Processus de Lévy en Finance - Laboratoire de Probabilités et ...

162 CHAPTER 4. DEPENDENCE OF LEVY PROCESSES
Suppose now that S is not an ordered set.
Then there exist two points u, v ∈ S such
that u m > v m and u n < v n for some m and n. Moreover, either u i ≥ 0 and v i ≥ 0 for all
i or u i ≤ 0 and v i ≤ 0 for all i. Suppose that u i ≥ 0 and v i ≥ 0, the other case being
analogous. Let x = u+v
2 . Since u, v ∈ S, we have ν({z ∈ Rd : z m < x m , z n ≥ x n }) > 0 and
ν({z ∈ R d : z m ≥ x m , z n < x n }) > 0. However
ν({z ∈ R d : z m < x m , z n ≥ x n }) = U n (x n ) − U {m,n} (x m , x n )
= U n (x n ) − min(U m (x m ), U n (x n ))
and
ν({z ∈ R d : z m ≥ x m , z n < x n }) = U m (x m ) − min(U m (x m ), U n (x n )),
which is a contradiction because these expressions cannot be simultaneously positive.
For the last assertion, we assume that the tail integrals U i of X i are continuous and satisfy
lim x→0 U i (x) = ∞, i = 1, . . . , d. It suffices to show that ν(S n ) = 0 for any n in decomposition
(4.28). If ξ(n) ≠ 0, then
ν(S n ) = lim
ε↓0
(U k(n) (ξ(n) − ε) − U k(n) (ξ(n))) = 0,
because U k(n) is continuous. Suppose now that ξ(n) = 0. Since S n does not reduce to a single
point, we must have either x m > 0 or x m < 0 for some x ∈ S n and some m. Suppose that
x m > 0, the other case being analogous. Since S is ordered, we have
ν({x ∈ R d : x k(n) ≥ ε} ∩ S) ≤ ν({ξ ∈ R d : ξ m ≥ x m } ∩ S) < ∞
uniformly in ε > 0. This implies lim x↓0 U k(n) (x) < ∞ in contradiction to lim x→0 U k(n) (x) = ∞.
Hence, ξ(n) > 0 for any n. Therefore, ν(R d \ S ∗ ) = 0 and the proof is completed.
A characterization of complete dependence of spectrally positive Lévy processes can be
obtained as a corollary of Theorem 4.11.
Corollary 4.2. Let X be a R d -valued spectrally positive Lévy process whose Lévy measure is
supported by an ordered set S ⊂ R d +. Then
F ‖ (x 1 , . . . , x d )| [0,∞] d ≡ min(x 1 , . . . , x d )
is a Lévy copula of X.