Therefore, we deduce from 39 that ν ≥ λ
Q
. On the other hand, the matrix Q
N
is finite so that, according to the Perron-Frobenius Theorem, its spectral radius is equal to its largest eigenvalue λ
Q
N
and is given by the formula λ
Q
N
= sup
x
1
,...,x
N
min
j
P
N i=
1
x
i
qi , j
x
j
where the supremum is taken over all N -dimensional vectors with strictly positive coefficients c.f. 1.1 p.4 of [19]. In view of 40, we deduce that λ
Q
N
≥ ν − ǫ. Furthermore, when Q
N
is irreducible, Theorem 6.8 of [19] states that λ
Q
N
≤ λ
Q
. We conclude that λ
Q
≤ ν ≤ λ
Q
+ ǫ.
7.1 Preliminaries
Recall the construction of the random variables ξ
i i
≥1
given in Definition 3.1 and set E
m ,n
def
= in the finite sequence ξ
1
, ξ
2
, . . . , ξ
M
, there are at least m terms equal to 0 and exactly n terms are equal to 1 before the m
th
E
′ m
,n
def
= in the finite sequence ξ
1
, ξ
2
, . . . , ξ
M
, there are exactly m terms equal to 0 and exactly n terms equal to 1
. Let us note that, for n + m M ,
P {E
m ,n
} = P{E
′ m
,n
} = 0. 41
In the rest of this section, we use the notation s
def
= q
q + 1 − qb
= P{ξ
M + 1
= 1 | ξ
M + 1
∈ {0, 1}}.
Lemma 7.3. For i , j ≥ 1, the coefficient pi, j of the matrix P associated with the cookie environment
C = p
1
, . . . , p
M
; q is given by pi
, j = P{E
i , j
} + X
0≤n≤ j 0≤m≤i−1
P {E
′ m
,n
} j + i − m − n − 1
j − n
s
j −n
1 − s
i −m
.
Proof. Recall that pi, j is equal to the probability of having j times 1 in the sequence ξ
l l
≥1
before the i
th
0. We decompose this event according to the number of 0’s and 1’s in the subsequence ξ
l l
≤M
. Let F
m ,n
be the event F
m ,n
def
= {in the sub-sequence ξ
i i
M
, n terms equal to 1 before the m
th
failure}.
1661
Thus we have pi
, j = P{E
i , j
} + X
0≤n≤ j 0≤m≤i−1
P {E
′ m
,n
}P{F
i −m, j−n
} 42
the first term of the r.h.s. of the equation comes from the case m = i which cannot be included in the sum. Since the sequence ξ
i i
M
is a sequence of i.i.d. random variables, it is easy to compute
P {F
m ,n
}. Indeed, noticing that,
P {F
m ,n
} = P{F
m ,n
| ξ
l
∈ {0, 1} for all l ∈ [M, M + n + m]}, we get
P {F
m ,n
} = n + m − 1
n s
n
1 − s
m
. 43
The combination of 42 and 43 completes the proof of the lemma. We can now compute the image
t
Y P of the exponential vector Y = s1 − s
i −1
i ≥1
. Let us first recall the notation
λ
sym def
= q
b 1 − q
M
Y
i= 1
1 − p
i
q b
1 − q +
b − 1p
i
b +
p
i
b q
b 1 − q
−1
.
We use the convention that P
v u
= 0 when u v.
Lemma 7.4. We have
∞
X
i= 1
pi , j
s
1 − s
i −1
= λ
sym
s
1 − s
j −1
+ A j, with
A j
def
=
M − j
X
i= 1
P {E
i , j
}
s 1 − s
i −1
−
M
X
n= j+ 1
M −n
X
m=
P {E
′ m
,n
}
s 1 − s
j+m −n
. In particular, A j =
0 for j ≥ M. Proof.
With the help of Lemma 7.3, and in view of 41, we have
∞
X
i= 1
pi , j
s
1 − s
i −1
=
∞
X
i= 1
P {E
i , j
}
s 1−s
i −1
+
∞
X
i= 1
X
0≤n≤ j 0≤m≤i−1
P {E
′ m
,n
} j+i−m−n−1
j − n
s
j+i −n−1
1−s
1−m
=
M − j
X
i= 1
P {E
i , j
}
s 1−s
i −1
+ X
0≤n≤ j∧M 0≤m≤M−n
P {E
′ m
,n
}
∞
X
i=
j + i − n i
s
j+i+m −n
1−s
1−m
. Using the relation
∞
X
i=
j + i − n i
s
j+i+m −n
1 − s
1−m
=
s 1 − s
j+m −n
, 1662
we deduce that
∞
X
i= 1
pi , j
s
1−s
i −1
=
M − j
X
i= 1
P {E
i , j
}
s 1−s
i −1
+
j ∧M
X
n= M
−n
X
m=
P {E
′ m
,n
}
s 1−s
j+m −n
. =
M
X
n= M
−n
X
m=
P {E
′ m
,n
}
s 1−s
j+m −n
+ A j. It simply remains to show that
λ
sym
=
M
X
n= M
−n
X
m=
P {E
′ m
,n
}
s 1 − s
m −n+1
. 44
Let us note that, λ
sym
= s
1 − s
M
Y
l= 1
s 1 − s
P {ξ
l
= 0} + P{ξ
l
≥ 2} + P{ξ
l
= 1} 1 − s
s .
Expanding the r.h.s. of this equation and using the definition of E
′ m
,n
, we get 44 which concludes the proof of the lemma.
We have already noticed that A j = 0 whenever j ≥ M. In fact, if some cookies have strength 0, the lower bound on j can be improved. Let M
denote the number of cookies with strength 0: M
def
= ♯{1 ≤ i ≤ M, p
i
= 0}.
