and ξ
n
s, r
i
= M
n,i [ns]
· e
1
+ ED · e
1
ζ
n
s, r
i
= M
n,i [ns]
· e
1
− E
D · e
1
E D · e
2
M
n,i [ns]
· e
2
+ O1. So recall the definiton of ˆ
w in 1,
N
X
i=1
θ
i
ξ
n
·, r
i
n
1 4
= n
−
1 4
N
X
i=1
θ
i
h E
ω r
i
p n
X
λ
[n·]
· e
1
− E
r
i
p n
X
λ
[n·]
· e
1
i
= ˆ
w ·
N
X
i=1
θ
i
M
n,i [n·]
n
1 4
+ On
−
1 4
⇒ ˆ w
· B h· .
Returning to equation 7, the proof of Theorem 2.1 will be complete if we show 1
p n
n
X
k=1
P
ω r
i
p n, r
j
p n
X
λ
k −1
= ˜ X
˜ λ
k −1
on level k − 1 −→
1 βσ
2
Z
σ2 c1
1 p
2 πv
exp −
r
i
− r
j 2
2v d v
in P-probability as n → ∞. We will first show the averaged statement with ω integrated away
P
ROPOSITION
3.2. 1
p n
n
X
k=1
P
r
i
p n, r
j
p n
X
λ
k −1
= ˜ X
˜ λ
k −1
on level k − 1 −→
1 βσ
2
Z
σ2 c1
1 p
2 πv
exp −
r
i
− r
j 2
2v d v
and then show that the difference of the two vanishes: P
ROPOSITION
3.3. 1
p n
n
X
k=1
P
ω r
i
p n, r
j
p n
X
λ
k −1
= ˜ X
˜ λ
k −1
on level k − 1
− 1
p n
n
X
k=1
P
r
i
p n, r
j
p n
X
λ
k −1
= ˜ X
˜ λ
k −1
on level k − 1 −→ 0 in P-probability.
3.1 Proof of Proposition 3.2
For simplicity of notation, let P
n i, j
denote P
r
i
p n,r
j
p n
and E
n i, j
denote E
r
i
p n,r
j
p n
. Let 0 = L L
1
· · · be the successive common levels of the independent walks in common environment X and ˜ X ;
that is L
j
= inf{l L
j −1
: X
λ
l
· e
2
= ˜ X
˜ λ
l
· e
2
= l}. Let Y
k
= X
λ
Lk
· e
1
− ˜ X
˜ λ
Lk
· e
1
. Now
1 p
n
n
X
k=1
P
n i, j
X
λ
k −1
= ˜ X
˜ λ
k −1
on level k − 1
1274
= 1
p n
n
X
k=1
P
n i, j
X
λ
Lk−1
= ˜ X
˜ λ
Lk−1
and L
k −1
≤ n − 1
= 1
p n
n
X
k=1
P
n i, j
Y
k −1
= 0 and L
k −1
≤ n − 1. 9
We would like to get rid of the inequality L
k −1
≤ n − 1 in the expression above so that we have something resembling a Green’s function. Denote by ∆L
j
= L
j+1
− L
j
. Call L
k
a meeting level m.l. if the two walks meet at that level, that is X
λ
Lk
= ˜ X
˜ λ
Lk
. Also let I = { j : L
j
is a meeting level }. Let
Q
1
, Q
2
, · · · be the consecutive ∆L
j
where j ∈ I and R
1
, R
2
, · · · be the consecutive ∆L
j
where j ∈ I.
We start with a simple observation. L
EMMA
3.4. Fix x, y ∈ Z. Under P
x, y
, Q
1
, Q
2
, · · · are i.i.d. with common distribution P
0,00,0
L
1
∈ · R
1
, R
2
, · · · are i.i.d. with common distribution P
0,0,1,0
L
1
∈ · Proof. We prove the second statement. The first statement can be proved similarly. Call a level a
non meeting common level n.m.c.l if the two walks hit the level but do not meet at that level. For positive integers k
1
, k
2
, · · · , k
n
, P
x, y
R
1
= k
1
, R
2
= k
2
, · · · , R
n
= k
n
= X
i
P
x, y
R
1
= k
1
, · · · , R
n
= k
n
, i is the n’th n.m.c.l. =
X
i, j 6=0
P
x, y
R
1
= k
1
, · · · , R
n −1
= k
n −1
, i is the n’th n.m.c.l., ˜ X
˜ λ
i
· e
1
− X
λ
i
· e
1
= j, R
n
= k
n
= X
i, j 6=0,l
E h
P
ω x, y
R
1
= k
1
, · · · , R
n −1
= k
n −1
, i is the n’th n.m.c.l., X
λ
i
· e
1
= l, ˜ X
˜ λ
i
· e
1
= j + l ×P
ω l,i,l+ j,i
∆L = k
n
i =
X
i, j 6=0,l
E h
P
ω x, y
R
1
= k
1
, · · · , R
n −1
= k
n −1
, i is the n’th n.m.c.l., X
λ
i
· e
1
= l, ˜ X
˜ λ
i
· e
1
= j + l i
×P
0,0,e
1
L
1
= k
n
= P
0,0,e
1
L
1
= k
n
P
x, y
R
1
= k
1
, R
2
= k
2
, · · · , R
n −1
= k
n −1
. The fourth equality is because P
ω l,i,l+ j,i
∆L = k
n
depends on {ω
m
: m ≥ i} whereas the previous
term depends on the part of the environment strictly below level i. Note also that P
0,0,e
1
L
1
= k
n
= P
0,0, je
1
L
1
= k
n
. This can be seen by a path by path decomposition of the two walks. The proof is complete by induction.
