Advanced Microeconomics - Tistory

Microeconomics Game Theory

Advanced Microeconomics

Pierre von Mouche

Wageningen University

2013


Outline

1 MicroeconomicsMotivationReminderThe corePareto efficiencyWalrasian equilibriumExistence of equilibriumWelfare theoremsLarge economiesQuiz

2 Game TheoryMotivationGames in strategic formGames in extensive form


Capitalism

According to Adam Smith: a laissez-faire approach toeconomics is the essential way to ensure prosperity for anation as a whole. Ultimately, when capitalism is allowed torun its course, the greed and self-interest of the capitalistswould produce results in the economy that benefit not onlythe individual, but society as well.Scientific problem: proof such claims with models.To prove: equilibria exist, are efficient and stable.


Overall notations

Space of goodsRn+ = x = (x1, . . . , xn) ∈ Rn | x1 ≥ 0, . . . , xn ≥ 0.

Utility function u : Rn+ → R.

Prices p = (p1, . . . ,pn). All pi > 0 (if not stated otherwise).Note: prices do not appear in the utility function.Budget (income) m ≥ 0.Marshallian demand correspondences xi(p; m). Oftenthese are functions.

Producer theory will be less important for what follows.


Concrete functions

Cobb-douglas:u(x) = Axα1

1 · · · xαnn ,

xi(p; m) =αi

α1 + · · ·+ αn

mpi.

Ces:u(x) = A(α1xρ1 + · · ·+ αnxρn )

h/ρ.


Concrete functions (cont.)

Leontief:u(x) = min (x1/α1, . . . , xn/αn).

ExampleWhat are the marshallian demand functions for this utilityfunction?

Answer:xi(p; m) =

αimα1p1 + · · ·+ αnpn

.



Solow (for n = 2):

u(x1, x2) = α1x1 + α2x2.

Here there are marshallian demand correspondences (insteadof functions).

Maximum (for n = 2):

u(x1, x2) = max (α1x1, α2x2).

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Special quasi-linear:

u(x1, x2) = α√

x1 + x2.

In general for quasi-linear functions

u(x1, . . . , xn) = v(x1, . . . , xn−1) + xn

the marshallian demand for the quasi-linear goods (i.e.1, . . . ,n − 1) is independent of the income (if this is not toosmall).

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Increasingness

Remember the notations ≤, <,. For instance:(3,4) ≤ (3,4), (3,4) ≤ (3,5).(3,4) < (3,5), (3,4) < (4,5).(3,4) (4,5).

Remember:u is increasing: x ≤ y ⇒ u(x) ≤ u(y).u is strongly increasing: x < y ⇒ u(x) < u(y).u is strictly increasing: u is increasing, andx y ⇒ u(x) < u(y).

ExampleIs the cobb-douglass utility function strongly increasing?Answer: no (but it is strictly increasing).

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Setting for core

Pure exchange economy. Specified by:N consumers and n goods.for each consumer h a good bundle

ωh = (ωh1 , . . . , ω

hn) > 0

(i.e. each consumer has something), to be calledinitial good bundle such that for each good k ,

Ok :=N∑

h=1

ωhk > 0

(i.e. each good is present).for each consumer h a continuous utility function

uh : Rn+ → R.

Note: there are (still) no prices!

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Allocations

Allocation: X := (x1, . . . ,xN) ∈ (Rn+)N .

Initial allocation: Ω := (ω1, . . . ,ωN).Feasible allocation:

∑Nh=1 xh

k = Ok (1 ≤ k ≤ n). Note: not≤-sign!Feasible allocation is interior if 0 < xh

k < Ok for all h and k .

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Pareto


Pareto (ctd.)

Vilfredo Pareto (1848-1923):Italian engineer, economist and sociologist.Very good knowledge of mathematics.For 20 years director of two Italian railway companies.Later, motivated by Walras to switch to economic research.After disenchantment in economics, switched to sociology.His articles are difficult to read.


