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)
.
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