Conditional limit theorems for ordered random walks

In a recent paper of Eichelsbacher and Koenig (2008) the model of ordered random walks has been considered. There it has been shown that, under certain moment conditions, one can construct a k-dimensional random walk conditioned to stay in a strict order at all times. Moreover, they have shown that the rescaled random walk converges to the Dyson Brownian motion. In the present paper we find the optimal moment assumptions for the construction of the conditional random walk and generalise the limit theorem for this conditional process.


Introduction, main results and discussion
1.1. Introduction. A number of important results have been recently proved relating the limiting distributions of random matrix theory with certain other models. These models include the longest increasing subsequence, the last passage percolation, non-colliding particles, the tandem queues, random tilings, growth models and many others. A thorough review of these results can be found in [12].
Apparently it was Dyson who first established a connection between random matrix theory and non-colliding particle systems. It was shown in his classical paper [7] that the process of eigenvalues of the Gaussian Unitary Ensemble of size k × k coincides in distribution with the k-dimensional diffusion, which can be represented as the evolution of k Brownian motions conditioned never to collide. Such conditional versions of random walks have attract a lot of attention in the recent past, see e.g. [14,11]. The approach in these papers is based on explicit formulas for nearest-neighbour random walks. However, it turns out that the results have a more general nature, that is, they remain valid for random walks with arbitrary jumps, see [1] and [9]. The main motivation for the present work was to find minimal conditions, under which one can define multidimensional random walks conditioned never to collide.
In this paper we study the asymptotic behaviour of the random walk S n conditioned to stay in W . Let τ x be the exit time from the Weyl chamber of the random walk with starting point x ∈ W , that is, One can attribute two different meanings to the words 'random walk conditioned to stay in W .' On the one hand, the statement could refer to the path (S 0 , S 1 , . . . , S n ) conditioned on {τ x > n}. On the other hand, one can construct a new Markov process, which never leaves W . There are two different ways of defining such a conditioned processes. First, one can determine its finite dimensional distributions via the following limit Second, one can use an appropriate Doob h-transform. If there exists a function h (which is usually called invariant function) such that h(x) > 0 for all x ∈ W and then one can make a change of measure As a result, one obtains a random walk S n under a new measure P (h) x . This transformed random walk is a Markov chain which lives on the state space W .
To realise the first approach one needs to know the asymptotic behaviour of P(τ x > n). And for the second approach one has to find a function satisfying (2). It turns out that these two problems are closely related to each other: The invariant function reflects the dependence of P(τ x > n) on the starting point x. Then both approaches give the same Markov chain. For one-dimensional random walks conditioned to stay positive it was shown by Bertoin and Doney [2]. They proved that if the first moment of a random walk is finite, then the function V (x) = x − E(x + S σx ) is invariant and that P(σ x > n) ∼ CV (x)P(σ 0 > n), where σ x = min{k ≥ 1 : x + S k ≤ 0}. The analogous program for random walks in the Weyl chamber was carried out by Eichelsbacher and König [9]. If we define the direct analogue of the invariant function used by Bertoin and Doney as follows where ∆(x) denotes the Vandermonde determinant, that is, Then it was shown in [9] that if E|ξ| r k < ∞ with some r k > ck 3 , then it can be concluded that V is a finite and strictly positive invariant function. Moreover, the authors determined the behaviour of P(τ x > n) and studied some asymptotic properties of the conditioned random walk. They also posed a question about minimal moment assumptions under which one can construct a conditioned random walk by using V . In the present paper we answer this question. We prove that the results of [9] remain valid under the following conditions: • Centering assumption: We assume that Eξ = 0.
