博弈论(复旦大学 王永钦)复旦大学研究生一年级博弈论课程讲义,英文
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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
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)}
4 / 33
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
博弈论讲义完整版
第一章 导论
注意两点： 1、是两个或两个以上参与者之间的对策论 当鲁滨逊遇到了“星期五”
石匠的决策与拳击手的决策的区别
第一章 导论
2、理性人假设 理性人是指一个很好定义的偏好，在面临定的约束条 件下最大化自己的偏好。 博弈论说起来有些绕嘴，但理解起来很好理解， 那就是每个对弈者在决定采取哪种行动时，不但要根 据自身的利益的利益和目的行事，而且要考虑到他的 决策行为对其他人可能的影响，通过选择最佳行动计 划，来寻求收益或效用的最大化。
不完全信息静态博弈-贝叶斯纳什均衡 海萨尼（1967-1968）
你 接受 求爱博弈： 品德优良者求爱 求爱者 求爱
100，100
不接受
-50，0 0，0
不求爱 0，0
100x+（-100）（1-x）=0 当x大于1/2时，接受求爱 求爱博弈： 品德恶劣者求爱 求爱者 接受 求爱 不求爱 0，0 你 不接受
问题：什么叫“完全而不完美信息博弈”？
第二章 完全信息静态博弈
一 博弈的基本概念及战略表述 二 占优战略（上策）均衡
三 重复剔除的占优均衡（严格下策反复消去法）
四 划线法
五 箭头法
六 纳什均衡
完全信息静态博弈
完全信息：每个参与人对所有其他参与人的特 征（包括战略空间、支付函数等）完全了解
同样的情形发生在： 公共产品的供给 美苏军备竞赛 经济改革 中小学生减负 ……
第一章 导论-囚徒困境
囚徒困境的性质：
个人理性和集体理性的矛盾； 个人的“最优策略”使整个“系统”处于不利 的状态。
思考：为什么会造成囚徒困境 是否由于“通讯”问题造成了囚徒困境？ “要害”是否在于“利己主义”即“个人理 性”？
研究生课程 博弈论 英文课件3
3. Since player 2 is intelligent, he will predict the reaction of player 1 and figure out the expected payoff m2 (a2 ) from playing a2 : m2 (a2 ) = = ∀ a 1 ∈ A1 .
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a1 ∈A1 a1 ∈A1
min u1 (a1 , a2 ) min {u2 (a3 , a1 ), u2 (a2 , a2 ), u2 (a1 , a3 )}, 1 2 1 2 1 2
1\2 a1 1 a2 1 a3 1 m2 (a2 )
a1 2 0, 0 4, −4 9, −9 −9
3. Since player 1 is intelligent, he will predict the reaction of player 2 and figure out the expected payoff m1 (a1 ) from playing a1 : m1 (a1 ) = = ∀ a 1 ∈ A1 .
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First argument of player 1: Player 2 moves second
1. Since player 2 is intelligent, he will predict any action a1 ∈ A1 that player 1 may choose. 2. Since player 2 is rational, he will choose the action a2 that maximizes his payoff (or, equivalently, minimizes the payoff of player 2). P1 : P1 : P1 : a1 1 a2 1 3 a1 ⇒ ⇒ ⇒ P2 : P2 : P2 : a1 2 a2 2 2 a2 ⇒ ⇒ ⇒ u1min u1min u1min = 0 = 2 = 0
博弈论英文课件 (3)
Ø If Player 2 chooses Head, r-(1-r)=2r-1 Ø If Player 2 chooses Tail, -r+(1-r)=1-2r
Solving matching pennies
Player 2
Head
Tail
Expected payoffs
Head Player 1
Static (or SimultaneousMove) Games of Complete Information
Matching pennies
Player 2
Head
Tail
Player 1
Head Tail
-1 , 1 1 , -1 1 , -1 -1 , 1
n Head is Player 1’s best response to Player 2’s strategy Tail n Tail is Player 2’s best response to Player 1’s strategy Tail
( 1(H)=0.5, 1(T)=0.5 ) is a Mixed strategy. That is, player 1 plays H and T with probabilities 0.5 and 0.5, respectively.
( 1(H)=0.3, 1(T)=0.7 ) is another Mixed strategy. That is, player 1 plays H and T with probabilities 0.3 and 0.7, respectively.
Ø Player 2’s expected payoff of playing s22:
EU2(s22, (r, 1-r))= r×u2(s11, s22)+(1-r)×u2(s12, s22)
Solving matching pennies
Player 2
Head
Tail
Expected payoffs
Head Player 1
Static (or SimultaneousMove) Games of Complete Information
Matching pennies
Player 2
Head
Tail
Player 1
Head Tail
-1 , 1 1 , -1 1 , -1 -1 , 1
n Head is Player 1’s best response to Player 2’s strategy Tail n Tail is Player 2’s best response to Player 1’s strategy Tail
( 1(H)=0.5, 1(T)=0.5 ) is a Mixed strategy. That is, player 1 plays H and T with probabilities 0.5 and 0.5, respectively.
