6948772 Game Theory NOTES MIT

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14.12 Game Theory Fall 2001 Announcement The final exam is on Friday, Dec. 21, 9:00-12:00 in Walker. The conflict exam will be on Thursday, December 20, 9:00-12:00 in E51-085. Faculty Professor: Muhamet Yildiz [email protected] (Office hours: M 4:00-5:30, E52-251a, 253-5331) TA: Kenichi Amaya [email protected] (Office hours: T 4:00-6:00, E52-303, 253-3591) TA: Astrid Dick [email protected] Schedule Class: Room E51-372 Recitation: F10 or F3, Room E51-085 Handouts syllabus Lecture notes 1 Lecture notes 2 Lecture notes 3 Lecture notes 4 Lecture notes 5 Lecture notes 6 Review notes 1 Review notes 2 Review notes 3 (11/30) 14.12 Game Theory, Fall 2001 http://web.mit.edu/14.12/www/ (1 of 4) [2002-05-11 23:27:55]

Transcript of 6948772 Game Theory NOTES MIT

Page 1: 6948772 Game Theory NOTES MIT

14.12 Game Theory

Fall 2001

AnnouncementThe final exam is on Friday, Dec. 21, 9:00-12:00 in Walker. The conflict exam will be on Thursday,December 20, 9:00-12:00 in E51-085.

Faculty

Professor: Muhamet Yildiz  [email protected]  (Office hours: M 4:00-5:30, E52-251a, 253-5331)

TA: Kenichi Amaya  [email protected]  (Office hours: T 4:00-6:00, E52-303, 253-3591)

TA: Astrid Dick  [email protected]

Schedule

Class: Room E51-372Recitation: F10 or F3, Room E51-085

Handouts syllabus●

 Lecture notes 1●

 Lecture notes 2●

 Lecture notes 3●

 Lecture notes 4●

 Lecture notes 5●

 Lecture notes 6●

 Review notes 1●

 Review notes 2●

 Review notes 3  (11/30)●

14.12 Game Theory, Fall 2001

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Lecture SlidesSome of the slides for lectures 1,2, and 6 are not included, as they are made by another program.

 Slides 1●

 Slides 2●

 Slides 3●

 Slides 4●

 Slides 5●

 Slides 6●

 Slides for lecture 8●

 Slides for lecture 9●

 Slides for lecture 10●

 Slides for lecture 12●

 Slides for lecture 13●

 Slides for lecture 15-18●

 Slides for lecture 19-21●

Homeworks Homework 1   (due 9/26)●

 Solutions for homework 1●

 Homework 2  (due 10/3)●

 Solutions for homework 2●

 Homework 3  (due 10/29)●

 Solutions for homework 3  Revised●

 Homework 4  (due 11/7)●

 Solutions for homework 4●

 Homework 5  (due 12/7)●

 Solutions for homework 5  Revised●

Homework Grading PoliciesYou may turn in assignments during the lecture on the day they are due. After the lecture,assignments may be placed in a designated box that will be set out outside E52-303 until 5:30 p.

14.12 Game Theory, Fall 2001

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m.  Do not leave assignments in the professor or T. A.'s office or mailbox.

You are permitted to discuss course material, including homework, with other students in theclass.  However, you must turn in your own individual solutions to each homework set.

You need to explain how you come up with the answers.Correct answers without properexplanations will receive no credit.

Exams Midterm 1  (10/10)●

 Midterm 1 solutions●

 Midterm 1 Grade Distribution●

 Midterm 2●

 Midterm 2 Solutions  New!●

 Midterm 2 Grade Distribution New!●

 Grade distribution of the average of midterm 1 and 2  New!●

 Mock Final  New!●

 Mock Final solutions  New!●

●  Final exam solutions  New!

●  Final exam grade distribution  New!

●  Final grade+quiz distribution  New!

●  Final grade distribution  New!

Past years' exams

Midterm 1 1995●

 1995 (Solutions)●

 1997●

 1999●

 2000●

 2000 (Solutions)●

Midterm 2

14.12 Game Theory, Fall 2001

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 1996●

 1997●

 1999  Mock Midterm●

 2000  (11/16)●

 2000  (11/17)●

 Some previous midterm questions  (including some questions from files above)●

 Solutions for Question 7 (or 8)●

Final 2000●

 2000 solutions●

 2000 make-up●

 

14.12 Game Theory, Fall 2001

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Page 71: 6948772 Game Theory NOTES MIT

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Page 72: 6948772 Game Theory NOTES MIT

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Page 73: 6948772 Game Theory NOTES MIT

14.12 Review Notes 1

Kenichi Amaya∗

September 14, 2001

1 Why do we learn game theory?

Economic agents’ (e.g., consumers, firms, government) behavior is describedas a maximization problem (utility maximization, profit maximization). Forexample, consider a firm’s profit maximization problem.

Perfect competition Under perfect competition, a firm takes the price asgiven, and choose the quantity to sell.

maxqpq − c(q)

⇒ FOC p = c′(q) = MC.

Monopoly A monopolist choose the price, and the quantity to sell is deter-mined by the demand curve D(p).

maxp

pD(p)− c(D(p))

⇒ FOC pD′(p) +D(p) = c′(D(p))D(p).

In the examples above, the firm is the only decision maker and everythingother than the choice variables (q in perfect competition and q in monopoly) areexogenously given. Therefore we can solve for the optimal value of the choicevariable.

Consider the Cournot Duopoly. Firm 1 chooses the quantity to sell q1, andFirm 2 chooses q2. Let P (q) be the inverse demand curve, where q = q1 + q2.Firm 1 maximizes its profit:

maxq1

P (q1 + q2)q1 − c1(q1).

∗E52-303, [email protected]

1

Page 74: 6948772 Game Theory NOTES MIT

Firm 2 maximizes its profit:

maxq2

P (q1 + q2)q2 − c2(q2).

To solve for the optimal q1, we need to know what q2 is. But to know theoptimal q2, we need to know what q1 is. This goes forever and we can neversolve it mathematically. We need some theory to solve it: Game theory.

Game theory is a tool to analyze people’s behavior under strategic environ-ment: An agent’s payoff depends not only on his decisions but also other agents’decisions.

Other examples includes:

• Employer — Employee (Wage scheme — Effort)

• Government — Firm (Regulation — Business strategy)

• U.S. Government — Japanese government (Trade policy)

2 Decision making under uncertainty

In the environment of games, economic agents often faces uncertainty. Theymust choose among alternatives which gives different probability distributionover outcomes. We often call outcomes of lotteries prizes We call these alterna-tives lotteries. For example:

Lottery 1 gives you beer with probability 12 and chocolate with probability 1

2 .

Lottery 2 gives you beer with probability 13 and coffee with probability 2

3 .

We want to have a utility function U(·) such that

U(lottery 1) ≥ U(lottery 2)

if and only if the agent prefers lottery 1 to lottery 2, i.e.,

lottery 1 � lottery 2.

The basic theory of choice implies there exists such a utility function if � is apreference relationship, i.e., it is complete and transitive

Completeness For all p, q ∈ P , p � q or q � p.

Transitivity If p � q and q � r, then p � r.

2

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But thinking directly about this function U is too messy! It is more conve-nient if we have a function u such that

U(lottery 1) =12u(beer) +

12u(chocolate),

U(lottery 2) =13u(beer) +

23u(coffee).

But can we really find such an convenient function u? Actually, it is known thatwe can find u if the agent’s preference � over lotteries satisfies the followingcondition, in addition to completeness and transitivity.

Independence For any p, q, r ∈ P , and any a ∈ (0, 1],

ap+ (1− a)r � aq + (1− a)r ⇔ p � q.

Continuity For any p, q, r ∈ P , if p � r, then there exist a, b ∈ (0, 1) such that

ap+ (1− p)r � q � bp+ (1− b)r.

Furthermore, if a von-Neuman Morgenstern utility function u represents �,then its affine transformation u′ = au+ b where a > 0 also represents �.

3 Attitudes toward risk

Suppose now the prizes of lotteries are money. Lottery A gives you 50 dollars forsure (a degenerated lottery). Lottery B gives you 100 dollars with probability12 and 0 with probability 1

2 . Which do you prefer? The two lotteries give youthe same expected amount of money, 50 dollars, so your choice reflects whetheryou like risk or not.

If you prefer Lottery A, then we can say you are risk averse. Your vNMutility function satisfies

u(50) >12u(0) +

12u(100).

This is true if your vNM utility function is concave (u′′ < 0 if u is twice differ-entiable). In general, if your vNM utility function is concave, you always prefergetting the avarage for sure to getting some risky lottery.

In contrast, if your vNM utility function is concave, you always prefer gettingsome risky lottery to getting the avarage for sure. In this case, you are calledrisk loving.

If your vNM utility function is linear, you are indifferent between gettingsome risky lottery and getting the avarage for sure. In this case, you are calledrisk neutral.

3

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4 Modelling games

When we write down a game which represents some economic environment,the payoffs of the game are vNM utilities of the players, not the actual moneyamount received. In this way, we can assume the players maximize their ex-pected payoffs. The only case where the payoffs of a player is equal to theactual money amount received is the case where the player is risk neutral. Ac-tually, in many textbook modelling we find the players are assumed to be riskneutral and therefore maximize expected revenue.

4

Page 77: 6948772 Game Theory NOTES MIT

14.12 Review Notes

Dynamic games with incomplete information

Kenichi Amaya∗

November 30, 2001

1 Perfect Bayesian equilibrium

• s = (s1, · · · , sn): strategy profile.

• A belief at an information set is a probability distribution over decisionnodes in the information set.

• A belief system µ is the collection of beliefs at all information sets.

Definition A pair of strategy profile and a belief system (s, µ) is a PerfectBayesian (Nash) equilibrium (PBE) if

1. Given their beliefs, the players’ strategies must be sequentially ratio-nal. That is, at each information set the action taken by the playerwith the move (and the player’s subsequent strategy) must be opti-mal given the player’s belief at that information set and the otherplayers’ subsequent strategies.

2. Beliefs are determined by Bayes’ rule and the players’ equilibriumstrategies wherever possible.

Note 1: When you are asked to describe a PBE, you need to show not onlythe strategy profile but also the belief system. (However, you usually don’t needto describe the belief at a singleton information set because it is obvious).

Note 2: PBE is a stronger concept than subgame perfect Nash equilibrium.Therefore, the strayegy profile of a PBE is a subgame perfect Nash equilibrium.∗E52-303, [email protected]

1

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2 How do we find PBE?

Here is an abstract procedure of looking for PBE.

1. Using sequential rationality and Bayes rule, determine strategies and be-liefs wherever possible. If you can determine strategies and beliefs every-where, that’s it.

2. When you can’t do more, make an assumption about a strategy at anyinformation set. Then, using sequential rationality and Bayes rule, deter-mine strategies and beliefs wherever possible.

3. You may be able to determine strategies and beliefs everywhere, i.e., tofind an equilibrium. Then, change your assumption and look for anotherequilibrium.

4. You may reach a contradiction. In this case, the assumption was wrong.Change your assumption.

5. When you can’t do more only with the assumption you made, make afurther assumption.

The idea we use in finding a mixed strategy Nash equilibrium applies heretoo: The mixing probability must be such that the other player is indifferentbetween the strategies he is mixing.

2

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1

14.12 Economic Applications of Game Theory

Professor: Muhamet YildizLecture: MW 2:30-4:00 @4-153?

Office Hours: M 4-5:30 @E52-251aTA: Kenichi Amaya

F 10,3 @E51-85Office Hours:TBA

Web: http://web.mit.edu/14.12/www/

Name of the game

Game Theory = Multi-person decision theory• The outcome is determined by the actions

independently taken by multiple decision makers.

• Strategic interaction.– Need to understand what the others will do– … what the others think that you will do– …

Page 80: 6948772 Game Theory NOTES MIT

2

Hawk-Dove game

(V/2,V/2)(0,V)

(V,0)

−−2

,2

cVcV

Chicken

(1/2,1/2)(0,1)

(1,0)(-1,-1)

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3

Stag Hunt

(6,6)(0,4)

(4,0)(2,2)

Quiz Problem

• Without discussing with anyone, each student is to write down a real number between 0 and 100 on a paper and submit it to the TA.

• The TA will then compute the average

• The students who submit the number that is closest to

.21

nxxxx n+++= �

ix

3/x will share 100 points equally; the others will get 0.

Page 82: 6948772 Game Theory NOTES MIT

1

14.12 Game Theory

Lecture 2: Decision TheoryMuhamet Yildiz

Road Map

1. Basic Concepts (Alternatives, preferences,…)2. Ordinal representation of preferences3. Cardinal representation – Expected utility

theory4. Applications: Risk sharing and Insurance5. Quiz

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2

Basic Concepts: Alternatives

• Agent chooses between the alternatives• X = The set of all alternatives• Alternatives are

– Mutually exclusive, and – Exhaustive

• Example: Options = {Tea, Coffee} X = {T, C, TC, NT} where T= Tea, C = Coffee, TC = Tea and Coffee, NT = Neither Tea nor Coffee

Basic Concepts: Preferences

• TeX • TeX –ordinal representation

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3

Examples

• Define a relation among the students in this class by – x T y iff x is at least as tall as y;– x M y iff x’s final grade in 14.04 is at least as

high as y’s final grade;– x H y iff x and y went to the same high school;– X Y y iff x is strictly younger than y;– x S y iff x is as old as y;

Exercises

• Imagine a group of students sitting around a round table. Define a relation R, by writing x R y iff x sits to the right of y. Can you represent R by a utility function?

• Consider a relation among positive real numbers represented by u with Can this relation be represented by What about

}

( ) .2xxu =( ) ?* xxu =

( ) ?/1** xxu =

Page 85: 6948772 Game Theory NOTES MIT

4

• TeX – OR Theorem• TeX – Cardinal representation• VNM Axioms• Theorem

A Lottery

Page 86: 6948772 Game Theory NOTES MIT

5

Two Lotteries

.3.3

.4

$1000

$10

$0

$1M

$0

.00001

.99999

Exercise

• Consider an agent with VNM utility function u with Can his preferences be represented by VNM utility function What about

( ) .2xxu =

( ) ?* xxu =( ) ?/1** xxu =

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6

Attitudes towards Risk

• A fair gamble: px+(1-p)y = 0.

• An agent is said to be risk neutral iff he is indifferent towards all fair gambles. He is said to be (strictly) risk averse iff he never wants to take any fair gamble, and (strictly) risk seekingiff he always wants to take fair gambles.

x

y

p

1-p

A utility functionEUu

u(pW1+(1- p)W2)

EU(Gamble)

W1 pW1+(1-p)W2 W2

B

C

A

Page 88: 6948772 Game Theory NOTES MIT

7

• An agent is risk-neutral iff he has a linear utility function, i.e., u(x) = ax + b.

• An agent is risk-averse iff his utility function is concave.

• An agent is risk-seeking iff his utility function is convex.

Risk Sharing• Two agents, each having a utility function u

with and an “asset:”

• For each agent, the value of the asset is

• Assume that the value of assets are independently distributed.

( ) xxu =$100

$0

.5

.5

Page 89: 6948772 Game Theory NOTES MIT

8

• If they form a mutual fund so that each agent owns half of each asset, each gets

$100

$0

$50

1/4

1/4

1/2

Insurance

• We have an agent with and

• And a risk-neutral insurance company with lots of money, selling full insurance for “premium” P.

$1M

$0

.5

.5

( ) xxu =

Page 90: 6948772 Game Theory NOTES MIT

1

Lecture 3 Decision Theory/Game Theory

14.12 Game Theory Muhamet Yildiz

Road Map1. Basic Concepts (Alternatives, preferences,…)2. Ordinal representation of preferences

3. Cardinal representation – Expected utility theory4. Applications: Risk sharing and Insurance5. Quiz6. Representation of games in strategic

and extensive forms7. Quiz?

Page 91: 6948772 Game Theory NOTES MIT

2

A Lottery

Two Lotteries

.3.3

.4

$1000

$10

$0

$1M

$0

.00001

.99999

Page 92: 6948772 Game Theory NOTES MIT

3

Exercise

• Consider an agent with VNM utility function u with Can his preferences be represented by VNM utility function What about

( ) .2xxu =

( ) ?* xxu =( ) ?/1** xxu =

Attitudes towards Risk

• A fair gamble: px+(1-p)y = 0.

• An agent is said to be risk neutral iff he is indifferent towards all fair gambles. He is said to be (strictly) risk averse iff he never wants to take any fair gamble, and (strictly) risk seekingiff he always wants to take fair gambles.

x

y

p

1-p

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4

A utility functionEUu

u(pW1+(1- p)W2)

EU(Gamble)

W1 pW1+(1-p)W2 W2

B

C

A

• An agent is risk-neutral iff he has a linear utility function, i.e., u(x) = ax + b.

• An agent is risk-averse iff his utility function is concave.

• An agent is risk-seeking iff his utility function is convex.

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5

Risk Sharing• Two agents, each having a utility function u

with and an “asset:”

• For each agent, the value of the asset is

• Assume that the value of assets are independently distributed.

( ) xxu =$100

$0

.5

.5

• If they form a mutual fund so that each agent owns half of each asset, each gets

$100

$0

$50

1/4

1/4

1/2

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6

Insurance

• We have an agent with and

• And a risk-neutral insurance company with lots of money, selling full insurance for “premium” P.

$1M

$0

.5

.5

( ) xxu =

Quiz Problem

• Without discussing with anyone, each student is to write down a real number between 0 and 100 on a paper and submit it to the TA.

• The TA will then compute the average

• The students who submit the number that is closest to

.21

nxxxx n+++= �

ix

3/x will share 100 points equally; the others will get 0.

Page 96: 6948772 Game Theory NOTES MIT

7

Multi-person Decision Theory

• Who are the players?• Who has which options?• Who knows what?• Who gets how much?

Knowledge

1. If I know something, it must be true.

2. If I know x, then I know that I know x.

3. If I don’t know x, then I know that I don’t know x.

4. If I know something, I know all its logical implications.

Common Knowledge: x is common knowledge iff

•Each player knows x

•Each player knows that each player knows x

• Each player knows that each player knows that each player knows x

•Each player knows that each player knows that each player knows that each player knows x

•… ad infinitum

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8

Representations of games

Normal-form representationDefinition (Normal form): A game is any list

where, for each • Si is the set of all strategies available to i,• is the VNM utility function of

player i.

},,,2,1{ nNi l=∈

( )nn uuSSG ,,;,, 11 ll=

ℜ→×× ni SSu m1:

Assumption: G is common knowledge.

Definition: A player i is rational iff he tries to maximize the expected value of ui given his beliefs.

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9

Chicken

(1/2,1/2)(0,1)

(1,0)(-1,-1)

Extensive-form representationDefinition: A tree is a set of nodes connected with

directed arcs such that 1. For each node, there is at most one incoming

arc; 2. each node can be reached through a unique path;

Page 99: 6948772 Game Theory NOTES MIT

10

A tree?

A tree??

A

B

C

A

B

C

D

Page 100: 6948772 Game Theory NOTES MIT

11

A tree

Terminal NodesNon-terminal

nodes

Extensive form – definition

Definition: A game consists of – a set of players– a tree– an allocation of each non-terminal node to a

player– an informational partition (to be made precise)– a payoff for each player at each terminal node.

Page 101: 6948772 Game Theory NOTES MIT

12

Information set

An information set is a collection of nodes such that

1. The same player is to move at each of these nodes;

2. The same moves are available at each of these nodes.

An informational partition is an allocation of each non-terminal node of the tree to an information set.

A game1

1 1

2(2,2)

(1,3) (3,1) (3,3) (1,1)

(0,0)

L R

l r

λρ Λ Ρ

u

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13

Another Game

1

BT

x

2

L R RL

The same game1 x

T B

2

L R L R

Page 103: 6948772 Game Theory NOTES MIT

14

Strategy

A strategy of a player is a complete contingent-plan, determining which action he will take at each information set he is to move (including the information sets that will not be reached according to this strategy).

Matching pennies with perfect information2’s Strategies:HH = Head if 1 plays Head,

Head if 1 plays Tail;HT = Head if 1 plays Head,

Tail if 1 plays Tail;TH = Tail if 1 plays Head,

Head if 1 plays Tail;TT = Tail if 1 plays Head,

Tail if 1 plays Tail.

1

22

Head Tail

head tail head tail

(-1,1) (1,-1) (1,-1) (-1,1)

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15

Matching pennies with perfect information

Tail

Head

TTTHHTHH12

Matching pennies with Imperfect information

1

2

Head Tail

head tail head tail

(-1,1) (1,-1) (1,-1) (-1,1)(-1,1)(1,-1)Tail

(1,-1)(-1,1)Head

TailHead12

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16

A game with nature

Nature

Head1

Left (5, 0)

(2, 2)Right

Tail 2Left

(3, 3)

Right(0, -5)

1/2

1/2

A centipede game1 2A

D

α

δ

(4,4) (5,2)

(1,-5)a

d

(3,3)

1

Page 106: 6948772 Game Theory NOTES MIT

1

Lecture 4 Representation of Games &

Rationalizability

14.12 Game Theory Muhamet Yildiz

Road Map1. Representation of games in strategic

and extensive forms2. Dominance3. Dominant-strategy equilibrium4. Rationalizability

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Normal-form representationDefinition (Normal form): A game is any list

where, for each • Si is the set of all strategies available to i,• is the VNM utility function of

player i.

},,,2,1{ nNi l=∈

( )nn uuSSG ,,;,, 11 ll=

ℜ→×× ni SSu m1:

Assumption: G is common knowledge.

Definition: A player i is rational iff he tries to maximize the expected value of ui given his beliefs.

Chicken

(1/2,1/2)(0,1)

(1,0)(-1,-1)

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Extensive-form representationDefinition: A tree is a set of nodes connected with

directed arcs such that 1. For each node, there is at most one incoming

arc; 2. each node can be reached through a unique path;

A tree

Terminal NodesNon-terminal

nodes

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Extensive form – definition

Definition: A game consists of – a set of players– a tree– an allocation of each non-terminal node to a

player– an informational partition (to be made precise)– a payoff for each player at each terminal node.

Information set

An information set is a collection of nodes such that

1. The same player is to move at each of these nodes;

2. The same moves are available at each of these nodes.

An informational partition is an allocation of each non-terminal node of the tree to an information set.

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A game1

1 1

2(2,2)

(1,3) (3,1) (3,3) (1,1)

(0,0)

L R

l r

λρ Λ Ρ

u

Another Game

1

BT

x

2

L R RL

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The same game1 x

T B

2

L R L R

Strategy

A strategy of a player is a complete contingent-plan, determining which action he will take at each information set he is to move (including the information sets that will not be reached according to this strategy).

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Matching pennies with perfect information2’s Strategies:HH = Head if 1 plays Head,

Head if 1 plays Tail;HT = Head if 1 plays Head,

Tail if 1 plays Tail;TH = Tail if 1 plays Head,

Head if 1 plays Tail;TT = Tail if 1 plays Head,

Tail if 1 plays Tail.

