Collusion - E-Cours - Université de la Réunion

, (1 ) (x) +On résoud pour le cas particulier où = 1 2 . (1 ) m m1 2 dp (x) , 2 (x) dp (x) (x)La première condition devient :1 m2 2 (x) d m 2 , 1 2(A c) 28(A 2x c) x!9 (A c)264(A c) 28, 1 2(A 2x c) x 9 (A c)264 = (A c) 2 4 2 x =2x =23 (A c)2323 (A c) 2, 2x 2 (A c) x +163 (A c)232= (A c) 2 34 (A c)2 = 1 (A c)24 1 1(A c)42 (A c) ; 1 (A c) + 1 42 (A c) 18 (A c) ; 3 8 (A c) 0La seconde condition devient :1 m2 2 (x) dp (x) (x) , 1 2"(A c) 28(A 2x c) x# 1 4 (A x c)2 (A 2x c) x,,(A c)2 1 16 4 (A x 1c)2 (A 2x c) x2(A c)2 1 h(A c) 2 2 (A c) x + x 2i 116 42 (A c) x + x2 , 0 5 4 x2 (A c) x + 3 (A c)216 = (A c) 2 4 5 4 316 (A c)2 = (A c) 2 1516 (A c)2 = 1 (A c)216x 2x 2 12 5 4(A c)14 (A c) ; 310 (A c) ; 1 2 (A c) 12 5 4(A c) + 1 4 (A c) D’où, la collusion parfaite peut être soutenue, pour = 1 2, en suivant la stratégie proposée par Abreu(1986) et en choisissant x dans l’intervalle 38 (A c) ; 1 2 (A c) .13