Utility Representation of Lower Separable Preferences

Transcription

1 Utility Representation of Lower Separable Preferences Özgür Yılmaz June 2008 Abstract Topological separability is crucial for the utility representation of a complete preference relation. When preferences are incomplete, this axiom has suitably defined counterparts: upper separability and lower separability (Ok (2002)). We consider the problem of representing an incomplete preference relation by means of a vector-valued utility function; we obtain representation results under the lower separability assumption. Our results extend the main representation theorems by Ok (2002) in terms of the separability axioms. Keywords: Incomplete Preferences, Utility Representation, Upper/Lower Separability, Near-completeness Journal of Economic Literature Classification Number: D11 1 Introduction Preference relations are usually assumed to satisfy completeness and transitivity. These two properties have been the core axioms, tying self-interest to rationality. They are also useful for analytical tractability. Particularly, the possibility of representation by real-valued functions owes much to the completeness axiom. However, for most problems of individual decision making, it seems unrealistic to assume completeness. Our focus is the utility representation without the completeness axiom. College of Administrative Science and Economics, Koç University, Sarıyer, İstanbul, Turkey 34450; Fax: ; address: ozyilmaz@ku.edu.tr 1

2 Obviously, an incomplete preorder can not be represented by a real-valued utility function. The scope of this limitation of analytical tractability is extensively discussed in the literature. 1 Basically, in the case of incomplete preferences, while utility representation is not possible, one may still find a utility function that represents the preference order in a weaker sense: as before, if x is preferred to y then u(x) > u(y), but the opposite implication is no longer true. This approach is explored by Richter (1966) and Peleg (1970). Clearly, this quasi-representation is very much problematic, because the preorder cannot be recovered from the utility function u. An alternative approach is representation by means of a vector-valued utility function. 2 The question is whether, given a preorder, it is possible to find a function u : X R n for some positive integer n such that, for each x, y X, x y if and only if u(x) u(y). As a natural extension of the classical utility representation, multi-criteria decision making is not only intuitive but also analytically convenient. Ok (2002) shows that, under certain topological separability assumptions, a not-too-incomplete preference relation, a preference relation such that any subset of mutually incomparable alternatives is finite, can be represented by a vector-valued utility function. First, he suggests two counterparts of the usual separability axiom, upper separability and lower separability, which are suitably defined for incomplete preferences. Then, he shows that upper separability is sufficient for the vector-valued utility representation; also, he shows that upper and lower separability together imply the existence of an upper semicontinuous vector-valued utility representation. We extend these results in terms of the separability axioms: we show that lower separability is sufficient for representability by means of a vector-valued utility function (Theorem 3); it follows from this result that lower separability is also the key axiom for the existence of such a function that is upper semicontinuous (Theorem 4). 1 See Aumann (1962), Richter (1966), Peleg (1970), and Sondermann (1980). 2 For a discussion of the multi-attribute utility representation, see Ok (2002) and Eliaz and Ok (2006). 2

