It seems to be a common view that von Neumann and Morgenstern (1944, 1947, 1953), and the host of axiomatic theories that have appeared in their

Transcription

1 PER-ERIK MALMNÄS AXIOMATIC JUSTIFICATIONS OF THE UTILITY PRINCIPLE A FORMAL INVESTIGATION 1. INTRODUCTION It seems to be a common view that von Neumann and Morgenstern (1944, 1947, 1953, and the host of axiomatic theories that have aeared in their aftermath (see, for instance, Fishburn (1981 rovide a formal foundation for the utility rincile or at least for the rincile of maximizing exected utility. The aim of the resent aer is to show that this oinion rests on a misunderstanding of the results obtained: existing axiomatic theories do not rovide any formal foundation for the utility rincile at all, and nor can they be made to do that without the addition of new axioms that will do most of the job themselves. To substantiate this claim I will in this aer discuss a few of the axiom systems roduced to date. To simlify the resentation, I will divide the systems into two grous: the first grou consists of the systems directly insired by von Neumann and Morgenstern (1944 and the second one of the systems insired by Ramsey (1931. Within each grou I will select one system for secial attention: in the first grou I will in articular consider the axiom system formulated by Herstein and Milnor (1953, which, at least from a mathematical oint of view, is the most satisfactory one I have seen, and in the second grou I will ay secial attention to the system of Savage (1954, 1972, which has been the most influential one. Hence most readers can content themselves by reading the discussion of these two systems.

2 2. FUNDAMENTAL CONCEPTS Let us consider an agent who is to choose between a finite number of courses of actions (alternatives each having a finite number of outcomes or consequences. Let us moreover assume that each alternative a can be identfied with a matrix ( 1,, n o 1,, o n. Here, as usual, o 1,, o n are suosed to be the ossible consequences of erforming a and i is to be the robability that o i occurs given that a is erformed, (1 i n. Then the utility rincile may be formulated as follows: If a = ( 1,, n o 1,,o n and V is a real -valued function on O, the set of all outcomes, then a has a value that equals 1 n i V(o i. An agent is then said to accet the utility rincile if he assigns the value n 1 i v i to the alternative ( 1,, n o 1,,o n given that he has assigned the value v i to o i. Now, given a function V: O R, E V (a can be defined for each alternative n a by setting E V (a = 1 i v i for a = ( 1,, n o 1,,o n. An ordering > of the alternatives is then said to be comatible with the rincile of maximizing exected utility if a > b imlies E V (a > E V (b. Finally, an agent is said to accet the rincile of maximizing exected utility if his value ordering of the alternatives is comatible with that rincile.

3 3. HERSTEIN AND MILNOR (1953. These authors consider a non emty set of outcomes O, a set of rosects S, and an oeration s such that S O and s(α, a, b S if a,b S and 0 α 1. s(α, a, b is to be the rosect that a occurs with robability α and b with robability 1 - α, and s is to satisfy the following conditions: (i s(1, a, b = a, (ii s(α, a, b = s(1 - α, b, a, (iii s(λ, s(µ, a, b, b = s(λ µ, a, b, From (i - (iii we can derive (iv s(α, s(β, a, b, s(γ, a, b = s(α β + (1 - α γ, a, b which will be needed below. They also consider the relation 'at least as good as' between rosects. This relation, labeled, is emloyed to define the relations 'better than' ( > and 'equally good' ( ~ in the following way: a > b if and only if a b and not b a, and a ~ b if and only if a b and b a. The following axioms are then assumed: Axiom 1 is a comlete semi-order on S. Axiom 2 { α s(α, a, b c } and { α c s(α, a, b } are closed, a,b,c S. Axiom 3 If a ~ a', then s(1/2, a, b ~ s(1/2, a', b. From these assumtions the following Reresentation Theorem is then roved. (a If and s satisfy (i - (iii and Axioms 1, 2, and 3, then there exists a function U: S R, which is comatible with and s. (b If U and U' are real-valued functions on S that are comatible with and s, then U = α U' + β, for some α > 0 and β.

4 Here, a function U: S R is comatible with and s if and only if (i (ii a b if and only if U(a U(b, and U(s(α, a, b = α U(a + (1 - α U(b. 3.1 PROOF OF THE REPRESENTATION THEOREM. The Rereentation Theorem is derived from the following lemmata: (1 If a b c, then b ~ s(α, a, c for some α. (2 If b ~ s(α i, a, c and lim α i = α when i, then b ~ s(α, a, c. (3 If a > b, then a > s(1/2, a, b > b. (4 If a > b and 0 < α <1, then a > s(α, a, b > b. (5 If a ~ a', then a ~ s(α, a, a'. (6 If a ~ a', then s(α, a, b ~ s(α, a', b. (7 If a > b, then s(α, a, b > s(β, a, b if and only if α > β. (8 If a > b > c, then there exists a unique α such that b ~ s(α, a, c. Proof of the theorem. Part (a. (i a ~ b, for all a,b S. Set U(x 1. Then U is a real-valued function on S that is comatible with and s. (ii I a,b = S, for some a > b S. Here I a,b = { x a x b }. Set U(x = α if and only if x ~ s(α, a, b. Then (1, (4 and (8 imly that U is a real-valued function on S, and (7 yields that U is comatible with. Assume next that x ~ s(β, a, b and y ~ s(γ, a, b. Then (iv of section 3 and (6 imly that s(α, x, y ~ s(α, s(β, a, b, y ~ s(α, s(β, a, b, s(γ, a, b = s(α β + (1 - α γ, a, b. Hence U(s(α, x, y = α β + (1 - α γ = α U(x + (1 - α U(y, which shows that U is comatible with s. (iii Neither (i nor (ii. Pick c,d S such that c > d. Consider then two intervals I a,b, I a',b' such that c,d I a,b I a',b'. Set µ a,b (x = α if and only if x ~ s(α, a, b and µ a',b' (x = β if and only if x ~ s(β, a', b'. Moreover, set

5 Μ a,b (x = (µ a,b (x - µ a,b (d/(µ a,b (c - µ a,b (d and Μ a',b' (x = (µ a',b' (x - µ a',b' (d/(µ a',b' (c - µ a',b' (d. Then Μ a,b and Μ a',b' are comatible with and s on I a,b and I a',b' resectively, and Μ a,b (c = Μ a',b' (c = 1 and Μ a,b (d = Μ a',b' (d = 0. Assume now that x I a,b I a',b'. Then Μ a,b (x = Μ a',b' (x. This can be seen as follows: Case 1. c x d. Then x ~ s(α, c, d for some α. Hence Μ a,b (x = α Μ a,b (c = α Μ a',b' (c = Μ a',b' (x. Case 2. x > c > d. Then c ~ s(α, x, d for some α > 0. This yields α Μ a,b (x = 1 = α Μ a',b' (x. Hence Μ a,b (x = Μ a',b' (x. Case 3. c > d > x. Then d ~ s(α, c, x for some α < 1. This yields 0 = α + (1 - α Μ a,b (x = α + (1 - α Μ a',b' (x. Hence Μ a,b (x = Μ a',b' (x. Hence we can set U(x = y if and only if Μ a,b (x = y for some a,b in S such that x,c,d I a,b. We then get a real-valued function on S that is comatible with and s. Part (b. Assume that U and U' are comatible with and s. Assume also that c> d in S. Set V(x = (U(x - U(d/(U(c - U(d and V'(x = (U'(x - U'(d/(U'(c - U'(d. Then the argument in (iii above yields V(x = V'(x. Hence U(x = α U'(x + β with α = (U(c - U(d/(U'(c - U'(d and β = U(d - α U'(d. Proof of lemma 1. Set I 1 = { α s(α, a, c b } and I 2 = { α b s(α, a, c }. Then I 1 I 2 = [0,1] since is a comlete semiorder on S. Moreover, I 1 and I 2 are closed and non-emty. Hence I 1 I 2. Proof of lemma 2. Assume first that a ~ s(1/2, a, b. Then axiom 3 imlies s(1/2, a, b ~ s(1/2, s(1/2, a, b, b = s(2-2, a, b. Hence a ~ s(2-2, a, b. Reeated use of axiom 3 then yields a ~ s(2 -n, a, b for all n 1. Hence lemma 2 imlies a ~ b.

