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Multiattribute Utility Functions Satisfying MutualPreferential IndependenceAli E. Abbas, Zhengwei Sun

To cite this article:Ali E. Abbas, Zhengwei Sun (2015) Multiattribute Utility Functions Satisfying Mutual Preferential Independence. OperationsResearch 62(2):378-393. http://dx.doi.org/10.1287/opre.2015.1350

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OPERATIONS RESEARCHVol. 63, No. 2, March–April 2015, pp. 378–393ISSN 0030-364X (print) � ISSN 1526-5463 (online) http://dx.doi.org/10.1287/opre.2015.1350

© 2015 INFORMS

Multiattribute Utility Functions Satisfying MutualPreferential Independence

Ali E. AbbasEpstein Department of Industrial and Systems Engineering and Department of Public Policy, Viterbi School of Engineering, and

Price School of Public Policy, University of Southern California, Los Angeles, California, 90089, [email protected]

Zhengwei SunDepartment of Management Science and Engineering, East China University of Science and Technology, Shanghai 200237, China,

[email protected]

The construction of a multiattribute utility function is an important step in decision analysis. One of the most widelyused conditions for constructing the utility function is the assumption of mutual preferential independence where trade-offsamong any subset of the attributes do not depend on the instantiations of the remaining attributes. Mutual preferentialindependence asserts that ordinal preferences can be represented by an additive function of the attributes. This paper derivesthe most general form of a multiattribute utility function that (i) exhibits mutual preferential independence and (ii) is strictlyincreasing with each argument at the maximum value of the complement attributes. We show that a multiattribute utilityfunction satisfies these two conditions if and only if it is an Archimedean combination of univariate utility assessments. Thisresult enables the construction of multiattribute utility functions that satisfy additive ordinal preferences using univariateutility assessments and a single generating function. We also provide a nonparametric approach for estimating the generatingfunction of the Archimedean form by iteration.

Subject classifications : multiattribute utility; Archimedean utility copula; preferential independence.Area of review : Decision Analysis.History : Received February 2014; revisions received November 2013, June 2014, August 2014; accepted December

2014. Published online in Articles in Advance March 4, 2015.

1. IntroductionIn decisions with multiple objectives, it is important tothink about the ordinal preferences for the consequences ofthe decision. One of the earliest conditions that specifiedsome forms of ordinal preferences is the notion of mutualpreferential independence, where trade-offs among any sub-set of the attributes do not depend on the instantiationsof the remaining attributes. In his classic work, Debreu(1960) showed that this condition on ordinal preferencescorresponds to a value function that is a monotone trans-formation of an additive function of the attributes when thenumber of attributes, n¾ 3, i.e., the value function can beexpressed as

V 4x11 0 0 0 1 xn5=m

( n∑

i=1

fi4xi5

)

1 (1)

where m is a monotone function, n¾ 3, and fi, i = 11 0 0 0 1 nare arbitrary univariate functions.

The condition of mutual preferential independence rep-resents an important class of value functions that hasbeen used extensively in the literature. For example, netpresent value functions, multiplicative value functions, andCobb-Douglas value functions all satisfy the conditionof mutual preferential independence. For more applica-tions of value functions satisfying the condition of mutualpreferential independence, see Keeney (1974, 1992), Dyerand Sarin (1979), Keelin (1981), Howard (1984), Barronand Schmidt (1988), Edwards and Barron (1994), Stewart

(1996), Kirkwood (1997), Greco et al. (2008), Eisenführet al. (2010), and Lichtendahl and Bodily (2012).

When uncertainty is present, it is also important tothink about the cardinal preferences for the consequencesof the decision. The classic work of von Neumann andMorgenstern (1947) shows that a utility function is neededto determine the best decision alternative in this case. Itwas soon realized, however, that the construction of amultiattribute utility function can be a tedious task unlesssome decomposition of the utility function is performed,and several methods have since been proposed to facilitatethis task.

Keeney and Raiffa’s (1976) classic work proposed sev-eral fundamental conditions for decomposing the utilityfunction. Of particular interest is the notion of mutual util-ity independence, where preferences for lotteries over anysubset of the attributes do not depend on the instantiationof the remaining attributes. The condition of mutual utilityindependence implies that the multiattribute utility func-tion is either an additive or a multiplicative combination ofsingle-attribute utility assessments,

U4x11 0 0 0 1 xn5=

n∑

i=1

kiUi4xi51 (2)

or

1 − kU4x11 0 0 0 1 xn5=

n∏

i=1

41 − kkiUi4xi551 (3)

where Ui4xi5, i = 11 0 0 0 1 n are univariate utility functions.

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mailto:[email protected]

Abbas and Sun: Multiattribute Utility Functions Satisfying Mutual Preferential IndependenceOperations Research 63(2), pp. 378–393, © 2015 INFORMS 379

Note that the functional forms (2) and (3) can be con-verted into an additive value function using a mono-tone transformation. This implies that a decision makerwho exhibits mutual utility independence over lotter-ies also exhibits mutual preferential independence overdeterministic consequences. However, the converse is notnecessarily true.

This paper derives the most general form of a multiat-tribute utility function that satisfies the condition of mutualpreferential independence over deterministic multiattributeconsequences but does not necessarily satisfy mutual util-ity independence. As we shall see, the functional form ofthe utility function can also be constructed using univari-ate utility assessments, but a univariate generating functionis also needed to reflect the decision maker’s preferencesfor lotteries. In particular, we show that (i) a decisionmaker exhibits mutual preferential independence among theattributes and (ii) his utility function is strictly increasingwith each argument at the maximum value of the com-plement attributes if and only if his utility function is anArchimedean combination of univariate utility assessments.This result completes the class of increasing utility func-tions that satisfy mutual preferential independence but donot necessarily satisfy mutual utility independence. Thisresult also sheds new light on the structure of the multiat-tribute utility function when the condition of mutual pref-erential independence is satisfied.

A natural question that arises with this result is howto assess the Archimedean utility form? Or equivalently,how to determine the generating function that should beused with the univariate assessments? This topic has beencovered extensively in the probability literature using bothparametric and nonparametric approaches for construct-ing an Archimedean probability copula (see for example,Genest and MacKay 1986, Genest and Rivest 1993, Sungurand Yang 1996, and for more information on probabilitycopula functions, see Nelsen 1999). Parametric approachesfor constructing probability copulas are relatively straight-forward; they assume a functional form and then assessits parameters. Nonparametric approaches usually requiremore assessments and computational effort, but they alsoenjoy the benefits of providing a copula that matches theexact assessments.

Sungur and Yang (1996) provide a nonparametric itera-tive approach to determine the surface of an Archimedeanprobability copula using probability assessments for pointson a diagonal path in the domain of the copula func-tion. There are, however, several fundamental differencesbetween Archimedean utility copulas (Abbas 2009) andArchimedean probability copulas: (i) utility copula func-tions need not be grounded (this implies that if an attributeis at its minimum value, the utility copula need not bezero), and (ii) the cross derivative of a utility copula func-tion can be positive, negative, or zero (see for exampleAbbas and Howard 2005). These conditions require severalmodifications to the work of Sungur and Yang (1996) if

an iterative approach for constructing the generating func-tion is to be applied to utility copulas. In this paper, weprovide an iterative procedure for estimating the generat-ing function of an Archimedean utility copula from directutility assessments. These assessments include (i) a util-ity assessment at the lower boundary value of the domainand (ii) utility assessments on a path in the domain of thecopula function.

In our search of the literature, we have found a wealth ofrelated work on constructing multiattribute utility functionsthat relax the conditions of mutual utility independence.Farquhar (1975) introduced a general decomposition the-orem to develop the functional form of a multiattributeutility function with particular preference structures alongthe vertices of a hypercube. Bell (1979a, b) generalizedutility independence to interpolation independence, wherethe conditional utility function is an interpolation of theconditional utility functions at the boundary values of thedomain. Abbas and Bell (2011, 2012) introduced one-switch independence for multiattribute utility functions,which is a weaker condition than utility independenceand allows preferences over lotteries to change, but onlyonce, as a parameter varies. In related work, Abbas (2013)defined double-sided utility copulas that match all bound-ary assessments of the attributes.

Other approaches for constructing multiattribute utilityfunctions that are relevant to our formulation have alsobeen proposed and use a deterministic value function andthen assign a single attribute utility function over value.For example, Dyer and Sarin (1982) assign a utility func-tion over a univariate value function and define the conceptof relative risk aversion, and Matheson and Abbas (2005)assign a utility function over a multivariate value functionand relate the trade-off assessments among the attributes tothe ratio of their relative risk-aversion functions.

