14 1. BASICS ON LARGE DEVIATIONS For any 0 δ , by Stirling formula there is c 0 such that θ[pm]+1 ([pm] + 1)! cp θpm (m!)p 1 + δ p(1 + ) pm . Thus, (1 + δ)−m E exp θbnYn 1/p 1/p c 1 m! θ p(1 + ) m bm n EY m n 1/p whenever m Q m bm n EY m n 1/p 1 . Summing up over m, 1 + δ δ E exp θbnYn 1/p 1/p c m∈Q 1 m! θ (1 + )p m bm n EY m n 1/p . Consequently, 1 + δ δ E exp θbnYn 1/p 1/p + c exp θ (1 + )p (1.2.15) c m=1 1 m! θ (1 + )p m bm n EY m n 1/p . Thus, max 0, 1 p lim inf n→∞ 1 bn log E exp θbnYn 1/p Ψ θ (1 + )p . Letting 0+ on the right hand side, by the lower semi-continuity of Ψ(·), we get max 0, 1 p lim inf n→∞ 1 bn log E exp θbnYn 1/p Ψ θ p . Since the right hand side is positive, we have proved (1.2.12). The same estimate can be used to establish “(1.2.14) =⇒ (1.2.13)”. To proceed we may assume that the left hand side of (1.2.13) is positive. Replacing θ by (1 + )pθ in (1.2.15) gives 1 + δ δ E exp (1 + )pθbnYn 1/p 1/p + ceθ c m=1 θm m! bm n EY m n 1/p . (1.2.16) By (1.2.14) (with θ being replaced by (1 + )pθ), we get max 0, Ψ ( (1 + ) lim sup n→∞ 1 bn log m=0 θm m! bn m EYn m 1/p . Since the right hand side is positive, we must have lim sup n→∞ 1 bn log m=0 θm m! bn m EYn m 1/p Ψ ( (1 + ) . To continue, we may assume that θ (for otherwise (1.2.13) is trivial). By continuity of Ψ(·), letting 0+ on the right hand side gives (1.2.13).
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