Pareto efficiency

DefinitionA feasible allocation X

is called weakly pareto efficient allocation if there is nofeasible allocation Y with uh(yh) > uh(xh) (1 ≤ h ≤ N).(strongly) pareto efficient if there is no feasible allocation Ywith uh(yh) ≥ uh(xh) (1 ≤ h ≤ N) with at least one of theseinequalities strict.

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Pareto efficient allocations

Each strongly pareto efficient allocation is also weaklypareto efficient. So each weakly pareto inefficientallocation is strongly pareto inefficient.In fact weakly and strongly pareto efficiency make sense inother contexts. (See next Example.)


Example

ExampleConsider N agents for which there are a finite number of ’statesof the world’. Denote by (a1,a2, . . . ,aN) a state where agent ihas ‘utility’ ai . Determine for the following situations whichstates are strongly pareto efficient and which are weakly paretoefficient.

a. A = (5,10), B = (6,9), C = (6,11), D = (4,12).Answer: Weak: B,C,D. Strong: C,D.

b. A = (6,6), B = (6,7), C = (3,2), D = (7,6), E =(5,6), F = (11,1).Answer: Weak: A,B,D,F . Strong: B,D,F .

c. A = (5,4), B = (9,1), C = (3,8).Answer: Weak: A,B,C. Strong: A,B,C.

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Example (cont.)

Example

d. A = (−4,8), B = (−4,3), C = (−5,−3), D = (6,0).Answer: Weak: A,B,D. Strong: A,D.

e. A = (1,2,6,4), B = (4,8,3,2), C = (1,8,1,2), D =(0,0,0,0).Answer: Weak: A,B,C. Strong: A,B.

f. A = (1,3,5), B = (1,3,5), C = (2,4,3).Answer: Weak: A,B,C. Strong: A,B,C.


Example (cont.)

Example

g. A = (1,3,5), B = (1,3,5), C = (2,4,3), D = (1,3,6).Answer: Weak: A,B,C,D. Strong: C,D.

h. A = (1), B = (−8), C = (137).Answer: Weak: C. Strong: C.


Strong versus weak pareto efficiency

TheoremIf each utility function is continuous and strongly increasing,then the set of weak and strong pareto efficient allocations isthe same.

Proof.This is a technical result. We omit here its proof. (Also seeexercise 5.44 in the text book).


Barter equilibrium

What would be a reasonable feasible allocationX = (x1, . . . ,xN) when the N consumers exchange goods?

1. X is individually rational.

This is defined as follows: X is individually rational for consumerh, if uh(xh) ≥ uh(ωh) and X is individually rational if X is foreach consumer individually rational.

Is that all?

If an allocation is strongly pareto inefficient, then there isanother feasible allocation making someone better off and noone worse off: then a trade can be arranged to which noconsumer will object. So:

2 X is strongly pareto efficient.

Is that all?


Core

NO:

3 X should belong to the core.

The core is the set of feasible allocations such that there is nocoalition (i.e. non-empty subset of consumers) that can improveon it.

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Definition

A feasible allocation X = (x1, . . . ,xN) belongs to the (strong)core if there is no coalition S and good bundles yh (h ∈ S) suchthat

1∑

h∈S yh =∑

h∈S ωh;2 uh(yh) ≥ uh(xh) for all h ∈ S with at least one inequality

strict.

The core depends on the initial allocation! The set of paretoefficient allocations does not.


Properties of core

TheoremEach element of the core is individually rational and paretoefficient.

Proof.Let X be in the core.X is individual rational: take S = i and yi = ωi . Thenui(ωi) > ui(xi) does not hold. So ui(ωi) ≤ ui(xi) holds.X is pareto efficient: take S = 1, . . . ,N and Y an arbitraryfeasible allocation.


Core for N = 2

TheoremFor N = 2 the core is the set of pareto efficient individuallyrational allocations.

