Computation of Moments for the Length of the One Dimensional ISE Support

: We consider in this paper the support [ L 0 ; R 0 ] of the one dimensional Integrated Super Brownian Excursion. We determine the distribution of ( R 0 ; L 0 ) through a modiﬂed Laplace transform. Then we give an explicit value for the ﬂrst two moments of R 0 as well as the covariance of R 0 and L 0


Introduction and results
The motivation of this work comes from the paper of Chassaing and Schaeffer [2]. They prove the rescaled radius of random quadrangulation converges (in law), as the number of faces goes to infinity, to the width r = R − L of the one dimensional Integrated Super Brownian Excursion (ISE) support [L , R ] (see [1] and the reference therein for the definition of the ISE). They also prove the convergence of moments. As pointed by Aldous in [1], little is known about the law of r . It is of particular interest to compute therefore the law of r = R − L as well as its moments. We recall that L ≤ 0 ≤ R a.s., and that by symmetry, R and −L are equally distributed. More precisely we give a sort of Laplace transform of R in the following proposition, which is proved in section 4. For b = 1, from the uniqueness of the Laplace transform, we deduce the function r → r −3/2 P(R > r −1/4 ), hence the law of R , is uniquely defined by the above equation. However, this does not allow us to give explicitly the law of R . We derive in section 4, the first two moments of R .

Corollary 2. We have
Multiplying equation (1) by b α−1 , with α > 2, and integrating w.r.t. b on (0, +∞) gives the following corollary. In section 5, we prove the following result. where u λ,r is the unique non-negative solution of the non-linear differential equation Then, one can derive the expectation of R |L | (see section 5).
To prove those results, we use the fact that the ISE has the same distribution (up to a constant scaling) as the total mass of an excursion of the Brownian snake conditioned to have a duration σ of length 1. Then we compute under the excursion measure of the Brownian snake the joint law of σ, R and L . In particular we use the special Markov property of the Brownian snake and the connection between Brownian snake and PDE The next section is devoted to the presentation of the Brownian snake and its link with ISE.

Brownian snake and ISE
Let W be the set of stopped continuous paths (w, ζ w ) defined on R + with values in R. ζ w ≥ 0 is called the lifetime of the path w, and w is a continuous path w = (w(t), t ≥ 0) defined on R + with values in R and constant for t ≥ ζ w . We sometimes write w for (w, ζ w ). We define It is easy to check that d is a distance on W, and that (W, d) is a Polish space. We shall denote by N x [dW ] the excursion measure on W of the Brownian snake W = (W s , s ≥ 0) started at x ∈ R with underlying process a linear Brownian motion. We refer to [5] for the definition and properties of the Brownian snake. We recall that under N x , the law of the lifetime process ζ = (ζ s , s ≥ 0) is the Itô measure, n + , on positive excursions of linear Brownian motion, where we take the normalization N x [sup s≥0 ζ s > ε] = 1 2ε . Under N x , conditionally on the lifetime process, W is a continuous W-valued Markov process started at the constant path (with lifetime zero) equal to x ∈ R. Conditionally, on the lifetime process and on (W u , u ∈ [0, s]), the law of W s , with s ≥ s is as follow: the two paths W s and W s coincide up to time m = inf u∈[s,s ] ζ u , and (W s (t + m), t ≥ 0) is a linear Brownian motion, constant after ζ s − m, which depends on (W u , u ∈ [0, s]) only through its starting point W s (m) = W s (m).
We define σ = inf{s > 0; ζ s = 0} the duration of the excursion. From the normalization of N x , we deduce that σ is distributed according to It is easy to check that for any x ∈ R, The Brownian snake enjoys a scaling property: if λ > 0, the law of the process W We now recall the connection between ISE and Brownian snake. There exists a unique collection N (2) For every λ > 0, r > 0, F , nonnegative measurable functional on C(R + , W), (3) For every nonnegative measurable functional F on C(R + , W), We take the opportunity to stress a misprint in [3], where 1/2 is missing in the right member of formula (3). The measurability of the mapping r → N Let R = {W s (t); 0 ≤ t ≤ ζ s , 0 ≤ s ≤ σ} be the range of the Brownian snake. Since we are in dimension one, using the continuity of the paths, we get that N x -a.e., the range is an interval which we denote by [ The law of the ISE is the law of the continuous tree associated to √ 2W , under N 0 (see corollary 4 in [4] and [1], see also [7] section IV.6). In particular the law of the support of ISE is the law of is the maximal non-negative solution of u = 4u 2 in (a, b). But,as we said, we need the law of R under N (1) x . Therefore, we need the joint law of R and σ. We shall compute N x [1 − 1 R⊂[a,b] e −λσ ] using Brownian snake techniques (see the key lemma 6 and its consequences lemmas 7 and 8). From this, thanks to scaling properties mentioned above, we will deduce a (kind of) Laplace transform for N (1) We prove in section 4 the following result: for λ > 0, b > 0, which determines the law of R. We also compute In section 5, lemma 8, we compute for