1

22

Head Tail

head tail head tail

(-1,1) (1,-1) (1,-1) (-1,1)

Matching pennies with perfect information

Tail

Head

TTTHHTHH12

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8

Matching pennies with Imperfect information

1

2

Head Tail

head tail head tail

(-1,1) (1,-1) (1,-1) (-1,1)(-1,1)(1,-1)Tail

(1,-1)(-1,1)Head

TailHead12

A game with nature

Nature

Head1

Left (5, 0)

(2, 2)Right

Tail 2Left

(3, 3)

Right(0, -5)

1/2

1/2

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Mixed StrategyDefinition: A mixed strategy of a player is a

probability distribution over the set of his strategies.

Pure strategies: Si = {si1,si2,…,sik}A mixed strategy: σi: S → [0,1] s.t.

σi(si1) + σi(si2) + … + σi(sik) = 1.

If the other players play s-i =(s1,…, si-1,si+1,…,sn), then the expected utility of playing σi is

σi(si1)ui(si1,s-i) + σi(si2) ui(si2,s-i) + … + σi(sik) ui(sik,s-i).

How to play

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Dominances-i =(s1,…, si-1,si+1,…,sn)

Definition: A pure strategy si* strictly dominates si if and only if

A mixed strategy σi* strictly dominates si iff

A rational player never plays a strictly dominated strategy.

.),(),( *iiiiiii sssussu −−− ∀>

iiiiiikiikiiiiii sssussusssus ∀>++ −−− ),(),()(),()( 11 σσ m

Prisoners’ Dilemma

(1,1)(6,0)Defect

(0,6)(5,5)Cooperate

DefectCooperate12

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11

A game

(3,0)(1,1)(0,3)B

(1,0)(0,10)(1,0)M

(0,3)(1,1)(3,0)T

RmL12

Weak DominanceDefinition: A pure strategy si* weakly dominates si

if and only if

and at least one of the inequalities is strict. A mixed strategy σi* weakly dominates si iff

and at least one of the inequalities is strict.If a player is rational and cautious (i.e., he assigns

positive probability to each of his opponents’ strategies), then he will not play a weakly dominated strategy.

.),(),( *iiiiiii sssussu −−− ∀≥

iiiiiikiikiiiiii sssussusssus ∀>++ −−− ),(),()(),()( 11 σσ �

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Dominant-strategy equilibriumDefinition: A strategy si* is a dominant

strategy iff si* weakly dominates every other strategy si.

Definition: A strategy profile s* is a dominant-strategy equilibrium iff si* is a dominant strategy for each player i.

If there is a dominant strategy, then it will be played, so long as the players are …

Prisoners’ Dilemma

(1,1)(6,0)Defect

(0,6)(5,5)Cooperate

DefectCooperate12

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13

Second-price auction

• N = {1,2} buyers;• The value of the house

for buyer i is vi;• Each buyer i

simultaneously bids bi;• i* with bi* = max bi gets

the house and pays the second highest bid

p = maxj≠ibj.

Question

What is the probability that an nxn game has a dominant strategy equilibrium given that the payoffs are independently drawn from the same (continuous) distribution on [0,1]?

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14

1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1(1/n)(2n-2)

A game

(3,0)(1,1)(0,3)B

(1,0)(0,10)(1,0)M

(0,3)(1,1)(3,0)T

RmL12

Assume: Players are rational and player 2 knows that 1 is rational.

1 is rational and 2 knows this:

(3,0)(1,1)(0,3)B

(0,3)(1,1)(3,0)TRmL

And 2 is rational:

(3,0)(0,3)B

(0,3)(3,0)TRL

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Rationalizability

The play is rationalizable, provided that …

Eliminate all the strictly dominated strategies.

Any dominated strategyIn the new game?

Yes

NoRationalizable strategies

Simplified price-competition

4,48,08,0Low

0,85,510,0Medium

0,80,106,6High

LowMediumHighFirm 2

Firm 1

Dutta

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A strategy profile is rationalizable when …

• Each player’s strategy is consistent with his rationality, i.e., maximizes his payoff with respect to a conjecture about other players’ strategies;

• These conjectures are consistent with the other players’ rationality, i.e., if i conjectures that j will play sj with positive probability, then sj maximizes j’s payoff with respect to a conjecture of j about other players’ strategies;

• These conjectures are also consistent with the other players’ rationality, i.e., …

• Ad infinitum

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Lecture 5 Nash equilibrium & Applications

14.12 Game Theory Muhamet Yildiz

Road Map1. Rationalizability – summary2. Nash Equilibrium3. Cournot Competition

1. Rationalizability in Cournot Duopoly4. Bertrand Competition5. Commons Problem6. Quiz7. Mixed-strategy Nash equilibrium

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Dominant-strategy equilibriums-i =(s1,…, si-1,si+1,…,sn)

Definition: si* strictly dominates si iff

si* weakly dominates si iffand at least one of the inequalities is strict. Definition: A strategy si* is a dominant strategy iff

si* weakly dominates every other strategy si.Definition: A strategy profile s* is a dominant-

strategy equilibrium iff si* is a dominant strategy for each player i.

Examples: Prisoners’ Dilemma; Second-Price auction.

;),(),( *iiiiiii sssussu −−− ∀>

),(),( *iiiiii ssussu −− ≥ is−∀

Question

What is the probability that an nxn game has a dominant strategy equilibrium given that the payoffs are independently drawn from the same (continuous) distribution on [0,1]?

Page 124: 6948772 Game Theory NOTES MIT

3

1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1(1/n)(2n-2)

Rationalizability

The play is rationalizable, provided that …

Eliminate all the strictly dominated strategies.

Any dominated strategyIn the new game?

Yes

NoRationalizable strategies

Page 125: 6948772 Game Theory NOTES MIT

4

Simplified price-competition

4,48,08,0Low

0,85,510,0Medium

0,80,106,6High

LowMediumHighFirm 2

Firm 1

Dutta

A strategy profile is rationalizable when …

• Each player’s strategy is consistent with his rationality, i.e., maximizes his payoff with respect to a conjecture about other players’ strategies;

• These conjectures are consistent with the other players’ rationality, i.e., if i conjectures that j will play sj with positive probability, then sj maximizes j’s payoff with respect to a conjecture of j about other players’ strategies;

• These conjectures are also consistent with the other players’ rationality, i.e., …

• Ad infinitum

Page 126: 6948772 Game Theory NOTES MIT

5

Stag Hunt

(6,6)(0,4)

(4,0)(2,2)

A summary

• If players are rational (and cautious), then they play the dominant-strategy equilibrium whenever it exists– But, typically, it does not exist

• If it is common knowledge that players are rational, then they will play a rationalizable strategy-profile– Typically, there are too many rationalizable strategies

• Now, a stronger assumption: The players are rational and their conjectures are mutually known.

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Nash EquilibriumDefinition: A strategy-profile s* =(s1*,…,sn*) is a

Nash Equilibrium iff, for each player i, and for each strategy si, we have

i.e., no player has any incentive to deviate if he knows what the others play.

??If players are rational, and their conjectures about what the others play are mutually known, then they must be playing a Nash equilibrium.

),,,,,,,(

),,,,,,(**

1*

1*1

**1

**1

*1

niiii

niiii

sssssusssssull

ll

+−

+−

Stag Hunt

(6,6)(0,4)

(4,0)(2,2)

Page 128: 6948772 Game Theory NOTES MIT

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Economic Applications

1. Cournot (quantity) Competition1. Nash Equilibrium in Cournot duopoly2. Nash Equilibrium in Cournot oligopoly3. Rationalizability in Cournot duopoly

2. Bertrand (price) Competition3. Commons Problem

Cournot Oligopoly• N = {1,2,…,n} firms;• Simultaneously, each firm i

produces qi units of a good at marginal cost c,

• and sells the good at priceP = max{0,1-Q}

where Q = q1+…+qn.• Game = (S1,…,Sn; π1,…,πn)

where Si = [0,∞∞),),

1

1

Q

P

πi(q1,…,qn) = qi[1-(q1+…+qn)-c] if q1+…+qn < 1, -qic otherwise.

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Cournot Duopoly -- profit

0 1-0.2

0

1-qj-c(1-qj-c)/2

-cqi

qi(1-qj-c)

Profitqj=0.2

c=0.2

C-D – best responses

• qiB(qj) = max{(1-qj-c)/2,0};

• Nash Equilibrium q*:q1* = (1-q2*-c)/2;q2* = (1-q1*-c)/2;

• q1* = q2* = (1-c)/3

q1

q2

q2=q2B(q1)

q1=q1B(q2)

q*

21 c−

1-c

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9

Cournot Oligopoly --Equilibrium • q>1-c is strictly dominated, so q ≤ 1-c. • πi(q1,…,qn) = qi[1-(q1+…+qn)-c] for each i.• FOC:

• That is,

• Therefore, q1*=…=qn*=(1-c)/(n+1).

.0)1(

)]1([),,(

***1

11

**

=−−−−−=

∂−−−−∂=

∂∂

==

in

qqi

ni

qqi

ni

qcqq

qcqqq

qqq

��π

cnqqq

cqqqcqqq

n

n

n

−=+++

−=+++

−=+++

1

12

12

**2

*1

**2

*1

**2

*1

Cournot oligopoly – comparative statics

1

1

Q

P

n = 1

n = 2

c

n = 3n = 4

Page 131: 6948772 Game Theory NOTES MIT

10

Rationalizability in Cournot Duopoly

q1

q2

21 c−

1-c2

1 c−

1-c

Assume that players are rational.

Players are rational:

q1

q2

21 c−

1-c2

1 c−

1-c

Assume that players know this.

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Players are rational and know that players are rational

q1

q2

21 c−

1-c2

1 c−

1-c

Assume that players know this.

Players are rational; players know that players are rational; players know that players know

that players are rational

q1

q2

21 c−

1-c2

1 c−

1-c

Assume that players know this.

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Rationalizability in Cournot duopoly

• If i knows that qj ≤ q, then qi ≥ (1-c-q)/2.• If i knows that qj ≥ q, then qi ≤ (1-c-q)/2.• We know that qj ≥ q0 = 0.• Then, qi ≤ q1 = (1-c-q0)/2 = (1-c)/2 for each i;• Then, qi ≥ q2 = (1-c-q1)/2 = (1-c)(1-1/2)/2 for each i;• …• Then, qn ≤ qi ≤ qn+1 or qn+1 ≤ qi ≤ qn where

qn+1 = (1-c-qn)/2 = (1-c)(1-1/2+1/4-…+(-1/2)n)/2.• As n→∞, qn → (1-c)/3.

Bertrand (price) competition• N = {1,2} firms.• Simultaneously, each firm i sets a price pi;• If pi < pj, firm i sells Q = max{1 – pi,0}

unit at price pi; the other firm gets 0.• If p1 = p2, each firm sells Q/2 units at price

p1, where Q = max{1 – p1,0}.• The marginal cost is 0.

( )otherwise.

if if

02/)1(

)1(, 21

21

11

11

211 pppp

pppp

pp =<

−−

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Bertrand duopoly -- EquilibriumTheorem: The only Nash equilibrium in the “Bertrand

game” is p* = (0,0).

Proof:1. p*=(0,0) is an equilibrium. 2. If p = (p1,p2) is an equilibrium, then p = p*.

1. If p = (p1,p2) is an equilibrium, then p1 = p2... • If pi > pj= 0, for sufficiently small ε>0, pj’ = ε is a better

response to pi for j. If pi > pj> 0, pi’ = pj is a better response for i.

2. Given any equilibrium p = (p1,p2) with p1 = p2, p = p*. • If p1 = p2>0, for sufficiently small ε>0, pj’ = pj - ε is a better

response to pj for i.

Commons Problem• N = {1,2,…,n} players, each with unlimited

money;• Simultaneously, each player i contributes xi

≥ 0 to produce y = x1+…xn unit of some public good, yielding payoff

Ui(xi,y) = y1/2 – xi.

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14

Quiz

Each student i is to submit a real number xi. We will pair the students randomly. For each pair (i,j), if xi ≠ xj, the student who submits the number that is closer to (xi+xj)/4 gets 100; the other student gets 20. If xi = xj, then each of i and j gets 50.

Page 136: 6948772 Game Theory NOTES MIT

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Lectures 6-7 Nash Equilibrium

&Backward Induction

14.12 Game Theory Muhamet Yildiz

Road Map1. Bertrand Competition2. Commons Problem3. Mixed-strategy Nash equilibrium4. Bertrand competition with costly search5. Backward Induction6. Stackelberg Competition7. Sequential Bargaining

Page 137: 6948772 Game Theory NOTES MIT

2

Bertrand (price) competition• N = {1,2} firms.• Simultaneously, each firm i sets a price pi;• If pi < pj, firm i sells Q = max{1 – pi,0}

unit at price pi; the other firm gets 0.• If p1 = p2, each firm sells Q/2 units at price

p1, where Q = max{1 – p1,0}.• The marginal cost is 0.

( )otherwise.

if if

02/)1(

)1(, 21

21

11

11

211 pppp

pppp

pp =<

−−

Bertrand duopoly -- EquilibriumTheorem: The only Nash equilibrium in the “Bertrand

game” is p* = (0,0).

Proof:1. p*=(0,0) is an equilibrium. 2. If p = (p1,p2) is an equilibrium, then p = p*.

1. If p = (p1,p2) is an equilibrium, then p1 = p2... • If pi > pj= 0, for sufficiently small ε>0, pj’ = ε is a better

response to pi for j. If pi > pj> 0, pi’ = pj is a better response for i.

2. Given any equilibrium p = (p1,p2) with p1 = p2, p = p*. • If p1 = p2>0, for sufficiently small ε>0, pj’ = pj - ε is a better

response to pj for i.

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Commons Problem• N = {1,2,…,n} players, each with unlimited

money;• Simultaneously, each player i contributes xi

≥ 0 to produce y = x1+…xn unit of some public good, yielding payoff

Ui(xi,y) = y1/2 – xi.

Stag Hunt

(5,5)(0,4)

(4,0)(2,2)

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Equilibrium in Mixed Strategies

What is a strategy?– A complete contingent-plan of a player.– What the others think the player might do under

various contingency.What do we mean by a mixed strategy?

– The player is randomly choosing his pure strategies.

– The other players are not certain about what he will do.

Stag Hunt

(5,5)(0,4)

(4,0)(2,2)

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Mixed-strategy equilibrium in Stag-Hunt game• Assume: Player 2 thinks that,

with probability p, Player 1 targets for Rabbit. What is the best probability q she wants to play Rabbit?

• His payoff from targeting Rabbit: U2(R;p) = 2p + 4(1-p)

= 4-2p.• From Stag:

U2(R;p) = 5(1-p) • She is indifferent iff

4-2p = 5(1-p) iff p = 1/3.( ) [ ]

1/3p if1/3p if1/3p if

11,0

0

>=<

∈= qpqBR

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

4 - 2p

5(1-p)

Best responses in Stag-Hunt game

1/3

p

1/3

q

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Bertrand Competition with costly search

• N = {F1,F2,B}; F1, F2 are firms; B is buyer

• B needs 1 unit of good, worth 6;

• Firms sell the good; Marginal cost = 0.

• Possible prices P = {1,5}.

• Buyer can check the prices with a small cost c > 0.

Game:1. Each firm i chooses price

pi;2. B decides whether to

check the prices;3. (Given) If he checks the

prices, and p1≠p2, he buys the cheaper one; otherwise, he buys from any of the firm with probability ½.

Bertrand Competition with costly search

F1F2

High

Low

High Low F1F2

High

Low

High Low

Check Don’t Check

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7

Mixed-strategy equilibrium• Symmetric equilibrium: Each firm charges

“High” with probability q;• Buyer Checks with probability r.• U(check;q) = q21 + (1-q2)5 – c = 5 - 4 q2 – c;• U(Don’t;q) = q1 + (1-q)5 = 5 - 4 q;• Indifference: 4q(1-q) = c; i.e.,• U(high;q,r) = 0.5(1-r(1-q))5;• U(low;q,r) = qr1 + 0.5(1-qr) • Indifference = r = 4/(5-4q).

Dynamic Games of Perfect Information

& Backward Induction

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Definitions

Perfect-Information game is a game in which all the information sets are singleton.

Sequential Rationality: A player is sequentially rational iff, at each node he is to move, he maximizes his expected utility conditional on that he is at the node – even if this node is precluded by his own strategy.

In a finite game of perfect information, the common knowledge of sequential rationality gives “Backward Induction” outcome.

A centipede game1 2A

D

α

δ

(4,4) (5,2)

(1,-5)a

d

(3,3)

1

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9

Backward InductionTake any pen-terminal node

Pick one of the payoff vectors (moves) that gives ‘the mover’ at the node the highest payoff

Assign this payoff to the node at the hand;

Eliminate all the moves and the terminal nodes following the node

Any non-terminalnode

Yes

No

The picked moves

Battle of The Sexes with perfect information

2

1

2

T B

L LR R

(2,1) (0,0) (0,0) (1,2)

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10

Note

• There are Nash equilibria that are different from the Backward Induction outcome.

• Backward Induction always yields a Nash Equilibrium.

• That is, Sequential rationality is stronger than rationality.

Matching Pennies (wpi)

1

22

Head Tail

head tail head tail

(-1,1) (1,-1) (1,-1) (-1,1)

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11

Stackelberg DuopolyGame:N = {1,2} firms w MC = 0;1. Firm 1 produces q1 units 2. Observing q1, Firm 2 produces

q2 units3. Each sells the good at price

P = max{0,1-(q1+q2)}.

1

1

Q

P

πi(q1, q2) = qi[1-(q1+q2)] if q1+ q2 < 1, 0 otherwise.

“Stackelberg equilibrium”

• If q1 > 1, q2*(q1) = 0. • If q1 ≤ 1, q2*(q1) = (1-q1)/2.• Given the function q2*, if q1 ≤ 1

π1(q1;q2*(q1)) = q1[1-(q1+ (1-q1)/2)]

= q1 (1-q1)/2;

0 otherwise.• q1* = ½.• q2*(q1*) = ¼.

1

1

P

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12

Sequential Bargaining

• N = {1,2}• X = feasible

expected-utility pairs (x,y ∈ X )

• Ui(x,t) = δitxi

• d = (0,0) ∈ D disagreement payoffs

1

1D

Timeline – 2 periodAt t = 1,

– Player 1 offers some (x1,y1),– Player 2 Accept or Rejects the

offer– If the offer is Accepted, the

game ends yielding (x1,y1),– Otherwise, we proceed to date

2.

At t = 2,– Player 2 offers some (x2,y2),– Player 1 Accept or Rejects the

offer– If the offer is Accepted, the

game ends yielding payoff δ(x2,y2).

– Otherwise, the game end yielding d = (0,0).

Accept

Reject(x1,y1)1

2

Accept

Reject(x2,y2)2

1

(x1,y1)(δx2,δy2)

(0,0)

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Accept

Reject(x1,y1)1

2

Accept

Reject(x2,y2)2

1

(x1,y1)(δx2,δy2)

(0,0)

At t = 2,•Accept iff y2 ≥ 0.•Offer (0,1).At t = 1,•Accept iff x2 ≥ δ.•Offer (1−δ,δ).

Timeline – 2n periodT = {1,2,…,2n-1,2n}If t is odd,

– Player 1 offers some (xt,yt),

– Player 2 Accept or Rejects the offer

– If the offer is Accepted, the game ends yielding δt(xt,yt),

– Otherwise, we proceed to date t+1.

If t is even– Player 2 offers some

(xt,yt),– Player 1 Accept or Rejects

the offer– If the offer is Accepted,

the game ends yielding payoff (xt,yt),

– Otherwise, we proceed to date t+1, except at t = 2n, when the game end yielding d = (0,0).

Page 149: 6948772 Game Theory NOTES MIT

14

Equilibrium

• Scientific Word

Page 150: 6948772 Game Theory NOTES MIT

1

Lectures 8 Subgame-perfect Equilibrium

&Applications

14.12 Game Theory Muhamet Yildiz

Road Map1. Subgame-perfect Equilibrium

1. Motivation2. What is a subgame?3. Definition4. Example

2. Applications1. Bank Runs2. Tariffs & Intra-industry trade

3. Quiz

Page 151: 6948772 Game Theory NOTES MIT

2

A game

1

B

X

2

L R RL

T

E

1

(2,6)

(0,1) (3,2) (-1,3) (1,5)

Sequential Bargaining

• N = {1,2}• X = feasible

expected-utility pairs (x,y ∈ X )

• Ui(x,t) = δitxi

• d = (0,0) ∈ D disagreement payoffs

1

1D

Page 152: 6948772 Game Theory NOTES MIT

3

Timeline – ∞ periodT = {1,2,…, n-1,n,…}If t is odd,

– Player 1 offers some (xt,yt),

– Player 2 Accept or Rejects the offer

– If the offer is Accepted, the game ends yielding δt(xt,yt),

– Otherwise, we proceed to date t+1.

If t is even– Player 2 offers some

(xt,yt),– Player 1 Accept or Rejects

the offer– If the offer is Accepted,

the game ends yielding payoff (xt,yt),

– Otherwise, we proceed to date t+1.

Backward induction

• Can be applied only in perfect information games of finite horizon.

How can we extend this notion to infinite horizon games, or to games with imperfect information?

Page 153: 6948772 Game Theory NOTES MIT

4

A subgame

A subgame is part of a game that can be considered as a game itself.

• It must have a unique starting point;• It must contain all the nodes that follow the

starting node;• If a node is in a subgame, the entire

information set that contains the node must be in the subgame.

A game1 2A

D

α

δ

(4,4) (5,2)

(1,-5)a

d

(3,3)

1

Page 154: 6948772 Game Theory NOTES MIT

5

And its subgames

(1,-5)a

d

(3,3)

1 2 α

δ

(5,2)

(1,-5)a

d

(3,3)

1

A game

1

B

X

2

L R RL

T

E

1

(2,6)

(0,1) (3,2) (-1,3) (1,5)

Page 155: 6948772 Game Theory NOTES MIT

6

Definitions

A substrategy is the restriction of a strategy to a subgame.

A subgame-perfect Nash equilibrium is a Nash equilibrium whose substrategy profile is a Nash equilibrium at each subgame.

Example

1

B

X

2

L R RL

T

E

1

(2,6)

(0,1) (3,2) (-1,3) (1,5)

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7

A “Backward-Induction-like” methodTake any subgame with no proper subgame

Compute a Nash equilibrium for this subgame

Assign the payoff of the Nash equilibrium to the starting node of the subgame

Eliminate the subgame

Any non-terminalnode

Yes

No

The moves computed as a part of any (subgame) Nash equilibrium

Theorem

In a finite, perfect-information game, the set of subgame-perfect equilibria is the set of strategy profiles that are computed via backward induction.

Page 157: 6948772 Game Theory NOTES MIT

8

A subgame-perfect equilibrium?

