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

Let b e (x) = ¯B(x)/m 1 = ∑ I ˜β i=1 i e −bix , where ˜β i = β i /m 1 , denote the ”stationary excess
density/integrated tail/ladder distribution”, and let b ∗ e (s) denote it’s Laplace transform.
Let l(s) = κ(s) = c −λ ¯B ∗ (s), (this is the reciprocal of the Wiener-Hopf factor φ − (s)) and
s
let −r i , r i > 0, i = 1, ..., I denote the non-zero (negative) roots of the ”simplified Cramér
Lundberg ” equation :
0 = l(s) ⇐⇒ c λ = ¯B ∗ (s) ⇐⇒ b ∗ e (s) = ρ−1 = 1 + θ ⇐⇒ c − λ
I∑
i=1
β i
1
b i + s = 0
Decomposing in simple fractions the Pollaczek-Khinchine formula (4.8) it follows that :
( I∑
)
ψ ∗ (s) = 1 s − l(0)
sl(s) = 1 s − C i
+ 1 I∑ C i
=
s + r
i=1 i s r
i=1 i + s
and therefore
Therefore,
ψ(x) =
I∑
C i e −r ix
i=1
ψ(x) =
where C i = − κ′ (0)
κ ′ (−r i )
I∑
C i e −r ix
i=1
(4.19) CLconst
(4.20) Ch
and thus finding the ruin probability reduces to solving the ”simplified Cramér Lundberg ”
equation (??), simple fractions decomposition and applying the explicit formula (4.20). Of
course, the first task can only be achieved analytically in particular cases § .
Exemple 4.7.3 Let c = 1, λ = 2, and ¯B ∗ (s) = 1 1
+ 3 1
. The ”simplified symbol” is :
4 2+s 4 4+s
( 1 1
l(s) = 1 − 2
4 2 + s + 3 )
1 4(1 + s)(3 + s)
=
4 4 + s 3(s + 2)(s + 4) =⇒ r 1 = 1, r 2 = 3
and the coefficients C i (obtained for example by partial fractions of ψ ∗ (s) = 1 − 3(s+2)(s+4) ,
s 8s(1+s)(3+s)
or by (4.19)) are : C 1 = 9/16, C 2 = 1/16.
( 1
) ( )
1 1 −1
Alternatively, the Cauchy matrix is
be obtained from the system
( 1 −1
1
3
1
) (C1
C 2
)
=
( 1
2 1
4)
2−1
1
4−1
2−3
1
4−3
=
=⇒
(
C1
C 2
)
= 3 4
1
1
3
and the coefficients C i may
( ) ( 1 1 1
− 1 2
1 1
3 4)
§ To get examples of this situation, consider the case when b i and the roots r i are integers (for example,
chose r i arbitrary under the necessary constraints 0 < r 1 < b 1 < r 2 ... < b I ).
1
Form next the ”Cauchy matrix” C A denote with elements
b i−r k
, i, k = 1, ..., I,, which intervenes naturally,
since the Cramér Lundberg equation may also be written in system form β C A = c λ 1 where β = (β 1, ..., β I ).
Note now that 1) the vector C = (C 1 , ..., C I ) satisfies the linear system C A C = b (−1) where b (−1) :=
(b −1
1
, ..., b−1
I
)), and thus is completely determined once b i and r i have been chosen.
2) Furthermore, we may determine ψ(0) = ∑ i C i and the safety loading θ. The equation c = λm 1 (1 + θ)
fixes then the quotient c/λ.
3) Finally, β may be computed from (??).