The remainder of this paper is structured as follows. Sec-tion 2 reviews the basic definitions and notation. Section 3characterizes utility functions satisfying mutual preferen-tial independence. Section 4 derives theoretical results todetermine the generating function using diagonal assess-ments and illustrates the approach with numerical exam-ples. Section 5 presents conclusions and summarizes themain results.

2. Basic Notation, Definitions, andReview of Previous Work

We assume that the decision maker follows the axioms ofexpected utility theory and has a multiattribute utility func-tion, U4x11 0 0 0 1 xn5, defined over n attributes, X11 0 0 0 1Xn.We use the lower case, xi, i ∈ 811 0 0 0 1 n9 to denote aninstantiation of attribute Xi, and use x0

i and x∗i to denote

the minimum and maximum of Xi, respectively. We use Xi

to denote the set of complement attributes to Xi and usexi to denote an instantiation of this complement. We alsouse the vector 4x11 0 0 0 1 xn5 to denote a consequence of the

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Abbas and Sun: Multiattribute Utility Functions Satisfying Mutual Preferential Independence380 Operations Research 63(2), pp. 378–393, © 2015 INFORMS

decision, and for notational convenience, we use 4xi1 xi5 torepresent a consequence.

We assume that the multiattribute utility function is(i) continuous, (ii) bounded, and (iii) strictly increasingin each argument at a reference value of the complementattributes. Our focus will be utility functions that are strictlyincreasing with each argument at the upper bound of thecomplement. We define a normalized conditional utilityfunction of xi at the maximum value of the complementattributes, x∗

i , as

Ui4xi � x∗

i 5=U4xi1 x

∗i 5−U4x0

i 1 x∗i 5

U4x∗i 1 x

∗i 5−U4x0

i 1 x∗i 50 (4)

Abbas (2009) introduced the notion of a utility copulafunction that expresses the multiattribute utility function interms of its conditional utility assessments at a referencevalue of the complement. A class 1 utility copula constructsa multiattribute utility function in terms of conditional util-ity assessments at the upper bound of the complementattributes—i.e., the utility function can be expressed as

U4x11 0 0 0 1 xn5=C(

U14x1 � x∗

151 0 0 0 1Un4xn � x∗

n5)

1 (5)

where C is a utility copula function.A class 1 utility copula is a linear function of each

attribute at the maximum values of its complementattributes, i.e.,

C411 0 0 0 111 vi111 0 0 0 115= aivi + 41 − ai51 i = 11 0 0 0 1 n0

The utility copula function is also normalized to range from0 to 1, i.e.,

C401 0 0 0 105= 0 and C411 0 0 0 115= 10

An important class of class 1 utility copulas is theArchimedean form

C4v11 0 0 0 1 vn5

= �−1

[ n∏

i=1

�4U4x0i 1 x

∗

i 5+ 41 −U4x0i 1 x

∗

i 55vi5

]

1 (6)

where �4v5 is a generating function that is continuous andstrictly increasing on the domain 60117, with �415= 1.

A multiattribute utility function U4x11 0 0 0 1 xn5 can alsobe constructed using a univariate utility assessment over anordinal value function V 4x11 0 0 0 1 xn5 through the relation

U4x11 0 0 0 1 xn5=UV 4V 4x11 0 0 0 1 xn551 (7)

where UV is the utility function over value.

3. Utility Functions Satisfying MutualPreference Independence

It is not surprising to see that a utility function of theArchimedean form (6) can be converted into an additive

function of the attributes using a monotone transformation.To illustrate, the functional form (6) can be converted intoa product form of univariate assessments by applying amonotone transformation � to both sides to get

�(

C4v11 v21 0 0 0 1 vn5)

=

n∏

i=1

�(

U4xi1 x∗

i 5+ 41−U4xi1 x∗

i 55vi)

0

A logarithmic transformation applied to this product formresults in an additive function of the attributes,

ln(

�4C4v11 v21 0 0 0 1 vn55)

=

n∑

i=1

ln(

�(

U4xi1 x∗

i 5+ 41 −U4xi1 x∗

i 55vi))

0 (8)

Since ordinal preferences are invariant to monotone trans-formations, the functional form (8) has the same ordinaltrade-offs as (6). What is not so obvious, however, is thatadditive ordinal preferences of the form

V 4x11 x21 0 0 0 1 xn5=m

( n∑

i=1

fi4x5

)

1

with arbitrary strictly increasing (and possibly different)functions fi must result in a multiattribute utility functionof the form (6), an Archimedean combination of functionsall having the same generating function, �. We prove thisbelow for the general case of n ¾ 3 but first assert thisresult for the case of two attributes having an additive valuefunction.

Proposition 1. If U4x11 x25 is continuous and strictlyincreasing with each argument at the upper bound of thecomplement attributes, then the following two statementsare equivalent:

(i) U4x11 x25=UV 4V 4x11 x255 withV 4x11 x25=m4f14x15+ f24x255,

where m is a continuous monotonic function and f1 and f2

are continuous and strictly increasing functions.(ii) U4x11 x25=C4U14x1 �x∗

251U24x2 �x∗155,

where C is an Archimedean utility copula of the form (6).

The following example illustrates this result.

Example 1 (Two-Attribute Archimedean Form). Con-sider the ordinal value function

V 4x1 y5= xy�1 0 < x1 y ¶ 11

that has been used to represent trade-offs for health andconsumption (Howard 1984). Note that this value functionis strictly increasing with each argument at the maximumvalue of the complement attribute. Moreover, a logarith-mic transformation converts this function into an additivefunction:

logV 4x1 y5= logx+� logy0

Proposition 1 asserts that any two-attribute utility functionsatisfying these ordinal preferences can be expressed as an

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Archimedean combination of the conditional utility func-tions at the maximum margin.

Consider the two-attribute utility function, obtained bytaking an exponential utility function over this value func-tion, as used in Howard (1984), and so

U4x1 y5=1 − e−�xy�

1 − e−�1 0 < x1 y ¶ 10 (9)

From (4), the conditional utility functions at the upperbounds are calculated as

u4

= U4x�y∗5=1−e−�x

1−e−�⇒ x4u5=−

1�

log41−41−e−�5u51

and

v4

=U4y �x∗5 =1 − e−�y�

1 − e−�

⇒ y4v5=

[

−1�

log41 − 41 − e−�5v5

]1/�

0

Substituting for x4u5 and y4v5 into (9) gives the utilitycopula function as

C4u1 v5=U4x4u51 y4v55

=1 − e−41/�5 log41−41−e−� 5u5·log41−41−e−� 5v5

1 − e−�0 (10)

Since ordinal preferences are additive, Proposition 1 assertsthat this copula function must be Archimedean. Indeed, thefunction (10) can be reduced to the Archimedean form

C4u1 v5= �−16�4u5�4v57

using the generating function

�4t5= −1�

log41 − 41 − e−�5t50

The following theorem extends the results of Proposition 1to multiple attributes, satisfying mutual preferential inde-pendence for n¾ 3.

Theorem 1. A multiattribute utility function U4x11 0 0 0 1 xn5that is continuous and strictly increasing with each argu-ment at the maximum value of the complement attributes,with n¾ 3, exhibits mutual preferential independence if andonly if its utility copula function is of the form (6).

Theorem 1 provides a fundamental method for con-structing multiattribute utility functions that satisfy mutualpreferential independence. Once we have asserted thatpreferential independence conditions exist, then the mul-tiattribute utility function must be Archimedean, and theassessment task reduces to the assessment of univariateutility functions for each attribute as well as a univariategenerating function and some corner values.

Methods for assessing single-attribute utility functionsfor each attribute are abundant in the literature (see forexample Keeney and Raiffa 1976). Our main focus willtherefore be the assessment of the generating function forthe Archimedean form.

4. Assessing a Multiattribute UtilityFunction Satisfying MutualPreferential Independence

Because a single generating function is sufficient to char-acterize the functional form of an Archimedean copula, itwill suffice to assess this generating function using a two-attribute formulation where the remaining attributes are setat their maximum values.

To illustrate, note that if C4v11 v21 0 0 0 1 vn5 is an Archi-medean utility copula, then the bivariate function

C4v11 v25=C4v11 v211111 0 0 0 115

is an Archimedean functional form having the same gener-ating function.

