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 Z2(n) 1 + · · · + Z2(n) l ≤ 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, xa k−1 + (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 t2 j,1 + · · · + t2 j,l = 1 j = 1, · · · , N, x2 1 + · · · + x2 l ≤ (1 + ) max 1≤j≤N l k=1 tj,kxk ∀x1, · · · , xl ∈ (R+)l. Hence, E exp θbn Z1(n) 2 + · · · + Z2(n) l ≤ 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 Z2(n) 1 + · · · + Z2(n) l ≤ C2(1 + )bθb max 1≤j≤N l k=1 tb j,k ≤ C2(1 + )bθb. Letting → 0+ on the right hand side proves part (b).
Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2010 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.