2010
DOI: 10.1016/j.crma.2010.01.026
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Un contre-exemple à une conjecture de Hutchinson et Lai

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Cited by 3 publications
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“…x,y (T x 1 ,y 1 ), T x 1 ,y 1 ) ≥ 0 (11) holds. Moreoever, we have equality in (11) if and only if ϕ x 1 x,y (T x 1 ,y 1 ) is a triangular Pickands dependence function too, i.e. if ϕ x 1…”
Section: A Sharp Inequality Between τ and ρmentioning
confidence: 99%
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“…x,y (T x 1 ,y 1 ), T x 1 ,y 1 ) ≥ 0 (11) holds. Moreoever, we have equality in (11) if and only if ϕ x 1 x,y (T x 1 ,y 1 ) is a triangular Pickands dependence function too, i.e. if ϕ x 1…”
Section: A Sharp Inequality Between τ and ρmentioning
confidence: 99%
“…, x n , x n+1 ) with x 0 := 0 < x 1 < x 2 < • • • < x n < x n+1 < 1 =: x n+2 , fix B ∈ A x and set y 1 = B(x 1 ). Since B can be represented as B = ϕ (x 2 ,...,x n+1 ) x 1 ,y 1 (A) for some piecewise linear A with at most n vertices, applying inequality (7) and inequality (11) yields The family of all piecewise linear Pickands functions is dense in (A, • ∞ ). The remaining assertion is a direct consequence of the identities τ (T x 1 ,y 1 ) = 1−y 1 y 1 and ρ(T x 1 ,y 1 ) = 3(1−y 1 ) 1+y 1 .…”
Section: Theorem 33 Every Pickands Dependence Function a Fulfills ρ(A...mentioning
confidence: 99%
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