Doob-Martin compactification of a Markov chain for growing random words sequentially
aa r X i v : . [ m a t h . P R ] D ec DOOB-MARTIN COMPACTIFICATION OF A MARKOVCHAIN FOR GROWING RANDOM WORDSSEQUENTIALLY
HYE SOO CHOI AND STEVEN N. EVANS
Abstract.
We consider a Markov chain that iteratively gener-ates a sequence of random finite words in such a way that the n th word is uniformly distributed over the set of words of length n in which n letters are a and n letters are b : at each step an a and a b are shuffled in uniformly at random among the letters ofthe current word. We obtain a concrete characterization of theDoob-Martin boundary of this Markov chain and thereby delin-eate all the ways in which the Markov chain can be conditionedto behave at large times. Writing N ( u ) for the number of letters a (equivalently, b ) in the finite word u , we show that a sequence ( u n ) n ∈ N of finite words converges to a point in the boundary if, foran arbitrary word v , there is convergence as n tends to infinity ofthe probability that the selection of N ( v ) letters a and N ( v ) let-ters b uniformly at random from u n and maintaining their relativeorder results in v . We exhibit a bijective correspondence betweenthe points in the boundary and ergodic random total orders onthe set { a , b , a , b , . . . } that have distributions which are sepa-rately invariant under finite permutations of the indices of the a ′ sand those of the b ′ s. We establish a further bijective correspon-dence between the set of such random total orders and the setof pairs ( µ, ν ) of diffuse probability measures on [0 , such that ( µ + ν ) is Lebesgue measure: the restriction of the random to-tal order to { a , b , . . . , a n , b n } is obtained by taking X , . . . , X n (resp. Y , . . . , Y n ) i.i.d. with common distribution µ (resp. ν ),letting ( Z , . . . , Z n ) be { X , Y , . . . , X n , Y n } in increasing order,and declaring that the k th smallest element in the restricted totalorder is a i (resp. b j ) if Z k = X i (resp. Z k = Y j ).2010 Mathematics Subject Classification.
Primary: 05A05, 60J10, 68R15.
Key words and phrases. harmonic function, exchangeability, bridge, shuffle, sub-word counting, binomial coefficient, Plackett-Luce model, vase model.S.N.E. was supported in part by NSF grants DMS-0907639 and DMS-1512933,and NIH grant 1R01GM109454-01. Introduction
There is a very simple way of producing a uniformly distributedrandom permutation of a set with n objects, say [ n ] := { , . . . , n } :we take the elements of [ n ] in order and lay them down successivelyso that the k th element goes into a uniformly chosen one of the k “slots” defined by the k − elements that have already been laid down(the slot before the first element, the slot after the last element, orone of the k − slots between elements). This sequential algorithmhas the attractive feature that when the first k elements have beenlaid down they are in uniform random order; that is, the algorithmbuilds uniformly distributed random permutations of [1] , [2] , . . . , [ n ] ina sequential manner.Suppose that we enumerate a standard deck of cards with the ele-ments of the set [52] . If the deck is in some order, then the colors ofthe successive cards ( R ed or B lack) define a word of length fromthe two-letter alphabet { R, B } in which letters are R and lettersare B (recall that a word of length k from a finite alphabet A is just anelement of the Cartesian product A k , although it is usual to write theword ( a , . . . , a k ) more succinctly as a · · · a k ). Moreover, if the orderof the deck is random and uniformly distributed, then the resultingword is uniformly distributed over the set of such words.Unfortunately, our sequential randomization algorithm doesn’t havethe feature that at the (2 k ) t h step for ≤ k ≤ we have a randomword from the alphabet { R, B } that is uniformly distributed over theset of (cid:0) kk (cid:1) words in which k letters are R and k letters are B .However, there is a simple way of modifying our algorithm to producethe latter type of random words sequentially. We begin at step withthe empty word. Suppose that we have completed k steps and a wordof length k has been produced. The first sub-step of step k + 1 insertsthe letter R uniformly at random into one of the k + 1 slots defined bythese k letters to produce a word of length k +1 . The second sub-stepinserts the letter B uniformly at random into one of the k + 2 slotsdefined by these k + 1 letters to produce a word of length k + 2 andthereby complete step k + 1 . It is not difficult to see that, despite theapparent dependence of this procedure on the ordering of the letters R and B , this procedure does indeed achieve what it is claimed to achieve.From now on we will replace the alphabet { R, B } by the alphabet { a, b } and write ( U n ) n ∈ N for the Markov chain that arises from ourrandom insertion procedure. Thus, U n ∈ W n , where W n is the setwords drawn from the alphabet { a, b } that consist of n letters a and n letters b . Set W := F n ∈ N W n and put N ( w ) = n for w ∈ W n , n ∈ N . ROWING RANDOM WORDS 3
We investigate the infinite bridges (equivalently, the Doob h -transforms) for the Markov chain ( U n ) n ∈ N ; that is, the Markov chainsthat have the same backwards-in-time transition dynamics as ( U n ) n ∈ N .We thereby identify the Doob-Martin compactification of the statespace W of the Markov chain. This enables us to characterize thenonnegative harmonic functions for the Markov chain and hence delin-eate all the ways that the Markov chain can be conditioned to “behaveat infinity”.More specifically, we show that a W -valued Markov chain is an infi-nite bridge for the Markov chain ( U n ) n ∈ N if and only if the backwardsdynamics are given by removing one letter a and one letter b uniformlyat random from the current word. We can enrich the state space ofthe Markov chain ( U n ) n ∈ N by replacing W n with the set ˜ W n that con-sists of words made up from the letters a , b , . . . , a n , b n written downin some order (each letter appearing once); that is, a word such as aababb will be associated with a word such as a a b a b b – a given w ∈ W n has ( n !) associated words in ˜ W n . We can then enhance aninfinite bridge ( U ∞ n ) n ∈ N to produce a Markov chain ( ˜ U ∞ n ) n ∈ N withvalues in ˜ W := F n ∈ N ˜ W n such that given U ∞ n = u the value of ˜ U ∞ n isuniformly distributed over all ways of “subscripting” the letters in u ;for example, if U ∞ = abba , then ˜ U ∞ is uniformly distributed over thefour words a b b a , a b b a , a b b a , a b b a . Moreover, in goingfrom ˜ U ∞ n to ˜ U ∞ n − the letters a n and b n are deleted. We may view ˜ U ∞ n as a random total (that is, linear) order on the set { a , b , . . . , a n , b n } .As n varies, these orders are consistent in the sense that the order ˜ U ∞ n induces on { a , b , . . . , a n − , b n − } is just the order given by ˜ U ∞ n − . Con-sequently, there is a total order on { a , b , a , b , . . . } that induces eachof the orders given by the ˜ U ∞ n . This total order is exchangeable in thesense that finite permutations of the subscripts of the a ’s and b ’s sep-arately leave its distribution unchanged. The infinite bridge ( U ∞ n ) n ∈ N is extremal (that is, not a mixture of infinite bridges or, equivalently,has an almost surely trivial tail σ -field) if and only if the exchange-able random total order on { a , b , a , b , . . . } is ergodic in the sensethat if an event is unchanged by finite permutations of the subscriptsof the a ’s and b ’s separately, then it has probability zero or one. Bygeneral Doob–Martin theory, extremal bridges correspond to extremalelements of the Doob–Martin boundary and, in general, some elementsof the Doob–Martin boundary may not be extremal. We show that thelatter phenomenon does not occur in our setting – all Doob–Martinboundary points are extremal. H.S. CHOI AND S.N. EVANS
We demonstrate that there is a bijective correspondence between er-godic exchangeable random total orders on { a , b , a , b , . . . } and pairs ( µ, ν ) of diffuse probability measures on the unit interval [0 , suchthat µ + ν = λ , where λ is Lebesgue measure on [0 , : let V , V , . . . be i.i.d. with distribution µ and W , W , . . . be independent and i.i.d.with distribution ν , then, writing ≺ for the total order, we have a i ≺ a j (resp. a i ≺ b j , b i ≺ a j , b i ≺ b j ) if V i < V j (resp. V i < W j , W i < V j , W i < W j ). Another way of describing this construction is the fol-lowing. We only need to describe the restriction of the random to-tal order to { a , b , . . . , a n , b n } for each n ∈ N . Let ( Z , . . . , Z n ) be { V , W , . . . , V n , W n } in increasing order and declare that the k th small-est element of { a , b , . . . , a n , b n } in the restricted total order is a i (resp. b j ) if Z k = X i (resp. Z k = Y j ).We remark that, due to the relationship µ + ν = λ , the probabilitymeasure ν is uniquely determined by the probability measure µ and vice versa and hence we could have said that the ergodic exchangeablerandom total orders are in bijective correspondence with the probabilitymeasures µ on [0 , that satisfy µ ≤ λ . However, we find the moresymmetric description to be preferable.In terms of the Doob–Martin topology, we show that a sequence ( y k ) k ∈ N with y k ∈ W N ( y k ) and N ( y k ) → ∞ as k → ∞ converges tothe point in the Doob–Martin boundary corresponding to the pair ofmeasures ( µ, ν ) if and only if for each m ∈ N the the sequence of ran-dom words obtained by selecting m letters a and m letters b uniformlyat random from y k and maintaining their relative order converges indistribution as k → ∞ to the random word that is obtained by writing V , . . . , V m , W , . . . , W m in increasing order to make a list ( Z , . . . , Z m ) as above and then putting a letter a (resp. b ) in position ℓ of the wordwhen Z ℓ ∈ { V , . . . , V m } (resp. Z ℓ ∈ { W , . . . , W m } ). Moreover, theconvergence of ( y k ) k ∈ N to y is equivalent to the weak convergence of µ k to µ and ν k to ν , where µ k (resp. ν k ) is the probability measure thatplaces mass N ( y k ) at the point ℓ N ( y k ) ≤ ℓ ≤ N ( y k ) , if the ℓ th letterof the word y k is the letter a (resp. b ).2. Background on the Doob–Martin compactification
The primary reference on the Doob–Martin compactification theoryfor discrete time Markov chains is [Doo59], but useful reviews may befound in [KSK76, Chapter 10], [Rev75, Chapter 7], [Saw97], [Woe00,Chapter IV], [RW00, Chapter III]. We restrict the following sketch tothe setting that is of interest to us.
ROWING RANDOM WORDS 5
Suppose that ( X n ) n ∈ N is a discrete time Markov chain with count-able state space E and transition matrix P . Suppose in addition that E can be partitioned as E = F n ∈ N E n , where E = { e } for some dis-tinguished state e , each set E n , n ∈ N is finite, and the transitionmatrix P is such that P ( k, ℓ ) = 0 unless k ∈ E n and ℓ ∈ E n +1 for some n ∈ N . Define the Green kernel or potential kernel G of P by G ( i, j ) := ∞ X n =0 P n ( i, j ) = P i { X n = j for some n ∈ N } =: P i { X hits j } ,i, j ∈ E , and assume that G ( e, j ) > for all j ∈ E , so that any statecan be reached with positive probability starting from e .The Doob–Martin kernel with reference state e is K ( i, j ) := G ( i, j ) G ( e, j ) = P i { X hits j } P e { X hits j } . If j, k ∈ E with j = k , then K ( · , j ) = K ( · , k ) and so E can be identifiedwith the collection of functions ( K ( · , j )) j ∈ E . Note that ≤ K ( i, j ) ≤ P e { X hits i } , and so the set of functions ( K ( · , j )) j ∈ E is a pre-compact subset of R E + .Its closure ¯ E is the Doob–Martin compactification of E . The set ∂E :=¯ E \ E is the Doob–Martin boundary of E .By definition, a sequence ( j n ) n ∈ N in E converges to a point in ¯ E if and only if the sequence of real numbers ( K ( i, j n )) n ∈ N convergesfor all i ∈ E . Each function K ( i, · ) extends continuously to ¯ E . Theresulting function K : E × ¯ E → R is the extended Martin kernel .For y ∈ ∂E the nonnegative function K ( · , y ) is harmonic and anynonnegative harmonic function can be represented as R K ( · , y ) µ ( dy ) for a suitable finite measure µ on ∂E .If Z is a P e -a.s. bounded random variable that is measurable withrespect to the tail σ -field of ( X n ) n ∈ N , then E e [ Z | X , . . . , X n ] = h ( X n ) for some bounded harmonic function h and, by the martingale con-vergence theorem, lim n →∞ h ( X n ) = Z P e -a.s. Conversely, if h is abounded harmonic function, then lim n →∞ h ( X n ) exists P e -a.s. and thelimit random variable is P e -a.s. equal to a random variable that ismeasurable with respect to the tail σ -field of ( X n ) n ∈ N .The limit X ∞ := lim n →∞ X n exists P e -almost surely in the topologyof ¯ E and the limit belongs to ∂E P e -almost surely. The tail σ -field of ( X n ) n ∈ N coincides P e -almost surely with the σ -field generated by X ∞ .Each j ∈ E = F n ∈ N E n belongs to a unique E n whose index n wedenote by N ( j ) . If the Markov chain starts in state e , then N ( j ) is the H.S. CHOI AND S.N. EVANS only time that there is positive probability the Markov chain will be instate j . Write ( X j , . . . , X jN ( j ) ) for