
<p>A set optimization approach to utility maximization under transaction costs </p><p>Andreas H. Hamel<sup style="top: -0.3615em;">∗</sup>, Sophie Qingzhen Wang<sup style="top: -0.3615em;">† </sup></p><p>Abstract </p><p>A set optimization approach to multi-utility maximization is presented, and duality results are obtained for discrete market models with proportional transaction costs. The novel approach admits to obtain results for non-complete preferences, where the formulas derived closely resemble but generalize the scalar case. </p><p>Keywords. utility maximization, non-complete preference, multi-utility representation, set optimization, duality theory, transaction costs </p><p>JEL classification. C61, G11 </p><p>1 Introduction </p><p>In this note, we propose a set-valued approach to utility maximization for market models with transaction costs. For finite probability spaces and a one-period set-up, we derive results which resemble very closely the scalar case as discussed in [4, Theorem 3.2.1]. This is far beyond other approaches in which only scalar utility functions are used as, for example, in [1, 3], where a complete preference for multivariate position is assumed. As far as we are aware of, there is no argument justifying such strong assumption, and it does not seem appropriate for market models with transaction costs. <br>On the other hand, recent results on multi-utility representations as given, among others, in [5] lead to the question how to formulate and solve an “expected multi-utility” maximization problem. The following optimistic goal formulated by Bosi and Herden in [2] does not seem achievable since, in particular, there is no satisfactory multi-objective duality which matches the power of the scalar version: ‘Moreover, as it reduces finding the maximal elements in a given subset of X with respect to ꢀ to a multi-objective optimization problem (cf. Evren and Ok, 2011), in applications, this approach is likely to be more useful than the Richter-Peleg approach.’ <br>The question remains how duality could work in a multi-utility framework. Using a simple model with finitely many utility functions as an exemplary case, our answer is that it works almost in the same way as in the scalar case if set optimization methods are used. </p><p><sup style="top: -0.3174em;">∗</sup>Free University Bozen, Faculty of Economics and Management, [email protected] <sup style="top: -0.3174em;">†</sup>Harvard University Department of Economics, [email protected] </p><p>1</p><p>Our approach is different, and it has several advantages. First, we demonstrate that the <br>“usual” duality can be established if one allows for a set-valued extension of the problem. This means that the vector-valued problem is replaced by a problem where the objective function maps into a well-defined complete lattice of sets. Secondly, the problem formulation allows to separate the market model, namely the time dependent solvency cone, from the the preference relation expressed by a (vector-valued) utility function and provides a link to no-arbitrage type results for market models with transaction costs as, for example, given in [16]. <br>Our model can be interpreted as follows: The decision maker has a complete preference for random positions of each asset in the market separately which is representable by a utility function (the classic von Neumann-Morgenstern set-up). These preferences work “component-wise,’ i.e. they are determined independently of the other assets. This may seem very restrictive and is indeed not the most general situation. However, there are two arguments in favor of studying it. First, it admits non-complete preferences which goes beyond the above quoted references and actually provides a path for a solution of the corresponding (multi-)utility maximization problem. Secondly, via the market model, an exchange mechanism