L
EMMA
3.5. There exists some a 0 such that
E
0,0,0,0
e
a L
1
∞ and
E
0,0,1,0
e
a L
1
∞.
1275
Proof. By the ellipticity assumption iii, we have P
0,0,0,0
L
1
k ≤ 1 − δ
[
k K
]
, P
0,0,1,0
L
1
k ≤ 1 − δ
[
k K
]
. Since L
1
is stochastically dominated by a geometric random variable, we are done. Let us denote by X
[0,n]
the set of points visited by the walk upto time n. It has been proved in Proposition 5.1 of [7] that for any starting points x, y
∈ Z
2
E
x, y
|X
[0,n]
∩ ˜ X
[0,n]
| ≤ C
p n.
10 This inequality is obtained by control on a Green’s function. The above lemma and the inequality
that follows tell us that common levels occur very frequently but the walks meet rarely. Let E
0,0,1,0
L
1
= c
1
, E
0,0,0,0
L
1
= c .
11 We will need the following lemma.
L
EMMA
3.6. For each ε 0, there exist constants C 0, bε 0, dε 0 such that
P
n i, j
L
n
n ≥ c
1
+ ε ≤ C exp
− nbε, P
n i, j
L
n
n ≤ c
1
− ε ≤ C exp
− ndε. Thus
L
n
n
→ c
1
P
0,0
a.s. Proof. We prove the first inequality. From Lemmas 3.4 and 3.5, we can find a
0 and some ν 0 such that for each n,
E
n i, j
expa L
n
≤ ν
n
. We thus have
P
n i, j
L
n
n ≥ C
1
≤ E
n i, j
expa L
n
expaC
1
n ≤ exp{nlog ν − aC
1
}. Choose C
1
large enough so that log ν − aC
1
0. Now P
n i, j
L
n
n ≥ c
1
+ ε ≤ exp{nlog ν − aC
1
} + P
n i, j
c
1
+ ε ≤ L
n
n ≤ C
1
. Let
γ =
ε 4c
. Denote by I
n
= { j : 0 ≤ j n, L
j
is a meeting level } and recall ∆L
j
= L
j+1
− L
j
. We have
P
n i, j
c
1
+ ε ≤ L
n
n ≤ C
1
≤ P
n i, j
|I
n
| ≥ γn, L
n
≤ C
1
n +
P
n i, j
|I
n
| γn, c
1
+ ε ≤ P
j ∈I
n
, j n
∆L
j
n +
P
j ∈I
n
∆L
j
n ≤ C
1
. 1276
Let T
1
, T
2
, . . . be the increments of the successive meeting levels. By an argument like the one given in Lemma 3.4,
{T
i
}
i ≥1
are i.i.d. Also from 10, it follows that E
0,0
T
1
= ∞. Let M = Mγ = 6
C
1
γ
and find K = K γ such that
E
0,0
T
1
∧ K ≥ M. Now,
P
n i, j
|I
n
| ≥ γn, L
n
≤ C
1
n ≤ P
n i, j
T
1
+ T
2
+ · · · + T
[γn]−1
≤ C
1
n ≤ P
n i, j
T
1
+ T
2
+ · · · T
[γn]−1
[γn] − 1 ≤ 4
C
1
γ ≤ P
n i, j
T
1
∧ K + T
2
∧ K + · · · T
[γn]−1
∧ K [γn] − 1
≤ 4 C
1
γ ≤ exp[−nb
2
]. for some b
2
0. The last inequality follows from standard large deviation theory. Also P
n i, j
|I
n
| γn, c
1
+ ε ≤ P
j ∈I
n
, j n
∆L
j
n +
P
j ∈I
n
∆L
j
n ≤ C
1
≤ P
n i, j
|I
n
| γn, c
1
+ ε
2 ≤
1 n
X
j ∈I
n
, j n
∆L
j
+ P
n i, j
|I
n
| γn, 1
n X
j ∈I
n
∆L
j
≥ ε
2 .
