Classnotes for the

Game Theory

course o�ered by

Professor Ferenc Szidarovszky

c© Ferenc Szidarovszky, 2009

Contents

1 Decision making 1

2 Examples of games 1

3 Discrete games with �nitely many strategies 42

4 Games Representable by Finite Rooted Trees 42

5 Continuous games 455.1 Some mathematical background . . . . . . . . . . . . . . . . . . . . . . . . 455.2 Brouwer��xed point theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 495.3 Kakutani �xed point theorem . . . . . . . . . . . . . . . . . . . . . . . . . 515.4 Banach��xed point theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 515.5 Conditions of contraction mappings . . . . . . . . . . . . . . . . . . . . . . 535.6 Relation of EP and �xed points . . . . . . . . . . . . . . . . . . . . . . . . 55

1. Based on payo� functions . . . . . . . . . . . . . . . . . . . . . . . . . . 552. Based on best responses . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

6 Nikaido�Isoda�theorem 576.1 Concave functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576.2 Counterexamples for Nikaido�Isoda theorem . . . . . . . . . . . . . . . . . 59

7 Applications 617.1 Matrix games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.2 Bimatrix games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627.3 Mixed �nite games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 637.4 Polyhedron games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647.5 Multiproduct oligopolies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647.6 Single-product oligopolies . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

8 Relation of EP problems and nonlinear programming 70

9 How to compute EP? 719.1 Lagrange method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 719.2 Kuhn�Tucker (K-T) conditions . . . . . . . . . . . . . . . . . . . . . . . . 729.3 Relation of Kuhn-Tucker conditions and Lagrange method . . . . . . . . . 739.4 Methodology to compute equilibria . . . . . . . . . . . . . . . . . . . . . . 75

10 Special games 7810.1 Bimatrix games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7810.2 Matrix games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8110.3 Oligopoly game (single-product model) . . . . . . . . . . . . . . . . . . . . 8310.4 Special matrix games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8410.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8610.6 Method of �ctitious play . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8910.7 von Neumann's method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

11 Uniqueness of Equilibrium 95

i

12 Leader�follower games (Stackelberg equilibrium) 10012.1 Application to duopolies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

1. Nash equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1012. Stackelberg equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . 1023. Optimum subsidy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

13 Games with incomplete information 103

14 Cooperative games 10714.1 Characteristic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10714.2 Core of game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11014.3 Stable sets (or von Neumann�Morgenstern solution) . . . . . . . . . . . . . 11114.4 The Shapley�value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

15 Social choice 11315.1 Equal players . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

1. Plurality voting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1132. Borda count . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1133. Hare system (successive deletions) . . . . . . . . . . . . . . . . . . . . . 1134. Pairwise comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1145. Dictatorship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

15.2 Nonsymmetric case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

16 Con�ict resolution 11716.1 Bargaining as a noncooperative game . . . . . . . . . . . . . . . . . . . . . 11716.2 Single-player decision problem . . . . . . . . . . . . . . . . . . . . . . . . . 11816.3 Axiomatic bargaining . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11916.4 Nonsymmetric Nash solution . . . . . . . . . . . . . . . . . . . . . . . . . . 12116.5 Area monotonic solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12216.6 Equal sacri�ce solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12316.7 Kalai�Smorodinsky solution . . . . . . . . . . . . . . . . . . . . . . . . . . 124

17 Multiobjective optimization, concepts and methods 12817.1 Existence of nondominated solutions . . . . . . . . . . . . . . . . . . . . . 13117.2 Method of sequential optimization . . . . . . . . . . . . . . . . . . . . . . . 13417.3 ε-constraint method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13717.4 Weighting method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14017.5 Distance�based methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14717.6 Direction�based methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

18 Dynamic games 15418.1 Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15418.2 Best response dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15818.3 Cournot oligopoly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16318.4 Best response dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

19 Controllability of dynamic economic systems 16619.1 Control of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 16619.2 Control in oligopolies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

20 Learning in oligopolies 171

ii

21 Partial cooperation 176

22 Certainty equivalent 177

23 Illustrative case studies 17823.1 Example 1. Selecting a restaurant for lunch . . . . . . . . . . . . . . . . . 17823.2 Example 2. Buying a family car . . . . . . . . . . . . . . . . . . . . . . . . 18523.3 Example 3. Restoration of a chemical land�ll . . . . . . . . . . . . . . . . . 190

iii

Game theory

1 Decision making

Three major elements:

Who is in charge to make the decision? The decision maker (DM):

• one or

• more

What choices the DM has? Alternatives:

• �nitely many (discrete problem), A1, A2, . . . , Am

• described by continuous variables (continuous problem), like

X ={x | x ∈ Rm, g(x) ≤ 0

}What are the consequences of the decision? Objective functions, ϕ1, ϕ2, . . . , ϕn.

Many cases:

1 DM with 1 objective: single objective optimization

1 DM with multiple objectives: multiobjective optimization

multiple DMs with 1 objective each: game

multiple DMs with multiple objectives each: Pareto game

Games:

DMs are called the players;decision alternatives are called the strategies;objective functions are called the payo� functions.

History:

• John von Neumann (1928)

• John Nash (1950-53)

• Nobel laureates Nash, Selten, Harsanyi (1996)

2 Examples of games

1. Prisoners' dilemma

Players: Two prisoners who robbed a jewellery store for hire, got arrested, but policedoes not have enough evidence to convict them with the full crime, only for a muchlesser crime of driving a stolen car

Strategies: Confess to the police (C) or not (NC)

1

Game theory

Payo�s:

• If only one confesses, then he recieves very light sentence (1 year), the othergets very harsh sentence (10 years)

• If both confess, they get medium long (5 year) sentence

• If none of them confesses, then they are convicted with a smaller crime, 2 yearssentence for each

1\ 2 NC CNC (-2, -2) (-10,-1)C (-1,-10) (-5,-5)

(ϕ1, ϕ2) in table

Question: What to do? Payo� of each depends on the choice of the other player.

Best response: Best choice of each player as a function of the choice of the other:

R1 =

{C if player 2 selects NCC if player 2 selects C

So player 1 always should confess. Same for player 2, who also should always confess.

⇒ C = dominant strategy

⇒ Solution is (C,C) = both confess

However, if the players cooperate, then they can choose (NC, NC) with better payo�for both.

2. Chicken game

Players: Two kids with motorcycle driving toward each other in a narrow alley

Strategies: Give way to the other (C=chicken) or not (C)

Payo�s: Being a chicken looks bad in the gang, but by having a crash both might die,even worse outcome

1\ 2 C CC (3, 3) (2,4)C (4,2) (1,1)(ϕ1, ϕ2) in table

R1(C) = C R1(C) = CR2(C) = C R2(C) = C

2

Game theory

Both points (C,C) and (C,C) are common in the two best responses, so theycalled the equilibria. No player wants to move away from his equilibrium strategyassuming that the other player keeps his corresponding equilibrium strategy.

Problem & di�culty: in a particular game which equilibrium is selected?

3. Hunting gameIn a forest, there are rabbits and deers, and two hunters.

Players: Two hunters

Strategies: Which animal they want to hunt down (D or R)

Payo�s:

1\ 2 D RD (2, 2) (-1, 1)R (1,-1) (0, 0)

(D, D) = cooperating to get the deer, lot of meat

(D, R) = shooting the rabbit the deer runs away, then player 1 has no success

(R, D) = same but player 2 gets nothing

(R, R) = they get the rabbit

Two equilibria: (D, D) and (R, R)= cooperation

4. Game of privilegeTwo families take care of a house, e.g. cleaning the common areas (basement, stair-

house, etc.)

Players: Two families

Strategies: Contribute or not in taking care of the house (C or C)

Payo�s:

3

Game theory

1\ 2 C CC (3, 3) (1, 2)C (2, 1) (0, 0)

In general:

1\ 2 C CC (b2 − c2, b2 − c2) (b1 − c1, b1)C (b1, b1 − c1) (0, 0)

Here, b1 = utility if only one contributes

b2 = utility if both contribute

c1 = cost if only one contributes

c2 = cost for one if both contribute

Same as contributing to public goods

In numerical example, equilibrium: (C, C) which is unique

5. Penalty in soccer

Players: Goalkeeper and the player kicking the penalty (G, K)

Strategies: Goalkeeper: moves to left or right (L, R)

Kicker: kicks the ball to the left or right (L, R)

Payo�s: Probability of saving for G

G\ K L RL 55 20R 10 80

ϕ1

ϕ2=100-ϕ1 ⇒ game is equivalent to a zero-sum game. No equilibrium exists

6. Penalty kick 2

Players: Goalkeeper and player kicking the penalty

Strategies: For both: left(L), right(R), middle(M)

Payo�s: If goalkeeper's strategy is same as that of the kicker, then he saves, otherwisekicker scores. Zero-sum game with payo� for player 1:

4

Game theory

G\ K L R ML 1 -1 -1R -1 1 -1M -1 -1 1

ϕ1

No equlibrium exists.

7. Battle of sexes

Players: Husband and wife, H prefers football game (F), W prefers a movie (M) for anevening. In the morning they leave home without decision and plan to call eachother in the afternoon to decide. They could not call each other, so in the eveningeach of them goes to F or to M.

Strategies: Selecting F or M

Payo�s:

H\ W F MF (2, 1) (0, 0)M (0, 0) (1, 2)

Two equilibria: (F, F) and (M, M).

Another interpretation:

Players: Two �rms with one product each, which complement each other

Strategies: Joining the patent of the competitor in order to have compatible products,or not

Payo�s: Similar, in the long run it is an advantage to accept the patent of the other�rm, since one sale generates the other, however it needs extra work in the shortrun.

8. Competition of gas stations

Players: Two gas stations next to each other

Strategies: Selling gas with low (L) or high (H) price

Payo�s:

1\ 2 H LH (40, 40) (10, 50)L (50, 10) (20, 20)

(H, H): both select high price, both get high pro�t

(H, L) or (L, H): the one with low price gets most of the customers, the other almostnobody

5

Game theory

(L, L): they equally share customers with low prices, so both get low pro�t

Best responses:

R1 =

{L if player 2 selects HL if player 2 selects L

⇒ L = dominant strategy for player 1

Similiar, L = dominant strategy for player 2

⇒ The unique equlibrium is (L, L) (same as prisoners' dilemma)

However if they form a coalition and charge high prices, pro�ts increase for bothplayers. This is an illegal act, they violate antitrust regulations.

9. Checking tax returns

Players: Internal Revenue Service (IRS) and a taxpayer (T)

Strategies: For IRS: checking the tax return of the taxpayer or not (C or C)

For T: cheating or not (C or C)

Payo�s: T should pay 5 thousand $ as tax. By cheating and being checked, his penaltyis also 5 thousand $; if not checked, then there is no penalty.

For IRS, checking T costs 1 thousand $. Payo�s:

IRS\ T C CC (9, -10) (4, -5)C (0, 0) (5, -5)

No equlibrium exists.

10. Checking a worker

Players: Supervisor (S) and a worker (W)

Strategies: Supervisor: checks the worker or not (C or C)

Worker: does his job well or not (J or J)

Payo�s: Checking costs amount c to supervisor. Payo�s:

S\ W J JC (8− c, 6) (6− c, 2)C (8, 6) (6, 8)

Equilibrium:

(C, J) no, since 8− c < 8

(C, J) no, since 6<8

(C, J) no, since 2<6

(C,J) yes, since 6 > 6− c and 8>6

6

Game theory

11. Driver and police

Players: Driver and a police o�cer

Strategies: Driver: speeding or not(S or S)

Police: giving ticket with penalty p or not (T or T )

Payo�s:

D\ P T TS (−p, 2) (5, -1)(bad feeling for not catching a speeder)S (0, 0) (0, 1)(no e�ort in checking drivers)

No equlibrium exists

12. Good citizen

Players: Two people witnessing a serious crime on the street

Strategies: Calling the police or not (C or C)

Payo�s:

1\ 2 C CC (7, 7) (7, 10)C (10, 7) (0, 0)

Value of letting the police come to arrest the criminal is 10, but by making the callit decreases by 3 (cost of call + possible revenge of criminal group).

Two equilibria: (C, C) and (C, C) meaning one makes the call and the otherwalks away. Public goods models are extensions of this one, some of them includemechanisms to avoid free ride (=giving nothing and bene�ting)

13. Waste managementA �rm wants to place dangerous waste on the border between two counties, which

would result in D1 and D2 damages to the counties, respectively. The only way to stopit is an intensive lobbying e�ort by at least one county which would cost C1 or C2.

Players: Two counties

Strategies: Lobbying (L) or not (L)

Payo�s:

1\ 2 L LL (−C1, −C2) (−C1, 0)L (0, −C2) (−D1,−D2)

(L, L) is never equlibrium

(L, L) is equlibrium if −C2 ≥ −D2 ⇔ C2 ≤ D2

(L, L) is equlibrium if −C1 ≥ −D1 ⇔ C1 ≤ D1

(L, L) is equlibrium if −D1 ≥ −C1 and −D2 ≥ −C2 ⇔ C1 ≥ D1 and C2 ≥ D2

7

Game theory

14. Matching pennies

Players: Two participants

Strategy: Each has a coin, and can show head (H) or tail (T)

Payo�: If the coins have identical sides, then player 1 wins $1, otherwise player 2 wins$1 from other player:

1\2 H TH 1 -1T -1 1

ϕ1

ϕ2 = −ϕ1

No equilibrium exists.

15. Cooperation in a job

Players: Two workers

Strategies: Working (W ) or not(W )

Payo�s: Cost of e�ort (if W )=c ∈ (0, 1)

Unit payment is given to both if job is done, and job can be done only if both work

1\ 2 W WW (1− c, 1− c) (−c, 0)W (0, −c) (0, 0)

Equilibria: (W , W ) and (W , W )

16. Chain store

Players: Chain store (C) & enterpreneur (E).

Strategies:

For C: soft or hard on E (S,H)

E: stay in business or out (I,O).

Payo�s:

8

Game theory

C\ E I OS (2, 2) (5,1)H (0,0) (5,1)

Normal form (n;S1, . . . , Sn, ϕ1, . . . , ϕn) giving

• number of players

• strategy sets

• payo� functions.

Extensive form, shows development & dynamism of game

RC(I) = S RC(O) = {S,H}RE(S) = I RE(H) = O

17.Two-person, zero-sum, discrete games

Two players

Strategies: {1, . . . ,m} and {1, . . . , n}.

9

Game theory

1\2 1 . . . j . . . n1 a11 . . . a1j . . . a1n...

......

...i ai1 . . . aij . . . ain...

......

...m am1 . . . amj . . . amn

ϕ1

Payo�s: ϕ1(i, j) = aij and ϕ2(i, j) = −aij

(aij) is equilibrium ⇔

aij is the largest among a1j, . . . , aij, . . . , amj

-aij is the largest among −ai1, . . . ,−aij, . . . ,−ain ⇔ aij is smallest amongai1, . . . , aij, . . . , ain

That is, aij is largest in its column and smallest in its row ⇒ saddle pointAssume that aij and akl are both equilibria. Then

1\2 j . . . li aij . . . ail...

......

k akj . . . akl

aij ≥ akj ≥ akl and aij ≤ ail ≤ akl.

So, aij = akl, that is, if multiple equilibria exists, then payo� values are the same.

What is the chance to have equilibrum?

Theorem 2.1 Assume that all aij are independent, identically distributed with a contin-uous distribution function. Then

P(equilibrium exists) = Pmn =m!n!

(m+ n− 1)!

Proof. Notice that

P(all elements aij are di�erent) = 1

P(aij is equilibrium) is same for all elements

P(equilibrium exists) = mnP(a11 is equilibrium)

a11 is equilibrium if a11 is largest in its column and smallest in its row. So if we orderthe elements of �rst row and �rst column in increasing order,

am1, . . . , a21, a11, a12, . . . , a1n,

10

Game theory

then a11 must not change position, only the elements before and after a11 can be inter-changed. So

P(a11 is equilibrium) =(m− 1)!(n− 1)!

(m+ n− 1)!

P(equilibrium exists) = mn(m− 1)!(n− 1)!

(m+ n− 1)!=

m!n!

(m+ n− 1)!

�

Example 2.1m = n = 2

P22 =2!2!

3!=

4

6=

2

3

m = 2, n = 5

P25 =2!5!

6!=

2 · 120

720=

1

3

m = 1, n = arbitrary

P1n =1!n!

(1 + n− 1)!= 1,

the smallest element in the only row is the equilibrium.m = 2, n ≥ 2

P2n =2!n!

(2 + n− 1)!=

2n!

(n+ 1)!=

2

n+ 1→ 0 as n→∞

What happens if m increases by 1:

Pm+1,n

Pm,n

=(m+ 1)!n!

(m+ 1 + n− 1)!· (m+ n− 1)!

m!n!

=(m+ 1)!n!(m+ n− 1)!

(m+ n)!m!n!=m+ 1

m+ n

which equals 1 if n = 1, and is less than 1 if n ≥ 2 ⇒ Pmn → 0 as m or n tends to ∞with other size ≥ 2.

O

Example 2.2 With discrete distribution theorem fails:

m = n = 2, P(aij = 0) = p, P(aij = 1) = 1− p = qWe have 24 = 16 possible matrices:1 is in 0 or 1 position:(

0 00 0

),

(1 00 0

),

(0 10 0

),

(0 01 0

),

(0 00 1

)1 is in 2 positions(

1 10 0

),

(1 01 0

),

(1 00 1

),

(0 11 0

),

(0 10 1

)(0 01 1

)11

Game theory

1 is in 3 or 4 positions:(1 11 0

),

(1 01 1

),

(0 11 1

),

(1 10 1

),

(1 11 1

)No equilibrium exists in cases(

1 00 1

)and

(0 11 0

)with probability 2p2q2, so

P(equilibrium exists) = 1− 2p2q2

O

18.Coin in pocketTwo players, each puts 0 or 1 coin into his pocket.

Step 1. Player 1 guesses total number of coins (no blu�ng, so with a coin in his pockethe cannot guess 0).

Step 2. Player 2 guesses total number of coins (no blu�ng and cannot repeat guess ofplayer 1).

Whoever's guess is correct wins $1 from other player.Extensive form:

12

Game theory

1\2 0 1(0, 0) 1 −1(0, 1) −1 1(1, 1) 1 −1(1, 2) −1 1

ϕ1

Normal form:ϕ2 = −ϕ1

Notice, guess of player 2 was always unique (no blu�, no repeat)

Strategy of player 1: (0, 0), (0, 1), (1, 1), (1, 2);

Strategy of player 2: 0 or 1 (No. of coins in pocket).

No equilibrium exists.

19. Sharing a pie

Players: 2 people to share a pie of unit size

Strategies: Requests from the pie, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

Payo�s: If the requests are feasible (x + y ≤ 1) then both get requested amount, ifinfeasible (x+ y > 1), then none of them receives anything:

ϕ1(x, y) =

{x if x+ y ≤ 10 otherwise

ϕ2(x, y) =

{y if x+ y ≤ 10 otherwise

13

Game theory

In�nitely many equilibria:

{(x, y) | 0 ≤ x, y ≤ 1, x+ y = 1}

20. War game

0 1

Players: Airplane (A) and a submarine (S).

Strategies:

For A: x ∈ [0, 1] where to drop a bomb

for S: y ∈ [0, 1] where to hide.

Payo�s:

ϕ1 = α e−β(x−y)2 damage to submarine

ϕ2 = −ϕ1

Zero sum game if∑ϕk = 0

R1(y) = drop bomb where submarine is hiding, so

R1(y) = y

R2(x) = hide as far as possible from x, so

R2(x) =

1 if x < 1

2

0 if x > 12

{0, 1} if x = 12

14

Game theory

No common point, no equilibrium exists.

21. Modi�ed war game

Players: Airplane (A) and a submarine (S)

Strategies:

for A: x ∈ [0, 1] where to drop a bomb

for S: y ∈ [0, 1] where to hide

Payo�s: if |x− y| < ε, then submarine is destroyed (ε > 0 small):

ϕ1 =

{1 if |x− y| < ε0 otherwise

ϕ2 = −ϕ1

0 1y

ε ε

R1(y) = {x | 0 ≤ x ≤ 1, y − ε < x < y + ε}R2(x) = {y | 0 ≤ y ≤ 1, y ≤ x− ε or y ≥ x+ ε}

No match, no equilibrium.

22. Selecting a numberThere are n students, each of them has to select a number from the set {1, 2, ...,m}.

Players: n students

Strategies: Selecting an integer from set {1, 2, ...,m}

15

Game theory

Payo�: If all select the same number, then all get 1$, otherwise none of them getsanything

Equlibrium: All selections, except when n-1 students select the same number and oneselects di�erent number

Proof:

(i) (i1, i2, ..., in) when i1 = i2 = ... = in is equilibrium, since if any student changeschoice, he will get nothing (as well as all others) ⇒ equilibrium

(ii) (i1, i2, ..., in) when at most n-2 students have identical choice, if any one changesstrategy, they still will get nothing ⇒ equilibrium

(iii) (i1, i2, ..., in) when n-1 students have identical choice, then if the nth studentchanges to the choice of others, then he can increase his payo� ⇒ not anequilibrium

23. Cournot oligopoly

Players: n �rms producing same product

Strategy: Produced amounts, x1, x2, . . . , xn, 0 ≤ xk ≤ Lk

Payo�s:

ϕk(x1, . . . , xn) = xkp

(n∑l=1

xl

)− Ck(xk)

where

p = price function, decreases in total supply

Ck = cost function of �rm k.

Example 2.3 n = 2, Ck(xk) = xk + 1, 0 ≤ xk ≤ 5

p(x1 + x2) = 10− (x1 + x2)

ϕ1 = x1(10− x1 − x2)− x1 − 1 = −x21 + 9x1 − x1x2 − 1 −→ max

Best response of player 1:

∂ϕ1

∂x1

= −2x1 + 9− x2 = 0

x1 =9− x2

2

always interior, so

R1(x2) =9− x2

2Similarly,

R2(x1) =9− x1

2Equilibrium:

x1 = 9−x22

2x1 + x2 = 9 x2 = 9− 2x1

x2 = 9−x12

2x2 + x1 = 918− 4x1 + x1 = 9

9 = 3x1

x1 = 3 x2 = 3O

16

Game theory

24. Commercial �shing

Players: n �rms �shing in an open sea

Strategies: Number of boats, hk, sent for �shing by �rm k

Payo�s: Pro�t per boat =A−B∑n

l=1 hl

Cost per boat=C (A > C)

⇒ pro�t of �rm k: ϕk = hk (A−B∑n

l=1 hl)− ChkBest response:∂ϕk

∂hk= A−B

∑l 6=k hl − 2Bhk − C = 0 (k=1, 2, ..., n)

Assuming nonnegative solution this is a system of linear equations. Let H =∑nl=1 hl, then

A−BH −Bhk − C = 0

So, hk = −H + A−CB

By adding for all k, H = −nH + n(A−C)B

(n+ 1)H=n(A−C)B

⇒ H=n(A−C)(n+1)B

By symmetry: hk = A−C(n+1)B

> 0

25. Single-product oligopolies with product di�erentiation

Players: n �rms producing related products

Strategies: Produced quatities, 0 ≤ xk ≤ Lk

Payo�s: Price functions, Pk(x1, ..., xn)

Cost functions, Ck(xk)

Pro�t of �rm k, ϕk = xkPk(x1, ..., xn)− Ck(xk)Checking conditions of Nikaido-Isoda theorem (see Chapter 6):

(i) Strategy set of player k is Sk = [0, Lk], which is convex, closed, bounded inone-dimensional space

(ii) ϕk is continuous if both Pk and Ck are continuous

(iii) ϕk is concave in xk if

∂2ϕk

∂xk2 =∂(Pk+xk

∂Pk∂xk−C′k

)∂xk

=2∂Pk

∂xk+ xk

∂2Pk

∂xk2 − C ′′k ≤ 0

⇒ under (i), (ii), (iii) there is at least one equilibrium.

26. Bertrand oligopoly

Players: n �rms producing similar products

Strategies: Setting prices for own products, 0 ≤ pk ≤ Pk

Payo�s:ϕk(p1, p2, . . . , pn) = pkdk(p1, . . . , pn)− Ck (dk(p1, . . . , pn))

where dk = demand of the product made by �rm k.

17

Game theory

27. Special duopoly

Players: 2 �rms, price setting.

Strategies: Prices but giving discounts to faithful customers.

Payo�s:

ϕ1 =

{p1 if p1 ≤ p2

p1 − c if p1 > p2

ϕ2 =

{p2 if p2 ≤ p1

p2 − c if p2 > p1

Assume max price, Pmax is large enough:

R1(p2) =

p2 if p2 > Pmax − cPmax if p2 < Pmax − c{p2, p

max} if p2 = Pmax − cR2 is the same

18

Game theory

In�nitely many equilibria:

{(p1, p2) | Pmax − c ≤ p1 = p2 ≤ Pmax}

28. Price war

Players: 2 �rms

Strategies: Selected prices p1, p2 ∈ [0, P ]

Payo�s: Demand D = A− p (p=min{p1,p2}, A ≥ 2P )

ϕ1 =

p1(A− p1) if p1 < p2 everybody buys for lower price12p1(A− p1) if p1 = p2 they share market

0 if p1 > p2 nobody buys for higher price

∂(p1(A−p1))∂p1

=A− 2p1 > 0, ∂2(p1(A−p1))

∂p12= −2 < 0

No best response, no equilibrium exists.

29. Sharing 100$

19

Game theory

Players: Two people

Strategies: Step 1. Player 1 gives an o�er x1 ∈ {0, 25, 50, 75, 100} to player 2, simulta-neously player 2 gives a minimum acceptable x2 amount

Payo�s: If x1 < x2, they get nothing, and if x1 ≥ x2, then player 2 gets x1 and player 1keeps 100− x1

1\ 2 0 25 50 75 1000 (100, 0) (0, 0) (0,0) (0,0) (0,0)25 (75, 25) (75, 25) (0,0) (0,0) (0,0)50 (50, 50) (50, 50) (50, 50) (0, 0) (0, 0)75 (25, 75) (25, 75) (25, 75) (25, 75) (0,0)100 (0, 100) (0, 100) (0,100) (0,100) (0,100)

Equilibria: (0,0), (25,25), (50,50),(75,75), (100,100)

30. Pick a number

Players: Two people

Strategies: x1, x2 positive integers ≤ 50

Payo�s: If x1 = x2, then both get 50-x1, ϕ1 = ϕ2 = 50− x1

If x1 > x2, then ϕ1 = 100− x1 and ϕ2 = 0

If x1 < x2, then ϕ1 = 0 and ϕ2 = 100− x2

20

Game theory

Discrete problem!!!

R1(x2) =

{all if x2 = 50x2 + 1 if x2 < 50

Three equilibria: (50,50), (49,50), (50,49)

31.Quality control A salesman sells an equipment to a customer with rules:

• If equipment is good, customer pays $α to salesman

• if equipment is defective, salesman pays $β to customer.

Equipment has 3 parts, they can be defective with equal probabilities.

Players: Salesman and equipment (S & E )

Strategies:

For S, how many parts to check before selling equipment: 0, 1, 2 or 3. Cost of eachchecking is $γ

For E, how many parts are defective: 0, 1, 2 or 3

Payo�s:

ϕ1 = expected pro�t of salesman

ϕ2 = −ϕ1

S\E 0 1 2 30 α −β −β −β1 α− γ −2

3β − γ −1

3β − γ −γ

2 α− 2γ −13β − 5

3γ −4

3γ −γ

3 α− 3γ −2γ −43γ −γ

A11: defective part is not found with probability 23

A12: defective part is not found with probability 13

21

Game theory

A21: defective part is found either in �rst or second checking, or not

1

3

(− γ)

+2

3

[1

2

(− 2γ

)+

1

2

(− 2γ − β

)]A22: same principle

2

3

(− γ)

+1

3

(− 2γ

)A31: defective part is found either in �rst, second or third checking

1

3

(− γ)

+2

3

[1

2

(− 2γ

)+

1

2

(− 3γ

)]A32: defective part is found either in �rst or second checking

2

3

(− γ)

+1

3

(− 2γ

)Equilibrium?

Row 0 has three smallest elements a01, a02, a03

a01 is equilibrium if

−β ≥ −23β − γ, −β ≥ −1

3β − 5

3γ, −β ≥ −2γ

β ≤ 3γ β ≤ 52γ β ≤ 2γ

a02 is equilibrium if

−β ≥ −13β − γ, −β ≥ −4

3γ, −β ≥ −4

3γ

β ≤ 32γ β ≤ 4

3γ

a03 is equilibrium if

−β ≥ −γ, β ≤ γ

Row 1 has one smallest element a11

a11 is equilibrium if

−23β − γ ≥ −β, −2

3β − γ ≥ −1

3β − 5

3γ, −2

3β − γ ≥ −2γ

β ≥ 3γ β ≤ 2γ β ≤ 32γ

contradiction

Row 2 has two potential smallest elements a20, a21

a20 is not equilibrium, it is not largest in column

a21 is equilibrium if

−13β − 5

3γ ≥ −β, −1

3β − 5

3γ ≥ −2

3β − γ, −1

3β − 5

3γ ≥ −2γ

β ≥ 52γ β ≥ 2γ β ≤ γ

contradiction

and

−13β − 5

3γ ≤ α− 2γ

irrelevant

22

Game theory

Row 3 has two possible smallest elements a30 and a31

a30 is not largest in column, it is not equilibrium

a31 is equilibrium if

−2γ ≥ −β, −2γ ≥ −23β − γ, −2γ ≥ −1

3β − 5

3γ

β ≥ 2γ β ≥ 32γ β ≥ γ

and−2γ ≤ α− 3γ

α ≥ γ

32.Timimg gameTwo players want to get an object valued as v1 and v2 by them (v1 6= v2). Both want

to wait as long as possible hoping that the other will give up �ghting for the object, sohe can get it. (Price war, isolating a community in war, etc.)

Players: Two agents

Strategies: When to give up, t1 and t2(≥ 0)

Payo�s:

ϕi =

−ti if ti < tj (he gives up)12vi − ti if ti = tj (he has

12chance to get the item)

vi − tj if ti > tj(other gives up earlier)

23

Game theory

Ri(tj) =

(tj,∞) if tj < vi{0} ∪ (tj,∞) if tj = vi0 if tj > vi

Equilibria are:

{tj ∈ [vi,∞) and ti = 0 } ∪ {tj = 0 and ti ∈ [vj,∞)}

33.Position gameTwo manufacturers produce one product each with a quality parameter x1 and x2.

Customers' expectation is about quality value M. That player wins who's quality is closerto customers' expectation

Players: Two manufacurers

Strategies: x1 and x2, quality parameters

Payo�s:

ϕi =

1 if |xi −M | < |xj −M |0 if |xi −M | = |xj −M |−1 if |xi −M | > |xj −M |

24

Game theory

Case 1 occurs if

xj > M :

2M − xj < xi < xj

xj < M :

xj < xi < 2M − xj

xj = M : never occurs

Ri(xj) =

(2M − xj, xj) if xj > MM if xj = M(xj, 2M − xj) if xj < M

25

Game theory

(vertically shaded region)

Unique equilibrium:xi = xj = M

34.Location gameTwo icecream sellers have to select locations for their shops on interval [0,1]. The

potential buyers are uniformly placed on the interval, and there are in�nitely many. Eachbuyer goes to the closer shop to buy icecream.