In principle, one can select from a library of functionsto determine the generating function of the Archimedeanform and then conduct some utility assessments on the sur-face to estimate the parameters of the chosen functionalform using a least-squares fit. This method of parameterestimation is widely used for utility functions, where theshape of the utility function is often assumed (such asan exponential function and the risk aversion coefficientis estimated to best match some utility assessments). Asshown in Abbas (2009), however, the generating functionof an Archimedean utility copula is strictly monotonic, butit does not need to be concave or convex on its entiredomain. In fact, it can even be S shaped to allow for furtherflexibility in the types of trade-offs that can be modeled.Therefore, the analyst must choose a functional form forthe generating function that allows for a wide variety ofshapes if the generating function is to accurately representthe assessments provided.

An alternate approach (that may be used for utilityfunctions) if we do not assume a particular form is toassess a few points on the curve and then fit those pointswith a smooth curve. Fritsch and Carlson (1980) pro-pose cubic polynomials, as an example, to connect orderedpoints using a differentiable path. This method could workwell for utility functions because each fitted point can beassessed directly as the utility of a consequence and can beinterpreted clearly in terms of lottery assessments. Unliketraditional utility function assessments, however, it is notpossible to immediately assess points on the generatingfunction of an Archimedean form because there is no clearinterpretation for the types of lottery questions one wouldask to determine points on the generating function directly.To remedy this problem, we provide a method to infer thegenerating function from utility assessments on the domainof the attributes and then provide proofs of convergence ofthese assessments to the generating function. This sectionexplains an iterative approach to infer the generating func-tion of the Archimedean form using direct utility assess-ments on the domain of the attributes.

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4.1. Assessments Needed for Constructing aTwo-Attribute Utility Function Using anArchimedean Utility Copula

Since the generating function of an Archimedean utilitycopula is strictly increasing on the interval 60117, its deriva-tive is positive (and possibly zero at some finite isolatedpoints). We consider the case where the derivative of thegenerating function can be zero at finite points but assumeit is strictly positive at 41115, i.e., we assume that �′415 > 0.

The utility assessments needed for constructing a two-attribute utility function using an Archimedean utility cop-ula can be divided into two steps:

(1) Step 1: Assess the boundary utility functions for eachattribute at the upper bound of the complement attributes.

(2) Step 2: Perform additional utility assessments todetermine the generating function. These additional assess-ments include

(a) a utility assessment at the lower bound of thedomain and

(b) a utility assessment on a path on the domain of thecopula, which we refer to as a skewed diagonal assessment.Define the corner values, kx = U4x∗1 y05 and ky =

U4x01 y∗5. Without loss of generality, we assume thatkx ¾ ky . If both corner values kx1 ky are zero, then theassessment task would be simplified and the lower bound,Step 2(a), would not be needed (the utility function wouldbe grounded and there would be no lower boundary assess-ments). We shall assume therefore that at least one of thecorner values kx > 0.

We now explain the assessments needed to construct theArchimedean utility function in more detail. Next we deriveconvergence results to determine the generating functionfrom the utility assessments of Steps 1 and 2.

Step 12 Assess two boundary utility functions U4x1 y∗5and U4x∗1 y5. This step requires a utility assessment foreach attribute at the upper bound of the complementattributes. For two attributes, we assess the two normalizedconditional utility functions U4x �y∗5 and U4x∗ �y5 as wellas the two corner values kx =U4x∗1 y05 and ky =U4x01 y∗5.The normalized assessments, U4x �y∗5 and U4y �x∗5, can

Figure 1. (a) The assessment U4x1 y∗5 is strictly increasing from ky to 1; (b) U4x∗1 y5 is strictly increasing from kx to 1.

x

U(x

,y* )

ky = 0.1

kx = 0.5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0(a)

U(x

* ,y)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0(b)

0 0.2 0.4 0.6 0.8 1.0x

0 0.2 0.4 0.6 0.8 1.0

be determined by fitting the individual assessments to someof the widely used functional forms of utility functionsor by assessing a few points and connecting them with asmooth path.

The utility function at the upper boundary values canthen be determined from these normalized conditional as-sessments and corner values using the relations

U4x1 y∗5= ky + 41 − ky5U4x �y∗51

U4x∗1 y5= kx + 41 − kx5U4y �x∗50(11)

Figure 1 plots an example of two boundary utility functionsfor attributes X and Y each on the domain 60117. The figureshows that U4x1 y∗5 and U4x∗1 y5 are strictly increasingand that ky = 001 and kx = 005.

Step 22 Additional utility assessments needed to deter-mine the generating function. To determine the generatingfunction when its functional form (or even shape) is notknown, we need to make some additional utility assess-ments.

Step 24a5: Utility Assessments at a Lower BoundU4x1 y05. To illustrate the intuition for assessing a lowerbound, note that the lower bound of an Archimedean utilityfunction is related to the upper bound using a transfor-mation that depends on the generating function. By directsubstitution into (6), we get

U4x1 y054

=C4v1105= �−1[

8�4U4x01 y∗5+ 41 −U4x01 y∗55

·U4x �y∗559�4U4x∗1 y055]

0

By assessing a lower boundary assessment and comparingit with the upper boundary assessment, we can infer someinformation about the shape of the generating function. Wehave assumed that kx ¾ ky , so we assess the utility of theattribute with the higher corner value, X, at the lower boundof the attribute with the lower corner value, Y , i.e., we needto assess the curve U4x1 y05.

We then define a general transformation, g, that relatesthe boundary assessments as

U4x1 y∗5= g4U4x1 y0550 (12)

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As attribute X spans its minimum to maximum values, thedomain of the function g spans U4x01y05=0 to U4x∗1y05=kx, and the range of g spans U4x01y∗5=ky to U4x∗1y∗5=1. Therefore,

g2 601 kx7→ 6ky117

The domain of the function g will be used to determine thegenerating function on the interval 601 kx7. We discuss theproperties of the transformation g in more detail in the nextsection. Because the domain of the generating function is60117, however, it is not sufficient to determine the gen-erating function from this assessment alone. This is whywe need a second assessment to characterize the generatingfunction on 6kx117.

Step 24b52 Utility Assessment on a Skewed DiagonalCurve. The second assessment is conducted across a pathon the domain of the attributes, which we refer to as askewed diagonal curve. The intuition behind this name isthat if both kx and ky were zero, this curve would be astraight (diagonal) line passing through the points 40105and 41115 in the domain of the copula function. Becauseboth kxandky need not be zero, however, and they neednot even be equal, the assessed curve in this case traces askewed and offset path in the domain of the consequences,as we illustrate below.

The skewed diagonal path is determined by first defininga parameter t that fills in the gap from kx to 1—i.e., wedefine t ∈ 6kx117. The values of x and y that determine thisskewed diagonal path are determined by

{

U4x1 y∗5= t1

U 4x∗1 y5= t0(13)

We therefore define x4t5 as the inverse function of thecurve U4x1 y∗5 and y4t5 as the inverse function of the curveU4x∗1 y5. The skewed diagonal path is traced by the points4x4t51 y4t55 on the interval t ∈ 6kx117. Denote the utilityvalues across this path as S4t5, i.e.,

S4t54

=U4x4t51 y4t551 t ∈ 6kx1170 (14)

Figure 2 illustrates the utility assessments on the lowerbound and the skewed diagonal path.

The following steps summarize the assessment procedurefor the skewed diagonal curve:

(i) define the parameter ton the interval 6kx117;(ii) define x4t5 using the equation U4x4t51 y∗5= t;(iii) define y4t5 using the equation U4x∗1 y4t55= t;(iv) trace the path 4x4t51 y4t55, which is shown using the

dashed line in Figure 2; and(v) conduct the utility assessments S4t5=U4x4t51 y4t55

using indifference assessments.The following example illustrates numerically the com-

plete set of utility assessments needed to construct anArchimedean utility function.

Figure 2. (Color online) Assessments along the skeweddiagonal curve and the lower bound.

