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tel-00703797, version 2 - 7 Jun 2012<br />

Chapitre 4. Asymptotiques <strong>de</strong> la probabilité <strong>de</strong> ruine<br />

The usual way to achieve this is to take a series expansion of f around x0 as g. f is said to<br />

take a series expansion at x0 if for all N ∈ N, we have<br />

f(x) = x0<br />

N<br />

anφn(x) + o(φN(x)) ,<br />

n=0<br />

where (φn)n is a sequence of so-called gauge functions such that ∀n ∈ N, φn+1(x) = o(φn(x))<br />

around x0. This condition is equivalent to<br />

N−1 <br />

f(x) − anφn(x) =<br />

x0<br />

n=0<br />

O(φN(x)) ,<br />

for all N ∈ N. In this paper, we also use the following notation for a series expansion of f at<br />

x0<br />

f(x) ∼ x0<br />

+∞<br />

n=0<br />

anφn(x).<br />

When x0 ∈ R, we generally choose φn(x) = (x − x0) n , whereas for x0 = +∞ we use<br />

φn(x) = x −n . Integration by part is a standard tool to study integral asymptotics and <strong>de</strong>rive<br />

asymptotics, as pointed in Olver et al. (2010). Integration by part will be extensively used in<br />

the next two subsections. In Appendix 4.6.5, we recall two integration by part theorems.<br />

4.3.2 Continuous time framework<br />

In this subsection, we present and show the A + B/u rule for the continuous time mo<strong>de</strong>l.<br />

Theorem 4.3.1. Let us consi<strong>de</strong>r the continuous time framework of Subsection 4.2.1 with a<br />

positive latent variable Θ and θ0 = λ/c.<br />

(i) For all u > 0, the ruin probability is boun<strong>de</strong>d<br />

182<br />

ψ(u) ≤ FΘ(θ0) + 1<br />

u<br />

FΘ(θ0)<br />

× .<br />

θ0<br />

(ii) If Θ has a continuous distribution with <strong>de</strong>nsity fΘ such that fΘ is almost everywhere<br />

differentiable on [θ0, +∞[ and f ′ Θ being a Lebesgue-integrable, then we have<br />

ψ(u) = FΘ(θ0) + fΘ(θ0)<br />

<br />

1<br />

+ o .<br />

u u<br />

(iii) If in addition fΘ is Ck-1 almost everywhere on [θ0, +∞[ and f (k)<br />

Θ<br />

and boun<strong>de</strong>d on [θ0, +∞[, then we have<br />

k−1<br />

ψ(u) = FΘ(θ0) +<br />

i=0<br />

where h(x) = θ0fΘ(x + θ0)/(x + θ0), so that<br />

h (i) (0) =<br />

i<br />

(−1) j<br />

j=0<br />

h (i) <br />

(0) 1<br />

+ o<br />

ui+1 uk <br />

,<br />

i!<br />

(i − j)!θ j f<br />

0<br />

(i−j)<br />

Θ (θ0).<br />

is Lebesgue integrable

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