6 Assume then that s(1/2, a, b > a. Then lemma 1 imlies a ~ s(α 1 1/2, a, b. Reeated use of axiom 3 and lemma 1 then yields a ~ s(α n 2 -n, a, b with 0 < α n < 1. Hence a ~ b. We have thus roved that a > s(1/2, a, b. To rove s(1/2, a, b > b, roceed in an analogous way. Proof of lemma 4. Assume a > b and 0 < α < 1. define two sequences as follows: (a r 1 = 0, s 1 = 1. (b If α > 1/2 (r n +s n, then r n+1 = 1/2 (r n +s n and s n+1 = s n, but if α 1/2 (r n +s n, then r n+1 = r n and s n+1 = 1/2 (r n +s n. Then r i α s i and r i - s i = 2 - i + 1. Hence lim r i = lim s i = α when i. Moreover, reeated use of lemma 3 yields s(r i, a, b s(r i+1, a, b < s(s i+1, a, b s(s i, a, b, and via axiom 1 we then arrive at the desired conclusion. Proof of lemma 5. I first show that a ~ s(α, a, a. Indeed, assume a > s(α, a, a. Then, in view of (i of section 3, 0 < α < 1. If we set 1 > α 2 > α 1 > α > 1 - α > 0, then lemma 4 yields s(α 2, a, a > s(α 1, a, a > s(α, a, a. But (ii of section 3 imlies s(α 1, a, a = s(1 - α 1, a, a. Hence, s(α 1, a, a < s(β, s(α 2, a, a, s(α 1, a, a = s(β, s(α 2, a, a, s(1 - α 1, a, a = s(β α 2 + (1 - β(1 - α 1, a, a. But β α 2 + (1 - β(1 - α 1 = α for some β, 0 < β < 1. Hence s(α 1, a, a < s(α, a, a. In the same way we can show that not s(α, a, a > a. We have thus roved that a ~ s(α, a, a. Assume now a ~ a'. Then axiom 3 yields a ~ s(1/2, a', a, and a roof by induction a ~ s(r n, a, b and a ~ s(s n, a, b, where r n and s n are as in the roof of lemma 4. Lemma 2 then enables us to conclude that a ~ s( α, a, a. Proof of lemma 6. Assume that a ~ a'. Then s(1/2, a, b ~ s(1/2, a', b (axiom 3. Assume thereafter that s(r n, a, b ~ s(r n, a', b and s(s n, a, b ~ s(s n, a', b. Then s(1/2(r n + s n, a, b = s(1/2, s(r n, a, b, s(s n, a, b ~ s(1/2, s(r n, a'

7 , b, s(s n, a', b. Hence s(r n+1, a, b ~ s(r n+1, a', b and s(s n+1, a, b ~ s(s n+1, a', b. Lemma 2 then yields s(α, a, b ~ s(α, a', b. Proof of lemma 7. Assume 1 > β > α and a > b. Then a > s(γ, a, s(α, a, b > s(α, a, b > b, if 0 < γ < 1. But β = γ + (1 - γ α for some such α. Hence (iv of section 3 imlies s(β, a, b > s(α, a, b. Proof of (iv (section 3. (1 γ = 1. Then s(α, s(β, a, b, s(γ, a, b = s(α, s(β, a, b, a = s(α, s(1 - β, b, a, a = s(α(1 - β, b, a = s(1 - α(1 - β, a, b = s(α β + (1 - α γ, a, b. (2 γ < 1. Then s( β, a, b = s(δ, a, s( γ, a, b with δ = (β - γ/(1 - γ. Hence s(α, s(β, a, b, s(γ, a, b = s(α, s(δ, a, s(γ, a, b, s(γ, a, b = s(α δ, a, s(γ, a, b = s(1 - α δ, s(1 - γ, b, a, a = s((1 - α δ(1 - γ, b, a = s(ε, a, b with ε = 1 - (1 - α δ(1 - γ = γ + α δ(1 - γ = γ + α β - α γ = α β + (1 - α γ. We thus get s(α, s(β, a, b, s(γ, a, b = s(α β + (1 - α γ, a, b. 3.2 THE JUSTFICATION OF THE UTILITY PRINCIPLE. The Reresentation Theorem stated above is to serve as a justification of the Utility Princile. To see how far this is ossible, we must first identify ( 1,, n o 1,, o n as an element of S. That is readily done as follows:

8 1, 0,, 0 (i ( o 1, o 2,, o n = o 1, (ii if 1 < 1, o 1 = b 1, and ( 2/1-1,, n/1-1 o 2,, o n = b 2, then ( 1,, n s( 1, b 1, b 2. o 1,, o n = From art (a of the Theorem we may then derive the following: If V: O R, a = ( 1,, n o 1,, o n, and Axioms 1, 2, and 3 hold, then (c there exists an extension U of V such that U: S R, U is comatible with and s, and U(a = 1 n i U(o i. Proof. Assume that O = { o 1,, o n } and that V(o 1 > > V(o n. Set o 1 > o n and o j ~ s(α j, o 1, o n with α j = (V(o j V(o n /(V(o 1 V(o n. Suose also that there exists an extension of ~ and > that satisfies Axioms 1, 2, and 3. The existence of such an extension may be said to follow from the assumtion that the axioms in question hold. It then follows from art (a of the Reresentation Theorem that there exists a function U': S R which is comatible with and s. Then U'(o 1 > U'(o n. Set U(a = (V(o 1 V(o n /(U'(o 1 U'(o n (U'(a U'(o n + V(o n, a S. Then U is the desired extension. Hence, the Utility Princile may be said to be consistent with the Herstein- Milnor Axioms, or to ut it differently, an agent who endorses these axioms is not contradicting himself if he also accets the Utility Princile. We then get a kind of weak justification of the Utility Princile, namely if it is the simlest rincile that is consistent with the Herstein-Milnor Axioms. But of course we demand more from the axiom system to be willing to seak of a justification of the Utility Princile; secifically, we require the following uniqueness result to

9 hold: If V: O R, a = ( 1,, n o 1,, o n, and Axioms 1, 2, and 3 hold, then (d it holds for all functions U: S R, which are extensions of V and comatible with the axioms that U(a = 1 n i U(o i. But does (d hold? Well that deends on how we interret the hrase 'comatible with the axioms'. We can interret it either as 'comatible with ' or as 'comatible with and s'. In the first case we get a version of (d that may be called (d' and in the second case we get a version that may be called (d''. Now (d' does not hold, since any strictly increasing function of a function U, whose existence is guaranteed by (c, is comatible with. On the other hand, it is easy to see that (d'' holds; we can even obtain the following somewhat stronger result: (e If V: O R, then it holds for all functions U: S R, which are n extensions of V and comatible with s that U(a = 1 i U(o i. Proof. A simle induction on the comlexity of s. Hence the contribution of the Herstein-Milnor Axioms to the justification of the Utility Princile is negligible.

