Uniform bounds for exponential moment of maximum of a Dyck paths

Let D be a Dyck path chosen uniformly from the set of Dyck paths with 2n steps. We prove that the sequence of the exponential moments of the maximum of D normalized by the square root of n converges in the limit of infinite n, and therefore is bounded uniformly in n. This result justifies corresponding assumption used to prove certain estimates of high moments of large random matrices.


Introduction
Let N = {0, 1, 2, 3, . . . } be the set of non-negative integers. For any n ∈ N, we denote by W n the set of Bernoulli chains with n steps : 2n be the uniform distributions on W n and D 2n . The expectations with respect to these measures will be denoted by E (w) n and E (d) 2n . The aim of this note is to prove the following statement.
Theorem 1 For any λ > 0, we have where (e(t), t ∈ [0, 1])) is the normalized Brownian excursion. In particular, for any λ > 0 We may notice that the right hand side of (2) is finite for every λ by using the computation of Chung [2]: and then max t∈[0,1] e(t) possesses all exponential moments.  [12,15]. In these settings, the Catalan numbers (1) represent the moments of the eigenvalue distribution of random matrices of Wigner ensemble A N in the limit of their infinite dimension N → ∞ [15].
Recent studies [12,14] of high moments of large Wigner random matrices have used the exponential moments of the maxima of the Dyck paths. More precisely, it was shown that [8,14] E Tr A 2n where Tr denotes the trace of the square matrix, C 1 , C 2 and C 3 are certain constants, and It was assumed in [8,12,14] that lim sup n Q n (λ) is bounded. Theorem 1 above shows that this assertion is true.

Proof of the Theorem
Before proving the Theorem, we first discuss the appearance of max t∈[0,1] e(t) and the non-triviality of the result. Let C[0, 1] be the set of continuous functions defined on [0, 1] with real values. For any S ∈ W n , denote by u n = u S n the function in C[0, 1] obtained from S by interpolation and rescaling: Kaigh [7] where this results is shown for general increment distributions). By Then the uniform integrability argument is sufficient: given λ > 0, in order to prove that (6) implies (2), it suffices to show that for some ε > 0 (see Billingsley [1,Section 16]). Hence to prove the Theorem, using (6), only the second assertion (3) (weaker in appearance) needs to be proved. This is what we will do.
Remark. Smith and Diaconis [13] proved that P 2n (max e ≤ y) + O(n 1/2 ), and the convergence of moments of max S √ 2n under P (d) 2n to those of max e is also known (see [5] and references therein, where this is stated in link with the convergence of the height of random trees). These convergence results are not strong enough to obtain Theorem 1.
The strategy will be at first to transform the question in terms of Bernoulli bridges, and then to transform the question in terms of simple random walks where the answer is easy. The steps follow some ideas developed by Janson and Marckert [6] in their proof of their Lemma 1.

From Dyck paths to Bernoulli bridges
Let us introduce the set B n of "Bernoulli bridges" with n steps The quotes around "Bernoulli bridges" are there to signal that often the terms "Bernoulli bridges" concerns walks ending at 0 instead at −1. Clearly, B n is empty for even n and it is easy to see that #B 2n+1 = 2n+1 The cycle Lemma introduced by Dvoretzky and Motzkin [3] (see also Raney [9] and also Pitman [11], Section 6.1) allows one to relate quantities on Dyck paths and on Bernoulli bridges, and among other explains why (2n + 1)#D 2n = #B 2n+1 .
Consider the set of Dyck paths with size 2n with an additional last step −1 : Obviously there is a canonical correspondence between D ⋆ 2n+1 and D 2n , and this correspondence conserves the value of the maximum of the paths. Now, the left hand side of (8) is viewed to be the cardinality of D ⋆ 2n+1 × L1, 2n + 1 ] ]. We state the Cycle Lemma as follows: There exists a one-to-one correspondence Ψ 2n+1 between D ⋆ 2n+1 × L1, 2n + 1 ] ] and B 2n+1 and such that if S ′ = Ψ 2n+1 (S, k) for some k ∈ L1, 2n + 1 ] ], then We provide a proof of this classical result for reader's convenience.
It remains to explain why each bridge belongs to a unique rotation class: take a bridge S that reaches its minimum for the first time at time k. The walk S ′ whose list of increments is (∆ (i+2n+1−k) mod (2n+1) (S), i = 0 . . . , 2n) is a Dyck path. Thus, Ψ 2n+1 (S ′ , k) = S, and then S belongs to the rotation class of S ′ (and only to this one). As a conclusion, each rotation class contains a unique element of D ⋆ 2n+1 , has cardinality 2n + 1, and of course each element of D ⋆ 2n+1 belongs to a rotation class. Now it is easy to see that for S in D ⋆ 2n+1 and for any k ∈ L0, 2n ] ], the bridge Hence, the uniform distribution on B 2n+1 is the push-forward measure of the uniform distribution on D ⋆ 2n+1 × L1, 2n + 1 ] ] by Ψ 2n (which amounts to first choosing a Dyck path uniformly, and then a rotation). It follows from all these considerations that < +∞ if and only if sup We now show that this second assertion holds. 2n+1] and the Cauchy-Schwarz inequality

From bridges to simple walks
The idea here is to work on the half of the trajectory where the conditioning S 2n+1 = −1 will appear to be "not so important". Since a time reversal of Bernoulli bridges with size 2n + 1, followed by a symmetry with respect to the x-axis send B 2n+1 onto B 2n+1 and exchange the "two halves" of the trajectory, we just have to prove that for a = 0 and a = 1 We will treat the case a = 0 the other one being very similar. We equip the space W 2n+1 with the filtration F := (F k ) where F k is generated by the random variables (S 1 , . . . , S k ). Lemma 3. Let A n be an F n -measurable event. We have for a constant C 0 valid for all n (and all A n ).
Proof . The equality in this formula is clear since under P (w) 2n+1 , S is a Markov chain. Only the existence of C 0 is needed to be proved. In the following computations, we will use that since both P (w) 2n+1 and P (b) 2n+1 are the uniform distributions on their respective set, we have We will also use that under both P (w) 2n+1 and P n (A|S n = k). This gives the following chain of equalities n (A|S n = k)P , is indeed finite (as one may check using the Stirling formula). Hence, the right hand side in (14) is bounded by k P (w) n (A|S n = k) · C 0 P (w) n (S n = k) = C 0 P (w) n (A).
This ends the proof of the Lemma.
To conclude the proof of Theorem 1, we explain why (12) holds true. Using Lemma 3, we have Since Y S [0,n] is F n -measurable. The right hand side of (15) is much simpler than the left one, since it deals with simple random walks under the uniform distribution. Then using again Cauchy-Schwarz, it suffices to show that sup n E (w) n e 4λ max 0≤k≤n S k √ 2n < +∞ (16) and the same thing for max replaced by min (which gives the same quantity). Now, by the André reflexion principle (see Feller [4], page 72), we have P (w) n ( max 0≤k≤n S k ≥ x) = 2P (w) n (S n > x) + P (w) n(S n = x) ≤ 2P (w) n (S n ≥ x).