Dual Random Utility Maximisation

impacts u in L 1 j , so that by Contraction Consistency it also impacts u in L1 k+1 ∪ {u} ⊂
L 1 j , i.e.
)
)
p
(u, L 1 k+1 ∪ {u} > p
(u, L 1 k+1 ∪ {u, v}
a contradiction with (7). A symmetric argument applies if u is a predecessor of v using
(8).
A straightforward adaptation of the argument above shows that cases (2.i) and (2.ii)
are exhaustive.
3. Showing that the algorithm retrieves the observed choice.
(
r 1 , r 2 ,
2
1
Let p 1 be the dRUM generated by
2
(weak) lower contour set in ranking r i , that is
L i x = {x} ∪ {s ∈ X : x ≻ i s}
)
. For any alternative x denote L i x its
and note that by construction p ( x, L i x) > 0. We examine the possible cases of failures
of the algorithm in succession.
3.1: p (a, A) = 0 and p 1 (a, A) > 0.
2
Then a ≻ i a ′ for some i, for all a ′ ∈ A \ {a}, hence A ⊆ L i a, and thus by Regularity
and p ( a, L i a) > 0 we have p (a, A) > 0, a contradiction.
3.2: p (a, A) = 1 and p 1 (a, A) < 1.
2
Then there exists b ∈ B such that b ≻ i a ′ for some i, for all a ′ ∈ A \ {b}, hence
A ⊆ L i b , and thus by Regularity and p ( b, L i a) > 0 we have p (b, A) > 0, a contradiction.
3.3: p (a, A) = 1 2 and p 2
1 (a, A) = 1.
)
Let I
(p 1 be the set of all pairs (a, A) satisfying the conditions of this case. Fix an
2 )
)
)
(a, A) ∈ I
(p 1 that is maximal in I
(p 1 in the sense that (b, B) ∈ I
(p 1 ⇒ a ≻ i b
2 2 2
for some i.
Because p 1 (a, A) = 1 we have A ⊆ L 1
2
a ∩ L 2 a. If p ( a, L 1 a ∩ L 2 a) = 1 we have an immediate
contradiction, since by Regularity p (a, A) = 1. So it must be p ( a, L 1 a ∩ L 2 a) =
1
2
.
We show that this also leads to a contradiction.
19