10 3.3 MAXIMIZATION OF EXPECTED UTILITY. In this section I will show that an agent may very well accet the Herstein- Milnor Axioms without his endorsing the rincile of maximizing exected utility. Indeed, let V: O R with O = { o 1, o 2, o 3, o 4 } and set a 1 = ( o 1 o 2 and a 2 = ( o 3 o 4. Assume moreover that we have an agent ψ who accets the Herstein-Milnor Axioms, who has set 1 > v 1 > v 3 > v 4 > v 2 > 0 with v i = V(o i and who has the following references : o 1 > o 3 > o 4 > o 2. Since ψ accets the axioms, he sets a 1 equal to s ( 1, o 1, o 2 and a 2 equal to s( 2, o 3, o 4. Moreover he should be willing to set o 3 ~ s(α 1, o 1, o 2 and o 4 ~ s(α 2, o 1, o 2 for some α 1, α 2 such that 1 > α 1 > α 2 > 0. We may then derive s( 1, o 1, o 2 > s( 2, o 3, o 4 if and only if 1 > 2 α 1 + (1-2 α 2. We also get E V (a 1 > E V (a 2 if and only if 1 > 2 (v 3 - v 2 /(v 1 -v 2 + (1-2 (v 4 -v 2 /(v 1 -v 2. Now these orderings coincide for all values of 1, 2 such that 0 < 1, 2 < 1 if and only if α 1 = (v 3 -v 2 /(v 1 -v 2 and α 2 = (v 4 -v 2 /(v 1 -v 2. Hence the Herstein-Milnor Axioms force ψ to accet the rincile of maximizing exected utility if and only if they force him to set α 1 = (v 3 -v 2 /(v 1 -v 2 and α 2 = (v 4 -v 2 /(v 1 -v 2. But they clearly cannot do that, since any choice of α 1, α 2 such that 1 > α 1 > α 2 > 0 is comatible with the Herstein-Milnor Axioms in the following sense : Set O = { o 1, o 2, o 3, o 4 } and S equal to the least set T such that T O and s(α, a, b T if a,b T and 0 α 1. Assume that s satisfies (i - (iii of section 3 above and set o 3 ~ s(α 1, o 1, o 2 and o 4 ~ s(α 2, o 1, o 2 for some α 1, α 2 such that 1 > α 1 > α 2 > 0. Let ~' be the least equivalence-relation that is an extension of ~ and such that a ~' b if a = b and c ~' d imlies s(α, a, c ~' s(α, a, d. Then for each a T there exists a unique α between 0 and 1 such that a ~' s(α, o 1, o 2. Define a b if and only if α β for all α,β such that a ~'

11 s(α, o 1, o 2 and b ~' s(β, o 1, o 2. Then satisfies the Herstein-Milnor Axioms as was forshadowed. Hence the Herstein-Milnor Axioms ut very mild constraints on the ossible choices of α 1 and α 2. This fact then leads to the following consequences for an agent who accets these axioms: Set V : O R, O {o 1,, o n } = O', n 2, V(o 1 > > V(o n, and A a set of matrices ( 1,, n o 1,, o n such that i 0 and 1 n i = 1. Assume a,b A with a = ( 1,, n q o 1,, o n and b = ( 1,, q n given o 1,, o n. Then a dominates b V if and only if 1 j i 1 j qi, for all j, 1 j n, and a strictly dominates b given V if and only if a dominates b given V and 1 j i > 1 j q i, for some j, 1 j n. Moreover, a is said to be an admissible alternative of A given V if and only if a is not srictly dominated given V by any element of A. Suose now that V and A are given. Then an agent who accets the Herstein - Milnor axioms should choose an alternative that is admissible given V. But since it holds for all ε > 0 and all j, 1 j n that there exists a strictly increasing function U : R R such that U(V(o k - U(V(o j < ε for k < j and U(V(o k - U(V(o n < ε for k > j, the construction above shows that an agent, who accets the Herstein - Milnor axioms, can choose any alternative that is admissible given V. Since any moderately rational agent knows that he should ot for an admissible alternative and certainly is not in need of an axiom system to tell him that, we can conclude that the Herstein - Milnor axioms imose no interesting constraints uon an agent in a decision context when he can base his decision on numerically recise robability - and value statements. Moreover, this also holds if we relax our assumtions like in Levi (1974 or Malmnäs

12 (1990. Hence these axioms have no interesting art to lay in a decision context. It should also be clear by now that if one were to add new axioms so as to force ψ to maximize exected utility these new axioms would have to carry most of the burden themselves. The results of the resent section also casts doubt on the ossibility of obtaining any interesting axiomatic foundation of the Utility Princile or the Princile of Maximizing Exected Utility that does not emloy a frequency interretation of robabilities and ought therefore to be of some concern to Bayesians. 3.4 NUMERICAL UTILITY Von Neumann and Morgenstern (1953, 617 claim that their axioms make utility a number u to a ositive linear transformation. In this section I will show that this contention is at best highly misleading. To see this, assume that o 0 > o 1 > o n. Then set o i ~ s(x i, o 0, o n for 1 i n-1 and assume that 1 > x 1 > x n-1 > 0. We can then construct a system S' that formally satisfies the Herstein-Milnor Axioms by alying the same method as in section 3.3. As in Herstein and Milnor (1953 it can then be roved that there exists a unique U: S R(x 1, x n-1 comatible with and s and such that U(o 0 = 1 and U(o n = 0. We then get U(o i = x i. Hence the construction of S' has not rovided us with a more secific value than we had at the outset. Thus it cannot be said of a erson who formally accets the Herstein-Milnor Axioms that he has a numerical utility function unique u to a ositive linear transformation, even if he is willing to let s(α, a, b stand for α a + (1-α b. Moreover, a erson's accetance of the Herstein-Milnor Axioms does not imly that he has such a utility function. Rather, as we saw in section 3.2, we have to let s(α, a, b stand for α a + (1-α b to get that result; but, as we saw in that section, from this asumtion alone we can deduce that our agent has a numerical utility

13 function modulo a ositive linear transformation. Thus the contribution of the axiom system is negligible even in this resect. 4. MARSCHAK (1950 Marschak considers the set A of vectors (x 1,, x n of R n such that 0 x i n 1 and 1 xi = 1 and the relation on A. A is to be viewed as the set of robability distributions of a finite set of consequences. He then assumes the following axioms. Axiom 1. is a comlete semi-order on A. Axiom 2. If a > b > c, then b ~ r a + (1 - r c for some r, 0 < r < 1. Axiom 3. For all a and r, 0 < r < 1, there exists a vector b such that not a ~ r a + (1 - r b. Axiom 4. If a ~ a ' and 0 < r < 1, then r a + (1 - r b ~ r a ' + (1 - r b. Rresentation Theorem. (1 If the axioms hold, then there exists a linear form U on A such that U(a U(b if and only if a b. (2 If U and U' are linear forms on A comatible with, then there exist α > 0 and β such that U = α U' + β. 4.1 PROOF OF THE REPRESENTATION THEOREM. The Reresentation Theorem can be roved from (1 - (8 below, using the same method as in the Herstein-Milnor case. (1 If a ~ b, then r a + (1 - r b ~ a. (2 If a ~ r a + (1 - r b and 1 > s > r, then a ~ s a + (1 - s b.

