Dual Random Utility Maximisation

Proof. Let p satisfy Constant Expansion, Regularity and Negative Expansion, let a ∈
B ⊆ A and b ∈ X and suppose by contradiction that p (a, A) > p (a, A ∪ {b}) but
p (a, B) ≤ p (a, B ∪ {b}). Since by Regularity p (a, B) ≥ p (a, B ∪ {b}), it can only be
p (a, B) = p (a, B ∪ {b}). Also, by Regularity it must be p (a, B) ≥ p (a, A). However
it cannot be p (a, B) = p (a, A), for in that case p (a, B ∪ {b}) = p (a, A) and Constant
Expansion would imply the contradiction p (a, A ∪ {b}) = p (a, (B ∪ {b}) ∪ A) =
p (a, A) > p (a, A ∪ {b}). So it must be p (a, B) > p (a, A), and the only possible configuration
of choice probabilities is p (a, B) = p (a, B ∪ {b}) > p (a, A) > p (a, A ∪ {b}).
Now w.l.o.g. let α > 1 − α. It cannot be that p (a, B ∪ {b}) = α and p (a, A) =
1 − α, since then Negative Expansion and p (a, B) ̸= p (a, A) imply the contradiction
p (a, A ∪ B) = p (a, A) = 0 > p (a, A ∪ {b}). Similarly, it cannot be p (a, A) =
α > p (a, A ∪ {b}) = 1 − α, for again Negative Expansion implies the contradiction
p (a, A ∪ {b}) = p (a, A ∪ (A ∪ {b})) = 0 ̸= 1 − α = p (a, A ∪ {b}). The only remaining
possibility is p (a, B ∪ {b}) = p (a, B) = 1, p (a, A) ∈ {α, 1 − α} and p (a, A ∪ {b}) =
0. By Regularity, p (a, B ∪ {b}) = 1 implies p (a, {a, b}) = 1. Moreover if p (a, A) >
p (a, A ∪ {b}) it must be p (b, A ∪ {b}) > 0 (if not, then there exist x, y ∈ A such
that p (x, A ∪ {b}) , p (y, A ∪ {b}) > 0 and p (x, A ∪ {b}) + p (y, A ∪ {b}) = 1, where
possibly x = y while a ̸= x, y since p (a, A ∪ {b}) = 0. But then by Regularity
p (x, A) ≥ p (x, A ∪ {b}) and p (y, A) ≥ p (y, A ∪ {b}), implying the contradiction
p (a, A) = 0). But if p (b, A ∪ {b}) > 0 the contradiction p (b, {a, b}) > 0 follows. We
conclude that Contraction Consistency holds.
□
Theorem 2 A stochastic choice rule is a dRUM if and only if it satisfies Regularity, Weak
Constant Expansion, Negative Expansion and Contraction Consistency.
Proof. Necessity is straightforward, so we show sufficiency.
Preliminaries and outline Let p satisfy Regularity, Weak Constant Expansion, Negative
Expansion and Contraction Consistency. The proof consists of three blocks: first
we define an algorithm to construct two rankings r 1 and r 2 explicitly (block 1), then we
show that the algorithm is well defined (block 2), and finally we show that r 1 and r 2 so
constructed retrieve p (block 3).
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