Proof.Because S = 1, S = 2 or S = 1,2.


Non-empty core?

Very fundamental question: is the core non-empty?We shall see:

TheoremEach pure exchange economy where each utility function iscontinuous, strictly quasi-concave and strongly increasing hasa non-empty core.

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Box of Edgeworth

Box of Edgeworth:

D := x ∈ R2 | 0 ≤ xk ≤ Ok (1 ≤ k ≤ 2).

One can identify a feasible allocation with the correspondingpoint in D.

The set of pareto efficient allocations in the box is thecontract curve.


Pareto efficient allocations: necessary condition

Theorem

If X = (x1, . . . ,xN) is an interior pareto efficient allocation, thenunder mild differentiability conditions, equality for eachconsumer of each specific marginal rate of substitution holds.

Proof.We omit the proof which can be given with the method ofLagrange and only illustrate with a figure the idea.


Pareto efficient allocations: sufficient condition

TheoremIf each utility function is quasi-concave, then under milddifferentiability conditions each interior feasible allocation wherefor each consumer each specific marginal rate of substitution isthe same, is pareto efficient.

Proof.We omit the proof and only illustrate with a figure the idea.


Example

ExampleDetermine the contract curve for:

uA = xα11 xα2

2 , uB = xγ11 xγ2

2 ,

where α1 + α2 = γ1 + γ2 = 1. Answer:

xA2 =

α2γ1O2xA1

α1γ2O1 + (γ1 − α1)xA1,

where 0 ≤ xA1 ≤ O1.

(In fact the points on the boundary of the box need specialinvestigation; we shall do that in an exercise.)


Walrasian equilibrium

Now allow for markets and prices: p ∈ Rn+.

Definition(p; X) is a Walrasian equilibrium if– for each h, xh is a maximiser of the utility function

uh : Rn+ → R under the restriction p · x ≤ p · ωh;

– X is feasible; i.e.∑N

h=1 xhk = Ok (1 ≤ k ≤ n).

p: equilibrium price vector.

X: equilibrium allocation.

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Equilibrium and aggregate excess demand

Suppose that each consumer has well-defined marshalliandemand functions. For this situation we can define the notion ofaggregate excess demands z1, . . . , zn as follows:

zi(p) :=N∑

h=1

ehi (p),

ehi (p) := xh

i (p; mh)−ωhi excess demand of good i for consumer h,

where mh := p · ωh.

Now: p ∈ Rn++ is an equilibrium price vector if and only if for the

aggregate excess demands zk of the goods one has

z1(p) = · · · = zn(p) = 0.


Fundamental observations

Each equilibrium allocation is individually rational.If p is an equilibrium price vector, then for each λ > 0 alsoλp is.Law of Walras: p1z1(p) + · · ·+ pnzn(p) = 0 holds for all p.Law of Walras implies that if n − 1 aggregate excessdemands are zero, then all are zero.


Existence of equilibrium

TheoremEach pure exchange economy where each utility function iscontinuous, strictly quasi-concave and strongly increasing hasa walrasian equilibrium with positive equilibrium prices.

Proof.Very complicated. (See text book.)


Notes on the proof

Strict quasi-concavity and continuity guarantee that thereare well-defined marshallian demand functions.Proving that there exists a p as above can be done withBrouwers’ fixed point theorem. In fact Debreu presented acomplete proof.The proof also allows for prices zero. A quite technicalproblem in the proof is to be sure that prices are not 0.

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Proof by not correct principle

Principle: n equations in n variables have a solution.

Applied to our equilibrium existence problem:zi(p) = 0 (1 ≤ i ≤ n) are n equations in n variables.

However: according to Law of Walras we only have n − 1equations, so principle does not apply.

But according to p equilibrium ⇒ λp equilibrium we haven − 1 equations. So principle applies.


Brouwer and his fixed point theorem


Brouwer and his fixed point theorem (cont.)