Players: Two icecream sellers

Strategies: Locations of shops, x, y ∈ [0, 1]

Payo�s:

ϕ1 =

x+y

2if x < y

12

if x = y1− x+y

2if x > y

For case 1:

For case 3:

ϕ2 =

1− x+y

2if x < y

12

if x = yx+y

2if x > y

26

Game theory

R1(y) = 12only if y = 1

2

Similiarly, R2(x) = 12only if x = 1

2

Unique equilibrium:x = y = 12

35.AdvertisementTwo �rms compete for m markets with number of potential customers a1 > a2 > · · · >

am.

Players: 2 �rms

Strategies: Which market is selected to conduct intensive advertisement (they can selectonly one), 1 ≤ i, j ≤ m

Payo�s: If they advertise in di�erent markets, then they get all customers, and if theyadvertise on the same market, they share customers:

1\2 1 2 . . . m1 p1a1 a1 . . . a1

2 a2 p2a2 . . . a2...

......

m am am . . . pmam

ϕ1

1\2 1 2 . . . m1 q1a1 a2 . . . am2 a1 q2a2 . . . am...

......

m a1 a2 . . . qmam

ϕ2

(i, j) is equilibrium if corresponding element in ϕ1 is largest in its column and thatin ϕ2 is largest in its row.

In column 1 the largest is

(1, 1) if p1a1 ≥ a2

(2, 1) if a2 ≥ p1a1.

27

Game theory

In columns 2, . . . ,m, elements (1, 2), . . . , (1,m) are the largest.

In row 1, the largest is

(1, 1) if q1a1 ≥ a2

(1, 2) if a2 ≥ q1a1.

In rows 2, . . . ,m, elements (2, 1), . . . , (m, 1) are the largest.

Only matches are

(1, 1) ⇔ p1a1 ≥ a2 and q1a1 ≥ a2

(2, 1) ⇔ a2 ≥ p1a1

(1, 2) ⇔ a2 ≥ q1a1.

Modi�ed game: ϕ1 is as before but �rm 1 believes that �rm 2 wants to damageit, so it believes ϕ2 = −ϕ1

Equilibrium in ϕ1 matrix is largest in its column but smallest in its row:

Smallest elements in rows are (1, 1), (2, 2), . . . , (m,m)

Largest elements in columns are:

in column 1,

(1, 1) if p1a1 ≥ a2

(2, 1) if a2 ≥ p1a1

in columns 2, . . . ,m, only (1, 2), . . . , (1,m)

Only match is (1, 1) ⇔ p1a1 ≥ a2 otherwise there is no equilibrium.

36. Advertisement budget allocation

Players: 2 competing �rms on m markets

Strategies: Amounts spent on the markets in advertisement

(x1, . . . , xm) and (y1, . . . , ym)

Payo�s:

ϕ1 =m∑i=1

xiaixi + yi + zi

ϕ2 =m∑j=1

yjajxj + yj + zj

where zi = total spending of others in market i.

37. Market share

Players: Two �rms compete for a business of unit value

Strategies: E�orts in order to get larger portions of the business (e.g. market) x, y ≥ 0

28

Game theory

Payo�s:

ϕ1 =x

x+ y− x (business share value � cost)

ϕ2 =y

x+ y− y

Best responses:

∂ϕ1

∂x=

1 · (x+ y)− x · 1(x+ y)2

− 1 =y

(x+ y)2− 1 = 0

(x+ y)2 = y

x+ y =√y

x =√y − y stationary point

∂2ϕ1

∂x2=−y · 2(x+ y)

(x+ y)4< 0 as y > 0.

ϕ1 is strictly concave, vertex can be negative or nonnegative:

R1(y) =

{0 if y ≥ 1√y − y if y ≤ 1

, R2(x) =

{0 if x ≥ 1√x− x if x ≤ 1

29

Game theory

Intercepts are (0, 0) and (14, 1

4)

x =√y − y x+ y =

√y 2x =

√x

y =√x− x x+ y =

√x 4x2 − x = 0

x = y x(4x− 1) = 0x = 0 or x = 1

4

Equilibrium is x = y = 14.

38. Inventory control

Players: A retailer and a wholesaler

Strategies: Inventories, y, z ≥ 0

Payo�s: Random demand x with pdf f(x)

ϕ1 = a1

∫ y

0

xf(x)dx+

∫ y+z

y

[a1y + a2(x− y)

]f(x)dx+

∫ ∞y+z

[a1y + a2z

]f(x)dx− b1y

1st term: a1 = unit pro�t from own inventory (x ≤ y)

2nd term: a2 = unit pro�t from back order (y < x ≤ y + z)

3rd term: same (x > y + z)

4th term: b1 = unit inventory cost of retailer

ϕ2 = a3

∫ y+z

y

(x− y)f(x)dx+

∫ ∞y+z

a3zf(x)dx− b2z

1st term: a3 = unit pro�t of wholesaler from back order (y < x ≤ y + z)

3rd term: b2 = unit inventory cost of wholesaler.

39. Price strategy

Players: n �rms producing similar goods

Strategy: Time varying price of own product, pk(t) ∈ [0, Pk]

Payo�s: Let Ψk(p1, . . . , pn) be demand of good k, then

ϕk =

∫ T

0

Ψk(p1(t), . . . , pn(t))pk(t)dt = total revenue.

40. Duel without sound- �1 1

Two duelists are placed 2 units from each other, each has a gun with 1 bullet in it. Fora signal they start walking toward each other, and can shoot at any time. Their speedsare equal, guns have silencers.

Players: Two participants

Strategies: Where to shoot, 0 ≤ x, y ≤ 1

30

Game theory

Payo�s: Hitting probabilities P1(x) and P2(y), then

ϕ1 =

P1(x) · 1− (1− P1(x)) · P2(y) if x < yP1(x)− P2(y) if x = yP2(y) · (−1) + (1− P2(y))P1(x) if x > y

ϕ2 =

P2(y) · 1− (1− P2(y)) · P1(x) if y < xP2(y)− P1(x) if y = x−P1(x) + (1− P1(x))P2(y) if y > x

Example 2.4 P1(x) = x, P2(y) = y

ϕ1 =

x · 1− (1− x) · y = x− y + xy if x < yx− y = x− y if x = y−y + (1− y)x = x− y − xy if x > y

R1(y) exists only if

y2 ≤ 1− 2y

y2 + 2y − 1 ≤ 0

y12 =−2±

√4 + 4

2= −1±

√2 = 0.4142 or − 2.4142

y ≤ 0.4142

R1(y) =

{1 if y ≤ 0.4142does not exist, otherwise

31

Game theory

No match, no equilibrium.

O

41. Duel with soundSame as previous problem, but guns have no silencers:

ϕ1 =

P1(x)− (1− P1(x)) = 2P1(x)− 1 if x < yP1(x)− P2(y) if x = y−P2(y) + (1− P2(y)) = 1− 2P2(y) if x > y

Example 2.5 P1(x) = x, P2(y) = y

ϕ1 =

2x− 1 if x < y0 if x = y1− 2y if x > y

Case 1: y < 12

32

Game theory

Case 2: y = 12and Case 3: y > 1

2

R1(y) =

{x | x > y} if y < 1

2

{x | x ≥ 12} if y = 1

2

does not exist if y > 12

33

Game theory

R2(x) is mirror image, only equilibrium is x = y = 12

⇒ do not shoot early and also do not shoot late. O

42. Spying game

Players: Spy and counterespionage

Strategies: E�orts, x, y ≥ 0

Payo�s: Let

P (x, y) = probability of arrest

V (x) = value of information collected by spy

U = value of spy.

Then

ϕ1 = P (x, y) · (−U) + (1− P (x, y)) · V (x)

ϕ2 = −ϕ1

Example 2.6 U = 4, V (x) = x, P (x, y) = A · (x+ y) (A > 0 is small)

ϕ1 = A · (x+ y)(−4) + [1− A · (x+ y)]x

= −4Ax− 4Ay + x− Ax2 − Axy strictly concave in x∂ϕ1

∂x= −4A+ 1− 2Ax− Ay = 0

x =1− 4A− Ay

2Astationary point

R1(y) =

{1−4A−Ay

2Aif y ≤ 1−4A

A

0 otherwise

ϕ2 = 4Ax+ 4Ay − x+ Ax2 + Axy

strictly inreases in y, so

R2(x) = ymax

34

Game theory

Equilibrium:y = ymax

x =

{0 if ymax ≥ 1−4A

A1−4A−Aymax

2A, otherwise

O

43. Hidden bomb in a cityCity with rectangular shape with m horizontal and n vertical blockrows and block-

columns respectively. Value of block (i, j) is aij.City map:

a11 a12 a13 . . . a1,n−2 a1,n−1 a1n

a21 a22 a23 . . . a2,n−2 a2,n−1 a2n

......

......

......

am1 am2 am3 . . . am,n−2 am,n−1 amn

A terrorist group places a bomb in one of the blocks, and requests release of criminalsfrom city prisons. City can check only one blockrow or blockcolumn to �nd the bomb.

Players: City (C) and terrorists (T)

Strategies:

For C: row i or column j to search

For T: where to place the bomb

Payo�s: ϕ1 = value of block if the bomb was there and became found:

35

Game theory

1\2 (1,1) . . . . . . (1,n) (2,1) . . . . . . (2,n) . . . (m,1) . . . . . . (m,n)rows

1 a11 . . . . . . a1n

2 a21 . . . . . . a2n...

. . .

m am1 . . . . . . amn

columns 1 a11 a21 am1

2 a12 a22 am2...

. . .. . .

. . .

n a1n a2n amn

ϕ1

ϕ2 = −ϕ1

Equilibrium: Matrix element is largest in its column and smallest in its row.

Facts:

largest elements in all columns are positive,

smallest elements in all rows are zeros

⇒ no element satis�es both ⇒ no equilibrium

44. First price auctionOne unit is sold in an auction

Players: n potential buyers with subjective valuation of the unit v1 > v2 > ... > vn(known to all)

Strategies: Each of them presents a bid, x1, ..., xn, simultaneously, bids are secret

Payo�s: Highest bidder wins the unit and pays his price, in case of more highest biddersthe one with highest valuation wins

ϕk =

vk − xk if xk = max{x1, ..., xn}, and maximum is unique

or k = min{l|xl = xk}0 otherwise

Fact: In any Nash equilibrium player 1 wins.

Proof : Assume not, if player i 6= 1 wins, then xi > x1. If xi > v2, then ϕi =vi − xi ≤ v2 − xi < 0, so player i can increase his payo� to zero by decreasing hisbid xi.

If xi ≤ v2, then player 1 can increase his 0 payo� to v1 − xi by increasing his bid toxi. (Note, v1 − xi > v2 − xi ≥ 0) �

Example 2.7 n=2, payo�s:

ϕ1 =

{v1 − x1 if x1 ≥ x2

0 otherwise

ϕ2 =

{v2 − x2 if x2 > x1

0 otherwise

36

Game theory

R1(x2) =

x2 if x2 < v1

[0, x2] if x2 = v1

[0, x2) if x2 > v1

R2(x1) =

{∅ if x1 < v2

[0, x1] if x1 ≥ v2

37

Game theory

Equilibrium set: {(x1, x2)|v2 ≤ x1 = x2 ≤ v1}

45. Second price auction

Players: n bidders with valuations v1 > v2 > ... > vn

Strategies: Bids x1, x2, ..., xn

Payo�s: Player k gets the item if he has the largest bid and in the case of a tie, hisvaluation is the largest; and the winner pays second largest bid

ϕk =

vk − xl if xk = max{x1, ..., xn}, xl = max{xi|i 6= k}

and k = min{i|xi = xk} or unique maximal xk0 otherwise

Example 2.8 n=2, payo�s:

ϕ1 =

{v1 − x2 if x1 ≥ x2

0 otherwise

ϕ2 =

{v2 − x1 if x2 > x1

0 otherwise

38

Game theory

R1(x2) =

[x2,∞) if x2 < v1

[0,∞) if x2 = v1

[0, x2) if x2 > v1

(horizontally shaded region)

R2(x1) =

(x1,∞) if x1 < v2

[0,∞) if x1 = v2

[0, x1] if x1 > v2

(vertically shaded region)

Set of equilibria= the two sets with both horizental and vertical shades.

39

Game theory

46. VotingTwo candidates (A and B) run for an o�ce which is decided by election. Among the

voters k support A, m = n − k support B. Each vote costs amount c (0<c<1) for thevoter, so each of them makes the decision to vote or not.

Players: n voters

Strategies: Votes (xi = 1) or not (xi = 0)

Payo�s: For voter, 1, 0, -1 if his candidate wins, the result is tie, or the opponent wins,but it decreases by c; for non voter, 1, 0 or -1 as above without voting cost

1 ≤ i ≤ k:

ϕi =

1− cxi if

∑kl=1 xl >

∑nj=k+1 xj

−cxi if∑k

l=1 xl =∑n

j=k+1 xj−1− cxi if

∑kl=1 xl <

∑nj=k+1 xj

k + 1 ≤ j ≤ n:

ϕj =

−1− cxj if

∑ki=1 xi >

∑nl=k+1 xl

−cxj if∑k

i=1 xi =∑n

l=k+1 xl1− cxj if

∑ki=1 xi <

∑nl=k+1 xl

Example 2.9 k = m=1

1\ 2 votes (x2 = 1) does not (x2 = 0)votes (x1 = 1) (-c, -c) (1-c, -1)

does not (x1 = 0) (-1,1-c) (0, 0)

Unique equilibrium (x1 = 1, x2 = 1)

In general: What is the equilibrium?

Fact: One candidate wins, not an equilibrium

Proof :

If at least one voted in losing group ⇒ if that voter does not vote, he increasespayo�

If nobody voted in losing group, then two cases:

(i) more than one voted in winning group ⇒ if one of them does not vote, he canincrease payo�;

(ii) only one voted in winning group ⇒ if one in losing group votes, then there is a tie,so he can increase his payo�.

⇒ At any equilibrium it has to be a tie. If somebody did not vote, then bychanging his mind and voting, group becomes winner, so this is not equilibrium;

⇒ Everybody has to vote, so equilibrium exists only if k = m and everybody votesfor his candidate. This is really an equilibrium, since if any player does not vote,his group becomes the loser.

47. Irrigation systemFarms use common water supply to irrigate.

40

Game theory

Players: n farms

Strategy: Amount of water used, x1, . . . , xn

Payo�s: Bene�t of irrigation � cost of water

ϕk = Bk(xk)− xk ·K(∑n

l=1 xl

)∑n

l=1 xl︸ ︷︷ ︸unit cost of water

Similar to oligopoly, p(s) = −K(s)s, Ck(xk) = −Bk(xk).

48.Waste water managementn �rms treat waste water in a common plant.

Players: n �rms

Strategy: Amounts of treated waste water, x1, . . . , xn

Payo�s: Bene�t (usage of treated water, not paying penalty, etc.) � cost of water treat-ment

ϕk = Bk(xk)− xk ·K(∑n

l=1 xl

)∑n

l=1 xl.

Same as previous example.49. Multipurpose water management systemn water users (industry, agriculture, domestic, recreation).

Players: Water users

Strategies: Amounts of allocated water to users, x1, . . . , xn

Payo�s: Bene�t � cost, same as above

50. Chess game

Players: 2 players controlling W and B �gures

Strategies: For all possible con�gurations on the board a selected next move

Payo�s:

ϕW =

1 if W wins−1 if B wins0 if tie.

,

ϕB = −ϕW .

41

Game theory

3 Discrete games with �nitely many strategies

Example 3.1 No equilibrium exists

1 2

2 0

2 1

4 5

ϕ1 and ϕ2 � no match

O

Example 3.2 Unique equilibrium

-2 -10

-1 -5-2 -1

-10 -5

ϕ1 and ϕ2 � one match

O

Example 3.3 Multiple equilibria

1 11 1

1 11 1

ϕ1 and ϕ2 � everything is equilibrium

O

Method: Do loop goes through every element (i, j) and checks if ϕ1(i, j) is largest in itscolumn and ϕ2(i, j) is largest in its row. If yes, then (i, j) is equilibrium, otherwise not.

4 Games Representable by Finite Rooted Trees

Assume: full information to all players;

(i) game starts at the root of the tree;

(ii) to each node of the tree a player is assigned and the game proceeds on an arcoriginating at this node to its end vertex (by the decision of the player which isassigned to the originating node);

(iii) each player knows the vertices at which he/she has to make decisions;

(iv) at each terminal node, each player has a given payo� value

Example 4.1

42

Game theory

O

Example 4.2 Chess-game O

Theorem 4.1 The game has at least one equilibrium pont (EP).

Proof. From root to each terminal node there is a unique path, since otherwise cyclewould arise:

Length of the path= number of arcs on itProof by induction with respect to the length h of the tree (which is the longest path

from root to any endpoint). If h = 0, only one point, no decision, that point is the EP.If h ≥ 1, then assume player l is assigned to the root. Let I1, I2, . . . , Im be vertices

connected directly to the root. With roots I1, I2, . . . , Im we have subgames with smallerlengths, each has EP with payo�s (ϕ

(1)1 , . . . , ϕ

(1)n ), . . . , (ϕ

(m)1 , . . . , ϕ

(m)n ).

The EP of original game is obtained as follows:Let ϕ

(k0)l = max

1≤k≤m{ϕ(k)

l }, then

• For player l, the EP of subgame k0 with the initial step from the root to Ik0

43

Game theory

• For players i 6= l, the EP of subgame k0

�Note. We got a method: backward induction.

Example 4.3

So EP is with payo�s (2, 2, 3) and path A→ B → C

O

Remark. Some vertices might not be assigned to players, they might be random withgiven discrete distributions de�ned on arcs originating from each random vertex. Sameproof applies to show existence of EP.

Example 4.4 Chain store

Equilibrium is (soft,in)Note. The other EP, (hard,out) is lost. O

44

Game theory

5 Continuous games

5.1 Some mathematical background

(i) Mean value theorems in Rn

Theorem 5.1 Let f : D 7→ R be di�erentiable, where D ⊆ Rn is convex. Then

f(x)− f(y) =∂f(z)

∂x1

(x1 − y1) + · · ·+ ∂f(z)

∂xn(xn − yn)

where z is a point in the linear segment connecting x and y.

Proof. Letg(t) = f(y + t(x− y)),

theng(0) = f(y), g(1) = f(x), g′(t) = 5f

(y + t(x− y)

)(x− y)

where 5f is the gradient of f as a row vector. So

g(1)− g(0) = g′(c)(1− 0) with some c ∈ (0, 1), so

f(x)− f(y) = 5f(z)(x− y),with z = y + c(x− y) being on the linear segment

�

Theorem 5.2 Let f : D 7→ Rn be di�erentiable, where D ⊆ Rm is convex. Then

f(x)− f(y) =

∫ 1

0

Of(y + t(x− y)

)(x− y)dt,

where Of is the Jacobian matrix of f .

Proof. Let g(t) = f(y + t(x− y)

), then

g(1) = f(x), g(0) = f(y), so

f(x)− f(y) = g(1)− g(0) =

∫ 1

0

g′(t)dt

=

∫ 1

0

Of(y + t(x− y)

)(x− y)dt

�

(ii) Cauchy�inequality

a1, a2, . . . , an; b1, b2, . . . , bn are real numbers, then(∑aibi

)2

≤(∑

a2i

)(∑b2i

)Proof. Let

f(x) = (a1 − xb1)2 + · · ·+ (an − xbn)2

=∑

a2i︸ ︷︷ ︸

A

−2x∑

aibi︸ ︷︷ ︸C

+x2∑

b2i︸ ︷︷ ︸

B

≥ 0

45

Game theory

This quadratic polynomial has 0 or 1 real root only ⇒ discriminant must not bepositive:

4C2 − 4AB ≤ 0

C2 ≤ AB.

�

(iii) Vector and matrix norms

Lengths of vectors in Rn (or Cn) are called norms and denoted as ‖x‖. Axioms ofnorms:

(a) ‖x‖ ≥ 0, and ‖x‖ = 0 ⇔ x = 0;

(b) ‖αx‖ = |α| · ‖x‖ with all vectors x and real (or complex) scalars α;

(c) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ (triangle inequality)

Example 5.1 Three particular norms are used most frequently:l1-norm:

‖x‖1 =n∑i=1

|xi| (1)

Proof. (a), (b) are trivial, (c) is proved as follows:

|xi + yi| ≤ |xi|+ |yi|

and by adding it for all i the result is obtained. �

l2-norm:

‖x‖2 =

√√√√ n∑i=1

|xi|2 (2)

Proof. (a), (b) are trivial, (c) is proved by using Cauchy-inequality

‖x+ y‖2 =n∑i=1

|xi + yi|2 =n∑i=1

|x2i + 2xiyi + y2

i |

≤n∑i=1

|xi|2 +n∑i=1

|yi|2 + 2n∑i=1

|xi| · |yi|

≤n∑i=1

|xi|2 +n∑i=1

|yi|2 + 2

√√√√ n∑i=1

|xi|2n∑i=1

|yi|2

=

√√√√ n∑i=1

|xi|2 +

√√√√ n∑i=1

|yi|2

2

=(‖x‖+ ‖y‖

)2

�

46

Game theory

l∞-norm:

‖x‖∞ = max |xi| (3)

Proof. (a), (b) are trivial, (c) can be proved as follows. Let maxi |xi + yi| = |xi0 + yi0|,then

‖x+ y‖∞ = |xi0 + yi0| ≤ |xi0|+ |yi0| ≤ maxi|xi|+ max

i|yi| = ‖x‖∞ + ‖y‖∞

�O

Let A be n× n matrix, norm of this matrix ‖A‖ should satisfy axioms:

(a) ‖A‖ ≥ 0, and ‖A‖ = 0 if and only if A = 0;

(b) ‖αA‖ = |α| · ‖A‖ for all matrices A and constants α;

(c) ‖A+B‖ ≤ ‖A‖+ ‖B‖ (triangle inequality).

De�nition 5.1 A matrix norm is generated from a vector norm, if

‖A‖ = max‖x‖=1

‖A x‖.

Example 5.2 It can be proved that the generated matrix norms are:

‖A‖1 = maxj

n∑i=1

|aij| (column-norm) (4)

‖A‖∞ = maxi

n∑j=1

|aij| (row-norm) (5)

‖A‖2 = max√λATA (Euclidean-norm) (6)

where λATA are the eigenvalues of matrix ATA. O

Lemma 5.1 If matrix norm ‖.‖ is generated from vector norm ‖.‖, then for all matricesA and vectors x,

‖A x‖ ≤ ‖A‖ · ‖x‖

Proof.

‖A x‖ =

∥∥∥∥A · x

‖x‖

∥∥∥∥ · ‖x‖ ≤ max‖x‖=1

‖A x‖ · ‖x‖ = ‖A‖ · ‖x‖.

�

Lemma 5.2 Any matrix-norm generated from a vector norm satis�es axioms (a), (b)and (c), furthermore

(d) ‖A B‖ ≤ ‖A‖ · ‖B‖

Proof.

(a) ‖A‖ ≥ 0 trivial,

‖A‖ = 0 ⇔ ‖A x‖ = 0 for all ‖x‖ = 1 ⇔ A x = 0

for ‖x‖ = 1 ⇔ A x = 0 for all x ⇒ A = 0;

47

Game theory

(b) Trivial;

(c)

‖A+B‖ = max‖x‖=1

‖(A+B)x‖ ≤ max‖x‖=1

(‖A x‖+ ‖B x‖)

≤ max‖x‖=1

‖A x‖+ max‖x‖=1

‖B x‖ = ‖A‖+ ‖B‖.

(d) With some ‖z‖ = 1,

‖A ·B‖ = ‖A ·B · z‖ ≤ ‖A‖ · ‖B · z‖ ≤ ‖A‖ · ‖B‖ · ‖z‖ = ‖A‖ · ‖B‖.

�

Example 5.3

‖A‖F =

√√√√ n∑i=1

n∑j=1

|aij|2 (Frobenius-norm) (7)

Proof. Axioms (a), (b) are trivial, (c) can be proved as axiom (c) for ‖x‖2. O

Lemma 5.3 For all vectors x and matrices A,

‖A x‖2 ≤ ‖A‖F · ‖x‖2

Proof. By Cauchy-inequality,

‖A x‖22 =

n∑i=1

∣∣∣ n∑j=1

aijxj

∣∣∣2 ≤ n∑i=1

(n∑j=1

|aij|2)·

(n∑j=1

|xj|2)

= ‖A‖2F · ‖x‖2

2

‖A‖F is a vector norm of the n2-element vector, so (a), (b), (c) are satis�ed. �

Lemma 5.4 The Frobenius norm satis�es property (d).

Proof.

‖A B‖2F =

∑i

∑j

∣∣∣∑l

ailblj

∣∣∣2 ≤∑i

∑j

(∑l

|ail|2∑l

|blj|2)

=∑i

∑l

|ail|2∑j

∑l

|blj|2 = ‖A‖2F · ‖B‖2

F

�

De�nition 5.2 Matrix norm ‖A‖ and vector norm ‖x‖ are compatible, if for all matricesand vectors, ‖A x‖ ≤ ‖A‖ · ‖x‖

Corollary. The norm pairs {‖A‖2, ‖x‖2}, {‖A‖1, ‖x‖1}, {‖A‖∞, ‖x‖∞} and {‖A‖F , ‖x‖2}are compatible.Remark. Not all matrix norms satisfying axioms (a), (b), (c) are compatible with avector norm, as the following example shows.

48

Game theory

Example 5.4 Let ‖A‖∗ = 12· ‖A‖1, clearly axioms are satis�ed, and

‖x‖ = ‖I x‖ ≤ ‖I‖∗ · ‖x‖ =1

2· ‖x‖,

contradiction.O

It is known that the vector norms in Rn are equivalent to each other, that is, if ‖ · ‖and ‖ · ‖∗ are two vector norms, then there are constants K1 and K2 such that for allx ∈ Rn,

K1‖x‖ ≤ ‖x‖∗ ≤ K2‖x‖

.

De�nition 5.3 Distance of vectors x and y can be de�ned as %(x, y) = ‖x−y‖ with somevector norm.

Sometimes distances of vectors are de�ned more generally by the following axioms:

(i) %(x, y) ≥ 0, and %(x, y) = 0 ⇔ x = y;

(ii) %(x, y) = %(y, x);

(iii) %(x, y) + %(y, z) ≥ %(x, z) (triangle inequality).

If distance is de�ned as %(x, y) = ‖x− y‖, then these axioms are satis�ed. Properties (i),(ii) are trival. For (iii),

%(x, y) + %(y, z) = ‖x− y‖+ ‖y − z‖ ≥

‖(x− y) + (y − z)‖ = ‖x− z‖ = %(x, z).

Note, not all vector distances satisfying (i), (ii), (iii) can be obtained from vector norm.

Example 5.5 De�ne

%(x, y) =

{0 if x = y1 if x 6= y

Clearly axioms are satis�ed, and if it were generated from a vector norm, then with x 6= 0,1 =%(1

2x, 0) = ‖1

2x − 0‖ = ‖1

2x‖ = 1

2‖x‖ =1

2‖x − 0‖ = 1

2%(x, 0) = 1

2· 1 = 1

2, which is

impossible. O

We will always assume that a vector distance is obtained from a vector norm.

5.2 Brouwer��xed point theorem

Theorem 5.3 (In one dimension) Let f : [a, b] 7→ [a, b] be continuous, then there ex-ists an x∗ ∈ [a, b] such that

x∗ = f(x∗)

Proof.

49

Game theory

Case 1 f(a) = a, then x∗ = a

Case 2 f(b) = b, then x∗ = b

Case 3 a < f(a) and b > f(b), then curve must intercept 45�degree line.

�

Theorem 5.4 (Generally) Let D ⊆ Rn be convex, closed, bounded, nonempty set, f :D 7→ D continuous function. Then there exists x∗ ∈ D such that x∗ = f(x∗).

Examples showing that without conditions theorem fails:D nonconvex:

D = [−2,−1] ∪ [1, 2], f(x) = −x

D not closed:D = (0, 1), f(x) =

x

2

D not bounded:D = R, f(x) = x+ 1

f not continuous:

D = [0, 1], f(x) =

{x2

if x 6= 01 if x = 0

O

50

Game theory

5.3 Kakutani �xed point theorem

Theorem 5.5 (Kakutani �xed point theorem) Let f be a point-to-set mapping suchthat for all x ∈ D, f(x) is a subset of D. Assume

(i) D is nonempty, convex, closed, bounded in Rn,

(ii) f(x) is nonempty, closed, convex subset of D for all x ∈ D,

(iii) the graph of mapping f, Gf = {(x, y) | x ∈ D, y ∈ f(x)} is closed.

Then there is an x∗ such that x∗ ∈ f(x∗).

Remark. Generalization of Brouwer's, since if f is one-to-one and continuous, then Gfis the curve of the function, which is a closed set.

5.4 Banach��xed point theorem

Theorem 5.6 (Bolzano�Weierstrass theorem) In Rn, every bounded sequence hasconvergent subsequence.

Proof. Since the vector norms are equivalent, convergence of vectors does not dependon the selection of norms, it means component-wise convergence. We give proof in onedimension, in Rn same by repeating it component-wise.

There are∞ many points in [A,B]; so one half also has∞ many points from sequence.Let

I0 = [A,B], select x0 ∈ I0

Let I1 be half of I0 having ∞ points, select x1 6= x0 and x1 ∈ I1

Let I2 be half of I1 having ∞ points, select x2 6= x0, x1 and x2 ∈ I2 and so on.

Intervals shrink to a single point, this is limit of sequence x0, x1, x2, . . . �Convergent sequence: xn → x∗ as n→∞ if %(xn, x

∗)→ 0Cauchy sequence: %(xn, xm)→ 0 as n,m→∞

• Convergent ⇒ Cauchy

0 ≤ %(xn, xm) ≤ %(xn, x∗)︸ ︷︷ ︸

→0

+ %(x∗, xm)︸ ︷︷ ︸→0

→ 0

51

Game theory

• Cauchy ⇒ Convergent

For any ε > 0 there exists N0, such that %(xm, xn) < ε if m,n ≥ N0. Consider theball B with center xN0

and radius ε.

Only x1, x2, . . . , xN0−1 can be outside the ball B ⇒ sequence is bounded ⇒ hasconvergent subsequence ⇒ since %(xn, xm)→ 0, whole sequence converges to samelimit.

Theorem 5.7 (Banach �xed-point theorem) Let D ⊆ Rn be a closed set, f : D 7→ Dbe a contraction:

%(f(x), f(y)

)≤ q · %(x, y)

with some �xed 0 ≤ q < 1. Then there is a unique x∗ ∈ D such that x∗ = f(x∗).

Proof. Let x0 ∈ D arbitrary, consider sequence

x1 = f(x0)

x2 = f(x1)

. . .

Since f(x) ∈ D for all x ∈ D, sequence is well de�nied.

(i) We �rst prove that {xk} is a Cauchy�sequence:

%(xk+1, xk) = %(f(xk), f(xk−1)

)≤ q · %(xk, xk−1) ≤ q · q · (xk−1, xk−2)

≤ · · · ≤ qk%(x1, x0) = qk%(x0, x1),

so for n < m,

%(xn, xm) ≤ %(xn, xn+1) + %(xn+1, xn+2) + · · ·+ %(xm−1, xm)

≤ qn%(x0, x1) + qn+1%(x0, x1) + · · ·+ qm−1%(x0, x1)

≤ %(x0, x1)[qn + qn+1 + . . . until ∞

]= %(x0, x1)

qn

1− q.

If n,m→∞, this converges to zero.

(ii) Since {xk} is a Cauchy�sequence it has a limit in Rn, and since D is closed, thislimit denoted by x∗, is also in D.

(iii) xn+1 = f(xn) for all n. Let n→∞ and since f is continuous,

x∗ = f(x∗) ⇒ x∗ is a �xed point.