00.010.020.030.040.050.060.070.080.091.00

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0x

y

Example 2 (Utility Assessments for the Archime-dean Form). Step 12 Upper Boundary Assessments. Thefirst step is to assess the upper boundary curves for eachattribute. Once again, this can be done by identifying afunctional form and assessing its parameters or by assess-ing several points and fitting them. Here we assume a par-ticular functional form. Suppose that the upper boundaryutility functions are

U4x1 y∗5= 1052 − 1042e−x1 x ∈ 60117 (15)

and

U4x∗1 y5= 1029 − 0079e−√y1 y ∈ 601170 (16)

By direct substitution, this implies that kx = U4x∗1 y05 =

005 and ky =U4x01 y∗5= 001.Step 24a52 Lower Boundary Assessment. Because the

highest corner value is kx, we need to assess the lowerboundary curve U4x1 y05. For this example, we use a hyper-bolic absolute risk aversion (HARA) utility function at thelower bound because of its generality. The analyst mayassess utility values on this lower bound and then use theseutility assessments to estimate the parameters of the HARAutility. Suppose that the resulting lower boundary assess-ment is

U4x1y05=00582−4009884−00041x546041 x∈ 601170 (17)

From (12), (15), and (17), the transformation g satisfies

1052 − 1042e−x= g400582 − 4009884 − 00041x5460450 (18)

To determine the transformation g1 define r = U4x1 y05.From (17),

r = 00582 − 4009884 − 00041x54604

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Figure 3. Transformation g4r52 601 kx7→ 6ky117.g

(r) kx

g (0) = ky

g (kx) = 1

0.6

0.7

0.8

0.9

1.0

0.5

0.4

0.3

0.2

0.1

00 0.05 0.10 0.15 0.20 0.25

r0.30 0.35 0.40 0.45 0.50

Note that r ∈ 601 kx7= 6010057. Rearranging gives

x = 240107 − 24039400582 − r50002161 r ∈ 60100570 (19)

Substituting from (19) into (18) gives

g4r5= 1052 − 1042e−4240107−24039400582−r500021651

r ∈ 60100570 (20)

Figure 3 plots the transformation g4r5 in (20). Note thatg2 601 kx7→ 6ky117 (as expected).

Step 24b52 Skewed Diagonal Assessment. To assess theutility values along the skewed diagonal curve, we firstdefine t ∈ 6kx117. We then determine x4t5 and y4t5 for dif-ferent values of t using (15) and (16), respectively. Table 1shows the assessments. The first column shows discrete val-ues of the parameter t. The second and third columns showthe corresponding values of x4t5 and y4t5. Note that fort = kx, y4kx5= y0, and for t = 1, x415= x∗ and y415= y∗.

Figure 4 plots this skewed diagonal path from Table1,which is the x–y plane of Figure 2.

The last column in Table 1 shows the utility assessmentsfor the points x4t5 and y4t5 defining the curve S4t5 obtainedusing indifference lottery assessments of (x4t51 y4t55 fora binary gamble that gives either 4x∗1 y∗5 with a prob-ability U4x4t51 y4t55 or 4x01 y05 with a probability 1 −

U4x4t51 y4t55.The utility assessments across the skewed diagonal path

can also be made by decomposing the assessment intomultiple steps using the utility tree decomposition (Abbas2011), where lotteries representing only one variation of

Table 1. Determine the skewed diagonal path4x4t51 y4t55 and the utility assessment S4t5.

t x4t5 y4t5 S4t5

kx = 005 0033 0 0027650.6 0043 0002 0039720.7 0055 0008 0052530.8 0068 0023 0066410.9 0083 0050 0081891 1 1 1

Figure 4. Skewed diagonal path on the X–Y domain.

x

y

(1, 1)

(0.83, 0.50)

(0.68, 0.23)

(0.55, 0.08)(0.43, 0.02)(0.33, 0)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

each attribute can be incorporated. For example, considerthe utility assessment at 4x11 y15. The utility tree decompo-sition is

U4x11 y15

=U4x11 y∗5U4y1 �x15+U4x11 y

05U 4y1 �x15 (21)

where U = 1 −U .The assessment U4x11 y15 can therefore be composed

into U4x11 y∗5, U4x11 y

05, and U4y1 �x15. Note that we havealready assessed the utility function on the upper and lowerbounds, U4x1 y∗5 and U4x1 y05. Therefore U4x11 y

∗5 andU4x11 y

05 are already determined. Furthermore, the termU4y1 �x15 is a single indifference assessment that can beobtained using indifference assessments of 4x11 y15 for abinary gamble that gives either 4x11 y

∗5 or 4x11 y05. This

gamble keeps the level x1 fixed and varies only y1 fromy0 to y∗. Figure 5 illustrates the six assessments for S4t5versus t in Table 1.

Figure 5. A utility assessment on the skewed diagonalcurve S4t5.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

S(t

)

t

S (0.5) = 0.2765

S (0.6) = 0.3972

S (0.7) = 0.5253

S (0.8) = 0.6641

S (0.8) = 0.8189

S (1) = 1

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We have now conducted all utility assessments needed todetermine the utility surface.

4.2. Determining the Generating Function on theInterval 6kx117

4.2.1. Relating S4t5 to the Generating Function onthe Interval 6kx117. We now determine the generatingfunction on the interval 6kx117 using the assessment S4t5.The following proposition relates S4t5 to the generatingfunction.

Proposition 2. Relating S4t5 to the Generating Function

S4t5= �−144�4t55251 t ∈ 6kx117

Proposition 2 shows that a portion of the generating func-tion on the interval t ∈ 6kx117 can be estimated if we solvethe functional equation �4S4t55 = 4�4t552. We provide aniterative solution to this functional equation using the fol-lowing steps:

Step 1: Determine the inverse function S−1 on the inter-val 6kx117.

Step 2: Determine the composite inverse function forany positive integer m as

S4−m54t54

= S−1� · · · � S−14t51 ∀ t ∈ 6kx1170

Step 3: For any S4−m54t5, define the exponential function�m4t5 such that

�m4t5= e2m4S4−m54t5−150

We prove in Theorem 2 that the iterations �m4t5 con-verge to the generating function on the interval 6kx117as mincreases. The following example illustrates the stepsneeded to solve this functional equation numerically and todetermine the generating function on the interval 6kx117.

Example 3 (Determining the Generating Functionon the Interval 6kx117). Step 12 Determine the Inversefunction S−14t5. To determine the inverse function S−14t5from the assessments in Table 1, we interchange the orderof the assessments of 4t1 S4t55. The six assessments of4t1 S−14t55 are as follows: 400276510055, 400397210065,400525310075, 400664110085, 400818910095, and 41115. Thenext step is to fit the assessments of S−14t5 using asmooth curve to help determine its composite functions.Appendix B derives the properties of S4t5 and its inverseS−14t5 and illustrates why the inverse function is clearlydefined. Appendix C provides a procedure to determine apiecewise polynomial fit for S−14t5 that may be used inpractice.

Step 22 Determine the Composite Functions S−4m54t5.Given S−14t5, the calculation of S−24t5 is obtained by itera-tion, where S−4254t5

4

= S−14S−14t55, and similarly for higherorders to get S−4m54t5. Figure 6 plots the inverse functionS−1 and its composite functions, S4−35 and S4−65 on 6kx117

Figure 6. Inverse function S−1 and its composite func-tions S4−35 and S4−65.

0.60

0.65

0.70

0.75

0.80

0.85

0.90

0.95

1.00

0.5 0.6 0.7 0.8 0.9 1.0

t

S (–6)(t )

S (–3)(t )

S–1(t)

as determined by the polynomial fit of S−1 in Appendix C.Appendix B, explains why S4−m54t5 ¾ S4−4m−1554t5 and,therefore, why the curves in Figure 6 are increasing with m.

Step 32 Determine the Iterations �m4t5. For any S4−m54t5,define the exponential function

�m4t5= e2m4S4−m54t5−150 (22)

The iterations �m4t5 are obtained by direct substitution.Figure 7 plots the curves �14t5 = e24S−14t5−15, �34t5 =

e84S4−354t5−15, and �64t5 = e644S4−654t5−15 computed directlyfrom S−1, S4−35, and S4−65. As we shall see, the iterations�m4t5 approximate the generating function on the interval6kx117.

4.2.2. Convergence of �m4t5 to the Generating Func-tion on the Interval 6kx117. Observe that the generatingfunction of an Archimedean copula is unique up to a powertransformation, i.e., if C4v11 0 0 0 1 vn5 is an Archimedeanutility copula with generating function �, then it is also thecopula formed by the generating function ��, �> 0.

Lemma 1. If the derivative of the generating function att = 1 is not equal to zero (i.e., �′415 6= 0), then therealways exists �> 0 such that �′415= 1, where � = ��.

Figure 7. Convergence of �m4t5, m= 11316 on 6kx117.

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00.50 0.55 0.60 0.65 0.70 0.75

t

m

0.80 0.85 0.90 0.95 1.00

�1(t)

�3(t)

�6(t)

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Lemma 1 implies that if �′415 6= 0 (as we have assumed)then we can further assume without loss of generality thatthe generating function satisfies �′415= 1.

Theorem 2 (Determining the Generating Functionon the Interval 6kx117). If the generating function of anArchimedean utility copula, �4t5, satisfies �′415= 1, then

�4t5= limm→�

�m4t51 ∀ t ∈ 6kx1170 (23)

Theorem 2 asserts that higher orders of �m4t5 convergeto the generating function, �4t5, on the interval t ∈ 6kx117.While any power of a generating function results in anequivalent Archimedean copula, the convergence of Theo-rem 2 results in the generating function that satisfies thecondition �′415= 1.

