Processus de Markov, de Levy, Files d'attente, Actuariat et Fiabilité ...

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FILES D’ATTENTE, FIABILITÉ, ACTUARIAT The Kolmogorov partial integro-differential equations above are seldom tractable analytically. Slightly easier to solve (and derive) is the perpetual version : G ψ(x) = 0, ψ(t, x) = 1 {x
(ρ is sometimes called ”geometric compounding parameter” (due to its role in the Benes- Pollaczek-Khinchin decomposition of ruin probabilities as geometric sums–see (?)) A00 and In conclusion ψ ∗ (s) = ∫ ∞ 0 ⇐⇒ ψ ∗ (s) = ψ(0) = P 0 [Ȳ = 0] = 1 − ρ = κ′ (0) . c e −su ψ(u)du = λm 1 − λ ¯F ∗ (s) = 1 κ(s) s − κ′ (0) κ(s) = 1 s − p κ(s) , ∫ ∞ 0 e −su ψ(u)du = κ′ (0) κ(s) = 1 − ρ s(1 − ρb ∗ e (s)) (4.8) PK Note also that ψ ∗ (s) = ∫ ∞ e −su P[Ȳ ≤ u]du = s ∫ ∞ e −su 1−ρ f 0 0 Ȳ (u)du = s(1−ρb ∗ e (s)). Finally, we obtain the famous Pollaczek-Khinchine formula for the Laplace transform of the density of the supremum of a spectrally positive Levy process : φ(s) = ∫ ∞ 0 e −su f Ȳ (u)du = 1 − ρ (4.9) 1 − ρb ∗ PKf e (s), where φ(s) is defined by ψ ∗ (s) = φ(s) ⇐⇒ ψ ∗ (s) = 1−φ(s) § . s s In conclusion, a careful treatment of the singularity at 0 finds simultaneously the Laplace transform and the ”auxiliary boundary unknown” ψ(0). 4.6 Exponential claims Exemple 4.6.1 With exponential claim sizes of intensity µ, the ultimate ruin probability is : ψ(x) = ρe −γx . (4.10) expruin Here −γ = −µ+λ/c > 0 is the unique negative root of the Cramér Lundberg equation : κ(s) = 0, (4.11) CL with κ(s) = log ( E 0 e sX(1)) = cs−λs/(µ+s) being the ”cumulant generating function/Laplace exponent/symbol” of the process X(t) –see below, and ρ = − κ′ (0) κ ′ (−γ) = λ < 1. (4.12) CLct µc It turns out that whenever a negative solution −γ of (4.24) exists, the expression − κ′ (0) κ ′ (−γ) e−γx is asymptotic to ψ(x)! This is the famous Cramér Lundberg approximation. The ruin probability killed at rate q is also exponential : ψ q (x) = (1 + s − µ )exs − = λ/c µ + s + e xs − , the exponent being the unique negative root of the Cramér Lundberg equation : κ(s) = q, (4.13) CLq which reduces here to the quadratic equation cs 2 + s(cµ − λ − q) − qµ = 0. § in fact, φ(s) has a probabilistic interpretation, being one of the two factors of the celebrated Wiener-Hopf decomposition –see (?), Ber and φ(s) = E[e −sȲ ] = P[Ȳ = 0] + ∫ ∞ 0 e −su f Ȳ (u)du = sψ ∗ (s) = κ′ (0)s κ(s)
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Processus de Markov, de Levy, Files
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1.5.4 Les distributions et transfor
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Définition 1.1.2 Soit X . = X t ,
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3. La fonction génératrice des mo
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et du fait que les distributions co
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B U A O Fig. 1.1 - Marche aléatoir
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Trouvez T n . b) Calculez T(z) =
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3. Quand p = q = 1/2, b x = P x [X(
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En demandant que c n = c p (n) + A(
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De lors, ψ ∗ (z) = 1 1 − z −
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et alors c = −2 et c 2 = 2c 1 + 1
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toujours à des sommes closes. Une
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Solution : 1. C’est une équation
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Théorème 1.3.1 Une fonction monot
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Il est facile de voir que le nombre
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continu par un processus de Bernoul
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Le processus (X t ) t≥0 peut alor
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1.4 Les semigroupes de Markov homog
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Or, on sait que p 11 (0) = 1 donc C
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(a) Donnez la formule exacte, ainsi
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et x 1 = 2 1 + 1 1 = 7 , x 3 5 3 8
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Par exemple, les systémes associé
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1.5.3 Les probabilités d’absorbt
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Exercice 1.5.5 a) Calculez l’espe
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Corollaire 1.5.5 Soit X t un proces
Page 51 and 52: ( q1 p 1 | ) 0 ser Exercice 1.5.15
Page 53 and 54: 1.6.1 L’existence de la matrice d
Page 55 and 56: Le calcul des probabilités d’abs
Page 57 and 58: 1.6.4 Le théorème de Perron-Frobe
Page 59 and 60: En conclusion, l’étude de l’ex
Page 61 and 62: distribution stationnaire π ∞ de
Page 63 and 64: On aperçoit par la structure de ma
Page 65 and 66: 1.7 Exercices 1. Considérez une pa
Page 67 and 68: permutation cyclique de ses éléme
Page 69 and 70: ii. E ˜T 0 = 1 + 1 4 (t 2 + t 3 +
Page 71 and 72: et finalement p H y 1 + p F x 1 = p
Page 73 and 74: (c) En résolvant le système d’a
Page 75 and 76: Chapitre 2 Processus de Levy : marc
Page 77 and 78: Définition 2.2.1 Le processus de r
Page 79 and 80: Chapitre 3 Processus de naissance-e
Page 81 and 82: Rémarque : Dans tous ces cas, le g
Page 83 and 84: Théorème 3.2.1 (admis) Soit (X t
Page 85 and 86: 3. Montrez que la transformée de L
Page 87 and 88: 3. Obtenez et resolvez l’équatio
Page 89 and 90: Comme Gz x = z x (λ(z −1)+µ(z
Page 91 and 92: Exercice 3.6.3 Montrez que B(c, ρ)
Page 93 and 94: 3.7 Exercices 1. Soit X t une file
Page 95 and 96: 3.8 Les probabilités transitoires
Page 97 and 98: Rémarquons aussi que la fraction c
Page 99 and 100: fig1 X(t): Risk process: skipfree u
Page 101: Note : For Cramer Lundberg processe
Page 105 and 106: The unconditional distribution of a
Page 107 and 108: Let b e (x) = ¯B(x)/m 1 = ∑ I ˜
Page 109 and 110: Notes : 1) This was first derived v
Page 111 and 112: 1. Les probabilités de ruine satis
Page 113 and 114: 2. Soit Y (t) un processus de Cram
Page 115 and 116: 5. a) La distribution stationnaire
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