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
+
xk)a
.
By the fact 0 a 1, xk−1 a + xk)a xk + xk−1)a 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).
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