4.3. Determining the Generating Function on theInterval 601 kx5

We have determined the generating function on the inter-val 6kx117. We now show how to determine the generatingfunction on the remaining interval using the estimated gen-erating function on the interval 6kx117 and the transforma-tion function g4r5.

4.3.1. Relating g4r5 to the Generating Function onthe Interval 6kx117. Define the pth composite functiong4p54r5 as

g4p54r54

= g � · · · � g4r50

Before we relate g4r5 to the generating function, we need toprovide some intuition about the shape of the compositionsg4p54r5. Figure 8(a) plots the functions g4r5, g4254r5, g4354r5,and g4454r5 for the assessments of Example 2.

Observe the following important features fromFigure 8(a):

(a) The domain of g4r5 is 601 kx7 and its range is kyto 1 (as expected). Moreover, it satisfies the inequality kx ¶g4r5 < 1 on the interval r ∈ 6002710055. No other compositefunction of g satisfies the inequality kx ¶ g4p54r5 < 1 on theinterval r ∈ 6002710055.

Figure 8. (a) Composite functions g4r5, g4254r5, g4354r5, and g4454r5; (b) integer valued function p4r5.

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0(a) (b)

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.500

1

2

3

4

r0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

r

p(r

)

kx = 0.5

r4 = 0.03r4 = 0.03 r3 = 0.13 r2 = 0.27

r3 = 0.13r2 = 0.27

r1 = kx = 0.5 r1 = kx = 0.5

g(4)(r )

g(3)(r )g(2)(r ) g(r )

p (r ) = 4

p (r ) = 3 p (r ) = 2 p (r ) = 1

(b) The function g4254r5 satisfies kx ¶ g4254r5 < 1 on theinterval r ∈ 60013100275. No other composite function of gsatisfies kx ¶ g4p54r5 < 1 on this interval.

(c) The function g4354r5 satisfies kx ¶ g4354r5 < 1 on theinterval r ∈ 60003100135. No other composite function of gsatisfies kx ¶ g4p54r5 < 1 on this interval.

(d) The function g4454r5 satisfies kx ¶ g4454r5 < 1 on theinterval r ∈ 60100035. No other composite function of gsatisfies kx ¶ g4p54r5 < 1 over this interval. The functiong4454r5 has the property that its value at zero is greater thankx. This is where we end the compositions of g4r5 for thepurposes of estimating the generating function.We can now define the integer-valued decreasing functionp4r5 as the smallest integer, p, for any r , that satisfies kx ¶g4p54r5 < 1. From Figure 8(a), we have

p4r5= 11 ∀ r ∈ 60027100551

p4r5= 21 ∀ r ∈ 600131002751

p4r5= 31 ∀ r ∈ 600031001351

p4r5= 41 ∀ r ∈ 60100035

Figure 8(b) plots the function p4r5 and shows that it is adecreasing step function. The domain of p4r5 is the interval401 kx5 and it is integer-valued.

The following proposition relates the functions g4p54r5 tothe generating function.

Proposition 3 (Relating g4p54r5 to the GeneratingFunction on r ∈ 601 kx5).

�4r5= 4�4kx55p�4g4p54r551 r ∈ 601 kx50 (24)

Equation (24) is a functional equation that shows that�4r5 on the interval 601 kx5 is equal to the product of theconstant4�4kx55

p and the composition �4g4p54r55 for anyvalue of p. This might suggest at first that we can determinethe generating function on the interval 601 kx5 by direct sub-stitution for r ∈ 601 kx5 into (24). The problem we encounterhowever is that we have only determined the value of �4t5on the interval 6kx117. If the value of r in (24) is suchthat g4p54r5 ∈ 6kx1171 then we can substitute directly into

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the right-hand side of (24) to determine the correspond-ing value of �4r5. It might be possible, however, that thecomposition g4p54r5, for a given value of p, lies outsidethe interval 6kx117, as we have seen in Figure 8(a). If thisis the case, then we cannot determine �4r5 by direct sub-stitution into the right-hand side for that value of p.

The question that arises now is whether we can alwaysfind a value of p such that the composition g4p54r5 belongsto the interval 6kx117 for any r ∈ 601 kx5? If this were thecase, then we can determine �4r5 over the whole inter-val 601 kx5 by direct substitution into the right-hand side of(24). The following lemma asserts this fact.

Lemma 2 (Existence of the Integer p). For any givenr ∈ 401 kx5, there exists a composite function g4p5 such that

kx ¶ g4p54r5 < 10

As a result of Lemma 2, for any r ∈ 401 kx5, we areguaranteed a value of p such that kx ¶ g4p54r5 < 1. Forease of calculation of the composite functions, we shalluse the lowest value of p that satisfies this condition,i.e., p4r5. This will enable us to determine the values of�4r5 over 401 kx5.

The Iteration �m4r5. Define the function

�m4r5= 4�m4kx55p4r5�m4g

4p4r554r551 ∀ r ∈ 401 kx51 (25)

where �m4t5 = e2m6S4−m54t5−157, t ∈ 6kx117 is the same itera-tion defined earlier and S4−m54t5 is the mth-order compos-ite function of the inverse function S−14t5 (also calculatedearlier).

To better understand this function (25), note that thefirst term is simply a constant term �m4kx5 raised to thepower of p4r5. The second term �m4g

4p4r554r55 is a com-posite function based on the iteration �m4t5 and the p4r5composition of g4r5.

We do not need to plot the full curves in Figure 8 everytime we compute �m4r5. To illustrate, suppose we wish tocalculate �340025. We first determine the value g40025 =

00377. Because g40025 < 005 = kx, we conduct anothercomposition to get

g42540025= g4g400255= g4003775= 00688 > 005 = kx0

Hence, the integer valued function p40025 = 2, and we donot need higher compositions at r = 002. Now we calculate�34g

425400255= �34006885 as

�34g425400255= e234S4−354006885−155

= e8400946−15= 00650 (26)

For the constant term,

�34kx5= e234S4−3540055−15= e84008955−15

= 00430 (27)

Substituting from (26) and (27) into (25) gives

�340025= 4�34kx55p40025�34g

4p400255400255= 00432× 0065

= 00120

Figure 9 plots the curve �m4r5 form= 11316 for the skeweddiagonal assessment of Table 1 and the boundary assess-ments of Example 3.

Figure 9. Convergence of �m4r5, m= 11316 on 401 kx5.

0.6

0.5

0.4

0.3

0.2

0.1

00 0.05 0.10 0.15 0.20 0.25

�1(r )

�3(r )

�6(r )

r

m

0.30 0.35 0.40 0.45 0.50

4.3.2. Determining the Generating Function on theInterval 401 kx5.

Theorem 3. If the generating function of an Archimedeanutility copula, �4t5, satisfies �′415= 1, then

�4r5= limm→�

�m4r51 ∀ r ∈ 401 kx50 (28)

Theorem 3 asserts that higher orders of �m converge tothe generating function on the interval 401 kx5. Once again,while any power of a generating function results in anequivalent Archimedean copula, the convergence of The-orem 3 results in a generating function that satisfies thecondition �′415= 1.

4.4. Summary of the Approach

We now summarize the steps needed to determine the gen-erating function from the utility assessments and then illus-trate the deviation between consecutive iterations.

(i) Assess two corner values: kx and ky . Assume thatkx ¾ ky .

(ii) Assess two utility functions at the upper bound:U4x �y∗5 and U4y �x∗5.

(iii) Assess the utility function at the lower bound,U4x1 y05.

(iv) Determine g4r5 on the interval 601 kx7 using (12)and its composites g4p54r5.

(v) Determine the integer-valued function p4r5 from thefunctions g4p54r5.

(vi) Assess the utility values along the skewed diagonalcurve, S4t5.

(vii) Determine the inverse function S−14t5 and its com-posite functions S4−m54t5.

(viii) Use Theorems 2 and 3 to determine the generatingfunction �4t5.

