Dual Random Utility Maximisation

and ˆp (a, B ∪ {b}) = 1 2
, so that in either case b impacts a in B.
Observe that the only role performed by Impact Consistency in the proof is that it
ensures, together with Modal Regularity that p is binary. So, we have immediately:
Corollary 1 A stochastic choice rule is an mdRUM if and only if it is binary and satisfies
Modal Regularity.
It should be easy to check whether a p is binary, and in this sense being binary can
serve as a good property for testing the model. However, Impact Consistency is behaviourally
a more interesting property because it offers indications on the ‘comparative
statics’ of the model across menus.
We conclude this section with a few remarks.
Remark 2 mdRUMs can accommodate violations of Weak Stochastic Transitivity as
well as of Regularity. For example consider p given by p (a, {a, b}) = p (a, {a, b, c}) =
p (b, {b, c}) = 2 3 , p (a, {a, c}) = 1 3
= p (b, {a, b, c}). Then p violates Weak Stochastic
Transitivity but it is an mdRUM generated by the rankings r 1 = acb and r 2 = bca with
˜α ({a, b}) = ˜α ({a, b, c}) = 2 3 and ˜α ({b, c}) = ˜α ({a, c}) = 1 3 .
Remark 3 The ordinal preference information contained in an mdRUM is entirely determined
by which choice of alternative, in each menu, is possible, impossible or certain,
not by the exact values of the choice probabilities. More precisely suppose that p
is an mdRUM generated by (r 1 , r 2 , ˜α) and that p ′ is an mdRUM generated by (r ′ 1 , r′ 2 , ˜α′ ).
If for all menus A, p (a, A) > 0 ⇔ p ′ (a, A) > 0, then it must be the case that r 1 = r ′ 1
and r 2 = r
2 ′ or r 2 = r
1 ′ and r 1 = r
2 ′ : that is, the rankings coincide up to a relabelling.
The exact probabilities of choice in each menu then serve to determine the probabilities
with which each ranking is maximised, but not to determine what those rankings look
like.
Remark 4 In an mdRUM, while the rankings can have arbitrarily small probability,
they always have positive probability. This is crucial to tell apart the case in which
the choice probability of an alternative a is zero because it is dominated by another
alternative in both rankings, from the case in which a is dominated say in r 1 and top in
r 2 , but r 2 occurs with zero probability in a given menu.
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