14 (3 If a ~ r a + (1 - r b for some r, 0 < r < 1, then a ~ r a + (1 - r b for all r, 0 < r < 1. (4 If a ~ r a + (1 - r b, 0 < r <1 and b is an interior oint of A, then a ~ b. (5 If a and b are interior oints and a > b, then a > r a + (1 - r b > b for all r, 0 < r <1. (6 If a > b, then a > r a + (1 - r b > b for all r, 0 < r <1. (7 If a > b, then r a + (1 - r b > s a + (1 - s b if and only if r > s. (8 If a > c > b, then c ~ r a + (1 - r b for a unique r, 0 < r < 1. Proof of 2. Assume a ~ r a + (1 - r b. Then a ~ t a + (1 - t a ~ t a + (1 - tr a + (1 - t(1 - r b = v a + (1 - v b if v = r + t - r t. Now v = s if and only if t = (s - r/(1 - r which is consistent with 1 > s > r > 0. Hence a ~ s a + (1 - s b. Proof of 3. Assume a ~ r a + (1 - r b. Then a ~ r (r a + (1 - r b + (1 - r b = r 2 a + (1 - r 2 b. Hence a ~ r n a + (1 - r n b for all n > 1. But then (2 imlies a ~ s a + (1 - s b for all s, 0 < s < 1. Proof of 4. If b is an interior oint of A, then there exists a oint c of A and a number s, 0 < s < 1, such that b = s a + (1 - s c. Then a ~ r a + (1 - r (s a + (1 - s c = (r + (1 - r s a + (1 - r(1 - s c. But then (3 imlies that a ~ b. Proof of 5.

15 (5 follows easily from (4 and axiom 2.

16 Proof of 6. (6 is a consequence of (5 and the following Lemma. If H is a hyerlane and x ~ y for all interior oints of H, then x ~ y for all oints of H. Proof of the lemma. Assume that it does not hold. Then there exists a hyerlane of maximum dimension that contradicts the lemma. Let H be such a hyerlane and a a limit oint of H such that a > x for all interior oints x of H. Then axiom 3 imlies that there exists a oint c H such that a > c or c > a. Assume a > c first and let x be an interior oint of H. Then x < c or x ~ c or c < x. (i x < c. But then c ~ r a + (1 - r x for some r, 0 < r < 1. Hence x ~ c. (ii x ~ c. But then there exists a hyerlane which has higher dimension than H and which contradicts the lemma. (iii c < x. Then x ~ r a + (1 - r c for some r, 0 < r < 1. Set y = r a + (1 - r x. Then y ~ r a + (1 - r (r a + (1 - r c = (2 r - r 2 a + (1 - r 2 c. But we have assumed x ~ y. Hence there are distinct interior oints b 1, b 2 on the linesegment (a, c such that x ~ b 1 ~ b 2. But then x ~ d, for all interior d (a, c, and at least one such oint must lie outside H. Let d 0 be such a oint. But then we have case (ii. Thus we must have c > a.then a ~ r c + (1 - r x for some r, 0 < r < 1. Set y = r a + (1 - r x. Then y ~ r (r c + (1 - r x + (1 - r x = r 2 c + (1 - r 2 x. But x ~ y. Hence x ~ z for all interior z (c, x. But then a ~ x. 4.2 THE JUSTFICATION OF THE UTILITY PRINCIPLE. We can first note that the Reresentation Theorem above imlies that Marschak's axioms are comatible with the Utility Princile: this follows from (a and (b below. We can then note that the function H is comatible with

17 if H(a = f(u(a, U comatible with and f a strictly increasing function from R to R and that any linear 'extension' of a function V: O R satisfies the Utility Princile; see (c below. Hence the contribution of Marschak's axioms to a formal justification of the Utility Princile is negligible. (a For all O = { o 1,, o n } and V: O R such that V(o 1 > > V(o n there exist A and such that (A, satisfies Marschak's axioms and V is comatible with (A,. (b If V: O R, a = 1,, n, (A, satisfies Marschak's axioms and V is comatible with (A,, then there exists an 'extension' U of V such that U: A R, U is comatible with, and U(a = 1 n i U(i for all a A. Here, i = q 1,, q n A is such that q i = 1, an 'extension' U to A of V is such that U(i = V(o i, and a function V: O R with V(o 1 > > V(o n is said to be comatible with a system (A, that satisfies Marschak's axioms if and only if 1 > n and there exists an a = 1,, n A such that (i 1 + n = 1, (ii 1 = (V(o i V(o n /(V(o 1 V(o n, and (iii i ~ a. (c If V: O R, then it holds for all linear functions U: A R, which are 'extensions' of V that U(a = 1 n i U(i for all a A. Proof of (a. See the construction of section 4.3. Proof of (b. Let U' be a linear form on A comatible with. Then U'(i = (V(o i V(o n /(V(o 1 V(o n U'(1 + (V(o 1 V(o i /(V(o 1 V(o n U'(n. We

18 thus get V(o i = α U'(i + β, for some α > 0 and some β. Hence there exists an 'extension' U of V with the desired roerties. Proof of (c. Obvious. 4.3 MAXIMIZATION OF EXPECTED UTILITY. Set A = { (x 1,x 2,x 3,x 4 x 1 + x 2 + x 3 + x 4 = 1, 0 x i 1} and U(x 1,x 2,x 3,x 4 = 1 x x 2 + α 1 x 3 + α 2 x 4 with α 1, α 2 as in section 3.3. Define a b if and only if U(a U(b. Then (A, satisfies Marschak's axioms. Hence these axioms also ut very mild constraints on the ossible choices of α 1 and α FRIEDMAN AND SAVAGE ( 1952 These authors the same set A as did Marschak and assume the following axioms: Axiom 1. is a comlete semi-order on A. Axiom 2. If r a + (1 - r b c for all r, 0 r < 1, then a c. Axiom 3. If 0 < r < 1, then r a + (1 - r c r b + (1 - r c if and only if a b. Rresentation Theorem. (1 If the axioms hold, then there exists a linear form U on A such that U(a U(b if and only if a b. (2 If U and U' are linear forms on A comatible with, then there exist α > 0 and β such that U = α U' + β.

19 5.1 PROOF OF THE REPRESENTATION THEOREM. This theorem can roved from (1 - (8 below, emloying the same method as in the Herstein- Milnor case. (1 r a + (1 - r c ~ r b + (1 - r c if and only if a ~ b. (2 r a + (1 - r c > r b + (1 - r c if and only if a > b. (3 If a > b, then a > r a + (1 - r b > b. ( 0 < r < 1 in (1 - (3 (4 If a > b, then r a + (1 - r b > s a + (1 - s b if and only if r > s. (5 If s a + (1 - s b c for all s, 0 s < r, then r a + (1 - r b c. (6 If s a + (1 - s b c for all s, 0 s < 1, then a c. (7 If s a + (1 - s b c for all s, 0 s < r, then r a + (1 - r b c. (8 If a > c and a b c, then b ~ r a + (1 - r c for a unique r, 0 r 1. Proofs of (1 - (5. (1 and (2 are easy consequences of axiom 3, and (3 follows from (2. (4 follows from (3 and axiom 1, and (5 is an easy consequence of axiom 2. Proof of (6. I consider only the case a b, since the other one is trivial. Now, a c holds if it holds that (a s c + (1 - s b s a + (1 - s b for some s, 0 < s < 1. To rove (a it is, in view of axiom 2, sufficient to show that (b r (s c + (1 - s b + (1 - r a s a + (1 - s b for all r, 0 r < 1. Assume now that 0 t < 1. Then r (s c + (1 - s b + (1 - r a r (s (t a + (1 - t b + (1 - s b + (1 - r a = u a + (1 - u b. Here u = r s t r. But a b. Hence u a + (1 - u b s a + (1 - s b if and only if u s. But u s if and only if t (s + r - 1/r s. Hence we need only rove that 1 > (s + r - 1/r s or that s r - r - s + 1 > 0 for 0 r <1. Set f(x = s x - x - s + 1.