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5. Convergence and Comparison withOther Approaches

5.1. Convergences of Successive Iterations

To provide some insights into the rate of conversion of theresults for examples, define the iteration �m4t5 as

�m4t5=

{

�m4t51 t ∈ 6kx1171

�m4t51 t ∈ 401 kx50(29)

Define the deviation between �m4t5 and �m+14t5 over theinterval 60117 as

em =

(

∫ 1

04�m+14t5−�m4t55

2 dt

)1/2

0

Figure 10 shows the deviation, em, plotted versus m, wherethe iterations of �m4t5 are obtained from the assessments ofTable 1 with the lower boundary assessment in (17). Fromthe convergence results of Theorems 2 and 3, this deviationconverges to 0, i.e.,

limm→�

em = 00

The iterations can be conducted to reach the acquired accu-racy of estimating the generating function as shown in Fig-ure 10. Since the generating function �4t5 is unknown inthe approximation procedure, we terminate the iterations on�m4t5 when em is sufficiently small. Note that these iter-ations do not require additional cognitive effort from thedecision maker; they are simply computations used to cal-culate additional composite functions.

5.2. Comparison with Mutual Utility Independence

We now compare the estimates of the transformation gand the skewed diagonal assessment S4t5 to those obtainedusing the assumption of mutual utility independence. If twoattributes are mutually utility independent, then U4x �y∗5=

U4x �y05 and so

g4r5= ky +1 − ky

kxr1 (30)

Figure 10. Deviation Between �m+14t5 and �m4t5.

0

0.005

0.010

0.015

0.020

0.025

0.030

0.035

0.040

1 2 3 4 5

e1 = 0.0389

e2 = 0.0232

e3 = 0.0137

e4 = 0.0078

e5 = 0.0045

m

Dev

iatio

n

a linear function. It is natural to consider what S4t5 wouldlook like if the attributes are mutually utility independent.The following proposition determines S4t5 in this case.

Proposition 4. Two attributes are mutually utility inde-pendent with kx ¾ ky if and only if the utility functionU4x1 y5 has an Archimedean utility copula, and the follow-ing two statements hold:

(i) U4x �y∗5=U4x �y05, x ∈ 6x01 x∗71 and(ii) S4t5 = kt2 + 241 − k5t + k − 1, t ∈ 6kx1171 where

k = 41 − kx − ky5/441 − kx541 − ky55.

Proposition 4 shows a new method to verify mutual util-ity independence between two attributes. First, we verifythat U4x �y∗5=U4x �y051 where X is the attribute with thegreater corner value. Next, we assert that S4t5 is quadratic.If these conditions hold then the attributes are mutuallyutility independent. For the special case where kx + ky = 1(the case of an additive utility function), then k = 0 andS4t5 is a linear function. The following example comparesthe accuracy of the Archimedean utility copula obtained bythe iterative approach for Example 2 to the utility functionobtained assuming mutual utility independence.

Example 4 (Comparison with Mutual Utility Inde-pendence). Consider again the two-attribute utility func-tion of Example 2, where

U4x1 y∗5= 1052 − 1042e−x1 x ∈ 601171 (31)

and

U4x∗1 y5= 1029 − 0079e−√y1 y ∈ 601170 (32)

This implies that kx = 005 and ky = 001. To compare ouriterative assessment approach with that of mutual utilityindependence, we first compute the constant k

k =1 − kx − ky

41 − kx541 − ky5=

1 − 001 − 00541 − 005541 − 0015

= 00890

Proposition 4 asserts that skewed diagonal assessmentmust be

S4t5= 0089t2+ 0022t − 00111 t ∈ 60051170

From (30), the transformation function has the form

g4r5= 001 + 108r1 r ∈ 60100570

Figure 11 shows the skewed diagonal assessment S4t5 andthe transformation function g4r5 for the utility function ofExample 2.

It is straightforward to see that a linear generating func-tion of the form

�4t5= kt + 41 − k51 t ∈ 60117 (33)

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Figure 11. Archimedean copula vs. mutual utility independence.

t

S(t

)

r

S (1) = 1

g (0) = ky

g (kx) = 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0(a)

g(r

)

(b)

0.5 0.6 0.7 0.8 0.9 1.00

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

Archimedeanutility copula

Mutual utilityindependence

Mutual utilityindependence

Archimedeanutility copula

Note. (a) Skewed diagonal assessment S4t5; (b) transformation function g4r5.

satisfies the condition of mutual utility independence. Butto compare the generating function obtained from the iter-ative approach to that of mutual utility independence, weneed take a power of (33) that makes �′415 = 1. Recallthat the condition of �′415 = 1 applies to the generatingfunctions obtained from the iterative approach. Direct sub-stitution shows that the function �4t5 = 4kt + 41 − k551/k

satisfies the condition �′415= 1. From here on, we denotethe generating function for mutual utility independence as

�UI4t5= 6kt + 41 − k571/k

as it will be used in the comparison with the generatingfunction obtained from the iterative approach.

Figure 12 plots the �3 and �6, �UI4t5 for Example 2. Thefigure also shows the actual generating function �4t5 thatwe used to determine the numerical values of the examplesin this paper. Of course, �4t5 is not known to the ana-lyst during this entire procedure. We simply included �4t5

Figure 12. Generating functions �UI4t5, �4t5, 3rd-order iteration �34t5 and 6th-order iteration�64t5.

�UI(t )

�3(t)�6(t)

� (t)

t

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

in the figure to illustrate the convergence of the results.The figure shows that �4t5 and �64t5 are indistinguishableover the entire interval 60117. The figure also shows theimprovement that �3 offers over �UI in this example.

6. ConclusionOrdinal preferences represented by additive value functionsare widely used in practice. When uncertainty is present,a multiattribute utility function is needed to determine thebest decision alternative. We completed the class of multi-attribute utility functions that correspond to additive ordinalpreferences that are strictly increasing with each attributeat the maximum value of the complement. We showed thatthis class of utility functions must be an Archimedean com-bination of the utility assessments.

The results of this paper shed some new light on thestructure of multiattribute utility functions satisfying mutualpreferential independence. As we have shown, the func-tional form of the utility function is highly constrained,even if mutual utility independence conditions are not sat-isfied. The main insight from this formulation is the asser-tion that if mutual preferential independence is verified,then preferences over lotteries can be decomposed into twoparts: (1) single attribute assessments at the upper boundand (2) a single generating function that combines thesesingle attribute assessments. The assumption of mutualutility independence focuses only on the upper boundaryassessments but ignores the second component: the gen-erating function. The inclusion of the generating functionallows for more general trade-offs and preferences over lot-teries and simultaneously satisfies the same utility valuesat the boundaries.

Another implication of the results of this work is thatonce preferential independence is verified, we do not needto determine the actual values of the ordinal functionsof the attributes when constructing the multiattribute util-ity function; assessing the boundary utility functions usingindifference assessments and constructing the generating

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function (also using indifference assessments) is sufficientto capture the whole structure of the utility function.

When constructing a multiattribute utility function, thenext step after preferential independence is to verifywhether the attributes can be formulated to achieve someforms of utility independence. If such independence condi-tions can be verified, then the functional form of the utilityfunction is highly simplified. When utility independenceconditions cannot be verified, it is important to capturethe utility dependence among the attributes if we wish toprovide an accurate representation of the decision maker’spreferences. The results of this paper provide the completeclass of utility functions and methods for their assessmentwhen decision makers have additive ordinal preferences forincreasing utility functions but when mutual utility inde-pendence conditions do not exist.

Acknowledgments

The authors thank the editor, associate editor, and three anony-mous referees for their comments on content and exposition. Thiswork was supported by the National Science Foundation awards[SES 08-46417, CMMI 12-58482, and CMMI 13-01150].

Appendix A

Proof of Proposition 1. Necessity: Given

U4x11 x25=UV 4m4f14x5+ f24x5551

where U4x11 x∗25 and U4x∗

11 x25 are strictly increasing. Thisimplies that f14x

∗151 f24x

∗25 < �; otherwise U4x11 x

∗25 would not

be strictly increasing and likewise for U4x∗11 x25. We have

U4x11 x254=UV

(

m4f14x15+ f24x255)

=UV

(

m4ln4ef14x15ef24x2555)

0

Define f14x15 = e4f14x15−f14x∗1 55 and f24x25 = e4f24x25−f14x

∗2 55. There-

fore,

U4x11 x25=UV

(

m(

ln4ef14x15ef24x255+ f14x∗

15+ f24x∗

25))

4= UV

(

f14x15f24x25)

1 (A1)

where UV 4v5=UV 4m4ln v+ f14x∗15+ f24x

∗2555, which is a strictly

increasing function.Note that f14x

∗15= f24x

∗25=1. When x2 = x∗

2 , (A1) gives

U4x11x∗

25= UV 4f14x15f24x∗

255= UV 4f14x15·15= UV 4f14x1550

Define �4= U−1

V , which is strictly increasing, and note that�415=1. Therefore,

f14x15= U−1V 4U4x11 x

∗

255= �4U4x11 x∗

2550 (A2)

