Asymptotics of Bernoulli random walks, bridges, excursions and meanders with a given number of peaks

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by  Jean-Maxime Labarbe

Asymptotics of Bernoulli random walks, bridges, excursions and
meanders with a given number of peaks
Jean-Maxime Labarbe Jean-Franc¸ois Marckert
Universit´e de Versailles CNRS, LaBRI
45 avenue des Etats-Unis Universit´e Bordeaux 1
6 78035 Versailles cedex 351 cours de la Lib´eration
0
[email protected] 33405 Talence cedex
0
2 [email protected]

n
a
J

5 Abstract
2
A Bernoulli random walk is a random trajectory starting from 0 and having i.i.d. in-

]
R crements, each of them being +1 or 1, equally likely. The other families cited in the title
−
P are Bernoulli random walks under various conditionings. A peak in a trajectory is a local
h. maximum. In this paper, we condition the families of trajectories to have a given number
t of peaks. We show that, asymptotically, the main effect of setting the number of peaks is
a
m to change the order of magnitude of the trajectories. The counting process of the peaks,
[ that encodes the repartition of the peaks in the trajectories, is also studied. It is shown

1 that suitably normalized, it converges to a Brownian bridge which is independent of the
v limiting trajectory. Applications in terms of plane trees and parallelogrampolyominoes are
4
also provided.
2
6
1
0 1 Introduction
6
0
/ Let N = 0,1,2,3,... be the set of non-negative integers. For any n N, we denote by
h n
{ } ∈ W
t the set of Bernoulli chains with n steps :
a
m
v: Wn = {S= (S(i))0≤i≤n :S(0) = 0,S(i+1)= S(i)±1 for any i ∈J0,n−1K}.
i
X
The sets of Bernoulli bridges , Bernoulli excursions , Bernoulli meanders with n steps
n n n
r B E M
a are defined by
= S : S ,S(n)= 0 ,
n n
B { ∈ W }
= S : S ,S(n)= 0,S(i) 0 for any i J0,nK ,
n n
E { ∈ W ≥ ∈ }
= S : S ,S(i) 0 for any i J0,nK .
n n
M { ∈ W ≥ ∈ }
The cardinalities of these sets are given by
2n 1 2n n
# = 2n, # = , # = , # = , (1)
n 2n 2n n
W B n E n+1 n M n/2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18)⌊ ⌋(cid:19)
1
and for every odd number n, = = ∅. The two first formulas are obvious, the third can be
n n
B E
proved for instance thanks to the cyclical lemma (see also the 66 examples of the appearance of
the Catalan numbers # in combinatorics in Stanley [21, ex. 6.19 p.219]), and the last one,
2n
E
may be proved iteratively or thanks to a bijection with Bernoulli bridges (see Section 4.4).
Let n N. For every S , the set of peaks of S, denoted by S , is defined by
n
∈ ∈ W ∧
S = x : x J1,n 1K, S(x 1) = S(x+1) =S(x) 1 .
∧ { ∈ − − − }
Theset ( S) is called the set of valleys of S : it is easy to check that for any S, #S #( S)
− ∧ ∧− − ∧
belongs to +1,0, 1 . The value of this difference depends only on the signs of the first and
{ − }
(k)
last steps of S. In this paper, we focus only on the number of peaks and we denote by
n
W
(k) (k) (k)
(resp. , and ) the subset of , (resp. , , ) of trajectories having exactly
n n n n n n n
B E M W B E M
k peaks (for any k > n/2 these sets are empty). We have
⌊ ⌋
Proposition 1 For any k 0 and any n 0,
≥ ≥
2
n+1 n 1 n n n/2 n/2
# (k) = , # (k) = , # (k) = , # (k) = ⌊ ⌋ ⌈ ⌉ ,
Wn n 2k B2n k E2n n k k 1 Mn k k
(cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) − (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
where, by convention
m!
m if p and m are non negative integers and p J0,mK,
= p!(m p)! ∈
p  −
(cid:18) (cid:19)  0 in any other cases.

