4 1. BASICS ON LARGE DEVIATIONS

Similarly, one can show that

lim sup

n→∞

1

bn

log P{Yn ≤ α } ≤ −l.

Summarizing our argument, we have

lim sup

n→∞

1

bn

log P{Yn ∈ F } ≤ −l.

Letting l → inf

λ∈F

Λ∗(λ)

on the right hand side leads to (1.1.4).

The proof of lower bound (1.1.5) is harder. To do this we claim that for any

closed (bounded or unbounded) interval I ⊂ R and θ ∈ DΛ,o

lim sup

n→∞

1

bn

log E exp θbnYn 1{Yn∈I} ≤ sup

λ∈I

θλ −

Λ∗(λ)

. (1.1.6)

Due to similarity, we only deal with the case θ ≥ 0. We first consider the case

when I = [a, b] is bounded. For 0, let a = x0 x1 · · · xm = b be a

partition of I such that xj − xj−1 . Thus,

E exp θbnYn 1{Yn∈I} ≤

m

j=1

exp θbnxj P{Yn ∈ [xj−1,xj]}.

Consequently, by (1.1.4) we get

lim sup

n→∞

1

bn

log E exp θbnYn 1{Yn∈I}

≤ max

1≤j≤m

θxj − inf

xj−1≤λ≤xj

Λ∗(λ)

≤ θ + sup

λ∈I

θλ −

Λ∗(λ)

.

Letting →

0+

leads to (1.1.6).

We now prove (1.1.6) in the case when I is unbounded. By duality,

Λ(θ) = sup

λ∈R

λθ −

Λ∗(λ)

θ ∈ R.

Consequently, (1.1.6) is equivalent to Assumption 1.1.1 when I = (−∞, ∞). The

remaining case is when I is a half line. For similarity we let I be unbounded from

above. Given a large M 0, write IM = I ∩ (−∞,M]. We have

E exp θbnYn 1{Yn∈I}

≤ E exp θbnYn 1{Yn∈IM

}

+ E exp θbnYn 1{Yn≥M} .

By what has been established,

lim sup

n→∞

1

bn

log E exp θbnYn 1{Yn∈I}

≤ max lim sup

n→∞

1

bn

log E exp θbnYn 1{Yn≥M} , sup

λ∈I

θλ −

Λ∗(λ)

.