Luitzen Jan Egbertus Brouwer (1881-1966).Dutch mathematician, Frisian and idealist.Brouwer proved a number of theorems that werebreakthroughs in the emerging field of topology. Mostfamous is his fixed point theorem.He died after he was strucked by a vehicle while crossingthe street in front of his house.Fixed point theorem of Brouwer: each continuous functionf from the unit ball in Rn to itself has a fixed point, i.e. thereis x such that f (x) = x.For one dimension the theorem is not so deep.


Debreu

Gérard Debreu (1921-2004):French economist and mathematician.Wrote ’Theory of Value’ for his PhD thesis.Nobel price for economics in 1983 together with Arrow.


Example

ExampleDetermine an equilibrium price vector for

uA(x1, x2) = x1x2, uB(x1, x2) = min (x1, x2),

ωA = (5,0), ωB = (0,6).

Answer: Equilibrium price vector p2/p1 = 5/7.


First welfare Theorem

Theorem(First welfare Theorem.) Consider a pure exchange economywhere each utility function is continuous and locallynon-satiated. Then: each equilibrium allocation is paretoefficient.

Proof.Proof, with additional assumptions, possible with Gossen’ssecond law.


Perfect proof of theorem

By contradiction: suppose X is an equilibrium allocation that ispareto inefficient.

Let (p; X) be a walrasian equilibrium.

Because X is pareto inefficient, there exists a feasible allocationY and a consumer, say j , such that for all h

uh(yh) ≥ uh(xh)

with strict inequality for h = j .

Because xj is a maximiser of uj under the restrictionp · x ≤ p · ωj , and by the above yj is not, it follows that

p · yj > p · ωj .


Perfect proof of theorem (cont.)

Also for each hp · yh ≥ p · ωh.

Indeed: because if not, then there is a k with p · yk < p · ωk .Because uk is in yk locally non-satiated, there exists a withp · a < p · ωk and uk (a) > uk (yk ). But then p · a < p · ωk anduk (a) > uk (xk ), a contradiction.Adding now the N inequalities

p · yh ≥ p · ωh

with strict inequality for h = j :

p ·N∑

h=1

yh > p ·N∑

h=1

ωh,

a contradiction with feasibility of Y.


Improvement of first welfare theorem

TheoremConsider a pure exchange economy where each utility functionis continuous and locally non-satiated. Then: each equilibriumallocation is in the core.

Proof.Adapt proof of First welfare Theorem.

Thus: core is not empty if walrasian equilibrium exists.


Second welfare Theorem

Theorem(Second welfare theorem.) Consider a pure exchange economywhere each utility function is continuous, strongly increasingand strictly quasi-concave. Let X be a pareto efficientallocation. Then the pure exchange economy where X is theinitial allocation has X as unique equilibrium allocation.


Perfect proof

Proof.Consider the pure exchange economy where X is the initialallocation.

Also here X is pareto efficient.

We know: there exists a walrasian equilibrium (p; Y). Thisimplies, for all h, uh(yh) ≥ uh(xh). Because X is pareto efficient,even uh(yh) = uh(xh) for all h.

We show by contradiction Y = X.


Perfect proof

Proof.So suppose yr 6= xr for some r . Then, as ur is strictlyquasi-concave

ur (12

(xr + yr )) > min (ur (xr ),ur (yr )) = ur (yr ).

Because p · yr ≤ p · xr , we have

p · 12

(xr + yr ) =12

(p · xr ) +12

(p · yr ) ≤ p · xr ,

a contradiction.


Summary

First results on general equilibrium obtained by Walras.Debreu made these results mathematical rigorous usingBrouwer’s fixed point theorem.(Under weak conditions) each equilibrium is in the core.Not each element of the core has to be an equilibrium.As the economy becomes larger and larger, the coreshrinks to include only equilibrium allocations.Method of proof by concept of ’replica economy’. (Section5.5 in text book.)Proof first provided by Debreu and Scarf.


