The Exact Asymptotic of the Time to Collision

In this note we consider the time of the collision τ for n independent copies of Markov processes X 1 t ,. .. , X n t , each starting from x i , where x 1 <. .. < x n. We show that for the continuous time random walk IP x (τ > t) = t −n(n−1)/4 (Ch(x) + o(1)), where C is known and h(x) is the Vandermonde determinant. From the proof one can see that the result also holds for X t being the Brownian motion or the Poisson process. An application to skew standard Young tableaux is given.


Introduction and the results
In this note X t is either a standard Brownian motion (SBM) W t or the standard symmetric continuous time random walk (CTRW).Recall that a compound Poisson process with intensity parameter λ = 1 and jump distribution 1  2 δ −1 + 1 2 δ 1 is called CTRW; see also Asmussen [2, page 99], with his µ = δ = 1  2 .We consider a sequence X 1 t , . . ., X n t of independent copies of X t , each starting from X i 0 = x i .We assume that x ∈ W = {y ∈ IR n : y 1 < y 2 < . . .< y n }.Two processes X i t and X j t collide at τ ij = min{t > 0 : X i t = X j t }.The time of the collision is τ = min 1≤i<j≤n τ ij .
Let x = (x 1 , . . ., x n ) and let be the Vandermonde determinant.Our aim is the proof of the following theorem.The Brownian part was first given by Grabiner [6], see also a new proof of Doumerc and O'Connell [4] which uses the representation of collision time obtained there, and an elementary proof of Pucha la [11].We also remark that from the theorem proof and Proposition 6.1 from Doumerc and O'Connell [4] we may immediately conclude that the theorem also holds for X t being the Poisson process with unit intensity.Following Mehta [10, page 354], we may rewrite the constant C in (1.2) in the following form: 3 The proof is based on the following recent result by Doumerc and O'Connell [4], which expresses IP x (τ > t) in terms of Pfaffians.Let P = (p ij ) n i,j=1 , where p ij = p ij (t) = IP x i ,x j (τ ij > t) for i ≤ j and p ij = −p ji .Then IP x (τ > t) = Pf(P ) if n is even, n l=1 (−1) l+1 Pf(P (l) ) if n is odd, (1.4) where P (l) = (p ij ) i,j =l .By Pf we denote the Pfaffian.To recall this notion, let for n even P 2 (n) be the set of partitions of {1, . . ., n} into n 2 pairs and c(π) is the number of crossings.For a given skew-symmetric matrix A = (a ij ) n i,j=1 we define the Pfaffian In particular we have the formula: The paper is organized as follows.In Section 2 we give some preliminary ideas and we state without proofs two key lemmas and a proposition.We work out special cases for the Brownian motion and the continuous time random walk in Section 3.An application to Young tableaux is given in Section 4.There is also mentioned some relationship of Theorem 1.1 with Markovian tandem queues.Proofs are given in Section 5.

Preliminaries
We need some technical facts.Suppose that x ∈ R n and n ∈ 2N.If , where then Pf(Q) can be written in the form for some polynomials ). Hence we conclude that all polynomials W k must be skew symmetric polynomials too.We will also use the generalized Vandermonde determinant where l = (l 1 , . . ., l n ).The special case is the Vandermonde determinant when l = (0, 1, . . ., n − 1).Since generalized Vandermonde determinants creates basis for skew polynomials (see Macdonald [8, page 24]) we can write W k as a linear combination of generalized Vandermonde determinants.
The following lemmas will be useful in calculating asymptotics, which proofs will be demonstrated in Section 5. Notice that in both the lemmas we suppose that n is even, because only for this case Pfaffian is defined. where . (2.7) In particular The next proposition will be the key to calculate asymptotics.
Notice that constants C even (n) and C odd (n) depend only on coefficients a k,2k+1 .The above lemmas and the proposition will be proved in Section 5.

Special cases
We find here details of expansions (2.5) for two special cases, from which with the use of Proposition 2.3 we may conclude the result of Theorem 1.1.

Brownian motion
To calculate p ij (t) we will use the reflection principle: where Proof.We have Setting q ij (t) = ψ 2t (x j − x i ), the assumptions of Proposition 2.3 are satisfied with We have Pf(Q) .
Using formula (1.4) and Proposition 2.3 we obtain: for n even: lim and for n odd: where the constants are defined in (2.7) and (2.8) respectively.We can rewrite above and for n odd: In Section 5.4 we find another expressions for C even (n) and C odd (n), which gives an alternative proof of Grabiner theorem (see Grabiner [6], Dourmerc and O'Connell [4], or Pucha la [11]).

