SOLUTION MANUAL Game Theory Basics 1st Edition By Bernhard von Stengel. Chapters 1 - 12

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,SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12




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,TABLE OF CONTENTS MB MB MB




1 - Nim and Combinatorial Games
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2 - Congestion Games
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3 - Games in Strategic Form
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4 - Game Trees with Perfect Information
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5 - Expected Utility
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6 - Mixed Equilibrium
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7 - Brouwer’s Fixed-Point Theorem
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8 - Zero-Sum Games
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9 - Geometry of Equilibria in Bimatrix Games
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10 - Game Trees with Imperfect Information
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11 - Bargaining
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12 - Correlated Equilibrium
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, Game Theory Basics MB MB




Solutions to Exercises M B M B




© M B Bernhard von Stengel 2022 MB MB MB




Solution to Exercise 1.1 MB MB MB




(a) Let ≤ be defined by (1.7). To show that ≤ is transitive, consider x, y, z with x ≤ y and y ≤
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z. If x = y then x ≤ z, and if y = z then also x ≤ z. So the only case left is x < y and y
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< z, which implies x < z because < is transitive, and hence x ≤ z.
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Clearly, ≤ is reflexive because x = x and therefore x ≤ x.
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To show that ≤ is antisymmetric, consider x and y with x ≤y and y ≤ x. If we had
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x ≠ y then x < y and y < x, and by transitivity x < x which contradicts (1.38). Hence x =
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y, as required. This shows that ≤ is a partial order.
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Finally, we show (1.6), so we have to show that x < y implies x y and
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≤ x ≠ y and vice v
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ersa. Let x < y, which implies x y by (1.7). If≤ we had x = y then x < x, contradicting (1.3
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8), so we also have x ≠ y. Conversely, x y and x ≠ y imply by
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≤ (1.7)x < y or x = y where
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M MB MB MB MB MB MB MB MB



the second case is excluded, hence x < y, as required.
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(b) Consider a partial order and ≤ assume (1.6) as a definition of <. To show that < is transiti
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ve, suppose x < y, that is, x y and x ≠≤y, and y < z, that is, y z and y ≠ z. ≤
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ransitive, ≤
x z. If we had ≤
x = z then x y and y
MB x≤and hence≤x = y by antisymmet
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ry of , which contradicts x ≠ y, so we have x z and x ≠ z, that is,x < z by (1.6),
≤ ≤
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as required. MB




Also, < is irreflexive, because x < x would by definition mean x
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≤ x ≠ x, but the la
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tter is not true.
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Finally, we show (1.7), so we have to show that x ≤ y implies x < y or x = y and vice v
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ersa, given that < is defined by (1.6). Let x ≤ y. Then if x = y, we are done, otherwise x
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≠ y and then by definition x < y. Hence, x ≤ y implies x < y or x = y. Conversely, suppos
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e x < y or x = y. If x < y then x ≤ y by (1.6), and if x = y then x ≤ y because ≤ i
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s reflexive. This completes the proof.
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Solution to Exercise 1.2 MB MB MB




(a) In analysing the games of three Nim heaps where one heap has size one, we first lookat
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some examples, and then use mathematical induction to prove what we conjecture to be the lo
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sing positions. A losing position is one where every move is to a winning position, becau
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se then the opponent will win. The point of this exercise is to formulate a precise stateme
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nt to be proved, and then to prove it.
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First, if there are only two heaps recall that they are losing if and only if the heaps are
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of equal size. If they are of unequal size, then the winning move is to reduce thelarger h
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eap so that both heaps have equal size.
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