16 1. BASICS ON LARGE DEVIATIONS
Theorem 1.2.8. Let Y 0 be a random variable such that
(1.2.21) lim
m→∞
1
m
log
1
(m!)γ
EY
m
= κ
for some γ 0 and κ R. Then
(1.2.22) lim
t→∞
t−1/γ
log P{Y t} =
−γe−κ/γ.
Proof. We check the condition posed in Theorem 1.2.6 with Yt = Y/t, bt = t1/γ
and p = 2γ. Indeed, for any θ 0,
log

m=0
θm
m!
tm/(2γ)
EY
m
1

log

m=0
θm
m!
tm/(2γ) (m!)γeκm
1

= log

m=0
1

m!
θt
1

e
κ

m
.
Consider the decomposition

m=0
1

m!
θt
1

e
κ

m
=

m=0
1
(2m)!
θt
1

e
κ

2m
+

m=0
1
(2m + 1)!
θt
1

e
κ

2m+1
.
By the Stirling formula, we get
(2m)! =
(
1 + o(1)
)m
(2mm!)2
and (2m + 1)! =
(
1 + o(1)
)m
(2mm!)2
as m ∞.
Thus,
log

m=0
1

m!
θt
1

e
κ

m
log

m=0
1
2mm!
θt
1

e
κ

2m
=
1
2
θ2t1/γeκ/γ
.
Summarizing our discussion,
lim
t→∞
t−1/γ
log

m=0
θm
m!
tm/(2γ)
EY
m
1

=
1
2
θ2eκ/γ.
Therefore, (1.2.22) follows from Theorem 1.2.6 and the fact that
IΨ(λ) = sup
θ0
θλ
1


1
2
θ2eκ/γ
=
λ1/γ γe−κ/γ
0).
(so IΨ(1) =
γeκ/γ
appearing on the right hand side of (1.2.22)).
The following theorem appears as an inverse to Theorem 1.2.7.
Theorem 1.2.9. Let I(λ) be a non-decreasing rate function I(λ) on
R+
with
I(0) = 0.
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