10 1. BASICS ON LARGE DEVIATIONS

(b) If there are constants C2 0 and b ≥ 2 such that

lim sup

n→∞

1

bn

log E exp θbnZj (n) ≤

C2θb

∀θ 0

for j = 1, · · · , l, then

lim sup

n→∞

1

bn

log E exp θbn Z1 2(n) + · · · + Zl2(n) ≤

C2θb

∀θ 0.

Proof. Clearly, part (a) needs only to be proved in the case l = 2. Given 0 δ λ,

let 0 = x0 x1 · · · xN = λ be a partition of [0,λ] such that xk − xk−1 δ.

Then

P Z1(n) + Z2(n) ≥ λ ≤

N

k=1

P{Z1(n) ≥ xk−1}P{Z2(n) ≥ λ − xk}.

Consequently,

lim sup

n→∞

1

bn

log P Z1(n) + Z2(n) ≥ λ ≤ −C1 min

1≤k≤N

xk−1

a

+ (λ −

xk)a

.

By the fact 0 a ≤ 1, xk−1 a + (λ − xk)a ≥ (λ − xk + xk−1)a ≥ (λ − δ)a. So we have

lim sup

n→∞

1

bn

log P Z1(n) + Z2(n) ≥ λ ≤ −C1(λ −

δ)a.

Letting δ → 0+ on the right hand side proves part (a).

We now come to part (b). For any 0, there are finitely many vectors

tj = (tj,1, · · · , tj,l) ∈

(R+)l

j = 1, · · · , N

such that

tj,1 2 + · · · + tj,l 2 = 1 j = 1, · · · , N,

x1

2

+ · · · + xl

2

≤ (1 + ) max

1≤j≤N

l

k=1

tj,kxk ∀x1, · · · , xl ∈

(R+)l.

Hence,

E exp θbn Z1

2(n)

+ · · · +

Zl2(n)

≤

N

j=1

E exp (1 + )θbn

l

k=1

tj,kZk(n)

=

N

j=1

l

k=1

E exp (1 + )θbntj,kZk(n) .

Consequently, by the fact that b ≥ 2,

lim sup

n→∞

1

bn

log E exp θbn Z1 2(n) + · · · + Zl2(n)

≤ C2(1 +

)bθb

max

1≤j≤N

l

k=1

tj,k

b

≤ C2(1 +

)bθb.

Letting →

0+

on the right hand side proves part (b).