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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).