Proof let π ρ by completeness of followsorequal for

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Proof. Let π ρ . By completeness of followsorequal , for all x i , x j X , δ x i followsorequal δ x j or δ x j followsorequal δ x i . We will prove this claim by contradiction. Let | X | = n , and suppose that there do not exist x, y X for which δ x δ y . Then, for all i ∈ { 1 , . . . , n } , we have δ x i δ x j . We can use independence (mixing δ x 1 and δ x 2 with weights π 1 π 1 + π 2 and π 2 π 1 + π 2 ) to show
Section 6: vNM Representation Theorem and Risk 6-14 that δ x 1 δ x 2 parenleftbigg π 1 π 1 + π 2 δ x 1 + π 2 π 1 + π 2 δ x 2 parenrightbigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright γ 2 δ x 2 γ 2 δ x 3 . Then we can mix γ 2 and δ x 3 with weights (1 π 3 π 1 + π 2 + π 3 ) and ( π 3 π 1 + π 2 + π 3 ) respectively and we get parenleftbigg π 1 + π 2 π 1 + π 2 + π 3 parenrightbigg γ 2 + parenleftbigg π 3 π 1 + π 2 + π 3 parenrightbigg δ x 3 δ x 3 parenleftbigg π 1 + π 2 π 1 + π 2 + π 3 parenrightbiggparenleftbigg π 1 π 1 + π 2 δ x 1 + π 2 π 1 + π 2 δ x 2 parenrightbigg + parenleftbigg π 3 π 1 + π 2 + π 3 parenrightbigg δ x 3 δ x 3 parenleftbigg π 1 π 1 + π 2 + π 3 parenrightbigg δ x 1 + parenleftbigg π 2 π 1 + π 2 + π 3 parenrightbigg δ x 2 parenleftbigg π 3 π 1 + π 2 + π 3 parenrightbigg δ x 3 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright γ 3 δ x 3 γ 3 δ x 4 . We repeat this procedure, and, at each step we use independence and the indifference of every δ x j with each other, to mix γ i with δ i +1 , with weights i j =1 π j i +1 j =1 π j on γ i and π i +1 i +1 j =1 π j on δ x i +1 , to draw indifference between γ i +1 and δ x i +1 . By iterating n 1 times we will get: parenleftBigg n 1 j =1 π j n j =1 π j parenrightBigg γ n 1 + π n n j =1 π j δ x n δ x n n 1 summationdisplay j =1 π j γ n 1 + π n δ x n bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = π δ x n since n summationdisplay j =1 π j = 1 . Thus π δ x n . A similar construction (changing π i for ρ i on every step) will work for ρ , and we can obtain ρ δ x n as well. Transitivity will imply ρ π , a contradiction. Therefore there must exist x, y X such that δ x δ y . Exercise 6.13 (9.4). Suppose | X | = 2. Prove that if followsorequal is an independent preference relation on Δ X , then followsorequal is Archimedean.
Section 6: vNM Representation Theorem and Risk 6-15 Proof. Note that any element in Δ X has the form ( p, 1 p ) for some p [0 , 1]. That is, the probability of x 2 is pinned down by the probability of x 1 . Therefore, we can use p [0 , 1] to denote ( p, 1 p ) Δ X . Let followsorequal be independent. We have to consider two cases: If 1 0 (i.e. if δ x 1 δ x 2 ), then we can show that independence implies π ρ for all π, ρ [0 , 1]. Let α [0 , 1] be given. Then independence implies that α (1) + (1 α )(0) α (0) + (1 α )(0), which means that α 0. This is true for all α [0 , 1], so (by transitivity) indifference holds for any two elements in Δ X . Then the Archimedean property holds vacuously since π ρ σ is never satisfied.

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