# Suppose that 1 0 ie thatδ x 1 δ x 2 first we show

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Suppose that 1 0 (i.e. that δ x 1 δ x 2 ). First we show that π ρ if and only if π > ρ . To begin, consider the case where ρ = 0. Independence and 1 0 imply that α = 1 α +0(1 α ) 0 α +0(1 α ) for all α (0 , 1). Thus π 0 for all π (0 , 1]. Obviously, if π 0, then π negationslash = 0, so we also have that π 0 implies that π > 0. Now suppose that π > ρ > 0. Then π > 0, and (from above) we have π 0. Let β = 1 ρ π (0 , 1). Note that ρ = β 0 + (1 β ) π . Independence and the fact that π 0 imply that π = βπ + (1 β ) π β 0 + (1 β ) π = ρ . Hence π ρ whenever π > ρ . Now, suppose that π ρ . If π < ρ , then we have π ρ by the preceding argument. If π = ρ , then clearly π ρ . In either case, π ρ implies π negationslash≻ ρ . Thus π ρ implies π > ρ , as claimed. Now let π ρ σ . Then, by the previous part, we have π > ρ > σ . Thus there exists a (0 , 1) such that ρ = + (1 a ) σ . 7 Then for any α (0 , a ) ρ > απ + (1 α ) σ , and thus ρ απ + (1 α ) ρ . Similarly, for any β ( a, 1) ρ < βπ + (1 β ) σ , and thus βπ + (1 β ) σ ρ . The case where 0 1 is symmetric to the previous one. Exercise 6.14 (9.5). Suppose | X | = 3. Find a preference relation on Δ X that is independent, but not Archimedean. Find a preference relation on Δ X that is Archimedean, but not independent. Proof. Let X = { x 1 , x 2 , x 3 } . 7 You can claim this by arguing that the function f ( a ) = + (1 a ) σ is continuous and strictly increasing, and f (0) = σ < ρ , f (1) = π > ρ . Thus, by the Intermediate Value Theorem, there exists some a (0 , 1) such that f ( a ) = ρ .
Section 6: vNM Representation Theorem and Risk 6-16 1. Let followsorequal be an analogue of the lexicographic preference, defined as π followsorequal ρ ⇐⇒ π ( x 1 ) > ρ ( x 1 ) or π ( x 1 ) = ρ ( x 1 ) and π ( x 2 ) > ρ ( x 2 ) or π ( x 1 ) = ρ ( x 1 ) and π ( x 2 ) = ρ ( x 2 ) and π ( x 3 ) ρ ( x 3 ) This is not Archimedean. Let π = (1 , 0 , 0), ρ = (0 , 1 , 0) and σ = (0 , 0 , 1), which gives us π ρ σ . Note that for any α (0 , 1), απ +(1 α ) σ = ( α, 0 , 1 α ) ρ = (0 , 1 , 0). This relation is independent. To see this, let π, ρ, σ Δ X and α (0 , 1). Then π followsorequal ρ ⇐⇒ π ( x 1 ) > ρ ( x 1 ) or π ( x 1 ) = ρ ( x 1 ) and π ( x 2 ) > ρ ( x 2 ) or π ( x 1 ) = ρ ( x 1 ) and π ( x 2 ) = ρ ( x 2 ) and π ( x 3 ) ρ ( x 3 ) ⇐⇒ απ ( x 1 ) > αρ ( x 1 ) or απ ( x 1 ) = αρ ( x 1 ) and απ ( x 2 ) > αρ ( x 2 ) or απ ( x 1 ) = αρ ( x 1 ) and απ ( x 2 ) = αρ ( x 2 ) and απ ( x 3 ) αρ ( x 3 ) ⇐⇒ απ ( x 1 ) + (1 α ) σ ( x 1 ) > αρ ( x 1 ) + (1 α ) σ ( x 1 ) or απ ( x 1 ) + (1 α ) σ ( x 1 ) = αρ ( x 1 ) + (1 α ) σ ( x 1 ) and απ ( x 2 ) + (1 α ) σ ( x 2 ) > αρ ( x 2 ) + (1 α ) σ ( x 2 ) or απ ( x 1 ) + (1 α ) σ ( x 1 ) = αρ ( x 1 ) + (1 α ) σ ( x 1 ) and απ ( x 2 ) + (1 α ) σ ( x 2 ) = αρ ( x 2 ) + (1 α ) σ ( x 2 ) and απ ( x 3 ) + (1 α ) σ ( x 3 ) αρ ( x 3 ) + (1 α ) σ ( x 3 ) ⇐⇒ απ + (1 α ) σ followsorequal αρ + (1 α ) σ.
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