18 1. BASICS ON LARGE DEVIATIONS we have m=0 θm m! bn m E ( Yn m 1{Y n l} ) 1/p N i=1 m=0 θm m! bn m λm/p i P{Yn Ai} 1/p N i=1 exp θbn(λi−1 + ) P{Yn λi−1} 1/p . Consequently, by (1.2.25), we get lim sup n→∞ 1 bn log m=0 θm m! bn m E ( Yn m 1{Y n l} ) 1/p θ + max 1≤i≤N θλ1/p i−1 p−1I(λi−1) θ + sup λ0 θλ1/p p−1I(λ) . Letting 0+ on the right hand side, we get lim sup n→∞ 1 bn log m=0 θm m! bm n E ( Y m n 1{Y n l} ) 1/p sup λ0 θλ1/p p−1I(λ) . By the decomposition (1.2.28), we get lim sup n→∞ 1 bn log m=0 θm m! bm n EY m n 1/p max sup λ0 θλ1/p p−1I(λ) , lim sup n→∞ 1 bn log m=0 θm m! bm n E ( Y m n 1{Y n ≥l} ) 1/p . In view of (1.2.25), letting l on the right hand side leads to (1.2.26). The condition (1.2.26) is called uniform exponential integrability. It can be examined through the following lemma. Lemma 1.2.10. Given θ 0, the assumption (1.2.26) holds if there is 0 such that either one of the following happens: (1.2.29) lim sup n→∞ 1 bn log E exp p(1 + )θbnYn 1/p ∞, (1.2.30) lim sup n→∞ 1 bn log m=0 (1 + )mθm m! bn m EYn m 1/p ∞. Proof. The conclusion follows from the second part of Lemma 1.2.6 and a standard application of Chebyshev inequality.
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