20 1. BASICS ON LARGE DEVIATIONS
is stochastically bounded. Then there is a θ 0 such that
sup
n
E exp θYn/cn ∞. (1.3.2)
Furthermore, if
lim inf
n→∞
cmn/cn 1 (1.3.3)
for some m 1, then (1.3.2) holds for all θ 0.
Proof. We first show that for any s, t 0 and n 1,
max
1≤k≤n
P Yk s + t + C P max
1≤k≤n
Yk s max
1≤k≤n
P Yk t . (1.3.4)
Set
Ts = inf j 1; Yj s .
Notice that for any j 1,
Yj Yj−1 Y1
d
= Y1 C a.s.
For any 1 k n,
P Yk s + t + C = P Ts k 1; Yk s + t + C

k−1
j=1
P Ts = j, Yk Yj−1 t + C
k−1
j=1
P Ts = j, Yk Yj t
where the second step follows from the fact that YTs−1 s. By sub-additivity,
Yk Yj Yk−j
d
= Yk−j and Yk−j is independent of {Ts = j}. Therefore,
P Yk s + t + C
k−1
j=1
P Ts = j P Yk−j t
P Ts n max
1≤k≤n
P Yk t = P max
1≤k≤n
Yk s max
1≤k≤n
P Yk t .
Taking maximum over k on the left hand side leads to (1.3.4).
For any N 0 and integers m, n 1, by (1.3.4),
max
1≤k≤n
P Yk m(Ncn + C) P max
1≤k≤n
Yk Ncn
m
.
In particular,
P Yn m(Ncn + C) P max
1≤k≤n
Yk Ncn
m
.
By assumption, one can make N 0 sufficiently large, so that
P max
1≤k≤n
Yk Ncn
e−2
n = 1, 2, · · · .
Therefore,
sup
n≥1
P Yn m(Ncn + C)
e−2m
m = 1, 2, · · · .
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