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 2γ ∼ log ∞ m=0 θm m! tm/(2γ) (m!)γeκm 1 2γ = log ∞ m=0 1 √ m! θt 1 2γ e κ 2γ m . Consider the decomposition ∞ m=0 1 √ m! θt 1 2γ e κ 2γ m = ∞ m=0 1 (2m)! θt 1 2γ e κ 2γ 2m + ∞ m=0 1 (2m + 1)! θt 1 2γ e κ 2γ 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 2γ e κ 2γ m ∼ log ∞ m=0 1 2mm! θt 1 2γ e κ 2γ 2m = 1 2 θ2t1/γeκ/γ. Summarizing our discussion, lim t→∞ t−1/γ log ∞ m=0 θm m! tm/(2γ) EY m 1 2γ = 1 2 θ2eκ/γ. Therefore, (1.2.22) follows from Theorem 1.2.6 and the fact that IΨ(λ) = 2γ sup θ0 θλ 1 2γ − 1 2 θ2eκ/γ = λ1/γγe−κ/γ (λ 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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