• Moment assumption: We assume that E|ξ| α < ∞ with α = k − 1 if k > 3 and some α > 2 if k = 3. Furthermore, we shall assume, without loss of generality, that Eξ 2 = 1. It is obvious, that this moment condition is the minimal one for the finiteness of the function V defined by (3). Indeed, from the definition of ∆ it is not difficult to see that the finiteness of the (k − 1)-th moment of ξ is necessary for the finiteness of ∆(x + S 1 ). Thus, this moment condition is also necessary for the integrability of ∆(x + S τx ), which is equivalent to the finiteness of V . In other words, if E|ξ| k−1 = ∞, then one has to define the invariant function in a different way. Moreover, we give an example, which shows that if the moment assumption does not hold, then P(τ x > n) has a different rate of divergence.
1.2. On the tail of τ x . Here is our main result: Theorem 1. Assume that k ≥ 3 and let the centering as well as the moment assumption hold. Then the function V is finite and strictly positive. Moreover, as n → ∞, where κ is an absolute constant.
All the claims in the theorem have been proved in [9] under more restrictive assumptions: As we have already mentioned, the authors have assumed that E|ξ| r k < ∞ with some r k such that r k ≥ ck 3 , c > 0. Furthermore, they needed some additional regularity conditions, which ensure the possibility to use an asymptotic expansion in the local central limit theorem. As our result shows, these regularity conditions are superfluous and one needs k − 1 moments only.
Under the condition that ξ (1) , . . . , ξ (k) are identically distributed, the centering assumption does not restrict the generality: One has only to change to the random walk S n − nEξ. But if the drifts are allowed to be unequal, then the asymptotic behaviour of τ x and that of the conditioned random walk might be different, see [15] for the case of the Brownian motion.
We now turn to the discussion of the moment condition in the theorem. We start with the following example.
Example 2. Assume that k ≥ 4 and consider the random walk, which satisfies with some α ∈ (k − 2, k − 1). Then, The CLT because is applicable due to the condition α > k − 2, which implies the finiteness of the variance.
For the second term in the product we need to analyse (k − 1) random walks under the condition E|ξ| k−2+ε < ∞. Using Theorem 1, we have As a result the following estimate holds true for sufficiently small ε, The right hand side of this inequality decreases slower than n −k(k−1)/4 for all sufficiently small ε. Moreover, using the same heuristic arguments, one can find a similar lower bound in case (5) We believe that the lower bounds constructed above are quite precise, and we conjecture that It is clear that E|ξ| k−1 < ∞ is necessary for the finiteness of V . Furthermore, the example shows that this condition is almost necessary for the validity of (4): One can not obtain the relation P(τ x > n) ∼ C(x)n −k(k−1)/4 assuming that E|ξ| k−1−ε < ∞ with some ε > 0.
If we have two random walks, i.e. k = 2, then τ x is the exit time from (0, ∞) of the random walk Z n := (x (2) − x (1) ) + (S (2) n − S (1) n ). It is well known that, for symmetrically distributed random walks, EZ τx < ∞ if and only if E(ξ However, the existence of EZ τx is not necessary for the relation P(τ x > n) ∼ C(x)n −1/2 , which holds for all symmetric random walks. This is contary to the high-dimensional case (k ≥ 4), where the integrability of ∆(x + S τx ) and the rate n −k(k−1)/4 are quite close to each other.
In case we have three random walks our moment condition is not optimal. We think that the existence of the variance is sufficient for the integrability of ∆(x + S τx ). But our approach requires more than two moments. Furthermore, we conjecture that, as in the case k = 2, the tail of the distribution of τ x is of order n −3/2 for all random walks.
