Etude des marchés d'assurance non-vie à l'aide d'équilibres de ...

tel-00703797, version 2 - 7 Jun 2012
Chapitre 4. Asymptotiques de la probabilité de ruine
Discrete case
If Y follows an exponential distribution E(θ), then W = ⌊Y ⌋ follows a geometric distribution
G(1 − e −θ ), where ⌊x⌋ denotes the floor function. Indeed, we have
P (W = k) = P (Y ∈ [k, k + 1[) = e −θk (1 − e −θ ).
Furthermore, P (W > x) = F Y (⌊x⌋ + 1) = e −θ(⌊x⌋+1) and P (W ≤ x) = FY (⌊x⌋ + 1) =
1 − e −θ(⌊x⌋+1) for all x ∈ R. Hence, the distribution function FW and FY only coincide for
integer values with a shift, i.e. FW (n) = FY (n + 1) for all n ∈ N.
Using the same assumption for Y1 and Y2 as in Subsection 4.4.1, we look at the dependence
of W1, W2 with Wi = ⌊Yi⌋, i = 1, 2. The distribution function is given by
FW1 (x) = P (W1 ≤ x) = P (⌊Y1⌋ ≤ x) = P (Y1 < ⌊x⌋ + 1) = P (Y1 ≤ ⌊x⌋ + 1) = 1 − LΘ(⌊x⌋ + 1),
for x ∈ R+. In this case, the random variable FWi (Wi) is not uniform on [0,1] since
P (FW1 (W1) ≤ u) = P (W1 ≤ L −1
Θ (1 − u) − 1) = 1 − LΘ
Their joint distribution function is given by
DW1,W2 (u, v) = P (FW1 (W1) ≤ u, FW2 (W2) ≤ v) = P (W1 ≤ L −1
=
∞
0
(1 − e θ(⌊L−1
Θ (1−u)−1⌋+1) )(1 − e θ(⌊L−1
Θ (1−v)−1⌋+1) )dFΘ(θ).
The distribution function of (FW1 (W1), FW2 (W2)) can be rewritten as
⌊L −1
Θ (1 − u)⌋ = u.
Θ (1 − u) − 1, W2 ≤ L −1
Θ
DW1,W2 (u, v) = 1 − LΘ (⌊lu⌋) − LΘ (⌊lv⌋) + LΘ (⌊lu⌋ + ⌊lv⌋) , (4.11)
where lp = L −1
Θ (1 − p). Equations (4.10) and (4.11) differ by the use of the floor function in
the arguments of LΘ.
Remark 4.4.1. If one uses another parametrization for Wi than the geometric distribution
G(1 − e −θ ), then we have to replace all the Laplace transform by an appropriate expectation.
For example, if we consider G(e −θ ), we use E((1 − e −Θ ) x ) instead of LΘ(x).
4.4.2 Differences between CY1,Y2 and DW1,W2
Before looking at the differences between CY1,Y2 and DW1,W2 , we check that these distribution
functions (defined in Equations (4.10) and (4.11)) are identical on the support of
(FW1 , FW2 ). Let Im(FWi ) be the inverse image by FWi of the integer set N. Let u ∈ Im(FW1 )
and v ∈ Im(FW2 ), i.e. there exist n, m ∈ N, such that u = FW1 (n) and v = FW2 (m). Firstly,
we have
u = FW1 (n) = FY1 (n + 1) and v = FW2 (m) = FY2 (m + 1).
Secondly, functions CY1,Y2
and DW1,W2 can be written as
CY1,Y2 (u, v) = P (FY1 (Y1) ≤ FY1 (n + 1), FY2 (Y2) ≤ FY2 (m + 1)) = P (Y1 ≤ n + 1, Y2 ≤ m + 1).
and
DW1,W2 (u, v) = P (FW1 (W1) ≤ FW1 (n), FW2 (W2) ≤ FW2 (m)) = P (W1 ≤ n, W2 ≤ m).
Hence, both functions CY1,Y2 (u, v) and DW1,W2 (u, v) are equal for u, v ∈ Im(FW1 ) × Im(FW2 ).
Now, we are going to determine the maximal distance between CY1,Y2 (u, v) and DW1,W2 (u, v).
For that, we begin by deriving two elementary lemmas.
198
(1 − v) − 1)