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, · · · .
Previous Page Next Page