(3) (2) (3) (4)
Figure 1: Trajectories from , , , and . Black dots correspond to peaks.
W12 B12 E12 M12
(k)
Theformulagiving# isduetoNarayana[16]computedinrelationwithpairsofk-compositions
E2n
of n satisfying some constraints (see also Stanley [21, ex. 6.36 p.237]).
Let Pw, Pb, Pe and Pm be the uniform law on , , , and and Pw,(k), Pb,(k), Pe,(k)
n n n n Wn Bn En Mn n n n
and Pm,(k) be the uniform law on (k), (k), (k) and (k). For x w,b,e,m , a random
n n n n n
variable under Px,(k) is then simplyWa ranBdom vEariable uMnder Px cond∈itio{ned to h}ave k peaks.
n n
We are interested in the asymptotic behavior of random chains under the distributions Px,(k),
n
when n and k = go to infinity.
n
K
2
Let C[0,1] be the set of continuous functions defined on [0,1] with real values. For any
S , denote by u the function in C[0,1] obtained from S by the following rescaling:
n n
∈ W
1
u (t) = S( nt )+ nt (S( nt ) S( nt )) for any t [0,1]. (2)
n
√n ⌊ ⌋ { } ⌈ ⌉ − ⌊ ⌋ ∈
(cid:0) (cid:1)
Wecall Brownianbridgeb,Brownian excursioneandBrownian meandermthe(normalized)
processes characterized as follows : let w be a 1-dimensional standard Brownian motion. Let
d = inf t : t 1,w = 0 and g = sup t : t 1,w = 0 . Almost surely, we have g d > 0,
t t
{ ≥ } { ≤ } −
g (0,1). The processes b,e and m have the following representations :
∈
(bt)t [0,1] (=d) wgt , (et)t [0,1] (=d) |wd+(g−d)t| , (mt)t [0,1] (=d) |w(1−g)t| .
∈ √g t [0,1] ∈ √d g t [0,1] ∈ √1 g t [0,1]
(cid:16) (cid:17)∈ (cid:16) − (cid:17) ∈ (cid:16) − (cid:17)∈
As a consequence of the Donsker [5] theorem (for x = w), Kaigh [12] (for x = b), Iglehart [13]
(for x = m) and Belkin [2] (for x = e),
Theorem 2 For any x w,b,e,m , under Px, u (d) x in C[0,1] endowed with the topology
∈ { } n n −−n→
of the uniform convergence.
In the case x b,e , even if non specified, it is understood that n + in 2N.
∈{ } → ∞
In fact, Theorem 2 can be proved directly, thanks to the elementary enumeration of paths
passingvia someprescribedpositions inthemodelof Bernoullipaths. Themethodusedtoshow
the tightnesses in our Theorem 4 may be used to prove the tightness in Theorem 2; thanks to
some probability tricks, this reduces to show the tightness under Pw, which is simple.
n
The finite dimensional distributions of w,e,b and m are recalled in Section 3.1. Numerous
relations exist between these processes, and their trajectories, and a lot of parameters have
been computed. We refer to Bertoin & Pitman [6], Biane & Yor [4], Pitman [18] to have an
overview of the subject. These convergences have also provided some discrete approaches to the
computation of values attached to these Brownian processes, and the literature about that is
considerable, see e.g. Csa´ki & Y. Hu [7], and reference therein.
We introduce the counting process of the number of peaks : for any S , denote by
n
∈ W
Λ(S) = (Λ (S)) the process :
l l J0,nK
∈
Λ (S) = #S J0,lK for any l J0,nK. (3)
l
∧∩ ∈
For S , Λ (S) = #S is simply the total number of peaks in S. We have
n n
∈ W ∧
Proposition 3 For any x w,b,e,m , under Px,
∈ { } n
Λn n/4 (d)
− (0,1/16),
√n −−n→ N
where (0,1/16) denotes the centered Gaussian distribution with variance 1/16.
