and q
∗
{a, b} × Z
d n
−2
= 1
n=2
papb ¯
G0 − 1
p
2
a + b + 1
n ≥3
papb ¯
G0 − 1
X
x
3
,...,x
n
∈Z
d
p
n
a + b + x
3
+ · · · + x
n n
Y
k=3
px
k
= papb
¯ G0
− 1 p
2n −2
a + b, 3.32
we find recall that ¯ G0
− 1 = P
∞ n=1
p
2n
Ψ
Q
∗
a, b = papb
h
∞
P
n=1
p
2n
ih
∞
P
n=1
np
2n
i
∞
X
n=1
p
2n −1
a
∞
X
n=1
p
2n −1
b +
∞
X
n=1
p
2n ∞
X
n=2
n − 1p
2n −2
a + b
. 3.33
Pick b = −a with pa 0. Then, shifting n to n − 1 in the last sum, we get
Ψ
Q
∗
a, −a pa
2
− 1 =
∞
P
n=1
p
2n −1
a
2
h
∞
P
n=1
p
2n
ih
∞
P
n=1
np
2n
i 0. 3.34
This shows that consecutive letters are not uncorrelated under Ψ
Q
∗
, and implies that 3.27 holds as claimed.
3.3 Proof of Theorem 1.6
The proof follows the line of argument in Section 3.2. The analogues of 3.4–3.7 are z
e V
=
∞
X
N =0
log z
N
e V
N
N ,
3.35 with
e V
N
N =
Z
∞
d t
1
· · · Z
∞ t
N −1
d t
N
1
{e S
t1
=e S
′ t1
,...,e S
tN
=e S
′ tN
}
, 3.36
and E
h z
e V
| e S
i =
∞
X
N =0
log z
N
F
1 N
e S,
E h
z
e V
i =
∞
X
N =0
log z
N
F
2 N
, 3.37
with F
1 N
e S :=
Z
∞
d t
1
· · · Z
∞ t
N −1
d t
N
P e
S
t
1
= e S
′ t
1
, . . . , e S
t
N
= e S
′ t
N
| e S
, F
2 N
:= E F
1 N
e S
, 3.38
567
where the conditioning in the first expression in 3.37 is on the full continuous-time path e S =
e S
t t
≥0
. Our task is to compute er
1
:= lim
N →∞
1 N
log F
1 N
e S
e S
− a.s., er
2
:= lim
N →∞
1 N
log F
2 N
, 3.39
and show that er
1
er
2
. In order to do so, we write e
S
t
= X
♮ J
t
, where X
♮
is the discrete-time random walk with transition kernel p
·, · and J
t t
≥0
is the rate-1 Poisson process on [0, ∞, and then average over the jump
times of J
t t
≥0
while keeping the jumps of X
♮
fixed. In this way we reduce the problem to the one for the discrete-time random walk treated in the proof of Theorem 1.6. For the first expression in
3.38 this partial annealing gives an upper bound, while for the second expression it is simply part of the averaging over e
S. Define
F
1 N
X
♮
:= Z
∞
d t
1
· · · Z
∞ t
N −1
d t
N
P e
S
t
1
= e S
′ t
1
, . . . , e S
t
N
= e S
′ t
N
| X
♮
, F
2 N
:= E F
1 N
X
♮
, 3.40
together with the critical values r
♮ 1
:= lim
N →∞
1 N
log F
1 N
X
♮
X
♮
− a.s., r
♮ 2
:= lim
N →∞
1 N
log F
2 N
. 3.41
Clearly, er
1
≤ r
♮ 1
and er
2
= r
♮ 2
, 3.42
which can be viewed as a result of “partial annealing”, and so it suffices to show that r
♮ 1
r
♮ 2
. To this end write out
P e
S
t
1
= e S
′ t
1
, . . . , e S
t
N
= e S
′ t
N
| X
♮
= X
≤ j
1
≤···≤ j
N
∞ N
Y
i=1
e
−t
i
−t
i −1
t
i
− t
i −1
j
i
− j
i −1
j
i
− j
i −1
X
≤ j
′ 1
≤···≤ j
′ N
∞ N
Y
i=1
e
−t
i
−t
i −1
t
i
− t
i −1
j
′ i
− j
′ i
−1
j
′ i
− j
′ i
−1
N
Y
i=1
p
j
′ i
− j
′ i
−1
j
i
− j
i −1
X
k=1
X
♮ j
i −1
+k
.
3.43
Integrating over 0 ≤ t
1
≤ · · · ≤ t
N
∞, we obtain F
1 N
X
♮
= X
≤ j
1
≤···≤ j
N
∞
X
≤ j
′ 1
≤···≤ j
′ N
∞ N
Y
i=1
2
− j
i
− j
i −1
− j
′ i
− j
′ i
−1
−1
[ j
i
− j
i −1
+ j
′ i
− j
′ i
−1
] j
i
− j
i −1
j
′ i
− j
′ i
−1
p
j
′ i
− j
′ i
−1
j
i
− j
i −1
X
k=1
X
♮ j
i −1
+k
.
3.44
Abbreviating Θ
n
u =
∞
X
m=0
p
m
u 2
−n−m−1
n + m m
, n
∈ N ∪ {0}, u ∈ Z
d
, 3.45
568
we may rewrite 3.44 as F
1 N
X
♮
= X
≤ j
1
≤···≤ j
N
∞ N
Y
i=1
Θ
j
i
− j
i −1
j
i
− j
i −1
X
k=1
X
♮ j
i −1
+k
.