Exit measure of the Brownian snake
We refer to [6] for general definition and properties of the exit measure of Brownian snake. Let −∞ ≤ a < x < b ≤ +∞, and consider X (a,b) the exit measure of the Brownian snake of (a, b) under N x . We recall that X (a,b) is a random measure on {a, b}, defined by : for any nonnegative measurable function ϕ defined on R, The continuous additive functional of the Brownian snake L increases when the path W s dies as it reaches for the first time the boundary of (a, b). It is well known that in The next lemma is proved in section 6 using the link between Brownian snake and non-linear PDE.
As an application of this lemma, we have the following result.
Since N x -a.e., for x < b, By translation and symmetry, it is clear that v µ,λ,−∞,b (0) = v µ,λ,0,∞ (b). As w λ (b) is the increasing limit of v µ,λ,0,∞ (b) as µ → ∞, and since the set of nonnegative solutions of (4) is closed under pointwise convergence (see proposition 9 (iii) in section V.3 of [7], stated for λ = 0, which can be extended to the case λ > 0), we deduce that w λ also solves (4) with (a, b) = (0, +∞). Notice that and that, since R is a compact set N 0 -a.e., Therefore w λ is solution of This ordinary differential equation has a unique nonnegative solution, which is given by

Proof of proposition 1 and corollary 2
Proof of proposition 1. By scaling, we get that the law of R under N (r) 0 is the law of Rr 1/4 under N (1) 0 . We deduce from (3) and the above scaling property, that and we get Since R is distributed as √ 2R, this proves proposition 1.
Proof of corollary 2. By monotone convergence, letting λ decrease to 0, we get : Set u = r −1/4 and b = 1, to get Since R is distributed as √ 2R, we get the second equality of corollary 2. We recall the following development near 0: Hence the function 1 On the other hand, by Fubini, we have we have Since R is distributed as √ 2R, we get the first equality of corollary 2.
By symmetry, we deduce that w λ,r (r) = 0. The ordinary differential equation has a unique nonnegative solution. However, we don't have an explicit formula for w λ,r . Arguing as in the proof of proposition 1, we get with c = −a > 0, We set for b > 0, c > 0, That is where r 0 = (c + b)/2. We have proved the following lemma, and thus proposition 4 (with the notation u λ,r (x) = w λ,r/ √ 2 (x/ √ 2)).
Proof of the second equality of corollary 5. In particular, taking c = b and letting λ decreases to 0 in (9), we get .
Solving the differential equation (7) for w 0,r , we get for all t ∈ (0, r), that is For t = r, we get 3 8 We define We get that We then deduce from (6) and (10), that If we set u = r −1/4 and b = 1, we get from (10) Therefore, we have N This prove the second equality of corollary 5.
Proof of the first equality of corollary 5. We look for a transformation of J λ which will give the expectation of R |L|. Notice first that for λ > 0, J λ is differentiable in the variable λ and that Our next task is to compute ∂ λ J λ (c, b). From (8), we deduce the following scaling property: for ρ > 0, Taking ρ = λ −1/4 , we get Of course, we have a similar scaling property for H. Differentiating with respect to λ, we get A similar computation yields Now we will study the limit of C ε,A = [ε,A] 2 dcdb (−λ∂ λ J λ (c, b)) as A → ∞ and ε → 0, since An integration by parts gives 4C ε,A = K 1 + K 2 + K 3 + K 4 , where Study of K 1 . We have In particular, we have (14) lim ε→0,A→∞ This implies From the study of K 1 , we deduce that (15) lim ε→0,A→∞ Study of K 3 . Set εt = b and use the scaling property of I and H (with ρ = ε) to get where we used the definition of I and H for the last equality. By monotone convergence, we get that −K 3 increases, as ε ↓ 0 and A ↑ ∞ to −K 3 , where Therefore, we deduce that Study of K 4 . We have for A ≥ 1, Thus we have Conclusion. Eventually, we deduce from (13), (14), (15), (18) and (19), that

Proof of lemma 6
We introduce the special Markov property for X (a,b) and we refer to [6] for the complete theory. We set where τ (w) = inf{t ∈ [0, ζ w ], w(t) ∈ (a, b)} is the first exit time of (a, b) for the path w ∈ W. We define the continuous process W s by W s = W ηs . By definition, the σ-field E (a,b) is generated by W = (W s , s ≥ 0) and all the N x -negligible sets.
From proposition 2.3 in [6], we get that X (a,b) is E (a,b) -measurable. Notice that N x -a.e., {s ≥ 0; ζ s = τ (W s )} is of Lebesgue measure zero and In particular the integral The random open set {s ∈ (0, σ); τ (W s ) < ζ s } is a countable union of open sets, say i∈I (a i , b i ). Since the set {s ∈ (0, σ); τ (W s ) = ζ s } is of Lebesgue measure zero N x -a.e., we get that N x -a.e., In particular, we deduce from the special Markov property, theorem 2.4 in [6], that From (2) we get that where Y (a,b) = X (a,b) (dx). This implies that for a < x < b, and λ ≥ 0, µ > 0, Now we replace exp(− σ s dL u ) by its predictable projection with respect to the filtration of the Brownian snake. Let E * w be the law of the Brownian snake started at the path (w, ζ w ), and whose lifetime is distributed according to a linear Brownian motion started at point ζ w and stopped as it reaches 0. The predictable projection of exp (− Let us recall the first moment formula for the Brownian snake: for F a nonnegative measurable function defined on W, we have From standard arguments on Brownian motion, we deduce lemma 6.