Chapter V: The Order-Dependent Luce Model
E.5 Completing revealed order R 0
Proposition 16 shows that R0 is almost complete. In particular, there are only one pair that is not comparable by R0. Now we will discuss how to complete R0 using the following observation.
Observation 13: Take any ODLM (u,R). For anya,b,c,d ∈ X withaRbRcRd, p(a,{a,b,d})
p(d,{a,b,d}) = u(a,1)
u(d,3) = p(a,{a,c,d}) p(d,{a,c,d}) and p(b,{a,b,d})
p(d,{a,b,d}) = u(b,2)
u(d,3) , u(b,1)
u(d,3) = p(b,{b,c,d}) p(d,{b,c,d}).
Observation 13 says that since the rankings of the highest ranking alternativea and the lowest ranking alternativeddoes not change, replacingbwithccannot affect the probability of choosinga relative to that of choosing d. However, replacing a withcaffects the probability of choosingbrelative to that of choosing das long as u(b,1) ,u(b,2).
Now we discuss how to completeR0using the idea of Observation 3. Suppose there is a pair (a∗,b∗) such that neither a∗R0b∗ nor b∗R0a∗. Take any c,d such that c,d ∈ X \ {a∗,b∗}. By Proposition 16, a∗,b∗R0c,d and either cR0d or dR0c. Without loss of generality, let us assumecR0d.
By Observation 13, if
p(a∗,{a∗,b∗,d})
p(d,{a∗,b∗,d}) = p(a∗,{a∗,c,d}) p(d,{a∗,c,d}),
then we completeR0in the way thata∗has a higher ranking thanb∗. However, if p(a∗,{a∗,b∗,d})
p(d,{a∗,b∗,d}) , p(a∗,{a∗,c,d}) p(d,{a∗,c,d}),
then we complete R0 in the way that b∗ has a higher ranking than a∗. As long as u(x,1), u(x,2)for anyx ∈ X, we can uniquely identify an unknownR.
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