the bridge obtained by starting theMarkov chain in state e and conditioning it to be in state j at time N ( j ) . This process is a Markov chain with transition probabilities P { X jn +1 = i ′′ | X jn = i ′ } = P e { X n = i ′ , X n +1 = i ′′ , X N ( j ) = j } P e { X n = i ′ , X N ( j ) = j } = P e { X hits i ′ } P ( i ′ , i ′′ ) P i ′′ { X hits j } P e { X hits i ′ } P i ′ { X hits j } = P ( i ′ , i ′′ ) P i ′′ { X hits j } / P e { X hits j } P i ′ { X hits j } / P e { X hits j } = K ( i ′ , j ) − P ( i ′ , i ′′ ) K ( i ′′ , j ) . The backward transition probabilities of ( X j , . . . , X jN ( j ) ) are given by P { X jn = i ′ | X jn +1 = i ′′ } = P e { X hits i ′ } P ( i ′ , i ′′ ) P i ′′ { X hits j } P e { X hits i ′′ } P i ′′ { X hits j } = P e { X hits i ′ } P ( i ′ , i ′′ ) P e { X hits i ′′ } , so that all bridges have the same backward transition probabilities.An infinite bridge for ( X n ) n ∈ N is a Markov chain ( X ∞ n ) n ∈ N with thesebackward transition probabilities. If ( X ∞ n ) n ∈ N is an infinite bridge,then P { X ∞ n +1 = i ′′ | X ∞ n = i ′ } = P e { X ∞ hits i ′′ } P { X ∞ n = i ′ | X ∞ n +1 = i ′′ } P e { X ∞ hits i ′ } = h ( i ′ ) − P ( i ′ , i ′′ ) h ( i ′′ ) , where h ( i ) = P e { X ∞ hits i } P e { X hits i } . Thus an infinite bridge is a Doob h -transform of ( X n ) n ∈ N with a par-ticular harmonic function h . Conversely, any Doob h -transform is aninfinite bridge.Suppose now that ( j k ) k ∈ N is a sequence of elements of the state space E such that N ( j k ) → ∞ as k → ∞ . As observed in [Föl75], such asequence ( j k ) k ∈ N converges in the Doob–Martin topology if and only iffinite initial segments of the corresponding bridges converge in distri-bution. Moreover, two sequences of states converge to the same limitif and only if the limiting distributions of finite initial segments arethe same. For a sequence ( j k ) k ∈ N that converges to a point in the ROWING RANDOM WORDS 7
Doob–Martin boundary, the limiting distributions of the initial seg-ments define the distribution of an E -valued Markov chain ( X ( h ) n ) n ∈ N with transition probabilities P ( h ) given by P ( h ) ( i, j ) := h ( i ) − P ( i, j ) h ( j ) , i, j ∈ E ( h ) , where h ( i ) = lim k →∞ K ( i, j k ) and E ( h ) := { i ∈ E : h ( i ) > } = { i ∈ E : lim k →∞ P { X N ( i ) = i | X N ( j k ) = j k } > } . This Markov chain ( X ( h ) n ) n ∈ N is an infinite bridge. A necessary condi-tion for an infinite bridge to be extremal (that is, having a distributionthat is not a nontrivial mixture of infinite bridge distributions) is thatit is of this form.3. Transition probabilities and the Doob–Martin kernelfor the growing word chain
Definition 3.1.
For n ∈ N write W n for the set of words from thealphabet { a, b } that have n letters a and n letters b and put W := F n ∈ N W n .By definition, the Markov chain ( U n ) n ∈ N has state space W andone-step transition probabilities P { U m +1 = w | U m = v } = M ( v, w )(2 m + 2)(2 m + 1) for v ∈ W n and w ∈ W n +1 , where M ( v, w ) is the number of ways towrite w = v xv yv in such a way that { x, y } = { a, b } and v , v , v are (possibly empty) words such that v = v v v . That is, M ( v, w ) isthe number of times that v appears inside w as a sub-word . (We recallthat, in general, a word c · · · c p is a sub-word of a word d · · · d q if thereis a map f : [ p ] → [ q ] such that f ( i ) < f ( j ) for ≤ i < j ≤ p and d f ( k ) = c k for ≤ k ≤ p .)In order to write down multi-step transition probabilities for theMarkov chain ( U n ) n ∈ N , it is convenient to introduce the following stan-dard notation (see, for example, [Lot97]). Definition 3.2.
Given two words w and v drawn from some finitealphabet, write (cid:0) wv (cid:1) for the number of times that v appears as a sub-word of w . Example . For example, (cid:0) abbababba (cid:1) = 4 because bba appears inside abbaba as a sub-word four times:
H.S. CHOI AND S.N. EVANS a bba ba a bb ab a a b ba ba ab b a ba . Remark . Note that if our alphabet has only one letter, then (cid:0) wv (cid:1) isjust the usual binomial coefficient (cid:0) | w || v | (cid:1) , where we use the notation | u | for the length of the word u .For a general finite alphabet A , (cid:0) wv (cid:1) is uniquely determined by thefollowing three properties, where we write A ∗ for the set of finite wordswith letters drawn from the alphabet A (see [Lot97, Proposition 6.3.3]): • (cid:0) w ∅ (cid:1) = 1 for all w ∈ A ∗ , where ∅ is the empty word, • (cid:0) wv (cid:1) = 0 for all v, w ∈ A ∗ with | w | < | v | , • (cid:0) wyvx (cid:1) = (cid:0) wvx (cid:1) + δ x,y (cid:0) wv (cid:1) , for all v, w ∈ A ∗ and x, y ∈ A , where δ isthe usual Kronecker delta.The counting involved in determining (cid:0) wv (cid:1) for general v, w ∈ A ∗ ishandled by the following result from [Cla15]. Define an infinite matrix P with entries indexed by A ∗ by setting the ( v, w ) entry to be (cid:0) wv (cid:1) . Ifthe row and column indices are ordered so that they are nondecreasingin word length, then P is an upper triangular matrix with in everyposition on the diagonal. Define another infinite matrix H indexedby A ∗ by setting the ( v, w ) entry to be (cid:0) wv (cid:1) if | w | = | v | + 1 and otherwise. With the same ordering of the indices as for P , the matrix H is upper triangular with in every position on the diagonal. Thematrix exponential exp( H ) is well-defined and is equal to P .Using the above notation, we can express the transition probabilitiesof ( U n ) n ∈ N as follows. Lemma 3.5.
For words v ∈ W m and w ∈ W m + n P { U m + n = w | U m = v } = (cid:18) wv (cid:19) n ! n !(2 m + 1)(2 m + 2) · · · (2( m + n )) . Proof.
We proceed by induction. The result is certainly true when n = 1 . Supposing it is true for some value of n , in order to show it istrue for n + 1 , we need to show that for u ∈ W m and w ∈ W m + n +1 wehave X v ∈ W m +1 (cid:18) vu (cid:19) m + 1)(2 m + 2) (cid:18) wv (cid:19) n ! n !(2 m + 3)(2 m + 4) · · · (2( m + n + 1))= (cid:18) wu (cid:19) ( n + 1)!( n + 1)!(2 m + 1)(2 m + 2) · · · (2( m + n + 1)) , ROWING RANDOM WORDS 9 or, equivalently, that X v ∈ W m +1 (cid:18) vu (cid:19)(cid:18) wv (cid:19) = (cid:18) wu (cid:19) ( n + 1) . This, however, is clear. The lefthand side counts the number of words v ∈ W m +1 such that u is subword of v and v is a subword of w . Anysuch v and its embedding in w arises by taking an embedding of u in w and then specifying which of the remaining n + 1 letters a in w and which of the remaining n + 1 letters b in w are used to build theword with its particular embedding, and this is what the righthand sidecounts. (cid:3) Corollary 3.6.
