Choice functions based on sets of strict partial orders: an axiomatic characterisation

03/25/2020 ∙ by Jasper De Bock, et al. ∙ Ghent University 0

Methods for choosing from a set of options are often based on a strict partial order on these options, or on a set of such partial orders. I here provide a very general axiomatic characterisation for choice functions of this form. It includes as special cases an axiomatic characterisation for choice functions based on (sets of) total orders, (sets of) weak orders, (sets of) coherent lower previsions and (sets of) probability measures.

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1 Introduction

A choice function is a simple mathematical model for representing choices: for any set of options , it returns a subset of options that are chosen from , meaning that the options in are rejected and that the options in are deemed incomparable (Seidenfeld et al., 2010; Van Camp et al., 2018; De Bock and de Cooman, 2019).

Such a choice function is often derived from a strict partial order on options, by choosing the options that are maximal—or undominated—with respect to this ordering, or equivalently, by rejecting the options that are dominated:

(1)

More generally, we can also associate a choice function with any set of such strict partial orders, by choosing the options in that are maximal with respect to at least one of the considered orderings:

(2)

This approach is conservative, since will only reject an option if it is rejected—or dominated—with respect to each of the partial orders in ; it can for example be used to represent conservative group decisions, by interpreting every in as the preferences of a different group member.

Choice functions of the form and appear in various settings, in all sorts of forms and variations, depending on what the options are and which kinds of properties are imposed on the partial orders involved. Maximising expected utility, for example is a well-known special case of a choice function of the type , where the options are utility functions and one chooses the option(s) whose expected utility with respect to some given probability measure is highest. If the probability measure involved is only know to belong to a set—for example because different group members assign different probabilities—this naturally extends to a choice function of the type .

The first main contribution of this paper are necessary and sufficient conditions for a general choice function to be of the form or

, for the case where options are elements of a real vector space. More generally, I provide generic necessary and sufficient conditions for the representing orders to satisfy additional properties, provided these properties are expressable in an abstract rule-based form. This leads in particular to representation theorems for choice functions that are based on (sets of) total orders, (sets of) weak orders, (sets of) coherent lower previsions, (sets of) probability measures, and various other special cases.

2 Choice functions based on (sets of) proper orderings

Let be a real vector space, the elements of which we call options, and let be the set of all—possibly infinite—subsets of , including the empty set. A choice function is a function that, for any option set , returns a subset consisting of the options in that are chosen—or rather, not rejected. The elements of are deemed incomparable when choosing from , but they are not necessarily indifferent.

We are particularly interested in choice functions of the form or , corresponding to a strict partial order on or to a set of such orders, respectively. Besides being strict partial orders, we will also require these orders to be compatible with the vector space operations of the real vector space , in the following sense. We will call such orders proper.

Definition 1

A binary relation on is a proper order if, for all and , it satisfies

  1. [label=.,ref=,leftmargin=*]

  2. irreflexivity

  3. if and , then also transitivity

  4. if then also

  5. if then also .

Since 4 implies that is equivalent to , every proper order is completely characterised by the set of options

(3)

In fact, as established in Proposition 1, proper orders are in one-to-one correspondence with convex cones in that are blunt, meaning that they do not include . We call any such a proper set of options.

Definition 2

A set of options is proper if it satisfies

  1. [label=.,ref=,leftmargin=*]

  2. if and , then

  3. if and , then .

Proposition 1

A binary relation on is a proper order if and only if there is a proper set of options such that

(4)

This is then necessarily unique and equal to .

It follows that choice functions of the form are completely determined by a single proper set of options , whereas choice functions of the form are characterised by a set of such proper sets of options.

3 Imposing additional properties through sets of rules

In most cases, rather than consider (sets of) arbitrary proper orders, one whishes to consider particular types of such orders, by imposing additional properties besides 14. Rather than treat each of these types seperately, we will instead consider an abstract axiom that includes a variety of them as special cases. We will characterise this axiom by means of a set of rules. Each such rule takes the form of a pair , with and . A set of rules , therefore, is a subset of , with the powerset of .

Definition 3

Let be a set of rules. For any proper set of options , we then say that is -compatible if it satisfies

  1. [label=.,ref=,leftmargin=*]

  2. if and for all , then also .

A proper order is called -compatible if is -compatible.