3 2 Preliminary Definitions and Existing Results Let be a binary relation on a set of alternatives X. Two alternatives x and y are -comparable if either x y or y x holds. They are -incomparable if they are not -comparable. For each -incomparable pair of alternatives x and y, we write x y. A binary relation is complete if and only if each x and y in X are -comparable. The strict (asymmetric) part of, denoted by, is the relation on X defined as x y if and only if x y and (y x). The symmetric part of, denoted by, is the relation on X defined as x y if and only if x y and y x. For each Y X, the relation induced by on Y, denoted Y, is defined as Y (Y Y ). The relation is an extension of, if and. A relation is a preorder if is reflexive and transitive, a partial order if it is an antisymmetric preorder, a linear order if it is a complete partial order. A preordered set is a pair (X, ) such that X is a nonempty set and is a preorder on X. Let (X, ) be a preordered set. A subset Y of X is -dense if, for each pair x, y X with x y, there is z Y such that x z y. The preorder is weakly separable if there is a countable -dense set in X. Clearly, if is weakly separable, then X is infinite. A preordered set (X, ) is a poset (partially ordered set) if is a partial order on X. Given a poset (X, ), the partial order is spacious if x y implies that {z : z X and x z} Closure{z : z X and y z}. Theorem 1 Peleg (1970) Let be a partial order on a nonempty set X. If for each x X, {y : y X and x y} is open and is weakly separable and spacious, then there exists a continuous function u : X [0, 1] such that for each x, y X, x y implies u(x) > u(y). A poset (X, ) is a chain if is a linear order, and an antichain if =. Clearly, if (X, ) is a chain, then =, and if it is an antichain, then = {(x, y) : x y}. A chain in a poset (X, ) is a set Y X such that the relation Y is complete. Similarly, a set Y X is an antichain in a poset (X, ) if Y =. An antichain Y in (X, ) is of maximal cardinality if its cardinality is at least as large as that of each antichain in (X, ). In this case, the cardinality of Y is the width of the poset (X, ). It is denoted by w(x, ). Clearly, the width of each chain 3

4 is zero, while the width of an antichain (X, ) is X. An important result which we will use extensively in the following analysis is Dilworth s Decomposition Theorem (1950): If (X, ) is a poset with finite width w, then X is the union of w chains in (X, ). For each poset (X, ), let L(X, ) be the set of all extensions of that are linear orders. By the classical Szpilrajn s Theorem (1930), each partial order can be extended to a linear order. Thus, for each poset (X, ), L(X, ). From this result, it follows that, for each poset (X, ), = R L(X, ) R, that is, each partial order is the intersection of all of its linear extensions. The order dimension of a poset (X, ), denoted as dim(x, ), is the minimum number of linear extensions of, the intersection of which is, provided that this number is finite, and is, otherwise. That is, dim(x, ) min{k N : R i L(X, ), i = 1,..., k, and = n i=1 R i}. For each poset (X, ), w(x, ) dim(x, ) (Hiragushi (1955)). A partial order is representable if there is a positive integer n and a function u : X R n such that, for each x, y X, x y u(x) u(y). Proposition 1 Ok (2002) Let be a partial order on a nonempty set X. If is representable, then dim(x, ) <. Moreover, if X is countable and dim(x, ) <, then is representable. Since w(x, ) dim(x, ), it implies, that if X is countable and w(x, ) is finite, then is representable. Our focus, on the other hand, is on representability without the restriction that X is a countable set. We analyze this problem in the following section. 3 Representation of Near-Complete Preferences A natural conception of the degree of incompleteness is near-completeness: A preorder is nearcomplete, if each subset A of X, with x y for each distinct x, y in A, is finite (Ok (2002)). 4

5 Thus, a partial order on X is near-complete if and only if w(x, ) <. Let (X, ) be any preordered set. A subset Y of X is upper -dense if, for each x, y X such that x y, there is z Y such that x z y. It is lower -dense if, for each x, y X such that x y, there is z Y such that x z y. A preorder is upper (lower) separable if there is a countable set in X which is both -dense and upper (lower) -dense. Finally, is separable if it is both upper and lower separable. Theorem 2 Ok (2002) Let X be any nonempty set and let be a near-complete and upper separable partial order on X. Then, is representable. Our first result is the counterpart of this theorem in terms of the separability assumption. We replace upper separability with lower separability and obtain a representability result. Theorem 3 Let X be any nonempty set and let be a near-complete and lower separable partial order on X. Then, is representable. Proof. Let (X, ) be a poset such that its width is n <, and is lower separable. By Dilworth s Decomposition Theorem, there is a partition of X, say X 1, X 2,..., X n such that, for each k, Xk is a linear order on X k. For each k, define R k and R k as follows: x R k y x X k and z X k s.t. x z y, x R k y x X k and x y. Then define the relation k = R k R k. Step 1: k is a partial order. Reflexivity of k : It follows from reflexivity of. Antisymmetry of k : Let x k y k x. Suppose x y. There are three cases: i) x y. Thus, (y x), and it implies (y R k x). Since is antisymmetric, y R k x. By definition of R k, there is z X k such that y z x. Since x y, transitivity of implies that x z. It contradicts with z x. 5