The Doob–Martin kernel of ( U n ) n ∈ N with distinguishedstate the empty word is, for v ∈ W m and w ∈ W m + n , K ( v, w ) = (cid:18) wv (cid:19) (cid:0) mm (cid:1)(cid:0) m + nm (cid:1) . Proof.
We have K ( v, w )= P { U m + n = w | U m = v } P { U m + n = w | U = ∅} = (cid:0) wv (cid:1) n ! n !(2 m +1)(2 m +2) ··· (2( m + n )) (cid:0) w ∅ (cid:1) ( m + n )!( m + n )!(2( m + n ))! = (cid:18) wv (cid:19) n ! n !(2( m + n ))!( m + n )!( m + n )!(2 m + 1)(2 m + 2) · · · (2( m + n ))= (cid:18) wv (cid:19) (cid:0) mm (cid:1)(cid:0) m + nn (cid:1)(cid:0) m + nn (cid:1) . (cid:3) Remark . Up to the factor (cid:0) mm (cid:1) , the Doob–Martin kernel K ( v, w ) isthe probability that if we select m of the letters a and m of the letters b uniformly at random from w and list these letters in the same relativeorder that they appear in w , then the resulting word is v . Therefore, asequence ( w k ) k ∈ N in W with N ( w k ) → ∞ as k → ∞ converges in theDoob–Martin topology if and only if for every m ∈ N the sequence ofrandom words in W m obtained by selecting m letters a and m letters b from w k (and maintaining their relative order) converges in distributionas k → ∞ . Definition 3.8.
For w ∈ W k , k ∈ N , let ( U w , . . . , U wk ) be the bridgefrom the empty word to w . Theorem 3.9.
The backward transition dynamics for all bridges fromthe empty word are the same and consist of removing at each step oneletter a and one letter b uniformly at random.Proof. Consider the bridge from the empty word to w ∈ W k .For ≤ m ≤ k − , v ∈ W m +1 , and u ∈ W m we have P { U wm = u | U wm +1 = v } = P { U m = u, U m +1 = v | U k = w } P { U m +1 = v | U k = w } = P { U m = u, U m +1 = v, U k = w } P { U m +1 = v, U k = w } = P { U m = u } P { U m +1 = v | U m = u } P { U k = w | U m +1 = v } P { U m +1 = v } P { U k = w | U m +1 = v } = (cid:18) vu (cid:19) m +1)(2 m +2) × m ! m !(2 m )!( m +1)!( m +1)!(2 m +2)! = (cid:0) vu (cid:1) ( m + 1) . In order to go backward from the word v of length m + 1) to theword u of length m , we have to remove one a and one b . There are (cid:0) vu (cid:1) pairs of a and b such that the removal of the pair from v results in u , and there are a total of ( m + 1) pairs of a and b in v , and so theresult follows from the calculation above. (cid:3) Labeled infinite bridges
Suppose that ( y n ) n ∈ N is a sequence of words in W := F n ∈ N W n thatconverges in the Doob–Martin topology and is such that N ( y n ) → ∞ as n → ∞ . Recall that ( U y n , . . . , U y n N ( y n ) ) , n ∈ N , is the associated bridgethat starts from the empty word and is tied to being in state y n attime N ( y n ) . The finite dimensional distributions of ( U y n , . . . , U y n N ( y n ) ) converge as n → ∞ . Thus, there exists a process ( U ∞ n ) n ∈ N such thatfor every k ∈ N the random ( k + 1) -tuple ( U y n , . . . , U y n k ) converges indistribution to ( U ∞ , . . . , U ∞ k ) .The forward evolution dynamics of the Markov chain ( U ∞ n ) n ∈ N de-pend on the sequence ( y n ) n ∈ N , whereas from Section 2 and Theorem 3.9 ROWING RANDOM WORDS 11 the backward evolution is Markovian and doesn’t depend on the se-quence ( y n ) n ∈ N ; given U ∞ k +1 , the word U ∞ k is obtained by removing oneletter a and one letter b uniformly at random from U ∞ k +1 .For each n ∈ N the distribution of U ∞ n defines the distributionof a random element ˜ U ∞ n,n of the set ˜ W n of words of length n drawnfrom the alphabet { a , b , . . . , a n , b n } with each letter appearing once byassigning the labels [ n ] uniformly at random to the letters a and to theletters b . More precisely, for U ∞ n = c . . . c n , let A n := { i ∈ [ n ] : c i = a } and B n := { j ∈ [ n ] : c j = b } , let Σ : A n → [ n ] and T : B n → [ n ] berandom bijections that are conditionally independent and uniformlydistributed given U ∞ n , and define ˜ U ∞ n,n := ˜ c . . . ˜ c n by ˜ c k := ( a Σ( k ) , k ∈ A n ,b T ( k ) , k ∈ B n . For ≤ p ≤ n , define ˜ U ∞ n,p to be the word obtained by deleting { a p +1 , b p +1 , . . . , a n , b n } from ˜ U ∞ n,n . Observe that if ≤ p ≤ m ∧ n ,then ˜ U ∞ m,p , ˜ U ∞ n,p and ˜ U ∞ p,p have the same distribution. Moreover, if for ≤ p ≤ n we let U ∞ n,p be the result of removing the labels from ˜ U ∞ n,p (that is, U ∞ n,p is the element of W p obtained by replacing the letters a k , ≤ k ≤ p , by the letter a and the letters b k , ≤ k ≤ p , by b ), then ( U ∞ n, , . . . , U ∞ n,n ) has the same distribution as ( U ∞ , . . . , U ∞ n ) .By Kolmogorov’s consistency theorem, there is a process ( ˜ U ∞ n ) n ∈ N such that ( ˜ U ∞ , . . . , ˜ U ∞ m ) has the same distribution as ( ˜ U ∞ n, , . . . , ˜ U ∞ n,m ) for any m ≤ n and the result of removing the labels from ( ˜ U ∞ n ) n ∈ N hasthe same distribution as ( U ∞ n ) n ∈ N . By the transfer theorem [Kal02,Theorem 6.10], we may even suppose that ( ˜ U ∞ n ) n ∈ N is defined on anextension of the probability space on which ( U ∞ n ) n ∈ N is defined in sucha way that ( U ∞ n ) n ∈ N is the result of removing the labels from ( ˜ U ∞ n ) n ∈ N .5. The exchangeable random total order associated withan infinite bridge