is built into the model which makes the components of the portfolio dependent on each other when one looks for best alternatives. The underlying idea is that one can exchange assets, but cannot exchange utility of one asset for utility of another. <br>The main tool is a set-valued Lagrangian, to which recent duality results for set-valued functions from [10] are applied (compare also the survey [8]). The resulting formulas look very much like the scalar ones, a feature that is nearly impossible to achieve if one would only apply multi-objective optimization duality. <br>The results are based on the senior thesis of the second author written in 2011 at Princeton <br>University and supervised by the first author and B. Rudloff (now at Vienna University for Economics and Business). </p><p>2 Problem formulation </p><p>Let (Ω, P) be a finite probability space where the σ-algebra is assumed to be the power set </p><p>of Ω = {ω<sub style="top: 0.1246em;">1</sub>, ω<sub style="top: 0.1246em;">2</sub>, . . . , ω<sub style="top: 0.1246em;">N </sub>} with p<sub style="top: 0.1246em;">n </sub>:= P ({ω<sub style="top: 0.1246em;">n</sub>}) > 0 for all n ∈ {1, . . . , N}. </p><p>A one-period conical model for a market with d assets is given by a pair (K<sub style="top: 0.1246em;">0</sub>, K<sub style="top: 0.1246em;">T </sub>= K<sub style="top: 0.1246em;">T </sub>(ω)) of finitely generated convex cones satisfying IR<sup style="top: -0.362em;">d</sup><sub style="top: 0.2053em;">+ </sub>⊆ K<sub style="top: 0.1246em;">0 </sub>= IR<sup style="top: -0.362em;">d </sup>and IR<sup style="top: -0.362em;">d</sup><sub style="top: 0.2053em;">+ </sub>⊆ K<sub style="top: 0.1246em;">T </sub>(ω) = IR<sup style="top: -0.362em;">d </sup>for all ω ∈ Ω. The cones K<sub style="top: 0.1245em;">0</sub>, K<sub style="top: 0.1245em;">T </sub>are called solvency cones and may arise, for example, due to explicit proportional transaction costs (see [13]) or bid-ask spreads (see [16]). <br>The linear space of IR<sup style="top: -0.3621em;">d</sup>-valued random variables X : Ω → IR<sup style="top: -0.3621em;">d </sup>is denoted by L<sup style="top: -0.3013em;">0</sup><sub style="top: 0.2351em;">d </sub>:= <br>L<sup style="top: -0.3012em;">0</sup><sub style="top: 0.2352em;">d</sub>(Ω, P) which can be identified with IR<sup style="top: -0.3621em;">dN </sup>. Further, denote </p><p>ꢀ</p><p>L<sup style="top: -0.3428em;">0</sup><sub style="top: 0.2053em;">d </sub>(K<sub style="top: 0.1245em;">T </sub>) = X ∈ L<sup style="top: -0.3428em;">0</sup><sub style="top: 0.2053em;">d </sub>| ∀ω ∈ Ω: X (ω) ∈ K<sub style="top: 0.1245em;">T </sub>(ω) , </p><p></p><ul style="display: flex;"><li style="flex:1">ꢃ</li><li style="flex:1">ꢄ</li></ul><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>and L<sup style="top: -0.3012em;">0</sup><sub style="top: 0.2352em;">d </sub>= L<sup style="top: -0.3012em;">0</sup><sub style="top: 0.2352em;">d </sub>IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>. Finally, 1I ∈ L<sup style="top: -0.3012em;">0</sup><sub style="top: 0.2352em;">d </sub>stands for the function with 1I (ω) = 1 for all ω ∈ Ω. </p><p>+</p><p>In this note, it is assumed that d utility functions U<sub style="top: 0.1246em;">i </sub>: IR → IR ∪ {−∞}, i = 1, . . . , d are given which are concave and non-decreasing. We define a vector-valued function U : IR<sup style="top: -0.3621em;">d </sup>→ </p><p>2</p><p>IR<sup style="top: -0.3621em;">d </sup>∪ {−∞} by </p><p>(</p><p>(U<sub style="top: 0.1246em;">1 </sub>(x<sub style="top: 0.1246em;">1</sub>) , U<sub style="top: 0.1246em;">2 </sub>(x<sub style="top: 0.1246em;">2</sub>) , . . . , U<sub style="top: 0.1246em;">d </sub>(x<sub style="top: 0.1246em;">d</sub>))<sup style="top: -0.4174em;">T </sup></p><p>::</p><p>∀i ∈ {1, 2, . . . , d} : x<sub style="top: 0.1246em;">i </sub>∈ dom U<sub style="top: 0.1246em;">i </sub></p><p>otherwise <br>U (x) = </p><p>−∞ </p><p>This can be interpreted as follows. The investor has a scalar utility function for each of the d assets which is independent from holdings in the others, and a portfolio is not acceptable for the investor if one asset produces a utility −∞. Of course, this means that the investor has a complete preference for random positions of each asset which admits a von NeumannMorgenstern representation. We note that even this might be a strong assumption and refer to [6, Chap. 2] for further details. However, this assumption is in some respect much less </p><p>d</p><p>ˆstrong than the existence of a function U : IR → IR ∪ {−∞} which represents a complete </p><p>preference (a total order) on the set of all multivariate positions. This is the starting point e.g. in the recent [1, 3] <br>Since the U<sub style="top: 0.1245em;">i</sub>’s are concave and non-decreasing, the function U is IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+</sub>-concave, that is </p><p>∀t ∈ (0, 1) , ∀x, x<sup style="top: -0.3428em;">0 </sup>∈ IR<sup style="top: -0.3621em;">d </sup>: U (tx + (1 − t) x<sup style="top: -0.3428em;">0</sup>) ∈ tU (x) + U ((1 − t) x<sup style="top: -0.3428em;">0</sup>) − IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+</sub>, and U is IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2052em;">+</sub>-monotone increasing, that is x ∈ x<sup style="top: -0.3428em;">0 </sup>+ IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>⇒ U (x) ∈ U (x<sup style="top: -0.3428em;">0</sup>) + IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>where we agree upon x+(−∞) = (−∞)+x = −∞, t (−∞) = −∞ for all t > 0 and −∞ ≤ x for all x ∈ IR<sup style="top: -0.3621em;">d</sup>. <br>The basic problem of this note is </p><p>maximize IE [U (X)] subject to X ∈ x1I − K<sub style="top: 0.1246em;">0</sub>1I − L<sup style="top: -0.3428em;">0</sup><sub style="top: 0.2052em;">d </sub>(K<sub style="top: 0.1246em;">T </sub>) where x ∈ IR<sup style="top: -0.3621em;">d </sup>is the given initial endowment and the expected value is understood componentwise with IE [−∞] = −∞. <br>The set x1I − K<sub style="top: 0.1246em;">0</sub>1I − L<sup style="top: -0.3012em;">0</sup><sub style="top: 0.2352em;">d </sub>(K<sub style="top: 0.1246em;">T </sub>) is precisely the set of (random) portfolios which can be obtained by trading at time t = 0 and time t = T starting with the initial endowment x ∈ IR<sup style="top: -0.362em;">d</sup>. The first question is how the maximization is understood. In contrast to (more popular) vector approaches, we will understand the above problem as a set-valued one. The advantage of this approach is that the notions of supremum and infimum make sense, and consequently, that a complete duality theory is available. One may compare the appendix and the survey [8] for details. <br>We define the set </p><p></p><ul style="display: flex;"><li style="flex:1">n</li><li style="flex:1">ꢃ</li><li style="flex:1">ꢄo </li></ul><p></p><p>G(IR<sup style="top: -0.3621em;">d</sup>, −IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+</sub>) = A ⊆ IR<sup style="top: -0.3621em;">d </sup>| A = cl co A − IR<sub style="top: 0.2053em;">+</sub><sup style="top: -0.3621em;">d </sup></p><p></p><ul style="display: flex;"><li style="flex:1">ꢃ</li><li style="flex:1">ꢄ</li></ul><p></p><p>and introduce an addition on G IR<sup style="top: -0.3621em;">d</sup>, −IR<sub style="top: 0.2053em;">+</sub><sup style="top: -0.3621em;">d </sup>by </p><p>(</p><p>cl {a<sub style="top: 0.1245em;">1 </sub>+ a<sub style="top: 0.1245em;">2 </sub>| a<sub style="top: 0.1245em;">1 </sub>∈ A<sub style="top: 0.1245em;">1</sub>, a<sub style="top: 0.1245em;">2 </sub>∈ A<sub style="top: 0.1245em;">2</sub>} </p><p>::</p><p>A<sub style="top: 0.1245em;">1</sub>, A<sub style="top: 0.1245em;">2 </sub>= ∅ </p><p>otherwise </p><p>A<sub