Little quiz (ctd.)

Consider a pure-exchange economy.

1. The assumption Ok > 0 means that every consumer has apositive amount of good type k . False

2. There is at least producer. False3. The consumers have market power. False4. Each weakly pareto inefficient allocation is strongly pareto

inefficient. True5. The initial allocation is pareto inefficient. False


Little quiz (ctd.)

6. If each utility function is continuous and stronglyincreasing, then there exists a walrasian equilibrium. False

7. It is possible that there exists a walrasian equilibriumwhere each consumer has utility 0. True

8. If each utility function is continuous and stronglyincreasing, then the set of weak and strong pareto efficientallocations is the same. True

9. The core depends on the initial allocation. True10. Every pareto efficient individual rational allocation belongs

to the core. False


Little quiz (ctd.)

11. If there exists a walrasian equilibrium, then there existmore then two equilibrium price vectors. True

Now suppose N = n = 2 and denoting the consumers by A andB, uB(x1, x2) = x1/3

1 x2/32 , ωA

1 = 2, ωA2 = 2, ωB

1 = 3, ωB2 = 1,

13. If uA(x1, x2) = x1x2, then the allocation X =(

(1,2); (1,4))

is pareto efficient. False14. If uA(x1, x2) = x1x2, then the allocation ((5,0), (0,3)) is

pareto efficient. False


What is game theory?

Traditional game theory deals with mathematicalmodels of conflict and cooperation in the real worldbetween at least two rational intelligent players.

Player: humans, organisations, nations, animals,computers,Situations with one player are studied by the classicaloptimisation theory.’Traditional’ because of rationality assumption.’Rationality’ and ’intelligence’ are completely differentconcepts.


Rationality

Because there is more than one player, especially rationalitybecomes a problematic notion.

For example, what would You as player 1 play in the followingbi-matrix-game: (

300; 400 600; 250200; 600 450; 500

).


Outcomes and payoffs

A game can have different outcomes. Each outcome hasits own payoffs for every player.Interpretation of payoff: ‘satisfaction’ at end of game.Nature of payoff: money, honour, activity, nothing at all,utility, real number, ... .


Tic-tac-toe

Notations:1 2 34 5 67 8 9

Player 1: X. Player 2: O.Many outcomes (more than three). Can, for player 1, beordered by player 1 wins, draw, player 1 looses. It is azero-sum game.Payoffs (example): winner obtains 13 Euro from looser.When draw, then each player cleans the shoes of the other.Example of a play of this game:


Tic-tac-toe (cont.)

X X

O

X X

OX X O

O

X X O

X O

X X OO

X OSo: player 2 is the winner.

Question: Is player 1 intelligent? Is player 1 rational?


Real-world types

all players are rational – players may be not rationalall players are intelligent – players who may be notintelligentbinding agreements – no binding agreementschance moves – no chance movescommunication – no communicationstatic game – dynamic gametransferable payoffs – no transferable payoffsinterconnected games – isolated games

perfect information – imperfect informationcomplete information – incomplete information


Perfect information

A player has perfect information if he knows at eachmoment when it is his turn to move how the game wasplayed untill that moment.A player has imperfect information if he does not haveperfect information.A game is with (im)perfect information if (not) all playershave perfect information.Chance moves are compatible with perfect information.Examples of games with perfect information: tic-tac-toe,chess, ...Examples of games with imperfect information: poker,monopoly (because of the cards, not because of the die).


Complete information

A player has complete information if he knows all payofffunctions.A player has incomplete information if he does not havecomplete information.A game is with (in)complete information if (not) all playershave complete information.Examples of games with complete information: tic-tac-toe,chess, poker, monopoly, ...Examples of games with imperfect information: auctions,oligopoly models where firms only know the own costfunctions, ...A game of incomplete information can be transformed (bythe Harsanyi transformation) into one with complete butimperfect inforamtionSolution concept: Baysian-Nash equilibrium.