A state of a labeled infinite bridge is a word of length n from thealphabet { a , b , . . . , a n , b n } in which each letter appears once. An-other way to think of such an object is as a total order on the set S nk =1 { a k , b k } . Because the labeled infinite bridge evolves by slotting inthe letters a n +1 and b n +1 at the ( n + 1) th step while leaving the relativepositions of { a , b , . . . , a n , b n } unchanged, these successive total ordersare consistent: the total order on { a , b , . . . , a n , b n } given by the stateof the infinite bridge at step n is the same as the total order obtained by taking the state of the infinite bridge at step n + 1 (a total or-der on { a , b , . . . , a n , b n , a n +1 , b n +1 } ) and looking at the correspondinginduced total order on { a , b , . . . , a n , b n } .This projective structure means that we can associate any path of alabeled infinite bridge with a unique total order on I := S n ∈ N { a n , b n } such that the induced total order on { a , b , . . . , a n , b n } coincides withthe state of the labeled infinite bridge at step n .We now introduce some general notions about random total orders. Definition 5.1. A random total order ≺ on I is a map from theunderlying probability space to the collection of total orders on I suchthat the indicator { x ≺ y } is a random variable for every x, y ∈ I .A random total order ≺ is exchangeable if for every n ∈ N the inducedtotal order ≺ n on S nk =1 { a k , b k } has the same distribution as the randomtotal order ≺ nσ,τ for any permutations σ, τ of { , , . . . , n } , where ≺ nσ,τ is defined as follows: • a σ ( i ) ≺ nσ,τ b τ ( j ) iff a i ≺ n b j • b τ ( i ) ≺ nσ,τ a σ ( j ) iff b i ≺ n a j • a σ ( i ) ≺ nσ,τ a σ ( j ) iff a i ≺ n a j • b τ ( i ) ≺ nσ,τ b τ ( j ) iff b i ≺ n b j . Remark . The distribution of a random total order ≺ is determinedby the joint distribution of the random variables { { x ≺ y } : x, y ∈ S nk =1 { a k , b k }} for arbitrary n ∈ N . Remark . If ≺ is an exchangeable random total order, then theinduced random total orders ≺ n , n ∈ N , are consistent in the sense thatif we take the random total order ≺ n +1 on S n +1 k =1 { a k , b k } and remove { a n +1 , b n +1 } , then the induced random total order on S nk =1 { a k , b k } is ≺ n .Conversely, suppose for each n ∈ N that there is a random totalorder ≺ n on S nk =1 { a k , b k } , these random total orders have the propertythat ≺ n has the same distribution as ≺ nσ,τ for any permutations σ, τ of [ n ] for all n ∈ N , and these total orders are consistent. Then there is anexchangeable random order ≺ on I such that ≺ n is the correspondinginduced total order on S nk =1 { a k , b k } .In terms of these general notions, if we let ≺ n , n ∈ N , be the randomtotal order on S nk =1 { a k , b k } corresponding to ˜ U ∞ n , then these total or-ders are consistent and there is an exchangeable random total order ≺ on I such that the restriction of ≺ to S nk =1 { a k , b k } is ≺ n . ROWING RANDOM WORDS 13 Characterization of exchangeable random totalorders
The results of the previous sections indicate that if we want to under-stand the Doob–Martin compactification, then we need to understandinfinite bridges, and this boils down to understanding exchangeablerandom total orders on I .A mixture of two exchangeable random total orders is also an ex-changeable random total order, so we are interested in exchangeablerandom total orders ≺ that are extremal in the sense that their distri-butions cannot be written as a nontrivial mixture of the distributionsof two other exchangeable random total orders. This is equivalent torequiring that if A is a measurable subset of the space of total orders on I with the property that ≺∈ A if and only if ≺ σ,τ ∈ A for all finite per-mutations σ, τ , then P {≺∈ A } ∈ { , } . We say that an exchangeablerandom total order with this property is ergodic .The following result can be established using essentially the sameargument as in Proposition 5.19 (see also the subsequent Remark 5.20)of [EGW15], and we omit the details. Lemma 6.1.
The tail σ -field of an infinite bridge ( U ∞ n ) n ∈ N is almostsurely trivial if and only if the exchangeable random total order inducedby the corresponding labeled infinite bridge ( ˜ U ∞ n ) n ∈ N is ergodic.Remark . There is one obvious way to produce an ergodic exchange-able random total order. Let ζ and η be two diffuse probability mea-sures on R . Let ( V n ) n ∈ N be i.i.d. with common distribution ζ , let ( W n ) n ∈ N be i.i.d. with common distribution η , and suppose that thesetwo sequences are independent. The total order ≺ on I defined bydeclaring that • a i ≺ a j if V i < V j , • b i ≺ b j if W i < W j , • a i ≺ b j if V i < W j , • b i ≺ a j if W i < V j ,is exchangeable and ergodic; exchangeability is obvious and ergodic-ity is immediate from the Hewitt–Savage zero–one law applied to thei.i.d. sequence (( V n , W n )) n ∈ N (indeed, it follows from the Hewitt–Savagezero–one law that if A is a measurable subset of the space of total or-ders on I with the property that ≺∈ A if and only if ≺ ρ,ρ ∈ A for allfinite permutations ρ , then P {≺∈ A } ∈ { , } ).We will show that all ergodic exchangeable random total orders arisethis way. Note that many pairs of probability measures can give rise torandom total orders with the same distribution: replacing ζ and η by their push-forwards by some common strictly increasing function doesnot change the distribution of the resulting random total order. Definition 6.3.
