Denition 3.2 For m 2 N and a measurable map f : C N0 ! C ;3 2nm 3 2nm ] with the property that f (~) depends only on ~j 0 2 212nm , we dene . 12nm+1 ; 2 212nm : 9w 2 C ;2nm 2nm ] such that
(24) t 2 0 2 m +1 Tf (~ ) := w
(25) f (~) and w
(26) f (t (~)) : +1 ) arranged in increasing order. Let t(1) < t(2) < be the elements of Tm f (~. m+1 (~ We dene the sequence Tfm+1(~) := Tfk ) k 1 by m+1 (~ Tfk ) :=. (. . . t(2 22nm+1 k) + 2 212nm if 2 22nm+1 k Tmf +1(~) . 212nm+1 otherwise.. m +1 Tf (~) is a sequence of G -adapted stopping times with values in 0 212nm+1 . the stopping times depend only on ~j 0 212nm+1 . We dene the event that a sequence of stopping times fulls the task of stopping the random walk \close" to the origin (on a rather rough scale). Denition 3.3 For n 2 N and a sequence = (k k 1) of G -adapted stopping n := times we dene the event Estop n 2\ . k=1. . k (~) < 212n jS (k (~))j 2n j (~) + 2 22n k (~) for j < k : 8.
(27) The next theorem states that given an appropriate partial reconstruction algorithm f , the stopping times Tfm+1 full their task with a high probability. By the denition of Tfm+1, we stop at time t + 2 212nm i f (~) and f (t (~)) agree on a large enough subpiece. Therefore, for the stopping times to stop the random walk close to the origin, it is necessary that f (~) is a correctly reconstructed piece of scenery around the origin. Since we apply f often to obtain enough stopping times, we need that given a scenery , there is a high enough chance for the random walk on to be stopped correctly, i.e. f must reconstruct correctly conditional on . This is why n with h high enough i probability o 1 m we need the event P Ereconstf j 2 in the following theorem.. Theorem 3.4 Let m 1, and let f : C N0 ! C ;3 2nm 3 2nm ] be a measurable map with the property that f (~) depends only on ~j 0 2 212nm . We have for all 2]0 1 P. . m+1 . nm+1 Tf m Ereconst f n Estop. . 1 m \ P Ereconst f j 2. .
(28) . e;nm+1 :. The next theorem shows that there exist partial reconstruction algorithms Algn (the reader should think of n = nm ) with the following properties: Given stopping times which stop the random walk close to the origin, nitely many observations with errors and a small piece of scenery close to the origin, Algn reconstructs with high probability a piece of scenery around the origin of length of order 2n. If the reconstruction is succesful, the output of Algn contains in the middle. The reader should think of as a piece of scenery that has been reconstructed before.. Theorem 3.5 For all n 2 N there exists a measurable map 12n Algn : 0 212n ]N C 2 2 C ;knkn] ! C ;3 2n3 2n] k c1 L. with the following property: There exist constants c3 3 c6 c7 > 0 such that for all n c3 , 2]0 3 and for any sequence = (k k 1) of G -adapted stopping times with values in 0 212n] ;. n where Ereconstruct :=. . n n E n ;c7 n P Estop reconstruct c6 e. .
(29). For all 2 C ;knkn] with k c1 L and
(30) j ;2n 2n ] we have : j ;2n 2n]
(31) Algn ( ~j 0 2 212n )
(32) j ;4 2n 4 2n].. Furthermore if j ;2n 2n]
(33) Algn ( ~j 0 2 212n )
(34) j ;4 2n 4 2n] holds, 2 C ;knkn] with k c1 L,
(35) j ;2n 2n ] and j ;2n 2n ] 6= (1) ;2n2n ] , then we conclude that Algn ( ~j 0 2 212n )j ;kn kn] = .. 9.
(36) To motivate the allowed range for the abstract arguments in this theorem, m (~ recall that the Tfk )'s in Denition 3.2 take their values in 0 212nm ]. We are now able to dene Am , m 1, which fulll the requirements of Theorem 3.2. m N0 ;3 2nm 3 2nm ] and sequences T m+1 = 3.4 ;Denition We dene A : C ! C m +1 Tk k 1 recursively for m 1 in the following way: ; 1 1 A1 (~) := Aninitial ~j 0 2 212n1 with n1 as in Section 2.1 and Aninitial as in Theorem 3.3, T m+1(~) := TAmm+1(~) with TAmm+1 as in Denition 3.2, ; Am+1 (~) := Algnm+1 T m+1(~) ~j 0 2 212nm+1 Am (~) with Algnm+1 as in Theorem 3.5. Theorem 3.6 There exists 1 > 0 such that the sequence (Am m 2 N ) dened in Denition 3.4 fulls (3.3) and (3.4) for all 2]0 1 . All theorems of this section together yield the proof of our main theorem: Proof of Theorem 1.1. By Theorem 3.6, the assumptions of Theorem 3.2 are satised. Hence the assumptions of Theorem 3.1 are satised and Theorem 1.1 follows.. 4 Proofs In this section, we prove the statements from Section 3 with the exception of Theorem 3.5 which will be proved in Sections 5 and 6. Lemma 4.1 The shift # : ! , ( S X Y ) 7! ( ( + S (1)) S ( + 1) ; S (1) X ( + 1) Y ( + 1)) is measure-preserving and ergodic with respect to P for all 2]0 1 . Proof. Let 2]0 1 . By assumption, Yk , k 0, is stationary and ergodic under P . Xk , k 0, is i.i.d., hence stationary and ergodic under P . By Lemma 4.1 of 21], ( S ) 7! ( ( + S (1)) S ( + 1) ; S (1)) is measure-preserving and ergodic with respect to P . The claim follows from these three observations and the fact that ( S X Y ) are independent. Proof of Theorem 3.1. Let 1 and A0 : C N0 ! C Z be as in the hypothesis of the theorem, and let 2]0 1 . We dene for k 2 N measurable maps A0k : C N0 ! C Z as follows: If there exists j 2 0 k such that i 2 0 k : A0 (i (~ )) A0 (j (~)) > i 2 0 k : A0 (i (~)) 6 A0 (j (~)) then let j0 be the smallest j with this property, and dene A0k (~) := A0 (j0 (~)). Otherwise dene A0k (~) to be the constant scenery (1)j2Z. Finally we dene A : C N0 ! C Z by ) if this limit exists pointwise, k!1 A0k (~ A(~) := lim (1)j2Z else. 10.