( 1(H)=0.3, 1(T)=0.7 ) is another Mixed strategy. That is, player 1 plays H and T with probabilities 0.3 and 0.7, respectively.
Ø Player 2’s expected payoff of playing s22:
EU2(s22, (r, 1-r))= r×u2(s11, s22)+(1-r)×u2(s12, s22)
lecture13-slides
() June 10, 2015 10 / 11
2
()
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Example
1
Consider a monopoly market. There are two types of consumers. A fraction λ is type 1 and the rest is type 2. The utility function of a type 1 consumer is θ i v ( qi ) pi where qi is the quality of a good and pi is the price. The cost of producing quality q is cq , c > 0.
3
Since θ 1 is indi¤erent between the two contracts and q2 > q1 , the single-crossing condition implies that θ 2 must prefer (q2 , p2 ) to ( q1 , p1 ) .
3
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Basic Properties
1
If the participation constraint for θ 1 is satis…ed, so is θ 2 ’ s. This follows from θ 2 v ( q2 ) p2 θ 2 v ( q1 ) p1 0. θ 1 v ( q1 ) p1 .
2
Hence, qi such that θ i dv dq = c for i = 1, 2 and pi = θ i v (qi ). Illustrate with a diagram.
2
()
June 10, 2015
1 / 11
Example
1
Consider a monopoly market. There are two types of consumers. A fraction λ is type 1 and the rest is type 2. The utility function of a type 1 consumer is θ i v ( qi ) pi where qi is the quality of a good and pi is the price. The cost of producing quality q is cq , c > 0.
3
Since θ 1 is indi¤erent between the two contracts and q2 > q1 , the single-crossing condition implies that θ 2 must prefer (q2 , p2 ) to ( q1 , p1 ) .
3
()
June 10, 2015
7 / 11
Basic Properties
1
If the participation constraint for θ 1 is satis…ed, so is θ 2 ’ s. This follows from θ 2 v ( q2 ) p2 θ 2 v ( q1 ) p1 0. θ 1 v ( q1 ) p1 .
2
Hence, qi such that θ i dv dq = c for i = 1, 2 and pi = θ i v (qi ). Illustrate with a diagram.
博弈论 SPE 复旦大学 王永钦PPT课件
Player H
1
T
HH -1 , 1
1 , -1
Player 2
HT
TH
-1 , 1 1 , -1
-1 , 1 1 , -1
TT 1 , -1 -1 rium
• The set of Nash equilibria in a dynamic game of complete information is the set of Nash equilibria of its normal-form.
8
Game tree
• If a node x is a
successor of another
node y then y is called a predecessor of x.
• In a game tree, any node other than the root has a unique predecessor.
Definition: extensive-form representation
• The extensive-form representation of a game specifies:
➢ the players in the game ➢ when each player has the move ➢ what each player can do at each of his or her
• Player 2’s strategies
➢H if player 1 plays H, H if player 1 plays T ➢H if player 1 plays H, T if player 1 plays T ➢T if player 1 plays H, H if player 1 plays T ➢T if player 1 plays H, T if player 1 plays T
博弈论讲义完整版
囚徒困境
第一章 导论-囚徒困境
案例1-囚徒困境-纳什均衡
囚徒A
坦白
囚徒 B
抵赖
坦白
-8，-8 -10，0
0，-10 -1，-1
-8大于-10 0大于-1
抵赖
（坦白，坦白）是纳什均衡
第一章 导论-囚徒困境
设定： （1）每个局中人都知道博弈规则和博弈结 果的支付矩阵； （2）每个局中人都是理性的（个人理性和 个人最优决策）； （3）不能“串通”
第一章 导论-基本概念
博弈论的基本概念包括： 参与人：博弈论中选择行动以最大化自己效用的决策主体； 行动：参与人的决策变量 战略：参与人选择行动的规则 信息：参与人在博弈中的知识，特别是有关其他参与人的特征和 行动的知识 支付函数：参与人从博弈中获得的效用水平 结果：博弈分析真正感兴趣的要素的集合 均衡：所有参与人的最优战略的组合 参与人、行动、结果称为博弈规则；博弈分析的目的是使用博弈 规则决定均衡。
第一章 导论
注意两点： 1、是两个或两个以上参与者之间的对策论 当鲁滨逊遇到了“星期五”
石匠的决策与拳击手的决策的区别
第一章 导论
2、理性人假设 理性人是指一个很好定义的偏好，在面临定的约束条 件下最大化自己的偏好。 博弈论说起来有些绕嘴，但理解起来很好理解， 那就是每个对弈者在决定采取哪种行动时，不但要根 据自身的利益的利益和目的行事，而且要考虑到他的 决策行为对其他人可能的影响，通过选择最佳行动计 划，来寻求收益或效用的最大化。
不开发
1000，0 0，0
开发商A
博弈的战略式表述
一 、博弈的基本概念及战略表述
2博弈论第二讲-Mixed-Strategies-复旦大学-王永钦
➢ v1((0.4, 0.6), (0.3, 0.7))=0.40.4+0.6(-0.4)=-0.08
Player 2:
➢ EU2(H, (0.4, 0.6)) = 0.4×1+0.6×(-1) = -0.2 ➢ EU2(T, (0.4, 0.6)) = 0.4×(-1)+0.6×1 = 0.2 ➢ v2((0.4, 0.6), (0.3, 0.7))=0.3×(-0.2)+0.7×0.2=0.08
Player 1’s expected payoffs
➢ If Player 1 chooses Head, -q+(1-q)=1-2q ➢ If Player 1 chooses Tail, q-(1-q)=2q-1