Given an exchangeable random total order ≺ on I ,define d : I × I → [0 , by requiring that d ( x, x ) = 0 for all x ∈ I , d ( x, y ) = d ( y, x ) for all x, y ∈ I , and d ( x, y ) := lim sup n →∞ n { ≤ k ≤ n : x ≺ a k ≺ y } + lim sup n →∞ n { ≤ ℓ ≤ n : x ≺ b ℓ ≺ y } for x ≺ y . It follows from exchangeability, de Finetti’s theorem, andthe strong law of large numbers that in the above the superior limitsare actually limits almost surely. Remark . It is clear that by redefining d on a P -null set we mayassume for every x, y, z ∈ I that • d ( x, y ) ≥ , • d ( x, y ) = d ( y, x ) , • d ( x, z ) ≤ d ( x, y ) + d ( y, z ) , • d ( x, y ) = 0 if x = y . Remark . For distinct x, y, z ∈ I the triangle inequality d ( x, z ) ≤ d ( x, y ) + d ( y, z ) can be sharpened to a statement that for all x, y, z • d ( x, z ) = d ( x, y ) + d ( y, z ) if x ≺ y ≺ z , • d ( x, z ) = d ( x, y ) − d ( y, z ) if x ≺ z ≺ y , • d ( x, z ) = d ( y, z ) − d ( x, y ) if y ≺ x ≺ z ,and three analogous equalities when z ≺ x . Proposition 6.6. If x, y ∈ I with x = y , then d ( x, y ) > almostsurely. Therefore almost surely d is a metric.Proof. We need to show for k, ℓ ∈ N with k = ℓ that d ( a k , a ℓ ) > and d ( b k , b ℓ ) > , and, furthermore, for arbitrary k, ℓ ∈ N that d ( a k , b ℓ ) > .Consider d ( a k , a ℓ ) . Set I m := ( { a k ≺ a m ≺ a ℓ } ∪ { a ℓ ≺ a m ≺ a k } ) , m / ∈ { k, ℓ } . ROWING RANDOM WORDS 15
Suppose that Π n , n ∈ N is a uniform random permutation of [ n ] . Byexchangeability of the total order, if k ∨ ℓ ≤ n , then P { I m = 0 , ≤ m ≤ n, m / ∈ { k, ℓ }} = P ( { Π n ( ℓ ) = Π n ( k ) + 1 } ∪ { Π n ( k ) = Π n ( ℓ ) + 1 } )= 2( n −
1) 1 n ( n − n and the random variables { I m : m ∈ N , m / ∈ { k, ℓ }} are exchangeable.It follows from de Finetti’s theorem and the strong law of large numbersthat lim n →∞ n { ≤ m ≤ n : a k ≺ a m ≺ a ℓ } = lim n →∞ n n X m =1 I m > almost surely and hence d ( a k , a ℓ ) > . A similar argument shows that d ( b k , b ℓ ) > .It remains to show that d ( a k , b ℓ ) > . Set M := { m ∈ N : a k ≺ b m } .It follows from exchangeability that on the event { M = ∅} ⊇ { a k ≺ b ℓ } we have M = ∞ almost surely and indeed that lim n →∞ n M ∩ [ n ]) > . Write M = { m , m , . . . } with m < m < . . . . Fix p ∈ N and set J q := { b m q ≺ b m p } , q = p. By exchangeability of the total order, if p ∨ q ≤ r , then P { J q = 0 , ≤ q ≤ r, q = p | M = ∅} = P { Π r ( p ) = 1 } = 1 r and the random variables { J q : q ∈ N , q = p } are conditionally ex-changeable given { M = ∅} . It follows from de Finetti’s theorem thaton the event { M = ∅} lim n →∞ n { q : m q ∈ [ n ] , a k ≺ b m q ≺ b m p } > almost surely and hence d ( a k , b ℓ ) > almost surely on the event { a k ≺ b ℓ } . A similar argument shows that d ( a k , b ℓ ) > almost surely on theevent { b ℓ ≺ a k } . (cid:3) Definition 6.7.
Given an ergodic exchangeable random total order ≺ on I , denote by I the completion of I with respect to the metric d . Definition 6.8.
Define f : I → [0 , by f ( y ) := sup { d ( x, y ) : x ∈ I , x ≺ y } . Remark . It follows from Remark 6.5 that f ( y ) = lim sup n →∞ n { ≤ k ≤ n : a k ≺ y } + lim sup n →∞ n { ≤ ℓ ≤ n : b ℓ ≺ y } , | f ( x ) − f ( y ) | = d ( x, y ) , x, y ∈ I , and f ( x ) < f ( y ) ⇐⇒ x ≺ y, x, y ∈ I , so that f is an order-preserving isometry from I into [0 , . Thus thefunction f extends by continuity to an isometry from I into [0 , andif ≺ is extended to I by declaring that x ≺ y ⇐⇒ f ( x ) < f ( y ) , then ≺ is a total order on I and f is an order-preserving isometry from I into [0 , and hence an order-preserving isometric bijection from I tothe image set J := f ( I ) ⊆ [0 , . Because I is complete, J is complete.Because J is a complete subset of [0 , it is closed and hence compact,and therefore I itself is compact. It follows from the ergodicity of ≺ that J is almost surely constant. We will see below that J = [0 , . Remark . Define a sequence (( X n , Y n )) n ∈ N of J -valued randomvariables by setting X n := f ( a n ) and Y n := f ( b n ) . The exchangeabilityof ≺ implies that if σ and τ are two finite permutations of N , then (( X σ ( n ) , Y τ ( n ) )) n ∈ N has the same distribution as (( X n , Y n )) n ∈ N . In par-ticular, the sequence (( X n , Y n )) n ∈ N is exchangeable. It is a consequenceof de Finetti’s theorem and the ergodicity of ≺ that this sequence isi.i.d. with common distribution some probability measure π on J . Itfollows from the next result that π = µ ⊗ ν for two probability measures µ and ν on J that we call the canonical pair . Because X m = X n and Y m = Y n almost surely for m = n , the probability measures µ and ν must be diffuse. Lemma 6.11.
Suppose that the random variables X ′ , Y ′ , X ′′ , Y ′′ aresuch that (1) ( X ′ , Y ′ ) d = ( X ′′ , Y ′′ ) (2) (( X ′ , Y ′ ) , ( X ′′ , Y ′′ )) d = (( X ′ , Y ′′ ) , ( X ′′ , Y ′ )) (3) ( X ′ , Y ′ ) ⊥⊥ ( X ′′ , Y ′′ ) .Then X ′ , X ′′ , Y ′ , Y ′′ are independent. ROWING RANDOM WORDS 17
Proof.
For Borel sets A ′ , A ′′ , B ′ , B ′′ we have P { X ′ ∈ A ′ , X ′′ ∈ A ′′ , Y ′ ∈ B ′ , Y ′′ ∈ B ′′ } = P { X ′ ∈ A ′ , Y ′ ∈ B ′ } P { X ′′ ∈ A ′′ , Y ′′ ∈ B ′′ } by (3) = P { X ′ ∈ A ′ , Y ′′ ∈ B ′ } P { X ′′ ∈ A ′′ , Y ′ ∈ B ′′ } by (2) = P { X ′ ∈ A ′ } P { Y ′′ ∈ B ′ } P { X ′′ ∈ A ′′ } P { Y ′ ∈ B ′′ } by (3) = P { X ′ ∈ A ′ } P { Y ′ ∈ B ′ } P { X ′′ ∈ A ′′ } P { Y ′′ ∈ B ′′ } by (1) . (cid:3) Theorem 6.12.
Any ergodic exchangeable random total order ≺ hasthe same distribution as one given by the construction in Remark 6.2for some pair of diffuse probability measures ( ζ , η ) on R . The canonicalpair of diffuse probability measures ( µ, ν ) on [0 , is uniquely deter-mined by the moment formulae Z [0 , x n µ ( dx )= (cid:18) (cid:19) n X c ∈ Q nk =1 { a k ,b k } P { c ≺ a n +1 , . . . , c n ≺ a n +1 } and Z [0 , y n ν ( dy )= (cid:18) (cid:19) n X c ∈ Q nk =1 { a k ,b k } P { c ≺ b n +1 , . . . , c n ≺ b n +1 } . The probability measure ( µ + ν ) is Lebesgue measure on [0 , and,in particular, J = [0 , . Moreover, µ and ν are the respective push-forwards of ζ and η by the function z ( ζ + η )(( −∞ , z ]) Proof.
