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 +
)
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 +
)
.
To continue, we may assume that θ (for otherwise (1.2.13) is trivial). By
continuity of Ψ(·), letting
0+
on the right hand side gives (1.2.13).
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