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A note on additive conjoint measurement
OF MATHEMATICAL
PSYCHOLOGY
A Note
two-factor solvability
489-494
(1971)
on Additive
Conjoint
ERIC
W. HOLMAN
Department
Additive unrestricted
8,
Measurement1
of Psychology, University of California, Los Angeles, California 90024
conjoint measurement is derived from axioms that do not include or a condition on interlocked standard sequences.
This note presents a weakened set of axioms for additive two-factor conjoint measurement. In order to discuss such measurement explicitly, the following notation will be used. The two factors will be denoted by the sets A, and A, ; levels of the factors will be denoted by elements a, b, c,... in A,, and p, 4, Y,... in A, . Joint effects of combining the two factors will be denoted by pairs of elements, such as ap, bq,..., in A, x A, . These effects will be assumed to be ordered by a relation 2 on A, x A,. Unless otherwise specified, all statements about elements such as a and p will be understood to apply to all a in A, and p in A, . In this situation, the representation for additive conjoint measurement states that there are functions $r on A, and $a on A, , such that up 2 bq if and only if +i(u) + 4,(p) 3 4,(b) + &(q). The usual uniqueness result states that the functions are interval scales with the same unit; that is, the functions & and &’ both satisfy the representation if and only if there are constants 01 > 0, /3i , and /?a , such that &’ = oz& + pi . Adams and Fagot (1959) demonstrated that the following three axioms are necessary for the representation:
AXIOM 1. up 2 bq or ap 5 bq or both; up 2 bq and bq 2 cr imply up 2 cr. AXIOM
2.
AXIOM 3.
ap 2 bp implies aq 2 bq; up 2 aq implies bp 2 bq. ax w fq and
fp
m bx imply up w bq.
Axioms 1 and 2 have usually been called, respectively, weak ordering dence. In the theory of webs, Axiom 3 is called the Thomsen condition; 1 This work was supported by National Science Foundation a Visiting Research Fellowship at Educational Testing Service. Kristof and R. Duncan Lute for their helpful suggestions.
489
Grant I would
and indepenin a stronger
GB-13588X like to thank
and by Walter
490
HOLMAN
form with * replaced by 2, it is called double cancellation in the measurement literature. Lute and Tukey (1964) stated sufficient conditions for additive conjoint measurement. They were weak ordering and double cancellation plus two additional axioms, of which one is necessary for the representation and the other is not. The necessary condition is an Archimedean axiom, which will be stated here in the improved version given by Krantz, Lute, Suppes, and Tversky (1971). AXIOM 4. If for some Z, _a in A, and p, q in A, , a sequence of elements a, in il, satisjies ~22 5 a,q < a$ a ai+1q 5 ag for all ai and ai+l , then the sequence must be Jinite. A similar statement holds with the roles of ‘4, and A, reversed. The nonnecessary condition of Lute and Tukey is a solvability axiom, which requires the sets A, and A, to be unbounded and is, consequently, too strong for many the following restricted empirical applications. Lute (1966), th erefore, substituted solvability axiom. AXIOM 5. For any a, b, _b in A, and p, q in A, such that bq 2 ap 2 _bq, there is a b in A, such that ap w bq; for any a, b in A, and p, 4, gin A, such that bq 2 ap 2 bq, there is a q in A, such that ap cz bq. In order to prove conjoint measurement under this weaker solvability condition, Lute added an axiom that assumes the existence of certain interlocked standard sequences. Fortunately, this last rather complicated axiom need not be repeated here, because the present note shows that it can be replaced by the following very weak condition, which is called essentialness by Debreu (1960) and Krantz, et al. (1971). AXIOM
6.
There are a, b, c in A, , andp,
q, r in A, , such that ar > br andpc > qc.
In other words, an additive conjoint representation, tions, will be derived from the necessary conditions Thomsen condition, and the Archimedean property, of restricted solvability and essentialness.
unique up to linear transformaof weak ordering, independence, and the nonnecessary conditions
THEOREM. [f the system (A, , A,, 2; a satisfies ,4xioms 1, 2, 3, 4, 5, and 6, then there are real-valued functions 4i on A, (i = I, 2), such that ap 2 bq if and only I? 41(a) + d,(p) > 4,(b) + &(q); moreover, the functions &’ also satisfy the representation if and only if& = a& + pi for some constants a: > 0, ,Q1, and p3 . Proof. By the ordered as follows. p, q in A, , let p 2 The theorem will a < a and p
ordering and independence axioms, the sets Ai can be weakly For a, b in A, , let a 2 b if and only if ap 2 bp for allp in d, ; for q if and only if ap 2 aq for all a in A, . first be proved for a square subset A,’ x AZ’, bounded by points such that _aP = ap, and Q 5 ap 5 Zp for all ap in A,’ x A,‘.