Similarly,

f24x25= �4U4x∗

11 x2550 (A3)

Substituting for (A2), (A3) into (A1) with �4= U−1

V gives

U4x11 x25= �−14�4U4x11 x∗

255 ·�4U4x∗

11 x25550 (A4)

Observe that

U4x11 x∗

25=U4x011 x

∗

25+ 41 −U4x011 x

∗

255U4x1 �x∗

25

and

U4x11 x∗

25=U4x∗

11 x025+ 41 −U4x∗

11 x0255U4x2 �x∗

150

Substituting for (A4) into (A2) with u = U4x1 �x∗25 and v =

U4x2 �x∗15, we get

C4u1 v5= �−1(

�(

U4x011 x

∗

25+(

1 −U4x011 x

∗

255u)

·�(

U4x∗

11 x025+ 41 −U4x∗

11 x0255v

)

1 (A5)

which is the Archimedean utility copula of (6).Sufficiency: If U4x11 x25 has an Archimedean utility copula,

then (A4) holds. Applying a monotone transformation ln4�4t55gives an additive form

ln(

�4U4x11 x∗

255)

+ ln(

�4U4x∗

11 x255)

0 (A6)

Because ordinal preferences are invariant to monotone transfor-mations, the form in (A6) is equivalent to the additive formm4f14x15+ f24x255, where

m4t5= t1 f24x15= ln(

�4U4x∗

11 x255)

1 and

f24x15= ln(

�4U4x∗

11 x255)

0

Proof of Theorem 1. Necessity: Following the steps in Propo-sition 1, for multiple attributes,

U4x11 0 0 0 1 xn5=UV

(

m

( n∑

i=1

fi4xi5

))

= UV

( n∏

i=1

fi4xi5

)

1 (A7)

where fi4xi5 = e4fi4xi5−fi4x∗i 55, i = 11 0 0 0 1 n, and UV 4v5 =

UV 4m4ln v+∑n

i=1 fi4x∗i 555, which is a strictly increasing function.

Note that fi4x∗i 5= 1, i = 11 0 0 0 1 n0 When xi = x∗

i , (A7) gives

U4xi1 x∗

i 5= UV

(

fi4xi5∏

j 6=i

fj4x∗

j 5

)

= UV

(

fi4xi5∏

j 6=i

1)

= UV 4fi4xi550

Define �4= U−1

V ; then �4v5 is strictly increasing with�415= 1 and

fi4xi5= U−1V 4U4xi1 x

∗

i 554= �4U4x0

i 1 x∗

i 5

+ 41 −U4x0i 1 x

∗

i 55U4xi � x∗

i 551 ∀ i = 11 0 0 0 1 n1 (A8)

Substituting for � = U−1V and (A8) into (A7) gives

U4x11 0 0 0 1 xn5

= �−1

( n∏

i=1

�4U4x0i 1 x

∗

i 5+ 41 −U4x0i 1 x

∗

i 55U4xi � x∗

i 55

)

Therefore, the Class 1 utility copula of U4x11 x25 is

C4v11 0 0 0 1 v5

= �−1

( n∏

i=1

�4U4x0i 1 x

∗

i 5+ 41 −U4x0i 1 x

∗

i 55vi5

)

1 (A9)

which is the Archimedean utility copula of (6).

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Sufficiency: This is straightforward by applying a transforma-tion ln4�4t55 to (A9).

Proof of Proposition 2 for S4t5. By definition, U4x4t51 y∗5= t

and U4x∗1 y4t55= t. Therefore,

�4U4x4t51 y∗55= �4U4x∗1 y4t555= �4t5

Substituting into (A4) gives,

S4t54=U4x4t51 y4t55= �−1

(

�4U4x4t51 y∗5�4U4x∗1 y4t555

= �−144�4t55251 ∀ t ∈ 6kx1170

Proof of Lemma 1. Because the generating function � is strictlyincreasing on the interval 60117, its derivative at 1, �′415¾ 0. If�′415 6= 0 (as we have assumed), then �′415 > 0. Because the gen-erating function of an Archimedean copula is invariant to a powertransformation, define the new generating function � = ��, with� = 1/�′415 > 0. Note that �415 = 1 and its derivative satisfies�′415= ��′415 · 6�4157= 1.

Proof of Theorem 2. Define �4t5 = − ln4�4t55, ∀ t ∈ 40117and its inverse �−14v5 = �−14e−v5, ∀v ∈ 601�5. Note that�4t5 is strict decreasing on the interval 60117 with �415 =

− ln 4�4155= 0. Moreover, �4t5= e−�4t5, ∀ t ∈ 40117, so

S4t54= �−144�4t5525= �−144e−�4t5525= �−14e−2�4t55

= �−142�4t551 ∀ t ∈ 6kx1170

Define qm4t5 = S4m541 − 2−mt5 and its inverse function q−1m 4t5 =

2m41 − S4−m54t55. Note that �′415 = 1, the derivative �′415 =

−4�′415/�4155= −1 6= 0.Sungur and Yang (1996) proved that for the eqnarray S4t5 =

�−142�4t55, the inverse function �−1 satisfies

�−14t5= limm→�

S4m5

(

1 +t

2m�′415

)

= limm→�

S4m541 − 2−mt5

= limm→�

qm4t51 ∀ t ∈ 6kx1170 (A10)

Hence, the function � satisfies �4t5 = limm→� q−1m 4t5, ∀ t ∈

6kx1170 Therefore,

�4t5=e−�4t5= lim

m→�e−q−1

m 4t5= lim

m→�e2m4S4−m54t5−15

= limm→�

�m4t50

Proof of Proposition 3. Taking y = y0 in (A4) gives

U4x1 y05= �−14�4U4x1 y∗55 ·�4U4x∗1 y0555

= �−14�4U4x1 y∗55 ·�4kx550 (A11)

Denote r = U4x1 y051 then g4r5 = g4U4x1 y055 = U4x1 y∗5. Sub-stituting g4r5 into (A11) gives

r = �−14�4g4r55 ·�4kx551 i.e., �4r5= �4g4r55 ·�4kx50 (A12)

Applying (A12) at g4r5 gives

�4g4r55= �4g4g4r555 ·�4kx5= �4g4254r55 ·�4kx50 (A13)

Substituting from (A13) into (A12) gives

�4r5= 4�4g4g4r555 ·�4kx55 ·�4kx5

= �4g4254r55 · 4�4kx5520 (A14)

Repeating the above iteration with g4r5 gives

�4r5= �4g4p54r554�4kx55p0

Proof of Lemma 2. For a given r ∈ 401 kx5, 0 ¶ �405 < �4r5 <�4kx5 < 1 since � is strictly increasing. Hence, limp→�4�4kx55

p =

0 <�4r50 Therefore, 4�4kx55p ¶ p4r5 for a sufficiently large inte-

ger p > 1. Define p0 as the smallest such integer—i.e., 4�4kx55p0

¶ �4r5 and 4�4kx55p0−1 >�4r5. Hence,

�4kx5¶�4r5

4�4kx55p0

·�4kx5

=�4r5

4�4kx55p0−1

< 1 = �4150 (A15)

Note that �4g4p0−154r55= �4r5/4�4kx55p0−1 due to Proposition 3.

Hence, inequality (A15) becomes �4kx5 ¶ �4g4p0−154r55 < �415.Since � is strictly increasing, �4kx5¶ �4g4p54r55 < �415, wherep = p0 − 1.

Proof of Theorem 3. Note that g4p4r554r5 ∈ 6kx115, ∀ r ∈ 401 kx5.Proposition 3 gives

�4r5= 6�4kx57p4r5

·�4g4p4r554r55

= limm→�

6�m4kx57p4r5

· limm→�

�m4g4p4r554r55

= limm→�

6�m4kx57p4r5

·�m4g4p4r554r55

= limm→�

�m4r50

Proof of Proposition 4. Necessity: If two attributes are mutu-ally utility independent, then

U4x �y∗5=U4x �y051 ∀x ∈ 4x01 x∗51

and

U4x1 y5= kxU4x �y∗5+ kyU4y �x∗5

+ 41 − kx − ky5U4x �y∗5U4y �x∗50 (A16)

According to the definition of the utility copula in (5), we get itsutility copula as

C4u1 v5= kxu+ kyv+ 41 − kx − ky5uv0 (A17)

Substituting the definitions of x4t5 and y4t5 in (13), (11), and(A16) into (14) gives

S4t5=U4x4t51 y4t55

= kxt − ky

1 − ky+ ky

t − kx1 − kx

+ 41 − kx − ky5t − ky

1 − ky

t − kx1 − kx

= kt2+ 241 − k5t + k− 10

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If kx + ky = 1, then U4x1 y5 = kxU4x �y∗5 + kyU4y �x∗5, whichhas the additive value function as kxU4x �y∗5 + kyU4y �x∗5. Ifkx + ky 6= 1, then

U4x1 y5= 41 − kx − ky5

(

U4x �y∗5−ky

41 − kx − ky5

)

·

(

U4x �y∗5−kx

41 − kx − ky5

)

−kxky

41 − kx − ky5

= 41 − kx − ky5

· eln4U4x �y∗5−ky/41−kx−ky 55+ln4U4x �y∗5−kx/41−kx−ky 55

−kxky

41 − kx − ky51

which has the additive value function as

ln(

U4x �y∗5−ky

41 − kx − ky5

)

+ ln(

U4x �y∗5−kx

41 − kx − ky5

)

0

Hence, U4x1 y5 has an Archimedean utility copula as a result ofProposition 1.