(37) As a limit of measurable maps, A is measurable. For k 2 N we dene kX ;1 1 Zk := k 1 A0 (i (~)) i=0. here 1B denotes the indicator function of the event B . It follows from Lemma 4.1 that the sequence 1 A0 (k (~)) , k 0, is stationary and ergodic because it can be written as a measurable function of the sequence #k ( S X Y ), k 0 note that ( + Sk ). Hence we can use the ergodic theorem and our assumption to obtain P -almost surely: lim Z = P (A0 (~) ) > 1=2: k!1 k. (4.1). Note that if Zk > 1=2, then A0k (~) . By (4.1) there exists a.s. a (random) k0 such that Zk > 1=2 for all k k0 , and hence A0k (~) = A0k0 (~) recall that we chose the smallest possible j0 in the denition of A0k . Thus a.s. A(~) . Proof of Theorem 3.2. We say a sequence (
(38) m m 2 N ) of pieces of sceneries converges pointwise to a scenery
(39) if lim inf m!1 domain(
(40) m ) = Z, and for every z 2 Z there is mz > 0 such that
(41) m (z ) =
(42) (z ) for all m mz . Let 1 and Am be as in the hypothesis of the theorem, and let 2]0 1 . We set A0 (~) := limm!1 Am (~) if this limit exists pointwise on Z otherwise we set A0 (~) := (1)j2Z. Being a pointwise limit of measurable maps, A0 : C N0 ! C Z m is measurable. We abbreviate E m := Ereconst Am , and dene the events m := f j ;2nm 2nm ] 41 j ;4 2nm+1 4 2nm+1 ]g : E1 t. We claim: m holds P -a.s., 1. lim inf m!1 E1 t m ) \ T1 E m holds, then A0 (~ 2. If the event (lim inf m!1 E1 t ) . m=1 mc Together with the assumption P 1 m=1 (E ) ] < 1=2 these two statements 0 imply that P (A (~) ) > 1=2 which yields the claim of the theorem. Proof of claim 1: We show for any integer intervals I1 6= I2 with jI1 j = jI2 j. P ( jI1 jI2 ) 2 C ;jIj j=3 :. (4.2). First we dene fj : 0 jIj j ! Ij for j = 1 2 to be the unique translation which maps 0 jIj j onto Ij . An argument similar to the proof of (6.26) below shows that there exists a subset J 0 jIj j of cardinality jJ j jIj j=3 with f1 (J ) \ f2 (J ) = . Since k , k 2 Z, are i.i.d. with a uniform distribution, we conclude. P ( jI1 jI2 ) P ( jf1 (J ) = jf2 (J )) = C ;jJ j C ;jIj j=3 : Since jI1 jI2 means jI1 jI2 or jI1 ( jI2 )$ with ( jI2 )$ denoting the piece of scenery obtained from jI2 by re
(43) ection, estimate (4.2) follows. 11.
(44) We apply (4.2) for I1 = ;2nm 2nm ] and all integer intervals I2 ;4 n m 2 +1 4 2nm+1 ], I1 6= I2 , of length jI1 j = jI2 j = 2 2nm + 1 there are not more than 8 2nm+1 choices for I2 . We obtain m )c ) 8 2nm+1 2 C ;(2 2nm +1)=3 16 22 nm ;2 2nm =3 P ((E1 t which is summable over m recall C 2 and (3.1). Hence by the Borel-Cantelli m )c occurs P -a.s. only nitely many times this proves claim 1. lemma (E1 t Proof of claim 2: By the assumption of this claim, there is a (random) M m and E m hold for all m M . By the assumption of such that the events E1 t m+1 holds for all m M , Theorem 3.2, M can be chosen in such a way that Ecenter m +1 n n m m m too. Consequently, A (~)j ;3 2 3 2 ] = A (~) for all m M and it p. follows that. A0 (~)j ;k k] = Am (~)j ;k k]. (4.3). for all k 1 and all m large enough. In particular, limm!1 Am (~) exists. m hold, Am (~ Since E m and E1 t )
(45) 1 j ;4 2nm 4 2nm ]. Hence there exists m a unique map h : Z ! Z of the form x 7! am + bm x with am 2 Z and bm 2 f;1 1g that maps Am (~) onto a subpiece of j ;4 2nm 4 2nm ]. It follows from (4.3) that hm is independent of m and maps A0 (~) to . This nishes the proof of claim 2. Proof of Theorem 3.3. By Theorem 1.1 of 26], we know that there exists > 0 and for innitely many n 2 N there exists a measurable map Anini : n ]c ) = 0, where C 02n7 +2 212n ! C ;5 2n 5 2n] such that limn!1 P ( Eini . . ;. . . n := j ;2n;1 2n;1 4 An j 0 2n7 + 2 212n 4 j ;10 2n 10 2n ] : Eini ini. Small modications in the proof of Theorem 1.1 in 26] prove our claim. We remark that alternatively, we could work directly with the maps Anini from 26] without adjusting the constants all proofs in the remainder of the article go through, but the notation becomes more cumbersome. Proof of Corollary 3.1. We estimate the probability under consideration by intersecting with the event B0 := Xk = 0 for all k 2 0 2 212n that there are no errors in the rst 2 212n observations: For any > 0 we have ;. ;. . . 1 ; P j ;2n 2n ]
(46) Aninitial ~j 0 2 212n
(47) j ;4 2n 4 2n ] ; ; 1 ; P j ;2n 2n]
(48) Aninitial ~j 0 2 212n
(49) j ;2n+2 2n+2] \ B0 ; ; = 1 ; (n)P j ;2n 2n]
(50) Aninitial j 0 2 212n
(51) j ;2n+2 2n+2] = 1 ; (n)(1 ; "3 (n)) with (n) := (1 ; )2 212n and "3 (n) as in Theorem 3.3. We choose 2 (n) > 0 such that the last expression is bounded above by 2"3 (n) for all 2]0 2 (n) . Proof of Theorem 3.4. The proof is very similar to the proof of Theorem 3.11 in section 7 of 21] (Our Theorem 3.4 is the analogon of their Theorem 3.11 12.
(52) for our setting). The errors in the observations do not require adaptations of their arguments note that the errors are independent of scenery and random walk and occurences of errors are i.i.d. Bernoulli. The rest of this section is dedicated to the proof of Theorem 3.6. Throughout we assume, Am , m 1, are as in Denition 3.4, and we set 1 := minf3 2 (n1 )g with 3 as in Theorem 3.5 and 2 (n1 ) as in Corollary 3.1. We set for m 1 m E m := Ereconst (4.4) Am : Denition 4.1 For 2]0 1 we dene events of sceneries n. $1 := $2 := =. 1 \. 1 \. m=2. $3 :=. . m=2. . . o. . 2 C Z : P (E 1 )c (2"3(n1 ))1=2 . 2 CZ : P. . 1 n \ m=2. m i. nm T e; 2 C Z : P E m;1 j 12 ) P E m;1 n Estop h. . h. m E m;1 n E nm T. stop. h. i. \. . P E m;1 j 12. m. nm T n E m 2 C Z : P E m;1 \ Estop. . i.
(53) . nm.
(54). 2.
(55). n e; 2m cn o. (c6 )1=2 e; 7 2 m . $ := $1 \ $2 \ $3 where "3 (n1 ) is as in Theorem 3.3 and c6 and c7 are as in Theorem 3.5. Note the similarity between these events and the bounds in Corollary 3.1, Theorems 3.4 and 3.5. The following lemma provides a link between bounds with and without conditioning on the scenery : Lemma 4.2 (
(56) 21], Lemma 4.6) Let A be an event, r 0, and let Q be a probability measure on . If Q(A) r2 , then Q (Q(Aj ) > r) r. Lemma; 4.3 For all n 2 N there exist "2 (n) > 0 with limn!1 "2 (n) = 0 such that P 2= $ "2 (n1 ) for all 2]0 1 . Proof. Let 2]0 1 . Using Corollary 3.1 and Lemma 4.2 for Q = P , we obtain ; P 2= $1 (2"3(n1 ))1=2 : (4.5) An application of Theorem 3.4 with f = Am yields for m 2
(57) m;1 m. 1 n T m ; 1 m P E n Estop \ P E j 2 e;nm : An application of Lemma 4.2 with Q = P yields ;. . P 2= $2 . 1 X. m=2. e;nm =2 e;c8n1 13. (4.6).
