3.1 Step 1
Recall 2.7 and set Y
n
= ǫ
1
+ · · ·+ǫ
n
for all n ≥ 1. The first step consists in proving that for all a b
in R lim
N →∞
P
δ,T
N
a Y
k
N
p v
δ
k
N
≤ b = Pa Z ≤ b ,
3.2 that is, under
P
δ,T
N
and as N → ∞ we have Y
k
N
p v
δ
k
N
=⇒ Z, where “=⇒” denotes convergence in distribution.
The random variables ǫ
1
, . . . , ǫ
N
, defined under P
δ,T
N
, are symmetric and i.i.d. . Moreover, they take their values in
{−1, 0, 1}, which together with A.6 entails E
δ,T
N
|ǫ
1
|
3
= E
δ,T
N
ǫ
1 2
−→ v
δ
as N → ∞.
3.3 Observe that k
N
→ ∞ as N → ∞ and E
δ,T
N
τ
1
= OT
3 N
, by 2.19. Thus, we can apply the Berry Esseèn Theorem that directly proves 3.2 and completes this step.
3.2 Step 2
Henceforth, we fix a sequence of integers V
N N
≥1
such that T
3 N
≪ V
N
≪ N. In this step we prove that, for all a
b ∈ R, the following convergence occurs, uniformly in u ∈ {0, . . . , 2V
N
}: lim
N →∞
P
δ,T
N
a
Y
L
N −u
p v
δ
k
N
≤ b
= Pa Z ≤ b . 3.4
To obtain 3.4, it is sufficient to prove that, as N → ∞ and under the law P
δ,T
N
, U
N
:= Y
k
N
p v
δ
k
N
=⇒ Z and
G
N
:= sup
u ∈{0,...,2V
N
}
Y
L
N −u
− Y
k
N
p v
δ
k
N
=⇒ 0 . 3.5
Step 1 gives directly the first relation in 3.5. To deal with the second relation, we must show that P
δ,T
N
G
N
≥ ǫ → 0 as N → ∞, for all ǫ 0. To this purpose, notice that {G
N
≥ ǫ} ⊆ A
N η
∪ B
N η,ǫ
, where for
η 0 we have set A
N η
: = L
N
− k
N
≥ ηk
N
∪ L
N −2V
N
− k
N
≤ −ηk
N
3.6 B
N η,ǫ
: = ¨
sup Y
k
N
+i
− Y
k
N
p v
δ
k
N
, i ∈ {−ηk
N
, . . . , ηk
N
} ≥ ǫ
« .
3.7 Let us focus on
P
δ,T
N
A
N η
. Introducing the centered variables f
τ
k
:= τ
k
− k · E
δ,T
N
τ
1
, for k ∈ N, by the Chebychev inequality we can write assuming that 1
− ηk
N
∈ N for notational convenience P
δ,T
N
L
N −2V
N
− k
N
−ηk
N
= P
δ,T
N
τ
1−ηk
N
N − 2V
N
= P
δ,T
N
e τ
1−ηk
N
N − 2V
N
− 1 − ηk
N
E
δ,T
N
τ
1
= ηN − 2V
N
≤ 1 − ηk
N
Var
δ,T
N
τ
1
ηN − 2V
N 2
≤ N Var
δ,T
N
τ
1
ηN − 2V
N 2
E
δ,T
N
τ
1
. 3.8
2049
With the help of the estimates in 2.19, 2.20, we can assert that Var
δ,T
N
τ
1
E
δ,T
N
τ
1
= OT
3 N
. Since N
≫ V
N
and N ≫ T
3 N
, the r.h.s. of 3.8 vanishes as N → ∞. With a similar technique, we
prove that P
δ,T
N
L
N
− k
N
ηk
N
→ 0 as well, and consequently P
δ,T
N
A
N η
→ 0 as N → ∞. At this stage it remains to show that, for every fixed
ǫ 0, the quantity P
δ,T
N
B
N η,ǫ
vanishes as η → 0, uniformly in N. This holds true because {Y
n
}
n
under P
δ,T
N
is a symmetric random walk, and therefore
{Y
k
N
+ j
− Y
k
N
2
}
j ≥0
is a submartingale and the same with j 7→ − j. Thus, the maximal
inequality yields P
δ,T
N
B
N η,ǫ
≤ 2
ǫ E
δ,T
N
Y
k
N
+ηk
N
− Y
k
N
2
v
δ
k
N
≤ 2
η E
δ,T
N
ǫ
2 1
ǫv
δ
≤ 2
η ǫ v
δ
. 3.9
We can therefore assert that the r.h.s in 3.9 tends to 0 as η → 0, uniformly in N. This completes
the step.
3.3 Step 3
Kategori