A first important special case are (sets of) rules of the type . Since makes the premise of 1 trivially true, such a rule allows one to express that should contain at least one option such that . If is the set of all gambles—bounded real functions—on some state space , coherence of for example correspond to the set of rules

with equal to or , depending on the specific type of coherence (Quaeghebeur, 2014). A second example are total orders. These correspond to the set of rules

(5)

Another special case are rules of the type . Since cannot be true for , rules of this type allow one to express that there should be at least one option set such that for all .

Besides total orders, weak orders can also be obtained as a special case of -compatibility. This can be achieved by considering any of the following three equivalent sets of rules,

with the set of finite positive linear combinations of options in , and where the in refers to the so-called mixingness property (De Bock and de Cooman, 2019).

If and is -compatible for , then partial orders that correspond to coherent lower previsions can be obtained by additionally imposing Archimedeanity (De Bock and de Cooman, 2019), using the set of rules

(6)

where we identify with the constant gamble on with value . Note that imposing and corresponds to imposing the single set of rules . And if we additionally impose —that is, if we impose —the representing lower previsions become linear expectations, resulting in orders that correspond to maximising expected utility with respect to a finitely additive probability measure (De Bock and de Cooman, 2019).

As a final example, orders that are evenly continuous (Cozman, 2018) can be obtained by imposing , with

4 Proper choice functions

In order to motivate the use of choice functions of the form and , without directly assuming that they must be of this type, we now proceed to provide an axiomatic characterisation for such types of choice functions. We start by introducing the notion of a proper choice function. As we will see in Theorem 4.3, these are exactly the choice functions of the form .

To state the defining axioms of a proper choice function, we require some additional notation. First, for any and , we let , and similarly for . One of the properties of a proper choice function—see 2—will be that rejecting from is equivalent to rejecting from . Similarly to how a proper order is completely determined by , a proper choice function will therefore be completely determined by the sets from which it rejects zero. In particular, as we will see in Proposition 2, the role of is now taken up by

(7)

Next, for any , we denote by the collection of all maps that, for all , select a single option . Furthermore, for any such selection map , we let be the corresponding set of selected options. We also let be the set of all maps with finite non-empty support, so for finitely many—but at least one— in and otherwise. Properness for choice functions is now defined as follows.

Definition 4

A choice function is proper if it satisfies

  1. [label=.,ref=,leftmargin=*,start=0]

  2. for all

  3. if , then also , for all and

  4. if and if, for all , , then also

  5. if and , then also , for all .

Each of these axioms can—but need not—be motivated by interpreting as ‘there is no in that is better than ’, where the ‘better than’ relation satisfies the defining properties of a proper order. For 12 and 4, this should be intuitively clear. For 3, the idea is that for any , there is at least one that is better than zero. Hence, for any , there is at least one such that every element of is better than zero. Combined with 24, or actually, 2 and 3, it follows that should be better than zero as well.

Every proper choice function is completely determined by a so-called proper set of option sets , which is furthermore unique and equal to .

Definition 5

A set of option sets is proper if it satisfies

  1. [label=.,ref=,leftmargin=*,start=0]

  2. ;

  3. if then also ;

  4. if and if, for all , , then also

  5. if and , then also .

Proposition 2

A choice function is proper if and only if there is a proper set of option sets such that

(8)

This is then necessarily unique and equal to .

Hence, proper choice functions correspond to proper sets of option sets . On the other hand, we know from Proposition 1 that proper orders correspond to proper sets of options . Relating proper choice functions with proper orders, therefore, amounts to relating proper sets of option sets with (sets of) proper sets of options . The first step consists in associating with any set of options a set of option sets

(9)

It is quite straightforward to show that is proper if and only if is.

Proposition 3

Consider a set of options and its corresponding set of option sets . Then is proper if and only if is.

What is perhaps more surprising though is that every proper set of option sets corresponds to a set of proper sets of options .

Theorem 4.1

A set of option sets is proper if and only if there is a non-empty set of proper sets of options such that .

To additionally guarantee that the option sets in are -compatible with a given set of rules , we introduce a notion of -compatibility for proper sets of option sets and choice functions.

Definition 6

Let be a set of rules. For any proper set of option sets , we then say that is -compatible if it satisfies

  1. [label=.,ref=,leftmargin=*]

  2. if and for all , then also .

A proper choice function is called -compatible if is -compatible.

Clearly, a set of options sets is -compatible if and only if is. What is far less obvious though, is that for sets of rules that are monotone a proper set of option sets is -compatible if and only if it corresponds to a set of proper -compatible sets of options .

