Dual Random Utility Maximisation

Proceed in an analogous way for the construction of r 1 , starting from y 1 , that is for
all i = 2, ...n define recursively
T i =
and as before 0 < |S i | ≤ 2.
L 2 i = X \
i−1 ⋃
j=1
{ ( )
t ∈ L 2 i : p t, L 2 i > p
y j
(
)}
t, L 2 i ∪ {y i−1}
If |T i | = {t}, then let y i = t, while if |T i | = 2 , then letting T i = {e, f }:
(2.i). p ( e, L 2 i
) > p
(
e, L
2
i
∪ { y j
})
(resp. p
(
e, L
2
i
) > p
(
e, L
2
i
∪ { y j
})
) for all j = 1, ...i − 1,
and p ( f , L 2 i
) = p
(
f , L
2
i
∪ { y j
})
(resp. p
(
e, L
2
i
) = p
(
e, L
2
i
∪ { y i−g
})
) for some j ∈
{1, ..., i − 2}.
In this case let y i = e (resp., f ).
(2.ii). p ( e, L 2 i
) > p
(
e, L
2
i
∪ { y j
})
and p
(
f , L
2
i
) > p
(
f , L
2
i
∪ { y j
})
for all j = 1, ...i − 1.
In this case let y i = f (resp., y i = e) whenever eP (r 1 ) f (resp., f P (r 1 ) e) (i.e. we
require consistency with the construction of the first order).
2. Showing that the algorithm is well-defined
We show that cases (1.i) and (1.ii) are exhaustive, which means showing that if
|S i | = 2 at least one alternative in S i is impacted by all its predecessors according
to r 1 . We proceed by induction on the index i. If i = 2 there is nothing to prove.
Now consider the step i = k + 1. If |S k+1 | = 1 again there is nothing to prove, so let
|S k+1 | = 2, with S k+1 = {c, d}. By construction p ( c, L 1 k+1) > p
(
c, L
1
k+1
∪ {x k } ) and
p ( d, L 1 k+1) > p
(
d, L
1
k+1
∪ {x k } ) , so that by Lemma 1 and Lemma 2 it must be
( )
p c, L 1 k+1 = 1 ( )
2 = p d, L 1 k+1
By contradiction, suppose that there exist u, v∈{x 1 , ...x k−1 } for which u does not
impact c in L 1 k+1 and v does not impact d in L1 k+1 , i.e.
) ( )
p
(c, L 1 k+1 ∪ {u} = p c, L 1 k+1 = 1 2
(1)
(2)
and
) ( )
p
(d, L 1 k+1 ∪ {v} = p d, L 1 k+1 = 1 2
(3)
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