Fall, 2007, Fudan
2021/2/4
5
Solving matching pennies
Fall, 2007, Fudan
2021/2/4
7
Solving matching pennies
Head Player 1
Tail
Player 2
Head
Tail
-1 , 1
1 , -1
1 , -1 -1 , 1
r 1-r
Expected payoffs
1-2q
2q-1
Expected payoffs
8
Solving matching pennies
Player 1’s best response B1(q):
➢ For q<0.5, Head (r=1)
➢ For q>0.5, Tail (r=0)
Player 1
➢ For q=0.5, indifferent (0r1)
Player 2:
➢ EU2(H, (0.4, 0.6)) = 0.4×1+0.6×(-1) = -0.2 ➢ EU2(T, (0.4, 0.6)) = 0.4×(-1)+0.6×1 = 0.2 ➢ v2((0.4, 0.6), (0.3, 0.7))=0.3×(-0.2)+0.7×0.2=0.08
Player 1’s expected payoffs
➢ If Player 1 chooses Head, -q+(1-q)=1-2q ➢ If Player 1 chooses Tail, q-(1-q)=2q-1
Fall, 2007, Fudan
2021/2/4
5
Solving matching pennies
Fall, 2007, Fudan
2021/2/4
7
Solving matching pennies
Head Player 1
Tail
Player 2
Head
Tail
-1 , 1
1 , -1
1 , -1 -1 , 1
r 1-r
Expected payoffs
1-2q
2q-1
Expected payoffs
8
Solving matching pennies
Player 1’s best response B1(q):
➢ For q<0.5, Head (r=1)
➢ For q>0.5, Tail (r=0)
Player 1
➢ For q=0.5, indifferent (0r1)
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are a NE, if for each player i,
si* is (at least tied for) player i’s best response to the strategies
specified for the n-1 other players,
( s1* , ...,
s* i 1
1.Static Game of Complete Information
1.3 Further Discussion on Nash Equilibrium (NE) 1.3.1 NE versus Iterated Elimination of Strict
Dominance Strategies
Proposition A In the n -player normal form game
G {S1,..., Sn ; u1,..., un}
if iterated elimination of strictly dominated strategies
eliminates all but the strategies (s1*,..., sn* ) , then these
iterated elimination of strictly dominated strategies.
1.3.2 Existence of NE
Theorem (Nash, 1950): In the n -player normal form game
G {S1,..., Sn;u1,...,un}
competing firms? (Convergence to Competitive Equilibrium)
1.4.2 The problem of Commons
David Hume (1739): if people respond only to private incentives, public goods will be underprovided and public resources overutilized.
if n is finite and S i is finite for every i , then there exist at
least one NE, possibly involving mixed strategies.
See Fudenberg and Tirole (1991) for a rigorous proof.
Hardin(1968) : The Tragedy of Commons
Cont’d
There are n farmers in a village. They all graze their goat on the
village green. Denote the number of goats the ith farmer owns
by gi , and the total number of goats in the village by G g1 ... gn
Buying and caring each goat cost c and value to a farmer of
grazing each goat is v(G) .
1.4 Applications 1.4.1 Cournot Model
Two firms A and B quantity compete.
Inverse demand function P a Q, a 0
They have the same constant marginal cost, and there is no fixed cost.
Game Theory (Microeconomic Theory (IV))
Instructor: Yongqin Wang Email: yongqin_wang@ School of Economics and CCES, Fudan University
December, 2004
goats to own (to choose gi ).
Cont’d
A maximum number of goats : Gmax : v(G) 0 ,
for G Gmax but v(G) 0 for G Gmax
Also v '(G) 0, v ''(G) 0
The villagers’ problem is simultaneously choosing how many
Cont’d
Firm A’s problem:
A PqA cqA (a qA qB )qA cqA
dA dqA
a 2qA
qB
c
0
qA
a
qB 2cຫໍສະໝຸດ d 2 A 2 0 dqA2
Cont’d
By symmetry, firm B’s problem. Figure Illustration: Response Function, Tatonnement Process Exercise: what will happens if there are n identical Cournot
strategies are the unique NE of the game.
A Formal Definition of NE
In the n-player normal form G {S1,..., Sn ;u1,...,un}
the strategies (s1*,..., sn* )
,
si* ,
s* i 1
,
...,
sn*
)
ui
( s1* , ...,
s* i 1
,
si
,
s* i 1
,
...,
sn* )
Cont’d
Proposition B In the n -player normal form game
G {S1,..., Sn;u1,...,un}
if the strategies (s1* ,..., sn* ) are a NE, then they survive