(58) for some constant c8 > 0, recall our choice of nm (3.1). Let m 2, and recall nm T m from Theorem 3.5. By Denition 3.4, the denition of the event Ereconstruct ; we have that Am (~) = Algnm T m(~) ~j 0 2 212nm with := Am;1 (~). By our choice of n1 , (j j ; 1)=2 = 3 2nm 1 c1 nm L. If E m;1 holds, then
(59) j ;2nm 2nm ]. Hence the inclusion ;. m. m. m. mT n E m E mT n E nm T E m;1 \ Estop stop reconstruct. holds. Together with Theorem 3.5 the last inclusion implies. m. nm T n E m P E m;1 \ Estop. (4.7). mT m n E nm T m ;c7nm : P Estop reconstruct c6 e. Another application of Lemma 4.2 yields for some constant c9 > 0 ;. . P 2= $3 . 1 X. m=2. (c6 )1=2 e;c7nm =2 e;c9 n1 :. (4.8). The claim of the lemma follows from (4.5), (4.6), and (4.8) recall "3 (n) ! 0 as. n ! 1.. Lemma 4.4 For all 2]0 1 , 2 $ , and m 2 the following holds for some constants c4 c5 > 0:. P (E m;1 j ) 1 ; (2"3 (n1 ))1=2 ;. mX ;1 k=2. c4 e;c5 nk 12 . (4.9). P (E m;1 n E m j ) c4 e;c5nm : (4.10) Proof. Let 2]0 1 and 2 $ . We prove (4.9) and (4.10) simultaneously by induction over m: For m = 2 it follows from 2 $1. h; i P (E 1 j ) = 1 ; P E 1 c j 1 ; (2"3 (n1 ))1=2 1=2 (4.11). recall our choice of n1 from Section 2.1. Thus (4.9) holds for m = 2. Suppose (4.9) holds for some m 2. Then we have m i. h. h. m i. mT + P E m;1 n E mT P E m;1 n E m j ] P (E m;1 n E m ) \ Estop stop c n m 7 (c6 )1=2 e; 2 + e;nm =2 c4 e;c5 nm (4.12) for some constants c4 c5 > 0 for the rst term we used 2 $3 and for the second term we used 2 $2 and our induction hypothesis (4.9). Using (4.12). and our induction hypothesis (4.9) we obtain. P (E m j ) P (E m;1 j ) ; P (E m;1 n E m j ). 1 ; (2"3 (n1 ))1=2 ; 14. m X. k=2. c4 e;c5 nk 21 .
(60) for the last inequality we used our choice of n1 . This completes the induction step. Proof of Theorem 3.6. Let 2]0 1 recall our choice 1 = minf3 2 (n1 )g. By Theorem 3.5 we know that whenever the events E m;1 and E m hold and m holds. Since P -a.s. 6= (1)Z, rela j ;2nm 2nm ] 6= (1) ;2nm 2nm ] , then Ecenter tion (3.3) holds. Using Lemma 4.3 we have. P. . 1 . (E m )c. !. m=1. . ;. P 2= $ + P. "2 (n1 ) +. Z. f2 g. f 2 $ g \. P. . 1 . 1 m=1. (E m )c. . !. !. (E m )c dP : (4.13). m=1. To bound the integrand, we use Lemma 4.4: For all 2 $ and k 1, we obtain. P. . k . . (E m )c . . m=1. !. ;. . P (E 1 )c j + (2"3(n1 ))1=2 +. and taking limits as k ! 1, we conclude. P. . 1 . . (E m )c . m=1. kX +1. m=2 kX +1. m=2. !. (2"3 (n1 ))1=2 + . P (E m;1 n E m j ). c4 e;c5 nm . 1 X m=2. (4.14). c4 e;c5nm :. Together with (4.13) the last estimate yields (3.4):. P. . 1 . m=1. (E m )c. !. "2 (n1 ) + (2"3(n1 ))1=2 +. 1 X m=2. c4 e;c5 nm < 21 . (4.15). for the last inequality we used that n1 is chosen as in Section 2.1.. 5 The key algorithm of the reconstruction. In this section, we dene algorithms Algn for which Theorem 3.5 holds. We x n 2 N. For two words w w0 2 C of the same length we dene their distance. d(w w0 ) := jfk 2 1 jwj] : wk 6= wk0 gj (5.1) d(w w0 ) is the number of places where w and w0 disagree. Clearly, d is a metric. When the random walk observes a piece of scenery and is small, the ob-. servations with errors dier \typically" from the errorfree observations in only a small proportion of the letters because the probability to see an error at a 15.
(61) particular time is small under P . Since the random walk observes a given piece of scenery very often, we are able to lter out the errors using a majority rule f . The following notions will be used in this context. For w = w1 w2 : : : wm 2 C m we dene Cut(w) := w2 : : : wm;1 Cut(w) is obtained from w by cutting o the rst and the last letter. Denition 5.1 Let W = (wj 1 j K ) 2 (C c1 n)K be a vector consisting of K words of length c1 n. For i 2 1 c1n] we dene fi (W ), the favorite letter at position i, to be the element in C which most of the rst 2n words in W have at position i. If there is no unique letter with this property, then we dene the favorite letter to be the smallest one. Formally, we set fi (W ) = k i jfj 2 1 2n] : wj (i) = kgj = max jfj 2 1 2n] : wj (i) = k0 gj k 2C 0. and k is the smallest element in C satisfying the last equality here wj (i) denotes the ith letter of the word wj . We set f (W ) := f1 (W )f2 (W ) : : : fc1 n (W ). Furthermore, we dene f (W ) := Cut(f (W )) if K 2n and maxj2 12n] d(Cut(wj ) Cut(f (W ))) "n (;1) 1c1n;2] otherwise. f (W ) equals the word Cut(f (W )) which is composed of the favorite letters i the vector W has suciently many components and each of the rst 2n words in W diers from f (W ) in not more than "n letters. In the proof of Lemma 6.9 below it will be essential that we use Cut(f (W )) and not f (W ) in the denition of f (W ). Note that ;1 2= C so that (;1) 1c1 n;2] diers from all words w 2 C c1 n;2 . The algorithm Algn which will be dened below takes input data. 2 0 212n N 2 C 2 212n and 2. . . k c1 L. C ;knkn] :. (5.2). First we dene the set of all observations of length 3c1n which are collected within a time horizon of length 22n after a time k k 2 1 2n ]: Denition 5.2 We dene Collectionn ( ) := n. o. (w1 w2 w3 ) 2 (C c1 n )3 : 9k 2 1 2n] such that w1 w2 w3 v j k k + 22n :. The set PrePuzzlen ( ) contains only (w1 w2 w3 ) 2 Collectionn ( ) with the following property: If (w10 w20 w30 ) 2 Collectionn ( ) and w10 and w30 are \not too dierent" from w1 and w3 respectively, then w20 is \not too dierent" from w2 . Formally: Denition 5.3 We dene PrePuzzlen( ) :=
(62) (w1 w2 w3 ) 2 Collectionn ( ) : If (w10 w20 w30 ) 2 Collectionn ( ) with : d(w1 w10 ) 2"n and d(w3 w30 ) 2"n, then d(w2 w20 ) 2"n. 16.