style="top: 0.1246em;">1 </sub>⊕ A<sub style="top: 0.1246em;">2 </sub>= </p><p>,</p><p>∅</p><p>and a multiplication · with non-negative real numbers by </p><p></p><p>−IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2052em;">+ </sub></p><p>{ta | a ∈ A} <br>∅</p><p>:::t = 0 </p><p></p><p></p><ul style="display: flex;"><li style="flex:1">t · A = </li><li style="flex:1">t > 0, A = ∅ </li></ul><p></p><p></p><p>t > 0, A = ∅ </p><p>3</p><p></p><ul style="display: flex;"><li style="flex:1">ꢃ</li><li style="flex:1">ꢄ</li></ul><p></p><p>On G IR<sup style="top: -0.3621em;">d</sup>, −IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>, the inclusion ⊆ is a partial order which is compatible with the two </p><p>ꢃ ꢃ </p><p></p><ul style="display: flex;"><li style="flex:1">ꢄ</li><li style="flex:1">ꢄ</li></ul><p></p><p>algebraic operations just defined, thus the quadruple G IR<sup style="top: -0.3621em;">d</sup>, −IR<sub style="top: 0.2053em;">+</sub><sup style="top: -0.3621em;">d </sup>, ⊕, ·, ⊆ is an ordered conlinear space in the sense of [7, 8]. We shall denote this structure by G<sup style="top: -0.3012em;">O </sup>and drop the · for multiplication in most cases. <br>The infimum and supremum of a set A ⊆ G<sup style="top: -0.3012em;">O </sup>are given by </p><p></p><ul style="display: flex;"><li style="flex:1">[</li><li style="flex:1">\</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">sup A = cl co </li><li style="flex:1">A, inf A = </li></ul><p></p><p>A</p><p></p><ul style="display: flex;"><li style="flex:1">A∈A </li><li style="flex:1">A∈A </li></ul><p></p><p>which are clearly elements of G<sup style="top: -0.3012em;">O</sup>, thus G<sup style="top: -0.3012em;">O </sup>even is a complete lattice (see [7, 8]). Here, we agree upon sup A = ∅ in G<sup style="top: -0.3012em;">O </sup>for A = ∅. <br>Our basic problem now is to find, in G<sup style="top: -0.3012em;">O</sup>, </p><p></p><ul style="display: flex;"><li style="flex:1">n</li><li style="flex:1">o</li></ul><p></p><p>sup IE [U (X)] − IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>| X ∈ x1I − K<sub style="top: 0.1245em;">0</sub>1I − L<sub style="top: 0.2053em;">d</sub><sup style="top: -0.3428em;">0 </sup>(K<sub style="top: 0.1245em;">T </sub>) <br>(UMAX) where we use the convention IE [U (X)] − IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>= ∅ whenever IE [U (X)] = −∞. <br>The plan of the remainder of the paper is as follows. In the next section, we will give a dual characterization of the constraint of (UMAX). Then, we introduce a Lagrangian function, define primal and dual value functions and establish a strong duality theorem which provides the essentials for a solution of (UMAX). The reader may compare [4, Chapter 3] for a concise presentation for the scalar case d = 1. It is the main purpose of this note to demonstrate that the vector-valued case d > 1 can be dealt with by means of set-valued duality as proposed in [7, 10], and that in doing so one obtains meaningful analogs to the well-known scalar formulas. Such a theory is not really possible in terms of the vector order since the supremum (or infimum) with respect to a vector order usually does not make sense since it produces “utopia” solutions which are not feasible in general. </p><p>3 Constraints transformation </p><p>Defining <br>C = K<sub style="top: 0.1246em;">0</sub>1I + L<sup style="top: -0.3427em;">0</sup><sub style="top: 0.2052em;">d </sub>(K<sub style="top: 0.1246em;">T </sub>) </p><p>we want to describe C by means of dual variables. Note that in our setting C is always closed since it can be identified with a finitely generated cone in IR<sup style="top: -0.3621em;">dN </sup>(see [15, Theorem 19.1]). In a more general situation, the closedness of the set of all terminal positions follows from a no-arbitrage type condition, see for example [16, Theorem 2.1]. </p><p></p><ul