5. Choice under Uncertainty Daisuke Oyama Microeconomics I May 23, 2018
Formulations von Neumann-Morgenstern (1944/1947) X: Set of prizes Π: Set of probability distributions on X : Preference relation on Π Representation: satisfies certain axioms if and only if there exists a function U : X R such that u(π) = x X U(x)π(x) represents. 1 / 22
Savage (1954) S: Set of states of nature X: Set of consequences/prizes An act is a function from S to X. A: Set of all acts (all functions from S to X) : Preference relation on A Representation: satisfies certain axioms if and only if there exist a probability distribution p on S and a function U : X R such that u(a) = s S U(a(s))p(s) represents. 2 / 22
Anscombe-Aumann (1963) S: Set of states of nature X: Set of consequences/prizes Π: Set of probability distributions on X H: Set of all functions from S to Π : Preference relation on H Representation: satisfies certain axioms if and only if there exist a probability distribution p on S and a function U : X R such that u(h) = [ ] U(x)h(s)(x) p(s) s S x X represents. 3 / 22
von Neumann-Morgenstern X = {x 1,..., x n }: Finite set of prizes (for simplicity) Π: Set of probability distributions on X, called lotteries δ x Π: Lottery that gives x X with probability 1 Compound lotteries: For π, ρ Π, the compound lottery aπ + (1 a)ρ is identified with the corresponding lottery. Example: X = {x, y, z, w} π = 0.3x + 0.1y + 0.6z + 0w ρ = 0.6x + 0y + 0z + 0.4w (1/3)π + (2/3)ρ is identified with the lottery 0.5x + (1/30)y + 0.2z + (4/15)w : Preference relation on Π 4 / 22
Axioms Order: is complete and transitive. Substitution (Independence): For all π, ρ, φ, φ Π and all a [0, 1], aπ+(1 a)φ aρ+(1 a)φ aπ+(1 a)φ aρ+(1 a)φ. Continuity: For all π, ρ, φ Π, π ρ = a (0, 1) : a (a, 1] : aπ + (1 a)φ ρ and π aρ + (1 a)φ. 5 / 22
Under Order, Substitution is equivalent to the following: Substitution : For all π, ρ, φ, φ Π and all a [0, 1], aπ+(1 a)φ aρ+(1 a)φ aπ+(1 a)φ aρ+(1 a)φ. Substitution implies the following: Indifference Substitution: For all π, ρ, φ, φ Π and all a [0, 1], aπ+(1 a)φ aρ+(1 a)φ aπ+(1 a)φ aρ+(1 a)φ. 6 / 22
Under Order, Continuity is equivalent to the following: Continuity : For all π, ρ, φ Π, {a [0, 1] aπ+(1 a)φ ρ} and {a [0, 1] ρ aπ+(1 a)φ} are open (relative to [0, 1]). Continuity : For all π, ρ, φ Π, {a [0, 1] aπ+(1 a)φ ρ} and {a [0, 1] ρ aπ+(1 a)φ} are closed (relative to [0, 1]). 7 / 22
Expected Utility Representation Proposition 5.3 For a preference relation on Π, the following statements are equivalent: 1. satisfies Order, Substitution, and Continuity. 2. There exists a function U : X R such that the utility function u: Π R defined by u(π) = x X U(x)π(x) represents. Moreover, U is unique up to positive affine transformations, i.e., x X V (x)π(x) represents if and only if there exist A > 0 and B such that V (x) = AU(x) + B for all x X. U is sometimes called a von Neumann-Morgenstern function. 8 / 22
Just to recall: Order: is complete and transitive. Substitution: For all π, ρ, φ, φ Π and all a [0, 1], aπ+(1 a)φ aρ+(1 a)φ aπ+(1 a)φ aρ+(1 a)φ. Continuity: For all π, ρ, φ Π, π ρ = a (0, 1) : a (a, 1] : aπ + (1 a)φ ρ and π aρ + (1 a)φ. 9 / 22
Proof of 1 2 In the following, we assume that satisfies Order. Step 1 (Lemmas 5.6 5.7) Strict Betweenness: For all π, ρ Π and all a (0, 1), π ρ = π aπ + (1 a)ρ ρ. Substitution + Continuity = Strict Betweenness Step 2 (Lemma 5.6 5.7) Indifference Betweenness: For all π, ρ Π and all a [0, 1], π ρ = π aπ + (1 a)ρ ρ. Substitution + Continuity = Indifference Betweenness 10 / 22
Step 3 (Lemma 5.7) Mixture Monotonicity: For all π, ρ Π, if π ρ, then aπ + (1 a)ρ bπ + (1 b)ρ a b. Strict Betweenness = Mixture Monotonicity 11 / 22