(63) n ( ) we denote by Denition 5.4 For an element (w1 w2 w3 ) 2 PrePuzzle n n 2 S (w1 w2 w3 ) the sequence of (random) times sn 2 k=1 k k + 22n ; 3c1 n such that w10 w20 w30 := j s s + 3c1n 2 PrePuzzle ( ), d(w1 w10 ) 2"n, and n (w1 w2 w3 ) d(w3 w30 ) 2"n we assume that the elements of the sequence S. are arranged in increasing order. We dene ; n (w1 w2 w3 ) Listn (w1 w2 w3 ) := j s + c1 n s + 2c1 n s 2 S. to be the sequence with components j s + c1 n s + 2c1 n indexed by the set n (w1 w2 w3 ). We set S . . PuzzleListsn ( ) := Listn (w1 w2 w3 ) : (w1 w2 w3 ) 2 PrePuzzlen ( ) : Clearly, w2 2 Listn (w1 w2 w3 ). Note that Listn (w1 w2 w3 ) is a sequence,. and not a set. If by coincidence observations j s + c1 n s + 2c1 n coincide for two dierent values of s, we want to keep them both. The components of Listn (w1 w2 w3 ) are close to w2 in d-distance because we assumed (w1 w2 w3 ) 2 PrePuzzlen ( ). Denition 5.5 We dene Puzzlen( ) := ff (W ) : W 2 PuzzleListsn( )g. Puzzlen ( ) is the set of all words of length c1 n ; 2 which are obtained by the majority rule f from the lists in PuzzleListsn ( ). We use the words in Puzzlen ( ) like the pieces in a puzzle game to reconstruct a piece of scenery. We want the piece of scenery reconstructed by Algn to contain in the middle the piece of scenery from the input data of the algorithm. Denition 5.6 For 2 C ;knkn] we dene SolutionPiecen( ) := .
(64). w 2 C ;3 2n3 2n] : wj ;kn kn] = and for all ladder intervals I : ;3 2n 3 2n ] with jI j = c1 n ; 2 we have (wjI )! 2 Puzzlen ( ). We will see in the proof of Lemma 6.4 below that under appropriate conditions, there is precisely one element in SolutionPiecen ( ). Denition 5.7 We dene 12n. Algn : 0 212n ]N C 2 2. . . k c1 L. C ;knkn] ! C ;3 2n3 2n]. as follows: If SolutionPiecen ( ) is not empty, then we dene Algn ( ) to be its lexicographically smallest element. Otherwise we dene Algn ( ) to be the constant scenery (1) ;3 2n 3 2n] .. 6 The key algorithm reconstructs correctly. In this section, we prove Theorem 3.5. Throughout we x n 2 N . We assume that 2 0 212n]N is a sequence of G -adapted stopping times. Recall that " was chosen in Section 2.1. 17.
(65) 6.1 Denition of the key events. In this subsection, we collect the denitions of all the \basic" events which we n will need to prove the correctness of Algn . The event Ball paths holds if the random walk traverses all paths of length 3c1 n in the region where we want n to do the reconstruction. Bfew mistakes makes sure that there are not too many n mistakes in the words in Collectionn ( ). Bladder di gives a lower bound for the d-distance of two dierent ladder words in the neighborhood of the origin. n Bmajority garanties that the majority decision f is not corrupted by the errors n in the observations. If Boutside out holds, then we can distinguish ladder words from the region where we want to reconstruct from observations which are read n further outside. Bsignals implies that there are "signal words" which can be read only left from a certain point z 2 Z or only right from a certain z 2 Z this event allows use to reconstruct all ladder words in a region around the origin. n Bstraight often guarantees that certain ladder paths are traversed often enough. We arranged the denitions of the events in alphabetical order so that the reader can easily nd them while following the proofs in the next two subsections. We suggest to have a quick look at the denitions, and then to skip ahead to the next subsection and look up denitions when needed.. Denition 6.1 For z 2 Z and n such that c1 n 2 N , we denote by wz!n the. ladder word of length c1 n starting at z read from left to right, and by wzn the word wz!n read from right to left:. wz!n := ( (z + kL) k 2 0 c1n )! and wzn := (wz!n ) : Note that wz;(c1 n;1)L!n is the ladder word of length c1 n ending at z .. Denition 6.2 We dene 8 <For. 9. c1n with starting= any admissible piece of path R 2 Z 03n n n n 2 Ball paths := :point in ;7 2 7 2 ] there exists t 2 k=1 k k + 22n ; : 3c1 n] such that R(i) = S (t + i) for all i 2 0 3c1n Denition 6.3 We dene (. Bn. few mistakes :=. t X k=t;c1 n+1. Xk "n for all t 2 c1. ). n ; 1 2 212n. :. Denition 6.4 We dene 8z1 z2 2 ;8 2n 8 2n ] and 8i1 i2 2 f !g with
(66) : n Bladder := di (z1 i1 ) = 6 (z2 i2 ) we have d(wz1 i1 n=3 wz2 i2 n=3 ) 10"n Denition 6.5 Let IL denote the set of ladder intervals I ;7 2n 7 2n] of. length c1 n. For w1 w3 2 C c1 n and I 2 IL , we denote by SwI 1 w3 := sIi i 1. 2n (SwI 1 w3 := sIi i 1 ) the sequence of all times s 2 2kn =1 k k + 2 ; 3c1 n !. . . 18. !.
(67) such that S j s + c1 n s + 2c1 n is a straight crossing from left to right (right to left) of I and d(~j s + (i ; 1)c1 n s + ic1 n wi ) 2"n for i = 1 3. We assume that the components of SwI 1 w3 and SwI 1 w3 are arranged in increasing order. We dene !. n Bmajority :=. \. \ . nI (w w ) \ B nI (w w ) with Bmaj 1 3 1 3 maj !. w1 w3 2C c1 n I 2IL ( If SwI 1 w3 . nI (w w ) := Bmaj 1 3 !. . !. . 2nP, then 8j 2 1 c1 n ; 1 the n. following holds:. 2 XI i=1 si +c1 n+j !. ). < 2n=2. nI (w w ) dened analogously. and Bmaj 1 3 . n Denition 6.6 We dene Boutside out := 8 9 <8z 2 ;5 2n 5 2n ], for any admissible piece of path R 2 ( ;2L 22n 2L = 22n ] n ;6 2n , 6 2n]) 0c1 n=2 and 8i 2 f !g we have that d( : : R wzin=2 ) 3"n n Denition 6.7 We dene Brecogn straight := 8 > For > <. 9. any admissible piece of path R1 2 ;7 2n 7 2n ] 0c1n which is> > not a ladder path there exists an admissible piece of path R2 2 ;8 = : > 2n 8 2n ] 0c1 n with R2 (0) = R1 (0), R2 (c1 n ; 1) = R1 (c1 n ; 1) and> > > : d( R1 R2 ) 5"n. Denition 6.8 We dene n n n n n Bsignals := Bsign l! \ Bsignr! \ Bsignl \ Bsignr with 8 9 <8z 2 ;6 2n 6 2n ] and for any admissible piece of path= n 2n 2n 0c n Bsign l! := :R 2 ;2L 2 2L 2 ] 1 with R(c1 n ; 1) > z we have that d( R wz;(c1 n;1)L!n) 5"n 8 9 <8z 2 ;6 2n 6 2n ] and for any admissible piece of path= n 2n 2n 0c n Bsign r! := :R 2 ;2L 2 2L 2 ] 1 with R(0) < z we have that d( R wz!n ) 5"n 8 9 <8z 2 ;6 2n 6 2n ] and for any admissible piece of path= 2n 2n 0c n n Bsign l := :R 2 ;2L 2 2L 2 ] 1 with R(0) > z we have that d( R wz;(c1 n;1)Ln) 5"n 8 9 <8z 2 ;6 2n 6 2n ] and for any admissible piece of path= n 2n 2n 0c n Bsign r := :R 2 ;2L 2 2L 2 ] 1 with R(c1 n ; 1) < z we : have that d( R wzn ) 5"n Denition 6.9 We denote the collection of ladder intervals I ;6 2n 6 2n] of length 3c1 n by JnL . For I 2 JL , we denote by S! (I ) (S (I )) the sequence of all times s 2 2k=1 k k + 22n ; 3c1n] such that S j s s + 3c1 n is a straight 19.