1

B

X

2

L R RL

T

(2,6)

(0,1) (3,2) (-1,3) (1,5)

Bank Run1

2

1

2

W DW

DWDW

DW

DWDW

W W

W

W

W

(r,r) (D,2r-D) (2r-D,D)

(R,R) (2R-D,D) (D,2R-D) (R,R)

R > D > r > D/2

Page 158: 6948772 Game Theory NOTES MIT

1

Lectures 9 Applications of Subgame-perfect

Equilibrium&

Forward Induction

14.12 Game Theory Muhamet Yildiz

Road Map

1. Applications 1. Tariffs & Intra-industry trade2. Infinite horizon bargaining – Single-

deviation principle2. Forward Induction – Examples3. Finitely Repeated Games4. Quiz

Page 159: 6948772 Game Theory NOTES MIT

2

Single-Deviation principleDefinition: An extensive-form game is continuous at

infinity iff, given any εε > 0, there exists some t such > 0, there exists some t such that, for any two path whose first t arcs are the same, that, for any two path whose first t arcs are the same, the payoff difference of each player is less than the payoff difference of each player is less than εε..

Theorem: Let G be a game that is continuous at infinity. A strategy profile s = (s1,s2,…,sn) is a subgame-perfect equilibrium of G iff, at any information set, where a player i moves, given the other players strategies and given i’s moves at the other information sets, player i cannot increase his conditional payoff at the information set by deviating from his strategy at the information set.

Sequential Bargaining

• N = {1,2}• D = feasible

expected-utility pairs (x,y ∈ D )

• Ui(x,t) = δitxi

• d = (0,0) ∈ D disagreement payoffs

1

1D

Page 160: 6948772 Game Theory NOTES MIT

3

Timeline – ∞ periodT = {1,2,…, n-1,n,…}If t is odd,

– Player 1 offers some (xt,yt),

– Player 2 Accept or Rejects the offer

– If the offer is Accepted, the game ends yielding δt(xt,yt),

– Otherwise, we proceed to date t+1.

If t is even– Player 2 offers some

(xt,yt),– Player 1 Accept or Rejects

the offer– If the offer is Accepted,

the game ends yielding payoff δt(xt,yt),

– Otherwise, we proceed to date t+1.

SPE of ∞-period bargainingTheorem: At any t, proposer offers the other

player δ/(1+δ), keeping himself 1/(1+δ), while the other player accept an offer iff he gets δ/(1+δ).

“Proof:” Single-deviation principle: Take any date t, at which i offers, j accepts/rejects. According to the strategies in the continuation game, at t+1, j will get 1/(1+δ). Hence, j accepts an offer iff she gets at least δ/(1+δ).i must offer δ/(1+δ).

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4

Forward Induction

Strong belief in rationality: At any history of the game, each agent is assumed to be rational if possible. (That is, if there are two strategies s and s’ of a player i that are consistent with a history of play, and if s is strictly dominated but s’ is not, at this history no player j believes that i plays s.)

Table for the bidding game

10060203

-80402

--601

321minbid

Ui = 20(2+2minjbidj – bidi)

Page 162: 6948772 Game Theory NOTES MIT

5

Nash equilibria of bidding game

• 3 equilibria: s1 = everybody plays 1; s2 = everybody plays 2; s3 = everybody plays 3.

• Assume each player trembles with probability ε < 1/2, and plays each unintended strategy w.p. ε/2, e.g., w.p. ε/2, he thinks that such other equilibrium is to be played.– s3 is an equilibrium iff

– s2 is an equilibrium iff

– s1 is an equilibrium iff

0 0.05 0.1 0.15 0.2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0.032

0.047

(1-ε/2)n-1/2

(1-ε)n+(1-ε/2)n-1

Page 163: 6948772 Game Theory NOTES MIT

6

Bidding game with entry feeEach player is first to decide

whether to play the bidding game (E or X); if he plays, he is to pay a fee p > 60.

10060203-80402--601

321Bid

min

For each m =1,2,3, ∃ SPE: (m,m,m) is played in the bidding game, and players play the game iff 20(2+m) ≥ p.Forward induction: when 20(2+m) < p, (Em) is strictly dominated by (Xk). After E, no player will assign positive probability to min bid ≤ m. FI-Equilibria: (Em,Em,Em) where 20(2+m) ≥ p.

What if an auction before the bidding game?

Burning Money

1,30,0S

0,03,1BSB

0,3-1,0S

-1,02,1BSB

10 D

REHTO

DS

DB

0S

0B

SSSBBSBB

Page 164: 6948772 Game Theory NOTES MIT

7

Repeated Games

Entry deterrence

1 2Enter

X

Acc.

Fight

(0,2) (-1,-1)

(1,1)

Page 165: 6948772 Game Theory NOTES MIT

8

Entry deterrence, repeated twice

1 2Enter

X

Acc.

Fight

1 2Enter

X

Acc.

Fight

(1,3) (0,0)

(2,2)

1 2Enter

X

Acc.

Fight

(-1,1) (-2,-2)

(0,0)

1 2 Enter

X

Acc.

Fight

(-1,1)

(1,3)

(0,4)

Page 166: 6948772 Game Theory NOTES MIT

Lectures 10 -11Repeated Games

14.12 Game Theory Muhamet Yildiz

Page 167: 6948772 Game Theory NOTES MIT

Road Map1. Forward Induction – Examples

2. Finitely Repeated Games with observable actions1. Entry-Deterrence/Chain-store paradox2. Repeated Prisoners’ Dilemma3. A general result4. When there are multiple equilibria

3. Infinitely repeated games with observable actions1. Discounting / Present value2. Examples3. The Folk Theorem4. Repeated Prisoners’ Dilemma, revisited –tit for tat5. Repeated Cournot oligopoly

4. Infinitely repeated games with unobservable actions

Page 168: 6948772 Game Theory NOTES MIT

Forward Induction

Strong belief in rationality: At any history of the game, each agent is assumed to be rational if possible. (That is, if there are two strategies s and s’ of a player i that are consistent with a history of play, and if s is strictly dominated but s’ is not, at this history no player j believes that i plays s.)

Page 169: 6948772 Game Theory NOTES MIT

Bidding game with entry fee

10060203-80402--601

321Bid

minEach player is first to decide whether to play the bidding game (E or X); if he plays, he is to pay a fee p > 60.

For each m =1,2,3, ∃SPE: (m,m,m) is played in the bidding game, and players play the game iff 20(2+m) ≥ p.Forward induction: when 20(2+m) < p, (Em) is strictly dominated by (Xk). After E, no player will assign positive probability to min bid ≤ m. FI-Equilibria: (Em,Em,Em) where 20(2+m) ≥ p.

What if an auction before the bidding game?

Page 170: 6948772 Game Theory NOTES MIT

Burning Money

1,30,0S

0,03,1BSB

0,3-1,0S

-1,02,1BSB

0 D1

REHTO

DS

DB

0S

0B

SSSBBSBB

Page 171: 6948772 Game Theory NOTES MIT

Repeated Games

Page 172: 6948772 Game Theory NOTES MIT

Entry deterrence

1 2Enter

X

Acc.

Fight

(0,2) (-1,-1)

(1,1)

Page 173: 6948772 Game Theory NOTES MIT

Entry deterrence, repeated twice, many times

1 2Enter

X

Acc.

Fight

1 2Enter

X

Acc.

Fight

(1,3) (0,0)

(2,2)

1 2Enter

X

Acc.

Fight

(-1,1) (-2,-2)

(0,0)

12 Enter

X

Acc.

Fight

(-1,1)

(1,3)

(0,4)

What would happen if repeated n times?

Page 174: 6948772 Game Theory NOTES MIT

Prisoners’ Dilemma, repeated twice, many times

• Two dates T = {0,1};• At each date the prisoners’ dilemma is played:

• At the beginning of 1 players observe the strategies at 0. Payoffs= sum of stage payoffs.

1,16,0D

0,65,5CDC

Page 175: 6948772 Game Theory NOTES MIT

Twice-repeated PD1

2

1

2

1

2

1

2

1

2

C D

C D C D

C D C D C D C D

C D C D C D C D C D C D C D C D

1010

511

115

66

511

012

66

17

115

66

120

71

66

17

71

22

What would happen if T = {0,1,2,…,n}?

Page 176: 6948772 Game Theory NOTES MIT

A general result

• G = “stage game” = a finite game• T = {0,1,…,n}• At each t in T, G is played, and players remember

which actions taken before t;• Payoffs = Sum of payoffs in the stage game.• Call this game G(T).Theorem: If G has a unique subgame-perfect

equilibrium s*, G(T) has a unique subgame-perfect equilibrium, in which s* is played at each stage.

Page 177: 6948772 Game Theory NOTES MIT

With multiple equilibria

T = {0,1} s* = •At t = 0, each i play Mi;•At t = 1, play (B,R) if (M1,M2) at t = 0, play (T,L) otherwise.

3,30,00,0B

0,04,40,5M1

0,05,01,1T

RM2L21

4,41,11,1B

1,17,71,6M1

1,16,12,2T

RM2L

Page 178: 6948772 Game Theory NOTES MIT

Infinitely repeated Games with observable actions

• T = {0,1,2,…,t,…}• G = “stage game” = a finite game• At each t in T, G is played, and players

remember which actions taken before t;• Payoffs = Discounted sum of payoffs in the

stage game.• Call this game G(T).

Page 179: 6948772 Game Theory NOTES MIT

Definitions

The Present Value of a given payoff stream π = (π0,π1,…,πt,…) is

PV(π;δ) = Σ∞t=1 δtπt = π0 + δπ1 + … + δtπt +…

The Average Value of a given payoff stream π is (1−δ)PV(π;δ) = (1−δ)Σ∞

t=1 δtπt

The Present Value of a given payoff stream π at t isPVt(π;δ) = Σ∞

s=t δs-t πs = πt + δπt+1 + … + δsπt+s +…

Page 180: 6948772 Game Theory NOTES MIT

Infinite-period entry deterrence

1 2 Enter

X

Acc.

Fight

(0,2) (-1,-1)

(1,1) Strategy of Entrant:Enter iff Accomodated before.

Strategy of Incumbent:Accommodate iff accomodated before.

Page 181: 6948772 Game Theory NOTES MIT

Incumbent:• V(Acc.) = VA = 1/(1−δ);• V(Fight) = VF = 2/(1−δ); • Case 1: Accommodated before.

– Fight => -1 + δVA

– Acc. => 1 + δVA.• Case 2: Not Accommodated

– Fight => -1 + δVF

– Acc. => 1 + δVA

– Fight -1 + δVF ≥ 1 + δVA

VF − VA = 1/(1−δ) ≥ 2/δ δ ≥ 2/3.

Entrant: • Accommodated

– Enter => 1+VAE

– X => 0 +VAE

• Not Acc.– Enter =>-1+VFE

– X => 0 +VFE

Page 182: 6948772 Game Theory NOTES MIT

• VD = 1/(1−δ);• VC = 5/(1−δ) = 5VD;• Defected before (easy)• Not defected

– D => – C =>– C

1,16,0D

0,65,5CDC

A Grimm Strategy:Defect iff someone defected before.

Infinitely-repeated PD

Page 183: 6948772 Game Theory NOTES MIT

Tit for Tat

• Start with C; thereafter, play what the other player played in the previous round.

• Is (Tit-for-tat,Tit-for-tat) a SPE?

• Modified: Start with C; if any player plays D when the previous play is (C,C), play D in the next period, then switch back to C.

Page 184: 6948772 Game Theory NOTES MIT

Folk TheoremDefinition: A payoff vector v = (v1,v2,…,vn) is feasible

iff v is a convex combination of some pure-strategy payoff-vectors, i.e.,

v = p1u(a1) + p2u(a2) +…+ pku(ak),where p1 + p2 +…+ pk = 1, and u(aj) is the payoff vector at strategy profile aj of the stage game.

Theorem: Let x = (x1,x2,…,xn) be s feasible payoff vector, and e = (e1,e2,…,en) be a payoff vector at some equilibrium of the stage game such that xi > eifor each i. Then, there exist δ < 1 and a strategy profile s such that s yields x as the expected average-payoff vector and is a SPE whenever δ > δ.

Page 185: 6948772 Game Theory NOTES MIT

Folk Theorem in PD

1,16,0D

0,65,5CDC • A SPE with PV

(1.1,1.1)?– With PV (1.1,5)?– With PV (6,0)?– With PV (5.9,0.1)?

Page 186: 6948772 Game Theory NOTES MIT

Infinitely-repeated Cournot oligopoly

• N firms, MC = 0; P = max{1-Q,0};• Strategy: Each is to produce q = 1/(2n); if any

firm defects produce q = 1/(1+n) forever.• VC =• VD = • V(D|C) = • Equilibrium

Page 187: 6948772 Game Theory NOTES MIT

0 20 40 60 80 1000.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

Page 188: 6948772 Game Theory NOTES MIT

0 200 400 600 800 10000.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

Page 189: 6948772 Game Theory NOTES MIT

IRCD (n=2)

• Strategy: Each firm is to produce q*; if any one deviates, each produce 1/(n+1) thereafter.

• VC = q*(1-2q*)/(1-δ);• VD = 1/(9(1-δ));• VD|C = max q(1-q*-q) +δVD

• Equilibrium iff

( ) ( )δδ−

+−=19

4/*1 2q

( ) ( )( ) 9/4/*11*21* 2 δδ +−−≥− qqq

)9(359*δδ

−−

≥q

Page 190: 6948772 Game Theory NOTES MIT

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

x

yx = δ, y = (3-5/3 δ)/(9-δ )

Page 191: 6948772 Game Theory NOTES MIT

Carrot and StickProduce ¼ at the beginning; at ant t > 0, produce ¼ if both

produced ¼ or both produced x at t-1; otherwise, produce x.

Two Phase: Cartel & PunishmentVC = 1/8(1-δ). Vx = x(1-2x) + δVC.VD|C = max q(1-1/4-q) + δVX = (3/8)2 + δVXVD|x = max q(1-x-q) + δVX = (1-x)2/4 + δVXVC ≥ VD|C VC ≥ (3/8)2 + δ2VC + δ x(1-2x)

(1-δ2) VC - (3/8)2 ≥ δ x(1-2x) (1+δ)/8 - (3/8)2 ≥ δ x(1-2x)VX ≥ VD|C (1-δ)Vx ≥ (1-x)2/4 (1-δ)(x(1-2x) + δ/8(1-δ)) ≥ (1-x)2/4

(1-δ)x(1-2x) + δ/8 ≥ (1-x)2/4 2x2 – x + 1/8 – 9/64δ ≥ 0

(9/4-2δ)x2 – (3-2δ)x +δ/8(1-δ) ≤ 0

Page 192: 6948772 Game Theory NOTES MIT

1

Lectures 12-13Incomplete Information

Static Case

14.12 Game Theory Muhamet Yildiz

Road Map1. Examples2. Bayes’ rule3. Definitions

1. Bayesian Game2. Bayesian Nash Equilibrium

4. Mixed strategies, revisited5. Economic Applications

1. Cournot Duopoly2. Auctions3. Double Auction

Page 193: 6948772 Game Theory NOTES MIT

2

Incomplete information

We have incomplete (or asymmetric) information if one player knows something (relevant) that some other player does not know.

An Example

N ature

High p

Low 1-p

Firm H ireW

W ork

Shirk

D o not hire

(1, 2)

(0, 1)

(0, 0)

D o nothire

H ireW W ork

Shirk

(1, 1)

(-1, 2)

(0, 0)

Page 194: 6948772 Game Theory NOTES MIT

3

The same example

NatureHigh p

Low1-p

Firm

hire

W

Work

Shirk

(1, 2)

(0, 1)

(0, 0)

Do not

Hire

WWork

Shirk

(1, 1)

(-1, 2)

Another Example

Nature

High 0.5

Low .5

Seller p

B

Buy

Buy

Don’t

(p, 2-p)

(0, 0)

(0, 0)

p’

pB Buy

Don’t

(p, 1-p)

(p’, 1-p’)

(0, 0)

p’

Don’t

Don’t

Buy

B

B

(p’,2-p’)

(0,0)What would you askif you were to choosep from [0,4]?

Page 195: 6948772 Game Theory NOTES MIT

4

Same “Another Example”

Nature

High 0.5

Low .5

Seller

p

B

Buy

Buy

Don’t

(p, 2-p)

(0, 0)

(0, 0)

p’B Buy

Don’t

(p, 1-p)

(p’, 1-p’)

(0, 0)

Don’t

Don’t

Buy

B

B

(p’,2-p’)

(0,0)What would you askif you were to choosep from [0,4]?

Nature

High 0.5

Low .5

Bayes’ Rule

Prob(A and B)• Prob(A|B) =

Prob(B)• Prob(A and B) = Prob(A|B)Prob(B) = Prob(B|A)Prob(A)

Prob(B|A)Prob(A)• Prob(A|B) =

Prob(B)

Page 196: 6948772 Game Theory NOTES MIT

5

Example

• Prob(Work|Success) = µp/[µp + (1−µ)(1-p)]

• Prob(Work|Failure) = (1-µ)p/[µ(1−p) + (1−µ)p]

Success

Failure

Workµ

Shirk1−µ

p

1-p

1-p

p

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

µ

P(w

|S),P

(W|F

)

P(W|S)

P(W|F)

Page 197: 6948772 Game Theory NOTES MIT

6

Bayesian Game (Normal Form)

A Bayesian game is a list G = {A1,…,An;T1,…,Tn;p1,…,pn;u1,…,un}where

• Ai is the action space of i (ai in Ai)• Ti is the type space of i (ti)• pi(t-i|ti) is i’s belief about the other players• ui(a1,…,an;t1,…,tn) is i’s payoff.

An Example

Nature

High p

Low 1-p

Firm HireW

Work

Shirk

Do not hire

(1, 2)

(0, 1)

(0, 0)

Do nothire

HireW Work

Shirk

(1, 1)

(-1, 2)

(0, 0)

TFirm={tf};TW = {High,Low}AFirm = {Hire, Don’t}AW = {Work,Shirk}pF(High) = ppF(Low) = 1-p

Page 198: 6948772 Game Theory NOTES MIT

7

Bayesian Nash equilibrium

A Bayesian Nash equilibrium is a Nash equilibrium of a Bayesian game.

Given any Bayesian game G = {A1,…,An;T1,…,Tn;p1,…,pn;u1,…,un}

a strategy of a player i in a is any function si:Ti → Ai;A strategy profile s* = (s1

*,…, s1*) is a Bayesian Nash

equilibrium iff si*(ti) solves

i.e., si* is a best response to s-i

*.

( ) ( ) ( ) ( )( ) ( )iiiTt

nniiiiiiAa

ttpttstsatstsuiiii

|;,...,,,,..., *1

*11

*11

*1max −

∈++−−

∈∑

−−

An Example

Nature

High p

Low 1-p

Firm HireW

Work

Shirk

Do not hire

(1, 2)

(0, 1)

(0, 0)

Do nothire

HireW Work

Shirk

(1, 1)

(-1, 2)

(0, 0)

TFirm={tf};TW = {High,Low}AFirm = {Hire, Don’t}AW = {Work,Shirk}pF(High) = p >1/2pF(Low) = 1-p

sF* = Hire, sF* (High) = WorksF* (Low) = Shirk

Another equilibrium?

Page 199: 6948772 Game Theory NOTES MIT

8

Stag Hunt, Mixed Strategy

(6,6)(0,4)

(4,0)(2,2)

Mixed Strategies• t and v are iid with uniform

distribution on [−ε,ε].• t and v are privately known by

1 and 2, respectively, i.e., are types of 1 and 2, respectively.

• Pure strategy: – s1(t) = Rabbit iff t > 0;– s2(v) = Rabbit iff t > 0.

• p = Prob(s1(t)=Rabbit|v) = Prob(t > 0) = 1/2.

• q = Prob(s2(v)=Rabbit|t) = 1/2.

6,60,4+v

4+t,02+t,2+v

U1(R|t) = t +2q+4(1-q) = t + 4 – 2qU1(S|t) = 6(1-q);U1(R|t) > U1(S|t) � t+4–2q > 6(1-q)� t > 6-6q+2q-4 = 2 – 4q = 0.

Page 200: 6948772 Game Theory NOTES MIT

Economic Applicationswith Incomplete

Information14.12 Game Theory

Muhamet Yildiz

Road Map1. Cournot duopoly with Incomplete Informa-

tion

2. A �rst price auction3. A double auction

4. Quiz

1

Page 201: 6948772 Game Theory NOTES MIT

1 Cournot duopoly withIncomplete Information

� Demand:� ��� � ���

where � � �� � ��.

� The marginal cost of Firm 1 = �� commonknowledge.

� Firm 2’s marginal cost:�� with probaility �,�� with probaility �� ��

its private information.

� Each �rm maximizes its expected pro�t.

2

Page 202: 6948772 Game Theory NOTES MIT

1.1 Bayesian Nash EquilibriumFirm 2 of high type:�����

�� � ����� � �����

�� ��� � �� � �� ��

������� ��� ��� � ��

�(*)

Firm 2 of low type:

�����

�� ��� � �� � �� ��,

������� ��� ��� � ��

� (**)

Firm 1:

�����

� �� �� � �������� � ��

� ��� �� �� �� � �������� � ��

��� �� �� �������� � � ��� �� �� �������� �

�(***)

3

Page 203: 6948772 Game Theory NOTES MIT

Solve *, **, and *** for ��� ������� �

������.�

� �����������������

�� �

�� � � �� �� � �� � �

�����

� �� ��� ���� ��

��

������� ��� ��� � �

��

��� ����� � ���

������� ��� ��� � �

�� ���� � ���

��� ��� �� � ��� � ��� ����

4

Page 204: 6948772 Game Theory NOTES MIT

2 A First-price Auction

� One object, two bidders

� �� � Valuation of bidder i� iid with uniformdistribution over � �.

� Simultaneously, each bidder � submits a bid �, then the highest bidder wins the objectand pays her bid.

� The payoffs:

��� � � �� ��� �

�� � � if � � ���� �� if � � �

� if � � �

� Objective of �:��� �

� ��� � � �� ���

where

� �� � ��� � �� ��� � � �������

���� � �� ��� � � ������

5

Page 205: 6948772 Game Theory NOTES MIT

2.1 Symmetric, linear equilib-rium

� � �� ���

Then,��� � � ������ � �

Equilibrium condition: � � � � ��.Hence,

� �� � ��� � �� ��� � � � � ����� ��� � �� ����� � � � �

��

� ��� � �� � � � �

FOC:

� �

������ if �� � �

� if �� � �(1)

Therefore,

� ��

���

6

Page 206: 6948772 Game Theory NOTES MIT

2.2 Any symmetric equilibrium

� � ����

Hence,

� �� � ��� � �� ��� � � ������ ��� � �� ����� � �� � ���� ��� � ��

�� � ��

FOC (��� � � �):

� �� � �� � ��� � ��� �� � ��

� �� �

��� � ��� � ������

����� �

���� �� � ���� � ��� ���� ��

���� ��

���� �� � ��� �� � �����

���� � ���� � ��������

��� � �, =� cons = 0. ���� � ����

7

Page 207: 6948772 Game Theory NOTES MIT

3 Double Auction1. Simultaneously, Seller names �� and Buyer

names � .

a. If � � ��, then no trade�

b. if � � ��, trade at price � � ������ .