1.3. Scaling limits of conditioned random walks. Theorem 1 allows us to construct the conditioned random walk via the distributional limit (1). In fact, if (4) is used, we obtain, as m → ∞, But this means that the distribution of S n is given by the Doob transform with function V . (This transformation is possible, because V is well-defined, strictly positive on W and satisfies E[V (x + S 1 ); τ x > 1] = V (x).) In other words, both ways of construction described above give the same process. We now turn to the asymptotic behaviour of S n . To state our results we introduce the limit process. For the k-dimensional Brownian motion with starting point x ∈ W one can change the measure using the Vandermonde determinant: The corresponding process is called Dyson's Brownian motion. Furthermore, one can define Dyson's Brownian motion with starting point 0 via the weak limit of P (∆) x , for details see Section 4 of O'Connell and Yor [14]. We will denote the corresponding probability measure as P (∆) 0 . Theorem 3. If k ≥ 3 and the centering as well as the moment assumption are valid, then where µ is the probability measure on W with density proportional to ∆(y)e −|y| 2 /2 . Furthermore, the process X n (t) = n , x ∈ W converges weakly to the Dyson Brownian motion under the measure P (∆) x . Finally, the process X n (t) = x , x ∈ W converges weakly to the Dyson Brownian motion under the measure P (∆) 0 . Relation (6) and the convergence of the rescaled process with starting point x √ n were proven in [9] under more restrictive conditions. Convergence towards P (∆) 0 was proven for nearest-neighbour random walks, see [14] and [16]. A comprehensive treatment of the case k = 2 can be found in [6].
One can guess that the convergence towards Dyson's Brownian motion holds even if we have finite variance only. However, it is not clear how to define an invariant function in that case.

1.4.
Description of the approach. The proof of finiteness and positivity of the function V is the most difficult part of the paper. To derive these properties of V we use martingale methods. It is well known that ∆(x + S n ) is a martingale. And in the case of a nearest-neighbour random walk, or in the case of the Brownian motion, we can define τ x as the first time of ∆(x + S n ) being non-positive. But in general it could happen that ∆(x + S τx ) > 0. In other words, the martingale ∆(x + S n ) does not 'feel' the stopping time τ x . So the stopping time T x = min{k ≥ 1 : ∆(x + S k ) ≤ 0} seems to be more natural for the martingale ∆(x + S n ). Moreover, it helps us to obtain the desired properties of V . We first show that ∆(x+S Tx ) is integrable, which yields the integrability of ∆(x+S τx ), see Subsection 2.1. Furthermore, it follows from the integrability of ∆(x + S Tx ) that the function To show that the function V is strictly positive, we use the interesting observation that the sequence V (T ) (x + S n )1{τ x > n} is a supermartingale, see Subsection 2.2.
It is worth mentioning that the detailed analysis of the martingale properties of the random walk S n allows one to keep the minimal moment conditions for positivity and finiteness of V . The authors of [9] used the Hölder inequality at many places in their proof. This explains the superfluous moment condition in their paper.
To prove the asymptotic relations in our theorems we use a version of the Komlos-Major-Tusnady coupling proposed in [10], see Section 3. A similar coupling has been used in [3] and [1]. In order to have a good control over the quality of the Gaussian approximation we need more than two moments of the random walk. This fact explains partially why we required the finiteness of E|ξ| 2+δ < ∞ in the case k = 3.

Finiteness and positivity of V
The main purpose of the present section is to prove the following statement.
Proposition 4. The function V has the following properties: As it was already mentioned in the introduction our approach relies on the investigation of properties of the stopping time T x defined by It is easy to see that T x ≥ τ x for every x ∈ W .
2.1. Integrability of ∆(x+S Tx ). We start by showing that E[∆(x+S Tx )] is finite under the conditions of Theorem 1. In this paragraph we omit the subscript x if there is no risk of confusion.
The statement of the lemma follows now from the facts that ∆(x + S n ) is a martingale and ∆(x + S T ) is non-positive.