(iv) Let x∗, y∗ be 2 �xed points:

%(f(x∗)︸ ︷︷ ︸x∗

, f(y∗)︸ ︷︷ ︸y∗

) ≤ q · %(x∗, y∗)

%(x∗, y∗)︸ ︷︷ ︸+

[1− q]︸ ︷︷ ︸+

≤ 0 ⇒ contradiction.

�

Examples showing that without conditions the theorem may fail:

52

Game theory

Drop that D is closed:

D = (0, 1], f(x) =x

2⇒ no �xed point

Drop contraction:

D = (−∞,∞), f(x) = x+ 1 ⇒ no �xed point

Remark. Comparing Brouwer and Banach theorems:

(i) in D, Banach is weaker (no condition on boundedness or convexity)

(ii) in f , Brouwer is weaker, contraction ⇒ continuity

%(f(xn), f(x∗)

)≤ q%(xn, x

∗),

so if xn → x∗, then %(xn, x∗)→ 0, so f(xn)→ f(x∗) as well.

5.5 Conditions of contraction mappings

In one dimension:f(x)− f(y) = f ′(c)(x− y)

x y

c

If |f ′(c)| ≤ q < 1 ⇒ contractionSo |f ′| has to be small:

xk+1 = f(xk)

g(x) = 0, how to rewrite?

x = x+ g(x)︸ ︷︷ ︸f(x)

f ′(x) = 1 + g′(x)︸︷︷︸maybe larger than zero

Instead select x = x+ g(x) · h(x)︸ ︷︷ ︸f(x)

(h(x) 6= 0)

f ′(x) = 1 + g′(x)h(x) + g(x)h′(x)

53

Game theory

We want this to be zero at the root x∗, then it will be small around x∗:

1 + g′(x∗)h(x∗) = 0

h(x∗) = − 1

g′(x∗)so select h(x) = − 1

g′(x)⇒ form of �xed point problem:

x = x− g(x)

g′(x), iteration xk+1 = xk −

g(xk)

g′(xk)Newton's method

Let g′(xk) ≈ g(xk)−g(xk−1)

xk−xk−1, then

xk+1 = xk −g(xk)(xk − xk−1)

g(xk)− g(xk−1)Secant method

In n dimension:From mean-value theorem

‖f(x)− f(y)|‖ ≤∫ 1

0

‖Of(y + t(x− y)

)‖ · ‖x− y‖dt

so if we assume that D ⊆ Rn is convex and f is di�erentiable on D, furthermore for allz ∈ D,

‖Of(z)‖ ≤ q < 1, (8)

then‖f(x)− f(y)‖ ≤ q · ‖x− y‖

⇒ f is contraction on D.Remark. (8) might hold with one norm but not with others.

Example 5.6 Consider

A1 =

(0.8 0.80 0

), A2 =

(0.8 00.8 0

), A3 =

(0.51 0.510.51 0

)with

‖A1‖1 = 0.8 < 1, ‖A1‖∞ = 1.6, ‖A1‖2 =√

1.28

‖A2‖1 = 1.6, ‖A2‖∞ = 0.8 < 1, ‖A2‖2 =√

1.28

‖A3‖1 = 1.02, ‖A3‖∞ = 1.02, ‖A3‖2 = 0.51√

3+√

52≈ 0.825 < 1

O

g(x) = 0, how to rewrite it into �xed point problem?

Rewrite

x = x+H(x)g(x)︸ ︷︷ ︸f(x)

, where H(x) is invertible n× n matrix

54

Game theory

Derivative of f(x) is

I +H ′(x)g(x) +H(x)

Jacobianof g(x)︷︸︸︷J(x)

If this is zero at x∗, then g(x∗) = 0 implying that

I +H(x∗)J(x∗) = 0

H(x∗) = −J−1(x∗)

So we select for all x,H(x) = −J−1(x),

so �xed�point equation:x = x− J−1(x)g(x),

with iteration method:xk+1 = xk − J−1(xk)g(xk)

It is the multivariable Newton's method.

5.6 Relation of EP and �xed points

We can see two di�erent ways.

1. Based on payo� functions

Consider gameG = {n;S1, . . . , Sn︸ ︷︷ ︸

strategy sets

; ϕ1, . . . , ϕn︸ ︷︷ ︸payo� functions

}

De�ne

Φ(x, y) =n∑k=1

ϕk(x1, . . . , xk−1, yk, xk+1, . . . , xn)

for all x, y ∈ S = S1 × S2 × · · · × Sn.

Lemma 5.5 (x∗1, . . . , x∗n) = x∗ is an EP ⇔

Φ(x∗, x∗) ≥ Φ(x∗, y) for all y ∈ S. (9)

Proof.⇒ for all k, ϕk(x

∗1, . . . , x

∗k, . . . , x

∗n) ≥ ϕk(x

∗1, . . . , yk, . . . , x

∗n) ⇒ adding up for k =

1, 2, . . . , n, (9) is obtained⇐ select y = (x∗1, . . . , yl, . . . , x

∗n), then from (9),

n∑k=1

ϕk(x∗1, . . . , x

∗n) ≥ ϕl(x

∗1, . . . , yl, . . . , x

∗n) +

∑k 6=l

ϕk(x∗1, . . . , x

∗n)

after cancellation

ϕl(x∗1, . . . , x

∗n) ≥ ϕl(x

∗1, . . . , yl, . . . , x

∗n) ⇒ x∗ is EP.

�De�ne

H(x) ={z ∈ S | Φ(x, z) = max{Φ(x, y) | y ∈ S}

}55

Game theory

Lemma 5.6 x∗ is EP ⇔ x∗ ∈ H(x∗).

Remark. EP is equivalent to a �xed point problem.

2. Based on best responses

Best reply is de�ned as

maxϕk(x∗1, . . . , x

∗k−1, xk, x

∗k+1, . . . , x

∗n)

subject to xk ∈ SkSet of optimal solutions Rk(x

∗):

Rk(x∗) = argmax

{ϕk(x

∗1, . . . , x

∗k−1, xk, x

∗k+1, . . . , x

∗n) | xk ∈ Sk

}Lemma 5.7 x∗ is EP ⇔ x∗ ∈ R(x∗) where

R(x∗) = (R1(x∗), . . . , Rn(x∗))

Remark. EP is equivalent to a �xed point problem, but this is di�erent than the problemwith mapping H(x).

Existence results from �xed-point theorems:

Theorem 5.8 Assume R(x) is unique (e.g ϕk is strictly concave in xk) and continuousin x, furthermore S = S1 × · · · × Sn is nonempty, convex, closed, bounded. Then there isat least one EP.

Proof. Trivial, from Brouwer �xed�point theorem. �

Theorem 5.9 Assume R(x) is unique and contraction, furthermore S = S1× · · ·×Sn isclosed. Then EP exists and is unique.

Proof. Trivial, from Banach's �xed�point theorem. �

Theorem 5.10 Assume that

(i) S is nonempty, convex, closed, bounded;

(ii) R(x) is nonempty, closed, convex in S for all x ∈ S;

(iii) GX = {(x, y) | x ∈ S, y ∈ R(x)} is closed.

Then there is an EP.

Proof. Trivial from Kakutani's �xed point theorem. �

In using Brouwer �xed�point theorem only continuity of best reply is needed, but inapplying Banach's �xed�point theorem it has to be a contraction.

56

Game theory

6 Nikaido�Isoda�theorem

Theorem 6.1 Assume that in an n-person game for all k,

(i) Sk is nonempty, convex, closed, bounded in �nite dimensional vector space;

(ii) ϕk is continuous in all variables;

(iii) ϕk is concave in xk with �xed x1, . . . , xk−1, xk+1, . . . , xn.

Then the game has at least one EP.

The proof can be based on Kakutani's �xed point theorem. Some simple facts areneeded.

6.1 Concave functions

D ⊆ Rn convex set, f : D 7→ R is called concave if

f(αx+ βy) ≥ αf(x) + βf(y)

for all 0 ≤ α, β ≤ 1, α + β = 1.

Assume f is di�erentiable, then with β = 1− α,

f(αx+ (1− α)y) ≥ αf(x) + (1− α)f(y)

f(y + α(x− y))− f(y) ≥ α[f(x)− f(y)

]f(y + α(x− y))− f(y)

α≥ f(x)− f(y)

If α→ 0 then the limit of left hand side is

∂

∂αf(y + α(x− y)

) ∣∣∣α=0

,

so

57

Game theory

5f(y)(x− y) ≥ f(x)− f(y)

and also x↔ y

5f(x)(y − x) ≥ f(y)− f(x)

In one dimension 5f(x) = f ′(x).

x < y ⇒ f ′(y) ≤ f(x)− f(y)

x− yand

f ′(x) ≥ f(y)− f(x)

y − x⇒ f ′(y) ≤ f ′(x)

⇒ f ′ decreases, so if f ′′ exists, then f ′′(x) ≤ 0.

Lemma 6.1 Let f : D 7→ R be a real valued concave, continuous function and D anonempty, convex, closed set in Rn, and there is at least one solution of problem

maximizex∈D f(x)

Then the optimal solutions form a convex, closed set.

Proof. Let f ∗ be the optimal objective function value.

(i) xk ∈ D, xk → x∗, xk is optimal solution ⇒ f(xk) = f ∗ for all k, k → ∞ and fcontinuous, f(x∗) = f ∗.

Since D is closed, x∗ ∈ D and so optimal solution.

(ii) x, y ∈ D and optimal solutions. Let z = αx + (1 − α)y (0 ≤ α ≤ 1). Since f isconcave

f(z) = f(αx+ (1− α)y) ≥ α f(x)︸︷︷︸f∗

+(1− α) f(y)︸︷︷︸f∗

= f ∗

and because D is convex, z ∈ D ⇒ f(z) must not be larger than f ∗, so f(z) =f ∗ ⇒ z is also optimal solution.

58

Game theory

�

Lemma 6.2 Let D be convex in Rn and f strictly concave. Then f cannot have multiplemaximum points.

Proof. Assume x, y are both maximum points, let z = αx + (1 − α)y ∈ D, 0 < α < 1,then

f(z) = f(αx+ (1− α)y) > αf(x) + (1− α)f(y) = αf ∗ + (1− α)f ∗ = f ∗

⇒ contradiction, f(z) cannot be better than maximum. �

Proof of Nikaido�Isoda theorem. Conditions of the Kakutani �xed-point theoremare veri�ed:

S = S1 × S2 × · · · × Sn is closed, convex, bounded by assumptions of the theorem.Consider

maxϕk(x1, . . . , xk−1, yk, xk+1, . . . , xn)

subject to yk ∈ SkBest response is the set of optimal solutions: Rk(x).It is nonempty since S is bounded and closed.ϕk is continuous and concave ⇒ Rk(x) is closed, convex, so

R(x) = R1(x)× · · · ×Rn(x) is also closed, convex

Only (iii) has to be shown. The set

G ={

(x, y) | x ∈ D, y ∈ R(x)}

has to be closed. Select x(l) → x∗, and y(l) ∈ R(x(l)) such that y(l) → y∗. Then

ϕk

(x

(l)1 , . . . , x

(l)k−1, y

(l)k , x

(l)k+1, . . . , x

(l)n

)≥ ϕk

(x

(l)1 , . . . , x

(l)k−1, yk, x

(l)k+1, . . . , x

(l)n

)for all l and yk ∈ Sk. Let l→∞, by continuity of ϕk,

ϕk(x∗1, . . . , x

∗k−1, y

∗k, x

∗k+1, . . . , x

∗n

)≥ ϕk

(x∗1, . . . , x

∗k−1, yk, x

∗k+1, . . . , x

∗n

)⇒ y∗k is optimal, y∗k ∈ Rk(x

∗), y∗ ∈ R(x∗) ⇒ (x∗, y∗) ∈ G. �

6.2 Counterexamples for Nikaido�Isoda theorem

Sk nonconvex: any discrete game without equilibrium

Sk not closed: S1 = S2 = [0, 1), ϕ1 = ϕ2 = x+ y

Sk not bounded: S1 = S2 = [0,∞), ϕ1 = ϕ2 = x+ y

ϕk not continuous: n = 2

59

Game theory

ϕ1 =

{x+ y if x < 1y if x = 1

, ϕ2 =

{x+ y if y < 1x if y = 1

⇒ no best responses exist

ϕk not concave: n = 2, S1 = S2 = [0, 1]

ϕ1 = (x− y)2 + 1, ϕ2 = −(x− y)2 + 1

0 112

y

R1(y) =

1 if y < 1

2

0 if y > 12

{0, 1} if y = 12

R2(x) = x same

⇒ no match ⇒ no EP exists.

O

Example 6.1 n = 2, S1 = S2 = [0, 10]

ϕ1 = ϕ2 = 2x+ 2y − (x+ y)2 strictly concave in x and in y

∂ϕ1

∂x= 2− 2(x+ y) = 0, 1− x− y = 0

x = 1− y

R1(y) =

{1− y if 0 ≤ y ≤ 10 if y > 1

Similarly,

R2(x) =

{1− x if 0 ≤ x ≤ 10 if x > 1

60

Game theory

In�nitely many equilibria⇒ Strict concavity of ϕk is not enough for uniqueness, but in optimization it is enough.

O

7 Applications

7.1 Matrix games

Players select strategies randomly according to chosen discrete distributions on their strat-egy sets.

Zero-sum 2-person game, payo� matrix A (m× n). De�ne

S1 =

{x1 | (x

(1)1 , . . . , x

(m)1 ) = x1, x

(i)1 ≥ 0,

∑i

x(i)1 = 1

}

S2 =

{x2 | (x

(1)2 , . . . , x

(n)2 ) = x2, x

(j)2 ≥ 0,

∑j

x(j)2 = 1

}x

(i)1 = P(strategy i is chosen by player 1)

x(j)2 = P(strategy j is chosen by player 2)

ϕ1(x1, x2) = E(payo�) =∑i

∑j

aijx(i)1 x

(j)2

= xT1A x2

ϕ2(x1, x2) = −ϕ1(x1, x2)

⇒ there is at least one EP.

Example 7.1 The two-person, zero-sum game with payo� table of player 1 has no equi-librium.The randomized extension is de�ned as

61

Game theory

1\2 1 21 2 12 0 2

or1\2 y 1− yx 2 1

1− x 0 2

S1 = {(x, 1− x) | 0 ≤ x ≤ 1} , S2 = {(y, 1− y) | 0 ≤ y ≤ 1} ,

and

ϕ1(x, y) = 2xy+1x(1−y)+0(1−x)y+2(1−x)(1−y) = 2xy+x−xy+2−2x−2y+2xy

= 3xy − x− 2y + 2 = x(3y − 1)− 2y + 2

So, best response of player 1:

R1(y) =

1 if 3y − 1 > 0 (or y > 1

3)

0 if 3y − 1 < 0 (or y < 13)

[0, 1] if 3y − 1 = 0 (or y = 13)

ϕ2 = −ϕ1, so ϕ2(x, y) = −3xy + x+ 2y − 2 = y(2− 3x) + x− 2,

R2(x) =

1 if 2− 3x > 0 (or x < 2

3)

0 if 2− 3x < 0 (or x > 23)

[0, 1] if 2− 3x = 0 (or x = 23)

Unique equilibrium is: x = 23, y = 1

3

7.2 Bimatrix games

Strategy sets as above, payo�s are expectations again. 2-person �nite game, payo� ma-trices: A,B. The random strategies S1 and S2 are as before, and

ϕ1(x1, x2) = xT1A x2, ϕ2(x1, x2) = xT1B x2

⇒ there is at least one EP.

Example 7.2 Consider the bimatrix game with matrices

A =

(1 22 0

)and B =

(2 14 5

),

62

Game theory

which has no equilibrium. However with mixed strategies

ϕ1(x, y) = 1xy + 2x(1− y) + 2(1− x)y + 0(1− x)(1− y) = xy + 2x− 2xy + 2y − 2xy

= −3xy + 2x+ 2y = x(2− 3y) + 2y

So,

R1(y) =

1 if y < 2

3

0 if y > 23

[0, 1] if y = 23

Similarly,

ϕ2(x, y) = 2xy+1x(1−y)+4(1−x)y+5(1−x)(1−y) = 2xy+x−xy+4y−4xy+5−5x−5y+5xy

= 2xy − y − 4x+ 5 = y(2x− 1)− 4x+ 5

So,

R2(x) =

1 if x > 1

2

0 if x < 12

[0, 1] if x = 12

unique equilibrium is: x = 12, y = 2

3

7.3 Mixed �nite games

n-players, �nite game, with original strategy sets

Sk = {1, 2, . . . ,mk} ,

Let payo� of player k be a(k)i1i2···in if simultaneous strategy vector is (i1, i2, · · · , in).

In mixed extension

Sk =

{xk | (x

(1)k , . . . , x

(mk)k ) = xk, x

(i)k ≥ 0,

∑i

x(i)k = 1

}expected payo�:

ϕk(x1, . . . , xn) =∑i1

∑i2

· · ·∑in

a(k)i1i2...in

x(i1)1 x

(i2)2 . . . x(in)

n

⇒ EP exists.

63

Game theory

7.4 Polyhedron games

n = 2Sk = {xk | Bkxk ≤ bk}

ϕ1(x1, x2) = xT1A1x2,

ϕ2(x1, x2) = −ϕ1(x1, x2)

⇒ EP exists.Select as a special case

Bk =

−I1T

−1T

, bk =

01−1

,

then

Bkxk ≤ bk ⇔

−I1T

−1T

xk ≤

01−1

⇔ −xk ≤ 0 ⇔ xk ≥ 0

1Txk ≤ 1 1Txk = 1−1Txk ≤ −1 ⇔ 1Txk ≥ 1

Strategy sets of matrix games.

7.5 Multiproduct oligopolies

n �rms, M products

xk = (x(1)k , . . . , x

(M)k ) = production vector of �rm k

p = p

(n∑l=1

xl

)= price vector

ϕk(x1, . . . , xn) = xTk p

(n∑l=1

xl

)︸ ︷︷ ︸

revenue

−Ck(xk)︸ ︷︷ ︸cost

Assumptions:

(i) Sk = nonempty, closed, convex, bounded in RM+

64

Game theory

(ii) p and Ck are continuous functions

(iii) xTk p (∑n

l=1 xl) is concave in xk

(iv) Ck is convex

⇒ EP exists.Question: When is (iii) satis�ed?

De�nition 7.1 Let f be a function, f : D 7→ RM , where D ⊆ RM such that for allx, y ∈ D,

(x− y)T (f(x)− f(x)) ≥ 0

Then f is called monotone.

Note. If M = 1, then this means that f is increasing.Before �nding conditions for monotonicity, a matrix-theoretical reminder.Let S be a real symmetric matrix (S = ST ).

De�nition 7.2 S is positive de�nite if either

(i) xTS x > 0 for all x 6= 0 (x is complex conjugate of x)

or

(ii) all eigenvalues of S are positive.

Theorem 7.1 (i) and (ii) and equivalent.

Proof. Fact, all eigenvalues of S have to be real:

xT/ S x = λx (multiplied by complex conjugate of xT )

xTS x = λxTx (10)

Notice, xTS x is real, since(xTS x

)=(xTS x

)T= xTS x = xTS x.

Similarly, xTx =∑xixi =

∑|xi|2 real and positive.

⇒ From (10) and (i),

λ =xTS x

xTx> 0

⇐ From diagonal form of S, with real invertible matrix U,

S = UT

λ1

. . .

λn

U,

so

xTS x = xTUT

λ1

. . .

λn

U x

65

Game theory

With z = U x,

xTS x = zT

λ1

. . .

λn

z =∑

λizizi =∑

λi|zi|2 > 0

if z 6= 0, which is the case if x 6= 0.

Similarly:

De�nition 7.3 S is negative de�nite if

(i) xTS x < 0 for all x 6= 0

or

(ii) all eigenvalues of S are negative.

De�nition 7.4 S is positive semide�nite if

(i) xTS x ≥ 0 for all x

or

(ii) all eigenvalues of S are nonnegative.

De�nition 7.5 S is negative semide�nite if

(i) xTS x ≤ 0 for all x

or

(ii) all eigenvalues of S are nonpositive.

De�nition 7.6 A symmetric matrix S is inde�nite, if

(i) xTS x is positive with some x, and also negative with another x;

or

(ii) among the eigenvalues of S there are positive and negative numbers.

Theorem 7.2 Assume D is convex in RM . Let J(x) denote the Jacobian of f , assume

J(x) is continuous on D. Function f is monotone ⇔ J(x) + J(x)T is positive semide�-nite.

Proof. ⇐Consider function

g(t) = f(y + t(x− y)

)with

g(0) = f(y), g(1) = f(x).

Theng′(t) = J

(y + t(x− y)

)(x− y),

66

Game theory

so

f(x)− f(y) = g(1)− g(0) =

∫ 1

0

g′(t)dt

=

∫ 1

0

J(y + t(x− y)

)(x− y)dt

Therefore

(x− y)T(f(x)− f(y)

)=

∫ 1

0

(x− y)TJ(y + t(x− y)

)(x− y)dt

=1

2

∫ 1

0

(x− y)T[J(y + t(x− y)

)+ JT

(y + t(x− y)

)](x− y)dt ≥ 0

since for any matrix A and vector u,

uTA u︸ ︷︷ ︸scalar

= (uTA u)T = uTATu.

⇒ If J(x) + J(x)T is not positive semide�nite, then there exist y0and u 6= 0 such that

uT(J(y

0) + J(y

0)T)u < 0,

so same holds in neighborhood of y0. Take u = 1

αu with large enough α such that

y0

+ tu

belongs to this neighborhood for 0 ≤ t ≤ 1. Then let x = y0

+ u , then from aboveequation with y = y0 we have

(x− y0)T(f(x)− f(y

0))

=1

2

∫ 1

0

uT[J(y

0+ tu

)+ JT

(y

0+ tu

)]udt

=1

2α2

∫ 1

0

uT[J(y

0+ tu

)+ JT

(y

0+ tu

)]udt < 0,

contradiction. �

De�nition 7.7 Let f : D 7→ Rn with D ⊆ Rn. This function is called strictly mono-

tone if for all x, y ∈ D and x 6= y,

(x− y)T(f(x)− f(y)

)> 0.

Theorem 7.3 Assume D ⊆ Rn is convex, the Jacobian J(x) of f is continuous, further-

more J(x) + J(x)T is negative de�nite for all x ∈ D. Then f is strictly monotone.

Proof. Same as �rst part of previous theorem. �Note. In one dimension J(x) = f ′(x), so we get back well-known monotonicity condition.

Lemma 7.1 Let f : D 7→ RM with D ⊆ RM+ being a convex set, −f is monotone and all

components of f are concave. Then g(x) = xTf(x) is concave on D.

67

Game theory

Proof. Let α, β ≥ 0, α + β = 1, and x, y ∈ D. Then

(x− y)T (f(x)− f(y)) ≤ 0 (/ · αβ)

αβxTf(x) + αβyTf(y) ≤ αβyTf(x) + αβxTf(y).

Since αβ = α(1− α) = α− α2, αβ = β(1− β) = β − β2, we get

(αx+ βy)T(αf(x) + βf(y)

)≥ αxTf(x) + βyTf(y),

so

g(αx+ βy) = (αx+ βy)T f(αx+ βy)︸ ︷︷ ︸≥αf(x)+βf(y)

≥ (αx+ βy)T(αf(x) + βf(y)

)≥ αxTf(x) + βyTf(y)

= αg(x) + βg(y) ⇒ g(x) is concave.

�Corollary. If −p is monotone and each component of p is concave, furthermore Ck isconvex, then

ϕk(x1, . . . , xn) = xTk p

(xk +

∑l 6=k

xl

)− Ck(xk)

is concave in xk ⇒ EP exists.

7.6 Single-product oligopolies

n �rms

x1, . . . , xn production levels, 0 ≤ xk ≤ Lk

Ck(xk) cost functions

p(s) price function with s = x1 + · · ·+ xn

Pro�t of �rm k:

ϕk = xkp

(n∑l=1

xl

)− Ck(xk)

Assumptions: Functions p and all Ck are twice continuously di�erentiable, and

(i) p′ < 0 (price decreases if supply becomes larger)

(ii) p′ + xkp′′ ≤ 0 (which holds if p is concave or slightly convex)

(iii) p′ − C ′′k < 0 (which holds if Ck is convex or slightly concave)

Best responses:xkp(xk + sk)− Ck(xk)→ max

where sk =∑

l 6=k xl = output of the rest of the industry

∂ϕk∂xk

= p(xk + sk) + xkp′(xk + sk)− C ′k(xk)

∂2ϕk∂x2

k

= 2p′ + xkp′′ − C ′′k < 0

68

Game theory

⇒ ϕk is strictly concave in xk, so

Rk(sk) =

0 if p(sk)− C ′k(0) ≤ 0Lk if p(Lk + sk) + Lkp

′(Lk + sk)− C ′k(Lk) ≥ 0x∗k otherwise,

where x∗k is the solution of the �rst order condition

g(xk) = p(xk + sk) + xkp′(xk + sk)− C ′k(xk) = 0

Note in case 3,

g(0) = p(sk)− C ′k(0) > 0

g(Lk) = p(Lk + sk) + Lkp′(Lk + sk)− C ′k(Lk) < 0

andg′(xk) = 2p′ + xkp

′′ − C ′′k < 0

⇒ unique solution.

In cases 1 and 2, R′k = 0. In case 3 by implicit di�erentiation: xk = Rk(sk),

p′(R′k + 1) + xkp′′ · (R′k + 1) +R′kp

′ − C ′′k ·R′k = 0

R′k(sk) = − p′ + xkp′′

2p′ + xkp′′ − C ′′k∈ (−1, 0] ,

which holds also in cases 1 and 2 except on the border lines between the cases.

Special case of duopoly:Assume n = 2, then equilibrium is the solution of equations

x = R1(y)

y = R2(x)

with Jacobian

J =

(0 R′1(y)

R′2(x) 0

),

and since |R′1(y)| ≤ q < 1, |R′2(x)| ≤ q < 1 with some q (because of continuous deriva-tives), best response mapping is contraction ⇒ unique EP.

General case of n:We can also prove existence and uniqueness of EP if n ≥ 2.We can rewrite best responses as functions of total output s as

Rk(s) =

0 if p(s)− C ′k(0) ≤ 0Lk if p(s) + Lkp

′(s)− C ′k(Lk) ≥ 0x∗k otherwise,

(11)

where x∗k is the solution of equation

g(xk) = p(s) + xkp′(s)− C ′k(xk) = 0

Note in case 3 of (11),

69

Game theory

g(0) = p(s)− C ′k(0) > 0

g(Lk) = p(s) + Lkp′(s)− C ′k(Lk) < 0

g′(xk) = p′(s)− C ′′k (xk) < 0

⇒ unique solution.

In cases 1 and 2 of (11), Rk = 0, in case 3 by implicit di�erentiation: xk = Rk(s),

p′ +R′kp′ + xkp

′′ − C ′′k ·R′k = 0

R′k(s) = −p

′ + xkp′′

p′ − C ′′k≤ 0

which also holds in cases 1 and 2 except on border lines ⇒ Rk decreases in s.

Equilibrium:Consider equation

h(s) =n∑k=1

Rk(s)− s = 0

h(s) strictly decreases in s, h(0) ≥ 0, h (∑n

k=1 Lk) ≤ 0⇒ unique equilibrium.

Note. We also found a simple method to compute EP.

8 Relation of EP problems and nonlinear programming

Consider optimization problem

maximize f(x) (x ∈ Rn)(P) s. to x ∈ X (X ⊆ Rn any set)

g(x) ≥ 0 (g(x) ∈ Rm)

De�nition 8.1 Lagrangean is

F (x, u) = f(x) + uTg(x) for all u ≥ 0.

De�ne the 2-person zero-sum game as

S1 = X,S2 = Rm+ , ϕ1 = F, ϕ2 = −F.

Theorem 8.1 If (x∗, u∗) is an EP, then x∗ is optimal solution for the original problem.

Proof.

f(x∗) + u∗Tg(x∗) ≥ f(x) + u∗Tg(x) (∀x) (12)

−[f(x∗) + u∗Tg(x∗)

]≥ −

[f(x∗) + uTg(x∗)

](∀u) (13)

(13) ⇒ with u = 0, u∗Tg(x∗) ≤ 0. (14)

Next we show that g(x∗) ≥ 0. Assume gi(x∗) < 0, and select su�ciently large ui to

violate (13) ⇒ x∗ is feasible solution of (P).Since u∗ ≥ 0 and g(x∗) ≥ 0, u∗Tg(x∗) ≥ 0.

70

Game theory

Together with (14) we conclude that

u∗Tg(x∗) = 0.

Finally we show that x∗ is optimal for (P):

(12) ⇒ f(x∗) = f(x∗) + u∗Tg(x∗)

≥ f(x) + u∗T︸︷︷︸≥0

g(x)︸︷︷︸≥ 0

for allfeasiblex

≥ f(x).

In summary: If we have a good general method for solving 2-person zero-sum games,then we can solve any nonlinear programming problem.

9 How to compute EP?

9.1 Lagrange method

maximize f(x) (15)

s. to g(x) = 0

Theorem 9.1 Under certain regularity conditions problem (15)

⇔ maximize f(x) + uTg(x)︸ ︷︷ ︸L(x,u) Lagrangean

= f(x) +∑

uigi(x)

At optimum

∂L∂xi

= 0 and ∂L∂uj

= 0

Of(x)︸ ︷︷ ︸gradient

+uT Og(x)︸ ︷︷ ︸Jacobian

= 0T

g(x) = 0

system of nonlinear equations

Note that the gradient is a row vector here.

Example 9.1

maximize 2x1 − (x1 − x2)2

s. to x1 + x2 = 1

Method 1.: x2 = 1− x1, so objective is

2x1 − (x1 − x2)2 = 2x1 − (x1 − 1 + x1)2 = 2x1 − (2x1 − 1)2

= −4x21 + 6x1 − 1

71

Game theory

−8x1 + 6 = 0

x1 =6

8=

3

4

x2 = 1− x1 =1

4

Method 2.: 2x1 − (x1 − x2)2 + u (x1 + x2 − 1) −→ max

∂

∂x1

: 2− 2(x1 − x2) + u = 0 (16)

∂

∂x2

: −2(x1 − x2)(−1) + u = 0 (17)

∂

∂u: x1 + x2 − 1 = 0 (18)

by adding (16) and (17) ⇒ 2 + 2u = 0 ⇒ u = −1 (19)

(17) ⇒ 2(x1 − x2)− 1 = 0 (20)

(18) ⇒ 2x1 + 2x2 − 2 = 0 (21)

by substracting ⇒ −4x2 + 1 = 0 ⇒ x2 =1

4(22)

(18) ⇒ x1 = 1− x2 =3

4(23)

O

9.2 Kuhn�Tucker (K-T) conditions

maximize f(x) (x ∈ Rn) (24)

s. to g(x) ≥ 0 (g(x) ∈ Rm)

Theorem 9.2 Under certain regularity conditions and di�erentiability of f and g let x∗

be an optimal solution. Then there exist u∗1, . . . , u∗m such that with vector u∗ = (u∗i )

u∗ ≥ 0

g(x∗) ≥ 0

Of(x∗) + u∗TOg(x∗) = 0T

u∗Tg(x∗) = 0

Kuhn-Tucker necessary conditions

Here Of is gradient of f as row vector and Og is the Jacobian matrix of g.If f and all components of g are concave, then K-T-conditions are also su�cient.Note: The last condition can be rewritten as u∗i gi(x

∗) = 0 for i = 1, 2, . . . ,m.