6 ii) x R k y. By definition of R k, x X k and there is z X k such that x z y. Suppose y R k x. Then, there is z X k such that x z y z x. Clearly, z = z is impossible. If z z, then transitivity of implies x z. It contradicts with z x. If z z, transitivity of implies y z. It contradicts with z y. Note that since z and z are in X k, and X k is a chain, z and z are -comparable. Suppose y x. It implies x z y x. This contradicts with transitivity of. Finally, since z X k, y X k and (y R k x). iii) x R k y. By definition of R k, x X k and x y. Since X k is a chain, y X k and (y R k x). Thus, y R k x, and there is z X k such that y z x. But, since x, z X k, it is not possible. Thus, x k y k x implies x = y. Thus, k is antisymmetric. Transitivity of k : Let x k y k z. If x y z, then, by transitivity of, x z. Note that, x R k y implies y X k and (y R k z) for each z X. Thus, x R k y R k z is not possible. Also, since X k is a chain and x a y for some a X k, y X k. Then, for each z X, (y R k z). Thus, x R k y R k z is not possible. There are six cases left to be considered: i) x y R k z By definition of R k, y X k and there is a X k such that x y a z. If x X k, then transitivity of implies x R k z. Suppose x X k. Since x y a z, either x z or x z. If x z, then x R k z. In either case, x k z. ii) x R k y z There is a X k such that x a y z, where x X k. Since a and y are -incomparable, either a z or a z. If a z, then x z. If a z, then, since x X k and a X k, x a z implies x R k z. Thus, x k z. iii) x R k y R k z There are a, b X k such that x a y b z where x, y X k. Since X k is a chain, a and b are -comparable. Since x a y b, a b. If b a, then y a. It is a contradiction. Thus, a b. By transitivity of, x b z. Thus, x k z. iv) x y R k z 6

7 If x X k, then, since y X k and x y z, x R k z. Suppose x X k. If z x, then z y. It is a contradiction. Thus, either x z or x z. If x z, then, by definition of R k, x R k z. In both cases, x k z. v) x R k y z Clearly, z x is not possible. Thus, either x z or x z. Since x X k, the latter implies x R k z. Thus, x k z. vi) x R k y R k z There is a X k such that x y a z, where x X k. Clearly, a x is not possible. Since X k is a chain, and both x and a are in X k, x a. If z x, then z a. It is a contradiction. Thus, either x z or x z. Since x X k, the latter implies x R k z. In both cases, x k z. Thus, k is transitive. Thus, k is a partial order. Step 2: k is weakly separable. Since is lower separable, there is a countable set Y in X which is both -dense and lower -dense. We claim that Y is k -dense in X. Let x, y X such that x k y, where k is the strict part of k. i) x y Since Y is -dense, there is z Y such that x z y. Thus, there is z Y such that x k z k y. ii) x R k y By definition of R k, x X k and x y. By lower -denseness of Y, there is z Y such that x z y. Since x X k, x R k z. Clearly, x, y, z are all distinct. Thus, x k z k y. iii) x R k y There is a X k such that x a y, where x X k. Since Y is lower -dense, there is z Y such that x a z y. Then, x R k z. Since x and z are distinct, x k z k y. We conclude that there is a countable set Y in X, which is k -dense in X. Thus, k is a weakly separable partial order on X. Step 3: Constructing multi-vector utility indices that represent. We have shown that (X, k ) is a poset and for each k = 1,..., n, k is weakly separable. By 7