For any ε > 0, define the following set Lemma 6. For any sufficiently small ε > 0 there exists γ > 0 such the following inequalities hold and Proof. We shall prove (7) only. The proof of (8) requires some minor changes, and we omit it. For a constant δ > 0, which we define later, let and split the expectation into 2 parts, It follows from the definition of the stopping time T that at least one of the differences (x (r) + S (r) − x (s) − S (s) ) changes the sign at time T , i.e. one of the following events occurs On the event A n ∩ B s,r , This implies that on the event A n ∩ B s,r , As is not difficult to see, where the sum is taken over all i 1 , i 2 , . . . , i k such that Combining Doob's and Rosenthal's inequalities, one has Then, where C 1 , C 2 , . . . are universal constants. Now note that since x ∈ W n,ε , we have a simple estimate for any j 1 < j 2 . Using (11) and (12), we obtain Thus, Now we estimate E 2 (x). Clearly, Then, using (10) once again, we get Applying the following estimate, which will be proved at the end of the lemma, we obtain This implies that Consequently, (15) Applying (13) and (15) to the right hand side of (9), and choosing ε and δ in an appropriate way, we arrive at the conclusion. Thus, it remains to show (14).
It is easy to see that, for any i r ∈ (0, α), Putting y = x/p in Corollary 1.11 of [13], we get the inequality As was shown in [5], this inequality remains valid for M (r) n , i.e.
Lemma 7. For every ε > 0 holds Proof. To shorten formulas in the proof we set S 0 = x. Also, set, for brevity, b n = [an 1/2−ε ]. The parameter a will be chosen at the end of the proof. First note that Then there exists at least one pair j, l such that for at least at [n ε /(a 2 k 2 )] points i | ≤ n 1/2−ε for i ∈ I. Without loss of generality we may assume that j = 1 and l = 2. There should exist at least [n ε /(2a 2 k 2 )] points with the distance less than 2k 2 b 2 n . To simplify notation assume that points i 1 , . . . i n ε /(2a 2 k 2 ) enjoy this property: n . In fact this means that i s − i s−1 can take only values {jn 1−2ε , 1 ≤ j ≤ 2k 2 }. The above considerations imply that Using the Stirling formula, we get ≤ a n ε (2k 2 ) n ε /a 2 .
By the Central Limit Theorem, Thus, for all sufficiently large n, Choosing a = 8(2k 2 ) 2k 2 , we complete the proof.
Lemma 8. For every ε > 0 the inequality Remark 9. If E|ξ| α < ∞ for some α > k − 1, then the claim in the lemma follows easily from the Hölder inequality and Lemma 7. But our moment assumption requires more detailed analysis. ⋄ Proof. We give the proof only for t = 0.
Noting that {ν n > n 1−ε } ⊂ G l,i , we get Therefore, we need to derive an upper bound for E[|∆(x + S n )|; G 1,2 ]. Let µ = µ 1,2 be the moment when |x (2) n | ≤ n 1/2−ε for the [n ε /(a 2 k 2 )] time. Then it follows from the proof of the previous lemma that Using the inequality |a + b| ≤ (1 + |a|)(1 + |b|) one can see that Making use of (10), one can verify that Recall that by the definition of µ we have |x (2) where the sum is taken over all i 1 , . . . , i k such that all there is at most one i j = k − 1, and the sum i r does not exceed k(k − 1)/2. Thus, Since i 1 ≤ k − 2 and i 2 ≤ k − 2, we can apply the Hölder inequality, which gives Consequently, Plugging (19) and (20) into (18), we arrived at the conclusion.

Lemma 10. There exists a constant C such that
for all n ≥ 1 and all x ∈ W .
By Lemma 8, the second term on the right hand side is bounded by Using Lemma 5, we have in the last step we used the fact that ∆(x+S n ) is a martingale. Then, by Lemma 6, Using Lemma 5 once again, we arrive at the bound As a result we have Iterating this procedure m times, we obtain Choosing m = m(n) such that n (1−ε) m+1 ≤ 10 and noting that the product and the sum remain uniformly bounded, we finish the proof of the lemma.

2.2.