2. Valuations are private information: � , ��iid w/ uniform on � �.

3. Payoffs:

� �

�� � �����

� if � � ��� otherwise

�� �

������� � �� if � � ��

� otherwise4. The buyer’s problem:

�����

� � � � ������

�� � � ������

5. The seller’s problem:

�����

�� � � �� �

�� �� � � �� � � ��

8

Page 208: 6948772 Game Theory NOTES MIT

An Equilibrium:

� �

�� if � � �� otherwise

�� �

�� if �� � �� otherwise

Trade

X

0

v bv b /v s

Efficient

not to trade

v S

Inefficientlack of trade

Trade

X

0

v bv b /v s

Efficient

not to trade

v S

Inefficientlack of trade

9

Page 209: 6948772 Game Theory NOTES MIT

Equilibrium with linear strategies:� � � � � � �� � �� � ����

� � �� ���� � �� � ���� � �� � � � ����

�� � � �� � � � � � � � � � �� � � �

10

Page 210: 6948772 Game Theory NOTES MIT

� � � �

� � � � ������

�� � � ������

� �������

� � � � ������

����

� �������

�� � � � �� � ����

����

�� � ��

��

�� � � � ��

�� ��

� �������

�����

�� � ��

��

�� � � � ��

�� ��

�� � ��

��

��

�� � ��

��

�� � � � ��

�� � � ��

�� � ��

��

�� � �� � ��

F.O.C. (����� � � ):�

��

�� � �� � ��

�� ��� � ���

���� �

i.e.,

� ��

�� �

��� (2)

11

Page 211: 6948772 Game Theory NOTES MIT

Similarly,

� �� � �

�� � � �� �

�� �� � � �� � � ��

� �

�������

��� � � � � � �

� ��

���

��� �� � �

���� � � �

� ��

��� �

� �

�������

� ��

��� �� � �

���� � � �

� ��

�� �

���

��� � �

���

��� �� � �

� �� � �

�� �� �

� ���� � �

��� �� � �

� ��� � �

�� �� �

� �

12

Page 212: 6948772 Game Theory NOTES MIT

F.O.C. (����� � ��):

� �

��� � �

�� �� �

��

��� �� � �

�� �

� ��� � �

�� �� �

� �

��

��� � ��� � � �� � �

����

� �� �� �� � �

��

��� � � � � �� �

� � � �

i.e.,

�� ��

��� �

� � � �

(3)

� � � ����

� �� � � �� � ���

� Hence, ��� � �� � �,

� �� � ���� � � ����.

� ��

�� �

��(4)

�� ��

��� �

� (5)

13

Page 213: 6948772 Game Theory NOTES MIT

We have trade iff� � ��

iff�

�� �

��� �

��� �

�iff

� � �� � �

��

�� �

��

��

��

pb

ps

1/4

1/12

3/4

pb

ps

1/4

1/12

3/4

14

Page 214: 6948772 Game Theory NOTES MIT

1

Lectures 15-18Dynamic Games with

Incomplete Information

14.12 Game Theory Muhamet Yildiz

Road Map

1. Examples2. Sequential Rationality3. Perfect Bayesian Nash Equilibrium4. Economic Applications

1. Sequential Bargaining with incomplete information

2. Reputation

Page 215: 6948772 Game Theory NOTES MIT

2

An Example

Nature

High 0.7

Low 0.3

Firm HireW

Work

Shirk

Do nothire

(1, 2)

(0, 1)

(0, 0)

Do nothire

HireW Work

Shirk

(1, 1)

(-1, 2)

(0, 0)

What is wrong with this equilibrium?

Nature

High .7

Low .3

Firm HireW

Work

Shirk

Do nothire

(1, 2)

(0, 1)

(0, 0)

Do nothire

HireW Work

Shirk

(1, 1)

(-1, 2)

(0, 0)

Page 216: 6948772 Game Theory NOTES MIT

3

What is wrong with this equilibrium?

1

B

X

2

L R RL

T

(2,6)

(0,1) (3,2) (-1,3) (1,5)

Beliefs

• Beliefs of an agent at a given information set is a probability distribution on the information set.

• For each information set, we must specify the beliefs of the agent who moves at that information set.

1

B

X

2

L R RL

T

(2,6)

(0,1) (3,2) (-1,3) (1,5)

µ 1−µ

Page 217: 6948772 Game Theory NOTES MIT

4

Sequential Rationality

A player is said to be sequentially rationaliff, at each information set he is to move, he maximizes his expected utility given his beliefs at the information set (and given that he is at the information set) – even if this information set is precluded by his own strategy.

An Example

Nature

High 0.7

Low 0.3

Firm HireW

Work

Shirk

Do nothire

(1, 2)

(0, 1)

(0, 0)

Do nothire

HireW Work

Shirk

(1, 1)

(-1, 2)

(0, 0)

Page 218: 6948772 Game Theory NOTES MIT

5

Another example

1

B

X

2

L R RL

T

(2,6)

(0,1) (3,2) (-1,3) (1,5)

Example

1

B

2

L R RL

T

(0,10) (3,2) (-1,3) (1,5)

.9.1

Page 219: 6948772 Game Theory NOTES MIT

6

“Consistency”

Definition: Given any (possibly mixed) strategy profile s, an information set is said to be on the path of play iff the information set is reached with positive probability if players stick to s.

Definition: Given any strategy profile s and any information set I on the path of play of s, a player’s beliefs at I is said to be consistent with s iff the beliefs are derived using the Bayes’ rule and s.

Example

1

B

2

L R RL

T

(0,10) (3,2) (-1,3) (1,5)

Page 220: 6948772 Game Theory NOTES MIT

7

Example

2

B

X

3

L R RL

T

E

1

200

121

333

012

011

“Consistency”

• Given s and an information set I, even if I is off the path of play, the beliefs must be derived using the Bayes’ rule and s “whenever possible,” e.g., if players tremble with very small probability so that I is on the path, the beliefs must be very close to the ones derived using the Bayes’ rule and s.

Page 221: 6948772 Game Theory NOTES MIT

8

Example

2

B

X

3

L R RL

T

E

1

200

121

333

012

011

Perfect Bayesian Nash Equilibrium

A Perfect Bayesian Nash Equilibrium is a pair (s,b) of strategy profile and a set of beliefs such that

1. Strategy profile s is sequentially rational given beliefs b, and

2. Beliefs b are consistent with s.

Nash Subgame-perfectBayesian Nash Perfect Bayesian

Page 222: 6948772 Game Theory NOTES MIT

9

Example

2

B

X

3

L R RL

T

E

1

200

121

333

012

011

Beer – Quiche11

01

beer

beer

quiche

quiche

duel

don’t

duel

don’t

don’t

duel

don’t

duel

30

00

21

20

10

31

tw

ts

{.1}

{.9}

Page 223: 6948772 Game Theory NOTES MIT

10

Example1 2 1

(4,4) (5,2) (3,3)

(1,-5)

1 2 1

(-1,4) (0,2) (-1,3)

(0,-5)

.9

.1

Sequential Bargaining

1. 1-period bargaining – 2 types2. 2-period bargaining – 2 types3. 1-period bargaining – continuum4. 2-period bargaining – continuum

Page 224: 6948772 Game Theory NOTES MIT

11

Sequential bargaining 1-p• A seller S with valuation 0• A buyer B with valuation v;

– B knows v, S does not– v = 2 with probability π– = 1 with probability 1-π

• S sets a price p ≥ 0;• B either

– buys, yielding (p,v-p)– or does not, yielding (0,0).

SH L

p p

Y N Y N

p2-p

00

p1-p

00

Solution

1. B buys iff v ≥ p;1. If p ≤ 1, both types buy:

S gets p.2. If 1 < p ≤ 2, only H-type

buys: S gets πp.3. If p > 2, no one buys.

2. S offers • 1 if π < ½,• 2 if π > ½.

1 2

1

Page 225: 6948772 Game Theory NOTES MIT

12

Sequential bargaining 2-period

• A seller S with valuation 0

• A buyer B with valuation v;– B knows v, S does not– v = 2 with probability π– = 1 with probability 1-π

1. At t = 0, S sets a price p0 ≥ 0;

2. B either – buys, yielding (p0,v-p0)– or does not, then

3. At t = 1, S sets a price p0 ≥ 0;

4. B either – buys, yielding (δp0,δ(v-p0))– or does not, yielding (0,0)

Solution, 2-period

1. Let µ = Pr(v = 2|history at t=1).2. At t = 1, buy iff v ≥ p;3. If µ > ½, p1 = 2 4. If µ < ½, p1 = 1.5. If µ = ½, mix between 1 and 2.6. B with v=1 buys at t=0 if p0 ≤ 1.7. If p0 > 1, µ = Pr(v = 2|p0,t=1) ≤ π.

Page 226: 6948772 Game Theory NOTES MIT

13

Solution, cont. π <1/2

1. µ = Pr(v = 2|p0,t=1) ≤ π <1/2. 2. At t = 1, buy iff v ≥ p;3. p1 = 1. 4. B with v=2 buys at t=0 if

(2-p0) ≥ δ(2−1) = δ p0 ≤ 2−δ.5. p0 = 1:π(2−δ) + (1−π)δ = 2π(1−δ) + δ < 1−δ+δ = 1.

Solution, cont. π >1/2

• If v=2 is buying at p0 > 2−δ, then– µ = Pr(v = 2|p0 > 2−δ,t=1) = 0;– p1 = 1;– v = 2 should not buy at p0 > 2−δ.

• If v=2 is not buying at 2> p0 > 2−δ, then– µ = Pr(v = 2|p0 > 2−δ,t=1) = π > 1/2;– p1 = 2;– v = 2 should buy at 2 > p0 > 2−δ.

• No pure-strategy equilibrium.

Page 227: 6948772 Game Theory NOTES MIT

14

Mixed-strategy equilibrium, π >1/2

1. For p0 > 2−δ, µ(p0) = ½;2. β(p0) = 1- Pr(v=2 buys at p0)

3. v = 2 is indifferent towards buying at p0:2- p0 = δγ(p0) γ(p0) = (2- p0)/δ

where γ(p0) = Pr(p1=1|p0).

.1)(1)(21

)1()()(

000

0

ππβππβ

ππβπβµ −

=⇔−=⇔=−+

= ppp

p

Sequential bargaining, v in [0,1]

• 1 period:– B buys at p iff v ≥ p;– S gets U(p) = p Pr(v ≥ p);– v in [0,a] => U(p) = p(a-p)/a;– p = a/2.

Page 228: 6948772 Game Theory NOTES MIT

15

Sequential bargaining, v in [0,1]• 2 periods: (p0,p1)

– At t = 0, B buys at p0 iff v ≥ a(p0);– p1 = a(p0)/2;– Type a(p0) is indifferent:

a(p0) – p0 = δ(a(p0) – p1 ) = δa(p0)/2a(p0) = p0/(1-δ/2)

• S gets

• FOC:

20

00

22/11

−+

−−

δδppp

( )( )4/312

2/1022

2/121

2

000

δδ

δδ −−

=⇒=−

+−

− ppp

Reputation

Page 229: 6948772 Game Theory NOTES MIT

16

Centipede Game

100100

98101

9999

97100

9898

11

03

22

1 2 1 1 2 1 2

Centipede Game – with doubt

100100

98101

9999

97100

9898

11

03

9699

1 2 2 1 2 1 2

0100

0101

-199

0100

-198

-11

03

099

1 2 2 1 2 1 2{.001}

{.999} 12345197 = n

µ1µ3µ5

Page 230: 6948772 Game Theory NOTES MIT

17

Facts about the Centipede

• Every information set of 2 is reached with positive probability.

• 2 always goes across with positive probability.• If 2 strictly prefers to go across at n, then

– she must strictly prefer to go across at n+1, – her posterior at n is her prior.

• For any n > 2, 1 goes across with positive probability. If 1 goes across w/p 1 at n, then 2’s posterior at n-1 is her prior.

If 2’s payoff at any n is x and 2 is mixing, then x = µn(x+1) + (1- µn)[(x-1)pn +(1-pn)(x+1)]

= µn(x+1) + (1- µn)[ (x+1) -2pn]= x+1 - 2pn(1- µn)

(1- µn) pn= 1/2

( )( ) ( ) ( ) nnnnn

n

nnn

nn pp

µµµµ

µµµµµ 2

11111 =−−−+

=−−+

=−

21−= n

nµµ

Page 231: 6948772 Game Theory NOTES MIT

18

0 20 40 60 80 1000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Go Across

Mix

Page 232: 6948772 Game Theory NOTES MIT

1

Lecture 19-21

14.12 Game TheoryMuhamet Yildiz

Road Map

• Market for Lemons (Adverse Selection)• Insurance Market (Screening)• Signaling – theory• Job Market Signaling

Page 233: 6948772 Game Theory NOTES MIT

2

Example for Adverse selection

• A buyer and a seller, who owns an object.• The value of the object is

– v for the seller,– v+b for the buyer.

• v is uniformly distributed on [0,1]. Seller knows v. b > 0 is a known constant.

• Buyers offers a price p; seller decides on whether to sell.

Solution

• Seller sells at p iff p ≥ v;• Buyer’s payoff:

• Buyer offers p = 2b.

( ) ( )

−=−+=−+= ∫ 22

1)( 2

0

pbppbppdvpbvpUp

Page 234: 6948772 Game Theory NOTES MIT

3

Example for Adverse selection

• A buyer and a seller, who owns an object.• The value of the object is

– v for the seller,– 3v/2 for the buyer.

• v is uniformly distributed on [0,1]. Seller knows v.

• Buyers offers a price p; seller decides on whether to sell.

Solution

• Seller sells at p iff p ≥ v;• Buyer’s payoff:

• Buyer offers p = 0.

4/22

323)( 2

0

ppppdvpvpUp

−=

−=

−= ∫

Page 235: 6948772 Game Theory NOTES MIT

4

Market for Lemons

• Two types of cars: Lemons and Peaches.– A Peach is worth $2500 to seller, $3000 to buyer;– A lemon is worth $1000 to seller, $2000 to buyer;

• Each seller knows whether his car is a Peach or Lemon; buyers cannot tell.

• There are 200 Lemons and 100 Peaches.• Equilibrium: a market clearing price p.

Demand/Supply – Market for Lemons

Q

P

1000

200 300

Supply

2000

Page 236: 6948772 Game Theory NOTES MIT

5

Quiz• The students are grouped in pairs;• In each pair, one is assigned to be seller, the other one

is buyer. The buyer is given a valuation v, an integer between 20 and 40 with uniform distribution.– Buyer offers a price p; – seller accepts or rejects;– If seller accepts, seller gets p, buyer gets 20+v-p.– If seller rejects, w/probability 0.1 each gets 20; w/p

0.95 we proceed to next date, and– Seller offers a price p;– If buyer accepts, seller gets p, buyer gets 20+v-p;

otherwise each gets 20.

Insurance with adverse selection• Two states: s1, s2.• A risk averse agent with

– risky endowment (Y1,Y2) where Y1 > Y2, and – Utility function u.

• Two types:– High risk: U(y1,y2) = (1−πH)u(y1) + πHu(y2)– Low risk: U(y1,y2) = (1−πL)u(y1) + πLu(y2), where πH > πL.

• A risk neutral insurance company offers a menu ((y1H,y2H), (y1L,y2L)) of insurance policies, and the risk averse agent either chooses one of the policies in the menu or rejects the offer.

• Agent knows his type, the company does not.

Page 237: 6948772 Game Theory NOTES MIT

6

High-risk Agent

Y1

Y2

Low-risk Agent

Y1

Y2

Page 238: 6948772 Game Theory NOTES MIT

7

High- and Low-risk Agent

Y1

Y2

Optimal menu

Y1

Y2

Page 239: 6948772 Game Theory NOTES MIT

8

Signaling Games

Beer – Quiche11

01

beer

beer

quiche

quiche

duel

don’t

duel

don’t

don’t

duel

don’t

duel

30

00

21

20

10

31

tw

ts

{.1}

{.9}

Page 240: 6948772 Game Theory NOTES MIT

9

Signaling Game -- Definition

• Two Players: (S)ender, (R)eceiver1. Nature selects a type ti from T = {t1,…,tI}

with probability p(ti);2. Sender observes ti, and then chooses a

message mj from M = {m1,…,mI};3. Receiver observes mj (but not ti), and then

chooses an action ak from A = {a1,…,aK};4. Payoffs are US(ti,mj,ak) and US(ti,mj,ak).

Beer – Quiche11

01

beer

beer

quiche

quiche

duel

don’t

duel

don’t

don’t

duel

don’t

duel

30

00

21

20

10

31

tw

ts

{.1}

{.9}

Page 241: 6948772 Game Theory NOTES MIT

10

Types of Equilibria

• A pooling equilibrium is an equilibrium in which all types of sender send the same message.

• A separating equilibrium is an equilibrium in which all types of sender send different messages.

• A partially separating/pooling equilibrium is an equilibrium in which some types of sender send the same message, while some others sends some other messages.

A Pooling equilibrium11

01

beer

beer

quiche

quiche

duel

don’t

duel

don’t

don’t

duel

don’t

duel

30

00

21

20

10

31

tw

ts

{.1}

{.9}

Page 242: 6948772 Game Theory NOTES MIT

11

A Separating equilibrium11

01

beer

beer

quiche

quiche

duel

don’t

duel

don’t

don’t

duel

don’t

duel

30

00

21

0.50

10

31

tw

ts

{.1}

{.9}

A Mixed equilibrium11

01

beer

beer

quiche

quiche

duel

don’t

duel

don’t

don’t

duel

don’t

duel

30

00

21

20

10

31

tw

ts

{.1}

{.9}

Page 243: 6948772 Game Theory NOTES MIT

12

Job Market Signaling

Model• A worker

– with ability t = H or t = L (his private information) – Pr(t = H) = q,

– obtains an observable education level e, – incurring cost c(t,e) where c(H,e) < c(L,e), and – finds a job with wage w(e), where he– produces y(t,e).

• Firms compete for the worker: in equilibrium,w(e) = µ(H|e)y(H,e) + (1– µ(H|e))y(L,e).

Page 244: 6948772 Game Theory NOTES MIT

13

Equilibrium

(eH, eL, w(e), µ(H|e)) where• et = argmaxe w(e) – c(t,e) for each t;• w(e) = µ(H|e)y(H,e) + (1– µ(H|e))y(L,e);

qPr(eH = e)• µ(H|e) =

qPr(eH = e) + (1-q)Pr(eL = e)whenever well-defined.

If t were common knowledge

w(e) = y(t,e)

e*(t)e

w

Page 245: 6948772 Game Theory NOTES MIT

14

No need to imitate

y(L,e)

e*(L) e*(H)

y(H,e)

e

w

No need to imitate

y(L,e)

e*(L) e*(H)

y(H,e)

w(.)

e

w

Page 246: 6948772 Game Theory NOTES MIT

15

want to imitate

y(L,e)

e*(L) e*(H)

y(H,e)

e

w

A pooling equilibrium

y(L,e)

e*(L) e*(H)

y(H,e)

e

w

qy(H,e)+(1-q)y(L,e)

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16

A separating equilibrium

y(L,e)

eL=e*(L) e*(H)

y(H,e)

e

w

eH

An intuitive separating equilibrium

y(L,e)

eL=e*(L)

y(H,e)

e

w

eH

Page 248: 6948772 Game Theory NOTES MIT

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Page 250: 6948772 Game Theory NOTES MIT

14.12 Solutions for Homework 1Kenichi Amaya1

September 28, 2001

Question 1(a)

Remember that a binary relation � is a preference relation if it is completeand transitive

Completeness For all x, y ∈ X, x � y or y � x.

Transitivity If x � y and y � z, then x � z.

The relation defined here is complete, but it is not transitive. We can show thisby a counterexample. Suppose x = 0, y = .5, and z = 1. Then, x � y becausex = 0 = y − 1

2 , and y � z because y = 12 = z − 1

2 , but we don’t have x � zbecause x = 0 < 1

2 = z − 12 . Therefore, � is not a preference relation.

(b)

x � y⇔ x ≥ y − 1/2 and y < x− 1/2

⇔ x > y + 1/2

Suppose x � y and y � z. Then, x > y + 1/2 and y > z + 1/2. This impliesthat

x > y + 1/2 > (z + 1/2) + 1/2 > z + 1/2.

Therefore � is transitive.We can show that ∼ is not transitive by using the same counterexample used

in (a). Suppose x = 0, y = .5, and z = 1. Since x = 0 = y − 12 , x � y and

y � x, therefore x ∼ y. Since y = 12 = z − 1

2 , y � z and z � y, therefore y ∼ z.However, we don’t have x � z because x = 0 < 1

2 = z − 12 . Therefore x ∼ z is

not satisfied.

(c)

If we had x, y ∈ X = {0, 1, 2, · · ·},

x � y ⇔ x ≥ y − 1/2⇔ x ≥ y.

Then both completeness and transitivity are satisfied, so � is a preference rela-tion.

1E52-303, [email protected], 253-3591

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Question 2

Since the agent is risk neutral, his von-Neuman Morgenstern utility function isrepresented by the amount of money he receives (or any affine transformationof it).

(a)

If he doesn’t buy the security, the amount of money he has at date 1 is 0 forsure, so the expected utility is 0. If he buys the security at price p, the amountof money he has at date 1 is 100− p, 50− p, or −p, each with probability 1/3,and thus the expected utility is

13

(100− p) +13

(50− p) +13

(−p).

He wants to buy this security if and only if this expected utility is higher thanor equal to 0, the expected utility from not buying.

13

(100− p) +13

(50− p) +13

(−p) ≥ 0⇔ p ≤ 50.

Therefore πS = 50.

(b)

If the agent exercises the option, he will pay K and receive the dividend d,so he will receive net payment of d −K. If he doesn’t exercise the option, hereceives 0. Therefore he will excercise if and only if d − K ≥ 0. His utility isrepresented by

u(d) = max{0, d−K} − p,where p is the price of the option.

(c)

Case 1 K > 100

In this case, the agent will never exercise the option, so he ends up withreceiving 0 no matter what the dividend is. Therefore he wants to pay forthis option no more than 0.

Case 2 50 < K ≤ 100

If d = 100, the agent will exercise the option and receive net payment of100−K. However, if d = 0 or 50, he won’t exercise the option and receive0. Therefore, he wants to buy this option if

13

(100−K − p) +23

(−p) ≥ 0

⇔ p ≤ 13

(100−K).

2

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Case 3 0 ≤ K ≤ 50

If d = 50 or 100, the agent will exercise the option and receive net paymentof d −K. However, if d = 0, he won’t exercise the option and receive 0.Therefore, he wants to buy this option if

13

(100−K − p) +13

(50−K − p) +13

(−p) ≥ 0

⇔ p ≤ 50− 23K.