N
3
We will now describe the main result of this paper. Its aim is to describe the influence of
the number of peaks on the shape of the families of trajectories introduced above. We will then
conditionthedifferentfamilies by#S = forageneralsequence( )satisfyingthefollowing
n n
∧ K K
constraints :
(H) = For any n, N, lim = + , limn/2 = + .
n n n
K ∈ n K ∞ n −K ∞
(cid:16) (cid:17)
Notice that for every S , #S J0, n/2 K and then (H) is as large as possible to avoid
n
∈ W ∧ ∈ ⌊ ⌋
that the sequences and n/2 have a finite accumulation point.
n n
K −K
We set p := 2 /n and
n n
K
β := n(1 p )/p , and γ := np (1 p )= p β . (4)
n n n n n n n n
− −
p p
Each peak can be viewed to be made by two consecutive steps; hence, if you pick at random one
step of a trajectory under Pnx,(Kn), the probability that this step belongs to a peak is pn.
We consider S and Λ(S) as two continuous processes on [0,n], the values between inte-
ger points being defined by linear interpolation. The normalized versions of S and Λ(S) are
respectively denoted by s and λ :
n n
S(nt) Λ t
nt n
s (t) := and λ (t):= 2 − K . (5)
n n
β γ
n n
Theorem 4 If (H) is satisfied, for any x w,b,e,m , under Pnx,(Kn),
∈ { }
(d)
(s ,λ ) (x,b) (6)
n n
−−n→
where bis a Brownian bridge independent of x and wherbe the weak convergence holds in C([0,1])2
endowed with the topology of uniform convergence.
b
Hence, under Pnx,(Kn), up to the scaling constant, the process sn behaves as under Pnx. The
normalizingfactor β , thatwillbeexplained later in thepaper,indicates theorderof magnitude
n
of the process S under Pnx,(Kn) (βn is a decreasing function of n). The normalizing constant γn
K
is smaller than n/4 whatever is p ; γ gives the asymptotic order of the “linearity defect” of
n n
t Λ . The fact that (λ ) converges to a Brownian bridge independent of the limit trajectory
nt p n
7→
is quite puzzling. For example, under Pe, one would expect that only few peaks appear in a
n
neighborhood of 0, this resulting in a negative bias in λ near 0. This must be true, but this
n
bias is not important enough to change the asymptotic behavior of λ .
n
A second direct corollary of Theorem 4 is stated below:
Corollary 5 For any x w,b,e,m , under Px, we have
∈ { } n
S(nt) Λnt tn/4 (d)
, 4 − (x ,w ) (7)
t t t [0,1]
√n √n −−n→ ∈
(cid:18) (cid:19)t [0,1]
∈
where w is a Brownian motion independent of x and wherebthe weak convergence holds in
C([0,1])2 endowed with the topology of uniform convergence.
b
4
Theorem 2 is of course a consequence of Corollary 5.
Proof. For any S , set q (S) = 2Λ (S)/n, β˜ (S) = n(1 q (S))/q (S), and γ˜(S) :=
n n n n n n n
∈ W −
nq (S)(1 q (S)). We have
n n p
−
p S(nt) Λ tn/4 S(nt) β˜ γ˜ Λ tΛ Λ n/4
nt n n nt n n
, 4 − = , 2 2 − +4t − . (8)
(cid:18) √n √n (cid:19) β˜n √n √n (cid:18) γ˜n (cid:19) √n !
By Proposition 3 and Theorem 4, under Px, the five-tuple
n
Λ n/4 S(nt) Λ tΛ γ˜ β˜
n nt n n n
4 − , ,2 − ,2 ,
√n (cid:18) β˜n (cid:19)t [0,1] (cid:18) γ˜n (cid:19)t [0,1] √n √n!