Step 4 (Lemma 5.8) Indifference Independence: For all π, ρ, φ Π and all a [0, 1], π ρ = aπ + (1 a)φ aρ + (1 a)φ. Indifference Betweenness + Substitution = Indifference Independence Step 5 Strict Independence: For all π, ρ, φ Π and all a (0, 1), π ρ = aπ + (1 a)φ aρ + (1 a)φ. Strict Betweenness + Substitution = Strict Independence 12 / 22
Step 6 (Lemma 5.9) Mixture Intermediate Value: For all π, ρ, φ Π, π ρ φ = a [0, 1] : aπ + (1 a)φ ρ. Continuity = Mixture Intermediate Value Step 7 (Lemma 5.9) Unique Mixture Intermediate Value: For all π, ρ, φ Π, π φ, π ρ φ =! a [0, 1] : aπ + (1 a)φ ρ. Mixture Intermediate Value + Mixture Monotonicity = Unique Mixture Intermediate Value 13 / 22
Step 8 X: finite + Indifference Betweenness + Strict Betweenness = π, π Π : π π π for all π Π Notes: In the textbook, X is possibly an infinite set, and Π is the set of simple probability distributions (probability distributions with finite support). In the proof there, the existence of such π and π is assumed, whereas this assumption is not necessary; see Problem 5.3. 14 / 22
Step 9 Unique Mixture Intermediate Value + Mixture Monotonicity = Utility representation Proof Assume that π π. By Unique Mixture Intermediate Value, for each π Π, there is a unique a π [0, 1] such that π a π π + (1 a π )π. Define the function u: Π R by u(π) = a π for each π Π. Then u represents : for any π, ρ Π, π ρ u(π)π + (1 u(π))π u(ρ)π + (1 u(ρ))π u(π) u(ρ) by Mixture Monotonicity. 15 / 22
Step 10 Utility representation + Indifference Independence = Linear utility representation Proof Take any π, ρ Π and a [0, 1]. By the construction of u, we have π u(π)π + (1 u(π))π and ρ u(ρ)π + (1 u(ρ))π. By Indifference Independence, aπ + (1 a)ρ a[u(π)π + (1 u(π))π] + (1 a)ρ a[u(π)π + (1 u(π))π] + (1 a)[u(ρ)π + (1 u(ρ))π] = [au(π) + (1 a)u(ρ)]π + [1 (u(π) + (1 a)u(ρ))]π. Therefore, by the construction of u, we have u(aπ + (1 a)ρ) = au(π) + (1 a)u(ρ). 16 / 22
Step 11 Linear utility representation = Expected utility representation Proof Define U : X R by U(x) = u(δ x ) for x X. For each π Π, which is written as x X π(x)δ x, by the linearity of u we have u(π) = x X π(x)u(δ x ) = x X π(x)u(x). 17 / 22
Step 12 Linear utility representation = Unique linear utility representation up to positive affine transformation i.e., if π ρ u(π) u(ρ) and π ρ v(π) v(ρ), and u and v are linear, then v( ) = Au( ) + B for some A > 0 and B. 18 / 22
Proof Assume that π π. Take any π Π, and define a π = u(π) u(π) [0, 1]. u(π) u(π) Since u is linear, u(a π π + (1 a π )π) = a π u(π) + (1 a π )u(π) = u(π). Since u is a representation of, this implies that a π π + (1 a π )π π. Since v is a representation of, this implies that v(a π π + (1 a π )π) = v(π), where v(a π π + (1 a π )π) = a π v(π) + (1 a π )v(π) since v is linear. Hence v(π) = Au(π) + B where A = v(π) v(π) u(π) u(π) > 0, which are independent of π. B = Au(π) + v(π), 19 / 22
Step 13 Unique linear utility representation up to positive affine transformation = Unique expected utility representation up to positive affine transformation Proof i.e., if π ρ x X U(x)π(x) x X U(x)ρ(x) and π ρ x X V (x)π(x) x X V (x)ρ(x), then V ( ) = AU( ) + B for some A > 0 and B. Let u(π) = x X U(x)π(x) and v(π) = x X V (x)π(x). By v( ) = Au( ) + B, A > 0, for all x X we have V (x) = v(δ x ) = Au(δ x ) + B = AU(x) + B. 20 / 22
Implications of Substitution + Continuity Assume that X = {x 1, x 2, x 3 } and δ x1 δ x2 δ x3. Indifference curves are straight lines: if π ρ, then aπ + (1 a)ρ ρ for all a [0, 1]. (Indifference Betweenness) Indifference curves are parallel straight lines: if π ρ, then aπ + (1 a)δ x1 aρ + (1 a)δ x1 and aπ + (1 a)δ x3 aρ + (1 a)δ x3 for all a [0, 1]. (Indifference Independence) 21 / 22
Critiques Allais Paradox Ellsberg Paradox Chapter 5.5 22 / 22