Example 9.2

maximize f = ln(x1 + x2)

s. to x1 + 2x2 ≤ 5x1, x2 ≥ 0

} g1 = −x1 − 2x2 + 5g2 = x1

g3 = x2

72

Game theory

n = 2,m = 3, we need u1, u2, u3 such that

u1, u2, u3 ≥ 0

−x1 − 2x2 + 5 ≥ 0

x1 ≥ 0

x2 ≥ 0

(1

x1 + x2

,1

x1 + x2

)+ (u1, u2, u3)

−1 −21 00 1

= (0, 0)

(u1, u2, u3) ·

−x1 − 2x2 + 5x1

x2

= 0

By components

+︷ ︸︸ ︷1

x1 + x2

−u1 +

≥0︷︸︸︷u2 = 0 ⇒ u1 6= 0 (25)

1

x1 + x2

− 2u1 + u3 = 0 (26)

u1(−x1 − 2x2 + 5) = 0 (27)

u2x1 = 0 (28)

u3x2 = 0 (29)

(25), (26) ⇒ u3 > u2 ⇒ u3 > 0 (30)

(29) ⇒ x2 = 0 (31)

u1 6= 0, (27) ⇒ −x1 − 2x2 + 5 = 0 (32)

hence ⇒ x1 = 5 (33)

O

9.3 Relation of Kuhn-Tucker conditions and Lagrange method

(i) Lagrange method is special case of Kuhn-Tucker conditions.

Problem:

maximize f(x)

s. to g(x) = 0

Rewrite:maximize f(x)

s. to g(x) ≥ 0−g(x) ≥ 0

}K-T conditions:

u, v ≥ 0

73

Game theory

g(x) ≥ 0−g(x) ≥ 0

}g(x) = 0

Of(x) + uTOg(x)− vTOg(x) = 0T

uTg(x)− vTg(x) = 0 (trivially satis�ed)

Introduce u− v = λ, without sign constraint:

g(x) = 0

Of(x) + λTOg(x) = 0T

}Lagrange method

(ii) We can also derive Kuhn-Tucker conditions from Lagrange method.

Problem:

maximize f(x)

s. to g(x) ≥ 0

Rewrite:maximize f(x)

s. to g(x)−

v2

1

v22...v2m

=0 (g(x) ∈ Rm)

Lagrangean (L):

maximize f(x) + uT

g(x)−

v2

1

v22...v2m

Di�erentiating:with respect to x:

Of(x)︸ ︷︷ ︸gradient as row

+uT Og(x)︸ ︷︷ ︸Jacobian matrix

= 0T

with respect to u:

g(x)−

v2

1

v22...v2m

= 0⇔ g(x) ≥ 0 and gi(x) = v2i (i = 1, 2, · · ·,m)

with respect to v:−2uivi = 0

Since gi(x) = 0 ⇔ vi = 0, the last condition can be rewritten as

uigi(x) = 0

The only thing left is to show that ui ≥ 0 for all i. We maximize Lagrangean, so itsHessian H matrix is negative semide�nite ⇒ eTi Hei = hii ≤ 0 (ei is the ith basis vector).Note

∂L

∂vi∂vi=∂(−2uivi)

∂vi= −2ui ≤ 0 ⇒ ui ≥ 0

74

Game theory

9.4 Methodology to compute equilibria

Assume thatSk =

{xk | gk(xk) ≥ 0

}If (x∗1, . . . , x

∗n) is an EP, then x∗k is optimal solution for

maximize ϕk(x∗1, . . . , x

∗k−1, xk, x

∗k+1, . . . , x

∗n)

s. to gk(xk) ≥ 0

where xk is the decision variable.Assume that regularity conditions for K-T�relations are satis�ed:

uk ≥ 0

gk(xk) ≥ 0

Ψk(x, uk) = Okϕk(x) + uTkOgk(xk) = 0T

uTk gk(xk) = 0

for all k = 1, . . . , n (34)

where Okϕk(x) is gradient of ϕk with respect to components of xkConsider problem:

minimizen∑k=1

uTk gk(xk)

s. to uk ≥ 0

gk(xk) ≥ 0 k = 1, 2, . . . , n

Ψk(x, uk) = 0T

(35)

Theorem 9.3 If (x∗1, . . . , x∗n) is an EP, then there exist u∗1, . . . , u

∗n such that

(x∗1, . . . , x∗n, u

∗1, . . . , u

∗n) is optimal solution for (35).

Proof. If (x∗1, . . . , x∗n) is EP, then objective function is zero, and at any feasible solution

it is ≥ 0. �

Theorem 9.4 Assume ϕk and all components of gkare concave. Then x∗ is equilibrium

⇔ it is optimal for (35) with some u∗1, . . . , u∗n.

Note. Optimal objective value is zero.

Method 1. Write up conditions (34) for all players simultaneously and solve the resultedsystem (�nd feasible solution).

Example 9.3 n = 2 oligopoly

0 ≤ x, y ≤ 10

C1(x) = x, C2(y) = y

P (x, y) = 20− (x+ y)

ϕ1(x, y) = x(20− x− y)− x = 19x− x2 − xyϕ2(x, y) = y(20− x− y)− y = 19y − y2 − xy

O

75

Game theory

Kuhn�Tucker conditions:

u1 x ≥ 0,v1 10− x ≥ 0u2 y ≥ 0v2 10− y ≥ 0

0 ≤ x, y ≤ 10

Player 1:

uk ≥ 0 u1, v1 ≥ 0gk

(xk) ≥ 0 x ≥ 0, 10− x ≥ 0

Okϕk(x1, . . . , xn) + uTkOkgk(xk) = 0T −2x+ 19− y + (u1, v1) ·(

1−1

)= 0

uTk gk(xk) = 0 u1 · x+ v1 · (10− x) = 0

⇒ u1 · x = 0, v1 · (10− x) = 0

Player 2:

uk ≥ 0 u2, v2 ≥ 0gk

(xk) ≥ 0 y ≥ 0, 10− y ≥ 0

Okϕk(x1, . . . , xn) + uTkOkgk(xk) = 0T −2y + 19− x+ (u2, v2) ·(

1−1

)= 0

uTk gk(xk) = 0 u2 · y + v2 · (10− y) = 0

⇒ u2 · y = 0, v2 · (10− y) = 0

14 constraints, 6 variables (x, y, u1, v1, u2, v2). We have to consider three cases:

(i) x = 0, then v1 = 0 ⇒ 19− y + u1 = 0 ⇒ y = 19 + u1 > 10, infeasible

(ii) x = 10, then u1 = 0 ⇒ −20 + 19− y − v1 = 0 ⇒ y = −1− v1 < 0, infeasible

(iii) 0 < x < 10, then u1 = v1 = 0, −2x+ 19− y = 0.

Because of symmetry between x and y, we also have

−2y + 19− x = 0 ⇒ x = 19− 2y

−38 + 4y + 19− y = 0 ⇒ 3y = 19 ⇒ y =19

3

⇒ x =19

3

Method 2. Solve problem (35)

Example 9.4 In the case of previous example, the Kuhn�Tucker conditions are equivalentto the optimization problem:

minimize {u1 · x+ v1(10− x) + u2 · y + v2 · (10− y)}subject to u1, v1, u2, v2 ≥ 0

0 ≤ x, y ≤ 1019− 2x− y + u1 − v1 = 019− 2y − x+ u2 − v2 = 0

If optimal objective value is zero, then we got solution of Kuhn�Tucker conditions (sowe got the equilibrium). Otherwise no solution exists, therefore there is no equilibrium. O

76

Game theory

Example 9.5 n = 2, x, y ≥ 0 ⇔ S1 = S2 = [0,∞),

ϕ1(x, y) = x+ y − (x+ y)2 = x+ y − x2 − 2xy − y2

ϕ2(x, y) = x+ y − 2(x+ y)2 = x+ y − 2x2 − 4xy − 2y2

K-T conditions:

u, x ≥ 0 v, y ≥ 01− 2x− 2y + u · 1 = 0 1− 4x− 4y + v · 1 = 0

u · x = 0 v · y = 0

In solving these conditions we have several cases:

(i) x = 0, y = 0 ⇒ 1 + u = 0 infeasible

(ii) x = 0, y > 0 ⇒ v = 0, 1− 4y = 0, y = 14, 1− 1

2+ u = 0 infeasible

(iii) x > 0, y = 0 ⇒ u = 0, 1− 2x = 0, x = 12, 1− 2 + v = 0, v = 1 ⇒ x = 1

2, y = 0

is solution

(iv) x > 0, y > 0 ⇒ u = v = 0, 1− 2x− 2y = 0, 1− 4x− 4y = 0, contradiction.

Optimization problem:

minimize {u · x+ v · y}subject to u, v ≥ 0

x, y ≥ 01− 2x− 2y + u = 01− 4x− 4y + v = 0

From constraints,

u = 2x+ 2y − 1 ≥ 0 ⇒ x+ y ≥ 1

2

v = 4x+ 4y − 1 ≥ 0 ⇒ x+ y ≥ 1

4Objective function:

(2x+ 2y − 1)x+ (4x+ 4y − 1)y = 2x2 + 2xy − x+ 4xy + 4y2 − y

= 2x2 + 6xy + 4y2 − x− y → minimize (optimal objective value is zero!)Trying vertices:

(i) (0, 12), objective = 1− 1

2= 1

2> 0

(ii) (12, 0), objective = 1

2− 1

2= 0, this is a solution.

77

Game theory

or,2x2 + 6xy + 4y2 − x− y = 2(x2 + 2xy + y2) + 2xy + 2y2 − (x+ y)

= (x+ y)(2x+ 2y + 2y − 1)

By requiring zero objective function value:

0 = 2x2 + 6xy + 4y2 − x− y = (x+ y)(2x+ 4y − 1)

so,2x+ 4y − 1 = 0

y = −1

2x+

1

4⇒ x =

1

2, y = 0 is only solution.

O

10 Special games

10.1 Bimatrix games

ϕ1 = xT1A x2, ϕ2 = xT1B x2

S1 =

{x1

∣∣∣ x(i)1 ≥ 0,

∑i

x(i)1 = 1

}

S2 =

{x2

∣∣∣ x(j)2 ≥ 0,

∑j

x(j)2 = 1

}⇒

g1(x1) =

x

(1)1...

x(m)1

x(1)1 + · · ·+ x

(m)1 − 1

−x(1)1 − · · · − x

(m)1 + 1

,Og1

=

I1T

−1T

g2(x2) =

x

(1)2...

x(n)2

x(1)2 + · · ·+ x

(n)2 − 1

−x(1)2 − · · · − x

(n)2 + 1

,Og2

=

I1T

−1T

78

Game theory

Let's formulate problem (35):Objective:

m∑i=1

u(i)1 x

(i)1 + u

(m+1)1

(∑i

x(i)1 − 1

)+ u

(m+2)1

(−∑i

x(i)1 + 1

)

+n∑j=1

u(j)2 x

(j)2 + u

(n+1)2

(∑j

x(j)2 − 1

)+ u

(n+2)2

(−∑j

x(j)2 + 1

)

Introduce u(m+2)1 − u(m+1)

1 = α, u(n+2)2 − u(n+1)

2 = β (without sign constraints), thenobjective becomes

uT1 x1 + uT2 x2 − α(1Tx1 − 1

)− β

(1Tx2 − 1

)Constraints:

u1 ≥ 0, u2 ≥ 0

x1 ≥ 0, x2 ≥ 0

1Tx1 = 1, 1Tx2 = 1

xT2AT + uT1 +

u(m+1)1 − u(m+2)

1︸ ︷︷ ︸−α

1T = 0T

xT1B + uT2 +

u(n+1)2 − u(n+2)

2︸ ︷︷ ︸−β

1T = 0T

From last two equations:

uT1 = α1T − xT2AT , uT2 = β1T − xT1B,

so objective:

(− xT2AT + α 1T

)x1︸ ︷︷ ︸

1

+(− xT1B + β 1T

)x2︸ ︷︷ ︸

1

−α

1Tx1 − 1︸ ︷︷ ︸0

− β1Tx2 − 1︸ ︷︷ ︸

0

= −xT1A x2 − xT1B x2 + α + β

⇒ Since u1, u2 must be nonnegative

maximize xT1 (A+B)x2 − α− βs. to x1 ≥ 0, x2 ≥ 0

1Tx1 = 1, 1Tx2 = 1,

A x2 ≤ α1

BTx1 ≤ β1

quadratic programming problem (36)

Theorem 10.1 EP ⇔ optimal solution

Example 10.1

A =

(2 −1−1 1

), B =

(1 −1−1 2

), A+B =

(3 −2−2 3

)79

Game theory

somaximize 3x

(1)1 x

(1)2 − 2x

(1)1 x

(2)2 − 2x

(2)1 x

(1)2 + 3x

(2)1 x

(2)2 − α− β

s. to x(1)1 , x

(2)1 , x

(1)2 , x

(2)2 ≥ 0

x(1)1 + x

(2)1 = 1, x

(1)2 + x

(2)2 = 1

2x(1)2 − x

(2)2 ≤ α

−x(1)2 + x

(2)2 ≤ α

x(1)1 − x

(2)1 ≤ β

−x(1)1 + 2x

(2)1 ≤ β

Optimal solutions:x1 (1, 0) (0, 1)

(35, 2

5

)x2 (1, 0) (0, 1)

(25, 3

5

)α 2 1 1

5

β 1 2 15

By best responses:

ϕ1 = 2xy − 1(1− x)y − 1x(1− y) + 1(1− x)(1− y)

= 2xy − y + xy − x+ xy + 1− x− y + xy

= 5xy − 2x− 2y + 1 = x(5y − 2) + 1− 2y

R1(y) =

1 if y > 2

5

0 if y < 25

[0, 1] if y = 25

ϕ2 = 1xy − 1(1− x)y − 1x(1− y) + 2(1− x)(1− y)

= xy − y + xy − x+ xy + 2− 2x− 2y + 2xy

= 5xy − 3y − 3x+ 2 = y(5x− 3)− 3x+ 2

R2(x) =

1 if x > 3

5

0 if x < 35

[0, 1] if x = 35

Three equilibria (0,0), (1,1) and (35, 2

5) or probability vectors (x1 = (0, 1), x2 = (0, 1));

(x1 = (1, 0), x2 = (1, 0)) and (x1 = (35, 2

5), x2 = (2

5, 3

5))

O

80

Game theory

10.2 Matrix games

A+B = 0

maximize −α− βx1 ≥ 0, x2 ≥ 01Tx1 = 1 1Tx2 = 1A x2 ≤ α1, BTx1 ≤ β1 ⇔ ATx1 ≥ −β1

Separate (x1, α) and (x2, β):

minimize α minimize β

s. to A x2 ≤ α1 s. to ATx1 ≥ −β1

1Tx2 = 1 1Tx1 = 1

x2 ≥ 0 x1 ≥ 0

Both are linear programming problems.

Theorem 10.2 (x∗1, x∗2) is EP ⇔ they are optimal solutions with some α and β.

Some remarks:

(i) Since α+β = 0 at optimum, a strategy pair x1, x2 is equilibrium⇔ they are feasiblesolutions of system

A x2 ≤ α1 ATx1 ≥ α1

1Tx2 = 1 1Tx1 = 1

x2 ≥ 0 x1 ≥ 0

(ii) if (x∗1, x∗2) and (x∗∗1 , x

∗∗2 ) are both equilibria, then (x∗1, x

∗∗2 ) and (x∗∗1 , x

∗2) are also

equilibria.

(iii) The optimal value of α is the payo� of player 1 at any equilibrium:

A x2 ≤ α1⇒ xT1A x2 ≤ αxT1 1 = α

AT x1 ≥ α1⇒ (ATx1)Tx2 ≥ (α1)Tx2,

that is,xT1A x2 ≥ α1Tx2 = α⇒ xT1A x2 = α.

This value is called the value of the matrix game.

Example 10.2

A =

2 1 02 0 3−1 3 3

81

Game theory

minimize α minimize β

s. to 2x(1)2 + x

(2)2 − α ≤ 0 s. to 2x

(1)1 + 2x

(2)1 − x

(3)1 + β ≥ 0

2x(1)2 + 3x

(3)2 − α ≤ 0 x

(1)1 + 3x

(3)1 + β ≥ 0

−x(1)2 + 3x

(2)2 + 3x

(3)2 − α ≤ 0 3x

(2)1 + 3x

(3)1 + β ≥ 0

x(1)2 + x

(2)2 + x

(3)2 = 1 x

(1)1 + x

(2)1 + x

(3)1 = 1

x(1)2 , x

(2)2 , x

(3)2 ≥ 0 x

(1)1 , x

(2)1 , x

(3)1 ≥ 0

Optimal solutions:

x1 =

(4

7,

4

21,

5

21

)T, β = −9

7

x2 =

(3

7,3

7,1

7

)T, α =

9

7

O

Example 10.3

A =

(2 10 2

)The corresponding discrete game has no EP. The associated matrix game has, and the pairof linear programming problems can be written as follows. By selecting x1 = (x, 1 − x)T

and x2 = (y, 1− y)T :

minimize α minimize β

s. to 2y + (1− y) ≤ α s. to 2x ≥ −β2(1− y) ≤ α x+ 2(1− x) ≥ −β0 ≤ y ≤ 1 0 ≤ x ≤ 1

The constraints are simpli�ed as:

−y + α ≥ 1 and 2x+ β ≥ 0

2y + α ≥ 2 −x+ β ≥ −2

0 ≤ y ≤ 1 0 ≤ x ≤ 1

82

Game theory

Solution is x = 23, y = 1

3, so x1 = (2

3, 1

3)T and x2 = (1

3, 2

3)T .

O

10.3 Oligopoly game (single-product model)

Sk = {xk | xk ≥ 0, Lk − xk ≥ 0}

where Lk is capacity limit:

gk(xk) =

(xk

Lk − xk

)

ϕk(x1, . . . , xn) = xkp(x1 + · · ·+ xn)− Ck(xk)

The K-T condition-based optimum problem:

minimizen∑k=1

(u

(1)k xk + u

(2)k (Lk − xk)

)(let αk = u

(1)k − u

(2)k︸ ︷︷ ︸

no sign constraint

, βk = u(2)k︸︷︷︸≥0

)

=n∑k=1

(αkxk + βkLk)

Constraints:u

(1)k , u

(2)k ≥ 0 ⇔ βk ≥ 0, αk + βk ≥ 0

0 ≤ xk ≤ Lk

p(∑

xl

)+ xkp

′(∑

xl

)− C ′k(xk) +

αk︷ ︸︸ ︷(u

(1)k , u

(2)k

)( 1−1

)= 0

⇒ αk = −[p(∑

xl

)+ xkp

′(∑

xl

)− C ′k(xk)

],

so problem is rewritten as follows.

minimize∑n

k=1 (−xk [p (∑xl) + xkp

′ (∑xl)− C ′k(xk)] + βkLk)

s. to βk ≥ 0, βk − [p (∑xl) + xkp

′(∑xl)− C ′k(xk)] ≥ 0

0 ≤ xk ≤ Lk

Example 10.4 n = 3, Ck(xk) = k · x3k + xk

Lk = 1

p(s) = 2− 2s− s2

83

Game theory

maximize3∑

k=1

[xk (2− 2s− s2 − 2xk − 2xks− 3kx2k − 1)]− β1 − β2 − β3

s. to 0 ≤ xk ≤ 1, β1, β2, β3 ≥ 0

βk − (2− 2s− s2 − 2xk − 2xks− 3kx2k − 1) ≥ 0 (k = 1, 2, 3)

x1 + x2 + x3 = s

Optimal solution:x∗1 = 0.1077x∗2 = 0.0986x∗3 = 0.0919

O

10.4 Special matrix games

(i)

A =

α . . . αα . . . α...

...α . . . α

ϕ1 =

∑i

∑j

αx(i)1 x

(j)2 = α

∑i

x(i)1︸ ︷︷ ︸

1

∑j

x(j)2︸ ︷︷ ︸

1

= α

⇒ arbitrary strategy pair is equilibrium.

(ii)

A =

a1 0 . . . 00 a2 . . . 0...

. . ....

0 0 . . . an

(diagonal game)

84

Game theory

Constraints:

akx(k)2 ≤ α

akx(k)1 ≥ α

x(k)1 , x

(k)2 ≥ 0∑

k

x(k)1 =

∑k

x(k)2 = 1

since we selected β = −α, all feasible solutions are EP

Case 1. ak > 0 for all k. Since x(k)2 > 0 with some k, α > 0. Therefore

1 =∑k

x(k)2 ≤

∑k

α

ak≤∑k

x(k)1 = 1,

so equality everywhere ⇒ x(k)1 = x

(k)2 = α

ak, but

∑x

(k)1 = 1 so α = 1∑

k1ak

.

Case 2. ak < 0 for all k, then

(−ak)x(k)2 ≥ −α

(−ak)x(k)1 ≤ −α

1 =∑k

x(k)2 ≥

∑k

−α−ak

≥∑k

x(k)1 = 1, so x

(k)1 = x

(k)2 =

α

ak,

and since∑

k x(k)1 = 1, we have α = 1∑

k1ak

.

Case 3. ai ≥ 0 and aj ≤ 0 with some i and j, then

α ≥ aix(i)2 ≥ 0 ≥ ajx

(j)1 ≥ α ⇒ α = 0,

akx(k)2 ≤ 0 ≤ akx

(k)1 (all k)

Set

uk

{= 0 if ak > 0≥ 0 if ak ≤ 0,

∑k uk = u

vk

{= 0 if ak < 0≥ 0 if ak ≥ 0,

∑k vk = v

⇒

x1 =1

v(v1, . . . , vn) ,

x2 =1

u(u1, . . . , un)

(iii) AT = −A symmetric matrix game

minimize α minimize βs. to A x2 ≤ α1 s. to Ax1 ≤ β1

1Tx2 = 1 1Tx1 = 1x2 ≥ 0 x1 ≥ 0

identical problems

x1, x2 ∈ X∗ = set of optimal solutions

Since α + β = 0 at optimum and α = β, we have α = β = 0, so the value of gameis zero and

X∗ ={x | x ≥ 0, 1Tx = 1, A x ≤ 0

}85

Game theory

10.5 Applications

(i) Consider primal�dual LP's

x ≥ 0A x ≤ b

max cTx(primal) &

y ≥ 0

ATy ≥ c

min bTy

(dual)

Lemma 10.1 Let x and y be feasible solutions of the primal and dual problems,respectively. Then

cTx ≤ bTy.

Proof.cTx ≤

(ATy

)Tx =

(yTA

)x = yT (A x) ≤ yT b = bTy.

�

Corollary. (Duality theorem) If x and y are feasible solutions of the primal and

dual problems, respectively, such that cTx = bTy, then x is an optimal solution ofthe primal problem and y is an optimal solution of the dual problem.

Construct next matrix

P =

0 A −b−AT 0 cbT −cT 0

which generates a symmetric matrix game.

Theorem 10.3 Let z = (u, v, λ) be an equilibrium of the symmetric matrix gamesuch that λ > 0. Then

x =1

λv and y =

1

λu

are optimal solutions of the primal and dual problems.

Proof. If z is equilibrium, then P z ≤ 0 ⇔

A v − bλ ≤ 0 (37)

−ATu+ cλ ≤ 0 (38)

bTu− cTv ≤ 0 (39)

Since λ > 0 and z ≥ 0,

x =1

λv ≥ 0 and y =

1

λu ≥ 0,

furthermore

(37) ⇒ A x ≤ b (40)

(38) ⇒ ATy ≥ c (41)

(39) ⇒ bTy ≤ cTx (42)

(40), (41) ⇒ x, y are feasible for primal and dual.

However from above lemma,bTy ≥ cTx

we concludebTy = cTx.

The duality theorem implies optimality. �

86

Game theory

Example 10.5 Consider LP:

maximize x1 + 2x2

s. to x1 ≥ 0, x2 no sign constraint−x1 + x2 ≥ 15x1 + 7x2 ≤ 25

Step 1: Rewrite to primal form:

x2 = x+2 − x−2 ,

x+2 =

{x2 if x2 ≥ 00 otherwise

x−2 =

{0 if x2 ≥ 0−x2 otherwise

Objective: x1 + 2x2 = x1 + 2x+2 − 2x−2

⇒ cT = (1, 2,−2)

Constraints:

− x1 + x+2 − x−2 ≥ 1 / · (−1)

x1 − x+2 + x−2 ≤ −1 (43)

5x1 + 7x+2 − 7x−2 ≤ 25 (44)

⇒ A =

(1 −1 15 7 −7

), b =

(−125

)

⇒ P =

0 0 1 −1 1 10 0 5 7 −7 −25−1 −5 0 0 0 11 −7 0 0 0 2−1 7 0 0 0 −2−1 25 −1 −2 2 0

O

(ii) Consider a matrix game A > 0 (adding the same constant to all elements does notchange equilibrium), and a symmetric matrix game

P =

0 A −1−AT 0 11T −1T 0

Theorem 10.4 Matrix games A and P are equivalent:

(a) If z = (u, v, λ) is EP of P , then with a = 1−λ2,

x =1

au and y =

1

av

give equilibrium of A, and λais the value of the game;

87

Game theory

(b) If (x, y) is equilibrium of A and v is the value of the matrix game, then

z =1

2 + v(x, y, v)

is equilibrium for P .

Proof.

(a) Let z = (u, v, λ) be an equilibrium of P , then

P z ≤ 0 0 A −1−AT 0 11T −1T 0

uvλ

≤

000

That is,

A v − 1λ ≤ 0

−ATu+ 1λ ≤ 0

1Tu− 1Tv ≤ 0

First we show that 0 < λ < 1.

If λ = 1, then (since z is probability vector), u = 0 and v = 0⇒ contradicts 2nd inequality.

If λ = 0, then 1Tu+ 1Tv = 1, then by 3rd inequality v must have positivecomponent, so contradiction to the �rst inequality.

Next we show that 1Tu = 1Tv. From 1st and 2nd relations

uTA v − λuT1 ≤ 0

−vTATu+ λvT1 ≤ 0

}(+) adding them

⇒ λ(vT1− uT1

)≤ 0 /÷ λ ⇒ vT1− uT1 ≤ 0

Compare it to 3rd inequality to see vT1 = uT1.

Select a = 1−λ2, then vT1 = uT1 = a, so

x =1

au and y =

1

av

are both probability vectors, furthermore

ATx =1

aATu ≥ λ

a1 (2nd inequality)

A y =1

aA v ≤ λ

a1 (1st inequality)

So choose α = λaand β = −λ

a, x1 = x and x2 = y, they satisfy the constraints

of the pair of LP's for matrix games with α+ β = 0. So they are optimal ⇒equilibrium of A.

(b) Similar, omitted

⇒ Su�cient to solve symmetric matrix games.

88

Game theory

10.6 Method of �ctitious play

Consider a matrix game A. The idea is an iteration process in which at each time periodeach player selects best choice against the last selected strategy of the other player. Theproblem is that best choices are usually basis vectors depending on which component ofA y or xTA is optimal. A sequence of basis vectors cannot converge to mixed strategies.So mixed strategy equilibrium cannot be obtained. In order to introduce mixed strategiesfor the players we consider averages of their choices.

Initial step, k = 1. Let x1 be initial strategy of player 1. If the basis vectors aredenoted by e1, e2, . . . , then de�ne

xT1A ej1 = minj{xT1A ej}

j1 is index of smallest element of row vector xT1A and choose y1

= ej1 .

In steps k ≥ 2, let yk−1

= 1k−1

∑k−1t=1 yt be the average of all previous choices of player

2 and select xk = eik such that

eTikA yk−1

= maxi

{eTi A y

k−1

}ik is index of largest element of column vectorA y

k−1

Let xk = 1k

∑kt=1 xt be the average of all previous choices of player 1, and select yk = ejk

such thatxTkA ejk = min

j

{xTkA ej

}jk is index of smallest element of row vector xTkAThen go back with next value of k.

Theorem 10.5 Any limit point of the sequences {xk} and {yk} gives equilibrium strate-gies.

Proof. Complicated, not presented. �Note. We got an iterative method for solving matrix games and so for solving LPs.

10.7 von Neumann's method

Consider a symmetric matrix game P .De�ne:

ui : Rn 7→ R, ui(y) = eTi P y (i = 1, 2, . . . , n) (ith element of P y)ϕ : R 7→ R, ϕ(u) = max {0, u} ≥ 0Φ : Rn 7→ R, Φ(y) =

∑ni=1 ϕ

(ui(y))≥ 0

Ψ : Rn 7→ R, Ψ(y) =∑n

i=1 ϕ2(ui(y))≥ 0

Lemma 10.2 √Ψ(y)≤ Φ

(y)≤√n ·Ψ

(y)

89

Game theory

Proof.

Ψ(y) =n∑i=1

ϕ2(ui(y)) ≤

[n∑i=1

ϕ(ui(y))]2

︸ ︷︷ ︸cross products are nonnegative

= Φ2(y)

Φ(y) =n∑i=1

1 · ϕ(ui(y))≤

(n∑i=1

12

) 12(

n∑i=1

ϕ2(ui(y))

) 12

︸ ︷︷ ︸Cauchy�Schwarz

=√n ·Ψ

(y).

�Assume that at a strategy y, ϕ(uj(y)) > 0. Then eTj P y > 0, however eTj P ej = 0, so

player 2 needs to increase yj to 1 (since at y its payo� is negative, and at ej it is zero), socomponent j has to increase. This is represented by the �rst term of the following ODEsystem. The second term is used to guarantee that solution is always probability vector.

Consider the system of ODE's:

yj(t) = ϕ(uj(y(t))

)− Φ

(y(t)

)yj(t)

yj(0) = y0j

}1 ≤ j ≤ n

where y0 is a probability vector.

Theorem 10.6 Let tk (k = 1, 2, . . . ) be a positive, strictly increasing sequence that con-verges to ∞. Then any limit point of sequence

{y(tk)

}is equilibrium strategy, and there

exists a constant c > 0 such that

eTi P y(tk) ≤√n

c+ tk.

Proof. Several steps.

(i) We show that y(t), t ≥ 0 is always a probability vector.

Assume �rst that yj(t1) < 0 with some j and t1 > 0.

Lett0 = sup {t | 0 < t < t1, yj(t) ≥ 0} .

By continuity, yj(t0) = 0 and for all t0 < τ ≤ t1, yj(τ) < 0. By de�nition

ϕ(uj) ≥ 0, Φ(y(t)

)≥ 0 so

yj(τ) = ϕ(uj(y(τ))

)︸ ︷︷ ︸≥0

−Φ(y(τ)

)︸ ︷︷ ︸≥0

yj(τ)︸ ︷︷ ︸<0

≥ 0.

By Lagrange's mean-value theorem

yj(t1) = yj(t0)︸ ︷︷ ︸0

+ yj(τ)︸ ︷︷ ︸≥0

(t1 − t0)︸ ︷︷ ︸≥0

≥ 0

contradiction to yj(t1) < 0 which was the assumption.

Next we show that for all t ≥ 0,∑n

j=1 yj(t) = 1, so y(t) is a probability vector.

90

Game theory

Notice that(1−

n∑j=1

yj(t)

)′= −

n∑j=1

yj(t) = −︸ ︷︷ ︸ODE's

n∑j=1

ϕ(uj(y(t))

)︸ ︷︷ ︸

Φ(y(t))

+Φ(y(t)

) n∑j=1

yj(t)

= −Φ(y(t)

) [1−

n∑j=1

yj(t)

].

So function f(t) = 1−∑n

j=1 yj(t) satis�es the initial�value problem

f(t) = −Φ(y(t)

)f(t), f(0) = 0,

and the unique solution is f(t) ≡ 0.