Definition 7

A set of rules is monotone if for all :

Theorem 4.2

Let be a monotone set of rules. A set of option sets is then proper and -compatible if and only if there is a non-empty set of proper -compatible sets of options such that .

Among the sets of rules that were discussed in Section 3, examples of monotone ones are , and .

For sets of rules that are not monotone, a characterisation similar to Theorem 4.2 can be obtained by replacing with its monotonification

which is the smallest monotone set of rules that includes . This still leads to a representation in terms of proper -compatible sets of options because - and -compatibility are equivalent for sets of options.

Proposition 4

Consider a set of rules and a set of options . Then is -compatible if and only if it is -compatible.

Putting the pieces together—combining Theorem 4.2 with Propositions 12 and 4—we arrive at our main result: an axiomatic characterisation for choice functions of the form , with a set of -compatible proper orders.

Theorem 4.3

Let be a set of rules. Then a choice function is proper and -compatible if and only if there is a non-empty set of proper -compatible orders such that .

A similar result holds without -compatibility as well. It corresponds to the special case , since - and -compatibility are then trivially true.

In order to obtain an axiomatic characterisation for choice functions of the form , with a single -compatible proper order, we additionally impose that is completely determined by its pairwise choices. Replacing by is not needed in this case.

Definition 8

A choice function is binary if for all and :

Theorem 4.4

Let be a set of rules. Then a choice function is proper, binary and -compatible if and only if there is a proper -compatible order such that . This order is unique and characterised by

(10)

5 Conclusions and future work

The main conclusion of this paper is that decision making based on (sets of) strict partial orders is completely characterised by specific properties of the resulting choice functions. That is, any choice function that satisfies these properties is guaranteed to correspond to a (set of) strict partial order(s). By additionally imposing a set of rules on , we can furthermore guaruantee that the representing orders are of a particular type. As discussed in Section 3, this includes total orders, weak orders, orders based on coherent lower previsions, orders based on probability measures (such as maximising expected utility) and evenly continuous orders, depending on the set of rules that is imposed.

In future work, I intend to study if there are other types of orders that correspond to a set of rules, as well as explain in more detail why the types of orders and sets of rules in Section 3 indeed correspond to one another. I would also like to further elaborate on the implications of Theorems 4.3 and 4.4, for example by demonstrating how they can be applied in various contexts, and establish connections with earlier axiomatic characterisations for decision methods that are based on orders.

Acknowledgements.
This work was funded by the BOF starting grant 01N04819. I also thank Gert de Cooman, with whom I am developing a theory of choice functions for finite option sets (De Bock and de Cooman, 2018, 2019). My continuous discussions with him have been very helpful in extending this theory to allow choosing from infinite option sets, as I do here.

Bibliography

  • F. G. Cozman (2018) Evenly convex credal sets. International Journal of Approximate Reasoning 103, pp. 124–138. Cited by: §3.
  • J. De Bock and G. de Cooman (2018) A desirability-based axiomatisation for coherent choice functions. In

    Uncertainty Modelling in Data Science (Proceedings of SMPS 2018)

    ,
    pp. 46–53. Cited by: §5.
  • J. De Bock and G. de Cooman (2019) Interpreting, axiomatising and representing coherent choice functions in terms of desirability. In Proceedings of ISIPTA 2019, PMLR, Vol. 103, pp. 125–134. Cited by: §1, §3, §3, §5.
  • E. Quaeghebeur (2014) Desirability. In Introduction to Imprecise Probabilities, Cited by: §3.
  • T. Seidenfeld, M. J. Schervish, and J. B. Kadane (2010) Coherent choice functions under uncertainty. Synthese 172 (1), pp. 157–176. Cited by: §1.
  • A. Van Camp, G. de Cooman, and E. Miranda (2018) Lexicographic choice functions. International Journal of Approximate Reasoning, pp. 97–119. Cited by: §1.

Proofs of our main results

Proof

of Proposition 1 For the ‘if’ part, we consider a binary relation on and a proper set of options that satisfy Equation (4). That is equal to (and hence unique) follows from the fact that, for any ,

using Equation (4) for the second equivalence. That satisfies 4 follows directly from Equation (4). That satisfies 1, 2 and 3 follows directly from Equation (4) and the fact that satisfies 12 and 3, respectively.

