20 1. BASICS ON LARGE DEVIATIONS is stochastically bounded. Then there is a θ 0 such that sup n E exp θYn/cn ∞. (1.3.2) Furthermore, if lim inf n→∞ cmn/cn 1 (1.3.3) for some m 1, then (1.3.2) holds for all θ 0. Proof. We first show that for any s, t 0 and n ≥ 1, max 1≤k≤n P Yk ≥ s + t + C ≤ P max 1≤k≤n Yk ≥ s max 1≤k≤n P Yk ≥ t . (1.3.4) Set Ts = inf j ≥ 1 Yj ≥ s . Notice that for any j ≥ 1, Yj − Yj−1 ≤ Y1 d = Y1 ≤ C a.s. For any 1 ≤ k ≤ n, P Yk ≥ s + t + C = P Ts ≤ k − 1 Yk ≥ s + t + C ≤ k−1 j=1 P Ts = j, Yk − Yj−1 ≥ t + C ≤ k−1 j=1 P Ts = j, Yk − Yj ≥ t where the second step follows from the fact that YT s −1 ≤ s. By sub-additivity, Yk − Yj ≤ Y k−j d = Yk−j and Y k−j is independent of {Ts = j}. Therefore, P Yk ≥ s + t + C ≤ k−1 j=1 P Ts = j P Yk−j ≥ t ≤ P Ts ≤ n max 1≤k≤n P Yk ≥ t = P max 1≤k≤n Yk ≥ s max 1≤k≤n P Yk ≥ t . Taking maximum over k on the left hand side leads to (1.3.4). For any N 0 and integers m, n ≥ 1, by (1.3.4), max 1≤k≤n P Yk ≥ m(Ncn + C) ≤ P max 1≤k≤n Yk ≥ Ncn m . In particular, P Yn ≥ m(Ncn + C) ≤ P max 1≤k≤n Yk ≥ Ncn m . By assumption, one can make N 0 suﬃciently large, so that P max 1≤k≤n Yk ≥ Ncn ≤ e−2 n = 1, 2, · · · . Therefore, sup n≥1 P Yn ≥ m(Ncn + C) ≤ e−2m m = 1, 2, · · · .

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