∈ ∈
converges in distribution to
(N,(s ) ,(λ ) ,A,B)
t t [0,1] t t [0,1]
∈ ∈
where N is a centered Gaussian random variable with variance 1, and where, conditionally on
(d)
N, (s,λ) = (x,b) where x and b are independent, b is a Brownian bridge, and A and B are
two random variables equal to 1 a.s.. By (8), S√(nnt),4 Λnt√−ntn/4 converges to (x,(bt+tN)t [0,1])
b b b ∈
whereN is independentof x and b, and thenth(cid:16)e resultfollows,(cid:17)since(b +tN) is astandard
t t [0,1]
∈ b
Brownian motion. (cid:3)
b b
Consequences in terms of plane trees
Consider the set T of plane trees (rooted ordered trees) with n edges (we refer to [1, 14] for
n
more information on these objects). There exists a well known bijection between T and
n 2n
E
which may be informally described as follows. Consider a plane tree τ T (see Figure 2), and
n
∈
a fly walking around the tree τ clockwise, starting from the root, at the speed 1 edge per unit of
time. Let V(t) be the distance from the root to the fly at time t. The process V(t) is called in
the literature, the contour process or the Harris’ walk associated with τ. The contour process
Figure 2: A plane tree and its contour process
is the very important tool for the study of plane trees and their asymptotics and we refer to
Aldous [1], Pitman [19, Section 6], Duquesne & Le Gall [10], Marckert & Mokkadem [14] for
considerations on the asymptotics of normalized trees. It is straightforward that the set of trees
(k) (k)
encoded by is the subset of T of trees having exactly k leaves (say T ), a leaf being a
E2n n n
5
node without any child. A corollary of Theorem 4, is that random plane tree with n edges and
leaves, converges, normalized by β /2, to the continuum random tree introduced by Aldous
n n
K
[1], which is encoded by 2e. The variable Λ gives the number of leaves visited at time nt.
nt
proba.
By Theorem 4, sup (Λ t )/n 0. This translates the fact that the leaves are
t∈[0,1]| nt − Kn | −→
asymptotically uniformly distributed on a random tree chosen equally likely in Tn(Kn).
Consequences in terms of parallelogram polyominoes
We refer to Delest & Viennot [8] for more information on parallelogram polyominoes. Unit
squareshavingtheirverticesatintegerpointsintheCartesianplanearecalledcells. Apolyomino
is a finite union of cells with connected interior. The number of cells is the area and the length
of the border is called the perimeter (see Figure 3). A polyomino is said to be convex if
P
the intersection of with any horizontal or vertical line is a convex segment. For any convex
P
polyomino there exists a minimal rectangle R( ) (that can be seen as a convex polyomino)
P P
containing . Then touches the border of R( ) along four connected segments. A convex
P P P
polyomino is said to be a parallelogram polyomino if the south-west point and the north-east
P
point of R( ) belongs to (see Figure 3). Let denote by H( ) and V( ) the horizontal and
P P P P
Figure 3: The first convex polyomino is not parallelogram, the second is. Their areas are 9
and 11, their perimeters equal that of their minimal rectangles, here 18. For both polyominoes
H( ) = 4 and V( ) = 5. The last picture represents the Bernoulli excursion associated by ρ
P P
with the parallelogram polyomino.
vertical length of the border of R( ), and let Pol be the set of parallelogram polyominoes with
n
P
perimeter n.
Proposition 6 (Delest & Viennot [8, Section 4]) For any integer N 1, there is a bijection ρ
≥
between and Pol , such that if = ρ(S), the area of is equal to the sum of the heights
2N 2N+2
E P P
of peaks of S, moreover #S = H( ), and V( ) = 2N +2 2#S (where 2N 2#S is the
∧ P P − ∧ − ∧
number of steps of S that do not belong to a peak).
By symmetry with respect to the first diagonal, the random variables V( ) and H( ) have the
P P
same distribution when is taken equally likely in Pol . Hence, the proposition says that
2N+2
P
under Pe, 2N +2 2#S and #S have the same distribution.
n − ∧ ∧
6
WedescribeinafewwordsDelest&Viennot’sbijection: thesuccessivelengthsofthecolumns
of the polyomino give the successive heights of the peaks of S. The difference between the
P
heights of the floor-cells of the ith and i+1th columns of plus one gives the number of down
P
steps between the ith and i+1th peaks of S.