Sufficiency: Theorems 2 and 3 show that the generating function�4t5 of the Archimedean utility copula in (6) satisfying �′415 =

1 is uniquely determined on the interval 40117 by the monotonetransformation g4r5 and the skewed diagonal assessment S4t50

Hence, g4r5 and S4t5 uniquely determine the Archimedean util-ity copula C4u1 v5 of (6). Note that, if U4x �y∗5= U4x �y051 theng4r5= ky + 441−ky5/kx5r0 Therefore, if (i) and (ii) hold, then theArchimedean utility copula C4u1 v5 is in the form of (6) and thentwo attributes are mutually utility independent.

Appendix B

Properties of S4t5.(1) S4t5 2 6kx117→ 6S4kx5117.(2) S4t5 is a continuous and strictly increasing function (refer

to Lemma 2).(3) S4t5 < t. This is because 4�4t552 <�4t5, so �−144�4t5525 <

�−14�4t55= t.(4) The minimum value is S4kx5¶ kx.(5) The maximum value is S415=U4x∗1 y∗5= 1.Because S4t5 is continuous and strictly increasing, we can

define the inverse function S−1 on the interval 6kx117.

Properties of S−14t5.(1) S−12 6kx117→ 6S−14kx5117.(2) S−14t5 > t, ∀ t ∈ 6kx115 because S4t5 < t.(3) Minimum value: S−14kx5 > kx (because S4kx5 < kx,

S−14kx5 > kx5.(4) Maximum value: S−1415= 1 (because S415= 15.(5) S4−m54t5 > S4−4m−1554t5, t ∈ 6kx115 (this is because

S−14t5 > t5.Note that S−14kx5 > kx, we know that the domain of S−1,6kx117 contains its range, 6S−14kx5117. Therefore, the composites,S4−m54t5, m= 1121 0 0 0, are well defined on 6kx117.

Appendix C. Piecewise CubicPolynomial Interpolation

We applied a piecewise cubic polynomial interpolation for curvefitting of S−14t5 (see Fritsch and Carlson 1980 for further details).Denote the assessed points as 4xi1 yi5, i = 11 0 0 0 1 n + 1. Theapproach assigns a cubic polynomial S−1

i 4x5 over each interval6xi1 xi+17, i = 11 0 0 0 1 n as the fitting curve, where

S−1i 4x5= h004r5yi +h104r54xi+1 − xi5mi +h014r5yi+1

+h114r54xi+1 − xi5mi+11 x ∈ 6xi1 xi+171 (C1)

where r = 4x−xi5/4xi+1 −xi5, and where mi, mi+1 are the deriva-tives at xi and xi+1, respectively. The functions h00, h01, h10, andh11 are cubic polynomials defined as

h004t5= 2r3− 3r2

+ 11 h104r5= r3− 2r2

+ r1

h014r5= −2r3+ 3r2 and h114r5= r3

− r20

The derivatives mi, mi+1 are the only unknown parameters in (C1)given the assessed points 4xi1 yi5, i = 11 0 0 0 1 n+ 10 We now showhow to calculate these derivatives.

Denote the length of the ith interval as li = xi+1 − xi and theslope of the line connecting its two endpoints as di = 4yi+1 − yi5/4xi+1 − xi5, i = 11 0 0 0 1 n− 1.

(i) The interior derivatives mi, i = 21 0 0 0 1 n− 1 are

mi =

di−1di

w1di−1 +w2di

1 if di−1di > 03

01 if di−1di ¶ 01(C2)

where the weights satisfy w1 = 42li−1 + li5/434li−1 + li55 and w2 =

1 −w1.(ii) The derivatives at two endpoints of the whole domain, m1

and mn are

m1 =

01 if q1d1<03

3d11 if q1d1¾01 d1d2 <0 and �q1�>3�d1�3

q11 otherwise1

(C3)

and

mn =

01 if q2dn−1 < 03

3dn1 if q2dn−1 ¾ 01 dn−1dn−2 < 0 and

�q2�> 3�dn−1�3

q21 otherwise,

(C4)

where

q1 =42l1 + l25d1 − l1d2

l1 + l2and

q2 =42ln−1 + ln−25dn−1 − ln−1dn−2

ln−1 + ln−20

Now we apply the curve fitting approach to determine theinverse function S−14t5 from the assessments. From Table 1, werelabel t and S4t5 to get the six points:

4002765100551 4003972100651 4005253100751

4006641100851 4008189100951 411150

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We calculate the interior derivatives m2, m3, m4, and m5

from (C2). For example,

l1 = x2 − x1 = 003972 − 002765 = 001207 and

l2 = x3 − x2 = 005253 − 003972 = 0012810

Furthermore, the slopes in these two intervals are

d1 =y2 − y1

x2 − x1=

006 − 005001207

= 008285 and

d2 =y3 − y2

x3 − x2=

007 − 006001281

= 0078060

From (C2), the weights for m2 are

w1 =2l1 + l2

34l1 + l25=

2 × 001207 + 0013813 × 4001207 + 0013815

= 004950 and

w2 = 1 −w1 = 0050500

Substituting w1, w2, d1, and d2 into (C2) gives

m2 =d1d2

w1d1 +w2d2=

008285 × 007806004950 × 008285 + 0005050 × 007806

= 0080410

Similarly, m3 = 007497, m4 = 006819, m5 = 005966.Calculate the derivatives at the two endpoints x1 = 002765

and x6 = 1. Note that q1 = 442l1 + l25d1 − l1d25/4l1 + l25 =

008517 < 3d1, m1 = q1 = 0085170 Similarly, m6 = q2 = 0050160Substituting the derivatives mi, i = 11 0 0 0 16 into (C1) and rear-ranging it gives the fitting curve of S−14t5 in the form of a piece-wise polynomial over 6002765117 as

S−14t5=

−000814t3−00115t2

+00934t+0025231t∈ 600241003673

−004533t3−204151t2

+006889t+0028931t∈ 60036100573

−004828t3−00617t2

+005013t+0033641t∈ 60051006573

−005626t3+009761t2

+001297t+0044811t∈ 600651008173

−001868t3+002472t2

+005675t+0037211t∈ 600811170

The values of composite function S4−254t5 are obtained by apply-ing S−14t5 twice. For example, S4−2540055 = S−14S−1400555 =

S−14008085= 008114.

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Comm. Statist.-Theory Meth. 25(7):1659–1676.von Neumann J, Morgenstern O (1947) Theory of Games and Economic

Behavior (Princeton University Press, Princeton, NJ).

Ali E. Abbas is professor of industrial and systems engineer-ing and public policy at the Viterbi School of Engineering andPrice School of Public Policy at the University of Southern Cal-ifornia. He is also director of the National Center for Risk andEconomic Analysis of Terrorism Events (CREATE). His researchfocuses on decision analysis, risk analysis, multiattribute utilitytheory, and data-based decision making. Dr. Abbas is a recip-ient of multiple awards from the National Science Foundationincluding the National Science Foundation CAREER Award in2008. He is also widely published in books, journals, and confer-ence publications, and has shared his expertise through televisionappearances, TEDx, and other invited talks.

Zhengwei Sun is a lecturer in the department of manage-ment science and engineering, East China University of Scienceand Technology. He received his Ph.D. in industrial engineeringfrom the University of Illinois at Urbana–Champaign, where hewas also a postdoctoral research associate. His research inter-ests include value of information, Bayesian updating, and utilitytheory.

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Multiattribute Utility Functions Satisfying Mutual Preferential … · 2017. 11. 1. · utility assessments and a single generating function. We also provide a nonparametric approach