CTRW
Let S t be the standard symmetric CTRW.Following Asmussen [2, page 99] (with his where I r (t) is the modified Bessel function of order r.As in the Brownian case to calculate p ij we will use the reflection principle.
We need the asymptotic of For this we define Recall that a function f t has the asymptotic expansion For details and basic facts on the asymptotic expansions we refer to Knopp [7].By Watson [13,page 203] we have Setting q ij (t) = ϕ 2t (x j − x i ), the assumptions of Proposition 2.3 are satisfied with Hence we have (remember about doubling t) Lemma 3.2 (see [14]) where B k are Bernoulli numbers.
In the next lemma we study polynomials A ϕ k (x).
is an odd polynomial (that is with even coefficient vanishing) of order 2k + 1 with the leading coefficient a 2k+1 defined in (3.9).That is where a k,2k+1 = a 2k+1 .
Proof.We have that is an odd polynomial of order 2m + 1 with the leading coefficient 2 2m+1 .This is because Hence the above equals to Using now Proposition 2.3 we get: and for n ∈ 2N + 1 we have where the constants are defined in (2.7) and (2.8) respectively.Therefore this case is identical to the Brownian case, which completes the proof of Theorem 1.1.We must notice that above considerations are valid for Poisson process N t , this is because difference of two independent Poisson processes with intensity 1 is a CTRW with intensity 2.
, so all calculations are identical.

Applications
Since the result of Theorem 1.1 is also valid in the case of independent Poisson processes, we can apply it to obtain an aymptotics for Young tableaux, which generalizes some earlier results of Regev [12].
Thus let X t = (X 1 t , . . ., X n t ) be vector of independent Poisson processes with intensity 1 starting from x ∈ W .Let σ m denote the time of the m-th transition of X t , and let where N t = max{m : σ m ≤ t} is a Poisson process with intensity n independent of T .
For integer partitions λ and µ with µ ≤ λ, let f λ/µ denote the number of skew standard tableaux with shape λ/µ.Set δ = (n − 1, n − 2, . . ., 1, 0).We denote by λ 1 the hight of the Young diagram defined by partition λ (the number of boxes in the first row of the conjugate diagram), see for definitions Fulton [5].Put The key observation relating our theorem with Young tableaux is that which together with (4.13) links the exit time theory with Young tableaux.The following corollary extends the asymptotics obtained by Regev [12] from Young tableaux to skew Young tableaux.
Proof.Let N t be a Poisson variable with intensity nt.We first show that for each a > 0, and g(t) ∼ ct −b , where c > 0 and 0 < b < 1/2 we have For the proof, without loss of generality, we may assume n = 1.The Fenchel-Legendre transform for random variable X − 1, where X is Poisson distributed with mean 1 is see Dembo and Zeitouni [3, page 35].Note that Λ * (x) = x 2 2 + o(x 2 ) for x → 0. Following Dembo and Zeitouni [3, page 27] we have the inequality: for all t nonnegative integer where F = (−∞, −g(t)] ∪ [g(t), ∞), which the inequality can be extended to all t ≥ 0. Since we have as t → ∞.Thus from (4.15) the proof of (4.14) follows.
Observe now that We relate t and k above by . By (4.14) as k → ∞.We also have by Theorem 1.1 where C is from (1.2).Hence by (4.16) we have .
In the similar way, using (4.17) with t(k) = (k + k 3/4 )/n, we prove which completes the proof of the corollary.
¿From the corollary we have that for k → ∞ Note that following Regev [12, (F.4.5.1)] Now by (1.2) and (1.3) the right hand side of (4.18) equals The exit time result of the type like in Theorem 1.1 has also an application to the series of queues M/M→ . . .→M/1 with n−1 stations, with service rate µ i on the i-th station and arrival rate µ 0 .Correspondingly we consider independent Poisson processes X 1 t , . . ., X n t with intensity µ i−1 respectively and τ is the collision time.We observe that for 0 ≤ t ≤ τ , the queue size at the i-th station is if there is at t = 0, q i = x i−1 −x i > 0 jobs at i-th station.Thus in terms of the theory of queues, τ is the moment for the first time a station is empty.Define γ = ( n−1 j=0 µ j ) 1/n and Massey [9] showed that, if No exact asymptotic is known.However for the case when µ 0 = µ 1 = . . .= µ n−1 , that is not fulfilling conditions of Massey [9], our Theorem 1.1 shows the right asymptotics.The exact asymptotics for IP x (τ > t) in the case of independent but not necessarily identically distributed process X 1 t , . . ., X n t is an open problem.