To summarise,

πO =

0 if K > 10013 (100−K) if 50 < K ≤ 10050− 2

3K if 0 ≤ K ≤ 50

Question 3

Here is an example.

L Rs 4,0 0,0s’ 0,0 6,0s” 1,1 1,1

The mixed strategy 12s + 1

2s′ gives player 1 expected utility of 2 if player 2

plays l and expected utility of 3 if player 2 plays r.

Question 4

The intuition tells us only 0 would be selected, but actually, all strategiesexcept 100 are rationalizable! The reason is that a player’s objective is justto win the game, i.e., to name closer number to one third of the average thaneveryone else, and not to name exactly one third of the average. For example,if everybody else is naming 100, you can win by naming any number below 100,even 99 wins. The proof is given in the answer 1.

But if, not like this game, a player’s objective is to name exactly one thirdof the average, our first intuition works. The proof for this case is given in theanswer 2.

Answer 1

If 0 ≤ x < 100, x is the best response to all other players’ playing x′, wherex < x′ < 100 and x′ is close enough to x. To see this, if a player plays x and allothers plays x′, x/3 < x (because x′ is close enough to x), therefore she is the

3

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only winner. Since she can do no better than being the only winner, x is thebest response.

For any 0 ≤ x < 100, we can find a sequence x < x1 < x2 < · · · < 100 suchthat x is a best response to everyone else’s playing x1, x1 is a best response toeveryone else’s playing x2, and so on.

Naming 100 is not a best response to anything. The only case a player winsis when everyone is naming 100, but even in this case, she can be the only winnerby naming a smaller number.

To conclude, x is rationalizable if and only if 0 ≤ x < 100.

Answer 2

Suppose the award is 100 − |x − x3 |, so that a player always want to name

exactly x3 .

Given that 0 ≤ xi ≤ 100 for all i, x3 always lies between 0 and 33.33 · · ·. So

x > 33.33 · · · is never best response, i.e., it is not rationalizable.As long as everyone is naming 0 ≤ xi ≤ 33.33 · · ·, x

3 always lies between 0and 11.11 · · ·. So x > 11.11 · · · is never best response, i.e., it is not rationalizable.

Repeating this procedure forever, any number greater than 0 is not ratio-nalizable.

If everyone else is naming 0, the best response is to name 0 as well. Therefore0 is rationalizable.

To conclude, x = 0 is the only rationalizable strategy.Bonus: In the discussion above, the number of players played no role at all.

Question 5(a)

l rLλΛ 2,2 2,2LλP 2,2 2,2Lλu 2,2 2,2LρΛ 2,2 2,2LρP 2,2 2,2Lρu 2,2 2,2RλΛ 1,3 3,3RλP 1,3 1,1Rλu 1,3 0,0RρΛ 3,1 3,3RρP 3,1 1,1Rρu 3,1 0,0

(b)

4

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First, RρΛ weakly dominates all other strategies of player 1. We eliminate themand then we have the following game.

l rRρΛ 3,1 3,3

Now, l is weakly dominated by r, so we eliminate l and we have

rRρΛ 3,3

5

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�� �������� ��� ��������� �����

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3� - ���# �� � ������� �� �� � ���������� 4� �� ������ ��������� ���� ���� ������ ���������������' ������ ��� ��� ����( %� ��� ������� ���� ����� ��� ����� %��� �+���'�4� � ������ ���� � ���� �� #���� � ��� ������%��� � ������� #�'��� ��� %���( ��� #�'�5���� %�

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Page 256: 6948772 Game Theory NOTES MIT

14.12 Solutions for Homework 2Kenichi Amaya1

October 5, 2001

Question 1(a)

L RT 1,1 1,0B 0,1 0,10000

Fact: The set of rationalizable strategies is same as the set of strategieswhich survives iterated elimination of strictly dominated strategies.

Step 1: For player 1, B is strictly dominated by T. Now we have

L RT 1,1 1,0

Step 2: For player 2, R is strictly dominated by L. Now we have

LT 1,1

Therefore, the rationalizable strategies are T for player 1 and L for player 2.

(b)

There was a typo in the problem. If player 2 intends to play L, L is playedwith probability 1 − ε, and R is played with probability ε. Of course, if youinterpreted the problem differently, you will get credit as long as you are solvingthe problem consistently with your interpretation.

We can write a new game which represents this situation as following: Welook for the players’ expected payoffs as functions of the intended play. If player1 intends to play T and player intends to play L, what are the expected payoffs?(T,L) is realized with probability (1 − ε)2, (T,R) and (B,L) are realized eachwith probability ε(1 − ε), and (B,R) is realized with probability ε2. Therefore,the expected payoff of player 1 is

(1− ε)2 · 1 + ε(1− ε) · 1 + ε(1− ε) · 0 + ε2 · 0 = 1− ε = .999.

In the same way, player 2’s expected payoff is

(1− ε)2 · 1 + ε(1− ε) · 0 + ε(1− ε) · 1 + ε2 · 10000 = (1− ε) + ε2 · 10000 = 1.009

Therefore we must have1E52-303, [email protected], 253-3591

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L RT .999, 1.009B

In the same way, we can calculate expected payoffs for the remaining of thepayoff matrix and we have

L RT .999, 1.009 .999, 9.991B .001, 10.999 .001, 9980.01

For player 1, B is strictly dominated by T, and for player 2, L is strictlydominated by R. Now we have

RT .999, 9.991

Therefore, the rationalizable strategies are T for player 1 and R for player2.

(c)

For player 1, the possibility that opponent trembles doesn’t affect her optimalchoice because T gives higher payoff than B no matter what the opponent isdoing.

For player 2, in the case of (a), the large payoff differnce between (B,L)and (B,R) didn’t matter because he was sure B is never played. However,when we introduce trembles, this large payoff difference matters with positiveprobability even though player 2 knows that player 1 never intends to play B.In this situation, the payoff difference between (T,L) and (T,R) is so small thatit is outweighed by the payoff differnce between (B,L) and (B,R).

Question 2

Fact All the Nash equilibrium strategies are rationalizable, and thereforesurvives iterated elimination of strictly dominated strategies. When we look forNash equilibria, we can first eliminate strictly dominated strategies and thenfind Nash equilibria of the game left.

L M RA 3,1 0,0 1,0B 0,0 1,3 1,1C 1,1 0,1 0,10

Step 1 For player 1, C is strictly dominated by 12A+ 1

2B. Now we have

L M RA 3,1 0,0 1,0B 0,0 1,3 1,1

2

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Step 2 For player 2, R is stricly dominated by 12L+ 1

2M . Now we have

L MA 3,1 0,0B 0,0 1,3

Step 3 In the game left, (A,L) and (B,M) are pure strategy Nash equilibria,because players are taking best response to each other.

Step 4 We want to find the mixed strategy equilibrium. Let (pA + (1 −p)B, qL + (1 − q)M) be the equilibrium. First, player 1 must be indifferentbetween playing A and B, which implies

3q = 1(1− q),

where the LHS is the expected payoff of playing A and the RHS is that of B,when player 2 is playing the equilibrium strategy. This implies

q =14.

Next, player 2 must be indifferent between playing L and M, implying

1p = (1− p)3

p =34.

To conclude, the Nash equilibria are; (A,L), (B,M), and ( 34A+ 1

4B,14L+ 3

4M).

Question 3

This question is hard!Notice that in the Cournot model with fixed cost, there can be equilibria

where not every firm is producing. For example, if there are two firms, therecan be an equilibrium where one firm is producing the monopoly quiantity andthe other firm is producing nothing. (Of course, whether this is actually a Nashequilibrium or not depends on the cost and demand structure.) To see why thiscan be an equilibium, consider each firm’s best response. If the opponent firmis producing the monopoly quantity, the revenue from selling some amount cannot cover the fixed cost, if the fixed cost is large enough, and thus producingnothing is the best response. On the other hand, if the opponent is producingnothing, you can get a revenue equal to the monopoly profit, which is higherthan the fixed cost.

It is hard to characterize all the equilibria, so we limit attention to purestrategy equilibria which are symmetric in the sense that all the producingfirms are choosing the same quantity. (Keep in mind that there might be otherequilibria! Actually, it is not very hard to show that all pure strategy equilibriaare symmetric. This would be a good exercise.)

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First, consider the equilibrium where all firms are producing q∗. What weneed is that q∗ is a best response to all other firms choosing q∗. The profit offirm i when it is producing qi and all other firms are producing q∗ is

πi(qi, q∗) = (max{1− (qi + (1− n)q∗), 0} − c)qi − F. (1)

In the equilibrium all firms must be getting positive revenue (because there isfixed cost), so it must be 1− (qi + (n− 1)q∗) > 0. Therefore

πi(qi, q∗) = (1− (qi + (n− 1)q∗)− c)qi − F. (2)

Taking first order condition,

qi =1− (n− 1)q∗ − c

2. (3)

In the symmetric equilibium, we must have qi = q∗, thus

q∗ =1− cn+ 1

. (4)

For this quantity to be actually construct equilibrium, each firm must be gettingnonnegative profit. Plugging this into (2),

πi =( 1− cn+ 1

)2

− F. (5)

Therefore there exists an equilibrium where all firms are producing q∗ = 1−cn+1 if

and only if ( 1−cn+1 )2 ≥ F and c ≤ 1.

Next consider an equilibrium where only k < n firms are producing q∗. Weneed to check incentives of firms which are producing and incentives of firmswhich are not producing.

For a firm which is producing, qi = q∗ must be a best response to k − 1 ofother firms producing q∗ and the rest producing zero. The firm’s profit whenproducing qi is

πi(qi, q∗) = (max{1− (qi + (1− k)q∗), 0} − c)qi − F. (6)

Notice this is same as equation (1), n being replaced by k. Therefore the sameargument as above holds.

q∗ =1− ck + 1

, (7)

For this to be an equilibrium, we need ( 1−ck+1 )2 ≥ F and c ≤ 1.

However, this is not a sufficient condition. It must also be true that for non-producing firms, not producing is actually the best response. From the point

4

Page 260: 6948772 Game Theory NOTES MIT

of view of a non-producing firm i, k other firms are producing q∗. By the sameargument above, if this firm does choose to produce, the optimal amount is

q∗i =1− kq∗ − c

2, (8)

and the payoff is

πi =(1− kq∗ − c

2

)2

− F. (9)

If this profit is positive, the firm wants to produce q∗i rather than producing

nothing. Therefore, we need(

1−kq∗−c2

)2

≤ F , i.e.,( 1− c2(k + 1)

)2

≤ F. (10)

Finally, consider an equilibrium where no firm produces. This is the casewhere c ≥ 1 or where even the monopoly profit is less than the fixed cost. Thatis, (1− c

2

)2

≤ F. (11)

Question 4

Each player i = 1, · · · , n chooses his price of meal pi. The payoff of player iis

ui =√pi −

∑nj=1 pj

n

=√pi −

pin−∑j 6=i pj

n.

Taking first order condition,

12√pi

=1n

pi =n2

4.

This gives the best response, but actually this optimum does not depend onwhat other players are doing, which means that it is the dominant strategy.

To conclude, the Nash equilibrium is

pi =n2

4∀i.

Notice if n = 1 then p = 12 . This is the efficient choice, because there is no

externality. If n → ∞, then pi → ∞. This is because every marginal increasein the price of the meal will splitted by all the people and the marginal increaseof the payment of the player herself converges to 0.

5

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14.12 Solutions for Homework 3Kenichi Amaya1

November 2, 2001

Question 1

We solve this problem by backward induction. First, we solve for the quan-tity choice (stage 2) given the entrance decision, and then we solve for theentrance decision (stage 1).

Stage 2 Suppose m firms has entered in stage 1. Assuming 1 − Q ≥ 0 inequilibrium (we check this is actually true later), each firm i maximizes

πi = qi(1−Q)

= qi(1−∑j 6=i

qj − qi).

Notice, since the fixed cost F was payed in stage 1 already, i.e., it isalready sunk, the firm doesn’t care about it any more. Taking first ordercondition,

qi =1−

∑j 6=i qj

2.

Assuming the equilibrium is symmetric (qi is same for all i),

qi =1− (m− 1)qi

2,

implying

qi =1

m+ 1.

(It is not hard to prove the equilibrium is actually symmetric.) Now wecan confirm that our assumption 1−Q ≥ 0 is actually satisfied because

Q = mqi =m

m+ 1< 1.

The equilibrium payoff of each firm is

π =1

(m+ 1)2.

Stage 1 Let π(m) be the profit (gained in stage 2) of each firm when m firmsare entering. As we solved above,

π(m) =1

(m+ 1)2.

There are three possible classes of equilibria:1E52-303, [email protected], 253-3591

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1. All n firms enter.

2. No firm enters

3. k firms enter, where 0 < k < n.

The first class of equilibrium exists if it is better to enter and gain π(n)−Fthan to exit and gain 0. That is,

π(n)− F ≥ 0,1

(n+ 1)2≥ F.

The second class of equilibrium exists if it is better to exit and get 0 thanto enter to be the only entering firm and gain π(1)− F . That is,

π(1)− F < 0,14≤ F.

To see when the third class of equilibrium exists, we need to check theincentives of the firms which are entering and the firms which are notentering.

Let’s consider the incentive of an entering firm first. It gets π(k)−F if itenters, and gets 0 if it exits. Therefore, there is no incentive to deviate if

π(k)− F ≥ 0,1

(k + 1)2≥ F.

Next consider the incentive of an exiting firm. It gets 0 if it exits. Ifit enters, it gets π(k + 1) − F because k other firms are also entering.Therefore, there is no incentive to deviate if

π(k + 1)− F ≤ 0,1

(k + 2)2≤ F.

Notice, the argument above says nothing about which k firms enter. Ifthe condition above is satisfied, any selection of k firms constitutes anequilibrium.

To summarize, let’s formally describe the equilibrium strategies. Remember,we need to describe all the contingent plans for stage 2, even for the informationsets which are not achieved on the equilibrium path.

The equilibrium strategies are as the following.

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1. • All firms enter in stage 1.

• If m firms enter in stage 1, each entering firm produces 1m+1 in stage

2.

The equilibrium payoff is 1(n+1)2−F for every firm. This equilibrium exists

if and only if 1(n+1)2 ≥ F .

2. • No firm enter in stage 1.

• If m firms enter in stage 1, each entering firm produces 1m+1 in stage

2.

The equilibrium payoff is 0 for every firm. This equilibrium exists if andonly if 1

4 ≤ F .

3. • k firms enter in stage 1, where 0 < k < n.

• If m firms enter in stage 1, each entering firm produces 1m+1 in stage

2.

The equilibrium payoff is 1(k+1)2 − F for every entering firm, and 0 for

every exiting firm. This equilibrium exists if and only if 1(k+1)2 ≥ F and

1(k+2)2 ≤ F .

Question 2(a)

�����

AAAAU�����

@@@@R����

@@@@R

�����

AAAAU

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AAAAU

1

2

1

2

X N

x n

L R

l r l r

(2,2)

(2,2)

(3,3) (0,0) (0,0) (1,1)

We can solve for subgame perfect equilibria by backward induction. Thisgame has three subgames: the whole game itself, a subgame starting from player

3

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2’s first decision node, and a subgame starting from player 1’s second decisionnode.

First we look at the subgame starting from player 1’s second decision node.This subgame is equivalent to the following nomal form game:

l rL 3,3 0,0R 0,0 1,1

Obviously, pure strategy equilibria are (L,l) and (R,r).If (L,l) is played in this subgame, player 2 chooses n at his first decision

node, because he gets payoff of 2 by playing x and 3 by playing n. Given this,player 1 chooses N at her first decision node because she gets payoff of 2 byplaying X and 3 by playing N. Therefore, player 1’s strategy (N,L) and player2’s strategy (n,l) constitute a subgame perfect Nash equilibrium.

If (R,r) is played in the subgame starting from player 1’s second decisionnode, player 2 chooses x at his first decision node, because he gets payoff of 2by playing x and 1 by playing n. Given this, player 1 is indifferent between Xand N at her first decision node because she gets payoff of 2. Therefore, player1’s strategy (X,R) and player 2’s strategy (x,r) constitute a subgame perfectNash equilibrium, and player 1’s strategy (N,r) and player 2’s strategy (x,r)constitute another subgame perfect Nash equilibrium.

To conclude, there are three subgame perfect Nash equilibria:

1. Player 1: (N,L), Player 2: (n,l)

2. Player 1: (X,R), Player 2: (x,r)

3. Player 1: (N,R), Player 2: (x,r)

(b)

Let’s apply the forward induction argument here.Suppose player 1’s second decision node is reached. This means that player

2 has chosen N at his first decision node, i.e., his strategy is either (n,l) or(n,r) (or any mixture of them). Should player 1 think that player 2 is choosing(n,r)? Player 2 can get payoff of 0 or 1 by playing (n,r), whereas he can get 2for sure by playing (x,l) or (x,r). In other words, (n,r) is a strictly dominatedstrategy. If player 1 knows player 2 is rational, she must know that there is noreason player 2 play (n,r). Therefore, given player 2 has played N at his firstdecision node, player 1 must conclude player 2 is choosing (X,l), and choose thebest response to it, namely L. To conclude, in any subgame perfect equilibriumwhich is consistent with common knowlegde of rationality, player 1’s choice athis second decision node must be L.

Therefore, only the first equilibrium (N,L), (n,l) is consistent.

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Question 3(a)

The monopoly price is pm = 12 . Construct the following “trigger strategy”.

• Start with playing pm, and play pm if everybody has played pm all thetime.

• Play 0 if anybody has played something different from pm at least once.

By the single deviation principle, it is sufficient to check incentives to devi-ate only once, in order to check when this is actually a subgabe perfect Nashequilibrium.

Firstly, check the incentive when nobody has deviated from pm before. If afirm plays pm, as suggested by the strategy, then everyone will be playing pm

all the time in future, and a firm’s profit per period is

pm(1− pm)2

=18,

and the present discounted sum of the profit sequence from today is

18

+ δ18

+ δ2 18

+ · · · = 11− δ

18.

If the firm is to deviate, the best it can do today is to set a price slightly lessthan pm and get the whole monopoly profit

pm(1− pm) =14.

If the firm does deviate, everyone will be playing 0 from the next period, yieldingper period payoff of zero. Therefore, the present discounted sum of the profitsequence from today is

14

+ δ0 + δ20 + · · · = 14.

The firm has no incentive to deviate if and only if

11− δ

18≥ 1

4

δ ≥ 12.

Secondly, check the incentive when somebody has ever deviated from pm

before. In this case, a firm’s action today doesn’t affect it’s future payoff becauseeveryone will be playing 0 in future no matter what happens today. Thereforeit is sufficient to check if the firm can increase the present period profit bydeviating. If it plays p = 0, as suggested by the strategy, it gets a payoff of 0.

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If it deviates and plays p > 0, it gets a payoff of 0 again because all consumersbuy from the other firm which is choosing p = 0. Therefore it can’t benefit fromdeviating.

To conclude, the trigger strategy constitutes a subgame perfect Nash equi-librium if δ ≥ 1

2 .

(b)

When there are n firms, we can use the trigger strategy constructed in part(a) again. As we did in part (a), we check incentives to deviate before and afterany deviation.

Firstly, check the incentive when nobody has deviated from pm before. If afirm plays pm, as suggested by the strategy, then everyone will be playing pm

all the time in future, and a firm’s profit per period is

pm(1− pm)n

=1

4n,

and the present discounted sum of the profit sequence from today is

14n

+ δ1

4n+ δ2 1

4n+ · · · = 1

1− δ1

4n.

If the firm is to deviate, the best it can do today is to set a price slightly lessthan pm and get the whole monopoly profit

pm(1− pm) =14.

If the firm does deviate, everyone will be playing 0 from the next period, yieldingper period payoff of zero. Therefore, the present discounted sum of the profitsequence from today is

14

+ δ0 + δ20 + · · · = 14.

The firm has no incentive to deviate if and only if

11− δ

14n≥ 1

4

δ ≥ n− 1n

.

To check the incentive when somebody has ever deviated from pm before,the same argument as in part (a) holds.

To conclude, the trigger strategy constitutes a subgame perfect Nash equi-librium if δ ≥ n−1

n .

Question 4(a)

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To begin with, let’s find the subgame perfect Nash equilibrium of the stagegame. We do this by backward induction. The long run firm, observing theshort run firm’s quantity xt, chooses its quantity yt to maximize its profit

πLt = yt(1− (xt + yt)).

Solving first order condition, the optimum is

y∗t (xt) =1− xt

2.

The short run firm chooses its quantity xt to maximize its profit

πSt = xt(1− (xt + yt)),

knowing that if it choose xt, the long run firm reacts with

y∗t (xt) =1− xt

2.

Therefore, the short run firm’s objective function can be rewritten as a functionof xt;

πSt = xt(1− (xt +1− xt

2)).

Solving first order condition, the optimum is

x∗t =12.

Therefore, the subgame perfect Nash equilibrium of the stage game is

x∗t =12, y∗t (xt) =

1− xt2

.

Now let’s solve for the subgame perfect Nash equilibrium of the finitelyrepeated game. Since it is finite, we can use backward induction. At the lastperiod, t = T , the players don’t care about future and they concern only aboutthe payoff of that period. Therefore they must play the subgame perfect Nashequilibrium of the stage game, regardless of what happened in the past. At timet = T − 1, the players know that the actions today doesn’t affect tomorrow’soutcome, so they concern only about the payoff of that period. Therefore theymust play the subgame perfect Nash equilibrium of the stage game, regardlessof what happened in the past. We can repeat the same argument until we reachthe very first period.

Therefore, the subgame perfect Nash equilibrium of the repeated game is

x∗t =12, y∗t (xt) =

1− xt2

for all t,

regardless of the history.

7

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(b)

Construct the following trigger strategy.

Long rum firm • Start with playing the following strategy:

(∗) yt(xt) ={

1/2 if xt ≤ 1/21− xt if xt ≥ 1/2

Keep playing this strategy as long it has not deviated from it.

• Play y∗t (xt) = 1−xt2 if it has deviated from (*) at least once.

Short run firms • Play xt = 1/4 if the long run firm has not deviated from(*) before.

• Play xt = 12 if the long run firm has deviated from (*) at least once.

To see this is actually a subgame perfect Nash equilibrium, let’s check in-centives to deviate.

Firstly, consider the long run firm’s incentive when it has never deviatedfrom (*) before.

Case 1: xt ≤ 1/2 If it follows the strategy and choose yt = 1/2, the presentperiod profit of the long run firm is

12

(1− (12

+ xt)).

Starting from the next period, the outcome will be xt = 1/4 and yt = 1/2every period, and the long run firm’s per period profit is 1/8 and thereforethe present discounted value of the profit sequence is

12

(1− (12

+ xt)) +δ

8(1− δ).

If it is to deviate, the best it can do today is to play

yt =1− xt

2

and get payoff of(1− xt)2

4.

However, starting from next period, the outcome will be xt = 1/2 andyt = 1/4 and the long run firm’s per period profit is 1/16. Therefore thepresent discounted value of the profit sequence is

(1− xt)2

4+

δ

16(1− δ).

