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)
.