(ii) Assume ϕ(ui(y(t))

)> 0 for some t ≥ 0, then

d

dtϕ(ui(y(t))

)=

d

dtui(y(t)

)= eTi P y(t) =

n∑j=1

pij yj(t)

=︸︷︷︸ODEs

n∑j=1

pijϕ(uj(y(t))

)−

n∑j=1

pijΦ(y(t)

)yj(t)︸ ︷︷ ︸

Φ(y(t))·∑

j pijyj(t)=Φ(y(t))ϕ(ui)

Multiply both sides by ϕ(ui) and add for all i,

n∑i=1

ϕ(ui)d

dtϕ(ui) =

n∑i=1

n∑j=1

pijϕ(ui)ϕ(uj)︸ ︷︷ ︸0

−Φ(y(t)

)Ψ(y(t)

)

Since P T = −P , the �rst term is zero:

ϕTP ϕ =(ϕTP ϕ

)T= ϕTP T ϕ = −ϕTP ϕ ⇒ ϕTPϕ = 0.

Thus1

2· ddt

Ψ(y(t)

)= −Φ

(y(t)

)Ψ(y(t)

)(Notice, if ϕ(ui) = 0, this equation remain valid, since zero terms are added to bothsides.)

(iii) Assume �rst that with some t0 > 0,Ψ(y(t0)) = 0. Then from the ODE, Ψ(y(t)) = 0for all t ≥ t0 (since ODE has zero solution with initial value 0), so for all i,

ϕi(y(t)

)= 0 ⇒ Py(t) ≤ 0 ⇒ y(t) is equilibrium stategy.

(iv) If Ψ(y(t)) > 0 for all t, then from lemma

1

2

d

dtΨ(y(t)

)≤ −Ψ

(y(t)

) 32

91

Game theory

or

1

2

d

dtΨ(y(t)

)Ψ(y(t)

)− 32 ≤ −1 /

∫ t

0

both sides

−Ψ(y(t)

)− 12 + c ≤ −t

where c = Ψ(y(0)

)− 12 . Hence

Ψ12

(y(t)

)≤ 1

c+ t.

By lemma

eTi P y(t) ≤ ϕ(ui(y(t))

)≤ Φ(y(t))

≤√nΨ(y(t)) ≤

√n

c+ t.

Taking an increasing sequence t1 < t2 < t3 < . . . with tk → ∞, for any limit pointy∗ and all i

eTi P y∗ ≤ 0 ⇒ P y∗ ≤ 0

⇒ y∗ is equilibrium strategy. �

Example 10.6

A =

2 1 02 0 3−1 3 3

∼(+2)

4 3 24 2 51 5 5

P =

0 0 0 4 3 2 −10 0 0 4 2 5 −10 0 0 1 5 5 −1−4 −4 −1 0 0 0 1−3 −2 −5 0 0 0 1−2 −5 −5 0 0 0 11 1 1 −1 −1 −1 0

The ODE system is solved by 4th order Runge-Kutta method on [0, 100] with h = 0.01 :

y(100) = (u, v, λ), λ ≈ 0.627163

a =1− λ

2≈ 0.1864185

x =1

au ≈ (0.563619, 0.232359, 0.241988)

y =1

av ≈ (0.485258, 0.361633, 0.115144)

The value of the game is

v =λ

a− 2 ≈ 1.364274

92

Game theory

In Example 10.2 we determined the equilibrium:

x =

(4

7,

4

21,

5

21

)T≈ (0.571429, 0.190476, 0.238095)T ,

y =

(3

7,3

7,1

7

)T≈ (0.428571, 0.428571, 0.142857)T ,

and the true value of the game is v = 97≈ 1.285714.

Comparing these true values to the results obtained by using the von Neumann methodwe can see that the maximum error in the equilibrium components is about 0.0669, andthe error in the value of v is 0.07856. This inaccuracy is probably due to the cumulativeerror in using the Runga�Kutta method with a relatively large (h = 0.01) stepsize. O

Note. We got another method for solving matrix games and so for LPs by solvingdi�erential equations (similar idea to interior point method).

Example 10.7 Matrix game A =

(1 22 1

)has no pure strategy equilibrium.

Fictitious play example

k = 1,

x1 =

(10

), xT1A =

(1, 0)(1 2

2 1

)=(1, 2)

minimum component is the �rst, so

y1

=

(10

).

k = 2,

y1

=1

1

(10

)=

(10

)

Ay1

=

(1 22 1

)(10

)=

(12

)maximum component is the second, so

x2 =

(01

), x2 =

1

2

[(10

)+

(01

)]=

(1212

)

xT2A =(

12, 1

2

)(1 22 1

)=(

32, 3

2

)same values, select y

2=

(10

).

k = 3,

y2

=1

2

[(10

)+

(10

)]=

(10

)

Ay2

=

(1 22 1

)(10

)=

(12

)

93

Game theory

maximum component is the second, so

x3 =

(01

), x3 =

1

3

[(10

)+

(01

)+

(01

)]=

(1323

)

xT3A =(

13, 2

3

)(1 22 1

)=(

53, 4

3

)minimum component is the second,

y3

=

(01

).

k = 4,

y3

=1

3

[(10

)+

(10

)+

(01

)]=

(2313

)Ay

3=

(1 22 1

)(2313

)=

(4353

)maximum component is the second so

x4 =

(01

), x4 =

1

4

[(10

)+

(01

)+

(01

)+

(01

)]=

(1434

)

xT4A =(

14, 3

4

)(1 22 1

)=(

74, 5

4

)minimum component is the second so

y4

=

(01

),

and so on for k = 5, 6, . . .

von Neumann�method example

P =

0 A −1−AT 0 11T −1T 0

=

0 0 1 2 −10 0 2 1 −1−1 −2 0 0 1−2 −1 0 0 11 1 −1 −1 0

Five-dimensional problem. Remember that

ui(y) = eTi P y = component i of P y

ϕ(ui) = max {0;ui} = larger of 0 and ui

Φ(y) =∑

i ϕ(ui)

94

Game theory

u1(y) = y3 + 2y4 − y5

u2(y) = 2y3 + y4 − y5

u3(y) = −y1 − 2y2 + y5

u4(y) = −2y1 − y2 + y5

u5(y) = y1 + y2 − y3 − y4

So the 5-dimensional system of di�erential equations is:

y1 = max {0; y3 + 2y4 − y5} − y1 ·[

max {0; y3 + 2y4 − y5}+ max {0; 2y3 + y4 − y5}+ max {0;−y1 − 2y2 + y5}+ max {0;−2y1 − y2 + y5}+ max {0; y1 + y2 − y3 − y4}

]y2 = max {0; 2y3 + y4 − y5} − y2 ·

[. . .]

y3 = max {0;−y1 − 2y2 + y5} − y3 ·[. . .]

y4 = max {0;−2y1 − y2 + y5} − y4 ·[. . .]

y5 = max {0; y1 + y2 − y3 − y4} − y5 ·[. . .]

O

11 Uniqueness of Equilibrium

Example 11.1 n = 2 oligopoly with S1 = S2 = [0, 1]

C1(x1) = 0.5x1, C2(x2) = 0.5x2,

p(s) =

{1.75− 0.5s if 0 ≤ s ≤ 1.52.5− s if 1.5 ≤ s ≤ 2

with s = x1 + x2. So pro�t of player i is

ϕi(x1, x2) = xip(x1 + x2)− Ci(xi).

It satis�es all conditions of Nikaido�Isoda theorem, and ϕi is strictly concave.In optimization, if objective function is strictly concave, maximal solution is unique.

But, in games, there might be multiple EP. In this case set of equilibria:

X∗ = {(x1, x2) | 0.5 ≤ x1 ≤ 1, 0.5 ≤ x2 ≤ 1, x1 + x2 = 1.5} .

O

Another such example was shown earlier in Example 6.1.⇒ Something else is needed for uniqueness.

Consider game {n, S1, . . . , Sn, ϕ1, . . . , ϕn} and let best reply of player k be Rk(x), sothe best reply mapping is R(x) = (R1(x), . . . , Rn(x)).

Lemma 11.1 Assume that the best reply mapping is point-to-point and either

(i) %(R(x), R(y)) < %(x, y) for all x, y ∈ S1 × · · · × Sn, x 6= y

or

95

Game theory

(ii) %(R(x), R(y)) > %(x, y) for all x, y ∈ S1 × · · · × Sn, x 6= y.

Then equilibrium cannot be multiple.More generally, it is su�cient to assume that for all x 6= y, %(R(x), R(y)) 6= %(x, y).

Proof. Assume that x∗ and y∗ are both equilibria, then R(x∗) = x∗ and R(y∗) = y∗, so

%(R(x∗), R(y∗)

)= %(x∗, y∗)

contradicting the assumption. �Note. Condition (i) is much weaker than the assumption that R is contraction. Existenceis not implied by (i).

Example 11.2 Consider R(x) = x+ 1x+1

on [0,∞].Clearly there is no �xed point,

R′(x) = 1− 1

(x+ 1)2

so0 ≤ R′(x) < 1,

so|R(x)−R(y)| = |R′(ξ)| · |x− y| < |x− y|

with ξ being between x and y, so (i) is satis�ed. Notice that if x, y →∞, then ξ →∞,so R′(ξ) becomes very close to 1. O

Similiarly, (ii) does not imply existence either.

Example 11.3 Let R(x) = 2x on [1,∞),then

|R(x)−R(y)| = |2x− 2y| = 2|x− y| > |x− y|

O

96

Game theory

Another way of guaranteeing uniqueness is based on monotonicity.Equation f(x) = 0

(i) In one-dimension, if f(x) strictly increases or strictly decreases then f(x) = 0 cannothave multiple solutions.

(ii) In multiple-dimension, more complicated, for f(x) = 0, component-wise monotonic-ity is not su�cient:

x+ y = 02x+ 2y = 0

}both components are strictly increasing in both variables,but ∞ many solutions: y = −x

−x− y = 0−2x− 2y = 0

}both components are strictly decreasing in both variables,but ∞ many solutions: y = −x

Fixed point problem x = f(x)

(i) In one-dimension, if f(x) decreases, then multiple �xed points are impossible.

(ii) In multiple dimension, more complicated:

x = −x− 2yy = −2x− y

}both components are strictly decreasing in both variables,but ∞ many solutions: y = −x

⇒ Di�erent kind of monotonicity is needed.

We already introduced monotonicity in Rn, a function f : D 7→ Rn (D ⊆ Rn) is calledmonotonic, if for all x and y ∈ D,

(x− y)T(f(x)− f(y)

)≥ 0

and it is called strictly monotonic if for all x 6= y ∈ D,

(x− y)T(f(x)− f(y)

)> 0.

Consider �rst equation f(x) = 0.

Theorem 11.1 If f or −f is strictly monotonic, then the solution cannot be multiple.

Proof. Assume x∗ and y∗ are both solutions, then f(x∗) = f(y∗) = 0, which contradictsthe de�nition of strict monotonicity:

(x∗ − y∗)T(f(x∗)− f(y∗)

)= 0.

�Consider next �xed point problem x = f(x).

Theorem 11.2 If −f is monotonic, then solution cannot be multiple.

Proof. Let x∗ and y∗ be solutions, then

(x∗ − y∗)T(f(x∗)− f(y∗)

)= (x∗ − y∗)T

(x∗ − y∗

)> 0,

contradiction. �Corollary. If the negative of the best reply mapping (−R(x)) is monotonic, then the EPcannot be multiple. This is true, even if R(x) is a set, then we have to assume that forall x, y ∈ S1 × · · · × Sn,

(x− y)T (u− v) ≤ 0

for all u ∈ R(x) and v ∈ R(y).

97

Game theory

6

-

y < x

R(y)

R(x)

We can also check uniqueness without determining the best response mappings.Consider next game {n, S1, . . . , Sn, ϕ1, . . . , ϕn}, and assume

(i) Sk = {xk | xk ∈ Rmk , gk(xk) ≥ 0} is nonempty, each component of g

kis concave

and continuously di�erentiable in an open set containing Sk;

(ii) Sk satis�es the Kuhn-Tucker regularity condition;

(iii) ϕk is twice continuously di�erentiable in an open set containing S = S1× · · · ×Sn.

De�ne

h(x, r) =

r1O1ϕ1(x)...

rnOnϕn(x)

,

where r ≥ 0 and Okϕk(x) is the gradient (now as column vector) of ϕk with respect toxk. If M = m1 + · · ·+mk, then

h : RM 7→ RM with any �xed r.

De�nition 11.1 The game is called strictly diagonally concave on S, if for all x(0) 6=x(1), x(0), x(1) ∈ S and with some r ≥ 0,

(x(1) − x(0))(h(x(1), r)− h(x(0), r)

)< 0.

Note. This inequality means that −h is strictly monotone in x, so we have the followingcondition:

Lemma 11.2 Assume S is convex, for all k, ϕk is twice continuously di�erentiable. Thenthe game is strictly diagonally concave, if

J(x, r) + J(x, r)T

is negative de�nite (where J is the Jacobian of h with respect to x)

98

Game theory

Theorem 11.3 (Theorem of Rosen) Assume conditions (i), (ii), (iii) are satis�edand the game is diagonally strictly concave. Then there is at most one equilibrium.

Proof. Assume x(0) =(x

(0)1 , . . . , x

(0)n

)and x(1) =

(x

(1)1 , . . . , x

(1)n

)are both EP. Then from

the Kuhn-Tucker conditions for l = 0, 1,

u(l)Tk g

k

(x

(l)k

)= 0 (45)

Okϕk(x(l))︸ ︷︷ ︸

as row vector

+u(l)Tk Ogk

(x

(l)k

)= 0

That is, by taking transpose,

Okϕk(x(l))︸ ︷︷ ︸

as column vector

+

pk∑j=1

u(l)kjOgkj

(x

(l)k

)︸ ︷︷ ︸ = 0

Here pk is the size of gk, and Ogkj(x

(l)k

)is the gradient of gkj as column vector.

For l = 0, multiply by rk

(x

(1)k − x

(0)k

)Tand

for l = 1, multiply by rk

(x

(0)k − x

(1)k

)Tand add for all k,

0 =[(x(1) − x(0)

)Th(x(0), r) +

(x(0) − x(1)

)Th(x(1), r

)]︸ ︷︷ ︸+

+n∑k=1

pk∑j=1

{rk[u

(0)kj

(x

(1)k − x

(0)k

)TOgkj

(x

(0)k

)︸ ︷︷ ︸

≥gkj(x(1)k

)−gkj

(x(0)k

)+ u

(1)kj

(x

(0)k − x

(1)k

)TOgkj

(x

(1)k

)︸ ︷︷ ︸

≥gkj(x(0)k

)−gkj

(x(1)k

)]}

From (45), ∑j

u(l)kjgkj

(x

(l)k

)= 0, so

0 >n∑k=1

pk∑j=1

{rk

[u

(0)kj gkj

(x

(1)k

)+ u

(1)kj gkj

(x

(0)k

)]}≥ 0, contradiction.

�

Example 11.4 Duopoly with n = 2, S1 = S2 = [0, 1], Ck(xk) = xk, p(s) = 2 − s withs = x1 + x2. Then

ϕ1(x1, x2) = x1(2− x1 − x2)− x1 = −x21 + x1(1− x2)

ϕ2(x1, x2) = x2(2− x1 − x2)− x2 = −x22 + x2(1− x1)

O1ϕ1(x1, x2) = −2x1 + 1− x2

O2ϕ2(x1, x2) = −2x2 + 1− x1

h(x, r) =

(r1 (−2x1 + 1− x2)r2 (−2x2 + 1− x1)

).

99

Game theory

So

J(x, r) =

(−2r1 −r1

−r2 −2r2

),

J(x, r) + J(x, r)T =

(−2r1 −r1

−r2 −2r2

)+

(−2r1 −r2

−r1 −2r2

)=

(−4r1 −r1 − r2

−r1 − r2 −4r2

).

This is negative de�nite, if eigenvalues are negative. Characteristic polynomial:

det

(−4r1 − λ −r1 − r2

−r1 − r2 −4r2 − λ

)= λ2 + λ(4r1 + 4r2) +

(16r1r2 − (r1 + r2)2

)Select r1 = r2 = 1, then the characteristic polinomial:

λ2 + 8λ+ 12 = 0

λ =−8±

√64− 48

2=−8± 4

2= −2 and − 6

Both negative ⇒ unique equilibrium. O

12 Leader�follower games (Stackelberg equilibrium)

n = 2,

Player 1 = leader,

Player 2 = follower

Step 1. Player 1 selects x1 and tells it to player 2

Step 2. Player 2 choses R2(x1) (follows player 1)

Player 1's payo� ϕ1(x1, R2(x1)) −→ max

Theorem 12.1 Assume both ϕ1 and R2(x1) are continuous, S1 is nonempty, closed andbounded. Then there is optimal solution.

Proof. From Weierstrass' theorem. �In general: n > 2,

Player 1 = leader,

Players 2, . . . , n = followers

For each x1 ∈ S1, others form Nash equilibrium:

(x2 = E2(x1), . . . , xn = En(x1))

then payo� of player 1:

max ϕ1(x1,E2(x1), . . . ,En(x1))

s. to x1 ∈ S1

Optimal solution x∗1, and for k = 2, . . . , n, x∗k = Ek(x∗1)

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Game theory

12.1 Application to duopolies

n = 2, oligopoly with 2 �rms

Player 1: home-�rm with subsidy from its government

Player 2: foreign-�rm

Price function:p(x+ y) = a− b(x+ y)

Pro�ts of the �rms:

ϕ1 = x(a− bx− by)− (c− s)x (s = subsidity for home �rm)

ϕ2 = y(a− bx− by)− cy

1. Nash equilibrium

Assuming interior optimum,

∂ϕ1

∂x= a− bx− by − bx− (c− s) = 0

x =a− by − c+ s

2b, y =

a− c+ s

b− 2x

∂ϕ2

∂y= a− bx− by − by − c = 0

y =a− bx− c

2b

Equating the two expressions

y =a− bx− c

2b=

a− c+ s

b− 2x (/ · 2b)

a− bx− c = 2a− 2c+ 2s− 4bx

3bx = a− c+ 2s ⇒ x = a−c+2s3b

so

y =a− c

2b− a− c+ 2s

6b=

2a− 2c− 2s

6b=a− c− s

3b

y = a−c−s3b

Here we assumed players move simultaneously.

101

Game theory

2. Stackelberg equilibrium

Player 1 is leader, player 2 follows whatever he/she does.y = a−bx−c

2bwhatever x is, so pro�t of player 1:

ϕ1 = x ·(a− bx− a− bx− c

2

)− (c− s)x

= x · 2a− 2bx− a+ bx+ c

2− (c− s)x

=x

2(a− bx+ c)− (c− s)x

Di�erentiate:

∂ϕ1

∂x=

a− bx+ c

2− bx

2− (c− s) = 0

a− 2bx+ c− 2c+ 2s = 0

x = a−c+2s2b

so from above

y =a− c

2b− a− c+ 2s

4b=a− c− 2s

4b

y = a−c−2s4b

Di�erent than Nash equilibrium, x is now larger and y is smaller as they should be.

3. Optimum subsidy

Welfare of home country = ϕ1 − subsidyWith Stackelberg equilibrium:

ϕ1 − subsidy =a− c+ 2s

2b·(a− a− c+ 2s

2− a− c− 2s

4

)− c · a− c+ 2s

2b

=a− c+ 2s

2b

(4a− 2a+ 2c− 4s− a+ c+ 2s− 4c

4

)=

a− c+ 2s

2b· a− c− 2s

4−→ max

2(a− c− 2s)− 2(a− c+ 2s) = 0

a− c− 2s− a+ c− 2s = 0

−4s = 0 ⇒ s = 0 , no subsidy

Stackelberg equilibrium without subsidy (with s = 0):

x =a− c+ 2s

2b=a− c

2b

y =a− c− 2s

4b=a− c

4b

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Game theory

With Cournot equilibrium:

ϕ1 − subsidy =a− c+ 2s

3b·(a− a− c+ 2s

3− a− c− s

3

)− c · a− c+ 2s

3b

=a− c+ 2s

3b·(

3a− a+ c− 2s− a+ c+ s− 3c

3

)=

a− c+ 2s

3b·(a− c− s

3

)−→ max

2(a− c− s)− (a− c+ 2s) = 0

2a− 2c− 2s− a+ c− 2s = 0

⇒ s = a−c4

Nash equilibrium with this optimal subsidy:

x =a− c+ a−c

2

3b=a− c

2b

y =a− c− a−c

4

3b=

3a− 3c

4 · 3b=a− c

4b

Two equilibria are the same, so optimal subsidy is equivalent to leader's advantage.

13 Games with incomplete information

Note. Incomplete information refers to amount of information the players have aboutthe game, while imperfect information refers to the amount of information they haveon others' and their own previous moves (and on previous chance moves).

Incomplete information results from

• lack of information on physical outcomes

• lack of information on own or others' utility function (which are the payo�s)

• lack of information on strategy sets

Problems:

• wrong physical outcome ⇒ wrong or uncertain payo�

• wrong or uncertain payo�

• by assigning −∞ (or large negative) payo�s for infeasible strategies, payo� becomesonly uncertain

⇒ all cases reduce to uncertain payo�s, so we will assume that strategy sets are knownfor every player, only payo�s are uncertain.

Example 13.1 Modi�ed prisoner's dilemma: DA's brotherPlayer 1 is the DA's brother, who will go free if none of the players confesses.

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Game theory

NC CNC (0, -2) (-10,-1)C (-1,-10) (-5,-5)

Type IProbability = µ

Assume that with probability µ this is the case (player 2 is of type I), but with probability1 − µ player 2 pays an emotional penalty or friends of his partner will take revenge(equivalent to 6 years in prison for giving up his partner), so payo� table is

NC CNC (0, -2) (-10,-7)C (-1,-10) (-5,-11)

Type IIProbability = 1− µ

Extensive form:

Prisoner 2 now has four possible pure strategies:

(i) C if Type I, C if Type II

(ii) C if Type I, NC if Type II

(iii) NC if Type I, C if Type II

(iv) NC if Type I, NC if Type II

104

Game theory

and player 1 still has his original two strategies: C and NC.O

Next we give formal de�nition of such games:

De�nition 13.1 (Bayesian game) Bayesian game, payo� of player i (i = 1, 2, . . . , n),

ϕi(si, s−i, θi) (θi ∈ Θi is random variable)

where si ∈ Si, s−i = (s1, . . . , si−1, si+1, . . . , sn), furthermore we assume that randomvariables θ1, . . . , θn are selected by nature according to a joint distribution functionF (θ1, . . . , θn), and the actual value of θi is known by only player i, unknown to all others.With normal form notation:

(n, S1, . . . , Sn, ϕ1, . . . , ϕn,Θ, F )

with Θ = Θ1 ×Θ2 × · · · ×Θn and known F by all players.

De�nition 13.2 (Pure strategy) Pure strategy in a Bayesian game for player i is adecision rule si(θi) for all θi ∈ Θi, where si : Θi 7→ Si.

De�nition 13.3 (Expected payo�) Player i's expected payo� given a pro�le of purestrategies (s1(.), . . . , sn(.)) is given as

ϕi (s1(.), . . . , sn(.)) = EΘ [ϕi (s1(θ1), . . . , sn(θn), θi)]

De�nition 13.4 (Pure strategy Bayesian equilibrium) Bayesian equilibrium is apro�le of decision rules (s∗1(.), . . . , s∗n(.)) such that

ϕi(s∗i (.), s

∗−i(.)

)≥ ϕi

(si(.), s

∗−i(.)

)for all functions si(.).

Theorem 13.1 A decision pro�le (s∗1(.), . . . , s∗n(.)) is Bayesian equilibrium ⇔ for all iand θi ∈ Θi accuring with positive probability,

Eθ−i

[ϕi(s∗i (θi), s

∗−i(θ−i), θi | θi

)]≥ Eθ−i

[ϕi(si, s

∗−i(θ−i), θi | θi

)]for all si ∈ Si, where Eθ−i

is the conditional expectation with respect to θ−i =

(θ1, . . . , θi−1, θi+1, . . . , θn) given θi = θi.

Example 13.2 (Continuation of DA's brother example)

In Type I, player 2 must play C (dominant strategy)

in Type II, player 2 must play NC (dominant strategy)

Sos∗2(I) = C, s∗2(II) = NC

and therefore

ϕ1(C, s2(.)) = µ(−5) + (1− µ)(−1) = −4µ− 1

ϕ1(NC, s2(.)) = µ(−10) + (1− µ)0 = −10µ.

105

Game theory

So,

s∗1 = C, if −4µ− 1 ≥ −10µ ⇔ µ ≥ 1

6

s∗1 = NC, if −4µ− 1 ≤ −10µ ⇔ µ ≤ 1

6

In Summary:

s∗1 =

C, if µ > 1

6

NC, if µ < 16

C and NC, if µ = 16

O

Example 13.3 2-player game, both can be in weak or strong position (for player 1, a1 =weak, a2 = strong; for player 2, b1 = weak, b2 = strong,). There are 4 possibilities forpayo� functions. Player 1's payo� tables:

z1 z2

y1 2 5y2 -1 20

z1 z2

y1 -24 -36

y2 0 24

(a1, b1) (a1, b2)

z1 z2

y1 28 15y2 40 4

z1 z2

y1 12 20y2 2 13

(a2, b1) (a2, b2)

For player 2, ϕ2 = −ϕ1 for all cases.Joint distribution:

b1 b2

a1 r11 = 0.40 r12 = 0.10 0.50a2 r21 = 0.20 r22 = 0.30 0.50

0.60 0.40

De�ne the following strategies for the players:

(i, j) = (pure strategy if weak, pure strategy if strong)

Payo� table for player 1:

(1, 1) (1, 2) (2, 1) (2, 2)(1, 1) 7.6 8.8 6.2 7.4(1, 2) 7.0 9.1 1.0 3.1

(2, 1) 8.8 13.6 14.6 19.4(2, 2) 8.2 13.9 9.4 15.1

For example: for pair ((1, 2), (1, 1)) we have

0.4(2) + 0.1(−24) + 0.2(40) + 0.3(2) = 7.0case (a1, b1) (a1, b2) (a2, b1) (a2, b2)strategy (y1, z1) (y1, z1) (y2, z1) (y2, z1)

106

Game theory

or for pair ((1, 2), (2, 1)) we have

0.4(5) + 0.1(−24) + 0.2(4) + 0.3(2) = 1.0case (a1, b1) (a1, b2) (a2, b1) (a2, b2)strategy (y1, z2) (y1, z1) (y2, z2) (y2, z1)

⇒ equilibrium is unique: (2, 1) and (1, 1) are equilibrium strategies. O

14 Cooperative games

14.1 Characteristic functions

De�nition 14.1 (Characteristic function) Let S ⊆ {1, 2, . . . , n} = N be a coalition,then

v(S) = maxi∈S,xi

minj 6∈S,xj

∑i∈S

ϕi(x1, . . . , xn) (46)

Meaning: maximal utility of coalition S without the cooperation of others or payo� valuefor coalition what they can get regardless what others are doing. Also v(∅) = 0 and

v({1, 2, . . . , n}) = maxn∑i=1

ϕi(x1, . . . , xn).

In the literature there are other de�nitions known as well, but this is the most popular.

De�nition 14.2 (Superadditive game) Game is superadditive, if S, T ⊂ N and S ∩T = ∅, then

v(S ∪ T ) ≥ v(S) + v(T ). (47)

By induction,

v(S1 ∪ S2 ∪ · · · ∪ Sk) ≥ v(S1) + v(S2) + · · ·+ v(Sk) (48)

if Si ∩ Sj = ∅ for all i 6= j.Note. v(S) de�ned by (46) is always superadditive.

De�nition 14.3 (Monotone game) Game is monotone, if S ⊂ T implies v(S) ≤ v(T ).

Convexity is de�ned by increasing �rst di�erences. De�ne

Di(S) =

{v(S ∪ {i})− v(S) if i 6∈ Sv(S)− v(S − {i}) if i ∈ S

De�nition 14.4 (Convex game) Game is convex, if for each i ∈ N, S ⊆ T impliesthat

Di(S) ≤ Di(T ). (49)

107

Game theory

De�nition 14.5 (Constant sum game) Game is constant sum, if for all coalitionsS ⊆ N,

v(S) + v(N − S) = v(N). (50)

De�nition 14.6 (Rational game) Game is rational, if

v(N) ≥∑i∈N

v({i}). (51)

In case of =, game is inessential, if > then it is called essential.

De�nition 14.7 (Weakly superadditive game) Game is weakly superadditive, if forall S ⊆ N,

v(N) ≥ v(S) +∑i∈N−S

v({i})

Note. Superadditive ⇒ weakly superadditive (use (48)).

De�nition 14.8 (Stragetically equivalent games) Games (N, v) and (N, v′) arestrategically equivalent, if there exist α > 0 and β1, . . . , βn such that for all S ⊆ N,

v(S) = αv′(S) +∑i∈S

βi

De�nition 14.9 (Normalized game) A game is normalized, if

v({i}) = 0 for all i ∈ N

andv(N) = 1.

Note. Normalized ⇒ essential.

De�nition 14.10 (Simple game) A superadditive normalized game is called simple, iffor all S ⊆ N, v(S) = 0 or v(S) = 1.

A coalition S is winning if v(S) = 1 and losing if v(S) = 0.

Theorem 14.1 If a function v is de�ned on all coalitions such that v(∅) = 0 and issuperadditive, then there is an n-person game such that v is its characteristic function.

The central question is what is given to the players in case of cooperation.Payo� vector x = (x1, . . . , xn) gives amounts to all players such that

n∑i=1

xi ≤ v(N)

(we cannot give out more than we might have). Here xi is payment to player i and it isnot strategy.

De�nition 14.11 (Individually rational payo� vector) A payo� vector is called in-dividually rational, if

n∑i=1

xi = v(N) and xi ≥ v({i}) for all i

(sometimes they are called imputations)

108

Game theory

Example 14.1 3-person oligopoly, 0 ≤ xk ≤ 3

p(s) = 10− sCk(xk) = xk + 1 (symmetric)

ϕk(x1, x2, x3) = xk(10− x1 − x2 − x3)− (xk + 1)

Construction of v(S) :

(i) S = {1} (or {2}, or {3})

v({1}) = maxx1

minx2,x3

x1(10− x1 − x2 − x3)− (x1 + 1)︸ ︷︷ ︸minimum occurs at x2 = x3 = 3︸ ︷︷ ︸

x1(4−x1)−(x1+1)

−x21 + 3x1 − 1 → max

−2x1 + 3 = 0

x1 =3

2(feasible)

v({1}) = v({2}) = v({3}) = −9

4+

9

2− 1 =

5

4

(ii) S = {1, 2} (or {1, 3}, or {2, 3})

v({1, 2} = maxx1,x2

minx3

(x1 + x2)(10− x1 − x2 − x3)− (x1 + x2 + 2)︸ ︷︷ ︸minimum occurs at x3 = 3︸ ︷︷ ︸

(x1+x2)(7−x1−x2)−(x1+x2+2)

u = x1 + x2 ∈ [0, 6]

u(7− u)− (u+ 2) = −u2 + 6u− 2

−2u+ 6 = 0

u = 3 (feasible)

v({1, 2}) = v({1, 3}) = v({2, 3}) = −9 + 18− 2 = 7

(iii) S = {1, 2, 3}

v({1, 2, 3}) = max(x1 + x2 + x3)(10− x1 − x2 − x3)− (x1 + x2 + x3 + 3)

u = x1 + x2 + x3 ∈ [0, 9]

u(10− u)− u− 3 = −u2 + 9u− 3

−2u+ 9 = 0

u = 4.5 (feasible)

v ({1, 2, 3}) = −81

4+

81

2− 3 =

−81 + 162− 12

4=

69

4= 17.25

109

Game theory

Imputations: {x = (x1, x2, x3) | xi ≥

5

4(∀i), x1 + x2 + x3 =

69

4

}Checking superadditivity:

v({1, 2})︸ ︷︷ ︸7

≥ v({1})︸ ︷︷ ︸54

+ v({2})︸ ︷︷ ︸54

v({1, 2, 3})︸ ︷︷ ︸694

≥ v({1})︸ ︷︷ ︸54

+ v({2, 3})︸ ︷︷ ︸7

this is su�cient because of symmetry

O

De�nition 14.12 (Dominating imputation) Imputation x dominates imputation yon coalition S, if

(i) xi > yi (∀i ∈ S) and

(ii)∑i∈S

xi ≤ v(S)

De�nition 14.13 We simply say x dominates y if x dominates y on a coalition of atleast 2 players.