style="display: flex;"><li style="flex:1">ꢅ</li><li style="flex:1">ꢆ</li></ul><p></p><p>Using the duality pairing (X, Y ) → IE Y <sup style="top: -0.3013em;">T </sup>X on L<sub style="top: 0.2351em;">d</sub><sup style="top: -0.3013em;">0 </sup>× L<sub style="top: 0.2351em;">d</sub><sup style="top: -0.3013em;">0</sup>, L<sup style="top: -0.3013em;">0</sup><sub style="top: 0.2351em;">d </sub>is turned into a Hilbert space which can be identified with its dual, and for this topological setting, we obtain </p><p></p><ul style="display: flex;"><li style="flex:1">ꢅ</li><li style="flex:1">ꢆ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li><li style="flex:1">ꢀ</li></ul><p></p><p>+</p><p>L<sup style="top: -0.3427em;">0</sup><sub style="top: 0.2053em;">d </sub>(K<sub style="top: 0.1246em;">T </sub>) = L<sup style="top: -0.3427em;">0</sup><sub style="top: 0.2053em;">d </sub>K<sub style="top: 0.244em;">T</sub><sup style="top: -0.3552em;">+ </sup>= Y ∈ L<sub style="top: 0.2053em;">d</sub><sup style="top: -0.3427em;">0 </sup>| ∀ω ∈ Ω: Y (ω) ∈ K<sub style="top: 0.244em;">T</sub><sup style="top: -0.3552em;">+ </sup>(ω) </p><p>+</p><p>where denotes the (positive) dual cone (see [17, p. 7]). Compare [9, Lemma 3.1]. </p><p>Lemma 3.1 For x ∈ IR<sup style="top: -0.362em;">d</sup>, X ∈ L<sub style="top: 0.2352em;">d</sub><sup style="top: -0.3012em;">0</sup>, </p><p></p><ul style="display: flex;"><li style="flex:1">ꢅ</li><li style="flex:1">ꢆ</li></ul><p></p><p>x1I − X ∈ C </p><p>⇔</p><p>∀ (Y, v) ∈ Y<sub style="top: 0.1245em;">d </sub>: IE Y <sup style="top: -0.3428em;">T </sup>X ≤ v<sup style="top: -0.3428em;">T </sup>x </p><p>where </p><p></p><ul style="display: flex;"><li style="flex:1">n</li><li style="flex:1">o</li></ul><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>Y<sub style="top: 0.1245em;">d </sub>= (Y, v) ∈ L<sup style="top: -0.3428em;">0</sup><sub style="top: 0.2053em;">d </sub>× IR<sup style="top: -0.3621em;">d </sup>| v ∈ K<sub style="top: 0.2213em;">0</sub><sup style="top: -0.3552em;">+</sup>\{0}, Y ∈ L<sub style="top: 0.2053em;">d</sub><sup style="top: -0.3428em;">0 </sup>K<sub style="top: 0.244em;">T</sub><sup style="top: -0.3552em;">+ </sup>, IE [Y ] = v . </p><p>4</p><p>Proof. By a separation argument, one may see that the dual cone of C is </p><p></p><ul style="display: flex;"><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>C<sup style="top: -0.3427em;">+ </sup>= Y ∈ L<sub style="top: 0.2053em;">d</sub><sup style="top: -0.3427em;">0 </sup>| Y ∈ L<sub style="top: 0.2053em;">d</sub><sup style="top: -0.3427em;">0 </sup>K<sub style="top: 0.244em;">T</sub><sup style="top: -0.3552em;">+ </sup>∧ IE [Y ] ∈ K<sub style="top: 0.2214em;">0</sub><sup style="top: -0.3552em;">+ </sup><br>.</p><p>Since C is a closed convex cone the bipolar theorem produces </p><p></p><ul style="display: flex;"><li style="flex:1">ꢅ</li><li style="flex:1">ꢆ</li><li style="flex:1">ꢅ</li><li style="flex:1">ꢆ</li></ul><p></p><p>x1I − X ∈ C </p><p>⇔</p><p>∀Y ∈ C<sup style="top: -0.3427em;">+ </sup>: IE Y <sup style="top: -0.3428em;">T </sup>(x1I − X) = x<sup style="top: -0.3427em;">T </sup>IE [Y ] − IE Y <sup style="top: -0.3427em;">T </sup>X ≥ 0. </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>The set Y<sub style="top: 0.1245em;">d </sub>is in one-to-one relationship with C<sup style="top: -0.3012em;">+</sup>\{0}. Indeed, Y ∈ L<sup style="top: -0.3012em;">0</sup><sub style="top: 0.2352em;">d </sub>K<sub style="top: 0.244em;">T</sub><sup style="top: -0.3552em;">+ </sup>implies Y ∈ </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>L<sub style="top: 0.2352em;">d</sub><sup style="top: -0.3012em;">0 </sup></p><p>and this in turn IE [Y ] ≥ 0 since IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>⊆ K<sub style="top: 0.1245em;">T </sub>(ω), hence K<sub style="top: 0.244em;">T</sub><sup style="top: -0.3552em;">+ </sup>(ω) ⊆ IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>for all ω ∈ Ω. </p><p>+</p><p>Thus, v = IE [Y ] = 0 implies Y = 0. The rest is obvious. </p><p>ꢀ</p><p>By means of Lemma 3.4 in [9] it can be shown that the set Y<sub style="top: 0.1245em;">d </sub>also is in one-to-one relationship with the set </p><p></p><ul style="display: flex;"><li style="flex:1">ꢇ</li><li style="flex:1">ꢈ</li></ul><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>dQ dP </p><p>W<sub style="top: 0.1245em;">d </sub>= (Q, w) ∈ M<sub style="top: 0.2052em;">1,d </sub>× IR<sup style="top: -0.3621em;">d </sup>| w ∈ K<sub style="top: 0.2214em;">0</sub><sup style="top: -0.3552em;">+</sup>\{0}, diag (w) </p><p>∈ L<sup style="top: -0.3428em;">0</sup><sub style="top: 0.2052em;">d </sub>K<sub style="top: 0.244em;">T</sub><sup style="top: -0.3552em;">+ </sup></p><p>.</p><p>P</p><ul style="display: flex;"><li style="flex:1">dQ </li><li style="flex:1">P</li></ul><p></p><p>This can be arranged by Y = diag (w) <sub style="top: 0.2862em;">dP </sub>, IE [Y ] = w. Here, M<sub style="top: 0.2351em;">1,d </sub>is the set of all vector probability measures with components which are absolutely continuous with respect to P. Thus, we can write the utility maximization problem as </p><p></p><ul style="display: flex;"><li style="flex:1">n</li><li style="flex:1">o</li></ul><p></p><p>sup IE [U (X)] − IR<sup style="top: -0.3621em;">d</sup><sub style="top: 0.2053em;">+ </sub>| ∀ (Q, w) ∈ W<sub style="top: 0.1245em;">d </sub>: w<sup style="top: -0.3428em;">T </sup>IE<sup style="top: -0.3621em;">Q </sup>[X] ≤ w<sup style="top: -0.3428em;">T </sup>x . <br>Under our assumptions, we can reduce the number of constraints in the above problem to a finite number. This can be done due to the fact that the cones K<sub style="top: 0.1245em;">0 </sub>and K<sub style="top: 0.1245em;">T </sub>(ω) are assumed to be finitely generated. Therefore, the cone C can be identified with a finitely generated cone in IR<sup style="top: -0.3621em;">dN </sup>. Hence (see [15, Corollary 19.2.2]) the cone C<sup style="top: -0.3012em;">+ </sup>in IR<sup style="top: -0.3621em;">dN </sup>is finitely generated by, </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢂ</li></ul><p></p><p>say, Y <sup style="top: -0.3012em;">1</sup>, Y <sup style="top: -0.3012em;">2</sup>, . . . , Y <sup style="top: -0.3012em;">M </sup>∈ Y<sub style="top: 0.1246em;">d</sub>. Let Q<sup style="top: -0.3012em;">1</sup>, w<sup style="top: -0.3012em;">1 </sup>, . . . , Q<sup style="top: -0.3012em;">M </sup>, w<sup style="top: -0.3012em;">M </sup>∈ W<sub style="top: 0.1245em;">d </sub>be the pairs corresponding </p><p></p><ul style="display: flex;"><li style="flex:1">ꢁ</li><li style="flex:1">ꢅ</li><li style="flex:1">ꢆꢂ </li><li style="flex:1">ꢁ</li><li style="flex:1">ꢅ</li><li style="flex:1">ꢆꢂ </li></ul><p></p><p>to Y <sup style="top: -0.3012em;">1</sup>, IE Y <sup style="top: -0.3012em;">1 </sup>, . . . , Y <sup style="top: -0.3012em;">M </sup>, IE Y <sup style="top: -0.3012em;">M </sup>. Then </p><p>Q<sup style="top: -0.2506em;">m </sup></p><p></p><ul style="display: flex;"><li style="flex:1">T</li><li style="flex:1">T</li></ul><p></p><p>x1I − X ∈ C </p><p>⇔</p><p>∀m ∈ {1, . . . , M} : (w<sup style="top: -0.3428em;">m</sup>) IE </p><p>[X] ≤ (w<sup style="top: -0.3428em;">m</sup>) x. <br>We can further simplify the problem by observing that, for (Q, w) ∈ W<sub style="top: 0.1246em;">d</sub>, </p><p>N</p><p>X</p><p>w<sup style="top: -0.3428em;">T </sup>IE<sup style="top: -0.3621em;">Q </sup>[X] ≤ w<sup style="top: -0.3428em;">T </sup>x </p><p>⇔</p><p>w<sup style="top: -0.3428em;">T </sup>qˆ<sub style="top: 0.2259em;">(</sub><sup style="top: -0.3428em;">T</sup><sub style="top: 0.2259em;">n)</sub>ξ<sup style="top: -0.3428em;">n </sup>≤ w<sup style="top: -0.3428em;">T </sup>x </p>
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