We have already shown that an ergodic exchangeable randomtotal order has the same distribution as one built from an arbitrarypair ( ζ , η ) of diffuse probability measures on R using the constructionin Remark 6.2.Define (( X n , Y n )) n ∈ N as in Remark 6.10. It follows from Remark 6.9that X n = 12 µ (( −∞ , X n ]) + 12 ν (( −∞ , X n ]) and Y n = 12 µ (( −∞ , Y n ]) + 12 ν (( −∞ , Y n ]) for any n ∈ N . Let ( I n ) n ∈ N be a sequence of i.i.d. random variablesthat is independent of (( X n , Y n )) n ∈ N with P { I n = 0 } = P { I n = 1 } = and set Z n := I n X n + (1 − I n ) Y n so that the sequence ( Z n ) n ∈ N is i.i.d.with common distribution ( µ + ν ) . We have Z n = 12 ( µ + ν )(( −∞ , Z n ]) , and so ( µ + ν ) is Lebesgue measure on [0 , . Thus, for any n ∈ N Z [0 , x n µ ( dx ) = P { Z < X n +1 , . . . , Z n < X n +1 } = (cid:18) (cid:19) n X c ∈ Q nk =1 { a k ,b k } P { c ≺ a n +1 , . . . , c n ≺ a n +1 } and Z [0 , y n ν ( dy ) = P { Z < Y n +1 , . . . , Z n < Y n +1 } = (cid:18) (cid:19) n X c ∈ Q nk =1 { a k ,b k } P { c ≺ b n +1 , . . . , c n ≺ b n +1 } , as claimed.The proof of the final claim is straightforward and we omit it. (cid:3) Remark . We haven’t shown that if ( y k ) k ∈ N is a sequence of pointsof W , where y k ∈ W N ( y k ) , N ( y k ) → ∞ as k → ∞ , and lim k →∞ y k = y in the Doob–Martin topology for some arbitrary y in the Doob–Martinboundary, then the harmonic function K ( · , y ) is extremal. This isequivalent to showing that if the infinite bridge ( U ∞ n ) n ∈ N is the limitof the bridges ( U y k , . . . , U y k N ( y k ) ) , then ( U ∞ n ) n ∈ N has an almost surelytrivial tail σ -field. This is, in turn, equivalent to showing that thecorresponding labeled infinite bridge induces an ergodic exchangeablerandom order. The latter, however, can be established along the linesof [EGW15, Corollary 5.21] and [EW16, Corollary 7.2], so we omit thedetails.7. Identification of extremal harmonic functions
Any extremal infinite bridge ( U ∞ n ) n ∈ N is the h -transform of our orig-inal Markov chain with an extreme harmonic function h . We knowfrom the above that such a process arises as follows in terms of thecanonical pair ( µ, ν ) of diffuse probability measures associated withthe corresponding point in the Doob–Martin boundary. ROWING RANDOM WORDS 19
We first require some notation. Given ( x , . . . , x n , y , . . . , y n ) ∈ R n with distinct entries, let z < · · · < z n be a listing of { x , . . . , x n , y , . . . , y n } in increasing order. Define W (( x , . . . , x n , y , . . . , y n )) = u . . . u n ∈ W n by u i = ( a, if z i ∈ { x , . . . , x n } ,b, if z i ∈ { y , . . . , y n } . Given v ∈ W n , set S ( v ) := W − ( { v } ) ⊂ R n . For example, S ( abba ) = G σ,τ { ( x , x , y , y ) ∈ R : x σ (1) < y τ (1) < y τ (2) < x σ (2) } , where the union is over all pairs of permutations σ, τ of the set { , } .In general, S ( v ) is the disjoint union of ( n !) connected open sets thatall have boundaries of zero Lebesgue measure.Now take independent sequences of real-valued random variables ( X k ) k ∈ N and ( Y k ) k ∈ N , where the X k are i.i.d. with common distribution µ and the Y k are i.i.d. with common distribution ν and set U ∞ n = W (( X , . . . , X n , Y , . . . , Y n )) . We have P { U ∞ n = u } = µ ⊗ n ⊗ ν ⊗ n ( S ( u )) We also know that P { U ∞ n = u | U ∞ n +1 = v } = (cid:0) vu (cid:1) ( n + 1) . It follows that P { U ∞ n +1 = v | U ∞ n = u } = µ ⊗ ( n +1) ⊗ ν ⊗ ( n +1) ( S ( v )) (cid:0) vu (cid:1) ( n + 1) (cid:30) µ ⊗ n ⊗ ν ⊗ n ( S ( u )) . On the other hand, P { U ∞ n +1 = v | U ∞ n = u } = 1 h ( u ) P { U n +1 = v | U n = u } h ( v )= h ( v ) h ( u ) (cid:0) vu (cid:1) (2 n + 2)(2 n + 1) . Thus, h ( v ) h ( u ) = µ ⊗ ( n +1) ⊗ ν ⊗ ( n +1) ( S ( v )) µ ⊗ n ⊗ ν ⊗ n ( S ( u )) (2 n + 2)(2 n + 1)( n + 1) and, up to an arbitrary multiplicative constant, h ( w ) = (cid:18) mm (cid:19) µ ⊗ m ⊗ ν ⊗ m ( S ( w )) for w ∈ W m .Since h ( ∅ ) = 1 , this normalization is the extended Doob–Martinkernel w K ( w, y ) , where y is the point in the Doob–Martin boundarythat corresponds to the pair of diffuse probability measures ( µ, ν ) . Remark . The constant harmonic function h ≡ arises from theabove construction with µ and ν both being the Lebesgue measure λ on [0 , . Therefore the process ( U n ) n ∈ N is itself the extremal bridge asso-ciated with the canonical pair ( λ, λ ) . In particular, ( U n ) n ∈ N convergesalmost surely to the point in the Doob–Martin boundary associatedwith this pair.We observed in Remark 3.7 that a sequence ( y k ) k ∈ N with N ( y k ) →∞ as k → ∞ converges in the Doob–Martin topology if and only iffor every m ∈ N the sequence of random words in W m obtained byselecting m letters a and m letters b uniformly at random from y k andmaintaining their relative order converges in distribution as k → ∞ .We can now enhance that result as follows. Proposition 7.2.