(68) crossing from left to right (right to left) of I we assume that the components of S! (I ) and S (I ) are arranged in increasing order. We dene \ n Bstraight := fjS! (I )j 2n and jS (I )j 2ng : often I 2JL. 6.2 Combinatorics. In this subsection, we prove that Algn reconstructs correctly in the sense that n n and all the \basic" the event Ereconstruct holds, under the assumption that Estop events dened in the previous subsection hold. We abbreviate . ~n := ~ 0 2 212n : The task is split in four parts: Lemma 6.1 states a property of the elements in the set PrePuzzlen ( ~n ). Lemma 6.2 shows that all words in Puzzlen ( ~n ) which are observed while the random walk is approximately in the region of the scenery which we want to reconstruct, are ladder words. Lemma 6.3 states that Puzzlen ( ~n ) contains all the ladder words we need. Finally Lemma 6.4 shows that the reconstruction works. Denition 6.10 We say (w1 w2 w3 ) 2 Collectionn(n~n ) is read while the random walk is walking on J Z if there exists t 2 2k=1 k k + 22n ; 3c1 n such that S (t + j ) 2 J for all j 2 0 3c1n and w1 w2 w3 = ~j t t + 3c1n . If we know the time t, we say that (w1 w2 w3 ) is read during t t + 3c1 n . n Denition 6.11 We dene Epre ladder := 8 > >If <. 9. (w1 w2 w3 ) 2 PrePuzzlen ( ~n ) and there exists t 2 2kn > =1 k k +> 22n ; 3c1 n] such that (w1 w2 w3 ) is read during t t + 3c1 n while the= : > random walk is walking on ;7 2n 7 2n ], then S j t + c1 n t + 2c1n is a> > > : ladder path. Lemma 6.1 For all n 2 N the following holds:. n n n n Epre ladder Ball paths \ Bfew mistakes \ Brecogn straight : n n n Proof. Suppose the events Ball paths , Bfew mistakes , and Brecogn straightnhold. Let n n (w1 w2 w3 ) 2 PrePuzzle ( ~ ), and suppose there exists t 2 2k=1 k k + 22n ; 3c1n] such that the triple (w1 w2 w3 ) is read during t t + 3c1 n while the random walk is walking on ;7 2n 7 2n ]. Let Ri (j ) := S (t + (i ; 1)c1 n + j ) for j 2 0 c1 n and i = 1 2 3. Then jRi (j )j 7 2n for all j 2 0 c1 n and d( Ri wi ) "n for i = 1 2 3 (6.1) n because Bfew mistakes holds. We have to show that R2 is a ladder path. Suppose n 0 not. Since Brecogn straight holds, there exists an admissible piece of path R2 2 ;8 2n 8 2n ] 0c1n with the same starting and endpoint as R2 and d( R2 R20 ) 5"n: (6.2). 20.
(69) n 0 Since Ball piece of paths holds and the concatenation R1 R2 R3 is an admissible 2n n n path with starting point in ;7 2 7 2 ], there exists t0 2 2kn. + =1 k k 2 ; 0 0 3c1n] such that R1 R2 R3 (i) = S (t + i) for all i 2 0 3c1n . Using the triangle inequality, we obtain. d(w2 ~j t0 + c1 n t0 + 2c1 n ) d(w2 j t0 + c1 n t0 + 2c1 n ) ; "n = d(w2 R20 ) ; "n d( R2 R20 ) ; d(w2 R2 ) ; "n 5"n ; "n ; "n = 3"n (6.3) n for the rst inequality we used that Bfew mistakes holds, and for the last inequaln ity we used (6.2) and (6.1). The fact that Bfew mistakes holds together with inequality (6.1) yields. d(w1 ~j t0 t0 + c1 n ) d(w1 j t0 t0 + c1 n ) + "n = d(w1 R1 ) + "n 2"n: By the same argument, d(w3 ~j t0 +2c1 n t0 +3c1 n ) 2"n. Together with (6.3) this contradicts (w1 w2 w3 ) 2 PrePuzzlen ( ~n ). Hence R2 is a ladder path.. Denition 6.12 We dene. 8 ; Listn ~n ( > > > <PrePuzzlen (. . w1 w2 w3 ) 2 C c1 n;2 : (w1 w2 w3 ) 29 > > > ~n ) and 9(w10 w20 w30 ) 2 = n n n n 0 Puzzle1 ( ~ ) := PrePuzzle ( ~ ) such that d(w1 w1 ) 2"n, > > 0 0 0 0 > > > :d(w3 w3 ) 2"n and (w1 w2 w3 ) is read while the> n n random walk is walking on Z n ;6 2 6 2 ].. Puzzlen2 ( ~n ) := Puzzlen ( ~n ) n Puzzlen1 ( ~n ) (;1) 1c1 n;2] : f. . . Note that Puzzleni ( ~n ), i = 1 2, together with (;1) 1c1n;2] , form a partition of the set Puzzlen ( ~n ). If we are given an element of Puzzlen ( ~n ), we cannot decide to which set of the partition it belongs. Nevertheless the sets Puzzleni ( ~n ), i = 1 2, will be useful in the following.. Denition 6.13 We dene n ( ~n ), then w2
(70) j ;7 2n 7 2n ]
(71) If w 2 Puzzle n 2 2 E := : only ladder. and w2 is a ladder word. Let c10 > 0 be chosen in such a way that for all n c10 3c1 nL 2n :. Lemma 6.2 For all n c10 the following holds: n n n n n Eonly ladder Epre ladder \ Bfew mistakes \ Bladder di \ Bmajority : 21. (6.4).
(72) n n n Proof. Let n c10, and suppose the events Epre ladder , Bfew mistakes , Bladder di n n n and Bmajority hold. Let (w1 w2 w3 ) 2 PrePuzzle ( ~ ) and abbreviate W := Listn ~n (w1 w2 w3 ). Suppose f (W ) 2 Puzzlen2 ( ~n ). Let w20 2 W . Then there exist w10 w30 such that (w10 w20 w30 ) 2 PrePuzzlen ( ~n ), d(w1 w10 ) 2"n, and d(w3 w30 ) 2"n. By denition of Puzzlen2 ( ~n ), at least once the random walk is in ;6 2n 6 2n] while it reads (w10 w20 w30 ). Since the random walk jumps at most a distance of L in each step, it can move in 3c1 n steps at most a distance of 3c1 nL 2n. Hence (w10 w20 w30 ) is observed while the random walk n 0 is walking on ;7 2n 7 2n ]. Using that Epre ladder holds, we obtain nthat w2 is observed while the random walk is walking on a ladder word. Since Bfew mistakes holds, there exists a ladder word wb2
(73) j ;7 2n 7 2n ] such that d(w20 wb2 ) "n: (6.5) Suppose w200 2 W . Then by the above argument, there exists a ladder word w2
(74) j ;7 2n 7 2n ] such that d(w200 w2 ) "n: (6.6) Since (w1 w2 w3 ) 2 PrePuzzlen ( ~n ), we have that d(w20 w2 ) 2"n and d(w2 w200 ) 2"n. Hence d(w20 w200 ) 4"n: (6.7). Using the triangle inequality, (6.5), (6.7) and (6.6) we obtain. d(wb2 w2 ) d(wb2 w20 ) + d(w20 w200 ) + d(w200 w2 ) "n + 4"n + "n = 6"n: (6.8) n If wb2 6= w2 , then it follows from Bladder di that d(wb2 w2 ) 10"n, which contradicts (6.8). Hence wb2 = w2 . We have shown that any w20 2 W is observed while the random walk reads the ladder word wb2 . Hence for j 2 0 c1 n , w20 (j ) equals wb2 (j ) or an error in the observations. Since by assumption, f (W ) 6= (;1) 1c1n;2] , W has at least 2n components recall the denition of f (Denition 5.1). An application of nI (w w ) with I equal to the ladder interval underlying wb shows that Bmaj 1 3 2 n th more than half of the rst 2 words in W have j letter equal to wb2 (j ). n Consequently, f (W ) = wb2 , and since Bfew mistakes holds, f (W ) = Cut(wb2 ). Denition 6.14 We dene Eallnladder := f8z 2 ;5 2n 5 2n ] : Cut(wz!n ) Cut(wzn ) 2 Puzzlen ( ~n )g : Lemma 6.3 For all n c10 the following holds: n n n n n Eall ladder Ball paths \ Bfew mistakes \ Bmajority \ Bsignals n n \Bstraight often \ Estop : 22.