Common knowledge

Something is common knowledge if everybody knows it and inaddition that everybody knows that everybody knows it and inaddition that everybody knows that everybody knows thateverybody knows it and ...


Common knowledge

A group of dwarfs with red and green caps are sitting in a circlearound their king who has a bell. In this group it is commonknowledge that every body is intelligent. They do notcommunicate with each other and each dwarf can only see thecolor of the caps of the others, but does not know the color ofthe own cap. The king says: ”Here is at least one dwarf with ared cap.”. Next he says: “I will ring the bell several times. Thosewho know their cap color should stand up when i ring the bell.”.Then the king does what he announced.

The spectacular thing is that there is a moment where a dwarfstands up. Even, when there are M dwarfs with red caps thatall these dwarfs simultaneously stand up when the king ringsthe bell for the M-th time.


Mathematical types

Game in strategic form.Game in extensive form.Game in characteristic function form.


Game in strategic form

Definition

Γ = (X1, . . . ,Xn; f1, . . . , fn)

n players: 1, . . . ,n.Xi : non-empty strategy (or action) set of player i .X := X1 × · · · × Xn: set of multi-strategies or strategyprofiles.fi : X→ R payoff function of player i .

Interpretation: players choose simultaneously a strategy.


Some concrete games in strategic form

Cournot-duopoly:n = 2, Xi = [0,mi ] or Xi = R+

fi(x1, x2) = p(x1 + x2)xi − ci(xi).

Transboundary pollution game: n arbitrary, Xi = [0,mi ]

fi(x1, . . . , xn) = Pi(xi)−Di(Ti1x1 + · · ·+ Tinxn).


Normalisation

Many games can be represented in a natural way (bynormalisation) as a game in strategic form.

For example, chess and tic-tac-toe: n = 2, Xi is set ofcompletely elaborated plans of playing of i ,

fi(x1, x2) ∈ −1,0,1.

Questions:1 Give for each player of tic-tac-toe a completely elaborated

plan of playing.2 How the game will be played?3 Give an optimal strategy for player 1.


Some concrete games.

0; 0 −1; 1 1;−11;−1 0; 0 −1; 1−1; 1 1;−1 0; 0

Stone-paper-scissors


Some types of strategic form games

A game in strategic form is calledfinite if each strategy set X i is finite;antagonistic if n = 2 and if f1 + f2 = 0 (i.e. a zero-sumgame).


Fundamental notions

Best reply correspondence Ri of player i :X1 × · · · × Xi−1 × Xi+1 × · · · × Xn ( Xi .(Strictly) dominant strategy of a player i : (the) best strategyof player i independently of strategies of the other players.Strictly dominant equilibrium: multi-strategy where eachplayer has a strictly dominant strategy.Strongly (or strictly) dominated strategy of a player: astrategy of a player for which there exists another strategythat independently of the strategies of the other playersalways gives a higher payoff.


Fundamental notions (cont.)

Procedure of iterative (simultaneous) elimination ofstrongly dominated strategies.Multi-strategy that survives this procedure.If there is a unique multi-strategy that survives the aboveprocedure this multi-strategy is called the iteratively notstrongly dominated equilibrium.Nash equilibrium: multi-strategy such that no player wantsto deviate from it.


Nash equilibria

A multi-strategy e = (e1, . . . ,en) is a nash equilibrium if andonly if for each player i one has

ei ∈ Ri(e1, . . . ,ei−1,ei+1, . . . ,en).

Sometimes can be determined by

∂fi∂xi

= 0 (i = 1, . . . ,n)


Example

Example 2; 4 1; 4 4; 3 3; 01; 1 1; 2 5; 2 6; 11; 2 0; 5 3; 4 7; 30; 6 0; 4 3; 4 1; 5

.