For the ‘only if’ part, we consider a binary relation on that is proper and prove that there is a proper option set that satisfies Equation (4). In particular, we will prove that this is the case for . That satisfies Equation (4) follows from the fact that, for any :

where the first equivalence follows from 4 (with and ). That satisfies 1 and 3 follows directly from Equation (3) and the fact that satisfies 1 and 3, respectively. It remains to show that satisfies 2. So consider any and in , implying that and . Since , 4 implies that also . Since and , 2 implies that and therefore, that , as desired.

Proof

of Proposition 2 For the ‘if’ part, we consider a choice function and a proper set of option sets that satisfy Equation (8). That is equal to (and hence unique) follows from the fact that, for any ,

where the first two equivalences follow from Equations (7) and (8) and the last equivalence follows from 2 and 4. Next, we prove that is proper. Since and satisfies 3 (since is proper), we immediately find that satisfies 4. Consider now any . It then follows 2 and 1 that and therefore, because of Equation (8), that . Hence, satisfies 1. That also satisfies 2 follows directly from Equation (8). Indeed, for any and :

To see that satisfies 4, it suffices to consider any and such that and . It then follows from Equation (8) that . Since , 4 therefore implies that . A final application of Equation (8) therefore yields , as desired.

For the ‘only if’ part, we consider a proper choice function and prove that there is a proper set of option sets that satisfies Equation (8). In particular, we will prove that this is the case for . That satisfies Equation (8) follows from the fact that, for any and :

where the first equivalence follows from 2 (with and ) and the second follows from the fact that . It therefore remains to prove that is a proper set of option sets. That satisfies 3 follows directly from the fact that satisfies 3. Since 1 implies that , we know that and therefore, that satisfies 1. That satisfies 2 follows from Equation (7) and the fact that . To see that satisfies 4, we consider any such that and . It then follows from Equation (7) that . Since , it therefore follows from 4 that , or equivalently, that .

Proof

of Proposition 3 Consider any set of options and let be its corresponding set of option sets.

For the ‘only if’ part of the statement, we assume that is proper and prove that is then proper as well. That satisfies 1 and 4 is immediate from Equation (9). That satisfies 2 follows from Equation (9) and 1. To see that also satisfies 3, we consider any and, for all , some . For any , since , there is some , which we denote by . The resulting map , we have that for all , and therefore, because of 2 and 3, that . Hence,

For the ‘if’ part of the statement, we assume that is proper and prove that is then proper as well. To prove that satisfies 1, we assume ex absurdo that . This implies that and therefore, because of 2, that , contradicting 1. To prove that satisfies 2, we consider any . It then follows from Equation (9) that and . Now let and, for all let and for all other . Since and for all , and since the axiom of choice implies that is non-empty, it then follows from 3 that , or equivalently, that . The proof for 3 is analogous; in this case, we let and let for all other .

Proof

of Theorem 4.1 For the ‘if’ part, consider any non-empty set of proper sets of options such that . Then for any , is proper because of Proposition 3. Since is non-empty, and because properness is clearly preserved under taking non-empty intersections of sets of option sets, it follows that is proper as well.

For the ‘only if’ part, we assume that is proper. We need to prove that there is a non-empty set of proper sets of options such that . Let be the set of all proper sets of options and let . We will prove the required properties for the set of proper option sets . That is, we will prove that is non-empty and that . To that end, we will prove that for any such that , there is a set of options such that . On the one hand, this implies that and therefore, since the converse set inclusion holds trivially, that . On the other hand, since 1 guarantees that there is at least one option set such that , it also implies that is non-empty.

So consider any such that and assume ex absurdo that for every . We will show that this leads to a contradiction.

Consider any and let

(11)

be the set of all finite positive linear combinations of options in . Then for any , we have that —just let and for all —and therefore, since , also that . Hence, . Furthermore, by construction, clearly satisfies 3 and 2.

We now consider two cases: and . If , then satisfies 1. Since it also satisfies 3 and 2, we therefore find that is a proper set of options. Since , this implies that . By our assumption, it then follows that , implying that and therefore also . If , then , implying once more that . Hence, in both cases, we find that . Now let be any element of . Since , it follows from Equation (11) that there is some such that .

In summary then, for any , we have found some such that . Now let . It then follows from 3 that . Since is by construction a subset of 4 therefore implies that . This is however impossible because . If , this contradiction is trivial. If , this contradiction follows from 2.

