4 Proofs
4.1 Proof of Theorem 2.1
We consider the set A
N
= k
1
, . . . , k
N
; k
1
= 1, for i ∈ {1, . . . , N − 1}, k
i+ 1
∈ k
i
, k
i
+ 1 .
Notice that PV
1
= k
1
, . . . , V
N
= k
N
0 if and only if k
1
, . . . , k
N
∈ A
N
. To prove the first part of Theorem 2.1, it is enough to show that, for N ≥ 2 and k
1
, . . . , k
N + 1
∈ A
N + 1
, P
V
N + 1
= k
N + 1
|V
N
= k
N
, . . . , V
1
= k
1
= 1 −
1+k
N
N + 1
if k
N + 1
= k
N
,
1+k
N
N + 1
if k
N + 1
= 1 + k
N
. 27
For p and q in N
∗
such that q p, we introduce the set: ∆
p ,q
= {α = α
1
, . . . , α
p
∈ {0, 1}
p
, α
1
= 1,
p
X
i= 1
α
i
= q}. Notice that Card ∆
p ,q
=
p− 1
q− 1
. Hence to prove the second part of Theorem 2.1, it is enough to show that: for all k
1
, . . . , k
N
∈ A
N
, and all α ∈ ∆
N ,k
N
, P
σ
N
= α|V
N
= k
N
, . . . , V
1
= k
1
= 1
N − 1
k
N
−1
· 28
We proceed by induction on N for the proof of 27 and 28. The result is obvious for N = 2. We suppose that 27 and 28 are true for a fixed N . We denote by I
N
and J
N
, 1 ≤ I
N
J
N
≤ N +1, the two levels involved in the look-down event at time s
N
. Notice that I
N
, J
N
and σ
N
are independent. This pair is chosen uniformly so that, for 1 ≤ i j ≤ N + 1,
P I
N
= i, J
N
= j = 2
N + 1N ,
P I
N
= i = 2N − i + 1
N + 1N ,
P J
N
= j = 2 j − 1
N + 1N ·
For α = α
1
, . . . , α
N + 1
∈ {0, 1}
N + 1
and j ∈ {1, . . . , N + 1}, we set α
j ×
= α
1
, . . . , α
j− 1
, α
j+ 1
, . . . , α
N + 1
∈ {0, 1}
N
. Let us fix k
1
, . . . , k
N + 1
∈ A
N + 1
, and α = α
1
, . . . , α
N + 1
∈ ∆
N + 1,k
N + 1
. Notice that {σ
N + 1
= α} ⊂ {V
N + 1
= k
N + 1
}. We first compute P
σ
N + 1
= α|V
N
= k
N
, . . . , V
1
= k
1
.
792
1st case: k
N + 1
= k
N
+ 1. We have:
P σ
N + 1
= α|V
N
= k
N
, . . . , V
1
= k
1
= X
1≤i j≤N +1
P I
N
= i, J
N
= j, σ
N + 1
= α|V
N
= k
N
, . . . , V
1
= k
1
= X
1≤i j≤N +1,α
i
=α
j
=1
P I
N
= i, J
N
= j, σ
N
= α
j ×
|V
N
= k
N
, . . . , V
1
= k
1
= X
1≤i j≤N +1,α
i
=α
j
=1
P I
N
= i, J
N
= jPσ
N
= α
j ×
|V
N
= k
N
, . . . , V
1
= k
1
= X
1≤i j≤N +1,α
i
=α
j
=1
2 N + 1N
1
N − 1
k
N
−1
= 2
N + 1N 1
N − 1
k
N
−1
k
N + 1
k
N + 1
− 1 2
= k
N
+ 1N − k
N
N + 1 ,
29 where we used the independence of I
N
, J
N
and σ
N
for the third equality, the uniform distribution of σ
N
conditionally on V
N
for the fourth, and that k
N + 1
= k
N
+ 1 for the sixth. Hence, we get P
V
N + 1
= k
N
+ 1|V
N
= k
N
, . . . , V
1
= k
1
= X
α∈∆
N + 1,kN+1
P σ
N + 1
= α|V
N
= k
N
, . . . , V
1
= k
1
= N
k
N + 1
− 1 k
N
+ 1N − k
N
N + 1 =
1 + k
N
N + 1
· 30
2nd case: k
N + 1
= k
N
. Similarly, we have:
P σ
N + 1
= α|V
N
= k
N
, . . . , V
1
= k
1
= X
1≤i j≤N +1,α
i
=α
j
=0
2 N + 1N
1
N − 1
k
N
−1
= 2
N + 1N 1
N − 1
k
N
−1
N + 1 − k
N
N − k
N
2 =
N − k
N
k
N
− 1N − k
N
+ 1 N + 1
. 31
Hence, we get P
V
N + 1
= k
N
|V
N
= k
N
, . . . , V
1
= k
1
= X
α∈∆
N + 1,kN+1
P σ
N + 1
= α|V
N
= k
N
, . . . , V
1
= k
1
= N
k
N + 1
− 1 N − k
N
k
N
− 1N − k
N
+ 1 N + 1
=1 − 1 + k
N
N + 1
· 32
793
Equalities 30 and 32 imply 27. Moreover, we deduce from 29 and 31 that, for k
N + 1
∈ {k
N
, k
N
+ 1}, P
σ
N + 1
= α|V
N + 1
= k
N + 1
, . . . , V
1
= k
1
= P
σ
N + 1
= α, V
N + 1
= k
N + 1
|V
N
= k
N
, . . . , V
1
= k
1
P V
N + 1
= k
N + 1
|V
N
= k
N
, . . . , V
1
= k
1
= 1
N k
N + 1
−1
, which proves that 28 with N replaced by N + 1 holds. This ends the proof.
4.2 Proof of Proposition 2.4
Kategori