20 Then f(0 = 1 - s > 0, f(1 = 0 and f'(x s - 1 < 0. Hence f(x > 0 for 0 x < 1.

21 Proof of (7. Assume that s a + (1 - s b c for all s, 0 s < r, and that 0 t < 1. Then t (r a + (1 - r b + (1 - t b = t r a + (1 - t r b c since t r < r. Hence (6 imlies that r a + (1 - r b c. Proof of (8. Assume a > b > c. Set I 1 = { r r a + (1 - r c b } and I 2 = { r r a + (1 - r c b }. Then axiom 1 entails that I 1 I 2 = [0, 1], and from (4, (5 and (7 it follows that I 1,I 2 are closed segments. Hence I 1 I 2 Ø. From (4 it then follows that I 1 I 2 is a singleton set. 5.2 JUSTIFICATION OF THE UTILITY PRINCIPLE We can roceed exactly as in sections 4.2 and 4.3. Hence the system of Friedman and Savage does not rovide a formal justification of the Utility Princile, nor does it entail the maximization of exected utility. 6. VON NEUMANN AND MORGENSTERN (1944, 1947, 1953 These authors assume a non-emty set S of rosects and an oeration s such that s(α, a, b S if a,b S and 0 < α < 1. They also consider a relation > on S and assume the following axioms: (3 : A > is a strict linear order on S. (3 : B : a If a < b, then a < s(α, a, b. (3 : B : b If a > b, then a > s(α, a, b. (3 : B : c If a < b < c, then s(α, a, c < b for some α. (3 : B : d If a > b > c, then s(α, a, c > b for some α. (3 : C : a s(α, a, b = s(1 - α, b, a. (3 : C : b s(α, s(β, a, b, b = s(α β, a, b.

22 Reresentation Theorem (a If > and s satisfy the axioms, then there exists a real-valued function on S comatible with them. (b If U and U' are real-valued functions on S that are comatible with > and s, then U = α U' + β for some α > 0 and β. This theorem can be roved from (1 - (2 below, using the same method as in the Herstein-Milnor case. (1 If a > b, then s(α, a, b > s(β, a, b if and only if α > β. (2 If a > b > c, then b = s(α, a, b for a unique α. Proof of (1. Assume a > b and α > β. Then (3 : B : a and (3 : B : b imly a > s(α, a, b > b and s(α, a, b > s(β/α, s(α, a, b, b > b. But (3 : C : b imlies s(β/α, s(α, a, b, b = s(β, a, b. Hence s(α, a, b > s(β, a, b if α > β. The other direction then follows from (3 : A. Proof of (2. Assume a > b > c. Set I 1 = { α s(α, a, c > b } and I 2 = { α s(α, a, c < b }. Then I 1 and I 2 are non-emty in view of (3 : B : c and (3 : B : d, and (1 imlies that they are oen segments of (0, 1. Hence I 1 = (α 2, 1 and I 2 = (0, α 1 for some α 1, α 2, 0 < α 1 α 2 < 1, and (1 imlies α 1 = α 2. From (3 : A we can then infer that b = s(α 1, a, c. 6.1 JUSTIFICATION OF THE UTILITY PRINCIPLE The Reresentation Theorem of section 6 shows that the Utility Princile is consistent with the axioms of von Neumann-Morgenstern, - this follows from (a and (b below -, but it cannot, of course, be formally justified by them, since we can obtain the same results as in section 3.2.

23 (a If O = { o 1,, o n }, V: O R and V(o 1 > > V(o n, then there exists a system (X, s, > which satisfies the axioms of von Neumann- Morgenstern and is comatible with V. (b If V: O R, a = ( 1,, n o 1,, o n and (X, s, > is a system which satisfies the axioms of von Neumann-Morgenstern and is comatible with V via f, then there exists an 'extension' U of V such that U: X R, U is comatible with > and s, and U(af = 1 n i U(o i f. Here (X, s, > is comatible with V via f if and only if f : O X such that (i f(o 1 > f(o n, and (ii f(o j = s((v(o j - V(o n /(V(o 1 - V(o n, f(o 1, f(o n for 1 < j < n. Moreover, a = ( 1,, n o 1,, o n element af X as follows: can in this case be identified with an (i If j = 1, for some j, then af = f(o j, and (ii If 0 < j < 1, for all j, and b = ( 2/1-1,, n/1-1 o 2,, o n, then af = s( 1, f(o 1, bf. Proof of (a. Set R = { o 1, o n } and let S be the least set T such that (i T R, and (ii a,b T and 0 < α <1 imlies t(α, a, b T. Assume t(α, a, b t(1 - α, a, b and t(α, t(β, a, b, b t(α β, a, b and let be the least equivalence relation satisfying these conditions. Suose a S. Then a = o 1 or a = o n or a t(α, o 1, o n for a unique α, 0 < α <1. Hence a 1 b 1 and a 2 b 2 imlies t(α, a 1, a 2 t(α, b 1, b 2.

24 Set S' = S/ and a' = { b S b a }. Define s(α, a', b' = t(α, a, b'. Suose a S'. Then a = o 1 ' or a = o n ' or a = s(α, o 1 ', o n ' for a unique α, 0 < α <1. Set a > b if and only if (i a = o 1 ' b or (ii a o n '= b or (iii a = s(α, o 1 ', o n ' and b = s(β, o 1 ', o n ' with α > β. Then the system (S', s, > satisfies the axioms of von Neumann-Morgenstern and is comatible with V. Proof of (b. Obvious. 6.2 MAXIMIZATION OF EXPECTED UTILITY If we combine the constructions of sections 3.3 and 6.2, we find that the axioms of von Neumann-Morgenstern do not ut sricter constraints on the ossible values of α 1 and α 2 than the axioms of Herstein-Milnor. Here α 1 and α 2 are as in section LUCE AND RAIFFA (1957 These authors consider a finite number of objects A 1,, A n and the set Λ of lotteries over A 1,, A n. Λ is defined inductively as follows: (i A 1,, A n Λ. (ii If L 1,, L k Λ, i 0 (1 i k and 1 k i = 1, then ( 1,, k L 1,, L k Λ. They then consider the relation on Λ and assume the following axioms: Axiom 1. A 1 A 2 A n, A i A i,1 i n, A 1 > A n.

25 Axiom 2. If L i = qi 1,, qi n A s, 1 i s and t j = 1,, A n i 1 i q j, 1 j n, then ( 1,, s t L 1,, L s ~ ( 1,, t n A 1,, A n. u 1-u Axiom 3. A i ~ ( A 1 A n for some u, 1 i n. u 1-u Axiom 4. If A i ~ ( A 1 A n ( 1,, i,, n = A i *, then A 1,, A i,, A n ~ ( 1,, i,, n A 1,, A i *,, A n. Axiom 5. If L L' and L' L'', then L L''. 1- ' 1-' Axiom 6. ( A 1 A n ( A 1 A n if and only if '. They then claim that the following holds: Theorem. If (A1 - (A6 hold, then there exists a function U : R such that (i L 1 L 2 imlies U(L 1 U(L 2 with U(L 1 > U(L 2 if L 1 > L 2, and (ii U( 1,, n n A 1,, A n = 1 i U(A i.

