# 8_mixture - 8 The Mixture Space Theorem Let be a convex...

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8 The Mixture Space Theorem Let Π be a convex subset of R n , i.e. if π, ρ Π, then απ + (1 - α ) ρ Π for all α (0 , 1). Definition 8.1. A binary relation % on Π is independent if, for all π, ρ, σ Π and α (0 , 1), π % ρ ⇐⇒ απ + (1 - α ) σ % αρ + (1 - α ) σ. Example 8.2. Suppose Π = R 2 and x % y if and only if x 2 1 + x 2 2 y 2 1 + y 2 2 , i.e if the norm of x is greater than the norm of y . Then (4 , 0) (0 , 4), but 1 2 (4 , 0) + 1 2 (2 , 0) = (3 , 0) (1 , 2) = 1 2 (0 , 4) + 1 2 (2 , 0). So % is not independent. Exercise 8.3. Prove that if % is independent, then its indifference classes are convex. Proposition 8.4. A binary relation % on Π is independent if and only if, for all π, ρ, σ Π and α (0 , 1) , π ρ ⇐⇒ απ + (1 - α ) σ αρ + (1 - α ) σ and π ρ ⇐⇒ απ + (1 - α ) σ αρ + (1 - α ) σ. Exercise 8.5. Prove Proposition 8.4. Definition 8.6. A binary relation % on Π is Archimedean if, for all π, ρ, σ Π, π ρ σ = α (0 , 1) such that απ + (1 - α ) σ ρ and β (0 , 1) such that ρ βπ + (1 - β ) σ Exercise 8.7. Recall that % is continuous if the upper and lower contour sets, { ρ Π : ρ % π } and { ρ Π : π % ρ } , are closed for all π Π. Prove that if % is continuous, then % is Archimedean. Exercise 8.8. A preference relation % is mixture continuous if the sets { α [0 , 1] : απ +(1 - α ) ρ % σ } and { α [0 , 1] : απ + (1 - α ) ρ - σ are closed (in [0 , 1]) for all π, ρ, σ Π. Prove that if % is mixture continuous, then % is Archimidean. Example 8.9. This example demonstrates that the converses of the previous exercises fails so the Archimedean condition is a weaker assumption than continuity or mixture continuity. Let Π = R and consider the preference relation % on R defined by the utility function U ( π ) = 1 if π > 0 0 if π = 0 - 1 if π < 0 . Verify that this preference relation is Archimedean, but is neither continuous nor mixture contin- uous. 44
Definition 8.10. A function f : Π R is affine if, for all π, ρ Π and α [0 , 1] f ( απ + (1 - α ) ρ ) = αf ( π ) + (1 - α ) f ( ρ ) . Proposition 8.11. A function f : Π R is affine if and only if, for all α 1 , α 2 , . . . , α m 0 such that m k =1 α k = 1 and π (1) , π (2) , . . . , π ( m ) Π , 19 f m X k =1 α k π ( k ) ! = m X k =1 α k f ( π ( k ) ) . Proof (optional). Necessity ( ) is immediate. The proof of sufficiency ( ) is by induction on m . The required conclusion is obviously true when m = 1. So, suppose the claim is true for m = M . Now consider α 1 , . . . , α M +1 [0 , 1] such that M +1 k =1 α k = 1 and π (1) , . . . , π ( M +1) Π. Let β k = α k 1 - α M +1 and verify that β 1 , . . . , β M 0 with M k =1 β k = 1. So, by the induction hypothesis, f M X k =1 β k π ( k ) ! = M X k =1 β k f ( π ( k ) ) . (A) Then: f M +1 X k =1 α k π ( k ) !