1 No strictly dominant strategies, thus no strictly dominantequilibrium.

2 The procedure gives(

2; 4 1; 4 4; 31; 1 1; 2 5; 2

). Thus the game

does not have an iteratively not strongly dominatedequilibrium.

3 The game has the following nash equilibria:(1,1), (1,2), (2,2) and (2,3) (i.e. row 2 and column 3).


Example

Example 6; 1 3; 1 1; 52; 4 4; 2 2; 35; 1 6; 1 5; 2

1 No player has as strictly dominant strategy, thus the game

does not have a strictly dominant equilibrium.2 The procedure of iterative elimination of strongly

dominated strategies gives (5,2). Thus the game has aniteratively not strongly dominated equilibrium: (3,3).

3 The game has one nash equilibrium: (3,3).


Solution concepts

Theorema. Each strictly dominant equilibrium is an iteratively not

strongly dominated equilibrium.

And if the game is finite:b. Each Nash equilibrium is an iteratively not strongly

dominated multi-stategy. (So each nash equilibriumsurvives the procedure.)

c. An iteratively not strongly dominated equilibrium is aunique nash equilibrium.

Proof.1. Already in first steps of procedure all strategies are removedwith the exception of strictly dominant ones.2, 3. One verifies that in each step of the procedure the set ofnash equilibria remains the same. (See the text book.)


Mixed strategies

Some games do not have a nash equilibrium.

Mixed strategy of player i : probability density over X i .With mixed strategies, payoffs have the interpretation ofexpected payoffs.Nash equilibrium in mixed strategies. Remark: each nashequilibrium is a nash equilibrium in mixed strategies. (Seetext book for formal proof.)


Bi-matrix-game with mixed strategies

Consider a 2× 2 bi-matrix-game

(A; B)

Strategies: (p,1− p) for player 1 and (q,1− q) for player B.

Expected payoffs:

f A(p,q) = (p,1− p) ∗ A ∗(

q1− q

),

f B(p,q) = (p,1− p) ∗ B ∗(

q1− q

).


Example

ExampleDetermine the nash equilibria in mixed strategies for(

0; 0 1;−12;−2 −1; 1

).

Answer:f A(p; q) = (−4q + 2)p + 3q − 1,

f B(p; q) = (4p − 3)q + 1− 2p.

This leads to the nash equilibrium

p = 3/4,q = 1/2.


Example

ExampleDetermine the nash equilibria in mixed strategies for(−1; 1 1;−11;−1 −1; 1

).

Answer:p = q = 1/2.


Existence of nash equilibria

Conditional payoff function: fi as a function of xi , givenstrategies of the other players.

Theorem(Nikaido-Isoda.) Each game in strategic form where

1 each strategy set is a convex compact subset of some Rn,2 each payoff function is continuous,3 each conditional payoff function is quasi-concave,

has a nash equilibrium.

Proof.This is a deep theoretical result. A proof can be based onBrouwer’s fixed point theorem. See text book for the proof of asimpler case (Theorem 7.2., i.e. the next theorem).


Theorem of Nash.

TheoremEach bi-matrix-game has a nash equilibrium in mixedstrategies.

Proof.Apply the Nikaido-Isoda result.


Appetizer

Antagonistic games:t-t-t chess 8× 8 checkers hex

value draw not known draw 1opt. strat. known not known known not known


Hex

1 http://www.mazeworks.com/hex7.2 Hex can not end in a draw. (’Equivalent’ with Brouwer’s

fixed point theorem in two dimensions.)3 If You can give a winning strategy for hex, then You solved

a ’1-million-dollar problem’.

http://www.mazeworks.com/hex7


Games in extensive form

Our setting is always non-cooperative with complete information(and for the moment) perfect information and no chance moves.Game tree:

Nodes (or histories): end nodes, decision nodes, uniqueinitial node.Directed branches.Payoffs at endnodes.Each non-initial node has exactly one predecessor.No path in tree connects a node with itself.Game is finite (i.e. a finite number of branches and nodes).Actual moves can be denoted by arrows.


