Correction des exercices du livre La Gestion des Risques Financiers
Correction des exercices du livre La Gestion des Risques Financiers
Correction des exercices du livre La Gestion des Risques Financiers
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1.17 Les fonctions copules (TR-GDR, page 540)1. (a) On a :C (u, 0) = C (0, u) = 0C (u, 1) = C (1, u) = u∂ 2 C (u 1 , u 2 )∂ u 1 ∂ u 2= 1 + θ (1 − 2u 1 ) (1 − 2u 2 )Comme −1 ≤ 1 + θ (1 − 2u 1 ) (1 − 2u 2 ) ≤ 1, on en dé<strong>du</strong>it que∂ 2 C (u 1 , u 2 )∂ u 1 ∂ u 2≥ 0On en dé<strong>du</strong>it que C est une fonction copule (TR-GDR, page 260).(b) On a (TR-GDR, page 289) :On a :On en dé<strong>du</strong>it que :1 − 2u + C (u, u)λ = limu→1 − 1 − u(1 − 2u + u 2 1 + θ (1 − u) 2)= limu→1 − 1 − u= lim (1 − u) ( 1 + θu 2)u→1 −= 0∂ u1 C (u 1 , u 2 ) = u 2 + θu 2 − 2θu 1 u 2 − θu 2 2 + 2θu 1 u 2 2∂ u1 C (u 1 , u 2 ) ∂ u2 C (u 1 , u 2 ) = u 1 u 2 + θu 1 u 2 − 2θu 1 u 2 2 − θu 2 1u 2 + 2θu 2 1u 2 2 +θu 1 u 2 + θ 2 u 1 u 2 − 2θ 2 u 1 u 2 2 − θ 2 u 2 1u 2 + 2θ 2 u 2 1u 2 2 −2θu 2 1u 2 − 2θ 2 u 2 1u 2 + 4θ 2 u 2 1u 2 2 + 2θ 2 u 3 1u 2 − 4θ 2 u 3 1u 2 2 −θu 1 u 2 2 − θ 2 u 1 u 2 2 + 2θ 2 u 1 u 3 2 + θ 2 u 2 1u 2 2 − 2θ 2 u 2 1u 3 2 +Comme on a :∫∫2θu 2 1u 2 2 + 2θ 2 u 2 1u 2 2 − 4θ 2 u 2 1u 3 2 − 2θ 2 u 3 1u 2 2 + 4θ 2 u 3 1u 3 2= u 1 u 2(1 + 2θ + θ2 ) +u 1 u 2 2(−3θ − 3θ2 ) + u 2 1u 2(−3θ − 3θ2 ) +u 2 1u 2 (2 4θ + 9θ2 ) + u 3 (1u 2 2θ2 ) + u 1 u 3 (2 2θ2 ) +u 3 1u 2 (2 −6θ2 ) + u 2 1u 3 (2 −6θ2 ) + u 3 1u 3 (2 4θ2 )u 1 u 2 <strong>du</strong> 1 <strong>du</strong> 2 = 1[0,1] 2 4∫∫∫∫u 2 1u 2 <strong>du</strong> 1 <strong>du</strong> 2 = u 1 u 2 2 <strong>du</strong> 1 <strong>du</strong> 2 = 1[0,1] 2 [0,1] 2 6∫∫u 2 1u 2 2 <strong>du</strong> 1 <strong>du</strong> 2 = 1[0,1] 2 9∫∫∫∫u 1 u 3 2 <strong>du</strong> 1 <strong>du</strong> 2 = u 3 1u 2 <strong>du</strong> 1 <strong>du</strong> 2 = 1[0,1] 2 [0,1] 2 8∫∫∫∫u 2 1u 3 2 <strong>du</strong> 1 <strong>du</strong> 2 = u 3 1u 2 2 <strong>du</strong> 1 <strong>du</strong> 2 = 1[0,1] 2 [0,1] 2 12∫∫u 3 1u 3 2 <strong>du</strong> 1 <strong>du</strong> 2 = 1[0,1] 2 1642