16 1. BASICS ON LARGE DEVIATIONS Theorem 1.2.8. Let Y 0 be a random variable such that (1.2.21) lim m→∞ 1 m log 1 (m!)γ EY m = κ for some γ 0 and κ R. Then (1.2.22) lim t→∞ t−1/γ log P{Y t} = −γe−κ/γ. Proof. We check the condition posed in Theorem 1.2.6 with Yt = Y/t, bt = t1/γ and p = 2γ. Indeed, for any θ 0, log m=0 θm m! tm/(2γ) EY m 1 log m=0 θm m! tm/(2γ) (m!)γeκm 1 = log m=0 1 m! θt 1 e κ m . Consider the decomposition m=0 1 m! θt 1 e κ m = m=0 1 (2m)! θt 1 e κ 2m + m=0 1 (2m + 1)! θt 1 e κ 2m+1 . By the Stirling formula, we get (2m)! = ( 1 + o(1) )m (2mm!)2 and (2m + 1)! = ( 1 + o(1) )m (2mm!)2 as m ∞. Thus, log m=0 1 m! θt 1 e κ m log m=0 1 2mm! θt 1 e κ 2m = 1 2 θ2t1/γeκ/γ. Summarizing our discussion, lim t→∞ t−1/γ log m=0 θm m! tm/(2γ) EY m 1 = 1 2 θ2eκ/γ. Therefore, (1.2.22) follows from Theorem 1.2.6 and the fact that IΨ(λ) = sup θ0 θλ 1 1 2 θ2eκ/γ = λ1/γγe−κ/γ 0). (so IΨ(1) = γeκ/γ appearing on the right hand side of (1.2.22)). The following theorem appears as an inverse to Theorem 1.2.7. Theorem 1.2.9. Let I(λ) be a non-decreasing rate function I(λ) on R+ with I(0) = 0.
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