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.