Proof of Lemma 2.1
The proof is partitioned into lemmas.

.19)
Proof.Using Newton coefficients we write the LHS of (5.19) Next using elementary properties of determinants the above is , which can be written as σ∈Sn m 1 ,...,mn l 1 ,...,ln=0 and finally m 1 ,...,mn l 1 ,...,ln=0 Lemma 5.3 Let n be an even number.Suppose that for x ∈ IR n we have and Proof.We write . (5.22)We now show that the first v 0 − 1 coefficients vanish.Then the inside sum in (5.22), for |k| = v and k 1 ≤ . . .≤ k n , with the use of Lemma 5.1, can be transformed as follows (note that s i 's are odd) We now analyze the sum s 1 ,...,sn l 1 ,...,ln n i=1 For h l (x)h s−l (−x) = 0, both the sequences l, s − l ∈ Z n + must have different elements respectively.By Lemma 5.2, the minimal possible case is when l and s−l are permutations of {0, 1, . . ., n − 1}.This corresponds to Recall that since k has components k 1 ≤ . . .≤ k n , then for components of m we have m 1 ≤ . . .≤ m n and they are odd.Moreover we have |m| = 2|k| + n.For |k| = v 0 , the only admissible splits are of the following form.Let S eo n be the set of all permutations σ of (1, 2, . . ., n) such that σ(i) is odd if and only if i is even.We may identify this family with S n/2 × S n/2 .We define for s ∈ Z n + , σ(s) = (s σ(1) , . . ., s σ(n) ).Let l = l * = (0, 1, . . ., n − 1).Then {(l * , σ(l * ) : σ ∈ S eo n } has the property that components of s = l * + σ(l * ) are odd.However the components of l * + σ(l * ) are not always nondecreasing components, and therefore we must introduce another permutation σ , defined for a given s ∈ Z + , which makes the components of σ (s) nondecreasing.Let σ be defined by l * + σ(l * ).Then the set of all admissible entries is Fortunately, if σ is defined by l + s, then ¿From these considerations we see that . Notice that sign(σ) = (−1) n/2 sign(η)sign(ξ), where (−1) n/2 is responsible for n/2 transpositions from odds to evens, and that (−1) n/2 = (−1) n(n−1)/2 .Hence the above can be rewritten in the form: Using standard properties of determinants we write above as: The proof of lemma 2.1 is completed.

Proof of Lemma 2.2
We want to calculate constant C odd (n) defined in (2.8).Recall that this is the one which stands at h (1,2,...,n) (x) in polynomial W n 2 /4 (x).Since Pfaffian of a matrix is the square root of the corresponding determinant, it suffices to look for the constant in the asymptotic expansion of the determinant standing at h (0,...,n−1) (x)h (1,...,n) (x) at t −n(n−1)/2 and divide it by the already known constant C even (n).The following argument provides the proof of the uniqueness for this procedure, that is h 0,...,n−1 h 1,...,n cannot be represented as a combination of other Vandermondes.This means that if Since Schur polynomials form a basis and µ, ν is not a subtableaux of (1, . . ., 1) the above equality cannot be satisfied.The method of calculating is similar to the one presented before in Section 2.1.Since v 0 = n(n−1)

2
, by Lemma 5.3 the coefficient standing with Since we are only interested in h l (x)h s−l (x), where l is a permutation of (0, 1, . . ., n − 1) and s − l is a permutation of (1, 2, . . ., n), for which we also have Therefore s i cannot be less than 2k + 1, and so we have Using the same argumentation as in the proof of Lemma 2.1 we can assume that l = (0, 1, . . ., n − 1) and 2k + 1 − l is a permutation of (1, 2, . . ., n), which places even numbers into even places.We denote the resulting set of permutations by S ee n .Thus we have As in the proof of Lemma 5.3 we identify S ee n with the product of permutations S n/2 ×S n/2 of even numbers and permutations of odd numbers respectively.Hence the above expression can be written as η,ξ∈S n/2 i∈{2,4,...,n} We now recognize in the expression above the product of two determinants, so we rewrite it in the form .
Thus we conclude that .

Calculating constants
Let n be even.We will work out alternative expressions for constant (2.7): We now consider the determinant in the product above (with the substitution K We now demonstrate that our constants are consistent with the ones in Grabiner theorem.Following Mehta [10, p.  (see e.g.Abramowitz and Stegun [1, formula 6.1.12]), we have