For i 1,...,H( ) , let v ( ) be the number of cells in the ith column of . The values
i
∈ { P } P P
(vi(P)i∈J1,H(P)K coincide with the ordered sequence (Si)i∈S∧. Let PP(Konl()2n+2) be the uniform law
on the set of parallelogram polyominos with perimeter 2n+2 and width (that is such that
n
K
H( ) = ). Assume that v is interpolated between integer points, and v(0) = 0. We have
n
P K
Proposition 7 If (Kn) satisfies (H), under PP(Konl()2n+2)
v( nt) (d)
K (e )
t t [0,1]
(cid:18) βn (cid:19)t [0,1] −→ ∈
∈
in C[0,1] endowed with the topology of uniform convergence.
Proof. Let (V ) be the successive height of the peaks in S. Assume also that
i i 1,..., n
∈{ K }
V(0) = 0 and that V is interpolated between integer points. By Delest & Viennot’s bijec-
tion, βn−1v(Kn.) under PP(Konl()2n+2) has the same distribution as βn−1V(Kn.) under Pne,(Kn). Since
(d)
(β 1S(nt)) (e ) , to conclude, it suffices to show that
n− t∈[0,1] −→ t t∈[0,1]
V( nt) S(nt) proba.
sup K − 0. (9)
β −→
t [0,1](cid:12) n (cid:12)
∈ (cid:12) (cid:12)
(cid:12) (cid:12)
Let J(i) be (abscissa of) the ith peak(cid:12)in S. We have, f(cid:12)or any t 0,1/ n,..., n/ n ,
∈{ K K K }
V( t) S(nt) = S(J( t)) S(nt). (10)
n n
K − K −
As one can see using the convergence of λ to b,
n
sup J(Knt)b−nt proba. 0. (11)
n −→
t (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) pr(cid:12)oba.
Indeed, sup J( t) nt /n sup Λ(cid:12) t /n (cid:12) 0. Since (s ) converges in C[0,1] under
Pne,(Kn), by at|simKpnle a−rgum|ent≤aboutt|itnstm−odKunlu|s of−c→ontinuity, usinng (10) and (11), formula (9)
holds true. (cid:3)
We would like to point out the work of de Sainte-Catherine & Viennot [9], who exhibit a
quiteunexpectedlinkbetweenthecardinalities ofexcursionshavingtheirpeaksinagivensubset
of N and the famous Tchebichev polynomials.
2 Combinatorial facts : decomposition of trajectories
The decomposition of the trajectories is determinant in our approach, since we will prove
directly the convergence of finite dimensional distributions under Px. An important difference
n
7
with the case where the peaks are not considered is that under Px,(k), S is not a Markov chain.
n
Indeed, the law of (S(j)) dependson S(l), on the number of peaks in J0,lK, and also on the
l j n
≤ ≤
step (S(l 1),S(l)). The computation of the distributions of vector (S(t ),...,S(t )) in these
1 k
−
various cases admits the complication that S may own a peak in some of the positions t ,...,t .
1 k
In order to handle the contribution of these peaks, we have to specify what the type u or d (+1
or 1) of the first and last steps of the studied parts (S(t ),...,S(t )) is.
i i+1
−
We set the following notation : ∆S = S(l) S(l 1), and write for convenience, ∆S = u
l l
− −
when ∆S = 1, and ∆S = d when ∆S = 1. In this paper, we deal only with discrete
l l l
−
trajectories S such that ∆S +1, 1 for any k. We will not recall this condition.
k
∈ { − }
For a and b in d,u , and l,x,y,j in Z, set
{ }
Tj (l,x,y) = S : S= (S(i)) , #S = j, ∆S = a, ∆S = b, S(0) = x, S(l)= y
ab { 0≤i≤l ∧ 1 l }
Tj,≥(l,x,y) = S : S Tj (l,x,y), S(i) 0 for any i J0,lK .
ab { ∈ ab ≥ ∈ }
For any l,j ,j ,x,y Z, set
1 2
∈
l+y x l y+x
− 1 − 1
l,j ,j ,x,y = 2 − 2 − .