Proof

of Theorem 4.2 For the ‘if’-part of the statement, we assume that there is a non-empty set of proper -compatible sets of options such that . It then follows from Theorem 4.1 that is proper. To show that it is also -compatible, we consider any such that for all . Fix any . Then for all , since , we know that . Since is -compatible, this implies that , which in turn implies that . Since this is true for every , we find that . Hence, is -compatible.

For the ‘only if’ part, we assume that is proper and -compatible. We need to prove that there is a non-empty set of proper -compatible sets of options such that . Let be the set of all proper -compatible sets of options and let . We will prove the required properties for the set of proper option sets . That is, we will prove that is non-empty and that . To that end, we will prove that for any such that , there is a set of desirable options such that . On the one hand, this implies that and therefore, since the converse set inclusion holds trivially, that . On the other hand, since 1 guarantees that there is at least one option set such that , it also implies that is non-empty. So consider any such that .

Since is proper, Theorem 4.1 implies that there is a non-empty set of proper sets of options such that . Since , this implies that there must be some such that and . Let be the set of all proper sets of options and consider the set , partially ordered by set inclusion. In particular, for any two , we say that dominates , denoted by , if ; so subsets dominate their supersets. We will use Zorn’s lemma to prove that this partially ordered set has a maximal—undominated—element. To do so, we must show that any chain in —any completely ordered subset of —has an upper bound in . So consider any chain in . Since the elements of are proper and since properness is preserved under intersections, the intersection is proper. Furthermore, since for all , we also have that . Hence, . Since is also by definition an upper bound for the chain , in the sense that for all , we conclude that has an upper bound in . Since this is true for every chain in , it follows from Zorn’s lemma that has a maximal element. Let be any such a maximal element.

Since , we know that is proper and that . Since , this also implies that . The only thing left to prove, therefore, is that is -compatible. Assume ex absurdo that it is not. We will show that this leads to a contradiction.

Since is not -compatible and is proper, it follows from Definition 3 that there is some such that for all but . There are now two possibilities: and . As we will see, they both lead to a contradiction. If , it follows from the -compatibility of that , contradicting the fact that . It remains to consider the case . In that case, let . Since and , the monotonicity of implies that . Consider now any and any . If , then , so . This is impossible because is a maximal element of with respect to and therefore a minimal element with respect to . Hence, we find that either or . If , then since and therefore , we find that . If , then and therefore also , which implies that . Hence, in all cases, we find that . Since this is true for every and , and since , it follows that for all . Since , the -compatibility of therefore implies that , so . Since , this implies that , which is clearly impossible because . Hence, we find that must indeed be -compatible.

Proof

of Proposition 4 Since is a subset of , we trivially have that -compatibility implies -compatibility. To prove the converse, asume that is -compatible and consider any such that for all . We need to prove that . If , this follows directly from the -compatibility of . Otherwise, since , there are and such that , and . Since for all , this implies that for all . We now consider two cases: and . If , then also , as desired. If , then for all , we infer from that . Since is -compatible and , this implies that . Hence, also in this second case, .

Proof

of Theorem 4.3 For the ‘if’-part of the statement, we assume that there is a non-empty set of proper -compatible orders such that . Let . Then is non-empty because is, and the sets of options in are proper and -compatible because of Proposition 1, Definition 3 and Proposition 4. Hence, is a non-empty set of proper -compatible sets of options. It therefore follows from Theorem 4.2 that is a proper -compatible set of option sets. Now observe that for any ,

where the first equality follows from Equation (2), the second from Proposition 1, the fifth from Equation (9) and the last from our choice of . Since is proper, it therefore follows from Proposition 2 that is proper and that . Since is -compatible, it furthermore follows from Definition 4 that is -compatible.

For the ‘only if’-part of the statement, we consider any proper -compatible choice function . It then follows from Proposition 2 and Definition 4 that is a proper -compatible set of option sets. Because of Theorem 4.2, this implies that there is a non-empty set of proper -compatible sets of options such that . Because of Proposition 4, each in is furthermore -compatible as well. With any in , we now associate a binary relation on , defined by

(12)

Since is proper, it follows from Proposition 1 that is a proper order and that . Since is -compatible, it therefore follows from Definition 3 that is -compatible as well. Hence, if we let , then is a set of proper -compatible orders and, since is non-empty, is non-empty as well. It remains to show that . This follows from the fact that, for all ,