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If δ = 0.99,

12

(1− (12

+ xt)) +δ

8(1− δ)>

(1− xt)2

4+

δ

16(1− δ),

(check it!), and therefore it is better to follow the equilibrium strategythan to deviate.

Case 2: xt ≥ 1/2 If it follows the strategy, pt = 0 and today’s payoff is 0, andthe outcome will be xt = 1/4 and yt = 1/2 every period, starting thenext period. The long run firm’s per period profit is 1/8 and therefore thepresent discounted value of the profit sequence is

δ

8(1− δ).

If it is to deviate, the best it can do today is to play yt = 1−xt2 and get

payoff of (1−xt)2

4 . However, starting from next period, the outcome will bext = 1/2 and yt = 1/4 and the long run firm’s per period profit is 1/16.Therefore the present discounted value of the profit sequence is

(1− xt)2

4+

δ

16(1− δ).

This value is the largest when xt = 1/2 and is equal to

116

16(1− δ).

If δ = 0.99, it is better to follow the equilibrium strategy than to deviate(check it!).

Secondly, consider the long run firm’s incentive when it has deviated from(*) before. According to the strategy strategy profile, future outcomes don’tdepend on today’s behavior. Therefore the long run firm cares only about itspayoff today. Actually, by following the strategy, it is taking best response tothe short run firm.

Finally, consider the short run firm’s incentives. Since they never care aboutfuture payoff, it must be playing a best response to the long run firm’s strategy,which is actually true. (Check it!)

(c)

In the equilibrium we saw in part (b), the per period profit on the equilibriumpath were 1/8 for the long run firm and 1/16 for the short run firms.

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If there is a subgame perfect Nash equilibrium where xt = yt − 1/4 on theequilibrium path, then the per period profit on the equilibrium path are 1/8for the long run firm and 1/8 for the short run firms.

Construct the following trigger strategy, which is different from (b) only in(*) where xt = 1/4:

Long rum firm • Start with playing the following strategy:

(∗) yt(xt) =

1/4 if xt = 1/41/2 if xt ≤ 1/2 and xt 6= 1/41− xt if xt ≥ 1/2

Keep playing this strategy as long it has not deviated from it.

• Play y∗t (xt) = 1−xt2 if it has deviated from (*) at least once.

Short run firms • Play xt = 1/4 if the long run firm has not deviated from(*) before.

• Play xt = 12 if the long run firm has deviated from (*) at least once.

The incentive of the long run firm when it is supposed to play (*) andxt = 1/4 is satisfied because it gets current payoff of 1/8, which is same as part(b), and what it can get by deviating is same as in part (b). The incentiveproblem of the long run firm is same as in (b).

The incentives of the short run firms are also satisfied because the payofffrom following the strategy is larger than part (b) and payoff when deviating issame as in (b).

10

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Page 274: 6948772 Game Theory NOTES MIT

14.12 Solutions for Homework 4Kenichi Amaya1

November 9, 2001

Question 1(a)

Action spaces A1 = A2 = {B,S}.

Type spaces T1 = {α}, T2 = {β1, β2}. Since player 1 has no private informa-tion, we can model it so that her type can take only one value. Player 2knows that the game above is played when his type is β1 and he knowsthat the game below is played when his type is β2.

Belief Player i’s belief µi(tj |ti) is the probability that player j’s type is tjconditional on that player i’s type is ti. In this model, since it is assumedthat the types are independent,

µ1(β1|α) = µ1(β2|α) =12,

µ2(α|β1) = µ2(α|β2) = 1.

vNM utility function ui(a1, a2, t1, t2) is the vNM utility when player 1’s ac-tion is a1, player 2’s action is a2, player 1’s type is t1 and player 2’s typeis t2.

u1(B,B, α, β1) = 2, u2(B,B, α, β1) = 1,u1(B,S, α, β1) = 0, u2(B,S, α, β1) = 0,u1(S,B, α, β1) = 0, u2(S,B, α, β1) = 0,u1(S, S, α, β1) = 1, u2(S, S, α, β1) = 2,

u1(B,B, α, β2) = 2, u2(B,B, α, β2) = 0,u1(B,S, α, β2) = 0, u2(B,S, α, β2) = 2,u1(S,B, α, β2) = 0, u2(S,B, α, β2) = 1,u1(S, S, α, β2) = 1, u2(S, S, α, β2) = 0.

(b)

First consider player 1’s incentive. Since she doesn’t know the game whichis being played, she maximizes her expected payoff.

If she plays B, with probability 1/2 the top game is being played and player2 chooses B and thus she gets a payoff of 2, and with probability 1/2 the bottomgame is being played and player 2 chooses S and thus she gets a payoff of 0.

1E52-303, [email protected], 253-3591

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Page 275: 6948772 Game Theory NOTES MIT

Therefore her expected payoff is 1. If she plays S, with probability 1/2 the topgame is being played and player 2 chooses B and thus she gets a payoff of 0,and with probability 1/2 the bottom game is being played and player 2 choosesS and thus she gets a payoff of 1. Therefore her expected payoff is 1/2.

Therefore B is actually player 1’s best response against player 2’s strategy.Next consider player 2’s incentive. When he knows that the top game is

being played, B is the best response given player 1 is choosing B. When heknows that the bottom game is being played, S is the best response given player1 is choosing B.

Therefore choosing B when the top game is being played and choosing S whenthe bottom game is being played is actually player 2’s best response againstplayer 1’s strategy.

Since both players are taking a best response to each other, the strategyprofile constitutes a Bayesian Nash equilibrium.

Question 2

Each player’s action is the choice of price. A price can take any nonnegativereal number. Therefore the action space is R+ for both players.

Player i’s type is her private information. In this model, bi is player i’s type.bi is either bH or bL. Therefore, the type space of each player is {bH , bL}.

Player i’s belief µi(bj |bi) is the probability that player j’s type is bj condi-tional on that player i’s type is bi. In this model, since it is assumed that thetypes are independent,

µi(bj |bi) ={θ if bj = bH1− θ if bj = bL

.

(von Neuman Morgenstern) utility in this model is the profit of each player(assuming the firms are risk neutral) as a function of the actions and types ofboth players:

ui(pi, pj , bi, bj) = pi(a− pi − bipj).Player i’s strategy specifies what action to take for any realization of her

type. In this model, it is a two dimensional vector (pi(bH), pi(bL)), where pi(bH)is the price when its type is bH and pi(bL) is the price when its type is bL. Thestrategy space is R2

+ for each i.A strategy profile {(p∗1(bH), p∗1(bL)), (p∗2(bH), p∗2(bL))} constitutes a Bayesian

Nash equilibrium if each p∗i (bi) is a best response, i.e., a maximizer of player i’sexpected payoff, conditional on that her type is bi and the opponent is choosingstrategy (p∗j (bH), p∗j (bL)). That is:

p∗1(bH) = argmaxp1θp1(a− p1 − bHp∗2(bH)) + (1− θ)p1(a− p1 − bHp∗2(bL)),

p∗1(bL) = argmaxp1θp1(a− p1 − bLp∗2(bH)) + (1− θ)p1(a− p1 − bLp∗2(bL)),

p∗2(bH) = argmaxp2θp2(a− p2 − bHp∗1(bH)) + (1− θ)p2(a− p2 − bHp∗1(bL)),

p∗2(bL) = argmaxp2θp2(a− p2 − bLp∗1(bH)) + (1− θ)p2(a− p2 − bLp∗1(bL)).

2

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Taking first order conditions,

p∗1(bH) =a− bH(θp∗2(bH) + (1− θ)p∗2(bL))

2,

p∗1(bL) =a− bL(θp∗2(bH) + (1− θ)p∗2(bL))

2,

p∗2(bH) =a− bH(θp∗1(bH) + (1− θ)p∗1(bL))

2,

p∗2(bL) =a− bL(θp∗1(bH) + (1− θ)p∗1(bL))

2.

Since the game is symmetric, let’s look for a symmetric equilibrium wherep∗1(bH) = p∗2(bH) = p∗H and p∗1(bL) = p∗2(bL) = p∗L. Then the conditions arereduced to

p∗H =a− bH(θp∗H + (1− θ)p∗L)

2,

p∗L =a− bL(θp∗H + (1− θ)p∗L)

2.

Solving these, we get

p∗H =a

2(1− bH

2 + θbH + (1− θ)bL),

p∗L =a

2(1− bL

2 + θbH + (1− θ)bL).

Question 3

Consider the following incomplete information game.

H TH 1 + εα, −1 + εβ −1 + εα, 1T −1, 1 + εβ 1, −1

Here, α is player 1’s type, β is player 2’s type, α and β are independent drawfrom uniform distribution in the interval [0, 1], and ε > 0 is a constant.

The interpretation is like this: Player 1 (2) can get an extra payoff of εα(εβ) by playing head, in addition to the payoff generated from the original game.This extra payoff is her private information.

Notice, as we converge ε to 0, the game converges to the original game.From the construction, player 1 (2) is more tempted to play H if she has a

higher value of α (β). Observing this, try to find the pure strategy BayesianNash equilibrium of the following form.

• Player 1 chooses H if α ≥ α and chooses L if α ≤ α, where α ∈ [0, 1] is aconstant.

3

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• Player 2 chooses H if β ≥ β and chooses L if β ≤ β, where β ∈ [0, 1] is aconstant.

This strategy profile is actually a Bayesian Nash equilibrium if player 1 withcutoff type α and player 2 with cutoff type β are indifferent between choosingH and T.

According to the strategy, the probability that player 2 plays H is 1− β, andthe probability that player 2 plays L is β. Therefore, if player 1 of type α playsH, her expected payoff is

1(1− β) + (−1)β + εα,

and if she plays T, she gets

(−1)(1− β) + 1β.

Therefore, we need

1(1− β) + (−1)β + εα = (−1)(1− β) + 1β. (1)

We do the same thing for player 2 with type β. According to the strategy,the probability that player 1 plays H is 1− α, and the probability that player 1plays L is α. Therefore, if player 2 of type β plays H, his expected payoff is

(−1)(1− α) + 1α+ εβ,

and if she plays T, she gets

1(1− α) + (−1)α.

Therefore, we need

(−1)(1− α) + 1α+ εβ = 1(1− α) + (−1)α. (2)

Solving (1) and (2), we get

α =8− 2ε16 + ε2

, (3)

β =8 + 2ε16 + ε2

. (4)

Now, as we converge ε to 0, i.e., as we converge the game to the completeinformation game, both α and β converge to 1/2. Therefore, in the limit, bothplayers are choosing H and T with probability 1/2, which is equivalent to themixed strategy equilibrium of the complete information game.

Question 4

4

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Let i = 1, · · · , n be the index of bidders, vi be bidder i’s valuation of thegood, and bi be player i’s bid. We denote player i’s strategy by a function xi(vi),meaning that player i bids bi = xi(vi) when her valuation is vi.

We want to show that the strategy profile

xi(vi) =(n− 1)vi

nfor all i

constitutes a Bayesian Nash equilibrium. Since the game is symmetric and strat-egy profile is symmetric, it is sufficient to check one player’s incentive becauseevery player is facing the same incentive problem.

We are going to show that if player i’s valuation is vi and all other playersare taking the strategy

xj(vj) =(n− 1)vj

n,

then the bid which maximizes her expected payoff is

bi = xi(vi) =(n− 1)vi

n.

First consider the probability of winning the auction by bidding bi. She winsif and only if all other players’ bids are less than bi, i.e.,

xj(vj) =(n− 1)vj

n≤ bi for all j 6= i.

This is equivalent to

vj ≤nbin− 1

for all j 6= i.

SinceProb(vj ≤

nbin− 1

) =nbin− 1

for all j because vj is uniformly distributed in [0.1],

Prob(win) = Prob(vj ≤nbin− 1

for all j 6= i)

= Prob(vj ≤nbin− 1

)n−1

=( nbin− 1

)n−1.

Therefore, the expected payoff from bidding bi is

Ui = (vi − bi)Prob(win) = (vi − bi)( nbin− 1

)n−1.

5

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Taking first order condition,

∂Ui∂bi

= (vi − bi)(n− 1)( nbin− 1

)n−2 n

n− 1−( nbin− 1

)n−1 = 0,

(vi − bi)n =nbin− 1

,

bi =(n− 1)vi

n.

Therefore, the strategy of player i

bi = xi(vi) =(n− 1)vi

n

is actually the best response to other player’s playing

xj(vj) =(n− 1)vj

n.

Q.E.D.

6

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Page 282: 6948772 Game Theory NOTES MIT

14.12 Game Theory

Fall 2001

Problem Set 5 Solutions

December 14, 2001

1. The possibilities are: Pooling on A, pooling on D, and hybrid equilibria (sincethere are only two possible actions for player 1, but three possible types ofplayer 1). Let us write the strategies in the form (t11, t21, t31; t12, t22, t32) forplayer 1, where tij indicates the action taken by type i at the jth move, and(a) for the strategy of player 2, which gets to move only once and has only oneinformation set. The only equilibrium here is (A,A,D; d, a, d) and (δ). Player2 moves δ since u2(α) = 5/6(3) + 1/6(−5) = 5/3 < 2 = u2(δ) – note howbeliefs are updated. Finally, each type of player 1 does not have an incentiveto deviate: top type is getting 5 as opposed to 4 (if deviates and plays D);middle type is getting 0 as opposed to -1 (if deviates and plays D); bottomtype is getting 0 as opposed to -1 (if deviates and plays A).

2. In the second stage of the game, it is clear that each type will always play thestatic best response: a = 2 type will fight and a = −1 type will accomodate.

The possibilities for the first stage are:

(a) Pooling on Fight: If π = 0.4, then at t = 2, given no updating of beliefs(so that µ(a = 2|F ) = π), the entrant stays out since EU(Enter) =2π − 1 = −0.2 < 0 = EU(Out). Then, in t = 1, incumbents pool onfight. The PBE equilibrium is [(F, F ;F,A), (O,O,E)], where we writethe strategy for player 1 as (t11, t21; t12, t22) where tij is the action takenby type i in stage j and for player 2 as (a1, a21, a22 where a1 refers to thefirst information set in t = 1, and the other two refer to the informationsets in t = 2: a21 being the information set after the incumbent fights inthe first period, and a22 being the information set after the incumbentfights in the first period. Beliefs are not updated on a21 and off theequilibrium belief on a22 is µ(a = −1|accomodate) = 1.

(b) Separating equilibrium: If π = 0.9, then entrant always enters sinceEU(Enter) = 2π − 1 = 0.8 > 0 = EU(Out), a=2 type fights and a=-1 type accomodates. The PBE is [(F,A;F,A), (E,O,E)]. Beliefs areupdated such that at a22 we have µ(a = −1|accomodate) = 1 and at a21

we have µ(a = 2|fight) = 1.

1

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If the game is repeated 100 times a similar structure applies.

3. In t = 1, buyer of type 1 will accept if and only if:

0.9(1− p1) ≥ 0, that is p1 ≤ 10.9(2− p1) ≥ 0, that is p1 ≤ 2

and seller will offer p1 ≥ 0.

If seller offers 1 ≤ p1 ≤ 2, then type 1 rejects the offer and type 2 accepts it:

EU(seller) = (1− π)(0) + p1π = p1π

or simply 2π, which is the value that maximizes the seller’s utility.

If seller offers 0 ≤ p1 ≤ 1 then both types accept the offer:

EU(seller) = p1(1− π) + p1π = p1

or simply 1, which is the value that maximizes the seller’s utility.

If π > 1/2, then seller offers p1 = 2 in t = 1 and she accepts offers in t = 0p0 ≥ 0.9 since if she waits until t = 1 she can get 0.9(1)=0.9 . If π < 1/2, thenseller offers p1 = 1 and she accepts offers in t = 0 p0 ≥ 1.8 since in t = 1 shecan get 0.9(2)=1.8.

4. • Proposed pooling strategy of Sender: Suppose there is an equilibrium inwhich the Sender’s strategy is (R,R), where (m′,m′′) means that type t1chooses m′ and type t2 chooses m′′.

• Receiver’s belief at R: Then the Receiver’s information set correspondingcorresponding to R is on the equilibrium path, so the Receiver’s belief(p, 1 − p) at this information set is determined by Bayes’rule and theSender’s strategy: p = .5, the prior distribution (since there is no updat-ing of belief in pooling).

• Receiver’s best response at R: Given this belief, the Receiver’s best re-sponse following R is to play d sinceu2(u|R) = 1(.5) + 0(.5) = .5 < u2(d|R) = 0(.5) + 2(.5) = 1

• Receiver’s belief and best response at information set for L: To determinewhether both Sender types are willing to choose R, we need to specifyhow the Receiver would react to L. In particular, we need to pin down thebeliefs such that u2(u|L) > u2(d|L) in order for the Sender’s willingnessto choose R. The Receiver’s belief at the information set correspondingto L needs to be, in order for u to be the best response, q ≥ 1/3. Thus, ifthere is an equilibrium in which the Sender’s strategy is (R,R) then theReceiver’s strategy must be (d, u) where (a′, a′′) means that the Receiverplays a′ following R and a′′ following L.

• Pooling perfect Bayesian equilibrium: [(R,R), (d, u), p = .5, q] for q ≥1/3.

5. Part a.

2

Page 284: 6948772 Game Theory NOTES MIT

• Pooling on R: Beliefs on the equilibrium path: µ(t1|R) = .5Receiver’s best response: u2(u|R) = .5(2) = 1 > u2(d|R) = .5(1) = .5Sender’s payoff: ut1(R) = 2,ut2(R) = 1.Any play by Receiver after observing L sustains (R,R).PBE: [(R,R), (u, u), q > 1/2], [(R,R), (u, d), q < 1/2], [(R,R), (u, αu +(1− α)d), q = 1/2]

• Pooling on L: Not an equilibrium, since t2 will always want to play R.

• t1 plays R, t2 plays L: Then both of the Receiver’s information sets areon the equilibrium path, so both beliefs are determined by Bayes’ ruleand the Sender’s strategy: µ(t1|R) = 1, µ(t2|L) = 1.Receiver’s best response: (u, d)t2, however, will always play R.

• t1 plays L, t2 plays R: µ(t1|L) = 1, µ(t2|R) = 1.Receiver’s best response: (u, d)Sender’s payoff: ut1(L) = 1,ut2(R) = 1.PBE: [(L,R), (u, d)]

Part b.

• Pooling on R: Beliefs on the equilibrium path: µ(t1|R) = .5Receiver’s best response: u2(u|R) = .5(2) = 1 > u2(d|R) = .5(1) = .5Sender’s payoff: ut1(R) = 0,ut2(R) = 1.However, this is not an equilibrium since t1 would always play L.

• Pooling on L: Beliefs on the equilibrium path: µ(t1|R) = .5Receiver’s best response: u2(u|L) = .5(3) = 1.5 > u2(d|L) = .5(1) +.5(1) = 1Sender’s payoff: ut1(L) = 3,ut2(L) = 3.Receiver’s belief and best response at information set for R: u2(u|R) =2(1− p) > u2(d|R) = p when p < 2/3.PBE: [(L,L), (u, u), p < 2/3]

• t1 plays R, t2 plays L: Then both of the Receiver’s information sets areon the equilibrium path, so both beliefs are determined by Bayes’ ruleand the Sender’s strategy: µ(t1|R) = 1, µ(t2|L) = 1.Receiver’s best response: (d, u)No deviations are profitable.PBE: [(R,L), (d, u)]

• t1 plays L, t2 plays R: µ(t1|L) = 1, µ(t2|R) = 1.Receiver’s best response: (d, u)No deviations are profitable.PBE: [(L,R), (d, u)]

3

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Page 288: 6948772 Game Theory NOTES MIT

8 0���� !� �� � "���� ��� !� �� � �� �������! ��� ��� !� �� �"���� ��� !� �� � �� �������!- �� ��� ��� �� � � ��� ��� �� � ,%�!�� ����- �������� ����� ��� � 0���� !� �� � �� �������!- �� ��� ����� � �- ������ 7� �� ���� ���

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Page 289: 6948772 Game Theory NOTES MIT

(2,1)

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8

Page 290: 6948772 Game Theory NOTES MIT

6 1� ���� #������� ��� �� ���� �� ������� ��� � ���� ��� ��� ��� ��)��� �� ���� � .���� ��/������� ��� �������� ������ 7� �������� ���� ���� ��� ������ ���� �� ��� ������!���������� �� �����- ��-

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Page 291: 6948772 Game Theory NOTES MIT

4�5 1��� ��!��� ������ ���) �� ��� �������� ����� ��� ��� ��)���- ��� �� � ��� ����� ���� ��� ��� ��� ��)� ��� <�� �� �4�? � �����

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Page 292: 6948772 Game Theory NOTES MIT

Grade Distribution For Midterm 1

0

10

20

30

40

50

60

70

80

90

100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46

rank

grad

e

if you got above 70, you will probably get A(+/-);if you got above 50, you will probably get B(+/-) or above;if you got below 40, you should reconsider ...

Average = 67.96 Std.Dev.= 18.05

Assuming that you will exhibit similar performance in the future,

Page 293: 6948772 Game Theory NOTES MIT
Page 294: 6948772 Game Theory NOTES MIT

14.12 Game Theory — Midterm II10/10/2001

Prof. Muhamet YildizInstructions. This is an open book exam; you can use any written material. You have onehour and 20 minutes. Each question is 25 points. Good luck!

1. Consider the following game:

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

Compute all the pure-strategy subgame-perfect equilibria. Use a forward inductionargument to eliminate one of these equilibria.

2. Below, there are pairs of stage games and strategy profiles. For each pair, check whetherthe strategy profile is a subgame-perfect equilibrium of the game in which the stagegame is repeated infinitely many times. Each agent tries to maximize the discountedsum of his expected payoffs in the stage game, and the discount rate is δ = 0.99.

(a) Stage Game:1\2 L M RT 2,-1 0,0 -1,2M 0,0 0,0 0,0B -1,2 0,0 2,-1

Strategy profile: Until some player deviates, player 1 plays T and player 2alternates between L and R. If anyone devites, then each play M thereafter.

(b) Stage Game:1\2 A BA 2,2 1,3B 3,1 0,0

Strategy profile: The play depends on three states. In state S0, each playerplays A; in states S1 and S2, each player plays B. The game start at state S0. Instate S0, if each player plays A or if each player plays B, we stay at S0, but if aplayer i plays B while the other is playing A, then we switch to state Si. At anySi, if player i plays B, we switch to state S0; otherwise we state at state Si.

1

Page 295: 6948772 Game Theory NOTES MIT

3. Consider the following first-price, sealed-bid auction where an indivisible good is sold.Thereare n ≥ 2 buyers indexed by i = 1, 2, . . . , n. Simultaneously, each buyer i submits abid bi ≥ 0. The agent who submits the highest bid wins. If there are k > 1 play-ers submitting the highest bid, then the winner is determined randomly among theseplayers – each has probability 1/k of winning. The winner i gets the object and payshis bid bi, obtaining payoff vi − bi, while the other buyers get 0, where v1, . . . , vn areindependently and identically distributed with probability density function f where

f (x) =

½3x2 x ∈ [0, 1]0 otherwise.

