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

4.6. EXAMPLES OF LEVY COPULAS 161
where λ denotes the Lebesgue measure on R. Then V F‖ ((a, b]) = µ((a, b]) for any a ≤ b, and
therefore F ‖ is d-increasing. The margins of F have the same form as F , namely
F‖ I ((x i) i∈I ) = min |x i|1 {(−1,...,−1),(1,...,1)} ((sgn x i ) i∈I ) ∏ sgn x i . (4.30)
i∈I
i∈I
Therefore, the one-dimensional margins satisfy F {i} (u) = u.
The first part. Let x ∈ (0, ∞) d . Clearly, U(x) ≤ U k (x k ) for any k. On the other hand, since
S is an ordered set, we have
{y ∈ R d : x k ≤ y k } ∩ S = {y ∈ R d : x ≤ y} ∩ S
for some k. Indeed, suppose that this is not so. Then there exist points z 1 , . . . , z d ∈ S and
indices j 1 , . . . , j d such that zk k ≥ x k and zj k k
< x jk for k = 1, . . . , d. Choose the greatest element
among z 1 , . . . , z d (this is possible because they all belong to an ordered set) and call it z k .
Then z k j k
< x jk . However, by construction of z 1 , . . . , z d we also have z j k
jk
contradiction to the fact that z k is the greatest element. Therefore,
U(x) = min(U 1 (x 1 ), . . . , U d (x d )).
≥ x jk , which is a
Similarly, it can be shown that for every x ∈ (−∞, 0) d ,
U(x) = (−1) d min(|U 1 (x 1 )|, . . . , |U d (x d )|).
Since U(x) = 0 for any x /∈ K, we have shown that
U(x) = F ‖ (U 1 (x 1 ), . . . , U d (x d ))
for any x ∈ (R \ {0}) d . Since the marginal Lévy measures of X are also supported by ordered
sets and the margins of F ‖ have the same form as F ‖ , we have
U I ((x i ) i∈I ) = F I ‖ ((U i(x i )) i∈I ) (4.31)
for any I ⊂ {1, . . . , d} and any (x i ) i∈I ∈ (R \ {0}) I .
The converse statement. Let S := supp ν. Let us first show that S ⊆ K. Suppose that
this is not so. Then there exists x ∈ S such that for some m and n, x m < 0 and x n > 0 and
for every neighborhood N of x, ν(N) > 0. This implies that U {m,n} (x m /2, x n /2) > 0, which
contradicts Equation (4.31).