26 Now that is not true as the following examle shows. Counterexamle. Consider only two rizes A 1, A 2 and assume A 1 > A 2, A 1 ~ 1/2 1/2 1/3 2/3 1- ( A 1 A 2 = A 1 *, A 2 ~ ( A 1 A 2 = A 2 * and ( ~ 1-1- ( A 1 * A 2 ~ ( A 1 A 2 * 1/2 1/2 function U such that U(A 1 > U(A 2, U(A 1 * = U ( 1/2 U(A 1 + 1/2 U(A 2. A 1 A 2. Then the axioms hold but there is no A 1 A 2 = The main defect of their system is axiom 2 which is said to imly that every lottery is equivalent to a lottery of the form ( 1,, n A 1,,A n. Now, to get such a result one needs rinciles that can be used reeatedly like the rinciles for the oeration s stated by von Neumann-Morgenstern and Herstein-Milnor. The simlest way of doing this seems to use the oeration s and the set S much as Herstein-Milnor did and then make a translation from Λ to S to get a reresentation theorem. This is at least how I will attemt to reair their system here. Assume that a 1,, a n S and that s(α, a, b S if a,b S and 0 α 1. Then assume the following axioms: Axiom 1'. a 1 a 2 a n with a 1 > a n. Axiom 2'. (i s(1, a, b = 1, (ii s(α, a, b = s(1- α, b, a, (iii s(α, s(β, a, b, b = s(α β, a, b. Axiom 3'. a i ~ s(α, a 1, a n for some α. Axiom 4'. If b ~ s(α, a 1, a n, then s(β, b, c ~ s(β, s(α, a 1, a n, c. Axiom 5'. is transitive in S. Axiom 6'. s(α, a 1, a n s(β, a 1, a n if and only if α β.

27 These axioms are meant to resemble the ones assumed by Luce-Raiffa as much as ossible. Note that my axiom 4' is slightly stronger than their axiom 4, and this increase in strength is robably necessary for a smooth ride to a reresentation theorem. From (A1' - (A6' we can easily derive the following consequences: (1 ~ is transitive in S. (2 s(α, s(β, a, b, s(γ, a, b = s(α β + (1- α γ, a, b. (3 For all b S there exists a unique α such that b ~ s(α, a 1, a n. (4 is a comlete semi-order on S. Proof of (3. Set b = s(γ, b 1, b 2. Assume b 1 ~ s(β 1, a 1, a n and b 2 ~ s(β 2, a 1, a n. Then s(γ, b 1, b 2 ~ s(γ, s(β 1, a 1, a n, b 2 = s(1- γ,b 2,s(β 1, a 1, a n ~ s(1- γ,s(β 2, a 1, a n,s(β 1, a 1, a n = s(γ, s(β 1, a 1, a n,s(β 2, a 1, a n = s(γ β 1 + (1- γ β 2, a 1, a n. One can then easily arrive at a Herstein-Milnor tye of reresentation theorem. Via such a result we can thereafter derive the following Reresentation Theorem If U is a real-valued function on S which is comatible with and s, V(L = U(T(L, L Λ, then V is comatible with on Λ and such that V ( 1,, n n L 1,, L n = 1 i V(L i.

28 Here: T(A i = a i, if 1 = 1, then T ( 1,, n L 1,, L n = T(L 1, if 1 < 1, T(L 1 = a and T ( 2 /(1-1,, n /(1-1 L 2,, L n = b, then T ( 1,, n L 1,, L n L L' if and only if T(L T(L'. = s( 1,a, b, and 7.1 MAXIMIZATION OF EXPCTED UTILITY Since we can reason as we did in sections , it is clear that the system of Luce and Raiffa does not give any formal justification of the Utility Princile or the Princile of Maximizing Exected Utility. 8. SAVAGE (1954, 1972 Savage considers a non-emty set X, called a world, another non-emty set K, called the set of consequences, and the set of all functions from X into K. This set is labelled F and its elements are called acts. They will be referred to by f, g, h etc. Similarly, x, y, z, will refer to elements of X, k, k', to elements of K and A, B, C, to subsets of X. He then considers the relations, >, ~ between acts and emloys these to define the following concets: Definition 1. f = g (B if and only if f(x = g(x for all x B. Definition 2. f g given B if and only if f' g' for all f', g' such that f = f' (B, g = g' (B and f' = g' (-B, and g' f' holds for all such airs or for none.

29 Definition 3. f = k if and only if f(x = k for all x X. Definition 4. k k' if and only if k k'. Definition 5. B is a null-set if and only if f g given B for all f, g F. Definition 6. f = (k, A, k' if and only if f(x = k for x A, and f(x = k' for x A. Definition 7. A B if and only if (k, A, k' (k, B, k' for all k, k' such that k > k'. Definition 8. f k given B ( k f given B if and only if f k given B ( k f given B. He can then state his axioms: Axiom 1. is a comlete semiorder on F. Axiom 2. f g given B or g f given B for all f, g F. Axiom 3. If f(x = k, f'(x = k' for all x B, and B is not a null-set, then f f' given B if and only if k k'. Axiom 4. A B or B A for all subsets A, B of X. Axiom 5. k < k' for some k, k' B. Axiom 6. If f < g, then there exists a artition {B i } n 1 of X such that f < g' and f' < g for all f', g' such that f'= f (-B i, g' = g (-B i and f' = g' (B i for some i, 1 i n. Axiom 7. If f g(x given B (g(x f given B for all x B, then f g given B (g f given B.

30 From these Savage then roves that there exists a unique finitely additive robability measure P on P(X such that P(A P(B if and only if A B. This measure is then emloyed to associate sets of acts with gambles as follows: Let k 1,, k n be consequences and assume that 1 n i = 1 with i 0. Then G = { ( i,k i 1 i n } is a gamble. With G we can associate the set F G that consists of all acts f such that there exists a artition {B i } 1 n P(B i = i and f(x = k i for x B i, 1 i n. of X with Savage then defines a utility as a real valued function U on K such that 1 n i U(k i 1 m qi U(k' i if and only if f g holds for all f F G, g F G'. Here G = { ( i,k i 1 i n } and G' = { (q i,k' i 1 i m }. Thereafter, he roves the following theorem: Reresentation Theorem. If (X,K, satisfies Savage's axioms, then there exists a utility on K. This theorem is to serve as a justification of the Utility Princile (see Savage To see how far Savage has succeded in this endeavour, we can first note that his axioms are consistent with it. This much follows from (a and (b below. But Savage's axioms can, of course, not imly or justify the Utility Princile, since they only contain the relations, >, as rimitive concets. Indeed: Let U be a utility in Savage's sense, and assume that f F G with G = n { ( i,k i 1 i n }. Set V(f = 1 i U(k i, and let V' be a strictly increasing function of V. Then V and V' are both comatible with the relation when it is restricted to acts associated with gambles.

31 (a For all O = { o 1,, o n } and V: O R such that V(o 1 > > V(o n axioms and there exist X and such that (X,O, satisfies Savage's V is comatible with (X,O,. (b If V: O R, a = ( 1,, n o 1,, o n, (X,O, satisfies Savage's axioms and V is comatible with (X,O,, then there exists an 'extension' U of V such that U: F R, U is comatible with, and U(f = 1 n i U(o i for all f T(a. Here, T(a = { f f F and P(f -1 (o i = i, 1 i n}, an 'extension' U to F of V is such that U(o i = V(o i, and a function V: O R with V(o 1 > > V(o n is said to be comatible with a system (X,O, that satisfies Savage's axioms if and only if there exists an f F such that (i f(x = o 1 or f(x = o 2, for all x X, (ii P({ x f(x = o 1 } = (V(o i V(o n /(V(o 1 V(o n, and (iii o i ~ f. Here P is the unique robability measure on P(X that is commensurate with (X,O,. Proof of (a. See the construction of section 8.1. Proof of (b. Let U' be a utility that exists according to the Reresentation Theorem above. Then U'(o i = (V(o i V(o n /(V(o 1 V(o n U'(o 1 + (V(o 1 V(o i /(V(o 1 V(o n U'(o n. We thus get V(o i = α U'(o i + β, for some α > 0 and some β. Hence there exists an 'extension' U of V with the desired roerties.

