14 1. BASICS ON LARGE DEVIATIONS

For any 0 δ , by Stirling formula there is c 0 such that

θ[pm]+1

([pm] + 1)!

≥

cp

θpm

(m!)p

1 + δ

p(1 + )

pm

.

Thus,

(1 +

δ)−m

E exp θbnYn

1/p

1/p

≥ c

1

m!

θ

p(1 + )

m

bn

m EYnm

1/p

whenever

m ∈ Q ≡ m; bn

m

EYn

m

1/p

≥ 1 .

Summing up over m,

1 + δ

δ

E exp θbnYn

1/p

1/p

≥ c

m∈Q

1

m!

θ

(1 + )p

m

bn

m

EYn

m

1/p

.

Consequently,

1 + δ

δ

E exp θbnYn

1/p

1/p

+ c exp

θ

(1 + )p

(1.2.15)

≥ c

∞

m=1

1

m!

θ

(1 + )p

m

bn

m

EYn

m

1/p

.

Thus,

max 0,

1

p

lim inf

n→∞

1

bn

log E exp θbnYn

1/p

≥ Ψ

θ

(1 + )p

.

Letting →

0+

on the right hand side, by the lower semi-continuity of Ψ(·), we get

max 0,

1

p

lim inf

n→∞

1

bn

log E exp θbnYn

1/p

≥ Ψ

θ

p

.

Since the right hand side is positive, we have proved (1.2.12).

The same estimate can be used to establish “(1.2.14) =⇒ (1.2.13)”. To proceed

we may assume that the left hand side of (1.2.13) is positive.

Replacing θ by (1 + )pθ in (1.2.15) gives

1 + δ

δ

E exp (1 + )pθbnYn

1/p

1/p

+

ceθ

≥ c

∞

m=1

θm

m!

bn

m

EYn

m

1/p

. (1.2.16)

By (1.2.14) (with θ being replaced by (1 + )pθ), we get

max 0, Ψ

(

(1 + )θ

)

≥ lim sup

n→∞

1

bn

log

∞

m=0

θm

m!

bn

m

EYn

m

1/p

.

Since the right hand side is positive, we must have

lim sup

n→∞

1

bn

log

∞

m=0

θm

m!

bn

m

EYn

m

1/p

≤ Ψ

(

(1 + )θ

)

.

To continue, we may assume that θ ∈ DΨ (for otherwise (1.2.13) is trivial). By

continuity of Ψ(·), letting →

0+

on the right hand side gives (1.2.13).