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
2γ
∼ log
∞
m=0
θm
m!
tm/(2γ) (m!)γeκm
1
2γ
= log
∞
m=0
1
√
m!
θt
1
2γ
e
κ
2γ
m
.
Consider the decomposition
∞
m=0
1
√
m!
θt
1
2γ
e
κ
2γ
m
=
∞
m=0
1
(2m)!
θt
1
2γ
e
κ
2γ
2m
+
∞
m=0
1
(2m + 1)!
θt
1
2γ
e
κ
2γ
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
2γ
e
κ
2γ
m
∼ log
∞
m=0
1
2mm!
θt
1
2γ
e
κ
2γ
2m
=
1
2
θ2t1/γeκ/γ
.
Summarizing our discussion,
lim
t→∞
t−1/γ
log
∞
m=0
θm
m!
tm/(2γ)
EY
m
1
2γ
=
1
2
θ2eκ/γ.
Therefore, (1.2.22) follows from Theorem 1.2.6 and the fact that
IΨ(λ) = 2γ sup
θ0
θλ
1
2γ
−
1
2
θ2eκ/γ
=
λ1/γ γe−κ/γ
(λ 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.
