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.