Dual Random Utility Maximisation

(note that by construction we have u, v ̸= x k ). We can rule out the case u = v. For suppose
u = v = x j for some j. Then, recalling (1), p ( d, L 1 k+1 ∪ { x j
}) = p
(
d, L
1
k+1
) =
1
2
=
p ( c, L 1 (
k+1) = p c, L
1
k+1
∪ { }) (
x j and thus p xj , L 1 k+1 ∪ { })
x j = 0, which contradicts
(
Regularity and p x j , L 1 j ∪ { } )
x j > 0 with L 1 k+1 ⊂ L1 j .
Since by construction and Regularity p ( v, L 1 k+1 ∪ {v}) > 0, Lemma 1, Lemma 2
and (3) imply
)
p
(c, L 1 k+1 ∪ {v} = 0 (4)
Similarly, p ( v, L 1 k+1 ∪ {v}) > 0 (by construction and Regularity), Lemma 1, Lemma 2
and (2) imply
)
p
(d, L 1 k+1 ∪ {u} = 0 (5)
(i.e. u impacts d in L 1 k+1 ∪ {u} and v impacts c in L1 k+1
). Now consider the menu
L 1 k+1
∪ {u, v}. By (4), (5) and Regularity it must be:
)
)
p
(c, L 1 k+1 ∪ {u, v} = 0 = p
(d, L 1 k+1 ∪ {u, v}
It cannot be that p ( u, L 1 k+1 ∪ {u, v}) = 1, for otherwise Regularity would imply p ( u, L 1 k+1 ∪ {u}) =
1, contradicting p ( c, L 1 k+1 ∪ {u}) = p ( c, L 1 k+1) =
1
2
. Similarly, it cannot be that p ( v, {u, v} ∪ L 1 k+1) =
1. Finally, if either p ( v, ∪L 1 k+1 ∪ {u, v}) = 0 or p ( u, L 1 k+1 ∪ {u, v}) = 0, then in view of
(6) it would have to be p ( w, L 1 k+1 ∪ {u, v}) > 0 for some w ∈ L 1 k+1
. But this is impossible
since c ̸= w ̸= d by (6), and then by Regularity the contradiction p ( w, L 1 k+1) > 0
would follow. Therefore it must be
)
p
(u, L 1 k+1 ∪ {u, v} = 1 )
(v,
2 = p L 1 k+1 ∪ {u, v}
(6)
It follows that both
)
)
p
(u, L 1 k+1 ∪ {u} = p
(u, L 1 k+1 ∪ {u, v}
(7)
and
)
)
p
(v, L 1 k+1 ∪ {v} = p
(v, L 1 k+1 ∪ {u, v}
(8)
i.e. neither does u impact v in L 1 k+1 ∪ {v}, nor does v impact u in L1 k+1
∪ {u}. Suppose
w.l.o.g. that v is a predecessor of u, and let u = x j . By the inductive hypothesis v
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