Proof of Proposition 4. We start by showing that Lemma 10 implies the integrability of ∆(x + S τx ). Indeed, setting τ x (n) := min{τ x , n} and T x (n) := min{T x , n}, and using the fact that |∆(x + S n )| is a submartingale, we have Since ∆(x + S n ) is a martingale, we have Therefore, we get This, together with Lemma 10, implies that the sequence E[|∆(x + S τ )|1{τ ≤ n}] is uniformly bounded. Then, the finiteness of the expectation E|∆(x + S τ )| follows from the monotone convergence.
To prove (a) note that since ∆(x + S n ) is a martingale, we have an equality Letting n to infinity we obtain (a) by the dominated convergence theorem. For (b) note that ∆(x + S n )1{τ x > n} ≤ ∆(y + S n )1{τ x > n} ≤ ∆(y + S n )1{τ y > n}.
Then letting n to infinity and applying (a) we obtain (b). (c) follows directly from Lemma 10. We now turn to the proof of (d). It follows from (24) and the inequality τ x ≤ T x that Thus, we need to get a lower bound of the form (1 + o(1))∆(x). We first note that Therefore, it is sufficient to show that under the condition min j<k (x (j+1) − x (j) ) → ∞. The sequence V (T ) (x + S n )1{T x > n} is a non-negative martingale. Then, arguing as in Lemma 5, one can easily see that V (T ) (x + S n )1{τ x > n} is a supermartingale.
We bound E[V (T ) (x + S n )1{τ x > n}] from below using its supermartingale property. This is similar to the Lemma 10, where an upper bound has been obtained using submartingale properties of ∆(x + S n )1{T x > n}. We have Then, applying (23) and (8), we obtain Using now Lemma 8, we have Starting from n 0 and iterating this procedure, we obtain for the sequence n m = Next we fix a constant δ > 0 and pick n 0 such that This is possible since both the series and the product converge. Together with the fact that V (T ) (x + S n )1{τ x > n} is a supermartingale and the with lower bound in (23) this gives us, Therefore, since δ > 0 is arbitrary we have a lower asymptotic bound provided that min 2≤j≤k (x (j) − x (j−1) ) → ∞.
Using the martingale property of V (T ) (x + S n )1{T x > n} and noting that Letting n → ∞, we obtain Combining (24), (26) and (27), ). Now (25) follows from the obvious bound Thus, the proof of (d) is finished.
To prove (e) note that it follows from (d) that there exists R and δ > 0 such that V (x) ≥ δ on the set S R = {x : min 2≤j≤k (x (j) − x (j−1) ) > R}. Then, with a positive probability p the random walk can reach this set after N steps if N is sufficiently large. Therefore, This completes the proof of the proposition.

Coupling
We start by formulating a classical result on the normal approximation of random walks.
Lemma 13. There exists a finite constant C such that Moreover, uniformly in y ∈ W satisfying |y| ≤ θ n √ n with some θ n → 0. Finally, the density b t (y, z) of the probability P(τ bm uniformly in y, z ∈ W satisfying |y| ≤ θ n √ n and |z| ≤ n/θ n with some θ n → 0. Here, Proof. (29) has been proved by Varopoulos [17], see Theorem 1 and formula (0.4.1) there. The proof of (30) and (31) can be found in Sections 5.1-5.2 of [8].
Using the coupling we can translate the results of Lemma 14 to the random walks setting when y ∈ W n,ε .

Thus, it remains to show that
for all sufficiently small ε > 0 and all y ∈ W n,ε . For that note that for y ∈ W n,ε , Therefore, (37) will be valid for all ε satisfying This proves (32). To prove (33) it is sufficient to substitute (29) in (35). The proof of (34) is similar. Define two sets, Clearly D − ⊂ D ⊂ D + . Then, arguing as above, we get P(τ y > n, y + S n ∈ √ nD) ≤ P(τ y > n, y + S n ∈ √ nD, A) + o n −r ≤ P(τ bm y + > n, y Similarly, P(τ y > n, y + S n ∈ √ nD) ≥ P(τ bm y − > n, y − + B n ∈ √ nD − ) + o n −r .