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Such a subset must exist by Axioms 5 and 6. For any a in Al’, _aP m tip 2 up 2 _a~; thus, using Axiom 5, let +a) be such that a n(a) m up. Let B be the set of all pairs (a, 6) in A,’ x A,’ such that a > CX, b > _b, and a n(b) 5 Z-JI For any (a, b) in B, $p 2 a r(b) 2 up; thus, using Axiom 5 again, let a 0 b be such that a m(b) GZ (a o b) p. It will now be shown that if B is nonempty, then the system (Al’, 2, B, 0) satisfies the axioms for bounded extensive measurement as stated by Krantz (1968) or Krantz, et al. (1971). To prove that 0 is commutative, suppose that (a, b) is in B. By definition, up ‘Q @n(a) and &b) m &; hence, by Axiom 3, en(b) * bn(a). It follows that (b, a) is in B, and (a o b)p e (b 0 a)p; thus, by independence, a o b m b o a. To prove that o ismonotonic, suppose that (a, c) is in B, and a 2 b. By independence, ar(c) 2 b-r(c). It follows that (b, c) is in B, and a 0 c ,k b 0 c. By commutativity, therefore, (c, a) and (c, b) are also in B, and c o a s a o c 2 b o c m c o b. To prove that o is associative, suppose that (a, b) and (a 0 b, c) are in B. Since a o b > b, it follows that (b, c) is in B. By definition and commutativity, bn(a)w (aab)p and (boc)pm bn(c); hence, by Axiom 3, (b 0 c) T(a) = (a o b) n-(c). By definition and commutativity, arr(b 0 c) =Z [ao(boc)]P = [(boc)oa]pw(boc)n(a); consequently, an(b 0 c) m (a 0 6) n(c). I t f o 11ows that (a, b 0 c) is in B, and a 0 (b o c) w (a 0 6) 0 c. The remaining conditions for extensive measurement can be verified immediately. Therefore, if B is nonempty, there is a function $r on Al’, such that +l(a 0 b) = &(a) + 4,(b), and &(a) > 4,(b) if and only if a 2 b. For any p in A, , (-ip a _aP 2 gp 2 Q.P; thus, let Q(P) be such that CX(~)$J m gp, and let &(p) = &[u(p)]. By the definitions of a(p) and n(a), gp * asp w _arr[cu(p)]; hence, p m n[ol(p)]. Thus, for up 5
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it follows from the definition that dp e bq, and thus a > d by independence; and the last sentence that Cl(u) 4 &(p) > &(d) $- &(p) =- 4,(b) -t &(q). If up a similar argument gives the reversed inequality. These results mean that ap 2 as claimed. In the remaining and only if d&4 + h(p) 3 Mb) i- Mq),
of & < bq, bq if case,
ap 2 c+ 2 bq if and only if A(a) + 4dp) 2 &(_a) + Cd P) 2 A(b) + h(q), by the results already proved. If B is empty, then _a and 8 must be the only elements in A,‘, and it follows from Axiom 5 that p and p are the only elements in d4,‘. By definition, _ap < gp =
0, fll , and /3? , let &(-a) = p1 , 41(a) = p1 -+ 01, 74(p) = A , and +2(P) = A + 01. These functions clearly satisfy the representation and uniqueness for conjoint measurement. Next, the representation will be generalized to the hypothesis that J1’ 1: -4,’ is rectangular and bounded. It will be assumed that tip > _a$; the proof is similar in the other case. Let a, be such that a,p a gp; if a, 1s defined and a,!ps ~rp, let anfl be defined such that u~+~P w a, p. It has already been shown possible to define conjoint scales & and & on _a 5 a 5 a, andp 5 p 5 p, respectively; for convenience, let +l(-a) = &( p) = 0 and A(q) = A( P) = 1. Th e scale $I can now be extended to the rest of A,’ by induction on n, since by the Archimedean axiom, there must be some m such that u,,,p 2 sp. Therefore, to perform the induction, it is assumed that &(a) has been defined witk the appropriate properties for all a ,< a, . Suppose that a,,, is defined; the proof is similar if a,$> 2~. For any a such that it follows that Q,-~$ hv a,~ 5 up 5 a,+l~ & a,~; thus, let a’ be a, 5 a 5 a,+, , such that a’p e up. Since a’ 5 a, by independence, +l(a’) is defined by the induction hypothesis; therefore, let #1(u) = (bl(a’) + I. For any a, 5 a 5 anil , and p in -g2’, a? 5 a’p 5 a’j5 w up; thus, let @up) be such that 6(ap)p c a’p. The definitions of a’ and 6(ap) imply by Axiom 3 that 6(ap) p q up. To compare up and bq, suppose first that a, 5 a 5 antI and a, 5 b 5 a,,, . Consequently, ap 2 bq if and only if 6(up)p 2 6(bq) P; which by m d e p en d ence is equivalent to S(up)p 2 6(bq)p, which by definition holds if and only if a’p 2 b’q, which by the induction hypoihesis occurs if and only if +,(a’) + &(p) 2 +,(b’) + &(q), which by definition is equivalent to &(a) + C,(p) 3 4,(b) + &(q). Next, suppose that a,, 5 a 5 a,iT1 and b 5 urL . If ap 2 a& than up 2 bq, and also &(a) -t &(p) -> +l(a,) + &(p) :: $,(b) + &(q). If ap 5 a& then because a,p 5 up, let p’ be such that ap P a,~‘; hence, ap * GP’ X bq if and only ifA + &(P) = $l(qJ + +2(~‘) 2 4,(b) -i- &(q). Since a’ is defined uniquely (up to w) for each a, the uniqueness of Al is the same as the uniqueness of ~&(a’), which is specified by the induction hypothesis. Finally, the representation can be extended to all A, :/ il, by defining an expanding sequence of bounded rectangular subsets that will eventually include any point in A, x A, . As has just been shown, the required functions can be defined on each subset in the series, and thus extended to the entire set. To fix the upper boundaries of these rectangles, an infinite sequence of points a;, will be defined, such that