(a) Compute the symmetric, linear Bayesian Nash equilibrium.

(b) What happens as n→∞?

[Hint: Since v1, v2, . . . , vn is independently distributed, for any w1, w2, . . . , wk, we have

Pr(v1 ≤ w1, v2 ≤ w2, . . . , vk ≤ wk) = Pr(v1 ≤ w1) Pr(v2 ≤ w2) . . .Pr(vk ≤ wk).]

4. This question is about a thief and a policeman. The thief has stolen an object. Hecan either hide the object INSIDE his car on in the TRUNK. The policeman stops thethief. He can either check INSIDE the car or the TRUNK, but not both. (He cannotlet the thief go without checking, either.) If the policeman checks the place where thethief hides the object, he catches the thief, in which case the thief gets -1 and the policegets 1; otherwise, he cannot catch the thief, and the thief gets 1, the police gets -1.

(a) Compute all the Nash equilibria.

(b) Now imagine that we have 100 thieves and 100 policemen, indexed by i =1, . . . , 100, and j = 1, . . . , 100. In addition to their payoffs above, each thief igets extra payoff bi form hiding the object in the TRUNK, and each policeman jgets extra payoff di from checking the TRUNK. We have

b1 < b2 < · · · < b50 < 0 < b51 < · · · < b100,d1 < d2 < · · · < d50 < 0 < d51 < · · · < d100.

Policemen cannot distinguish the thieves from each other, nor can the thievesdistinguish the policemen from each other. Each thief has stolen an object, hidingit either in the TRUNK or INSIDE the car. Then, each of them is randomlymatched to a policeman. Each matching is equally likely. Again, a policemancan either check INSIDE the car or the TRUNK, but not both. Compute apure-strategy Bayesian Nash equilibrium of this game.

2

Page 296: 6948772 Game Theory NOTES MIT

The game for problem 1.

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

3

Page 297: 6948772 Game Theory NOTES MIT

14.12 Game Theory — Midterm II10/10/2001

Prof. Muhamet YildizInstructions. This is an open book exam; you can use any written material. You have onehour and 20 minutes. Each question is 25 points. Good luck!

1. Consider the following game:

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

Compute all the pure-strategy subgame-perfect equilibria. Use a forward inductionargument to eliminate one of these equilibria.

Answer: There are two pure strategy Nash equilibria in the proper subgame, yieldingsubgame-perfect equilibria:

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

5/25/2

33

02

20

22

X E

L R

l r lr

5/25/2

33

02

20

22

X E

L R

l r lr

2

1

2

For player 2, Er is strictly dominated by Xr, while El is not dominated. Hence, if player1 keeps believing that 2 is rational whenever it is possible, then when he sees that 2played E, he ought to believe that 2 plays strategy El – not the dominated strategyEx. In that case, 1 would play L, and 2 would play E. Therefore, the equilibrium onthe left is eliminated.

1

Page 298: 6948772 Game Theory NOTES MIT

2. Below, there are pairs of stage games and strategy profiles. For each pair, check whetherthe strategy profile is a subgame-perfect equilibrium of the game in which the stagegame is repeated infinitely many times. Each agent tries to maximize the discountedsum of his expected payoffs in the stage game, and the discount rate is δ = 0.99.

(a) Stage Game:1\2 L M RT 2,-1 0,0 -1,2M 0,0 0,0 0,0B -1,2 0,0 2,-1

Strategy profile: Until some player deviates, player 1 plays T and player 2alternates between L and R. If anyone deviates, then each play M thereafter.Answer: It is subgame perfect. Since (M,M) is a Nash equilibrium of the stagegame, we only need to check if any player wants to deviate when player 1 playsT and player 2 alternates between L and R. In this regime, the present value ofplayer 1’s payoffs is

V1L =2

1− δ− δ

1− δ=2− δ

1− δ> 0

when 2 is to play L and

V1R =2δ

1− δ− 1

1− δ=2δ − 11− δ

= 98

when 2 is to play R. When 2 plays L, 1 cannot gain by deviating. When 2 playsR, the best 1 gets by deviating is

2 + 0 < 98

(when he plays B). The only possible profitable deviation for player 2 is to playR when he is supposed to play left. In that contingency, if he follows the strategyhe gets V1R = 98; if he deviates, he gets 2 + 0 < V1R.

(b) Stage Game:1\2 A BA 2,2 1,3B 3,1 0,0

Strategy profile: The play depends on three states. In state S0, each playerplays A; in states S1 and S2, each player plays B. The game start at state S0. Instate S0, if each player plays A or if each player plays B, we stay at S0, but if aplayer i plays B while the other is playing A, then we switch to state Si. At anySi, if player i plays B, we switch to state S0; otherwise we state at state Si.Answer: It is not subgame-perfect. At state S2, player 2 is to play B, and wewill switch to state S0 no matter what 1 plays. In that case, 1 would gain bydeviating and playing A (in state S2).

2

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3. Consider the following first-price, sealed-bid auction where an indivisible good is sold.There are n ≥ 2 buyers indexed by i = 1, 2, . . . , n. Simultaneously, each buyer isubmits a bid bi ≥ 0. The agent who submits the highest bid wins. If there are k > 1players submitting the highest bid, then the winner is determined randomly amongthese players – each has probability 1/k of winning. The winner i gets the object andpays his bid bi, obtaining payoff vi − bi, while the other buyers get 0, where v1, . . . , vnare independently and identically distributed with probability density function f where

f (x) =

½3x2 x ∈ [0, 1]0 otherwise.

(a) Compute the symmetric, linear Bayesian Nash equilibrium.Answer: We look for an equilibrium of the form

bi = a+ cvi

where c > 0. Then, the expected payoff from bidding bi with type vi is

U (bi; vi) = (vi − bi) Pr (bi > a+ cvj ∀j 6= i)= (vi − bi)

Yj 6=iPr (bi > a+ cvj)

= (vi − bi)Yj 6=iPr

µvj <

bi − ac

= (vi − bi)Yj 6=i

µbi − ac

¶3= (vi − bi)

µbi − ac

¶3(n−1)for bi ∈ [a, a+ c]. The first order condition is

∂U (bi; vi)

∂bi= −

µbi − ac

¶3(n−1)+ 3 (n− 1) 1

c(vi − bi)

µbi − ac

¶3(n−1)−1= 0;

i.e.,

−µbi − ac

¶+ 3 (n− 1) 1

c(vi − bi) = 0;

i.e.,

bi =a+ 3 (n− 1) vi3 (n− 1) + 1 .

Since this is an identity, we must have

a =a

3 (n− 1) + 1 =⇒ a = 0,

and

c =3 (n− 1)

3 (n− 1) + 1 .

3

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(b) What happens as n→∞?Answer: As n→∞,

bi → vi.

In the limit, each bidder bids his valuation, and the seller extracts all the gainsfrom trade.

[Hint: Since v1, v2, . . . , vn is independently distributed, for any w1, w2, . . . , wk, we have

Pr(v1 ≤ w1, v2 ≤ w2, . . . , vk ≤ wk) = Pr(v1 ≤ w1) Pr(v2 ≤ w2) . . .Pr(vk ≤ wk).]

4. This question is about a thief and a policeman. The thief has stolen an object. Hecan either hide the object INSIDE his car on in the TRUNK. The policeman stops thethief. He can either check INSIDE the car or the TRUNK, but not both. (He cannotlet the thief go without checking, either.) If the policeman checks the place where thethief hides the object, he catches the thief, in which case the thief gets -1 and the policegets 1; otherwise, he cannot catch the thief, and the thief gets 1, the police gets -1.

(a) Compute all the Nash equilibria.Answer: This is a matching-pennies game. There is a unique Nash equilibrium,in which Thief hides the object INSIDE or the TRUNK with equal probabilities,and the Policeman checks INSIDE or the TRUNK with equal probabilities.

(b) Now imagine that we have 100 thieves and 100 policemen, indexed by i =1, . . . , 100, and j = 1, . . . , 100. In addition to their payoffs above, each thief igets extra payoff bi form hiding the object in the TRUNK, and each policeman jgets extra payoff dj from checking the TRUNK. We have

b1 < b2 < · · · < b50 < 0 < b51 < · · · < b100,d1 < d2 < · · · < d50 < 0 < d51 < · · · < d100.

Policemen cannot distinguish the thieves from each other, nor can the thievesdistinguish the policemen from each other. Each thief has stolen an object, hidingit either in the TRUNK or INSIDE the car. Then, each of them is randomlymatched to a policeman. Each matching is equally likely. Again, a policemancan either check INSIDE the car or the TRUNK, but not both. Compute apure-strategy Bayesian Nash equilibrium of this game.Answer: A Bayesian Nash equilibrium: A thief i hides the object in

INSIDE if bi < 0TRUNK if bi > 0;

a policeman j checksINSIDE if dj < 0TRUNK if dj > 0.

This is a Bayesian Nash equilibrium, because, from the thief’s point of view thepoliceman is equally likely to to check TRUNK or INSIDE the car, hence it is thebest response for him to hide in the trunk iff the extra benefit from hiding in thetrunk is positive. Similar for the policemen.

4

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14.12 Midterm 2 Grade Distribution

0

10

20

30

40

50

60

70

80

90

1001 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45

Rank

grad

e

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14.12 Grade Distribution (Midterm 1 + Midterm 2)/2

20

30

40

50

60

70

80

90

1001 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45

rank

grad

e

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14.12 Game Theory — Final12/10/2001

Prof. Muhamet YildizInstructions. This is an open book exam; you can use any written material. You have 2hours 50 minutes. Each question is 20 points. Good luck!

1. Consider the following extensive form game.

1

2

1

2 2

3 1

00

00

13

IO

L

L L

R

RR

(a) Find the normal form representation of this game.

(b) Find all rationalizable pure strategies.

(c) Find all pure strategy Nash equilibria.

(d) Which strategies are consistent with all of the following assumptions?

(i) 1 is rational.(ii) 2 is sequentially rational.(iii) at the node she moves, 2 knows (i).(iv) 1 knows (ii) and (iii).

2. This question is about a milkman and a customer. At any day, with the given order,

• Milkman puts m ∈ [0, 1] liter of milk and 1−m liter of water in a container andcloses the container, incurring cost cm for some c > 0;

• Customer, without knowing m, decides on whether or not to buy the liquid atsome price p. If she buys, her payoff is vm−p and the milkman’s payoff is p−cm.If she does not buy, she gets 0, and the milkman gets −cm. If she buys, then shelearns m.

1

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(a) Assume that this is repeated for 100 days, and each player tries to maximize thesum of his or her stage payoffs. Find all subgame-perfect equilibria of this game.

(b) Now assume that this is repeated infinitely many times and each player tries tomaximize the discounted sum of his or her stage payoffs, where discount rate isδ ∈ (0, 1). What is the range of prices p for which there exists a subgame perfectequilibrium such that, everyday, the milkman chooses m = 1, and the customerbuys on the path of equilibrium play?

3. For the game in question 3.a, assume that with probability 0.001, milkman stronglybelieves that there is some entity who knows what the milkman does and will punishhim severely on the day 101 for each day the milkman dilutes the milk (by choosingm < 1). Call this type irrational. Assume that this is common knowledge. Forv > p > c, find a perfect Bayesian equilibrium of this game.

Bonus: [10 points] Discuss what would happen if the irrational type were known todilute the milk by accident with some small but positive probability.

4. Find a perfect Bayesian equilibrium of the following game.

11

-11

beer

beer

quiche

quiche

duel

don’tdu

el

don’t

don’tduel

don’t

duel

30

00

21

00

10

31

tw

ts

{.1}

{.9}

1111

-11

beer

beer

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quiche

duel

don’tdu

el

don’t

don’tduel

don’t

duel

30

00

21

00

10

31

tw

ts

{.1}

{.9}

5. A risk-neutral entrepreneur has a project that requires $100,000 as an investment, andwill yield $300,000 with probability 1/2, $0 with probability 1/2. There are two typesof entrepreneurs: rich who has a wealth of $1,000,000, and poor who has $0. For somereason, the wealthy entrepreneur cannot use his wealth as an investment towards thisproject. There is also a bank that can lend money with interest rate π. That is, ifthe entrepreneur borrows $100,000 to invest, after the project is completed he will payback $100, 000 (1 + π) – if he has that much money. If his wealth is less than thisamount at the end of the project, he will pay all he has. The order of the events is asfollows:

• First, bank posts π.

2

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• Then, entrepreneur decides whether to borrow ($100,000) and invest.• Then, uncertainty is resolved.

(a) Compute the subgame perfect equilibrium for the case when the wealth is commonknowledge.

(b) Now assume that the bank does not know the wealth of the entrepreneur. Theprobability that the entrepreneur is rich is 1/4. Compute the perfect Bayesianequilibrium.

3

Page 306: 6948772 Game Theory NOTES MIT

14.12 Game Theory — Final12/10/2001

Prof. Muhamet YildizInstructions. This is an open book exam; you can use any written material. You have 2hours 50 minutes. Each question is 20 points. Good luck!

1. Consider the following extensive form game.

1

2

1

2 2

3 1

00

00

13

IO

L

L L

R

RR

(a) Find the normal form representation of this game.A:

L ROL 2,2 2,2OR 2,2 2,2IL 3,1 0,0IR 0,0 1,3

(b) Find all rationalizable pure strategies.

L ROL 2,2 2,2OR 2,2 2,2IL 3,1 0,0

(c) Find all pure strategy Nash equilibria.

L ROL 2,2 2,2OR 2,2 2,2IL 3,1 0,0

1

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(d) Which strategies are consistent with all of the following assumptions?

(i) 1 is rational.(ii) 2 is sequentially rational.(iii) at the node she moves, 2 knows (i).(iv) 1 knows (ii) and (iii).

ANSWER: By (i) 1 does not play IR. Hence, by (iii), at the node she moves, 2knows that 1 does not play IR, hence he knows IL. Then, by (ii), 2 must play L.Therefore, by (i) and (iv), 1 must play IL. The answer is (IL,L).

2. This question is about a milkman and a customer. At any day, with the given order,

• Milkman puts m ∈ [0, 1] liter of milk and 1−m liter of water in a container andcloses the container, incurring cost cm for some c > 0;

• Customer, without knowing m, decides on whether or not to buy the liquid atsome price p. If she buys, her payoff is vm−p and the milkman’s payoff is p−cm.If she does not buy, she gets 0, and the milkman gets −cm. If she buys, then shelearns m.

(a) Assume that this is repeated for 100 days, and each player tries to maximize thesum of his or her stage payoffs. Find all subgame-perfect equilibria of this game.ANSWER: The stage game has a unique Nash equilibrium, in which m = 0 andthe customer does not buy. Therefore, this finitely repeated game has a uniquesubgame-perfect equilibrium, in which the stage equilibrium is repeated.

(b) Now assume that this is repeated infinitely many times and each player tries tomaximize the discounted sum of his or her stage payoffs, where discount rate isδ ∈ (0, 1). What is the range of prices p for which there exists a subgame perfectequilibrium such that, everyday, the milkman chooses m = 1, and the customerbuys on the path of equilibrium play?ANSWER: The milkman can guarantee himself 0 by always choosing m = 0.Hence, his continuation value at any history must be at least 0. Hence, in theworst equilibrium, if he deviates customer should not buy milk forever, giving themilkman exactly 0 as the continuation value. Hence, the SPE we are looking foris the milkman always chooses m=1 and the customer buys until anyone deviates,and the milkman chooses m=0 and the customer does not buy thereafter. If themilkman does not deviate, his continuation value will be

V =p− c1− δ

.

The best deviation for him (at any history on the path of equilibrium play) is tochoose m = 0 (and not being able to sell thereafter). In that case, he will get

Vd = p+ δ0 = p.

In order this to be an equilibrium, we must have V ≥ Vd; i.e.,p− c1− δ

≥ p,

2

Page 308: 6948772 Game Theory NOTES MIT

i.e.,p ≥ c

δ.

In order that the customer buy on the equilibrium path, we must also have p ≤ v.Therefore,

v ≥ p ≥ cδ.

3. For the game in question 2.a, assume that with probability 0.001, milkman stronglybelieves that there is some entity who knows what the milkman does and will punishhim severely on the day 101 for each day the milkman dilutes the milk (by choosingm < 1). Call this type irrational. Assume that this is common knowledge. For somev > p > c, find a perfect Bayesian equilibrium of this game. [If you find it easier,take the customers at different dates different, but assume that each customer knowswhatever the previous customers knew.]

ANSWER: [It is very difficult to give a rigorous answer to this question,so you would get a big partial grade for an informal answer that showsthat you understand the reputation from an incomplete-information pointof view.] Irrational type always sets m = 1. Since he will be detected whenever hesets m < 1 and the customer buys, the rational type will set either m = 1 or m = 0.We are looking for an equilibrium in which early in the relation the rational milkmanwill always setm = 1 and the customer will always buy, but near the end of the relationthe rational milkman will mix between m = 1 and m = 0, and the customer will mixbetween buy and not buy.

In this equilibrium, if the milkman setsm < 1 or the customer does not buy at anyt, then the rational milkman sets m = 0 at each s > t. In that case, if in addition thecostumer buys at some dates in the period {t+ 1, t+ 2, . . . , s− 1} and if the milkmanchooses m = 1 at each of those days, then the costumer will assign probability 1 tothat the milkman is irrational and buy the milk at s; otherwise, he will not buy themilk. On the path of such play, if the milkman sets m < 1 or the customer does notbuy at any t, then the rational milkman sets m = 0 and the costumer does not buy ateach s > t. In order this to be an equilibrium, the probability µt that the milkman isirrational at such history must satisfy

µt (100− t) (v − p)− (1− µt) p ≤ 0,

where the first term is the expected benefit from experimenting (if the milkman happensto be irrational) and the second term is the cost (if he is rational). That is,

µt ≤p

p+ (100− t) (v − p) .

Now we determine what happens if the milkman has always been setting m = 1,and the customer has been buying. In the last date, the rational type will set m = 0,and the rational type will set m = 1; hence, the buyer will buy iff

µ100 (v − p)− (1− µ100) p ≥ 0,

3

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i.e.,µ100 ≥

p

v.

Since we want him to mix, we setµ100 =

p

v.

We derive µt for previous dates using the Bayes’ rule and the indifference conditionnecessary for the customer’s mixing. Let’s write αt for the probability that the rationalmilkman sets m = 1 at t, and at = µt+(1− µt)αt for the total probability that m = 1at date t. Since the customer will be indifferent between buying and not buying att+1, his expected payoff at t+1 will be 0. Hence, his expected payoff from buying att is

at (v − p) + (1− at) (−p) .For indifference, this must be equal to zero, thus

at =p

v.

On the other hand, by Bayes’ rule,

µt+1 =µtat.

Therefore,µt = atµt+1 =

p

vµt+1.

That is,

µ100 =p

v

µ99 =³pv

´2µ98 =

³pv

´3...

Note that

at =p

v= µt + (1− µt)αt ⇒ αt =

pv− µt1− µt

.

Assume that (p/v)100 < 0.001. Then, we will have a date t∗ such that³pv

´101−t∗< 0.001 <

³pv

´100−t∗.

At each date t > t∗, we will have µt = (p/v)101−t and the players will mix so thatat =

pv. At each date t < t∗, the milkman will set m = 1 and the customer will buy.

At date t∗, the rational milkman will mix so that³pv

´100−t∗= µt∗+1 =

µt∗

at∗=0.001

at∗,

4

Page 310: 6948772 Game Theory NOTES MIT

henceat∗ =

0.001¡pv

¢100−t∗ .Note that at∗ > p/v, hence the customer will certainly buy at t∗.

Let’s write βt for the probability that the customer will buy at day t. In the day99, if the rational milkman sets m = 1, he will get

U = β99 (p− c) + β99β100p+ (1− β99) (−c) ,

where the first term is the profit from selling at day 99, the second term is the profitfrom day 100 (when he will set m = 0), and the last term is the loss if the customerdoes not buy at day 99. If he sets m = 0, he will get β99p (from the sale at 99, andwill get zero thereafter). Hence, he will set m = 1 iff

β99 (p− c) + β99β100p+ (1− β99) (−c) ≥ β99p

i.e.,β99β100 ≥

c

p.

We are looking for an indifference, hence we set

β99β100 =c

p.

Similarly, at day 98 the rational milkman will set m = 1 iff

β98 (p− c) + β98β99p+ (1− β98) (−c) ≥ β98p,

where the second term is due to the fact that at date 99 he will be indifferent betweenchoosing m = 0 and m = 1. For indifference, we set

β98β99 =c

p.

We will continue on like this as long as we need the milkman to mix. That is, we willhave

βt∗βt∗+1 =c

p,

βt∗+1βt∗+2 =c

p,

...

β98β99 =c

p,

β99β100 =c

p.

5

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As we noted before, βt∗ = 1. Hence, βt∗+1 =cp. Hence, βt∗+2 = 1, ... That is,

βt∗ = 1,

βt∗+1 =c

p,

βt∗+2 = 1,

βt∗+3 =c

p,

...

Bonus: [10 points] Discuss what would happen if the irrational type were known todilute the milk by accident with some small but positive probability.

4. Find a perfect Bayesian equilibrium of the following game.

11

-11

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beer

quiche

quiche

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don’t

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don’t

don’t

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don’t

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30

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{.9}

1111

-11

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don’t

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30

00

21

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{.9}

ANSWER:

6

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30

00

21

00

10

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{.1}

{.9}

{1}

{0}{1}

{0}

11

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{1}

{0}{1}

{0}

5. A risk-neutral entrepreneur has a project that requires $100,000 as an investment, andwill yield $300,000 with probability 1/2, $0 with probability 1/2. There are two typesof entrepreneurs: rich who has a wealth of $1,000,000, and poor who has $0. For somereason, the wealthy entrepreneur cannot use his wealth as an investment towards thisproject. There is also a bank that can lend money with interest rate π. That is, ifthe entrepreneur borrows $100,000 to invest, after the project is completed he will payback $100, 000 (1 + π) – if he has that much money. If his wealth is less than thisamount at the end of the project, he will pay all he has. The order of the events is asfollows:

• First, bank posts π.• Then, entrepreneur decides whether to borrow ($100,000) and invest.• Then, uncertainty is resolved.

(a) Compute the subgame perfect equilibrium for the case when the wealth is commonknowledge.ANSWER: The rich entrepreneur is always going to pay back the loan in fullamount, hence his expected payoff from investing (as a change from not investing)is

(0.5)(300, 000)− 100, 000 (1 + π) .

Hence, he will invest iff this amount is non-negative, i.e.,

π ≤ 1/2.

Thus, the bank will set the interest rate at

πR = 1/2.

7

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The poor entrepreneur is going to pay back the loan only when the project suc-ceeds. Hence, his expected payoff from investing is

(0.5)(300, 000− 100, 000 (1 + π)).

He will invest iff this amount is non-negative, i.e.,

π ≤ 2.