Perfect information (ctd.)

Theoretically:Imperfect information can be dealt with by usinginformation sets. The information sets form a partition ofthe decision nodes. (Example: Figure 7.10.)Perfect information: all information sets are singletons.Solution concept: Nash equilibrium.Games in strategic form are games with imperfectinformation.


Normalisation

Strategy: specification at each decision node how to move.(This may be much more than a completely elaborated plan ofplay.)

Normalisation: make out (in natural way) of game in extensiveform a game in strategic form.

So normalisation destroys the perfect information.

All terminology and results for games in strategic form now alsoapplies to games in extensive forms.


Solving from the end to the beginning

ExampleConsider the following game between two (rational andintelligent) players. There is a pillow with 100 matches. Theyalternately remove 1,3 or 4 matches from it. (Player 1 begins.)The player who makes the last move wins. Who will win?

Answer: the loosing positions are 0,2,7,9,14,16,21, . . ., i.e.the numbers that have remainder 0 or 2 when divided by 7.Because 100/7 has remainder 2, 100 is a loosing position andplayer 2 has a winning strategy.


Procedure of backward induction (explained at the blackboard)leads to a non-empty set of backward induction multi-strategies.

Theorem(Kuhn.) Each backward induction multi-strategy of a finite gamein extensive form with perfect information is a nash equilibrium.

Proof.See text book.

But a nash equilibrium not necessarily is a backward inductionmulti-strategy.


Subgame perfection

Subgame: game starts at a decision node.Subgame perfect nash equilibrium: a nash equilibrium thatremains for each subgame a nash equilibrium.

TheoremFor every finite extensive form game with perfect informationthe set of backward induction multi-strategies coincides with theset of subgame perfect nash equilibria.

Proof.See text book.


Games in extensive form: extensions

Three extensions:

Imperfect information.Incomplete information: the solution concept here is that ofBayesian equilibrium (7.2.3.). [Next part of course.]Randomization.


Imperfect information

Imperfect information.Can be dealt with by using information sets. Theinformation sets form a partition of the decision nodes.(Example: Figure 7.10.)Perfect information: all information sets are singletons.Strategy: specification at each information set how tomove.The procedure of backward induction cannot be appliedanymore, but the notion of subgame perfect Nashequilibria still makes sense (when ’subgame’ is properlydefined). [Next part of course.]Subgame: not all decision nodes define anymore asubgame. (Example: Figure 7.20.) [Next part of course.]Nash equilibria need not always exist. (Example: Figure7.23.) [Next part of course.]


Randomization

Three types of strategies: pure, mixed and behaviouralstrategies.


Randomization

[Next part of the course.]A pure strategy of player i is a book with instructions wherethere is for each decision node for i a page with thecontent which move to make at that node. So the set of allpure strategies of player i is a library of such books.A mixed strategy of player i is a probability density on hislibrary. Playing a mixed strategy now comes down tochoosing a book from this library by using a chance devicewith the prescribed probability density.


Randomization (ctd.)

A behavioural strategy, is like a pure strategy also a book,but of a different kind. Each page in the book still refers toa decision node, but now the content is not which move tomake but a probability density between the possiblemoves.For many games (for instance those with perfect recall) itmakes no difference whatever if players employ mixed orbehavioural strategies.


Nash

John Nash (1928 –) at the right-hand side.

Mathematician. (Economist ?)Nobel price for economics in 1994, together with Harsanyiand Selten.Got this price for his PhD dissertation (27 pages) in 1950.http://topdocumentaryfilms.com/a-brilliant-madness-john-nash .

http://topdocumentaryfilms.com/a-brilliant-madness-john-nash

http://topdocumentaryfilms.com/a-brilliant-madness-john-nash

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