Document information
- University
- Politecnico di Milano
- Degree programme
- Computer Engineering
- Subject
- Game Theory
- Academic year
- 2022-2023
- Classification
- Exam · Full exam
- Content
- Exam paper only
- Original format
- Text
- Searchable text
Full exam for Game Theory in the Computer Engineering degree programme at Politecnico di Milano. The document covers: GAME THEORY - July 10, 2023 Last name: First name: ID #: SOLVE THE EXERCISES AND ANSWER THE QUESTIONS ON THESE SHEETS Exercise 1 5 points Two players I and II have to split one asset, so that P1 gets x ≥ 0 and P2 gets y ≥ 0, with x + y ∈ [0, 1]. Their utilities are u(x) = x2 and
Full exam for Game Theory in the Computer Engineering degree programme at Politecnico di Milano. The document covers: GAME THEORY - July 10, 2023 Last name: First name: ID #: SOLVE THE EXERCISES AND ANSWER THE QUESTIONS ON THESE SHEETS Exercise 1 5 points Two players I and II have to split one asset, so that P1 gets x ≥ 0 and P2 gets y ≥ 0, with x + y ∈ [0, 1]. Their utilities are u(x) = x2 and
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GAME THEORY - July 10, 2023 Last name: First name: ID #: SOLVE THE EXERCISES AND ANSWER THE QUESTIONS ON THESE SHEETS Exercise 1 5 points Two players I and II have to split one asset, so that P1 gets x ≥ 0 and P2 gets y ≥ 0, with x + y ∈ [0, 1]. Their utilities are u(x) = x2 and v(y) = y + a respectively, and the disagreement point is d = (0 , 0). Find a ∈ R such that the Nash solution is x = y = 1 2. Answer of exercise 1 The bargaining set is characterised by the relation √u + v − a ≤ 1. Finding the maximum of uv on this set is equivalent to finding the maximum of v(1 − v + a)2, which is achieved at v = a+1 3 . Since we want y = v − a = 1 2, we have a = − 1 4. 1 Exercise 2 5 points Given the zero-sum game described by the following matrix: A = 1 4 6 2 2 5 5 6 4 find a pair of optimal strategies for the players and the value of the game. Answer of exercise 2 First of all, notice that 4 = v1 ̸= v2 = 5, therefore there are no optimal pure strategies. Then we observe that the second row is strictly dominated by a convex combination of the first one and the third one. Once the second row is removed, the first column strictly dominates the second one. Thus the game can be reduced to A = 1 6 5 4 . By the indifference principle we obtain the optimal strategies of P1 (1/6, 0, 5/6) and P2 (1/3, 0, 2/3). The value is 13/3. 2 Theory Questions Answer one and only one question (7 points). Only the question 2 may lead to the top grade 30 e lode. 1. State Arrow’s theorem explaining assumptions and consequences. 2. Let A be a matrix defining a symmetric game, and let u(x, y) = ( x, Ay) the utility for the “invading type” x entering the population y. Prove that a strategy y is an ESS if and only if for all x ̸= y either u(x, y) < u(y, y) or u(x, y) = u(y, y) and u(x, x) <…
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