32 8.1 MAXIMIZATION OF EXPECTED UTILITY. Set X = [0,1] and K = { o 1, o 2, o 3, o 4 }. Let P be a finitely additive measure on P(X that is an extension of the ordinary Lebesgue measure on [0,1]. Set U(f = 1 P(B P(B 2 + α 1 P(B 3 + α 2 P(B 4 with B i = {x f(x = o i } and α 1, α 2 as in section 3.3. Finally, define f g if and only if U(f U(g. Then (X, K, satisfies Savage's axioms. Hence these axioms also ut very mild constraints on the ossible choices of α 1 and α DAVIDSON, SUPPES AND SIEGEL (1957 In the monograh Decision Making, the authors mentioned above consider various systems for measuring subjective robability and utility. Some of these are quite comlicated and inelegant, and luckily for us, not relevant for our investigation. However, on age 31 ff they consider a simle axiom system that can be viewed as an attemt to justify the Utility Princile, albeit in a somewhat restricted context. They there consider a finite non-emty set K of consequences, a subject s and an event E which has robability 1/2 for s. They also consider the set K K of rosects. Here (x,y is the rosect that x will hold if E occurs and y will hold if -E occurs. > is then considered as a relation between consequences and ~ as a relation between rosects. The following axioms are then assumed: Axiom 1. > is a strict linear order on K. Axiom 2. Axiom 3. ~ is an equivalence relation on K K. (x,y ~ (y,x. Axiom 4. If (x,y ~ (u,v and x > u, then v > y. Axiom 5. If x > 1 y and u > 1 v, then (x,v ~ (u,y.

33 Here x > 1 y if and only if x > y and it holds for all z K that y = z or y > z if x > z. We can then define x > k+1 y if and only if x > k z and z > 1 y for some z K and derive the following easy consequences of the axioms: (a If x > k y and u > k v, then (x,v ~ (y,u. (b If (x,y ~ (u,v and x > k u, then v > k y. Via these we can thereafter derive the following Reresentation Theorem: (i If (A1 - (A5 hold, then there exists a function f: K R such that x > y if and only if f(x > f(y and (x,y ~ (u,v if and only if f(x + f(y = f(u + f(v. (ii If f and f' are functions on K that satisfy the conditions of (i, then there exist numbers a > 0 and b such that f = a f' + b. Proof of (a. An easy induction on k. (i k = 1. This is axiom 5. (ii k = n + 1. Assume x > n+1 y and u > n+1 v. Let z and t be such that x > 1 z and t > 1 v. Then (x,v ~ (z,t. Moreover, z > n y and u > n t, which imlies (z,t ~ (u,y. Hence (x,v ~ (u,y. Proof of (b. (i k = 1. Assume (x,y ~ (u,v and x > 1 u. Then axiom 4 yields v > y. Assume v > r+1 y. Then v > r z and z > 1 y for some z. Hence (u,z ~ (x,y ~ (v,u. Now v > z, which imlies u > u and a contradiction. (ii k > 1. Analogous to (i.

34 Proof of the theorem. (i Assume K = { x 1,,x n } and x n > x n-1,, x 2 > x 1. Set f(x i = i. Then x > y if and only if f(x > f(y. Assume then that f(x + f(y = f(u + f(v. (a f(x = f(y. Then f(y = f(v. Hence x = u and y = v. But then (x,y ~ (u,v. (b f(x - f(u = k > 0. Then x > k u and v > k y. Hence (x,y ~ (u,v. Assume now on the other hand that (x,y ~ (u,v. (a x = u. Then y = v. Hence f(x + f(y = f(u + f(v. (b x > k u. Then v > k y. Hence f(x - f(u = k = f(v - f(y. But this imlies f(x + f(y = f(u + f(v. (ii Assume that f and f' satisfy the conditions of art (i. Now (x j+1,x j-1 ~ (x j,x j if 1 < i < n. Hence f(x j+1 - f(x j = f(x j - f(x j-1 and f'(x j+1 - f'(x j = f'(x j - f'(x j-1. We thus get, for 1 j n, f(x j = (j - 1 (f(x 2 - f(x 1 + f(x 1, and f'(x j = (j - 1 (f'(x 2 - f'(x 1 + f'(x 1. Hence f(x j = a f'(x j + b if a = (f(x 2 - f(x 1 /(f'(x 2 - f'(x 1 and b = f(x 1 + a f'(x 1. Remark. The axioms 1-5 do not suffice to justify the Utility Princile even in this restricted context, since there are functions f comatible with the axioms and such that f(x,y (f(x + f(y/2. To see this, notice first that for a real-valued function f on K (K K to be comatible with the axioms above it is sufficient that f satisfies the following conditions: (1 x > y if and only if f(x > f(y, (2 (x,y ~ (u,v if and only if f(x,y = f(u,v, (3 f(x,y = f(x if x = y, and (4 f(x > f(x,y > f(y if x > y.

35 We can then notice that ~ gives rise to 2 n - 1 equivalence classes contains either an element (z,z or an element (x,y with x > 1 y. If the latter holds, we can choose f(x,y as we lease in the interval (f(y, f(x. 10. PRATT, RAIFFA AND SCHLAIFER (1964 The system develoed by these authors can be characterized as the result of a marriage between Luce and Raiffa (1957 and Savage (1954, and consequently has many features in common with the off-sring of Ouranos and Gaia. But the intuitive ideas are simle enough, and their system also merits attention for the reason that they exlicitly aim at a justification of the Utility Princile. In the resentation of it below I will retain the fundamental concets and axioms but I will deviate from Pratt et al (1964 to some extent in the formal develoment and there emloy as much as ossible the concets used before in this study. The reason for this is simly that I find some of their rinciles uncommonly awkward FUNDAMENTAL CONCEPTS Pratt et al consider a finite world E and a finite set of consequences C. C contains two elements c* and c * such that c* > c * and c* c c * for all c C. As in Savage (1954 acts are functions from E to C, with the set of acts labeled A. Let a be an act with codomain { c 1,, c n }. Then a can be identified with { (a -1 {c i }, c i } 1 n. Conversely, if {E i} 1 n is a artition of E, then { (E i, c i } n 1 can be viewed as an act. To facilitate the formulation of the axioms below, I will emloy subsets of such acts and will call them artial acts with a secified domain. For technical reasons I will define these as follows: (i If E 1,E 2 are disjoint, then { (E 1, c 1, (E 2, c 2 } is a artial act with domain E 1 E 2.