Thus, the bank will set the interest rate at

πP = 2.

(b) Now assume that the bank does not know the wealth of the entrepreneur. Theprobability that the entrepreneur is rich is 1/4. Compute the perfect Bayesianequilibrium.ANSWER: As in part (a), the rich type will invest iff π ≤ πR = .5, and the poortype will invest iff π ≤ πP = 2. Now, if π ≤ πR, the bank’s payoff is

U (π) =1

4100, 000 (1 + π) +

3

4

·1

2100, 000 (1 + π) +

1

20

¸− 100, 000

=5

8100, 000 (1 + π)− 100, 000

≤ 5

8100, 000 (1 + πR)− 100, 000

=5

8100, 000 (1 + 1/2)− 100, 000 = − 1

16100, 000 < 0.

If πR < π ≤ πP , the bank’s payoff is

U (π) =3

4

·1

2100, 000 (1 + π) +

1

20− 100, 000

¸=

3

8100, 000 (π − 1) ,

which is maximized at πP , yielding 38100, 000. If π > πP , U (π) = 0. Hence, the

bank will choose π = πP .

8

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14.12 Game Theory — Final12/21/2001

Prof. Muhamet YildizInstructions. This is an open book exam; you can use any written material. You have 2hours 50 minutes. Each question is 20 points. Good luck!

1. Consider the following extensive form game.

1 2A

D

α

δ

(1,1) (0,3)

(1,4) a

d

(2,2)

1

(a) Find the normal form representation of this game.

1\2 α δAa 1,4 0,3Ad 2,2 0,3Da 1,1 1,1Dd 1,1 1,1

(b) Find all rationalizable pure strategies.

1\2 α δAd 2,2 0,3Da 1,1 1,1Dd 1,1 1,1

(c) Find all pure strategy Nash equilibria.

1\2 α δAd 2,2 0,3Da 1,1 1,1Dd 1,1 1,1

(d) Which strategies are consistent with all of the following assumptions?

(i) 1 is rational.(ii) 2 is sequentially rational.(iii) at the node she moves, 2 knows (i).(iv) 1 knows (ii) and (iii).

1

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ANSWER: By (i) 1 does not play Aa. Hence, by (iii), at the node she moves, 2knows that 1 does not play Aa, hence he knows that 1 plays Ad. Then, by (ii), 2must play δ. Therefore, by (i) and (iv), 1 must play Ad or Aa. The answer is 1plays A, given chance 2 would play δ.

2. This question is about a game between a possible applicant (henceforth student) toa Ph.D. program in Economics and the Admission Committee. Ex-ante, AdmissionCommittee believes that with probability .9 the student hates economics and withprobability .1 he loves economics. After Nature decides whether student loves or hateseconomics with the above probabilities and reveals it to the student, the student decideswhether or not to apply to the Ph.D. program. If the student does not apply, both thestudent and the committee get 0. If student applies, then the committee is to decidewhether to accept or reject the student. If the committee rejects, then committee gets0, and student gets -1. If the committee accepts the student, the payoffs depend onwhether the student loves or hates economics. If the student loves economics, he will besuccessful and the payoffs will be 20 for each player. If he hates economics, the payoffsfor both the committee and the student will be -10. Find a separating equilibrium anda pooling equilibrium of this game.

ANSWER: A separating equilibrium:

Hate

Love

.9

.1

Apply

Don’t

(0,0)

Accept

Reject

(-10,-10)

(-1,0)

Apply

Don’t

(0,0)

Accept

Reject

(20,20)

(-1,0)

{1}

{0}

A pooling equilibrium:

2

Page 316: 6948772 Game Theory NOTES MIT

Hate

Love

.9

.1

Apply

Don’t

(0,0)

Accept

Reject

(-10,-10)

(-1,0)

Apply

Don’t

(0,0)

Accept

Reject

(20,20)

(-1,0)

{.1}

{.9}

3. Consider a bargaining problem where two risk-neutral players are trying to divide adollar they own, which they cannot use until they reach an agreement. The playersdo not discount the future, but at the end of each rejection of an offer the bargainingbreaks down with probability 1− δ ∈ (0, 1) and each player gets 0.

(a) Consider the following bargaining procedure. Player 1 makes an offer (x, 1− x),where x is player 1’s share. Then, player 2 decides whether or not to accept theoffer. If she accepts, they implement the offer, yielding division (x, 1− x). If sherejects the offer, then with probability 1− δ, the bargaining breaks down an eachgets 0; with probability δ, player 1 makes another offer, which will be accepted orrejected by player 2 as above. (If player 2 rejects the offer, bargaining will breakdown with probability 1 − δ again.) If the offer is rejected and the bargainingdid not break down, now player 2 makes a counter offer, and player 1 acceptsor rejects this counter offer as above. If the offer is rejected, this time the gamewill end, and each will get 0. Find the subgame-perfect Nash equilibrium of thisgame. Compute the expected payoff of each player at the beginning of the gamein this equilibrium.

ANSWER: On the last day, 1 accepts any offer, so 2 offers (0,1). Hence, on theprevious day, 2 accepts an offer iff she gets at least δ. Hence, 1 offers (1− δ, δ)–accepted. Thus, in the first day, 2 accepts an offer iff she gets at least δ2. Hence,1 offers

¡1− δ2, δ2

¢– accepted. The expected payoffs are

¡1− δ2, δ2

¢.

(b) Compute the subgame-perfect equilibrium of the game in which the procedure inpart (a) is repeated 2 times. (The probability of bargaining breakdown after eachrejection is 1− δ, except for the end of the game.)

ANSWER: The last period as above. Let’s look at the first period. On the lastday of the first period, 1 accepts an offer iff he gets at least δ

¡1− δ2

¢, so 2 offers

3

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¡δ¡1− δ2

¢, 1− δ

¡1− δ2

¢¢. Hence, on the previous day, 2 accepts an offer iff she

gets at least δ¡1− δ

¡1− δ2

¢¢. Hence, 1 offers¡

1− δ¡1− δ

¡1− δ2

¢¢, δ¡1− δ

¡1− δ2

¢¢¢=¡1− δ + δ2

¡1− δ2

¢, δ¡1− δ

¡1− δ2

¢¢¢– accepted. Thus, in the first day, 2 accepts an offer iff she gets at leastδ2¡1− δ

¡1− δ2

¢¢. Hence, 1 offers¡

1− δ2¡1− δ

¡1− δ2

¢¢, δ¡1− δ

¡1− δ2

¢¢¢=¡1− δ2 + δ3

¡1− δ2

¢, δ2¡1− δ

¡1− δ2

¢¢¢– accepted.

(c) Find the subgame-perfect equilibrium of the game in which this procedure isrepeated until they reach an agreement. Note that player 1 makes two offers,then 2 makes one offer, then 1 makes two offers, and so on. You need to showthat the proposed strategy profile is in fact a subgame-perfect equilibrium. (Theprobability of bargaining breakdown after each rejection is 1− δ.)[Hint: One way is to compute the SPE for the game in which the procedureis repeated n times and let n → ∞. A somewhat easier way is to consideran alternating offer bargaining procedure with some effective discount rates –different for a different player.]

ANSWER: If you compare the calculations above with the calculations with thealternating offer case with asymmetric discount rates, you should realize thatthe first offer player 1 makes and the offer player 2 makes are identical to theoffers players 1 and 2 make, respectively, if the discount rates were δ1 = δ andδ2 = δ2. Intuitively, in his second offer player 1 makes player 2 indifferent betweenaccepting 1’s second offer and making an offer next day, and in his first offer hemakes her indifferent between accepting the offer and waiting for the second offer.Therefore, 2 is indifferent between accepting 1’s first offer and waiting two daystwo make an offer, as in the the alternating offer case when her discount rateis δ2. Now conjecture that the subgame-perfect equilibrium would be as in thealternating offer game with above discount rates. That is,

• in his first offer, player 1 offersµ1− δ21− δ1δ2

, 1− 1− δ21− δ1δ2

¶≡

µ1− δ21− δ1δ2

,δ2 (1− δ1)

1− δ1δ2

¶≡µ1− δ2

1− δ3,δ2 (1− δ)

1− δ3

¶≡

µ1 + δ

1 + δ + δ2,

δ2

1 + δ + δ2

¶;

• in his secon offer, he will offerµ1− δ (1− δ1)

1− δ1δ2,δ (1− δ1)

1− δ1δ2

¶≡µ

1 + δ2

1 + δ + δ2,

δ

1 + δ + δ2

¶;

• player 2 will offerµ1− 1− δ1

1− δ1δ2,1− δ11− δ1δ2

¶≡µδ1 (1− δ2)

1− δ1δ2,1− δ11− δ1δ2

¶≡µ

δ + δ2

1 + δ + δ2,

1

1 + δ + δ2

¶.

4

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Player 1’s first offer and player 2’s offer are by formula for alternating offer, player1’s second offer is calculated by backward induction using the player 2’s offer inthe next period. Using single deviation property, you need to check that this isan equilibrium.

4. We have an employer and a worker, who will work as a salesman. The worker may bea good salesman or a bad one. In expectation, if he is a good salesman, he will make$200,000 worth of sales, and if he is bad, he will make only $100,000. The employergets 10% of the sales as profit. The employer offers a wage w. Then, the worker acceptsor rejects the offer. If he accepts, he will be hired at wage w. If he rejects the offer,he will not be hired. In that case, the employer will get 0, the worker will get hisoutside option, which will pay $15,000 if he is good, $8,000 if he is bad. Assume thatall players are risk-neutral.

(a) Assume that the worker’s type is common knowledge, and compute the subgame-perfect equilibrium.

ANSWER: A worker will accept a wage iff it is at least as high as his outsideoption, and the employer will offer the outside option – as he still makes profit.That is, 15,000 for the good worker 8,000 for the bad.

(b) Assume that the worker knows his type, but the employer does not. Employerbelieves that the worker is good with probability 1/4. Find the perfect BayesianNash equilibrium.

ANSWER: Again a worker will accepts an offer iff his wage at least as high ashis outside option. Hence if w ≥ 15, 000 the offer will be accepted by both types,yielding

U (w) = (1/4) (.1) 200, 000 + (3/4) (.1) 100, 000− w = 12, 500− w < 0

as the profit for the employer. If 8, 000 ≤ w < 15, 000, then only the bad workerwill accept the offer, yielding

U (w) = (3/4) [(.1) 100, 000− w] = (3/4) [10, 000− w]

as profit. If w < 0, no worker will accept the offer, and the employer will get 0. Inthat case, the employer will offer w = 8, 000, hiring the bad worker at his outsideoption.

(c) Under the information structure in part (b), now consider the case that the em-ployer offers a share s in the sales rather than the fixed wage w. Compute theperfect Bayesian Nash equilibrium.

ANSWER: Again a worker will accept the share s iff his income is at least as highas his outside option. That is, a bad worker will accept s iff

100, 000s ≥ 8, 000

5

Page 319: 6948772 Game Theory NOTES MIT

i.e.,

s ≥ sB = 8, 000

100, 000= 8%.

A good worker will accept s iff

s ≥ sG = 15, 000

200, 000= 7.5%.

In that case, if s < sG no one will accept the offer, and the employer will get 0; ifsG ≤ s < sB, the good worker will accept the offer and the employer will get

(1/4) (10%− s) 200, 000 = 50, 000 (10%− s) ,and if s ≥ sB, each type will accept the offer and the employer will get

(10%− s) [(1/4) 200, 000 + (3/4) 100, 000] = 125, 000 (10%− s) .Since 125, 000 (10%− sB) = 2%125, 000 = 2, 500 is larger than 50, 000 (10%− sG) =2.5%50, 000 = 1, 250, he will offer s = sB, hiring both types.

5. As in question 4, We have an employer and a worker, who will work as a salesman.Now the market might be good or bad. In expectation, if the market is good, theworker will make $200,000 worth of sales, and if the market is bad, he will make only$100,000 worth of sales. The employer gets 10% of the sales as profit. The employeroffers a wage w. Then, the worker accepts or rejects the offer. If he accepts, he will behired at wage w. If he rejects the offer, he will not be hired. In that case, the employerwill get 0, the worker will get his outside option, which will pay $12,000. Assume thatall players are risk-neutral.

(a) Assume that whether the market is good or bad is common knowledge, and com-pute the subgame-perfect equilibrium.

ANSWER: A worker will accept a wage iff it is at least as high as his outsideoption 12,000. If the market is good, the employer will offer the outside optionw = 12, 000, and make 20, 000− 12, 000 = 8, 000 profit. If the market is bad, thereturn 10,000 is lower than the worker’s outside option, and the worker will notbe hired.

(b) Assume that the employer knows whether the market is good or bad, but theworker does not. The worker believes that the market is good with probability1/4. Find the perfect Bayesian Nash equilibrium.

ANSWER: As in part (a). [We will have a separating equilibrium.]

(c) Under the information structure in part (b), now consider the case that the em-ployer offers a share s in the sales rather than the fixed wage w. Compute aperfect Bayesian Nash equilibrium.

ANSWER: Note that, since the return is 10% independent of whether the marketis good or bad, the employer will make positive profit iff s < 10%. Hence, except

6

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for s = 10%, we must have a pooling equilibrium. Hence, at any s, the worker’sincome is

[(1/4) 200, 000 + (3/4) 100, 000] s = 125, 000s.

This will be at least as high as his outside option iff

s ≥ s∗ = 12, 000

125, 000= 9.6% < 10%.

Hence an equilibrium: the worker will accept an offer s iff s ≥ s∗, and the employerwill offer s∗. The worker’s beliefs at any offer s is that the market is good withprobability 1/4. [Note that this is an inefficient equilibrium. When the market isbad, the gains from trade is less than the outside option.]

There are other inefficient equilibria where there is no trade (i.e., worker isnever hired). In any such equilibrium, worker take any high offer as a sign that themarket is bad, and does not accept an offer s unless s ≥ 12, 000/100, 000 = 12%,and the employer offers less than 12%. When the market is good, in any suchpure strategy equilibrium, he must in fact be offering less than s∗. (why?) Forinstance, employer offers s = 0 independent of the market, and the worker accepts iff s > 12%.

7

Page 321: 6948772 Game Theory NOTES MIT

14.12 Final Exam Grade Distribution

0

20

40

60

80

100

1201 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45

rank

grad

e

Average = 73.82 Std. Dev = 20.74

Page 322: 6948772 Game Theory NOTES MIT

Final Grade, Final Grade + Quiz

30

40

50

60

70

80

90

100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Series1 Series2

88

76

A+

A

A-

B+

52

45

B

B-

C

Page 323: 6948772 Game Theory NOTES MIT

Final Grade (.25M1+.25M2+.4F+.1PS)

30

40

50

60

70

80

90

100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

rank

88

76

A+

A

A-

B+52

45

B

B-

C

Page 324: 6948772 Game Theory NOTES MIT

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Page 327: 6948772 Game Theory NOTES MIT

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Page 328: 6948772 Game Theory NOTES MIT

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Page 329: 6948772 Game Theory NOTES MIT

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Page 332: 6948772 Game Theory NOTES MIT

14.12 Game Theory – Midterm I10/19/2000

Prof. Muhamet YildizInstructions. This is an open book exam; you can use any written material. You have one hour and20 minutes. Each question is 33 points. Good luck!

1. Consider the following game.

1\2 L M R

T 3,2 4,0 1,1

M 2,0 3,3 0,0

B 1,1 0,2 2,3

a. Iteratively eliminate all the strictly dominated strategies.b. State the rationality/knowledge assumptions corresponding to each elimination.c. What are the rationalizable strategies?d. Find all the Nash equilibria. (Don’t forget the mixed-strategy equilibrium!)

2. Consider the following extensive form game.

1

B

X

2

L R RL

T

(2,2)

(3,1) (0,0) (5,0) (0,1)

2

a. Find the normal form representation of this game.b. Find all pure strategy Nash equilibria.c. Which of these equilibria are subgame perfect?

3. Consider two agents �1,2� owning one dollar which they can use only after they divide it. Eachplayer’s utility of getting x dollar at t is �tx for � � �0,1�. Given any n � 0, consider thefollowing n-period symmetric, random bargaining model. Given any date t � �0,1,� ,n � 1�,we toss a fair coin; if it comes Head (which comes with probability 1/2), we select player 1; if itcomes Tail, we select player 2. The selected player makes an offer �x,y� � �0,1�2 such thatx � y � 1. Knowing what has been offered, the other player accepts or rejects the offer. If theoffer �x,y� is accepted, the game ends, yielding payoff vector ��tx,�ty�. If the offer is rejected,we proceed to the next date, when the same procedure is repeated, except for t � n � 1, afterwhich the game ends, yielding (0,0). The coin tosses at different dates are stochasticallyindependent. And everything described up to here is common knowledge.

a. Compute the subgame perfect equilibrium for n � 1. What is the value of playing this gamefor a player? (That is, compute the expected utility of each player before the coin-toss,given that they will play the subgame-perfect equilibrium.)

b. Compute the subgame perfect equilibrium for n � 2. Compute the expected utility of each

Page 333: 6948772 Game Theory NOTES MIT

player before the first coin-toss, given that they will play the subgame-perfect equilibrium.

c. What is the subgame perfect equilibrium for n � 3.

Page 334: 6948772 Game Theory NOTES MIT

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Page 349: 6948772 Game Theory NOTES MIT

14.12 Solutions for Question 7 (or 8)Kenichi Amaya1

November 9, 2001

(a)

Let’s denote firm 2’s by c2, i.e., c2 ∈ {cL, cH}. Let p1 and p2 be each firm’sprice (action). Then, each firm’s payoff functions can be written as

π1(p1, p2) =

p1 − c ifp1 < p2p1−c

2 ifp1 = p2

0 ifp1 > p2

,

π2(p1, p2; c2) =

p2 − c2 ifp2 < p1p2−c2

2 ifp2 = p1

0 ifp2 > p1

(b)

A pure strategy profile {p∗1, (p∗2L, p∗2H)} is a Bayesian Nash equilibrium if

p∗1 = argmaxp1θπ1(p1, p

∗2L) + (1− θ)π1(p1, p

∗2H),

p∗2L = argmaxp2π2(p∗1, p2; cL),

p∗2H = argmaxp2π2(p∗1, p2; cH),

(c)

If there exists a pure strategy Bayesia Nash equilibrium, it must be true that

p∗1 = min{p∗2L, p∗2H}.

To see this, if p∗1 < min{p∗2L, p∗2H}, firm 1 is always selling to the whole market.If it increases the price slightly, it still sells the whole market. So it can increasethe payoff. If p∗2k < p∗1, where k is either L or H, firm 2 is selling to the wholemarket when its type is k. If it increases the price slightly, it still sells the wholemarket. So it can increase the payoff.

Next, if there exists a pure strategy Bayesia Nash equilibrium, it must betrue that

p∗1 = min{p∗2L, p∗2H} = c.

To see this, if p∗1 < c, firm 1 is selling positive amount and losing money.Therefore it’s better to charge a price higher than R and not to sell to get zeroprofit. On the other hand, if p∗1 < c, firm 1 can increase it’s profit by slightlyreducing its price.

Finally, p∗1 = min{p∗2L, p∗2H} = c can not be an equilibrium because firm 2of type L can increase its profit by choosing a price slightly smaller than c.

1E52-303, [email protected], 253-3591

1

Page 350: 6948772 Game Theory NOTES MIT

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14.12 Game Theory

Fall 2000

Final Exam Solutions

December 14, 2001

1. (a) Given perfect competition among firms, wages will be w1 = 1,w2 = 2and w3 = 3 for types t = 1, 2, 3 respectively. The incentive compatibilityconstraints give us the conditions that the education levels must fulfillin equilibrium: e∗2 ≥ 1 + e∗1 and e∗3 ≥ 2 + e∗1 for t = 1, e∗2 ≤ 2 + e∗1 ande∗3 ≥ 2 + e∗2 for t = 2, and e∗3 ≤ 3 + e∗2 and e∗3 ≤ 6 + e∗1 for t = 3.If we let, for example, e∗1 = 0, then e∗2 = 1 and e∗3 = 3. Beliefs areµ(t = 1|e′) = 1, where e′ is in [0, inf), µ(t = 2|e = 1) = 1 and µ(t = 3|e =3) = 1.

(b) Let w1 = 1 as before, since type 1 is separating. Given that types 2and 3 are pooling, their wage is the expected marginal productivity or1/2(2) + 1/2(3) = 5/2. Letting e∗1 = 0, the compatibility constraintsprovide the conditions that the education levels must fulfill in equilibrium:e∗ ≥ 3/2 and e∗ ≤ 3. For instance, we choose e∗ = 2. Beliefs areµ(t = 1|e 6= e∗) = 1, µ(t = 2|e∗) = 1/2 and µ(t = 3|e∗) = 1/2.

2. (a) Let the separating equilibrium be one where top plays R, bottom plays L.The beliefs are µ(top|R) = 1 and µ(bottom|R) = 1. Player 2, the receiver,plays up in the information set on the right-hand side, and down, on theinformation set on the left-hand side. Now it is easy to check that player1’s types do not want to deviate: if top type plays L instead, she gets0 as opposed to 3, while bottom type, if she plays R instead, gets 1 asopposed to 2.

(b) Pooling on R: player 2 plays down when sees R since EU(up|R) =0.4(1) + 0.6(0) = 0.4 < EU(down|R) = 0.4(0) + 0.6(1) = 0.6. Anybeliefs about types given L actually support this equilibrium.

(c) Mixed strategies equilibrium: Bottom mixes αR + (1 − α)L. Player 2mixes 1/2up+1/2down when sees R. By applying Bayes’ rule and forcingit to be equal to 1/2, because this is the value that would make thebottom type want to mix, we have µ(top|R) = 0.4(1)/(0.4+0.6(α)) = 1/2.In particular, for player 2, EU(up|R) = P (top|R) and EU(down|R) =1− P (top|R). Setting these two equal, because we need player 2 to mix

1

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after observing R in order for bottom type to want to mix, we obtain thatP (top|R) = 1/2. From the Bayes’ rule equation we solve for α above, suchthat bottom type’s strategy is 2/3R + 1/3L. Top type always plays R.Player 2 plays 1/2up+ 1/2down when sees R and plays down when seesL. Beliefs are µ(bottom|L) = 1 and µ(top|R) = 1/2.

3. Given no discounting, and the fact that the plaintiff gets to make the first offer,the game ends in the first day, with the plaintiff making an offer w and thedefendant accepting the offer. In particular, taking into account the legal feesboth parties must pay until court day, and the amount the defendant must paythen, which is certain and common knowledge, the plaintiff will offer 1, 110Kand the defendant will accept all offers less than or equal to this amount, andreject all others.

4. One pooling equilibrium of this game would be for the firm to reject all offersw0 > 5 and accept all others, for both types of workers to offer w0 = 5, and tooffer w1 = 5 in the following period if reached. Beliefs: µ(type = 0|w 6= 5) = 1.Workers do not want to deviate because that would lead to a rejection and awage of 5 in the following period, which is what they can get now (so they areindifferent).

2

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