8 Theorem 1 above, there is a function ϕ k : X [0, 1] such that, for each x, y X and for each k, x k y implies ϕ k (x) > ϕ k (y). Note that, for each k, x y implies x k y and ϕ k (x) > ϕ k (y). Suppose there are x, y X such that x y. Since X 1,..., X n is a partition of X, and for each k, Xk is complete, there are i, j with i j, such that x X i and y X j. By definition of R k, x R k y and y R k x. Thus, x i y and y j x. Thus, ϕ i (x) > ϕ i (y) and ϕ j (y) > ϕ j (x). Finally, for each x X, define u(x) = (ϕ 1 (x),..., ϕ n (x)). This completes the proof. Our next result is on the upper semicontinuous representation of. This is an important result because upper semicontinuity is often sufficient for maximization of utility functions. Theorem 4 Let X be a topological space and let be a near-complete and lower separable partial order on X such that {y : x y} is open for each x X. Then there exists an upper semicontinuous mapping u :X [0, 1] n with n being the width of (X, ), such that x y if and only if u(x) u(y) for each x, y X. Proof. Following Ok (2002), we define the mapping u k : X [0, 1] as ϕ k (x) u k (a) = Inf b U(a) Sup t L(b) ϕ k (t) if U(a) = otherwise where ϕ k is the utility function in Theorem 3, for each a X, U(a) = {b : b a} and L(a) = {t : a t}. Let x, y X. If x y, then, by weak separability, there is z U(y) such that x z. By Theorem 1 above, ϕ k (x) > ϕ k (z). Since U(y) and u k (y) = Inf b U(y) Sup t L(b) ϕ k (t) Sup t L(z) ϕ k (t), where z U(y), Sup t L(z) ϕ k (t) u k (y). Thus, by definition of u k, u k (x) ϕ k (x) > ϕ k (z) Sup t L(z) ϕ k (t) u k (y). Thus, x y implies that, for each k, u k (x) > u k (y). Now, suppose x y. By lower separability, there is z X such that x z y. Since X 1,..., X n is a partition of X and each set in this 8

9 partition is a chain, x X k and z X k for some k. Thus, x k z and ϕ k (x) > ϕ k (z). Since z y, the inequalities above apply. Thus, u k (x) ϕ k (x) > ϕ k (z) Sup t L(z) ϕ k (t) u k (y). Since is symmetric, for some l k, u l (y) > u l (x). Thus, x (=)y holds if and only if, for each k, u k (x) > (=)u k (y) Thus is represented by u = (u 1,..., u n ). Finally, we need to show that, for each k, u k is upper semicontinuous. To this end, we repeat the argument in the proof of Theorem 2 in Ok (2002): First, note u 1 k [θ, 1] is closed for θ 0 or θ 1. For upper semicontinuity, let θ (0, 1). Let (x α ) α A be a sequence in X such that, for each α A, x α a and u k (x α ) θ. If, for some α A, a x α, then u k (a) u k (x α ) θ. Now, assume that, for each α A, either x α a or x α a. Then, lower separability of implies U(a). Let b U(a), and observe that, since L(b) is open, there is α(b) A such that x α(b) L(b). Thus, for each b U(a), Sup t L(b) ϕ k (t) u k (x α(b) ) θ. It implies u k (a) = Inf b U(a) Sup t L(b) ϕ k (t) θ. Thus, u k is upper semicontinuous. Since k is chosen arbitrarily, u is upper semicontinuous. This completes the proof. Ok (2002) proves a corollary of this theorem; the difference is that he assumes separability instead of lower separability. Thus, Theorem 4 extends his result to lower separable partial orders. 4 Conclusion We extend some of the existing results on the multi-vector utility representation by Ok (2002) in terms of separability assumptions. In its technicality, it is restricted to near-complete preferences. We showed that a lower separable partial order is representable by means of a multi-vector utility function. We also show that a lower separable partial order can be represented by an upper semicontinuous vector-valued utility function. 9

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