(75) n Proof. Let n c10 and z 2 ;5 2n 5 2n ]. Suppose the events Ball paths ,. n n n n n Bfew , Bsignals , Bstraight mistakes , Bmajority often , and Estop hold. We will prove n n Cut(wz!n ) 2 Puzzle ( ~ ). The proof for wzn is similar. We dene w1 := wz;c1 nL!n w2 := wz!n w3 := wz+c1 nL!n : Clearly, w1 w2 w3 is the ladder word of length 3c1 n starting at z ; c1 nL and ending at z + (2c1 n ; 1)L. We dene R : 0 3c1n ! Z by R(i) = z ; c1 nL + iL. Then R is a ladder path with starting point z ; c1 nL ;6 2n and endpoint z + (2c1 n ; 1)L 6 2nnby our choice of z and n recalln (6.4). Furthermore R = w1 w2 w3 . Since Ball paths holds, there exists t 2 2k=1 k k + 22n ; 3c1n] such that R = S j t t + 3c1 n . We set wbit := ej t + (i ; 1)c1 n t + ic1n for i = 1 2 3: (6.9) n n Since Bstraight t's with this propoften holds, there are at least 2 dierent erty. Fix t. Clearly, (wb1t wb2t wb3t ) 2 Collectionn ( ~n ). We want to show (wb1t wb2t wb3t ) 2 PrePuzzlen ( ~n ). The word wbit diers from wi only by n errors in the observations. Since Bfew mistakes holds, d(wi wbit ) "n for i = 1 2 3: (6.10) n ( ~n ) and d(w0 wbit ) 2"n for i = 1 3. Suppose (w10 w20 w30 ) 2 Collection i n 0 2 Then there exists t 2 k=1 k k + 22n ; 3c1 n] such that w10 w20 w30 = ~j t0 t0 + 3c1n . Using (6.10) and the triangle inequality, we obtain d(wi0 wi ) d(wi0 wbit ) + d(wbit wi ) d(wi0 wbit ) + "n 3"n for i = 1 3: n We set I1 := t0 t0 + c1 n , I3 := t0 + 2c1 n t0 + 3c1 n . Since Bfew mistakes holds, d( S jIi wi ) d( S jIi wi0 ) + d(wi0 wi ) "n + d(wi0 wi ) 4"n for i = 1 3: (6.11) n holds, jS ( )j 2n , and for all i 2 0 22n , jS ( + i)j 2n + L 22n Since Estop k k 2 n 2L 2 because each jump of the random walk has length L. Hence we can n n use that Bsign l! holds for w1 = wz;0 c1 nL!n (note that jz ; Lj 6 2 ) and S jI1 to conclude from (6.11) that S (t + c1 n ; 1) z ; L. Similarly, we can use n n that Bsign r! holds for w0 3 = wz+c1 nL!n (note that jz + c1 nLj 6 2 ) and S jI3 to conclude that S (t + 2c1 n) z + c1 nL. The only path of length c1 n + 2 from z ; L to z + c1nL is the ladder path which visits precisely the points z + iL, 0 i c1 n ; 1. Hence w20 is observed with errors by the random walk walking n on the ladder word w2 . Using the fact that Bfew mistakes holds and (6.10), we. obtain. d(w20 wb2t ) d(w20 w2 ) + d(w2 wb2t ) "n + "n = 2"n: Consequently, (wb1t wb2t wb3t ) 2 PrePuzzlen ( ~n ). We set W := Listn ~n (wb1t wb2t wb3t ): 23.
(76) Clearly, W 2 PuzzleListsn ( ~n ). Consider wbis for s 6= t. Recall that there are at least 2n ; 1 dierent s with this property. By the triangle inequality and (6.10), d(wbis wbit ) d(wbis wi )+d(wi wbit ) 2"n for i = 1 2 3. Consequently, (wb1s wb2s wb3s ) 2 W , and we conclude that W has at least 2n components. Suppose w20 2 W . Then there exist w10 w30 with d(wi0 wbit ) 2"n for i = 1 3 and (w10 w20 w30 ) 2 PrePuzzlen ( ~n ). We have shown above (after (6.10)) that under these conditions, w20 must be observed while the random walk reads the ladder word w2 . In particular, for j 2 0 c1n , w20 (j ) = w2 (j ) or w20 (j ) is an nI (wb wb ) holds for the ladder interval error in the observations. Since Bmaj 1t 3t I = fz + iL i 2 0 c1 n g, in more than half of the words in W the j th letter equals w2 (j ). Consequently, the j th letter of f (W ) equals w2 (j ), and we have proved that Cut(w2 ) 2 Puzzlen ( ~n ). n Recall the denition of Ereconstruct from Theorem 3.5.. Lemma 6.4 For all n c10 with c10 as in (6.4) the following holds: n n n n n Ereconstruct Eonly ladder \ Eall ladder \ Bfew mistakes \ Bladder di n n \Boutside out \ Estop : n n n Proof. Let n c10, and suppose all the events Eonly ladder , Eall ladder, Bladder di , n n n. ; knkn ] Bfew mistakes , Boutside out , and Estop hold. Let 2 C for some k c1 nL, and suppose
(77) j ;2n 2n ]. There exist a 2 ;2n 2n] and b 2 f;1 1g such that for all j 2 ;kn kn] (j ) = (a + bj ) and a + bj 2 ;2n 2n ] : (6.12) First we show w := ( (a + bj ) j 2 ;3 2n 3 2n ]) 2 SolutionPiecen ( ~n ). By (6.12), = wj ;kn kn]. Let I ;3 2n 3 2n ] be a ladder interval of length c1 n ; 2. The image of I under the map j 7! a + bj is a ladder interval n which is contained in ;4 2n 4 2n] because jaj 2n. Since Eall ladder holds, n n n (wjI )! 2 Puzzle ( ~ ). Consequently, w 2 SolutionPiece ( ~n ), and in particular, SolutionPiecen ( ~n ) is not empty. It remains to show that j ;2n 2n]
(78) w
(79) j ;4 2n 4 2n ] for any element w 2 SolutionPiecen ( ~n ). Let w 2 SolutionPiecen ( ~n ). Then wj ;kn kn] = , and it follows from (6.12) that for all j 2 ;kn kn] w(j ) = (a + bj ):. (6.13). Suppose we prove (6.13) for all j 2 ;3 2n 3 2n ]. Then we know there is precisely one element in SolutionPiecen ( ~n ). Since
(80) j ;2n 2n ], there are more than 22n letters to the left and to the right of in w, and consequently j ;2n 2n ]
(81) w. On the other hand, in w, there are less than 3 2n letters to the left and to the right of . Hence w
(82) j ;4 2n 4 2n ]. Thus, to nish the proof, it suces to verify (6.13) for all j 2 ;3 2n 3 2n ]. We have already seen that (6.13) holds for all j 2 ;kn kn]. Suppose we know 24.