1 2
j j
(cid:18) 1 (cid:19)(cid:18) 2 (cid:19)
(cid:2) (cid:3)
We have
Proposition 8 For any y Z, l 0, j 0,
∈ ≥ ≥
j
#T (l,0, y)= l, j , j , 0, y . (12)
ab −1b=d −1a=u
(cid:2) (cid:3)
For any x 0, y 1, l 0, j 0,
≥ ≥ ≥ ≥
#Tj, (l, x, y) = l, j, j 1, x, y l, j 1, j, x, y ,
uu≥ − − − −
j,
#T ≥(l, x, y) = (cid:2)l, j 1, j 1,(cid:3)x, y(cid:2) l, j 2, j, (cid:3)x, y ,
ud − − − − −
(cid:2) (cid:3) (cid:2) (cid:3)
For any x 1, y 1, l 0, j 0,
≥ ≥ ≥ ≥
j,
#T ≥(l, x, y) = l, j, j, x, y l, j, j, x, y ,
du − −
j,
#T ≥(l, x, y) = (cid:2)l, j 1, j,(cid:3)x, y(cid:2) l, j 1, (cid:3)j, x, y
dd − − − −
(cid:2) (cid:3) (cid:2) (cid:3)
j, j,
(notice that #Tdb≥(l,0,y) = 0 and #Tau≥(l,x,0) = 0). In other words
j,
#Tab≥(l,x, y)= l, j −1b=d, j −1a=u, x, y − l, j −1a=u −1b=d, j, −x, y . (13)
(cid:2) (cid:3) (cid:2) (cid:3)
Proof. Let n,k N. A composition of n in k parts is an ordered sequence x ,...,x of non
1 k
∈
negative integers, such that x + +x = n. The number of compositions of n in k parts (or
1 k
···
of compositions of n+k in k positive integer parts) is n+k 1 .
k −1
−
(cid:0) (cid:1)
8
We call run of the chain S= (S(i)) , a maximal non-empty interval I of J1,lK such that
0 i l
≤≤
(∆S ) is constant. The trajectories S of Tj (l,y) are composed by j + runs of u and
i i∈I ab 1b=u
j + runs of d. The u-runs form a compositions of (l +y)/2 (this is the number of steps
a=d
1
u) in positive integer parts, and the d-runs form a composition of (l y)/2 in positive integer
−
parts. Hence,
(l+y)/2 1 (l y)/2 1
j
#T (l,y) = − − − ,
ab j + 1 j + 1
(cid:18) 1b=u− (cid:19)(cid:18) 1a=d− (cid:19)
and Formula (12) holds true.
The proofs of the other formulas are more tricky; the main reason for it is that the reflexion
principle does not conserve the number of peaks. What still holds is, for any x 0, y 0,
≥ ≥
j 0, l 0
≥ ≥
j, j j,(cid:8)
#T ≥(l,x,y) = #T (l,x,y) #T (l,x,y)
ab ab − ab
j,(cid:8) j
whereT (l,x,y) is theset of trajectories belongingtoT (l,x,y) that reach thelevel 1. Since
ab ab −
j j j,(cid:8)
#T (l,x,y) = #T (l,0,y x) is known, it remains to determine T (l,x,y).
ab ab − ab
We define two actions on the set of chains :
(cid:4) letS= (S(i))i∈J0,lK ∈ Wl. Foranyt ∈J0,lKwedenotebyS′ = Ref(S,t)thepathS′ = (Si′)0≤i≤l
obtained from S by a reflexion from the abscissa t; formally :
S (i) = S(i) for any 0 i t,
′
≤ ≤ .
