By D.R. Cox

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**Read or Download Applied Statistics: Principles and Examples (Chapman & Hall CRC Texts in Statistical Science) PDF**

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13). Proof. 13). 14. 14 to find a constant a ≥ 0 so that ut ≤ a and vt ≤ a for all 0 ≤ t ≤ T . 11 we have t ht ≤ l(t)Px |φ(ξs , ut−s ) − φ(ξs , vt−s )|K(ds) 0 t ≤ La l(t) sup Px x∈E ht−s K(ds) . 2 Integral Evolution Equations 37 Then it is easy to get hs ≤ La k(t)l(t) sup sup 0≤s≤t hs , 0 ≤ t ≤ T. 0≤s≤t Take 0 < δ ≤ T so that La k(δ)l(δ) < 1. The above inequality implies ht = 0 and hence ut = vt for 0 ≤ t ≤ δ. 13). 11 with the constants L and La independent of n ≥ 1. Suppose that limn→∞ φn (x, f ) = φ(x, f ) uniformly on E × Ba (E)+ for every a ≥ 0 and for fn ∈ B(E)+ there is a unique locally bounded positive solution t → vn (t) = vn (t, x) to the equation vn (t, x) = Px e−Kt (β) fn (ξt ) − Px t e−Ks (β) φn (ξs , vn (t − s))K(ds) .

40) 0 Proof. 39). 39). 42) 0 where ∞ b=a− 0 u3 m(du). 41) holds. Proof. 39) as ∞ φ(λ) = b1 λ + cλ2 + e−λu − 1 + λu1{u≤1} m(du), 0 where ∞ b1 = a + 0 u − u1{u≤1} m(du). 43), for each λ > 0 we have u 1 − e−λu m(du) − φ (λ) = b1 + 2cλ + (0,1] ue−λu m(du). (1,∞) Then we use monotone convergence to the two integrals to get φ (0+) = b1 − um(du). (1,∞) If φ is locally Lipschitz, we have φ (0+) > −∞ and the integral on the right-hand side is finite. 41). 41) holds, then φ is bounded on each bounded interval and so φ is locally Lipschitz.

27) 0 where β ≥ 0 and (1 ∧ u)l(du) is a finite measure on (0, ∞). 39 The relation ψ = − log Lμ establishes a one-to-one correspondence between the functions ψ ∈ I and infinitely divisible probability measures μ on [0, ∞). 3 Let b > 0 and α > 0. The Gamma distribution γ on [0, ∞) with parameters (b, α) is defined by γ(B) = αb Γ (b) xb−1 e−αx dx, B ∈ B([0, ∞)). B This reduces to the exponential distribution when b = 1. The Gamma distribution has Laplace transform α α+λ Lγ (λ) = b λ ≥ 0. , It is easily seen that γ is infinitely divisible and its n-th root is the Gamma distribution with parameters (b/n, α).