(83) that (6.13) holds for all j 2 ;s s] for some s 2 kn 3 2n ; 1]. We set. wl := (wjIl )! with Il := (;s ; 1 + iL i 2 0 c1 n ; 2 ) wr := (wjIr )! with Ir := (s + 1 + (i ; c1 n + 3)L i 2 0 c1 n ; 2 ) note that Il denotes the ladder interval of length c1 n ; 2 which contains ;s ; 1 as leftmost point, and Ir denotes the ladder interval of length c1 n ; 2 which contains s + 1 as rightmost point. The words wl and wr are well dened because c1 nL j j = 2kn + 1. Since w 2 SolutionPiecen ( ~n ), we have wl wr 2 Puzzlen ( ~n ). Note that wl and wr have both precisely c1 n ; 3 points in common with wj ;s s] wl extends wj ;s s] one letter to the left, and wr extends wj ;s s] one letter to the right. Suppose wl 2 Puzzlen1 ( ~n ). Then we have wl = f (W ) for some W = Listn ~n (w1 w2 w3 ) and there exists (w10 w20 w30 ) 2 PrePuzzlen ( ~n ) such that d(wi wi0 ) 2"n, for i = 1 3 and (w10 w20 w30 ) is read while walk is the random 2n ; 3c1 n walking on Z n ;6 2n 6 2n ]. Thus, there exists t 2 2kn + 2 k k =1 such that jS (t + j )j > 6 2n for all j 2 0 3c1n and w20 = ~jJ with J = n holds, we know that jS ( )j 2n for t + c1 n t + 2c1 n . Using that Estop k all k. Since the random walk jumps a distance L in each step, it follows that jS (t + j )j 2n + L 22n 2L 22n for all j 2 0 3c1n . For a word w = w1 w2 : : : wm 2 C m of length m c1 n=2, we dene Last(w) := wm;c1 n=2+1 : : : wm to be the word consisting of the last c1 n=2 letters of w. Let z 2 ;5 2n 5 2n ] n n and i 2 f !g. Since Bfew mistakes and Boutside out hold, we obtain d(Last(Cut(w20 )) wzin=2 ) = d(Last(Cut(~jJ )) wzin=2 ) (6.14) d(Last(Cut(jJ )) wzin=2 ) ; "n 3"n ; "n = 2"n: By denition of f (W ), d(Cut(f (W )) Cut(w)) "n for all w 2 W . Hence d(Last(wl ) Last(Cut(w20 ))) "n: (6.15) Combining (6.14) and (6.15), we obtain. d(Last(wl ) wzin=2 ) d(Last(Cut(w20 )) wzin=2 ) ; d(Last(wl ) Last(Cut(w20 ))) 2"n ; "n = "n: (6.16) Recall that wl is a ladder word of w of length c1 n ; 2 and the c1 n ; 3 right-most letters of wl overlap with wj ;s s]. Using that (6.13) holds for all j 2 ;s s] together with jaj 2n and jsj 3 2n , yields Last(wl )
(84) j ;4 2n 4 2n ]. This contradicts (6.16), which implies that Last(wl ) is dierent from any ladder word n of j ;4 2n 4 2n ]. We conclude wl 2 Puzzlen2 ( ~n ). Since Eonly ladder holds, n n wl
(85) j ;7 2 7 2 ], and wl is a ladder word of . Suppose (6.13) does not hold for j = ;s ; 1. Let Il denote the image of Il under the map j 7! a+bj . Then jIl 6= wl more precisely, jIl and wl disagree in precisely one point, namely the leftmost point (a + b(;s ; 1)) 6= wl (0). Thus we found two ladder words of length c1 n ; 2 in j ;7 2n 7 2n] which disagree in precisely one point. Consequently, there exist z z 0 2 ;8 2n 8 2n], i i0 2 f 25.
(86) !g with (z i) 6= (z 0 i0 ) such that jIl = Cut(wzin ) and wl = Cut(wz i n ). Consequently, there exist z1 z2 2 ;8 2n 8 2n ], i1 i2 2 f !g with (z1 i1 ) 6= (z2 i2 ) such that the two ladder words consisting of the last c1 n=3 letters of jIl n and wl respectively, equal wz1 i1 n=3 , wz2 i2 n=3 , respectively. Since Bladder di holds, wz1 i1 n=3 6= wz2 i2 n=3 which is a contradiction. We conclude that (6.13) holds for j = ;s ; 1. To see that (6.13) holds for j = s +1, one applies the above argument with w dened by w(j ) := w(;j ) for j 2 ;3 2n 3 2n ] in place of w. By the induction principle, (6.13) holds for all j 2 ;3 2n 3 2n ]. 0. 0. 6.3 The basic events have high probabilities. In this subsection, we prove that the events B:::n dened in Subsection 6.1 have a probability which is exponentially small in n. For some events B:::n this is n holds, i.e. if the stopping times stop only true under the assumption that Estop correctly. We treat the events from Subsection 6.1 in alphabetical order. Recall that unless otherwise stated, constants depend only on the distribution of the random walk increments and the number of colors of the scenery. In particular, the constants ci in this section do not depend on n.. Lemma 6.5 There exists a constant c11 > 0 such that for all n c11 ,. n n B n ;n P Estop all paths e :. Proof. We have P (S0 = S2 = 0) > 0 because the random walk has a positive. probability to make rst a step of maximal length L to the right and then a step of maximal length L to the left. Hence 2 divides the period of the random walk, and the period must be 1 or 2. Therefore there exists c12 > 0 such that for all n c12 and for all x z 2 ;7 2n 7 2n], the random walk starting at x can reach z with positive probability in 22n;1 or 22n;1 + 1 steps: ;. . Px S (22n;1) = z or S (22n;1 + 1) = z > 0:. (6.17). We denote by R the set of all admissible pieces of path R 2 Z 03c1n with starting point in ;7 2n 7 2n ]. For R 2 R and t 2 N 0 , we dene the event. E (t R) := fS (t + i) = R(i) 8i 2 0 3c1n or S (t + 1 + i) = R(i) 8i 2 0 3c1n g : Let n maxfc12 c10 g with c10 as in (6.4), and let k 2 1 2n ]. We set tkn := k +22n;1 and we dene random variables Yk (R) as follows: If jS (k )j 2n and E (tkn R) does not hold, then we set Y (R) = 0. Otherwise we set Yk (R) = 1. n and B n k , we see that Using the denitions of Estop all paths E n. stop. n B n. all paths. . R2R. E n. stop \. ( 2n X. k=1. 26. ). Yk (R ) = 0. . R2R. E2n (R). (6.18).