Applying now Lemmas 14, we obtain P(τ x > n; ν n ≤ n 1−ε ) We now show that the first expectation converges to V (x) and that the second expectation is negligibly small.

Lemma 15. Under the assumptions of Theorem 1,
Proof. Rearranging, we have According to Lemma 8, Further, here we have used the martingale property of ∆(x+S n ). Noting that ν n ∧n 1−ε → ∞ almost surely, we have Then, using the integrability of ∆(x + S τx ) and the dominated convergence, we obtain Proof. We first note that where we used the submartingale property of ∆(x + S j )1{T x > j}, see Lemma 5.
We note that Since the conditioned distribution of S n given Σ is exchangeable, we may apply Theorem 2.1 of [11], which says that Therefore, Using this equality and conditioning on F l , we have Consequently, by the dominated convergence, since Σ n → 0. This implies that Combining (48)-(50), we see that the left hand side of (48) converges to zero. Then, taking into account (47), we get (46). Thus, the proof is finished.

Weak convergence results
Lemma 17. For any x ∈ W , the distribution P x+Sn √ n ∈ ·|τ x > n weakly converges to the distribution with the density 1 Z1 e −|y| 2 /2 ∆(y), where Z 1 is the norming constant.
Using the coupling and arguing as in Lemma 14, one can show that P(τ y > n − k, y + S n−k ∈ √ nA) ∼ P(τ bm y > n, y + B n ∈ √ nA) uniformly in k ≤ n 1−ε and y ∈ W n,ε . Next we apply asymptotics (31) and obtain that uniformly in y ∈ W n,ε , |y| ≤ θ n √ n. As a result we obtain where the latter equivalence holds due to Lemma 15. Substituting the latter equivalence in (51) and using the asymptotics for P(τ x > n), we arrive at the conclusion. Now we change slightly notation. Let n be the family of processes with the probability measure P (V ) x √ n , x ∈ W . Then X n weakly converges in C(0, ∞) to the Dyson Brownian motion with starting point x, i.e. to the process distributed according to the probability measure P (∆) x .
Proof. The proof is given via coupling from Lemma 12. To prove the claim we need to show that the convergence take place in C[0, l] for every l. The proof is identical for l, so we let l = 1 to simplify notation. Thus it sufficient to show that for every function f : 0 ≤ f ≤ 1 uniformly continuous on C[0, 1], x f (B) as n → ∞. By Lemma 12 one can define B n and S n on the same probability in such a way that the complement of the event for some a > 0 and γ > 0. Let B n t = B nt / √ n. By the scaling property of the Brownian motion E (∆) Split the expectation into two parts, Since the function f is uniformly continuous, tends to 0 as n → ∞. Therefore, Moreover, on the event A n hold the following inequalities . Arguing as in Lemma 14 and using monotonicity of V , we obtain 1 where we used (d) of Proposition 4 in the second and the third lines. Replacing x + with x − , one can easily obtain the following lower bound Note also that Therefore, Thus, if we show that E ∆ x ± n [f (B n ); A n ] = o(1), and E 2 = o(1), we are done. Since the proofs of these statements are almost identical we concentrate on showing that E 2 = o(1). We have, since f ≤ 1, Put y n = (2n 1/2+δ , . . . , 2(k − 1)n 1/2+δ ). Then, if we pick δ sufficiently small. Next, using the bounds V (x) ≤ V (T ) (x) ≤ ∆ 1 (x), we get Arguing similarly to the second part of Lemma 6, one can see that The expectation of the product can be estimated exactly as in Lemma 6 using the Fuk-Nagaev inequality. This gives us Thus, the proof is finished. Now we consider start from a fixed point x.
Lemma 19. Let X n (t) = S [nt] √ n be the family of processes with the probability measure P (V ) x , x ∈ W . Then X n converges weakly to the Dyson Brownian motion with starting point 0.