Notations: x �S y or x � y

Note.

(i) means all members of S are better o� with (x1, . . . , xn)

(ii) coalition is able to provide xi to players i ∈ S

Note. No dominance is possible if |S| = 1 or n:

|S| = 1 , S = {i}, thenv(S) ≥ xi > yi ≥ v({i}) = v(S)

|S| = n

v(S) ≥∑i

xi >∑i

yi = v(N) = v(S)

⇒ contradiction in both cases.

14.2 Core of game

Set of x vectors such that

x(S) =∑i∈S

xi ≥ v(S) for all S ⊆ N

x(N) =n∑i=1

xi = v(N)

110

Game theory

Example 14.2 Previous example:

x1, x2, x3 ≥ 54

x1 + x2, x1 + x3, x2 + x3 ≥ 7x1 + x2 + x3 = 69

4

convex polyhedron

O

Theorem 14.2 For weakly superadditive games, the core is exactly the set of nondomi-nated imputations.

14.3 Stable sets (or von Neumann�Morgenstern solution)

V = set of imputations such that

(i) there is no x, y ∈ V such that x �S y for some S (internal stability);

(ii) if y 6∈ V then there exists x ∈ V such that x �S y with some S (external stability).

Problem: No general description, no general method.Not unique and not necessarily exists.(Remember! Imputation:

∑ni=1 xi = v(N) and xi ≥ v({i}) for all i)

14.4 The Shapley�value

Let di(S) = v(S)− v(S − {i}) be the marginal contribution of player i ∈ S after he joinscoalition S. In this case we de�ne di(S) = 0 if i 6∈ S.

Assume players form the grand coalition N by entering one by one in a random order.The average contribution of player i is given as

xi =∑S⊆N

(s− 1)!(n− s)!n!

di(S),

where s = |S|, since there are (s−1)! orderings of other players already in S, and (n− s)!ordering of the players outside S.

xi = Shapley value for player i

Note: If i /∈ S, then di(S) = 0. So in computing xi we need to consider all coalitionsS such that i ∈ S.

Theorem 14.3 For any superadditive game, x = (xi) is an imputation.

Example 14.3 n = 2, coalitions ∅, {1}, {2}, {1, 2}

x1 =(1− 1)!(2− 1)!

2!

(v ({1})− v(∅)

)+

(2− 1)!(2− 2)!

2!

(v({1, 2})− v({2})

)=

v({1}) + v({1, 2})− v({2})2

and similarly,

111

Game theory

x2 =(1− 1)!(2− 1)!

2!

(v ({2})− v(∅)

)+

(2− 1)!(2− 2)!

2!

(v({1, 2})− v({1})

)=

v({2}) + v({1, 2})− v({1})2

Notice thatx1 + x2 = v({1, 2}).

O

Example 14.4 n = 3, coalitions: ∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}

x1 =(1− 1)!(3− 1)!

3!

(v ({1})− v (∅)

)+

(2− 1)!(3− 2)!

3!

(v ({1, 2})− v ({2})

)+

(2− 1)!(3− 2)!

3!

(v ({1, 3})− v ({3})

)+

(3− 1)!(3− 3)!

3!

(v ({1, 2, 3})− v ({2, 3})

)=

1

6

(2 · v ({1}) + v ({1, 2})− v ({2}) + v ({1, 3})− v ({3}) + 2 · v ({1, 2, 3})− 2 · v ({2, 3})

)Similarly

x2 =1

6

(2 · v ({2}) + v ({2, 1})− v ({1}) + v ({2, 3})− v ({3}) + 2 · v ({1, 2, 3})− 2 · v ({1, 3})

)x3 =

1

6

(2 · v ({3}) + v ({3, 1})− v ({1}) + v ({3, 2})− v ({2}) + 2 · v ({1, 2, 3})− 2 · v ({1, 2})

)Notice that

x1 + x2 + x3 =1

6· 6 · v ({1, 2, 3}) = v ({1, 2, 3})

as it is always the case. O

Example 14.5 In the case of symmetric games

xi =1

nv({1, 2, . . . , n}),

since the players have to get equal payments. So in the case of the symmetric oligopoly ofExamples 14.1 and 14.2,

x1 = x2 = x3 =23

4.

O

Example 14.6 Simple game, where v(S) = 0 or v(S) = 1 for all coalitions. In thisspecial case

xi =∑ (s− 1)!(n− s)!

n!

where summation is over all pivotal coalitions such that S is winning (v(S)=1), and S−{i}is losing (v(S − {i}) = 0) ⇒ xi = average contribution when player i turns losingcoalitions into winning ones.⇒ xi = power index of player i O

112

Game theory

15 Social choice

15.1 Equal players

Based on rankings of alternatives, no payo� functions are needed.

Example 15.1 Practical methods are introduced in a forest treatment problem

Data: Clear cut Uniform Strip cut Controlthinning and thinning

Water users 1 2 3 4Wildlife advocates 4 2 1 3Livestock producers 1 2 3 4Wood producers 3 4 2 1Environmentalists 4 3 2 1Managers 4 2 3 1

i = criteria, j = alternatives, aij = rankings

1. Plurality voting

f(aij) =

{1 if aij = 10 otherwise

Aj =∑i

f(aij) = number of times alternative j is the best

Aj∗ = max{Aj}

The alternative with the most ranking 1 is selected.

A1 = 2, A2 = 0, A3 = 1, A4 = 3

⇒ control is the social choice

2. Borda count

Bj =∑i

aij (total points), Bj∗ = min{Bj}

The alternative with the smallest total point is selected.

B1 = 17, B2 = 15, B3 = 14, B4 = 14

⇒ control & (strip cut and thinning) are best

3. Hare system (successive deletions)

• if any alternative is considered the best by more than half of the players, then stop,it is the social choice

• delete the alternative that is considered the best by the least number of players, andadjust table as

anewij =

{aij if aij < aij∗aij − 1 otherwise

if alternative j∗ is deleted

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Game theory

• go back to �rst step

In previous example

A1 = 2, A2 = 0, A3 = 1, A4 = 3 ⇒ Uniform thinning is deleted

Clear cut Strip cut Controland thinning

1 2 33 1 21 2 33 2 13 2 13 2 1

A1 = 2 A3 = 1 A4 = 3 ⇒ strip cut and thinning is deleted

Clear cut Control1 22 11 22 12 12 1

A1 = 2 A4 = 4

⇒ control is the social choice

4. Pairwise comparisons

For a pair of alternatives de�ne

N(j1, j2) = No. of players where j1 is more preferred then j2.

Then j1 � j2 ⇔ N(j1, j2) > N(j2, j1)

The pairwise comparisons can be used in two di�erent ways:

1. An agenda is de�ned, an order of alternatives is given, i1, i2, · · · , im. First i1 and i2are compared. The better is compared to i3, then the winner to i4, etc., and the winnerof the last comparison is the social choice.

If agenda is 1, 2, 3, 4 thenN(1, 2) = 3, N(2, 1) = 3, tie, for example, select 1N(1, 3) = 2, N(3, 1) = 4, select 3N(3, 4) = 3, N(4, 3) = 3, tie again, so both 3 and 4 are social choice.If at the �rst step alternative 2 is chosen, thenN(2, 3) = 3, N(3, 2) = 3, tie, for example, select 2N(2, 4) = 3, N(4, 2) = 3, both 2 and 4 are social choice.If in comparing 2 and 3, 3 is chosen, thenN(3, 4) = 3, N(4, 3) = 3, both 3 and 4 are social choice.

2. All pairs are compared and a preference graph is constructed.

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Game theory

N(j1, j2) 1 2 3 41 − 3 2 22 3 − 3 33 4 3 − 34 4 3 3 −

Preference graph:

Only conclusion: clear cut is eliminated.

5. Dictatorship

Dictator = player j∗, then ai∗j∗ = 1 decides on social choice (what is best for dictator)

15.2 Nonsymmetric case

Example 15.2 The players may have di�erent importance weights:

A1 A2 A3 A4 weights1 2 3 4 22 1 3 4 11 2 4 3 14 3 2 1 23 4 1 2 14 3 1 2 1

total 8

O

Voting: 3 1 2 2 Best = A1

Borda: 20 20 19 21 Best = A3

Hare: A2 outA1 A3 A4

1 2 3 21 2 3 11 3 2 13 2 1 23 1 2 13 1 2 14 2 2

115

Game theory

Either A3 or A4 can be eliminated.

If A3 is eliminated, then new table

A1 A4

1 2 21 2 11 2 12 1 22 1 12 1 14 4

equally good, A1 and A4 are the social choice.

If A4 is eliminated, then new table

A1 A3

1 2 21 2 11 2 12 1 22 1 12 1 14 4

equally good, A1 and A3 are the social choice.

Pair-wise comparisons

A1 � A2 2 + 1 + 1 = 4

A1 � A3 2 + 1 + 1 = 4

A1 � A4 2 + 1 + 1 = 4

A2 � A3 2 + 1 + 1 = 4

A2 � A4 2 + 1 + 1 = 4

A3 � A4 2 + 1 + 1 + 1 = 5

1 2 3 41 − 4 4 42 4 − 4 43 4 4 − 54 4 4 3 −

����

����

"!#

������

��

�

1

5

3

2

4

116

Game theory

16 Con�ict resolution

Con�ict: (H, f ∗), whereH ⊆ R2 convex, closed, comprehensive (f ≤ f ′ ∈ H ⇒ f ∈ H),and f ∗ = status quo point, gives payo�s if no agreement is reached.

Assume at the solution, f1 ≥ f ∗1 , and f2 ≥ f ∗2 (rational players). Assume thereforethat solution is chosen in set

H∗ = {(f1, f2) | (f1, f2) ∈ H, f1 ≥ f ∗1 , f2 ≥ f ∗2}

Rational players will not agree on a solution which can be improved for both of them.

De�nition 16.1 A point (f1, f2) ∈ H∗ is called Pareto optimal if there is no otherpoint (f ′1, f

′2) ∈ H∗ such that (f ′1, f

′2) 6= (f1, f2) and (f ′1, f

′2) ≥ (f1, f2). The set of Pareto

points is called Pareto frontier.

The Pareto frontier of H∗ is assumed to be a function

f2 = g(f1), f ∗1 ≤ f1 ≤ F ∗1

which is strictly decreasing and concave. Assume that H∗ is nonempty.

16.1 Bargaining as a noncooperative game

Two players,S1 = [f ∗1 , F

∗1 ] , S2 = [f ∗2 , F

∗2 ]

ϕ1(f1, f2) =

{f1 if (f1, f2) ∈ H∗f ∗1 otherwise

ϕ2(f1, f2) =

{f2 if (f1, f2) ∈ H∗f ∗2 otherwise

Theorem 16.1 A point is equilibrium ⇔ Pareto in H∗.

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Game theory

16.2 Single-player decision problem

Player 1 assumes that player 2 will select f2 in interval [f ∗2 , F∗2 ] uniformly, and wants to

maximize his own expected payo�:

maximize f1P (f2 ≤ g(f1)) + f ∗1 P (f2 > g(f1))︸ ︷︷ ︸1−P(f2≤g(f1))

= (f1 − f ∗1 )g(f1)− f ∗2F ∗2 − f ∗2

+ f ∗1

is maximal if and only if(f1 − f ∗1 )(f2 − f ∗2 )

is maximal on the Pareto frontier. Product is called the Nash product. Solution is:

maximize (f1 − f ∗1 )(f2 − f ∗2 )

s. to f2 = g(f1)

f ∗1 ≤ f1 ≤ F ∗1

For player 2 we have the same problem, so optimal solution is best for both players ⇒reasonable solution.

Assume g is di�erentiable, and since at f1 = f ∗1 , and f1 = F ∗1 , objective is zero, wemust have local optimum:

h(f1) = [(f1 − f ∗1 )(g(f1)− f ∗2 )]′ = g(f1)− f ∗2 + (f1 − f ∗1 )g′(f1)︸ ︷︷ ︸strictly decreases, 2g′ + (f1 − f∗1 )g′′ < 0

= 0

h(f ∗1 ) = g(f ∗1 )− f ∗2 > 0h(F ∗1 ) = (F ∗1 − f ∗1 )g′(F ∗1 ) < 0

}unique solution

Example 16.1 Sharing a pie

g(f1) = 1− f1

f ∗1 = 0

f ∗2 = 0

maximize (f1 − 0)(1− f1 − 0)

By di�erentiation

1− f1 − 0 + (f1 − 0)(−1) = 0

1− f1 − f1 = 0

1 = 2f1

f1 =1

2⇒ f2 =

1

2fair share

118

Game theory

O

16.3 Axiomatic bargaining

Solution f = Ψ(H, f ∗) which satis�es certain axioms:

1. Ψ(H, f ∗) ∈ H (feasibility)

2. Ψ(H, f ∗) ≥ f ∗ (rationality)

3. f ∈ H, f ≥ Ψ(H, f ∗) ⇒ f = Ψ(H, f ∗) (Pareto optimality)

4. H1 ⊂ H, Ψ(H, f ∗) ∈ H1 ⇒ Ψ(H, f ∗) = Ψ(H1, f∗) (independence from unfavor-

able alternatives)

5. Let α1, α2 > 0, β1, β2 some constants,

f ∗′

= (α1f∗1 + β1, α2f

∗2 + β2)

H ′ = {(α1f1 + β1, α2f2 + β2) | (f1, f2) ∈ H} .

Then Ψ(H, f ∗) = (f 1, f 2) ⇒ Ψ(H ′, f ∗′) =

(α1f1 + β1, α2f2 + β2

)(independence

from increasing linear transformations)

6. (f ∗1 = f ∗2 and (f1, f2) ∈ H ⇔ (f2, f1) ∈ H) ⇒ f1 = f2 where Ψ(H, f ∗) = (f1, f2)(symmetry)

Theorem 16.2 (Theorem of Nash) There is a unique solution, which maximizes theNash product on H∗:

maximize (f1 − f ∗1 )(f2 − f ∗2 )

s. to f ≥ f ∗

f ∈ H

Lemma 16.1 Consider hyperbola y = cxand its tangent line at x = a. Then this point

is in the middle of the segment of the tangent line in the �rst quartal.

Proof.Equation of hyperbola y = c

xwith derivative y′ = − c

x2

119

Game theory

Targent line: y − b = − ca2

(x− a)

x-intercept: y = 0,a2b = cx− ca

x =a2b+ ac

c=

(ab)a+ ac

c= 2a

y-intercept: x = 0,

y = b+ca

a2= b+

ab · aa2

= 2b

�Idea of proof:

A Solution of the above optimization problem satis�es axioms. This is obvious.

B A solution that satis�es axioms is necessarily the optimal solution. Several steps:

(i)

Consider triangle with vertices (0,0), (1,0), (0,1). By 3. and 6., Ψ(H, f ∗) =(12, 1

2

)≡ optimal solution

(ii) By 5., Same for any triangle

(iii) H∗ is a convex set. Decrease the value of c of hyperbola until there is a commonpoint of hyperbola and H∗. This point f is a boundary point. Separatinghyperplane theorem ⇒ the tangent line of hyperbola at this point is theseparating hyperplane. Let H∗

′be the triangle de�ned by the tangent line and

the two coordinate lines, then Ψ(H∗′, f ∗) = f , since f is midpoint of segment.

Since f ∈ H∗ ⇒ Ψ(H∗, f ∗) = Ψ(H∗′, f ∗) = f.

If f ∗ 6= 0 then shifting the f1 and f2 lines will shift the solution in the sameway (Axiom 5 with αi = 0, βi 6= 0)

120

Game theory

Example 16.2 In the case of sharing a pie, solution is f1 = f2 = 12because of symmetry

(see previous example). O

Axiom 4 is critized by many authors, as given in �gure. Solution Ψ(H, 0) is fair, butΨ(H1, 0) is not at all.

16.4 Nonsymmetric Nash solution

By dropping 6., for all solutions there exists an α ∈ (0, 1) such that solution can beobtained:

maximize (f1 − f ∗1 )α(f2 − f ∗2 )1−α

s. to f ≥ f ∗

f ∈ H

(Harsányi and Selten)

Example 16.3 For previous example, the nonsymmetric Nash solution is:

maximize fα1 (1− f1)1−α

s. to 0 ≤ f1 ≤ 1,

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Game theory

which is equivalent to

maximize f p1 (1− f1) where p =α

1− αs. to 0 ≤ f1 ≤ 1

By di�erentiationpfp−1

1 (1− f1)− fp1 = 0

fp−11 (p− pf1 − f1) = 0

f1 6= 0, so f1 = pp+1

= α, f2 = 1− f1 = 1p+1

= 1− α. O

16.5 Area monotonic solution

Arc from status quo point divides region to equal areas, the solution is the intercept ofthis arc with Pareto frontier.

A1 =

f∫f∗1

g(t)dt− 1

2(f − f ∗1 ) (f ∗2 + g(f)) increasing in f

A2 =1

2(f − f ∗1 )(f ∗2 + g(f)) +

F ∗1∫f

g(t)dt− (F ∗1 − f ∗1 )f ∗2 decreasing in f

A1 = A2 or A1 = αA2

α = relative power of player 1 above player 2

122

Game theory

Solve equationαA2 − A1︸ ︷︷ ︸

strictly decreasing

= 0

at f = f ∗1 , A1 = 0, αA2 − A1 > 0at f = F ∗1 , A2 = 0, αA2 − A1 < 0

}⇒ unique solution

Example 16.4 Because of symmetry, f1 = f2 = 12.

O

16.6 Equal sacri�ce solution

Both players decrease their maximum payo�s, F ∗1 and F ∗2 with equal speed until feasiblesolution is found.

F ∗1 − f = F ∗2 − g(f)

or

α(F ∗1 − f) = F ∗2 − g(f) ⇔h(f) = αf − g(f)︸ ︷︷ ︸

strictly increasing

−αF ∗1 + F ∗2 = 0

123

Game theory

at f = f ∗1 , g(f ∗1 ) = F ∗2 , h = αf ∗1 − αF ∗1 < 0at f = F ∗1 , g(F ∗1 ) = f ∗2 , h = −f ∗2 + F ∗2 > 0

}⇒ unique solution

Example 16.5 Equal sacri�ce solution for previous example

1− f1 = 1− (1− f1)

1− f1 = f1 ⇒ f1 =1

2⇒ f2 = 1− 1

2=

1

2O

16.7 Kalai�Smorodinsky solution

Arc starts at status quo point and moves toward ideal point, then the last feasible pointis accepted as the solution.

g(f)− f ∗2 =F ∗2 − f ∗2F ∗1 − f ∗1︸ ︷︷ ︸S=slope

(f − f ∗1 )

h = Sf − g(f) + f ∗2 − Sf ∗1︸ ︷︷ ︸strictly increasing

= 0

at f = f ∗1 , h = f ∗2 − g(f ∗1 ) < 0at f = F ∗1 , h = SF ∗1 − Sf ∗1 > 0

}⇒ unique solution

Example 16.6 For previous problem the Kalai-Smorodinsky solution is f1 = f2 = 12

124

Game theory

O

Example 16.7 Consider con�ict with disagrement payo� vector (0, 0) and feasible set

H ={

(f1, f2) | f1, f2 ≥ 0, f2 ≤ 1− f 21

}1. Nash solution

max (1− f 21 − 0)(f1 − 0)

s.to 0 ≤ f1 ≤ 1

Objective f1 − f 31 is zero at endpoints f1 = 0 and f1 = 1.

Derivative:

1− 3f 21 = 0

f 21 =

1

3

f1 =

√3

3≈ 0.5774, f2 = g(f1) =

2

3≈ 0.6667

So solution is (0.5774,0.6667).

125

Game theory

2. Nonsymmetric Nash solution

maximize fα1 (1− f 21 )1−α

s. to 0 ≤ f1 ≤ 1.

By di�erentiation

αfα−11 (1− f 2

1 )1−α + fα1 (1− α)(1− f 21 )−α(−2f1) = 0

Divide by fα−11 (1− f 2

1 )−α to have

α(1− f 21 )− 2f 2

1 (1− α) = 0

α− αf 21 − 2f 2

1 + 2f 21α = 0

α = f 21 (2− α)

f 21 =

α

2− α, f1 =

√α

2− α, f2 = 1− f 2

1 =2− 2α

2− αIn the special case of α=1

2, the symmetric Nash solution is obtained:

f1 =

√12

2− 12

=

√1232

=1√3

=

√3

3,

f2 = 1− f 21 = 1− 1

3=

2

3.

3. Area monotonic solution

Total area: ∫ 1

0

(1− f 2)df =

[f − f 3

3

]1

0

=2

3,

126

Game theory

so

1

3=f1(1− f 2

1 )

2+

∫ 1

f1

(1− f 2)df =f1 − f 3

1

2+

[f − f 3

3

]1

f1

=f1 − f 3

1

2+

2

3− f1 +

f 31

3=

1

33f1 − 3f 3

1 + 4− 6f1 + 2f 31 = 2

h = f 3 + 3f − 2 = 0, h′ = 3f 2 + 3

f0 = 1, Newton's method:

f h h′

1 2 60.6667 0.2964 4.33350.5983 0.0091 4.07390.5961 0.0001 4.0660

So solution is (0.60,0.64).

4. Equal sacri�ce solution

1− (1− f 21 ) = 1− f1

f 21 + f1 − 1 = 0

f1 =−1 +

√1 + 4

2=−1 +

√5

2≈ 0.62

f2 = 1− f 21 = 0.62

So solution is (0.62,0.62).

5. Kalai-Smorodinsky solution

S = 1, so f1 − (1− f 21 ) + 0− 0 = 0

f 21 + f1 − 1 = 0, same as above

O

127

Game theory

17 Multiobjective optimization, concepts and methods

In the case of one objective

maximize f(x)

s. to x ∈ X

All values of f(x) when x runs through X are on the real line

Maximal solution has properties:

(i) maximal solution is at least as good as any other solution

(ii) there is no better solution

(iii) all maximal solutions are equivalent, i.e., they have the same objective value

In the case of several objectives no �optimal� solution existsIn the case of one objective, any two decisions x(1) and x(2) can be compared since

either f(x(1)) > f(x(2)), or f(x(1)) = f(x(2)), or f(x(1)) < f(x(2)).

In the case of multiple objectives, this is not true: for example(1

2

)and

(2

1

)cannot be compared. Ideal point is usually infeasible ⇒ there is no optimal solution.New concept is needed.

De�nition 17.1 A solution x∗ ∈ X is weakly nondominated, if there is no x ∈ X suchthat

fi(x) > fi(x∗) for all i

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Game theory

De�nition 17.2 A solution x∗ ∈ X is (strongly) nondominated, if there is no x ∈ Xsuch that

fi(x) ≥ fi(x∗)

for all i with strict inequality for at least one i. I.e., no objective can be improved withoutworthening another one.

Nondominated solutions satisfy only (ii), properties (i) and (iii) are not true anymore:f ∗ is worse than f in second objective, and both objective values are di�erent at thesepoints.

De�nition 17.3 (Payo� set or objective space)

H ={f | such that there exists x ∈ X with fi = fi(x),∀i

}.

Example 17.1

maximize x1 + x2, x1 − x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

Decision space:

129

Game theory

Letf1 = x1 + x2

f2 = x1 − x2

(+) x1 = f1+f22

(−) x2 = f1−f22

Constraints:

f1 + f2

2≥ 0 ⇒ f2 ≥ −f1

f1 − f2

2≥ 0 ⇒ f2 ≤ f1

3f1 + f2

2+f1 − f2

2≤ 4 ⇒ 3f1 + 3f2 + f1 − f2 ≤ 8 ⇒ 4f1 + 2f2 ≤ 8 ⇒ 2f1 + f2 ≤ 4

f1 + f2

2+ 3

f1 − f2

2≤ 4 ⇒ f1 + f2 + 3f1 − 3f2 ≤ 8 ⇒ 4f1 − 2f2 ≤ 8 ⇒ 2f1 − f2 ≤ 4

Payo� set:

130

Game theory

O

De�nition 17.4 A point f ∗ ∈ H is weakly nondominated, if there is no f ∈ H such thatfi > f ∗i for all i.

De�nition 17.5 A point f ∗ ∈ H is (strongly) nondominated, if there is no f ∈ H suchthat fi ≥ f ∗i for all i, with strict inequality for at least one i.

Example 17.2 In previous example all points of the linear segment connecting points(2, 0) and (4

3, 4

3) are weakly and strongly nondominated. O

Note. If f is strongly nondominated, then it is also weakly nondominated.

17.1 Existence of nondominated solutions

Example 17.3 No nondominated solution exists:

maximize x1, x2

s. to x1, x2 ≥ 0

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Game theory

O

Example 17.4 There is a unique nondominated solution, f1 = f2 = 1.

O

Example 17.5 There are multiple nondominated solutions

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Game theory

O

In Example 17.3, H was not bounded. Even if H is bounded there might be nonondominated solution:

Example 17.6

maximize x1, x2

s. to x21 + x2

2 < 1

O

Lemma 17.1 Let f ∈ H be an interior point of H. Then f must not be nondominated.

Proof. Take ε > 0 very small, then f + ε · 1 ∈ H but f + ε · 1 > f. �

Theorem 17.1 Assume H is nonempty, closed, and for all i and x ∈ X,

fi(x) ≤ F ∗i .

Then there is at least one nondominated solution.

Proof. Consider:

maximize g(f) = f1 + f2 + · · ·+ fI (52)

subject to f ∈ H.

(i) The setH(4) =

{f | f ∈ H, g(f) ≥ 4

}is bounded:

fi ≤ F ∗i

and

f1 + f2 + · · ·+ fI ≥ 4 ⇒fi ≥ 4−

∑j 6=i

fj ≥ 4−∑j 6=i

F ∗j

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Game theory

(ii) Let f be a feasible solution, then (52) is equivalent to

maximize g(f)

s. to f ∈ Hg(f) ≥ g(f)

where new feasible set is compact (closed and bounded) ⇒ there is optimal solution,that is obviously strongly nondominated. �

Note. All conditions are essential, as earlier examples 17.3, 17.6 show.

Corollary 1. Assume X is nonempty, closed, bounded and all objectives functions fi arecontinuous ⇒ there exists nondominated solution.

Corollary 2. Assume X consists of �nitely many points ⇒ there is at least onenondominated solution.

Example 17.7objectives

alternatives 2 3 2

1 4 33 3 24 2 5

First alternative is dominated:

(2, 3, 2) ≤ (3, 3, 2),

all others are nondominated. O

Since there are multiple nondominated solutions, additional preference information isneeded to �nd a special one. Each method is based on a particular way how preferencesare expressed.

17.2 Method of sequential optimization

Given: Ordinal preferences of objectives (I = number of objectives):

(f1 � f2 � · · · � fI)

Step 1.max f1(x)s.to x ∈ X

}optimum f ∗1

Step 2.max f2(x)s.to x ∈ X

f1(x) = f ∗1

optimum f ∗2

...

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Game theory

Step k.max fk(x)s.to x ∈ X

f1(x) = f ∗1...fk−1(x) = f ∗k−1

Process terminates if either a unique optimal solution is found in any of the steps, orwe proceeded I steps

Theorem 17.2 The solution is always nondominated. (Trivial)

Example 17.8 In Example 17.1 we determinated and illustrated the payo� set of problem

maximize x1 + x2, x1 − x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

If f1 � f2, then optimal solution is f1 = 2, f2 = 0, so

x1 = f1+f22

= 1 and x2 = f1−f22

= 1

If f2 � f1, then optimal solution is f1 = f2 = 43, so

x1 = f1+f22

= 43and x2 = f1−f2

2= 0

O

Example 17.9

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Game theory

O

No point (except the two endpoints) on the nondominated curve can be obtained assolution ⇒ we lose most of the nondominated solutions by the method selection.

Example 17.10objectives

alternatives2 3 2

1 4 3

3 3 2

⇒ 4 2 5

f1 � f2 � f3

alternative 4 is selected O

Problem: If there is a unique solution in an earlier step, the later objectives are notconsidered at all.

Modi�cation. Relaxing optimality conditions in steps:

Step k:

maximize fk(x)

s. to x ∈ Xf1(x) ≥ f ∗1 − ε1

...

fk−1(x) ≥ f ∗k−1 − εk−1

Theorem 17.3 Solution is weakly nondominated, and if solution in step I is unique, thenit is strongly nondominated.

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Game theory

17.3 ε-constraint method

Given: Most important objective and minimal acceptable lower bounds for all otherobjectives. If f1 is most important, then we solve:

maximize f1(x)

s. to x ∈ Xf2(x) ≥ ε2 (53)...

fI(x) ≥ εI

Theorem 17.4 Solution is always weakly nondominated.

137

Game theory

Proof. If not, there is an x ∈ X such that fi(x) > fi(x∗) (∀i) where x∗ is the solution.

Then x satis�es all constraints and gives better value for f1. Contradicts to the optimalityof x∗. �

Example 17.11 Solution is not necessarily strongly nondominated

O

Theorem 17.5 Let x∗ be a strongly nondominated solution. Then the bounds ε2, . . . , εIcan be selected so that x∗ is an optimal solution for problem (53).

Proof. Take εi = fi(x∗) for i ≥ 2. �

Note. By method selection we do not lose (strongly) nondominated solutions.

Example 17.12

maximize x1 + x2, x1 − x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

138

Game theory

Assume f1 � f2, ε2 = 1

maximize x1 + x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

x1 − x2 ≥ 1 ⇒ x2 ≤ x1 − 1

Solution is intercept of two lines:x2 = x1 − 1

3x1 + x2 = 43x1 + x1 − 1 = 4

4x1 = 5x1 = 5

4

x2 = 14

With payo� set it is easier:

139

Game theory

2f1 + f2 = 4

f2 = 1

f1 =3

2

x1 =f1 + f2

2=

32

+ 1

2=

5

4

x2 =f1 − f2

2=

32− 1

2=

1

4

O

Example 17.13objectives

alternatives 2 3 2

1 4 3

3 3 2

��4 �2 �5

Assume: f1 is most important, ε2 = 3, ε3 = 2. Alternative 4 is infeasible, and alter-native 3 is the choice. O

17.4 Weighting method

Given: c1, c2, . . . , cI positive weights,∑I

i=1 ci = 1, relative importances of objectives.Then we solve:

maximize c1f1(x) + · · ·+ cIfI(x) (54)

s. to x ∈ X

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Game theory

Theorem 17.6 Solution is always strongly nondominated.

Proof. If not, there is an x ∈ X such that fi(x) ≥ fi(x∗) for all i with strict inequality

for at least one i. Then ∑i

cifi(x) >∑i

cifi(x∗)

contradicting the optimality of x∗. �

Example 17.14 By the method selection we might lose nondominated solutions:

O

Theorem 17.7 Assume H is convex and x∗ is strongly nondominated Then there arenonnegative weights such that x∗ is optimal solution for problem (54).

