18 1. BASICS ON LARGE DEVIATIONS
we have

m=0
θm
m!
bn
m
E
(
Yn
m1{Ynl}
)
1/p

N
i=1

m=0
θm
m!
bn
mλi
m/p
P{Yn Ai}
1/p

N
i=1
exp θbn(λi−1 + ) P{Yn λi−1}
1/p
.
Consequently, by (1.2.25), we get
lim sup
n→∞
1
bn
log

m=0
θm
m!
bn
m
E
(
Yn
m1{Ynl}
)
1/p
θ + max
1≤i≤N
θλi−1
1/p

p−1I(λi−1)
θ + sup
λ0
θλ1/p

p−1I(λ)
.
Letting
0+
on the right hand side, we get
lim sup
n→∞
1
bn
log

m=0
θm
m!
bn
m
E
(
Yn
m1{Ynl}
)
1/p
sup
λ0
θλ1/p

p−1I(λ)
.
By the decomposition (1.2.28), we get
lim sup
n→∞
1
bn
log

m=0
θm
m!
bn
m EYnm
1/p
max sup
λ0
θλ1/p

p−1I(λ)
,
lim sup
n→∞
1
bn
log

m=0
θm
m!
bn
m
E
(
Yn
m1{Yn≥l}
)
1/p
.
In view of (1.2.25), letting l on the right hand side leads to (1.2.26).
The condition (1.2.26) is called uniform exponential integrability. It can be
examined through the following lemma.
Lemma 1.2.10. Given θ 0, the assumption (1.2.26) holds if there is 0
such that either one of the following happens:
(1.2.29) lim sup
n→∞
1
bn
log E exp p(1 + )θbnYn
1/p
∞,
(1.2.30) lim sup
n→∞
1
bn
log

m=0
(1 + )mθm
m!
bn
m
EYn
m
1/p
∞.
Proof. The conclusion follows from the second part of Lemma 1.2.6 and a standard
application of Chebyshev inequality.
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