5.2 Proof of positive recurrence
Proposition 5.4. Assume that the cookie environment C = p
1
, . . . , p
M
; q is such that q
b b +
1 and
λ 1
b .
Then, all the return times of the walk to the root of the tree have finite expectation. Proof.
Let σ
i
denote the time of the i
th
crossing of the edge joining the root of the tree to itself for the cookie random walk:
σ
i
def
= inf n
n 0,
n
X
j= 1
1
{X
j
=X
j −1
=o}
= i o
.
We prove that E[σ
i
] ∞ for all i. Recalling the construction of the branching Markov chain L in section 3.1 and the definition of Z, we have
E[
σ
i
] = i + 2E
i
h X
x ∈T\{o}
ℓx i
= i + 2
∞
X
n= 1
b
n
E
i
[Z
n
]. Let us for the time being admit that
lim sup
n →∞
P
i
{Z
n
0}
1n
≤ λ for any i.
20 Then, using Hölder’s inequality and b of Lemma 4.6, choosing α, β, ˜
λ such that ˜ λ λ, b ˜
λ
1α
1 and
1 α
+
1 β
= 1, we get
∞
X
n= 1
b
n
E
i
[Z
n
] ≤
∞
X
n= 1
b
n
P
i
{Z
n
0}
1α
E
i
[Z
β n
]
1β
≤ C
β ∞
X
n= 1
b ˜ λ
1α n
∞. It remains to prove 20. Recall that {0}, [l
1
, r
1
], . . . , [l
k
, ∞ denote the irreducible classes of P and that Z can only move from a class [l
k
, r
k
] to another class [l
k
′
, r
k
′
] with k
′
k. Thus, for i ∈ [l
k
, r
k
], we have
P
i
{Z
n
≥ l
k
} = P
i
{Z
n
∈ [l
k
, r
k
]} =
r
k
X
j=l
k
P
i
{Z
n
= j} =
r
k
X
j=l
k
p
n
i, j. For k K, the sum above is taken over a finite set. Recalling the definition of λ
k
, we get lim
n →∞
P
i
{Z
n
≥ l
k
}
1n
= λ
k
for all i ∈ [l
k
, r
k
]. Using the Markov property of Z, we conclude by induction that, for any i l
K
, lim sup
n →∞
P
i
{Z
n
0}
1n
≤ maxλ
1
, . . . , λ
K −1
≤ λ. 21
It remains to prove the result for i ≥ l
K
. In view of 21 and using the Markov property of Z, it is sufficient to show that, for i ≥ l
K
, lim sup
n →∞
P
i
{Z
n
≥ l
K
}
1n
≤ λ
K
. 22
1649
Let us fix i ≥ l
K
. We write
P
i
{Z
n
≥ l
K
} = P
i
{∃m ≥ n, Z
m
= l
K
} +
∞
X
j=l
K
P
i
{Z
n
= j}P
j
{∄m ≥ 0, Z
m
= l
K
}. According to lemma 4.7, there exists c 0 such that, for all j ≥ l
K
, P
j
{∄m ≥ 0, Z
m
= l
K
} ≤ 1 − c. Therefore, we deduce that
P
i
{Z
n
≥ l
K
} ≤ 1
c
P
i
{∃m ≥ n, Z
m
= l
K
} ≤ 1
c
∞
X
m=n
p
m
i, l
K
. 23
Moreover, we have lim
m →∞
p
m
i, l
K 1m
= λ
K
1 hence lim
n →∞
∞
X
m=n
p
m
i, l
K
1n
= λ
K
. 24
The combination of 23 and 24 yields 22 which completes the proof of the proposition.
5.3 Proof of transience when
Kategori