3.46 This expression is similar in form as the first line of 3.8, except that the order of the j
i
’s is not strict. However, defining
b F
1 N
X
♮
= X
j
1
··· j
N
∞ N
Y
i=1
Θ
j
i
− j
i −1
j
i
− j
i −1
X
k=1
X
♮ j
i −1
+k
,
3.47 we have
F
1 N
X
♮
=
N
X
M =0
N M
[Θ 0]
M
b F
1 N
−M
X
♮
, 3.48
with the convention b F
1
X
♮
≡ 1. Letting r
♮ 1
= lim
N →∞
1 N
log b F
1 N
X
♮
, X
♮
− a.s., 3.49
and recalling 3.41, we therefore have the relation r
♮ 1
= log h
Θ 0 + e
br
♮ 1
i ,
3.50 and so it suffices to compute
br
♮ 1
. Write
F
1 N
X
♮
= E
exp
N Z
Ý Z
d
π
1
R
N
d y log f
♮
y
X
♮
,
3.51 where f
♮
: Ý Z
d
→ [0, ∞ is defined by f
♮
x
1
, . . . , x
n
= Θ
n
x
1
+ · · · + x
n
p
2 ⌊n2⌋
[2 ¯ G0
− 1], n
∈ N, x
1
, . . . , x
n
∈ Z
d
. 3.52
Equations 3.51–3.52 replace 3.8–3.9. We can now repeat the same argument as in 3.16– 3.22, with the sole difference that f in 3.9 is replaced by f
♮
in 3.52, and this, combined with Lemma 3.3 below, yields the gap r
♮ 1
r
♮ 2
. We first check that f
♮
is bounded from above, which is necessary for the application of Varadhan’s lemma. To that end, we insert the Fourier representation 3.14 into 3.45 to obtain
Θ
n
u = 1
2π
d
Z
[−π,π
d
d k e
−ik·u
[2 − bpk]
−n−1
, u
∈ Z
d
, 3.53
from which we see that Θ
n
u ≤ Θ
n
0, u ∈ Z
d
. Consequently, f
♮ n
x
1
, · · · , x
n
≤ Θ
n
p
2 ⌊n2⌋
[2 ¯ G0
− 1], n
∈ N, x
1
, . . . , x
n
∈ Z
d
. 3.54
569
Next we note that lim
n →∞
1 n
log 2
−a+bn−1
a + bn an
¨ = 0,
if a = b, 0,
if a 6= b.
3.55 From 1.1, 3.45 and 3.55 it follows that Θ
n
0p
2 ⌊n2⌋
0 ≤ C ∞ for all n ∈ N, so that f
♮
indeed is bounded from above. Note that X
♮
is the discrete-time random walk with transition kernel p ·, ·. The key ingredient
behind br
♮ 1
br
♮ 2
is the analogue of Lemma 3.2, this time with Q
∗
= q
∗ ⊗N
and q
∗
given by q
∗
x
1
, . . . , x
n
= Θ
n
x
1
+ · · · + x
n 1
2
G0 − Θ
n
Y
k=1
px
k
, 3.56
replacing 3.19. The proof is deferred to the end.
Lemma 3.3. Assume 1.1. Let Q
∗
= q
∗ ⊗N
with q
∗
as in 3.56. If m
Q
∗
∞, then I
que
Q
∗
I
ann
Q
∗
. This shows that
br
♮ 1
br
♮ 2
via the same computation as in 3.21–3.23. The analogue of 3.18 reads
Z
♮
= X
n ∈N
X
x
1
,...,x
n
∈Z
d
Θ
n
x
1
+ · · · + x
n n
Y
k=1
p x
k
= X
n ∈N
∞
X
m=0
¨ X
x
1
,...,x
n
∈Z
d
p
m
x
1
+ · · · + x
n n
Y
k=1
p x
k
« 2
−n−m−1
n + m m
= −Θ 0 +
∞
X
n,m=0
p
n+m
0 2
−n−m−1
n + m m
= −Θ 0 +
1 2
∞
X
k=0
p
k
0 = −Θ 0 +
1 2
G0. 3.57
Consequently, log
ez
2
= e
−er
2
= e
−r
♮ 2
= 1
Θ 0 + e
br
♮ 2
= 1
Θ 0 + Z
♮
= 2
G0 ,
3.58 where we use 3.37, 3.39, 3.42, 3.50 and 3.57.
We close by proving Lemma 3.3. Proof. We must adapt the proof in Section 3.2 to the fact that q
∗
has a slightly different form, namely, p
n
x
1
+ · · · + x
n
is replaced by Θ
n
x
1
+ · · · + x
n
, which averages transition kernels. The computations are straightforward and are left to the reader. The analogues of 3.24 and 3.26 are
q
∗
Z
d n
= 1
1 2
G0 − Θ
∞
X
m=0
p
n+m
0 2
−n−m−1
n + m m
, m
Q
∗
= X
n ∈N
nq
∗
Z
d n
= 1
4 1
1 2
G0 − Θ
∞
X
k=0
kp
k
0, 3.59
570
while the analogues of 3.31–3.32 are q
∗
x
1
= a = pa
1 2
G0 − Θ
1 2
∞
X
k=0
p
k
a[1 − 2
−k−1
] = 1
2 pa
Ga − Θ
a
1 2
G0 − Θ
, q
∗
{a, b} × Z
d n
−2
= papb
1 2
G0 − Θ
∞
X
m=0
p
n −2+m
a + b 2
−n−m−1
n + m m
. 3.60
Recalling 3.30, we find Ψ
Q
∗
a, −a − pa
2
0, 3.61
implying that Ψ
Q
∗
6= ν
⊗N
recall 3.2, and hence HΨ
Q
∗
| ν
⊗N
0, implying the claim.
4 Proof of Theorem 1.2
This section uses techniques from [6]. The proof of Theorem 1.2 is based on two approximation lemmas, which are stated in Section 4.1. The proof of these lemmas is given in Sections 4.2–4.3.
4.1 Two approximation lemmas
