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Therefore i 1 i bi i showing that b is convex valued

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Unformatted text preview: ⊂ Σ is convex if and only if Bi (σ−i ) is convex for all i . Let σi� , σi�� ∈ Bi (σ−i ). Then, for all λ ∈ [0, 1] ∈ Bi (σ−i ), we have ui (σi� , σ−i ) ≥ ui (τ i , σ−i ) for all τ i ∈ Σi , ui (σi�� , σ−i ) ≥ ui (τ i , σ−i ) for all τ i ∈ Σi . The preceding relations imply that for all λ ∈ [0, 1], we have λui (σi� , σ−i ) + (1 − λ)ui (σi�� , σ−i ) ≥ ui (τ i , σ−i ) for all τ i ∈ Σi . By the linearity of ui , ui (λσi� + (1 − λ)σi�� , σ−i ) ≥ ui (τ i , σ−i ) for all τ i ∈ Σi . Therefore, λσi� + (1 − λ)σi�� ∈ Bi (σ−i ), showing that B (σ ) is convex-valued. 17 Game Theory: Lecture 5 Existence Results Proof (continued) 4. B (σ) has a closed graph. Suppose to obtain a contradiction, that B (σ ) does not have a closed graph. ˆ ˆ ˆ Then, there exists a sequence (σn , σn ) → (σ, σ ) with σn ∈ B (σn ), but ˆ/ ˆ/ σ ∈ B (σ), i.e., there exists some i such that σi ∈ Bi (σ−i ). This implies that there exists some σi� ∈ Σi and some � > 0 such that ˆ ui (σi� , σ−i ) > ui (σi , σ−i ) + 3�. By the continuity of ui and the fact that σn i → σ−i , we have for − sufficiently large n, ui (σi� , σn i ) ≥ ui (σi� , σ−i ) − �. − 18 Game Theory: Lecture 5 Existence Results Proof (continued) [step 4 continued] Combining the preceding two relations, we obtain ˆ ˆi − ui (σi� , σn i ) > ui (σi , σ−i ) + 2� ≥ ui (σn , σn i ) + �, − where the second relation follows from the continuity of ui . This ˆi contradicts the assumption that σn ∈ Bi (σn i ), and completes the − proof. The existence of the fixed point then follows from Kakutani’s theorem. If σ∗ ∈ B (σ∗ ), then by definition σ∗ is a mixed strategy equilibrium. 19 Game Theory: Lecture 5 Existence Results Existence of Equilibria for Infinite Games A similar theorem to Nash’s existence theorem applies for pure...
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