(87) with. M \. EM (R) :=. fjSk j 2n k;1 + 2 2n k Yk (R) = 0g. k=1. for M 2 1 2n]. Let R 2 R. Since n c10 , we have 3c1n 2n by (6.4). Hence tkn + 1 + 3c1 n = k + 1 + 22n;1 + 3c1 n k + 22n. Consequently, fk +2 22n < k+1 g\ E (tkn R) 2 Fkn+1 here Fk := (Si ~i i 2 0 k]) denotes the natural ltration of random walk and observations with errors. Using the strong Markov property at time M , we obtain. P EM (R)] P EM ;1 (R) \ jSM j 2n M ;1 + 2n+1 M YM (R) = 0. =P EM ;1 (R) \ jS (Mn )j 2n M ;1 + 2n+1 M \ E (tMn R)c h i P EM ;1 (R) \ fjS (Mn )j 2ng PS(Mn ) (E (22n;1 R)c ) P EM ;1 (R)] x2 max P E (22n;1 R)c ]: ;2n 2n] x An induction argument yields. 2n. . P (E2n (R)) x2 max P (E (22n;1 R)c ) : (6.19) ;2n 2n] x To estimate the right-hand side of (6.19), let b 2 N be minimal and let h 2 N be maximal such that Pp (S1 ; S0 2 b + hZ) = 1. We set 2 := E (S1 ; S0 )2 ], and Lm := f(mb + hy)= m : y 2 Zg. By the local central limit theorem ( 6],. page 132, Theorem (5.2)), . pm. . . . . Sm = y ; p 1 exp ; y2 = 0: p P 22 m y2Lm h 22 p We apply this with m 2 f22n;1 22n;1 + 1g, y := (R0 ; x)= m and p R0 equal to the starting point of R. Note that jR0 j 7 2n so that jR0 ; xj= m 16 for all ;x2)2 > 0. We conclude that x 2 ;2n 2n ]. Hence minx2 ;2n2n ]R2R exp ; (R20m
(88) there exist constants c13 > 0 and c14 maxfc12 c10 g such that for all n c14 ; min P S (22n;1 ) = R0 or S (22n;1 + 1) = R0 x2 ;2n 2n ]R2R x 2n;1 ) 2n;1 + 1) S (2 R ; x S (2 R ; x 0 0 = min P p =p or p =p lim sup m!1. 22n;1 + 1 (6.20) We set min := minf(j ) : j 2 Mg recall that is the distribution of the random walk increments Sk+1 ; Sk . The probability that the random walk starting at R0 follows the path R for the next 3c1 n ; 1 steps is bounded below c1 n;1 . Thus, (6.20) yields by 3min c1 n;1 = c15 2;n 3c1 n min Px (E (22n;1 R)) c13 2;n3min min n n x2 ;2n 2n ]R2R c13 2;n. 22n;1. 22n;1. x2 ;2 2 ]R2R. 27. 22n;1 + 1.
(89) 1 . Combining the last inequality with (6.18) and (6.19), we with c15 := c13 ;min obtain. ; n n B n ;n 3c1 n 2n P Estop all paths jRj 1 ; c15 2 min ; ; c1 n : (6.21) (14 2n + 1)jMj3c1 n;1 exp 2n ln 1 ; c15 2;n3min Note that choosing a path in R one has 14 2n + 1 possible starting points and jsupp()j = jMj possibilities for each step of the path. Using the estimate ln(1 ; x) ;x, we obtain h. i. c1 n = 2n+4 jMj3c1 n exp ;c15 ec16 n ] (6:21) 2n+4jMj3c1 n exp ;c15 2(;1)n 3min. and the last expression is e;n for all n suciently large because c16 = ( ; 1) ln 2 + 3c1 ln min > 0 by our choice of . Lemma 6.6 There exist 4 > 0 such that for all n 2 N and 2]0 4 c n ;n P ((Bfew mistakes ) ) e : Proof. Using Denition 6.3 and our convention " = c1" we obtain (B n. few mistakes. )c =. (. . t X. k=t;c1 n+1. t2 c1 n;12 212n . ). Xk > c1 "n :. (6.22). Recall that Xk , k h0, are i.i.d. Bernoulli random variables with parameter i Pt under P . Hence E k=t;c1 n+1 Xk = c1 n. By the large deviation principle (see e.g. 3]), we have for all 2]0 ". P. . with rate function. t X. k=t;c1 n+1. !. Xk > c1 "n exp (;I (" ; )c1 n) . (6.23). . x + x log x x 2]0 1 : I (x) = (1 ; x) log 11 ; ; Combining (6.22) with (6.23) we obtain for all 2]0 " c n P ((Bfew mistakes ) ) exp ( 1 + 12n] ln 2 ; I (" ; )c1 n) : Since. . . . . (6.24). lim I (" ; ) = lim (1 ; " + ) log 1 ;1 ;" + + (" ; ) log " ; = +1 !0 !0 there exists 4 2]0 " such that 1+12] ln 2 ; I (" ; )c1 < ;1 for all 2]0 4 . The assertion of the lemma follows. We will need the following lemma in the proofs of Lemmas 6.8, 6.10, and 6.13. 28.
(90) Lemma 6.7 There exist "1 c17 ("0) > 0 such that for all m with c1 m 2 N , "0 2]0 "1 , w 2 C 0c1m , and for any admissible piece of path R 2 Z 0c1m the following holds:. P (d( R w) < c1 "0 m) c17 ("0 )(c2 )c1 m max P (( R)jJ = wjJ ) J where the maximum is taken over all subsets J 0 c1 m with cardinality jJ j =. c1 m ; bc1 "0 mc and c2 is as in Section 2.1.. Proof. Let m be such that c1m 2 N, let w 2 C 0c1m , and let R 2 Z 0c1m be an admissible piece of path. If d( R w;) < c1 "0 m, then c1 m ; bc1 "0 mc letters of R and w agree. Since there are bcc11"mmc possibilities of choosing c1 m ; bc1 "0 mc out of c1 m letters, we have 0. . . P (d( R w) < c1 "0 m) bcc1"m max P (( R)jJ = wjJ ) 1 0 mc J where the maximum is taken over all subsets J 0 c1 m with cardinality c1 m ; bc1 "0 m pc. By Stirling's formula ( 1], p.24, formula (3.9)) we have for k 2 N , k! = 2kk+1=2 e;k+(k) with (k) 2]0 1 and limk!1 (k) = 0. Thus 0 mc c1 m c1 m b c " 1 0 bc1 "0 mc c17 (" )' c1 m with '(x) = x;x (1 ; x);(1;x) and some constant c17 ("0 ) > 0 independent of m. Note that ' is continuous at 0 with '(0) = 1, and recall that c2 2]1 C=(C ; 1) . There exists "1 such that '(x) < c2 for all x 2]0 "1 . Note that bc1 "0 mc=(c1 m) "0 . The claim follows.. Lemma 6.8 There exists a constant c18 > 0 such that for all n 2 N c n ;n P ((Bladder di ) ) c18 e : Proof. Let n o J := (z1 i1 z2 i2 ) 2 ( ;8 2n 8 2n ] f !g)2 : (z1 i1 ) =6 (z2 i2 ) : By Denition 6.4, c n (Bladder di ) =. . (z1 i1 z2 i2 )2J. fd(wz1 i1 n=3 wz2 i2 n=3 ) < 10"ng:. (6.25). Let (z1 i1 z2 i2) 2 J . For k = 1 2 we set ok := +1 if ik =!, ok := ;1 if ik =, and we set fk (j ) := zk + ok jL for j 2 0 c1 n=3 . First we prove that there exists a subset J 0 c1 n=3 of cardinality jJ j c1 n=9 such that f1 (J ) \ f2 (J ) = : (6.26) 29.
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