( S′(i+t)= 2S(i) S(i+t) for any 0 i l t
− ≤ ≤ −
Wheng isafunctionfrom takingitsvaluesinJ0,lK,wewritesimplyRef(.,g) forthereflexion
l
W
at abscissa g(S).
(cid:4) letS= (S(i))i J0,lK l. ForanycanddinN,0 c d l,wedenotebyS′ = Cont(S,[c,d])
∈ ∈ W ≤ ≤ ≤
the path S = (S (i)) obtained from S by a contraction of the interval [c,d] :
′ ′ 0 i l d
≤≤−
S (i) = S(i) for any 0 i c,
′
≤ ≤ .
( S′(c+i)= S(d+i) S(d)+S(c) for any 0 i l (d c)
− ≤ ≤ − −
As before, we write Cont(.,[g ,g ]) for the contraction of the interval [g (S),g (S)].
1 2 1 2
We denote by (S) = inf j : S(j) = 1 the hitting time of 1 by S. We proceed to a
1
T− { − } −
classification of the paths S from Tj,(cid:8)(l,x,y) according to the two first steps following (S),
ab T−1
that exist since y is taken positive. We encode these two first steps following above the
1
T−
symbol T : for any α,β u,d , set
∈ { }
j,(cid:8)
(αβ)
T ab (l,x,y) = {S :S∈ Tajb,(cid:8)(l,x,y), ∆ST−1(S)+1 = α, ∆ST−1(S)+2 = β}.
(αβ)
j,(cid:8)
Hence, T (l,x,y) is the union of four elements of that type. Let us compute # T for any
ab ab
α,β.
9
t
c d l
t c l (d c)
− −
Figure 4: On the first column S and Ref(S,t), on the second column S and Cont(S,[c,d])
For any a u,d , x 0, y 1, the following bijections (denoted by ) hold
∈ { } ≥ ≥ ↔
j,(cid:8) j,(cid:8)
(du) (ud)
T ad (l,x,y) ↔ T au(l,x,−2−y) ↔ Taju−1(l−2,x,−2−y)
j,(cid:8) j+1,(cid:8)
(du) (ud)
j
T (l,x,y) T (l,x, 2 y) T (l 2,x, 2 y).
au ↔ ad − − ↔ ad − − −
On each line, the first bijection is Ref(., ), the second one is Cont(.,[ , +2]). Notice
1 1 1
T− T− T−
that this last operation does not create any peak because ∆S −1(S) = d. The cardinalities of
T
(du) (du) (du) (du)
the sets in the right hand side are known, hence, T dd, T ud, T du, T uu are determined. Set
j,(cid:8) j,(cid:8) j,(cid:8)
(α) (αu) (αd)
T (l,x,y) = T (l,x,y) T (l,x,y). We have
ab ab ∪ ab
j,(cid:8) j 1,(cid:8)
(u) (d) −
T ad (l,x,y) ↔ Tau (l,x,−y−2) ↔ Taju−1(l−1,x,−y−1)
j,(cid:8) j,(cid:8)
(u) (d)
j
T (l,x,y) T (l,x, y 2) T (l 1,x, y 1)
au ↔ ad − − ↔ ad − − −
The first bijection is Ref(., ), the second one Cont(.,[ , +1]). Now,
1 1 1
T− T− T−
j,(cid:8) j+1,(cid:8)
(dd) (ud)
# T = # T (l,x,y+2)
ab ab
j+1,(cid:8) j+1,(cid:8)
(u) (uu)
= #T (l,x,y+2) # T (l,x,y+2)
ab − ab
j+1,(cid:8) j+1,(cid:8)
(u) (du)
= #T (l,x,y+2) # T (l,x,y)
ab − ab
in the first line, we have replaced ∆S = d by u, in the third line we have replaced uu by
−1
T
du. (cid:3)
2.1 Proof of Proposition 1
(i) To build a path with k peaks, dispose k peaks, that is k pairs ud. Take a composition
x ,...,x of n 2k in 2(k+1) non negative parts. Fill in now the k+1 intervals between
1 2(k+1)
−
10