Proof. LetH∗ =

{g | there exists f ∈ H such that g ≤ f

}

141

Game theory

which is also convex, and has the same nondominated points as H. Let f ∗ = f(x∗), whichis nondominated ⇒ boundary point. Theorem of separating hyperplanes implies theexistence of an I-dimensional vector c = (ci) such that

cT (f ∗ − f) ≥ 0 for all f ∈ H∗.

We next show that c ≥ 0. Assume next, ci < 0 for some i. Then for arbitrary ε > 0,

f = f ∗ − εei ∈ H∗(ei = (0, . . . , 0, 1, 0, . . . , 0)T

)and

cT (f ∗ − f) = cT εei = εci < 0,

contradiction. Notice: H ⊆ H∗, f ∗ ∈ H, and for all f ∈ H∗, cTf ∗ ≥ cTf ⇒ vectorf ∗ is an optimal solution of (54). �

Example 17.15 We do not have always positive weights:

O

Questions:

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Game theory

1. How can we guarantee that H is convex?

2. When do positive weights exist?

For Question 1:

Theorem 17.8 Assume X is convex and all fi are linear. Then H is convex.

Proof. f1, f

2∈ H ⇒ there exist x1, x2 ∈ X such that

f1

= f(x1) and f2

= f(x2).

Then for α ∈ [0, 1],

αf1

+ (1− α)f2

= αf(x1) + (1− α)f(x2) (f is linear)

= f

αx1 + (1− α)x2︸ ︷︷ ︸∈X

∈ H�

Example 17.16 If X is convex, and all fi are concave then H still might be nonconvex:X = [0, 1], f1(x) = x1, f2(x) = −x2

1

O

Theorem 17.9 Assume that X is convex and for all x ∈ X and g ≤ f(x), there existsan x ∈ X such that g = f(x) (that is, H is comprehensive). Assume furthermore that allfi are concave. Then H is convex.

Proof. f1, f

2∈ H ⇒ there exist x1, x2 ∈ X such that

f1

= f(x1) and f2

= f(x2).

Then for α ∈ [0, 1],

143

Game theory

αf1

+ (1− α)f2

= αf(x1) + (1− α)f(x2) (concavity of fi's)

≤ f

αx1 + (1− α)x2︸ ︷︷ ︸∈X

∈ H,so from assumption, there exists x ∈ X such that

α1f 1+ (1− α)f

2= f(x).

�

For Question 2:

Theorem 17.10 If X is a polyhedron, all fi are linear, and x∗ is a strongly nondominated

solution, then there are positive weights ci such that x∗ is optimal solution of (54).

Proof. Uses duality theorem twice, complicated. �

Example 17.17

maximize x1 + x2, x1 − x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

144

Game theory

Assume: c1 = c2 = 12

c1(x1 + x2) + c2(x1 − x2) = x1

So

maximize x1

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

Solution is x1 = 43, x2 = 0. O

Di�culties:

• Composite objectiveI∑i=1

cifi(x) has no meaning

• Solution changes if we change units of objectives

Solution for this problem: normalizing objectives to satisfaction levels.If a utility function is given for objective i with values in [0,1], then

f i(x) = Ui(fi(x)) = satisfaction level with value fi(x)

or by linear transformation into unit interval [0, 1].

Mi = max {fi(x) | x ∈ X}mi = min {fi(x) | x ∈ X}

f i(x) =fi(x)−mi

Mi −mi

∈ [0, 1] without units

I∑i=1

cif i(x) means average satisfaction level on the scale between worst and best case

scenarios.

Example 17.18

maximize x1 + x2, x1 − x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

We saw earlier that (from payo� set)

M1 = 2, m1 = 0, M2 =4

3, m2 = −4

3so

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Game theory

f 1(x) =x1 + x2 − 0

2− 0=

1

2(x1 + x2)

f 2(x) =x1 − x2 + 4

343− (−4

3)

=3

8(x1 − x2) +

1

2

Assume c1 = c2 = 12, then composite objective:

1

4(x1 + x2) +

3

16(x1 − x2) +

���1

4∼ 7x1 + x2

16

7x1 + x2 = 0

x2 = −7x1

7x1 + x2 = 1

x2 = −7x1 + 1

Optimal solution: x1 = 43, x2 = 0 O

Example 17.19 c1 = 1, c2 = 3, c3 = 2 without normalizing:

objectives∑cifi

alternatives 2 3 2 2 + 9 + 4 = 15

1 4 3 1 + 12 + 6 = 193 3 2 3 + 9 + 4 = 164 2 5 4 + 6 + 10 = 20

Alternative 4 is selectedWhen objectives are normalized:

i mi Mi

1 1 42 2 43 2 5

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Game theory

c1 = 1, c2 = 3, c3 = 2

objectives∑cif i

alternatives 1

312

0 13

+ 32

+ 0 = 116

0 1 13

0 + 3 + 23

= 113

23

12

0 23

+ 32

+ 0 = 136

1 0 1 1 + 0 + 2 = 3

Alternative 2 is selected. O

17.5 Distance�based methods

Given: Ideal point (computed or subjectively given) and a distance function % of I-dimensional vectors

We look for closest point to idea point.

In decision space:minimize %(f ∗, f(x)

)s. to x ∈ X

or (55)

in objective space:minimize %(f ∗, f

)s. to f ∈ H

Distance types between u = (ui) and v = (vi).

%∞(u, v) = maxi{ci|ui − vi|} (weighted %∞-distance)

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Game theory

Points with equal %∞ distance from f ∗

%1(u, v) =I∑i=1

ci|ui − vi| (weighted %1-distance)

Points with equal %1 distance from f ∗

%2(u, v) =

{I∑i=1

ci|ui − vi|2} 1

2

(weighted %2-distance)

Point with equal %2 distance from f ∗

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Game theory

%G(u, v) =I∏i=1

|ui − vi|ci (weighted geometric distance)

Points with equal %G distance from f ∗

Distance axioms may be violated. Let I = 2, c1 = c2 = 1. Then

1. % ((2, 3), (2, 2)) = 0 even if (2, 3) 6= (2, 2)

3. % ((1, 1), (1, 3)) + % ((1, 3), (3, 3)) < % ((1, 1), (3, 3))

0 + 0 < |3− 1| · |3− 1|0 < 4 contradiction

Assume ideal point satis�es:

f ∗i ≥ max {fi(x) | x ∈ X} (∀i)

Theorem 17.11 The optimal solution of (55) is always weakly nondominated, and ifoptimal solution is unique then it is strongly nondominated.

Proof. Obvious. �

Example 17.20

maximize x1 + x2, x1 − x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

149

Game theory

Ideal point:(2, 4

3

), c1 = c2

Minimizing %∞ distance: intercept of diagonal of square and Pareto frontier:

f2 −4

3= f1 − 2 ⇒ f2 = f1 −

2

3

f2 − 0 =0− 4

3

2− 43

(f1 − 2) ⇒ f2 = −2f1 + 4

intercept:

f1 −2

3= −2f1 + 4 ⇒

f1 =14

9, f2 =

8

9, x1 =

11

9, x2 =

3

9

Minimizing %1 distance:

f1 = f2 =4

3, x1 =

4

3, x2 = 0

Minimizing %2 distance: radius is perpendicular to Pareto frontier:

f2 −4

3=

1

2(f1 − 2) ⇒ f2 =

1

2f1 +

1

3

f2 = −2f1 + 4

1

2f1 +

1

3= −2f1 + 4

f1 =22

15, f2 =

16

15, x1 =

38

30, x2 =

6

30

Minimizing %G distance: zero distance is at

f1 = 2, f2 = 0, x1 = 1, x2 = 1

and

f1 = f2 =4

3, x1 =

4

3, x2 = 0

O

150

Game theory

Theorem 17.12 Assume that for all i there exist εi>0 such that for all x,

f ∗i > fi(x) + εi

and ci > 0. Then all solutions by the geometric distance are strongly nondominated.

Proof. Obvious. �

Special case: compromise programming, when all objectives are normalized

f i(x) =fi(x)−min{fi(x)}

max{fi(x)} −min{fi(x)}∈ [0, 1].

Example 17.21objectives

alternatives 2 3 2

1 4 33 3 24 2 5

Select c1 = c2 = 1, c3 = 2, f ∗ = (5, 5, 7) and % = %1

For alternative 1: %1 = |2− 5|+ |3− 5|+ 2|2− 7| = 15

For alternative 2: %1 = |1− 5|+ |4− 5|+ 2|3− 7| = 13

For alternative 3: %1 = |3− 5|+ |3− 5|+ 2|2− 7| = 14

For alternative 4: %1 = |4− 5|+ |2− 5|+ 2|5− 7| = 8

⇒ Alternative 4 is the choice O

Other variant of the method:Given: Nadir (worst case scenario computed or subjectively given) and a distance

function % of I-dimensional vectors.

We look for a feasible point with largest distance from nadir.

151

Game theory

Example 17.22 Best design, 5 alternatives to choose from:

AlternativesCost Reliability Resiliency -Vulnerabilitysaving above 95% % (scale)

1 1.4× 106 2.4 87 342 1.8× 106 2.2 76 413 2.0× 106 1.9 91 294 2.1× 106 1.6 89 405 2.1× 106 1.5 77 44

where

reliability: P(works until given time period T )

resiliency: how quickly a system is likely to recover or bounces back from failure oncefailure has occured

P(works at time period t+ 1 | failure at time period t)

vulnerability: for each failure event let s(X) denote a numerical indicator how severeis the consequence of a failure, and p(X) the probabilty that it occurs. Then

v =∑

X∈all failure modes

s(X)p(X)

Worst�case, if all objectives are on minimal value:

nadir = (1.4× 106, 1.5, 76, 29)

Geometric distance with equal c′is from the nadir is maximized

A1 : 0× 0.9× 11× 5 = 0

A2 : (0.4× 106)× 0.7× 0× 12 = 0

A3 : (0.6× 106)× 0.4× 15× 0 = 0

A4 : (0.7× 106)× 0.1× 13× 11 = 1.001× 107

A5 : (0.7× 106)× 0× 1× 15 = 0

⇒ A4 is selected O

Note. Same as Nash's bargaining solution

17.6 Direction�based methods

Given: A status quo point (worst case scenario or current situation) and a vector ofimprovement

152

Game theory

Starting from status quo point we improve objectives in given direction as much as wecan.

Similar to Kalai�Smorodinsky solutionLet f ∗ be status quo point, v the direction of improvement:

maximize t

s. to f ∗ + tv ∈ H

Vector v represents preference, larger vi value means that fi increases faster.Or alternative approach:

We relax objectives in the given direction until feasible solution is obtained.This method is mathematically represented by the following:

153

Game theory

minimize t

s. to f ∗ − tv ∈ H

where f ∗ is now the ideal point.

Example 17.23

maximize x1 + x2, x1 − x2

s. to x1, x2 ≥ 0

3x1 + x2 ≤ 4

x1 + 3x2 ≤ 4

Select

f ∗ = (0,−4

3), v = (1, 1)

f2 = f1 −4

3

f2 − 0 =−4

323

(f1 − 2)

f2 = f1 −4

3= −2f1 + 4

f1 =16

9, f2 =

4

9, x1 =

10

9, x2 =

6

9O

18 Dynamic games

18.1 Dynamic Systems

Discrete: x(t+ 1) = f(x(t), t) next state is given

154

Game theory

Continuous: x(t) = f(x(t), t) direction of change

Steady state: If state will remain the same for all future times:

x = f(x, t) or 0 = f(x, t)

Time invariant system:

x(t+ 1) = f(x(t)) or x(t) = f(x(t))

Linearization: around, steady state:

f(x) ≈ f(x) + f ′(x)︸ ︷︷ ︸Jacobian J(x)

(x− x)

Example 18.1 x = (x+ y)2 − 4, y = x2 + 3y − 4

(x+ y)2 = 4, x+ y = ±2

(i) x+ y = 2, y = 2− xx2 + 6− 3x− 4 = 0

x2 − 3x+ 2 = 0

x =

{12

y =

{10

(ii) x+ y = −2, y = −x− 2x2 − 3x− 6− 4 = 0

x2 − 3x− 10 = 0

x =

{−25

y =

{0−7

Linearizing around (1, 1):

f(x, y) ≈ f(x, y) + f ′x(x, y)(x− x) + f ′y(x, y)(y − y)

(x+y)2−4 ≈ 0+2(x+y)(x−1)+2(x+y)(y−1) = 4(x−1)+4(y−1) = 4x+4y−8

x2 + 3y − 4 ≈ 0 + (2x)(x− 1) + 3(y − 1) = 2(x− 1) + 3(y − 1) = 2x+ 3y − 5

O

Stability:Steady state x is locally asymptotically stable if there is an ε > 0 such that

%(x(0), x) < ε ⇒ limt→∞

x(t) = x (LAS in brief)

155

Game theory

Steady state x is globally asymptotically stable if regardless of x(0), limt→∞

x(t) = x

(AS in brief)

Theorem 18.1 Equilibrium x of nonlinear system is LAS, if the same holds for thelinearized system.

Note. Linearized system has the form

x(t+ 1) = x+ J(x)(x(t)− x) or x(t) = J(x)(x− x)

Note. Opposite of theorem is not true: x(t+ 1) = x(t)e−x(t)2

Proof.If x(0) > 0, then x(t) > 0 for all t > 0; if x(0) < 0, then x(t) < 0 for all t > 0;

|x(t+ 1)| = |x(t)|e−x(t)2 < |x(t)|

⇒ {x(t)} either decreases with zero lower bound, or increases with zero upper bound⇒ converges. Let x∗ be the limit. Then

x(t+ 1) = x(t)e−x(t)2

if t→∞,x∗ = x∗e−x

∗2

(1− e−x∗2)x∗ = 0 ⇒ x∗ = 0

Steady state:x = xe−x

2 ⇒ x = 0 ⇒ System is AS

Linearized system:

(xe−x2

)′ = e−x2

+ xe−x2

(−2x)∣∣∣x=0

= 1

156

Game theory

x(t+ 1) = x(t),

which is not A.S., since starting with any x(0) 6= x, x(t) never converges to x. �Question: How can I see that a linear system is asymptotically stable?

Discrete systems:x(t+ 1) = A x(t) + b

x = A x+ b

Subtractingx(t+ 1)− x = A(x(t)− x)

Let z(t) = x(t)− x, then z(t+ 1) = A z(t)Fact: System is asymptotically stable if and only if solution of homogeneous system

converges to zero.

z(1) = A z0

z(2) = A z(1) = A2z0

z(3) = A z(2) = A3z0

· · ·

z(t) = Atz0, so we need that At → 0 as t→∞

Assume:

A = T−1

λ1

λ2

. . .

λn

T

At = T−1Λ T T−1Λ T · · ·T−1Λ T = T−1ΛtT

= T−1

λ1

t

λ2t

. . .

λnt

T

This converges to zero ⇔ |λi| < 1 for all eigenvaluesNote. For linear systems local and global stability is the same.

Theorem 18.2 Linear discrete system is AS ⇔ for all eigenvalues |λi| < 1.

Continuous systems:x(t) = A x(t) + b

0 = A x+ b

Let z(t) = x(t)− x, then z(t) = A z(t) by subtracting these equations.Fact: System is asymptotically stable if and only if solution of homogeneous system

converges to zero.

A = T−1

λ1

λ2

. . .

λn

T

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Game theory

Solution of homogeneous equation is:

z(t) = T−1

eλ1t

eλ2t

. . .

eλnt

T z0

Proof. By substitution,

z(0) = T−1I T z0 = T−1T z0 = z0

z(t) = T−1

λ1e

λ1t

λ2eλ2t

. . .

λneλnt

T z0

A z(t) = T−1

λ1

λ2

. . .

λn

T T−1

eλ1t

eλ2t

. . .

eλnt

T z0,

which are the same. So, x is AS, if and only if eλt → 0 as t→∞ with all eigenvalues ofA. Let λ = α + iβ,

eλt = e(α+iβ)t = eαteiβt = eαt(cos βt+ i sin βt︸ ︷︷ ︸unit absolute value

)

So eλt → 0 as t→∞⇔ α < 0

Theorem 18.3 Linear continuous system is AS ⇔for all eigenvalues, Reλi < 0.

18.2 Best response dynamics

Discrete case: Moving into the direction toward best response:

xk(t+ 1) = xk(t) +Kk(Rk(x(t))− xk(t)) (1 ≤ k ≤ n)

where 1 ≥ Kk > 0 speed of adjustment (to avoid overshooting over steady state)Continuous case: Same idea:

xk(t) = Kk(Rk(x(t))− xk(t)) (1 ≤ k ≤ n)

Note. Steady state of these systems is always

Rk(x) = xk

⇔ equilibrium of n-person gameQuestion: How to check stability condition for n=2 ?

λ2 + pλ+ q = 0 (p, q real)

Lemma 18.1Reλ < 0⇔ p, q > 0

158

Game theory

Proof. ⇒ Complex roots:λ1 = α + iβ, λ2 = α− iβ

p = −(λ1 + λ2) = −2α > 0, q = α2 + β2

or real roots:λ1, λ2 < 0, p = −(λ1 + λ2) > 0, q = λ1λ2 > 0

⇐

λ =−p±

√p2 − 4q

2

Complex roots:

Re λ = −p2< 0

Real roots:both are negative, since p2 − 4q < p2

Lemma 18.2 |λ| < 1⇔ q < 1, ±p+ q + 1 > 0

Proof.

λ =−p±

√p2 − 4q

2

Real roots:

−2 < −p±√p2 − 4q < 2, (p2 − 4q ≥ 0 or q ≤ p2

4)√

p2 − 4q < 2− p,√p2 − 4q < 2 + p ⇒ −2 < p < 2

p2 − 4q < 4 + p2 − 4p p2 − 4q < 4 + p2 + 4p

0 < 1 + q − p 0 < 1 + p+ q

Complex roots:

λ = −p2± i

2

√4q − p2 (4q − p2 > 0 or q >

p2

4)

|λ|2 =p2

4+

4q − p2

4< 1 ⇔ q < 1

159

Game theory

Gradient adjustment, another dynamism:

xk = Kk ·∂ϕk∂xk

or

xk(t+ 1) = xk(t) +Kk∂ϕk∂xk

Only local information is needed. Only interior equilibrium are steady states.Comparison of discrete and continuous 2× 2 systems

Asymptotically stable continuous systems:

x = Kxf(x, y)

y = Kyg(x, y)

Jacobian:

J =

(Kxfx KxfyKygx Kygy

)Characteristic polynomial:

ϕ = (Kxfx − λ)(Kygy − λ)−KxfyKygx= λ2 − λ(Kxfx +Kygy︸ ︷︷ ︸

<0

) +KxKy((fxgy − fygx)︸ ︷︷ ︸+

)

Asymptotically stable discrete systems:

x(t+ 1) = x(t) +Kxf(x(t), y(t))

y(t+ 1) = y(t) +Kyg(x(t), y(t))

Jacobian:

J =

(1 +Kxfx KxfyKygx 1 +Kygy

)Characteristic polynomial:

ϕ = (1 +Kxfx − λ)(1 +Kygy − λ)−KxKygxfy= λ2 − λ(2 +Kxfx +Kygy) + (1 +Kxfx +Kygy +KxKyfxgy −KxKygxfy) = 0

Stability conditions:

q < 1,

Kxfx +Kygy +KxKy(fxgy − gxfy) < 0

p+ q + 1 > 0,

−2−Kxfx −Kygy + 1 +Kxfx +Kygy +KxKy(fxgy − gxfy) + 1 > 0⇔ fxgy − fygx > 0

−p+ q + 1 > 0,

2 +Kxfx +Kygy + 1 +Kxfx +Kygy +KxKy(fxgy − gxfy) + 1 > 0

4 + 2(Kxfx +Kygy) +KxKy(fxgy − gxfy) > 0

De�ne

Kxfx +Kygy = A, (fxgy − gxfy)KxKy = B,

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Game theory

then conditions areA+B < 0 B < −A

B > 0 B > 0

4 + 2A+B > 0 B > −2A− 4

Recall, for continuous systems: A < 0 and B > 0.Stability region for continuous system:

Stability region for discrete system:

Conclusion: Discrete system is much less stable, since stability region is much smaller.Jacobians of two cases satisfy

JD = JC + I,

so for eigenvalues

λD = λC + 1 or λC = λD − 1

AS of discrete system ⇒ same for continuous system|λD| < 1 ⇒ λD − 1 is in circle with center -1 and radius 1 ⇒ Re(λD − 1) < 0

161

Game theory

AS of continuous system ; same for discrete system

ReλC < 0 ; |λC + 1| < 1,

Notice that ReλC < 0 implies Re(λD) = Re(λC + 1) < 1, but ImλD can be anything,so |λC + 1| can be very large.

Consider discrete system

xk(t+ 1) = xk(t) +Kkfk(x)

consider smaller stepsize 1n, then

xk(t+1

n) = xk(t) +

Kk

nfk(x)

xk(t+1

n)− xk(t) =

Kk

nfk(x)

xk(t+ 1n)− xk(t)1n

= Kkfk(x)

n→∞xk(t) = Kkfk(x)

corresponding continuous system.

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Game theory

18.3 Cournot oligopoly

N = number of �rms;

xk = output of �rm k, k = 1, . . . , N ;

p

(N∑l=1

xl

)= unit price of product;

Ck(xk) = cost of �rm k, k = 1, . . . , N

Set of strategies for �rm k:Xk = [0, Lk] ,

Lk = capacity of �rm k, k = 1, . . . , N

Pro�t of �rm k:

ϕk(x1, . . . , xN) = xk · p

(N∑l=1

xl

)− Ck(xk)

18.4 Best response dynamics

We already saw that best response of �rm k is

Rk(sk) =

0 if p(sk)− C ′k(0) ≤ 0Lk if p(Lk + sk) + Lkp

′(Lk + sk)− C ′k(Lk) ≥ 0z∗k otherwise

with sk =∑

l 6=k xl, where z∗k is the unique solution of equation

p(xk + sk) + xkp′(xk + sk)− C ′k(xk) = 0.

We also saw that under assumptions

(i) p′ < 0;

(ii) p′ + xkp′′ ≤ 0;

(iii) p′ − C ′′k < 0

the best response functions are di�erentiable (except the boundary points between thethree cases) and

−1 < R′k(sk) ≤ 0.

In the case of discrete time scales the dynamic model is:

xk(t+ 1) = xk(t) +Kk ·

(Rk

(∑l 6=k

xl(t)

)− xk(t)

)(1 ≤ k ≤ N) (56)

where Kk > 0 is the speed of adjusment of �rm k.In the case of continuous time scales the model is:

xk(t) = Kk ·

(Rk

(∑l 6=k

xl(t)

)− xk(t)

)(1 ≤ k ≤ N) (57)

where Kk > 0 is the speed of adjustment of �rm k.In both cases �rm k moves into the direction of its best response.Before analysing stability, a matrix theoretical result is shown.

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Game theory

Lemma 18.3 If a, b ∈ RN , then det(I + a bT

)= 1 + bTa

Proof. By induction, if N = 1,

det(1 + ab) = 1 + ab = 1 + ba

N − 1 ⇒ N

DN = det

1 + a1b1 a1b2 . . . a1bN−1 a1bNa2b1 1 + a2b2 . . . a2bN−1 a2bN...

......

...aN−1b1 aN−1b2 . . . 1 + aN−1bN−1 aN−1bNaNb1 aNb2 . . . aNbN−1 1 + aNbN

Substract aN

aN−1� multiple of row N − 1 from row N , then

substract aN−1

aN−2� multiple of row N − 2 from row N − 1, then etc.

...substract a2

a1� multiple of row 1 from row 2.

Determinant does not change:

DN = det

1 + a1b1 a1b2 . . . a1bN−1 a1bN−a2a1

1 . . . 0 0...

......

...0 0 . . . 1 00 0 . . . − aN

aN−11

expanding with respect to last row

= DN−1 + (−1)N−1a1bN

(−a2

a1

)(−a3

a2

)· · ·(− aNaN−1

)= DN−1 + aNbN = 1 +

N−1∑k=1

akbk + aNbN = 1 +N∑k=1

akbk

= 1 + bTa.

�Corollary. Consider matrix

a1 b1 . . . b1

b2 a2 . . . b2...

......

bN bN . . . aN

,

its characteristic polynomial is:

ϕ(λ) = det

a1 − λ b1 . . . b1

b2 a2 − λ . . . b2...

......

bN bN . . . aN − λ

= det(D − λI + a 1T )

with

D =

a1 − b1

a2 − b2

. . .

aN − bN

, a =

b1

b2...bN

164

Game theory

and 1T = (1, 1, . . . , 1). Then

ϕ(λ) = det(D − λI) det(I + (D − λI)−1a 1T )

=N∏k=1

(ak − bk − λ) ·

[1 +

N∑k=1

bkak − bk − λ

].

The Jacobian of the discrete system (56) has the form

JD =

1−K1 K1r1 . . . K1r1

K2r2 1−K2 . . . K2r2...

... · · ·KNrN KNrN . . . 1−KN

with rk = R′k at equilibrium. Its characteristic polynomial (using above Corollary) is

ϕ(λ) =N∏k=1

(1−Kk(1 + rk)− λ)

[1 +

N∑k=1

Kkrk1−Kk(1 + rk)− λ

](58)

Eigenvalues are λ = 1−Kk(1 + rk) and/or solutions of equation

g(λ) = 1 +N∑k=1

Kkrk1−Kk(1 + rk)− λ

= 0 (59)

The values 1−Kk(1 + rk) are inside the unit circle if

Kk <2

1 + rk,

Notice thatlim

λ→±∞g(λ) = 1, lim

λ→1−Kk(1+rk)±0g(λ) = ±∞,

g′(λ) =N∑k=1

Kkrk

(1−Kk(1 + rk)− λ)2 < 0

(If rk = 0, then term is zero, same equation with smaller value of N .)

Figure. Shape of g(λ)

165

Game theory

So there are N − 1 roots between the poles and one before the �rst pole. All roots arefound, they are real and between −1 and 1 if and only if g(−1) > 0 ⇔

N∑k=1

Kkrk2−Kk(1 + rk)

> −1. (60)

In the linear case p(s) = A−Bs, Ck(xk) = ckxk + dk, so

Rk(sk) =

0 if A−Bsk − ck ≤ 0Lk if A−B(sk + Lk)−BLk − ck ≥ 0z∗k otherwise,

where z∗k is the solution of equation

A−Bsk − 2Bxk − ck = 0,

so

z∗k = −1

2sk +

A− ck2B

.

Therefore rk = −12in the case of interior EP. So asymptotical stability condition is:∑N

k=1

− 12Kk

2− 12Kk

> −1 or∑N

k=1Kk

4−Kk< 1

Assume Kk ≡ K, then this holds if

NK

4−K< 1

or

K <4

N + 1.

The Jacobian of the continuous system (57) has the form

JC =

−K1 K1r1 . . . K1r1

K2r2 −K2 . . . K2r2...

... · · ·KNrN KNrN . . . −KN

= JD − I,

and since eigenvalues of JD are less than 1, all eigenvalues of JC are real and negative.So continuous system (57) is always asymptotically stable.

19 Controllability of dynamic economic systems

19.1 Control of Linear Systems

Discrete system:x(t+ 1) = A x(t) +B u(t), x(0) = x0

where u(t) is the control at time period t.Continuous system:

x(t) = A x(t) +B u(t), x(0) = x0

where u(t) is the control at time period t.

166

Game theory

State is controllable to �nal state x∗ at t = T if there is a control function u(t) suchthat the solution satis�es x(T ) = x∗. State is completely controllable, if it is controllableto all x∗.

Question: How can we see that system is controllable or not?We discuss discrete system, but �nal conclusion remains true for continuous system

as well.

x(1) = A x0 +B u(0)

x(2) = A x(1) +B u(1)=A2x0 + A B u(0) +B u(1)

x(3) = A x(2) +B u(2)=A3x0 + A2B u(0) + A B u(1) +B u(2)

· · ·In general

x(t)=Atx0 +B u(t− 1) + A B u(t− 2) + · · ·+ At−1B u(0),

it can be rewritten as

(B,A B,A2B, · · · , At−1B)

u(t− 1)u(t− 2)

...u(0)

= x(t)− Atx0

So with condition x(T ) = x∗,

(B,A B,A2B, · · · , AT−1B)

u(T − 1)u(T − 2)

...u(0)

= x∗ − ATx0

Fact: This system has solution with all x∗ ⇔ rank (B,A B, · · · , AT−1B)︸ ︷︷ ︸K(T )

= n (size of

system)Fact:

rank(K(1)) ≤ rank(K(2)) ≤ · · · ≤ rank(K(n))︸ ︷︷ ︸adding more columns rank cannot become smaller

= rank(K(n+ 1)) = · · ·

Lemma 19.1 (Cayley-Hamilton theorem) Let

p(λ) = λn + an−1λn−1 + · · ·+ a1λ+ a0

be the characteristic polynomial of matrix A, then

p(A) = An + an−1An−1 + · · ·+ a1A+ a0I = 0

167

Game theory

Proof. If A = T−1

λ1

λ2

. . .

λn

T , then

Ak = T−1

λ1

λ2

. . .

λn

T T−1

λ1

λ2

. . .

λn

T · · ·T−1

λ1

λ2

. . .

λn

T

= T−1

λk1

λk2. . .

λkn

T .

So,

p(A) =∑n

k=0 akAk =

∑nk=0 akT

−1

λk1

λk2. . .

λkn

T =

T−1

∑nk=0 ak

λk1

λk2. . .

λkn

T = T−1

∑n

k=0 akλk1 ∑n

k=0 akλk2

. . . ∑nk=0 akλ

kn

T

= T−1

p(λ1)

p(λ2). . .

p(λn)

T = 0, since p(λk) = 0

Proof of Fact: From lemma,

An = −an−1An−1 − an−2A

n−2 − · · · − a1A− a0I

AnB = −an−1An−1B − an−2A

n−2B − · · · − a1A B − a0B

An+1B = −an−1AnB − an−2A

n−1B − · · · − a1A2B − a0A B,

where AnB is a linear combination of An−1B, · · · , A B and B. So AnB,An+1B, · · · areall linear combinations of columns of K(n). Adding dependent columns does not changerank. So if system does not become controllable for t = n, then it will not becomecontrollable later.

Theorem 19.1 System becomes controllable at t = n ⇔

rank(B,A B, · · · , An−1B︸ ︷︷ ︸Kalman matrix

) = n

For continuous systems (Since between 0 and any T > 0 there are ∞ many timeperiods):

Theorem 19.2 System is controllable at any T > 0 ⇔

rank(B,A B, · · · , An−1B) = n

168

Game theory

19.2 Control in oligopolies

Only single-product oligopolies will be examined. Let n denote the number of �rms, letp(s) = A − Bs (A,B > 0) be the price function and for k = 1, 2, . . . , n, let Ck(xk) =ckxk + dk (ck, dk > 0) denote the cost function of �rm k. Assume furthermore that themarket is controlled with the cost function of the �rms, which can be interpreted assubsidies, tax breaks, etc. Under this assumption the pro�t of �rm k can be expressed as

ϕk(x1, . . . , xn) = xk

(A−B

n∑l=1

xl

)− u(ckxk + dk), (61)

where u is the control variable. If tax and subsidy are imposed on unit output, thenequation (61) has to be modi�ed as follows:

ϕk(x1, . . . , xn) = xk

(A−B

n∑l=1

xk

)− (ckxku+ dk). (62)

The resulting dynamic models will be the same regardless which of the controls (61) or(62) is assumed. So in the following discussion, only the control (61) will be considered.