36 (ii If F,E 1 are disjoint and a is a artial act with domain F, then { (E 1, c 1, (F, a } is a artial act with domain E 1 F. Partial acts with domain E are then called acts tout court. The set of objects so defined is labeled A', and the objects of A' should be viewed as descritions of artial act and by by no means as artial acts with acts as consequences. The following observation can erhas clarify this oint. Consider, for instance, the set { (E 1, c 1, (E 2, c 2, (E 3, c 3 }. This set reresents the act that has c 1 as a consequence if one of the elements of E 1 occurs, c 2 as a consequence if one of the elements of E 2 occurs and c 3 as a consequence if one of the elements of E 3 occurs. But the same act can also be described in the following somewhat cumbersome way: if one of the elements of E 1 occurs, then c 1 will occur, and if one of the elements of E 2 E 3 occurs, then c 2 will occur if this element lies in E 2 and c 3 will occur if it lies in E 3. This descrition can then be said to corresond to the object { (E 1, c 1, (E 2 E 3, {(E 2, c 2, (E 3, c 3 }}. Pratt et al then consider the set of functions from E [0, 1] 2 into C calling its members lotteries. I will not emloy this set but will instead use the set S of rosects over C A'. As before this is the least set that contains C A' and is such that s(α, a, b S if 0 α 1 and a,b S. They thereafter take as a relation on C A L, whereas I will take it as a relation on S, and assume the following axioms: Axiom 1. There exists a function π: C [0, 1] such that c ~ { (E [0, π(c] [0, 1], c*, (E -[0, π(c] [0, 1], c * }. Axiom 2. There exists a function P: P(E [0, 1] such that {(E 1, c*, (-E 1, c * } ~ {(E [0, 1] [0, P(E 1 ], c*, (E [0, 1] [0, P(E 1 ], c * }. Axiom 3. is transitive in C A L. Axiom 1 says in effect that every consequence has a definite value, and that this value is equal to the value of a certain lottery. - True, they do not assume there

37 that π is unique, but that is only because they add other axioms to ensure that π and P are unique.- The first art of this assumtion is of course a natural starting-oint when the Utility Princile is to be justified, whereas the second art must be merely accidental to such an enterrise. The idea behind axiom 2 is that if we only consider acts with two consequences, then the value of an act is determined by the robabilities of its consequences. This is of course almost tantamount to an assumtion that the Utility Princile holds for acts with only two consequences. Hence we need only slightly generalize axiom 2 and add a few obvious axioms concerning rosects to derive what amounts to the Utility Princile. Such a revised system may for instance be as follows: Axiom 1. There exists a function π: C [0, 1] such that c ~ s(π(c, c*, c *. Axiom 2. There exists a function P: P(E [0, 1] such that {(E 1, a 1, (E 2, a 2 } ~ s(p(e 1 /P(E 1 E 2, a 1, a 2 if 0 < P(E 1 E 2, P(E = 1 and P(E 1 P(E 2 if E 1 E 2. Axiom 3. is transitive in S. Axiom 4. c* > c *. Axiom 5. s(1, a, b = a. Axiom 6. Axiom 7. s(α, a, b = s(1- α, b, a. s(α, s(β, a, b, b = s(α β, a, b. Axiom 8. If a > b, then s(α, a, b > s(β, a, b if and only if α > β. Axiom 9. If a ~ b, then s(α, a, c ~ s(α, b, c. Consequences of the axioms: (1 ~ is an equivalence relation in S. (2 π is unique. (3 P is unique. (4 P(E 1 + P(-E 1 = 1. (5 P(E 1 E 2 = P(E 1 + P(E 2 if E 1 E 2 = Ø.

38 (6 { (E i, c i } n 1 ~ s(( 1 n π(ci P(E i /P( 1 n Ei, c*, c * if P( 1 n Ei > 0. From (6 we can then conclude that the value of the act { (E i, c i } n 1 equals 1 n π(ci P(E i. Proof of (4. {(-E 1, c*, (E 1, c * } ~ s(p(-e 1, c*, c * and {(E 1, c *, (-E 1, c* } ~ s(p(e 1, c *, c* = s(1-p(e 1, c*, c *. Hence P(-E 1 = 1-P(E 1. Proof of (5. (a P(E 1 = 1. Then P(E 1 E 2 = 1 and 0 = P(-E 1 P(E 2. Hence P(E 1 E 2 = P(E 1 + P(E 2. (b P(E 1 < 1. Then (i {(E 1 E 2, c*, (-(E 1 E 2, c * } ~ s(p(e 1 E 2, c*, c *. On the other hand, {(E 1 E 2, c*, (-(E 1 E 2, c * }= {(E 1, c*, (E 2, c*, (-(E 1 E 2, c * }= {(E 1, c*, (-E 1, { (E 2, c*, (-(E 1 E 2, c * } }~ s(p(e 1, c*, { (E 2, c*, (-(E 1 E 2, c * } ~ s(p(e 1, c*, s(p(e 2 /P(-E 1, c*, c * since { (E 2, c*, (-(E 1 E 2, c * }~ s(p(e 2 /P(-E 1, c*, c *. But s(p(e 1, c*, s(p(e 2 /P(-E 1, c*, c * = s(p(e 1 + (1-P(E 1 P(E 2 /P(-E 1, c*, c * = s(p(e 1 + P(E 2, c*, c *. Hence (ii {(E 1 E 2, c*, (-(E 1 E 2, c * }~ s(p(e 1 + P(E 2, c*, c *. (5 then follows from (i and (ii.

39 Proof of (6. By induction on n. (i n = 1. Then { (E 1, c 1 } = { (E 1, c 1,(Ø, c 1 }~ s(1, c 1, c 1 = c 1 ~ s( π(c, c*, c *. (ii n = k + 1. { (E i, c i } n 1 = {(E 1, c 1, ( n 2 Ei, {(E i, c i } n 2 }~ s(p(e n 1/P( 1 Ei, c 1, {(E i, c i } n 2. (a P(E 1 = P( 1 n Ei. Then s(p(e 1 /P( 1 n Ei, c 1, {(E i, c i } n 2 = c 1 ~ s( π(c 1, c*, c * = s(( 1 n π(ci P(E i /P( 1 n Ei, c*, c *. (b P(E 1 < P( 1 n Ei. Then s(p(e 1 /P( 1 n Ei, c 1, {(E i, c i } n 2 ~ s(p(e 1 /P( 1 n Ei, s( π(c 1, c*, c *, s(( 2 n π(ci P(E i /P( 2 n Ei, c*, c * = s(p(e 1 /P( 1 n Ei π(c 1 + (1 - P(E 1 /P( 1 n Ei ( 2 n π(ci P(E i /P( 2 n Ei, c*, c * = s(( 1 n π(ci P(E i /P( 1 n Ei, c*, c *, since 1 - P(E 1 /P( 1 n Ei = P( 2 n Ei /P( 1 n Ei.

40 11. KRANTZ ET AL. (1971 Krantz et al. construct a conditional version of Savage (1954. In this way, they are able to rove a Reresentation Theorem that is almost as natural as the one roved by von Neumann-Morgenstern, but at the exense of a remarkably comlicated axiom system FUNDAMENTAL CONCEPTS AND REPRESENTATION THEOREM They consider a non-emty set X, an algebra E of subsets of X, and a subset N of E. The elements of N are called null-sets. They then consider a nonemty set K of consequences, a set D of functions from elements of E -N into D, and a relation on D. Finally, they assume the following axioms: 1. (i If A B =, then f A g B D. (ii If A B, then the restriction of f A to B is in D. 2. is a weak ordering of D. 3. If A B = and f A ~ g B, then f A g B ~ f A. 4. If A B =, then f (1 A f(2 A 5. If A B =, f (i A ~ g(i B if and only if f(1 A g B f (2 A g B., i = 1, 2, 3, 4, f(1 and h (1 A g(1 B ~ h(2 A g(2 B, then and only if h (1 A g(3 B h(2 A g(4 B. A k(1 B ~ f(2 A k(2 B, f(3 A k(1 B f(4 A k(2 B if 6. If A B =, N is a sequence of consecutive integers, not g (0 B ~ g(1 B, and f (i A g(1 B ~ finite or { f (i A f(i+1 A g(0 B i Ν } is unbounded. for all i, i + 1 Ν, then either Ν is