Let {M
j
}
j ≥1
be i.i.d. with the distribution of L
1
under P
0,e
1
and {N
j
}
j ≥1
be i.i.d. with the distribution of L
1
under P
0,0
. We thus have that the above expression is less than P
1 n
n
X
j=1
M
j
≥ c
1
+ ε
2 + P
1 n
[γn]
X
j=1
N
j
≥ ε
2 ≤ exp−nb
3
ε + P 1
[γn]
[γn]
X
j=1
N
j
≥ ε
2 γ
≤ exp−nb
3
ε + exp−nb
4
ε. for some b
3
ε, b
4
ε 0. Recall that
ε 2
γ
c by our choice of
γ. Combining all the inequalities, we have
P
n i, j
L
n
n ≥ c
1
+ ε ≤ 3 exp−nbε.
for some b ε 0. The proof of the second inequality is similar.
Returning to 9, let us separate the sum into two parts as 1
p n
[
n1+ ε
c1
]
X
k=0
P
n i, j
Y
k
= 0 and L
k
≤ n − 1
+ 1
p n
n −1
X
k=[
n1+ ε
c1
]+1
P
n i, j
Y
k
= 0 and L
k
≤ n − 1. Now the second term above is
1 p
n
n −1
X
k=[
n1+ ε
c1
]+1
P
n i, j
X
λ
Lk
= ˜ X
˜ λ
Lk
and L
k
≤ n − 1 ≤ n
p n
P
n i, j
L
[
n1+ ε
c1
]
≤ n
1277
which goes to 0 as n tends to infinity by Lemma 3.6. Similarly 1
p n
h
[
n1 −ε
c1
]
X
k=0
P
n i, j
Y
k
= 0 −
[
n1 −ε
c1
]
X
k=0
P
n i, j
Y
k
= 0, L
k
≤ n − 1 i
≤ 1
p n
[
n1 −ε
c1
]
X
k=0
P
n i, j
L
k
≥ n ≤
n p
n P
n i, j
L
[
n1 −ε
c1
]
≥ n also goes to 0 as n tends to infinity. Thus
1 p
n
n −1
X
k=0
P
n i, j
Y
k
= 0, L
k
≤ n − 1
= 1
p n
[
n1 −ε
c1
]
X
k=0
P
n i, j
Y
k
= 0 + 1
p n
[
n1+ ε
c1
]
X
k=[
n1 −ε
c1
]+1
P
n i, j
Y
k
= 0, L
k
≤ n − 1 + a
n
ε where a
n
ε → 0 as n → ∞. Now we will show the second term in in the right hand side of the above equation is negligible. Let
τ = min{ j ≥ [
n1 −ε
c
1
] + 1 : Y
j
= 0}. Using the Markov property for the second line below, we get
1 p
n
[
n1+ ε
c1
]
X
k=[
n1 −ε
c1
]+1
P
n i, j
Y
k
= 0, L
k
≤ n − 1 =
1 p
n E
n i, j
h
[
n1+ ε
c1
]
X
k= τ
I
{Y
k
=0,L
k
≤n−1}
i
≤ 1
p n
E
0,0
h
[
2n ε
c1
]
X
k=0
I
{Y
k
=0}
i
= 1
p n
E
0,0
h I
{L
[ 2nε c1
]
≤4nε} [
2n ε
c1
]
X
k=0
I
{Y
k
=0}
i
+ 1
p n
E
0,0
h I
{L
[ 2nε c1
]
4nε} [
2n ε
c1
]
X
k=0
I
{Y
k
=0}
i .
In the expression after the last equality, using 10 we have First term
≤ 1
p n
E
0,0
X
[0,4nε]
\ ˜ X
[0,4nε]
≤ C
p n
p 4n
ε ≤ C p
ε.
Second term ≤
C n p
n P
0,0
L
[
2n ε
c1
]
4nε −→ 0.
by Lemma 3.6. This gives us 1
p n
n −1
X
k=0
P
n i, j
Y
k
= 0, L
k
≤ n − 1 = 1
p n
[
n1 −ε
c1
]
X
k=0
P
n i, j
Y
k
= 0 + O p
ε + b
n
ε 1278
where b
n
ε → 0 as n → ∞. By the Markov property again and arguments similar to above 1
p n
n c1
X
k=
n1 −ε
c1
+1
P
n i, j
Y
k
= 0 ≤ 1
p n
n ε
c1
X
k=0
P
0,0
Y
k
= 0 ≤ O
p ε + c
n
ε where c
n
ε → 0 as n → ∞. So what we finally have is 1
p n
n −1
X
k=0
P
n i, j
Y
k
= 0, L
k
≤ n − 1 = 1
p n
[
n c1
]
X
k=0
P
n i, j
Y
k
= 0 + O p
ε + d
n
ε 12
where d
n
ε → 0 as n → ∞.
3.2 Control on the Green’s function