Consider a sequence ( y k ) k ∈ N in W , where y k ∈ W N ( y k ) , k ∈ N , and N ( y k ) → ∞ as k → ∞ . If y is the point in theDoob–Martin boundary that corresponds to the pair of (diffuse) prob-ability measures ( µ, ν ) with ( µ + ν ) = λ , then lim k →∞ y k = y in theDoob–Martin topology if and only if lim k →∞ (cid:0) y k w (cid:1)(cid:0) N ( y k ) m (cid:1) = µ ⊗ m ⊗ ν ⊗ m ( S ( w )) for all w ∈ W m for all m ∈ N . That is, lim k →∞ y k = y if and onlyif for each m ∈ N the sequence of random words in W m obtained byselecting m letters a and m letters b uniformly at random from y k andmaintaining their relative order converges in distribution as k → ∞ tothe random word U ∞ m = W ( X , . . . , X m , Y , . . . , Y m ) defined above. Given a sequence ( y k ) k ∈ N in W , where y k ∈ W N ( y k ) , k ∈ N , and N ( y k ) → ∞ as k → ∞ , define a sequence of pairs of discrete prob-ability measures (( µ k , ν k )) k ∈ N on [0 , as follows. For k ∈ N the twoprobability measure µ k and ν k both assign all of their mass to the set ROWING RANDOM WORDS 21 { ℓ N ( y k ) : 1 ≤ ℓ ≤ N ( y k ) } . For ≤ i ≤ N ( y k ) , µ k ( i N ( y k ) ) = N ( y k ) ifthe i th letter of y k is the letter a , otherwise µ k ( i N ( y k ) ) = 0 . Similarly,for ≤ j ≤ N ( y k ) , ν k ( j N ( y k ) ) = N ( y k ) if the j th letter of y k is the letter b , otherwise ν k ( j N ( y k ) ) = 0 . In particular, ( µ k + ν k ) is the uniformprobability measure on { ℓ N ( y k ) : 1 ≤ ℓ ≤ N ( y k ) } . Observe that if w ∈ W m , then, for w ∈ W m , ( N ( y k ) m ) µ ⊗ mk ⊗ ν ⊗ mk ( S ( w )) = ( m !) (cid:18) y k w (cid:19) so that (cid:0) y k w (cid:1)(cid:0) N ( y k ) m (cid:1) = (cid:18) N ( y k ) m N ( y k )( N ( y k ) − · · · ( N ( y k ) − m + 1) (cid:19) µ ⊗ mk ⊗ ν ⊗ mk ( S ( w )) . One direction of the following corollary is now immediate.
Corollary 7.3.
Suppose that ( y k ) k ∈ N and (( µ k , ν k )) k ∈ N are as above.If ( y k ) k ∈ N converges in the Doob–Martin topology to the point y in theDoob–Martin boundary that corresponds to the pair of probability mea-sures ( µ, ν ) , then ( µ k ) k ∈ N converges weakly to µ and ( ν k ) k ∈ N convergesweakly to ν . Conversely, if ( µ k ) k ∈ N converges weakly to µ and ( ν k ) k ∈ N converges weakly to ν , then ( µ + ν ) = λ , and if y is the point in theDoob–Martin boundary that corresponds to the pair ( µ, ν ) , then ( y k ) k ∈ N converges in the Doob–Martin topology to y .Proof. As we have already remarked, if ( µ k ) k ∈ N converges weakly to µ and ( ν k ) k ∈ N converges weakly to ν then, since the boundary of S ( w ) is Lebesgue null for any word w ∈ W m , m ∈ N , we have that µ ⊗ mk ⊗ ν ⊗ mk ( S ( w )) converges to µ ⊗ m ⊗ ν ⊗ m ( S ( w )) so that lim k →∞ (cid:0) y k w (cid:1)(cid:0) N ( y k ) m (cid:1) = µ ⊗ m ⊗ ν ⊗ m ( S ( w )) , and it follows from Proposition 7.2 that ( y k ) k ∈ N converges to the point y in the Doob–Martin boundary that corresponds to the pair ( µ, ν ) .Conversely, suppose that ( y k ) k ∈ N converges to the point y in theDoob–Martin boundary corresponding to the pair ( µ, ν ) . Given anysubsequence of N there is, by the compactness in the weak topology ofprobability measures on [0 , , a further subsequence such that alongthis further subsequence µ k converges weakly to some probability mea-sure µ ′ and ν k converges weakly to some probability measure ν ′ . Notethat ( µ ′ + ν ′ ) = λ . From the other direection of the corollary, thisimplies that along the subsubsequence y k converges to the point y ′ inthe Doob–Martin boundary corresponding to the pair of probability measures ( µ ′ , ν ′ ) . Because y ′ = y it must be the case ( µ ′ , ν ′ ) = ( µ, ν ) .Thus, from any subsequence of N we can extract a further subsequencealong which µ k converges weakly to µ and ν k converges weakly to ν , andthis implies that ( µ k ) k ∈ N converges weakly to µ and ( ν k ) k ∈ N convergesweakly to ν . (cid:3) Example: the Plackett-Luce chain
In general, there is no simple closed form expression for the transitionprobabilities of an infinite bridge ( U ∞ n ) n ∈ N associated with a pair of(not necessarily canonical) diffuse probability measures ζ , η and hencethe associated harmonic function h . However, it is possible to obtainsuch expressions in the special case where ζ is the exponential distribu-tion with rate parameter α and η is the exponential distribution withrate parameter β . Given u ∈ W n and ≤ i ≤ n , set A ni ( u ) := { i ≤ j ≤ n : u j = a } and B ni ( u ) := { i ≤ j ≤ n : u j = b } . By the reasoning that goes into the analysis of the Plackett-Luce orvase model of random permutations (see, for example, [Mar95]), P { U ∞ n = u } = ( n !) α n β n n Y i =1 A ni ( u ) α + B ni ( u ) β – this is essentially just repeated applications of the elementary resultusually called competing exponentials : if S and T are independent ex-ponentially distributed random variables with rate parameters λ and θ , then the probability of the event { S < T } is λλ + θ and conditionalon this event the random variables S and T − S are independent andexponentially distributed with rate parameters λ + θ and θ . (As acheck, note that when α = β = γ , say, this probability is, as expected, / (cid:0) nn (cid:1) .) We also know that P { U ∞ n = u | U ∞ n +1 = v } = (cid:0) vu (cid:1) ( n + 1) . ROWING RANDOM WORDS 23
It follows that P { U ∞ n +1 = v | U ∞ n = u } = (cid:0) vu (cid:1) ( n + 1) (( n + 1)!) α n +1 β n +1 2( n +1) Y i =1 A n +1 i ( v ) α + B n +1 i ( v ) β (cid:30) ( n !) α n β n n Y i =1 A ni ( u ) α + B ni ( u ) β = (cid:18) vu (cid:19) αβ Q ni =1 ( A ni ( u ) α + B ni ( u ) β ) Q n +1) i =1 ( A n +1 i ( v ) α + B n +1 i ( v ) β ) . As a check, when α = β = γ , say, this transition probability is (cid:18) vu (cid:19) (2 n )!(2( n + 1))! = (cid:0) vu (cid:1) (2 n + 2)(2 n + 1) , as expected.The corresponding harmonic function h satisfies (cid:18) vu (cid:19) αβ Q ni =1 ( A ni ( u ) α + B ni ( u ) β ) Q n +1) i =1 ( A n +1 i ( v ) α + B n +1 i ( v ) β )= h ( v ) h ( u ) (cid:0) vu (cid:1) (2 n + 2)(2 n + 1) . We conclude from this that, up to an arbitrary positive constant, h ( w ) = (2 m )! α m β m Q mi =1 ( A mi ( w ) α + B mi ( w ) β ) for w ∈ W n . Acknowledgments.
We thank two anonymous referees for a numberof helpful suggestions that improved the presentation considerably.
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