Lemma 7.5. Let C = p
1
, . . . , p
M
; q be a cookie environment with p
M
6= 0. We have, A j =
0 for all j ≥ M − M .
Proof. Since M
cookies have strength 0, there are at most M − M terms equal to 1 in the sequence
ξ
1
, . . . , ξ
M
. Keeping in mind the definitions of E
m ,n
and E
′ m
,n
, we see that
P {E
m ,n
} = P{E
′ m
,n
} = 0 for n M − M
. Moreover, recall that p
M
6= 0. Thus, if exactly M − M terms are equal to 1, the last one, ξ
M
, must also be equal to 1. Therefore, we have
P {E
m ,M −M
} = 0. Let us now fix j ≥ M − M
, and look at the expression of A j. A j
def
=
M − j
X
i= 1
P {E
i , j
}
s 1 − s
i −1
−
M
X
n= j+ 1
M −n
X
m=
P {E
′ m
,n
}
s 1 − s
j+m −n
. The terms in the first sum
P
M − j
i= 1
are all zero since j ≥ M − M . Similarly, all the terms in the sum
P
M −n
m=
are also zero since n ≥ j + 1 M − M .
1663
Proposition 7.6. Let C = p
1
, . . . , p
M
; q be a cookie environment such that q
b b +
1 and M
≥ M
2 .
If M is an odd integer, assume further that p
M
6= 0. Then, the spectral radius λ
K
of the infinite irreducible sub-matrix P
K
= pi, j
i , j≥l
K
is equal to λ
sym
. Proof.
Let us note that, when M is an even integer and p
M
= 0, we can consider C as the M + 1 cookie environment p
1
, . . . , p
M
, q ; q and this M + 1 cookie environment still possesses, at least, half of its cookies with zero strength because ⌊M + 12⌋ = ⌊M2⌋. Thus, we can assume, without
loss of generality that the cookie environment is such that p
M
6= 0. In order to prove the proposition, we shall prove that
Y
def
=
s 1 − s
i −1
i ≥l
K
is a left eigenvector of P
K
for the eigenvalue λ
sym
fulfilling the assumptions of Proposition 7.1. Since the cookie environment has M
≥ ⌊M2⌋ cookies with strength 0, there are, in the 2⌊M2⌋ first cookies, at most ⌊M2⌋ random variables taking value 1 in the sequence ξ
i i
≥1
before the ⌊M2⌋
th
failure i.e. pi
, j = 0 for i ≤ ⌊M2⌋ j.
45 This implies, in particular, that l
K
≥ ⌊M2⌋ + 1 ≥ M − M . Using Lemma 7.5, we deduce that
A j = for all j ≥ l
k
. 46
Combining 46 and Lemma 7.4, we conclude that Y is indeed a left eigenvector:
t
Y P
K
= λ
sym
t
Y .
Let ǫ 0. We consider the sub-vector Y
N
def
=
s 1 − s
i −1
l
K
≤il
K
+N
. It remains to show that, for N large enough,
t
Y
N
P
K ,N
≥ λ
K
− ǫ
t
Y
N
i.e.
l
K
+N −1
X
i=l
K
pi , j
s
1 − s
i −1
≥ λ
K
− ǫ
s 1 − s
j −1
for all j ∈ {l
K
, . . . , l
K
+ N − 1}. 47
Keeping in mind that, for j ≥ l
K ∞
X
i=l
K
pi , j
s
1 − s
i −1
= λ
K
s
1 − s
j −1
, we see that 47 is equivalent to proving that,
∞
X
i=l
K
+N
pi , j
s
1 − s
i −1
≤ ǫ
s 1 − s
j −1
for j ∈ {l
K
, . . . , l
K
+ N − 1}. 48
1664
Choosing N such that l
K
+ N ≥ M + 1, and using the expression of pi, j stated in Lemma 7.3, we get, for any j ∈ {l
K
, . . . , l
K
+ N − 1},
∞
X
i=l
K
+N
pi , j
s
1−s
i −1
= X
0≤n≤ j 0≤m≤M
P {E
′ m
,n
}
∞
X
i=l
K
+N
j+i−m−n−1 j
−n s
j+i −1−n
1 − s
1−m
where we used that P{E
′ m
,n
} = P{E
m ,n
} = 0 when either n or m is strictly larger than M. We now write
j + i − m − n − 1 j
− n s
j+i −1−n
1 − s
1−m
=
s 1 − s
j+m −n
j + i − m − n − 1 i
− m − 1 s
i −m−1
1 − s
j −n+1
and we interpret the term j + i − m − n − 1
i − m − 1
s
i −m−1
1 − s
j −n+1
as the probability of having i − m − 1 successes before having j − n + 1 failures in a sequence B
r r
≥1
of i.i.d. Bernoulli random variables with distribution P{B
r
= 1} = 1 − P{B
r
= 0} = s. Therefore, we deduce that
∞
X
i=l
K
+N
j + i − m − n − 1 i
− m − 1 s
i −m−1
1 − s
j −n+1
= P there are at least l
K
+N−m−1 successes before the j−n+1
th
failure in B
r r
≥1
≤ P there are at least l
K
+N−M−1 successes before the l
K
+N+1
th
failure in B
r r
≥1
. Noticing that s 12 since q bb+1, the law of large numbers for the biased Bernoulli sequence
B
r r
≥1
implies that the above probability converges to 0 as N tends to infinity. Thus, for all ǫ 0, we can find N ≥ 1 such that 48 holds.
7.2 Proofs of Theorem 1.4 and Proposition 1.8