A NOTE
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MEASUREMENT
493
a,,, 2 a;, for all n, and given any a in A, , there is an m such that a;, 2 a. If A, is bounded by an z such that a 2 a for all a, then let Z~ = a for all n; clearly, this sequence has the required properties. Next, suppose that there is no such upper bound. Let three points a,, q0 , and q be given, with q > q,, . The points (z,,, and qn+l will be defined recursively, given an and qn, with q > qrL. If there is an a such that uq, 2 i&q, then using Axiom 5, let a,+, be such that il,+,q, e a,q; also let q,+l = qn . Otherwise, because A, is assumed to be unbounded from above, let &+i be any point such that &+i > a,; since in this case zn+lqn < &q < %+s, let qn+l be such that %n+lq,+l e anq. Th ese definitions immediately imply that for all n, a;l+i 2 a;, and q > qn+l 2 qn . To show that the sequence a;, will eventually reach any point in A, , let any a in A, be given. By the hypothesis of unboundedness, there must also be a b such that b > a. If aq > bq, , then by Axiom 5, there is a q’ such that uq’ e bq,,; ootherwise, let q’ = q,, . Comparing q’ to the qTL,two cases are possible. First, suppose that qn 2 q’ for all n. Let b, = z,,; for any n > 0, if b, 5 ir, , let b,,, be such that bnq * bn+d, using Axiom 5 because b,,q’ < b,q 5 if,q GZ an+lqn+l 5 iin+lq’; thus, bn+1 5 G+1 Y so the sequence 6, can be continued indefinitely. Since the Archimedean axiom can be applied to the b, , it follows that a,,, 2 b,,, 2 a for some finite m, as claimed. In the second case, suppose that qc > 4’ for some finite 1; by definition, qn == qLfor n > 1 as long as Z~ 5 b. Hence, by the Archimedean axiom, there must be a finite m > 1 such that ii,q > bq,,, e bq, > bq’ 2 uq, and therefore a,, > a, as claimed. To define the other boundaries of the rectangles, Axiom 6 can be used to find _a, and g0 such that a,, < a,, and p, and P, such that p, < &. By a slight modification of the previous paragraph, it is possible to define a nonincreasing sequence a, , such that given any a in A r , a,, 5 a for some m; it is similarly possible to define a nonincreasing sequence p, and a nondecreasing sequence pn, such that given any p or some m Thus, let A?’ be the set of all a in A, such that in4,pm5P5Fmf and let A?) be the set of all p in A, such that pa 5 p 5 pn . Consea, 5 a 5 %l, x kp+l’ for all n, and for any ap in A, x A, quently, A?’ x Ap’ is included in A?+‘) there is an m such that up is in Ai”’ x Aim’. Since all the A?’ x A?’ are included in A, x A, , they must satisfy Axioms l-5. Since the A?’ x Ap) all include A?’ x A;“‘, they must also satisfy Axiom 6. Hence, each A?) :< Ap’ is a bounded rectangular subset, on which it is possible, as already shown, to construct functions #J?’ and Moreover, the arbitrary $2‘n) with the properties required for conjoint measurement. constants in these functions can be set so that for any n, +T+“(u) = +?‘(a) for all a in A?‘, and +F+l’ (p) = +c’(p) for all p in A p’. Thus, for any a in A, y&(u) can be defined as equal to +?‘(a), where n is such that a is in A?‘; similarly, for anyp in A, , &(p) can be defined as equal to $p’( p), where n is such that p is in A?‘. Q.E.D.
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HOLMAN REFERENCES
E., AND FGOT, R. A model of riskless choice. Behauiorul Science, 1959, 4, I-IO. G. Topological methods in cardinal utility theory. In K. J. Arrow, S. Karlin, and P. Suppes (Eds.), Mathematical methods ill the social sciemes. Sttlnford University Press, 1960. Pp. 16-26. KRANTZ, D. H. A survey of measurement theory. In G. B. Dantzig and A. F. Veinott, Jr. (Eds.), Fifth Summer Seminar on the Mathematics of the Decision Sciences, 1968, part 2. Providence, R. I.: American Mathematical Society. Pp. 3 14-350. KRANTZ, D. H., LUCE, R. D., SUPPES, P., AND TVERSKY, rl. Foundations of measurement, New York; Academic Press, 1971. LUCE, R. D. Two extensions of conjoint measurement. Journal of Mathematical Psychology, I, 66, ADAMS,
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Journal
J. Simultaneous of Mathematical
July 28, 1970
conjoint Psychology,
measurement: a new 1964, 1, l-27.
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of fundamental