研究生课程 博弈论 英文课件4

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Figure: Maximin strategies of player 1
u1 u1 (p, T ) = 1 − 2 p u1 (p, H) = 2 p − 1
p∗
p
m1 (p) = {u1 (p, T ), u1 (p, H)}
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The thick (kinked) line represents m1 (p), i.e. the worst outcome for player 1 for each value of p ∈ [0, 1]. In order to maximize m1 (p), we have to compute the intersection of u1 (p, H) and u1 (p, T ):
u1 (H, q) and u1 (T, q)
u1 (H, q) = q u1 (H, H) + (1 − q) u1 (H, T ) ⇔ u1 (H, q) = 2 q − 1 u1 (T, q) = q u1 (T, H) + (1 − q) u1 (T, T ) ⇔ u1 (T, q) = −2 q + 1
u1 (p, T ) and u1 (p, H)
u1 (p, H) = p u1 (H, H) + (1 − p) u1 (T, H) ⇔ ⇔ u1 (p, T ) = 2 p + (1 − p) (−1) u1 (p, T ) = 3 p − 1 u1 (p, T ) = p u1 (H, T ) + (1 − p) u1 (T, T ) ⇔ ⇔ u1 (p, T ) = p (−1) + (1 − p) u1 (p, T ) = −2 p + 1
u1 (T, q) = u1 (H, q) ⇔ 1 − 2 q = −1 + 2 q 1 ⇔ q∗ = . 2 Thus, the Minimax value of player 1 is
M1 = u1 (H, q ∗ ) ⇔ M1 = (1/2) u1 (T, H) + (1/2) u1 (T, T ) ⇔ M1 = (1/2)(−1) + (1/2) (1) ⇔ M1 = 0.
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Figure: Minimax strategies of player 2
u1
M1 (q) = {u1 (T, q), u1 (H, q)} u1 (H, q) = 3 q − 1 u1 (T, q) = 1 − 2 q
q∗
q
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The thick (kinked) line represents M1 (q), i.e. the best outcome for player 1 for each value of q ∈ [0, 1]. In order to minimize M1 (q), we have to compute the intersection of u1 (H, q) and u1 (T, q):
u1 (H, q) and u1 (T, q)
u1 (H, q) = q u1 (H, H) + (1 − q) u1 (H, T ) ⇔ u1 (H, q) = 3 q − 1 u1 (T, q) = q u1 (T, H) + (1 − q) u1 (T, T ) ⇔ u1 (T, q) = −2 q + 1
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Solution
Thus, the equilibrium of the game is thus (p∗ , q ∗ ) = and the value of the game is 1 m1 (p∗ , q ∗ ) = M1 (p∗ , q ∗ ) = u1 (p∗ , q ∗ ) = . 5 According to definition 2.7 the last game is ‘unfair’. If participation would be voluntary, agent 2 would only participate in this game, if (in 1 every round) he would get a sidepayment σ, with σ ≥ 5 . 2 3 , 5 5
Microeconomic Analysis
Lecture in the Graduate Program in International Economics and Finance OTTO-VON-GUERICKE-UNIVERSITY MAGDEBURG
———–
Part 4
Dr. Magnus Hoffmann
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Similarly, we compute the minimax value M1 = min[M1 (q)]
q
and the corresponding probability q ∗ = arg min [M1 (q)].
q
We do so by graphing the functions u1 (H, q), u1 (T, q) and, consequently M1 (q) = max{u1 (H, q), u1 (T, q)}.
According to definition 2.7 the last game is a fair game. Is this also true for our next example?
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Example 2.20 (Variation of ‘matching pennies’)
Find the minimax equilibrium of the following game! Is this game a fair one? 1\2 H T H 2 −1 T −1 1
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H 1, −1 −1, 1
T −1, 1 1, −1
First, we compute the maxmin value m1 = max [m1 (p)]
p
and the corresponding probability p∗ = arg max [m1 (p)].
p
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We do so by graphing the functions u1 (p, H), u1 (p, T ) and, consequently, m1 (p) = min{u1 (p, H)(p, H) ⇔ 1 − 2 p = −1 + 2 p 1 ⇔ p∗ = . 2 Thus, the Maximin value of player 1 is
m1 = u1 (p∗ , H) ⇔ m1 = (1/2) u1 (H, H) + (1/2) u1 (T, H) ⇔ m1 = (1/2)(1) + (1/2) (−1) ⇔ m1 = 0.
u1 (p, T ) = u1 (p, H) ⇔ 1 − 2 p = −1 + 3 p 2 ⇔ p∗ = . 5 Thus, the Maximin value of player 1 is m1 = u1 (p∗ , H) ⇔ m1 = (2/5) u1 (H, H) + (3/5) u1 (T, H) ⇔ m1 = (2/5)(2) + (3/5) (−1) 1 . ⇔ m1 = 5
magnus.hoffmann@ovgu.de
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Chapter 2.2 - Two-Player Zero-Sum Games (cont’d)
Example 2.1 (‘Matching Pennies’ again)
Find the minimax equilibrium for the “matching pennies” game! 1\2 H T
u1 (T, q) = u1 (H, q) ⇔ 1 − 2 q = −1 + 3 q 2 ⇔ q∗ = . 5 Thus, the Minimax value of player 1 is M1 = u1 (H, q ∗ ) ⇔ M1 = (2/5) u1 (T, H) + (2/5) u1 (T, T ) ⇔ M1 = (2/5)(−1) + (3/5) (1) 1 . ⇔ M1 = 5
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Again, first we compute the maxmin value m1 = max [m1 (p)]
p
and the corresponding probability p∗ = arg max m1 (p).
p
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We do so by graphing the functions u1 (p, H), u1 (p, T ) and m1 (p) = min{u1 (p, H), u1 (p, T )}.
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Figure: Minimax strategies of player 2
u1 u1 (T, q) = 1 − 2 q M1 (q) = {u1 (T, q), u1 (H, q)} u1 (H, q) = 2 q − 1
q∗
q
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The thick (kinked) line represents M1 (q), i.e. the best outcome for player 1 for each value of q ∈ [0, 1]. In order to minimize M1 (q), we have to compute the intersection of u1 (H, q) and u1 (T, q):
u1 (p, T ) and u1 (p, H)
u1 (p, H) = p u1 (H, H) + (1 − p) u1 (T, H) ⇔ ⇔ u1 (p, T ) = p + (1 − p) (−1) u1 (p, T ) = 2 p − 1 u1 (p, T ) = p u1 (H, T ) + (1 − p) u1 (T, T ) ⇔ ⇔ u1 (p, T ) = p (−1) + (1 − p) u1 (p, T ) = −2 p + 1