Assuming �rst discrete time scales and best response dynamics (Kk = 1), at eachtime period t ≥ 1, each �rm maximizes its expected pro�t

xk

(A−Bxk −B

∑l 6=k

xl(t− 1)

)− (ckxk + dk)u(t− 1)

Assuming interior optimum, the best response is

xk(t) = −1

2

∑l 6=k

xl(t− 1) +A− u(t− 1)ck

2B(k = 1, 2, . . . , n).

Introduce the new state variables

zk(t) = xk(t)−A

(n+ 1)B

to have a discrete control system based on best response dynamics

zk(t) = −1

2

∑l 6=k

zl(t− 1)− ck2B

u(t− 1) (63)

for k = 1, 2, . . . , n. Notice that this system can be written as

z(t) = A z(t− 1) + bu(t− 1) (64)

with

z =

z1

z2...zn

, A =

0 −1

2. . . −1

2

−12

0 . . . −12

......

...−1

2−1

2. . . 0

, and b =

− c1

2B

− c22B...− cn

2B

.

169

Game theory

It is well known from the theory of linear systems (see Theorem 19.1) that system (63)is completely controllable if and only if the rank of the Kalman-matrix

K = (b, A b, A2b, . . . , An−1b)

is n.Consider �rst the special case of a duopoly (that is when n = 2). In this case

K = (b, A b) =

(− c1

2Bc24B

− c22B

c14B

) (det(K) = − c2

1

8B2+

c22

8B2

)which has full rank if and only if c1 6= c2.

Assume next that n ≥ 3. We will now verify that always rank(K) < n, that is thesystem is not controllable. Observe �rst that

A = 12(I − E), or E = I − 2A

where I is the n × n identity matrix and E is the n × n real matrix where all elementsequal one. Since E2 = nE,

A2 =1

4

(I − 2E + E2

)=

1

4(I + (−2 + n)E) =

n− 1

4I +

2− n2

A

and

A2b =n− 1

4b+

2− n2

A b

By induction it follows that for all k ≥ 2, Akb is a linear combination of b and A b showingthat the columns of K are linearly dependent.

Let us modify the above model by assuming that di�erent �rms are controlled bydi�erent control variables. The modi�ed model can be written as

z(t) = A z(t− 1) +B u(t− 1) (65)

where A is as before,

B = diag(− c1

2B,− c2

2B, . . . ,− cn

2B

),

and u is an n-element vector. Since B itself is nonsingular, the �rst n columns of theKalman matrix K are linearly independent. Therefore rank(K) = n showing that thesystem is completely controllable.

Assume next continuous time scales. The dynamic system based on best responsedynamics now has the form

xk(t) = Kk

(−1

2

∑l 6=k

xl(t) +A− cku(t)

2B− xk(t)

)(k = 1, 2, . . . , n)

By introducing

Kk =Kk

2B(k = 1, 2, . . . , n)

it can be rewritten as

xk(t) = Kk

(−2Bxk(t)−B

∑l 6=k

xl(t) + A− cku(t)

)(k = 1, 2, . . . , n),

170

Game theory

where Kk is a positive constant for all k. Introduce the new state variables zk(t) =xk(t)− A

(n+1)Bto have the continuous control system

z(t) = A z(t) + bu(t)

where

A = K ·

−2B −B . . . −B−B −2B . . . −B...

......

−B −B . . . −2B

, b = K ·

−c1

−c2...−cn

with

K = diag(K1, K2, . . . , Kn)

Consider �rst the special case of a duopoly (that is when n = 2). Then the Kalmanmatrix is the following:

K = (b, A b) =

(−K1c1 B(2K

2

1c1 +K1K2c2)

−K2c2 B(K1K2c1 + 2K2

2c2)

)

which has full rank if and only if

K1c21 + 2(K2 −K1)c1c2 −K2c

22 6= 0.

For example, if K1 = K2, then this condition is equivalent to the assumption thatc1 6= c2.

If n ≥ 3, then the su�cient and necessary controllability conditions are even morecomplicated. However, if K1 = K2 = · · · = Kn = K, then

A = −BK(I + E)

and

A2 = B2K2(I + 2E + E2) = B2K

2(I + (n+ 2)E)

= B2K2(I + (n+ 2)

(−I − 1

BKA

))= −(n+ 1)B2K

2 · I − (n+ 2)BK · A.

showing that the columns of the Kalman matrix are linearly dependent. Similiarly to thediscrete case, it is easy to show by induction that for all k, Ak is a linear combination ofI and A.

The case of di�erent control for di�erent �rms as well as the cases of adaptive andextrapolative expectations can be discussed analogously to the discrete case.

20 Learning in oligopolies

A. n-�rms, price function p(S) = B−AS (S =∑n

l=1 xl), cost functions Ck(xk) = ck+dkxk.Assume A is known, B is not. At t=0, each �rm has initial estimate about B, say Bk(0).

At t ≥ 1, based on its estimate Bk(t), �rm k computes equilibrium, his productionlevel and equilibrium price based on his belief of the price function. The market gives theprice, what the �rm compares with his believed price and adjusts its belief based on this

171

Game theory

discrepancy. If believed price > actual price, then decreases his belief, if believed price <actual price then increases it and if they are equal, then he thinks his estimate was good,so does not change his belief.

(1) So player k does the following: He believes

ϕl = xl(Bk(t)− Axl − A∑i 6=l

xi)− (cl + dlxl)

is maximized,

Bk(t)− 2Axl − A∑i 6=l

xi − dl = 0

xl = −1

2

∑i 6=l

xi +Bk(t)− dl

2A

is best response of player l believed by �rm k. Notice with S =∑n

l=1 xl

xl = −1

2(S − xl) +

Bk(t)− dl2A

2xl = −S + xl +Bk(t)− dl

A

xl = −S +Bk(t)− dl

A

By adding

S = −nS +nBk(t)−

∑nl=1 dl

A

S =nBk(t)−

∑nl=1 dl

(n+ 1)A

is the total production believed by player k, so he thinks the price is

Bk(t)− AS = Bk(t)−nBk(t)−

∑nl=1 dl

n+ 1=Bk(t) +

∑nl=1 dl

n+ 1

and he produces

xk = −S +Bk(t)− dk

A= −nBk(t)−

∑nl=1 dl

(n+ 1)A+Bk(t)− dk

A

=Bk(t) +

∑nl=1 dl − (n+ 1)dk

(n+ 1)A

(2) In reality, all �rms think the same way, each produces his believed equilibriumoutput, so actual total output becomes

n∑k=1

Bk(t) +∑n

l=1 dl − (n+ 1)dk(n+ 1)A

=

∑nk=1 Bk(t)−

∑nk=1 dk

(n+ 1)A

So actual market price is

B − AS = B −∑n

k=1Bk(t)−∑n

k=1 dkn+ 1

=(n+ 1)B −

∑nk=1Bk(t) +

∑nk=1 dk

n+ 1

172

Game theory

(3) For �rm k, actual price-believed price:

∆pk(t) =(n+ 1)B −

∑nl=1Bl(t) +

∑nl=1 dl

n+ 1− Bk(t) +

∑nl=1 dl

n+ 1

=1

n+ 1[(n+ 1)B −

n∑l=1

Bl(t)−Bk(t)]

So dynamic models:Continuous:

Bk(t) =Kk

n+ 1[(n+ 1)B −

n∑l=1

Bl(t)−Bk(t)]

Discrete:

Bk(t+ 1) = Bk(t) +Kk

n+ 1[(n+ 1)B −

n∑l=1

Bl(t)−Bk(t)]

(4) Steady states:

(n+ 1)B −n∑l=1

Bl −Bk = 0

Bk = identical, (n+ 1)B − (n+ 1)B = 0

B = B true knowledge.

(5) Stability: Linear modelsContinuous case:

J =

− 2K1

n+1− K1

n+1. . . − K1

n+1

− K2

n+1− 2K2

n+1. . . − K2

n+1...

......

− Kn

n+1− Kn

n+1. . . −2Kn

n+1

= D + a 1T

with

D =

− K1

n+1

− K2

n+1. . .

− Kn

n+1

, a =

− K1

n+1

− K2

n+1...

− Kn

n+1

From Corollary of Lemma 18.3 with ak = −2Kk

n+1, bk = − Kk

n+1,

ϕ(λ) =n∏k=1

(− Kk

n+ 1− λ)(1 +

n∑k=1

− Kk

n+1

− Kk

n+1− λ

)

Eigenvalues are

λ = − Kk

n+ 1< 0

or roots of

g(λ) =∑ Kk

Kk + (n+ 1)λ= −1

g(±∞) = 0, limλ→− Kk

n+1±0

g(λ) = ±∞

g′(λ) =∑ −Kk(n+ 1)

(Kk + (n+ 1)λ)2< 0

173

Game theory

All roots are negative ⇒ system is always A.S.Discrete case:

J =

1− 2K1

n+1− K1

n+1. . . − K1

n+1

− K2

n+11− 2K2

n+1. . . − K2

n+1...

......

− Kn

n+1− Kn

n+1. . . 1− 2Kn

n+1

with ak = 1− 2Kk

n+ 1, bk = − Kk

n+ 1

ϕ(λ) =n∏k=1

(1− Kk

n+ 1− λ)(1 +

n∑k=1

− Kk

n+1

1− Kk

n+1− λ

)

Eigenvalues are

λ = 1− Kk

n+ 1

or roots of

g(λ) =n∑k=1

Kk

n+1

1− Kk

n+1− λ

= 1

Note, |1− Kk

n+1| < 1⇔ 1− Kk

n+1> −1⇔ Kk < 2(n+ 1) always holds.

g(±∞) = 0, limλ→1− Kk

n+1±0

g(λ) = ∓∞

g′(λ) =n∑k=1

Kk

n+1

(1− Kk

n+1− λ)2

> 0

174

Game theory

All roots are real and less than 1. They are greater than -1 ⇔ g(−1) < 1 or

n∑k=1

Kk

n+1

2− Kk

n+1

< 1 or

n∑k=1

Kk

2(n+ 1)−Kk

< 1 or

all speeds of adjustment are su�ciently small.B. Same as before, but in price function p(S) = B−AS, B is known but A is unknown.

Learning scheme is also the same.(1) Player k does the following:It believes that total production will be

S =nB −

∑nl=1 dl

(n+ 1)Ak(t),

his believed price isB +

∑nl=1 dl

n+ 1,

he produces his believed equilibrium output

xk =B +

∑nl=1 dl − (n+ 1)dk

(n+ 1)Ak(t)

so total output of industry becomes

n∑k=1

1

Ak(t)(n+ 1)[B +

n∑l=1

dl − (n+ 1)dk].

Then actual price-believed price becomes

∆pk(t) =

(B − A

n∑k=1

1

Ak(t)(n+ 1)[B +

n∑l=1

dl − (n+ 1)dk]

)− B +

∑nl=1 dl

n+ 1,

Notice that all players have equal ∆pk(t) values = ∆p(t).

175

Game theory

Dynamic systems:xk(t) = Kk∆pk(t)

orxk(t+ 1) = xk(t) +Kk∆pk(t)

Steady state ⇔ ∆p(t) = 0, one equation for n unknown steady state Ak values ⇒in�nitely many steady states ⇒ no learning is possible.

21 Partial cooperation

n = 2 symmetic oligopoly,p(x+ y) = B − A(x+ y)

C1(x) = αx, C2(y) = αy

Pro�t functions:ϕ1 = x(B − Ax− Ay)− αxϕ2 = y(B − Ax− Ay)− αy

γ=cooperation level, 0 < γ < 1.Payo� functions:

ψ1 = x(B − Ax− Ay)− αx+ γ(y(B − Ax− Ay)− αy)

∂ψ1

∂x= B − 2Ax− Ay − α− γyA = 0

x =B − Ay − α− γyA

2A=B − α− Ay(1 + γ)

2A

y =B − Ax− α− γxA

2A=B − α− Ax(1 + γ)

2ASymmetry:

x =B − α− Ax(1 + γ)

2A2Ax = B − α− Ax− Axγ

x =B − αA(3 + γ)

= y, x+ y =2(B − α)

A(3 + γ)

Both are decresing as γ increases ⇒ price increases. Untitrust authorities watch increasedprice! Pro�ts:

ϕ1 =B − αA(3 + γ)

(B − 2(B − α)

3 + γ− α)︸ ︷︷ ︸

3B+Bγ−2B+2α−3α−αγ=(B−α)(1+γ)

ϕ1 =(B − α)2(1 + γ)

A(3 + γ)2

(1 + γ

(3 + γ)2)′ =

1(3 + γ)2 − (1 + γ)2(3 + γ)

(3 + γ)4=

3 + γ − 2(1 + γ)

(3 + γ)3

numerator: 3 + γ − 2− 2γ = 1− γ > 0

So increasing value of γ increases pro�t!! In the expense of the consumer.Special cases. γ = 0, noncooperativeγ = 1, complete cooperation.

176

Game theory

22 Certainty equivalent

x= random variable, E(x) = µ, V ar(x) = σ2

u(x)=utility function

u(x) ≈ u(x) + u′(x)(x− x) +u′′(x)

2(x− x)2 + R3(x)︸ ︷︷ ︸

small, neglected

(66)

E(u(x)) = u(x) + u′(x) (E(x)− x)︸ ︷︷ ︸0

+u′′(x)

2V ar(x)

For any x,

u(x) = u(x) + u′(x)(x− x) + R2(x)︸ ︷︷ ︸small, neglected

(67)

For certainty equivalent we need u(x) = E(u(x)),

u(x) +u′′(x)

2V ar(x) = u(x) + u′(x)(x− x)

x− x =u′′(x)

2u′(x)V ar(x)

x = x+u′′(x)

2u′(x)︸ ︷︷ ︸α

V ar(x)

where α shows risk acceptance.Problems: 1. In (66) quadratic, in (67) linear Taylor's polynomial is used, so error

in (67) has the same order of magnitude as last considered term in (66).2. x︸︷︷︸+αV ar(x)︸ ︷︷ ︸: terms have di�erent units.

Modi�cation : New (67):

u(x) = u(x) + u′(x)(x− x) +u′′(x)

2(x− x)2 +R3(x) (68)

So

u(x) +u′′(x)

2V ar(x) = u(x) + u′(x)(x− x) +

u′′(x)

2(x− x)2

Let x− x = ∆,u′′(x)∆2 + 2u′(x)∆− u′′(x)V ar(x) = 0

∆ =−2u′(x)±

√4u′(x)2 + 4u′′(x)2V ar(x)

2u′′(x)

or by dividing equation by 2u′(x),

α∆2 + ∆− αV ar(x) = 0

∆ =−1±

√1 + 4α2V ar(x)

2α

177

Game theory

Since if V ar(x) = 0, we need ∆ = 0, so the only reasonable solution is

∆ =

√1 + 4α2V ar(x)− 1

2α,

and so x = x+ ∆Application to oligopoly: If price function is uncertain,

ϕk = xk[p(s) + ξk]− Ck(xk), E(ξk) = 0, V ar(ξk) = σ2k

E(ϕk) = xkp(s)− Ck(xk), V ar(ϕk) = x2kσ

2k,

Since we maximize, certainty equivalent is

ϕk = xkp(s)− Ck(xk)− αkx2kσ

2k

= xkp(s)− (Ck(xk) + αkx2kσ

2k︸ ︷︷ ︸

new cost function

)

or

ϕk = xkp(s)− (Ck(xk) +

√1 + 4α2

kx2kσ

2k − 1

2αk︸ ︷︷ ︸new cost function again

)

23 Illustrative case studies

23.1 Example 1. Selecting a restaurant for lunch

Decision makers: 4 people to go together from di�erent locations

Alternatives: Chinese (C), American (A), Mexican (M), Italian (I), no food (N)

Attributes:

A1 Quality/taste

A2 Quantity/amount

A3 Price

A4 Service/speed

A5 Distance to restaurant

Problem: Find collective decision

Step 1: Individual preferences assessed

Step 2: Collective choice is made

Individual preferences:

Individual ratings of Decision maker No. 1Attributes C A M I N ImportanceA1 (0-100) 75 80 95 65 0 0.3A2 (0-100) 80 60 70 70 0 0.3A3 ($2-12) 5 8 7 11 0 0.2A4 (0-40 mins) 15 25 20 35 0 0.1A5 (0-8 miles) 0.7 0.8 1.5 2.75 0 0.1

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Game theory

A1, A2 are given in satisfaction levels. A3, A4 and A5 have to be given in a similarway: How satis�ed one is with an attribute.

We cannot use the numbers directly from the table; money, minutes, miles cannot becompared and added directly. We need common measures of satisfaction to be introducedand used.

Value function for price (A3)

Value in (0,100) scale

V (0) = 100V (12) = 0

}Linear function: V (P ) = 100− 100

12· P

Therefore:

V (5) = 100− 100

12(5) = 58.3%

V (8) = 100− 100

12(8) = 33.3%

V (7) = 100− 100

12(7) = 41.7%

V (11) = 100− 100

12(11) = 8.3%

V (0) = 100− 100

12(0) = 100%

Value function for service (A4)

Value in (0,100) scale

179

Game theory

If people are in a hurry, then 20 minutes is considered a slow service, so 25% satisfactionis given there. Between 0 and 20 minutes, and between 20 and 40 minutes value functionlinearly decreases:

V1(0) = 100V1(20) = 25

}Linear function: V1(S) = 100− 300

80(S)

V2(20) = 25V2(40) = 0

}Linear function: V2(S) = 50− 100

80(S)

Therefore

V (15) = V1(15) = 43.8%

V (25) = V2(25) = 18.8%

V (20) = V1(20) = 25%

V (35) = V2(35) = 6.3%

V (0) = V1(0) = 100%

Value function for distance (A5)

Value in (0,100) scale

180

Game theory

Similar reason for distance as for service time, 4 miles already considered a largedistance:

V1(0) = 100V1(4) = 25

}Linear function: V1(D) = 100− 300

16(D)

V2(4) = 25V2(8) = 0

}Linear function: V2(D) = 50− 100

16(D)

Therefore

V (0.7) = V1(0.7) = 86.9%

V (0.8) = V1(0.8) = 85%

V (1.5) = V1(1.5) = 71.9%

V (2.75) = V1(2.75) = 48.4%

V (0) = V1(0) = 100%

Table with satisfaction levels for Decision maker No. 1Attributes C A M I N ImportanceA1 75 80 95 65 0 0.3A2 80 60 70 70 0 0.3A3 58.3 33.3 41.7 8.3 100 0.2A4 43.8 18.8 25 6.3 100 0.1A5 86.9 85 71.9 48.4 100 0.1

Individual average satisfaction level for each alternative is the weighted sum:

For C: 75(0.3) + 80(0.3) + 58.3(0.2) + 43.8(0.1) + 86.9(0.1) = 71.23

For A: 80(0.3) + 60(0.3) + 33.3(0.2) + 18.8(0.1) + 85(0.1) = 59.04

For M: 95(0.3) + 70(0.3) + 41.7(0.2) + 25(0.1) + 71.9(0.1) = 67.53

For I: 65(0.3) + 70(0.3) + 8.3(0.2) + 6.3(0.1) + 48.4(0.1) = 47.63

For N: 0(0.3) + 0(0.3) + 100(0.2) + 100(0.1) + 100(0.1) = 40

In summary: Preference for Decision maker 1:

C �M � A � I � N

Collective choice for Restaurant Example:After individual preferences are assessed for the other three persons as well (similar

to Person 1 shown above) the following preference table is established:

C A M I N PowerPerson 1 1 3 2 4 5 5Person 2 3 1 4 5 2 8Person 3 1 4 5 2 3 6Person 4 4 5 2 1 3 10

Total: 29

Majority Rule (voting)

181

Game theory

Persons 1 & 3 choose C as their �rst choice. Using the power factors, the total for Cis 5+6=11

Person 2 was the only one choosing A as the �rst choice, hence A : 8

Person 4 selects I, so I : 10

Nobody selects M and N, so M,N : 0

Most votes were given to C, so Chinese restaurant is the collective choice.The drawback with this technique is that it only considers �rst choices, later rankings

are ignored.

Weighting (Borda count)

Weighted sum of rankings is computed for each alternative (weight · ranking), the onewith the lowest sum is the �nal choice.

For:

C: 1 · 5 + 3 · 8 + 1 · 6 + 4 · 10 = 75

A: 3 · 5 + 1 · 8 + 4 · 6 + 5 · 10 = 97

M: 2 · 5 + 4 · 8 + 5 · 6 + 2 · 10 = 92

I: 4 · 5 + 5 · 8 + 2 · 6 + 1 · 10 = 82

N: 5 · 5 + 2 · 8 + 3 · 6 + 3 · 10 = 89

Smallest number is at C, so Chinese restaurant is the collective choice

Hare system

First we can delete M or N , both got no �rst ranking. For example, deleting N, newtable

C A M I Power1 3 2 4 52 1 3 4 81 3 4 2 63 4 2 1 10

Deleting M, new table

C A I Power1 2 3 52 1 3 81 3 2 62 3 1 10

Deleting A, new table

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Game theory

C I Power1 2 51 2 81 2 62 1 10

Total weights

for C = 19

for I = 10

So, C is the social choice.

Pair-wise comparison

This technique utilizes a pair-wise comparison by choosing the better between two optionsfor all individuals. This comparison continues with the others until we arrive at the �nalchoice.

Starting from the top

Choice between C and A

C � A by total power 5 + 6 + 10 = 21

A � C 8

C is the winner

Choice between C and M

C �M by total power 5 + 8 + 6 = 19

M � C 10

C is the winner again

Choice between C and I

C � I by total power 5 + 8 + 6 = 19

I � C 10

C is the winner again

Choice between C and N

C � N by total power 5 + 6 = 11

N � C 8 + 10 = 18

N is the winner

183

Game theory

So N (no lunch) is the collective social choice.

The drawback with this technique is that the choice may depend on the order in whichcomparisons are made.

By reversing the order of pair-wise comparisons for the example starting from thebottom with N then:

Choice between N and I

N � I by 8

I � N 5 + 6 + 10 = 21

I is the winner, previous winner is out

Choice between I and M

I �M by 6 + 10 = 16

M � I 5 + 8 = 13

I is the winner

Choice between I and A

I � A by 6 + 10 = 16

A � I 5 + 8 = 13

I remains winner

Choice between I and C

I � C by 10

C � I 5 + 8 + 6 = 19

C is the winner

So C (Chinese restaurant) is the collective choice, di�erent than before.Contradictory preferences may occur at the collective level.By comparing all pairs the preference graph becomes:

184

Game theory

Consider:

C � I � AC � AConsistent

However:

I � N � CbutC � IContradiction! We usually delete such cycles from preference graphs.

23.2 Example 2. Buying a family car

Decision makers: Father, mother, child

Alternatives: Volvo (V), Geo Metro (GM), Dodge Status (DS), Corvette (C)

Attributes:

A1 Style/look

A2 Speed

A3 Safety

A4 Maintenance

A5 Gas usage

A6 Price

A7 Comfort

Problem: Find collective family decision

Step 1: Individual preferences assessed

Step 2: Collective choice is made

Individual preferences:

185

Game theory

Individual ratings of father

Attributes V GM DS C ImportanceA1 (0-100) 90 55 75 70 0.05A2 (mph) 150 80 120 150 0.25A3 (0-100) 100 50 90 75 0.25A4 (0-100) 90 70 80 50 0.15A5 (mpg) 20 35 25 10 0.13A6 ($1000s) 40 10 20 55 0.10A7 (0-100) 95 60 80 70 0.07

A1, A3, A4 and A7 are given in satisfaction levels (0-100) while A2, A5 and A6 arenot. Therefore value functions should be assessed for A2, A5 and A6.

Value function for speed (A2)

Value in (0,100) scale

Father does not want a fast car, since teenage child would endanger himself:

V2(75) = 100V2(100) = 0

}Linear function V2(S) = 400− 4S

Therefore:

V (150) = 0

V (80) = 80

V (120) = 0

V (150) = 0

Value function for mileage (A5)

Value in (0,100) scale

186

Game theory

Better gas mileage means better satisfaction uniformly:

V (0) = 0V (40) = 100

}Linear function V (M) =

100

40(M)

Therefore

V (20) = 50

V (35) = 87.5

V (25) = 62.5

V (10) = 25

Value function for price (A6)

Value in (0,100) scale

V2(20) = 100V2(30) = 50

}Linear function V2(P ) = 200− 5(P )

V3(30) = 50V3(60) = 0

}Linear function V3(P ) = 100− 100

60(P )

Therefore:

V (40) = V3(40) = 33.3

V (10) = 100

V (20) = 100

V (55) = 8.3

187

Game theory

Table with satisfaction levels:Attributes V GM DS C ImportanceA1 (0-100) 90 55 75 70 0.05A2 (mpg) 0 80 0 0 0.25A3 (0-100) 100 50 90 75 0.25A4 (0-100) 90 70 80 50 0.15A5 (mpg) 50 87.5 62.5 25 0.13A6 ($1000s) 33.3 100 100 8.3 0.10A7 (0-100) 95 60 80 70 0.07

Average satisfaction level is computed by weighted sums:

For V: 90 · 0.05 + 0 · 0.25 + 100 · 0.25 + 90 · 0.15 + 50 · 0.13 + 33.3 · 0.10 + 95 · 0.07 = 59.48

For GM: 55·0.05+80·0.25+50·0.25++70·0.15+87.5·0.13+100·0.10+60·0.07 = 71.33

For DS: 75 ·0.05 + 0 ·0.25 + 90 ·0.25 + 80 ·0.15 + 62.5 ·0.13 + 100 ·0.10 + 80 ·0.07 = 61.98

For C: 70 · 0.05 + 0 · 0.25 + 75 · 0.25 + 50 · 0.15 + 25 · 0.13 + 8.3 · 0.10 + 70 · 0.07 = 38.73

In summary: Preference of father as single decision maker:

GM � DS � V � C

Collective choice for family car example:After individual preferences are assessed for all members of the family, we obtain the

following preference table:

V GM DS C PowerFather 3 1 2 4 4Mother 2 3 1 4 3Child 3 4 2 1 3

Total 10

In this example, the Father pays for the car, so he has slightly higher power than othersin the family

Majority rule (voting)

Total vote for

V: 0

GM: 4

DS: 3

C: 3

The collective choice is GM.

Weighting (Borda Count)

V: 3 · 4 + 2 · 3 + 3 · 3 = 27

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Game theory

GM: 1 · 4 + 3 · 3 + 4 · 3 = 25

DS: 2 · 4 + 1 · 3 + 2 · 3 = 17

C: 4 · 4 + 4 · 3 + 1 · 3 = 31

The collective choice is DS.

Hare system

Deleting V, new table

GM DS C Power1 2 3 42 1 3 33 2 1 3

DS and C are with lowest vote, deleting DS new table

GM C Power1 2 41 2 32 1 3

Total vote

For GM=7

For C=3

So GM is the social choice.If we delete C instead of DS, then new table

GM DS Power1 2 42 1 32 1 3

Total vote

For GM=4

For DS=6

So DS is the social choice.

Pair-wise comparison

V or GM:

V � GM 3+3=6

GM � V 4

V is the winner

189

Game theory

V or DS:

V � DS 0

DS � V 4 + 3 + 3 = 10

DS is the winner

DS or C:

DS � C 4+3=7

C � DS 3

The collective choice is DS

By comparing all pairs, the preference graph becomes:

C is less preferred than all the others. After C is eliminated, GM is the second leastpreferred since both V and DS are more preferred than GM. After GM is eliminated, thenbetween V and DS, DS is collectively more preferred.

Very consistent preferences.

23.3 Example 3. Restoration of a chemical land�ll

Alternatives:

1. Digging and disposing

2. In situ remediation with established technology

3. In situ remediation with new technology

4. Ex situ remediation with established technology

5. Ex situ remediation with new technology

Criteria:

190

Game theory

C1: human health (death rate)

C2: clean-up level

C3: life cycle cost (106$)

C4: timing

C5: social and political issues

Alternatives versus criteria

A1 A2 A3 A4 A5

C110−6 10−6 2 · 10−6 10−6 2 · 10−6

10% 10% 10% 10% 10%

C20 10−3 10−2 10−3 5 · 10−3

0% 30% 30% 20% 20%

C37.6 3.32 2.9 4.8 3.8

10% 20% 40% 20% 40%

C40.15 0.6 0.6 0.1 0.110% 20% 30% 15% 20%

C50.8 1.2 1.2 1 15% 10% 30% 10% 30%

For each entry

{mean valuestandard deviation in % of mean value

Decision makers (interest groups):

Importances of criteria for decision makersC1 C2 C3 C4 C5 Weights

site owners 0.25 0.20 0.70 0.25 0.20 0.50stake holders 0.50 0.60 0.25 0.05 0.60 0.25regulators 0.25 0.20 0.05 0.70 0.20 0.65

Last column gives weights for the decision makers. The combined weights of thecriteria are as follows:

W1 = 0.25(0.50) + 0.50(0.25) + 0.25(0.65) = 0.4125

W2 = 0.20(0.50) + 0.60(0.25) + 0.20(0.65) = 0.3800

W3 = 0.70(0.50) + 0.25(0.25) + 0.05(0.65) = 0.4450

W4 = 0.25(0.50) + 0.05(0.25) + 0.70(0.65) = 0.5925

W5 = 0.20(0.50) + 0.60(0.25) + 0.20(0.65) = 0.3800

Normalized table:

A1 A2 A3 A4 A5 WeightsC1 1 1 0 1 0 0.4125C2 1 0.9 0 0.9 0.5 0.3800C3 0 0.91 1 0.6 0.81 0.4450C4 0.9 0 0 1 1 0.5925C5 1 0 0 0.5 0.5 0.3800

191

Game theory

Weighting method with deterministic data:

A1: 1.706

A2: 1.159

A3: 0.445

A4: 1.804

A5: 1.332

A4 � A1︸ ︷︷ ︸small di�erence

� A5 � A2 � A3

Simulation results (N = 100, 000):

Normally distributed simulated data were used with six distance based methods,and probabilities of being optimal for all alternatives were obtained by relativefrequencies:

Minimize Maximizel∞ l1 l2 l∞ l1 l2

A1 0.10 0.22 0.30 0.25 0.33 0.11A2 0.01 0.02 0.04 0.01 0.01 0.01A3 0.00 0.01 0.01 0.00 0.01 0.01A4 0.82 0.75 0.63 0.71 0.54 0.82A5 0.07 0.00 0.02 0.03 0.11 0.05

Clearly A4 is best, largest probability with all methods.

Accuracy of relative frequencies:Markov inequality

X ≥ 0 ⇒ P (X ≥ a) ≤ E(X)

a

Proof.

E(X) =

∫ ∞0

xf(x)dx ≥∫ ∞a

xf(x)dx ≥∫ ∞a

af(x)dx

= a

∫ ∞a

f(x)dx = aP (X ≥ a)

Chebyshev inequality

E(X) = µ, V ar(X) = σ2

Let X ⇐ (X − µ)2, a⇐ ε2

P ((X − µ)2 ≥ ε2) ≤ E((X − µ)2)

ε2

P (|X − µ| ≥ ε) ≤ σ2

ε2

or

1− P (|X − µ| ≥ ε) ≥ 1− σ2

ε2

192

Game theory

P (|X − µ| < ε) ≥ 1− σ2

ε2

ε = kσ,

P (|X − µ| < kσ) ≥ 1− 1

k2

X=binomial, E(X) = Np, V ar(X) = Npq

P (|X −Np| < Nε) ≥ 1− Npq

N2ε2

P (|XN− p| < ε) ≥ 1− pq

Nε2≥ 1− 1

4Nε2

Since

pq = p(1− p) ≤ 1

4⇒ −pq ≥ −1

4

In our case ε = 0.01, N = 105

P(|error| ≤ 0.01) ≥ 1− 1

4 · 105 · 10−4= 1− 1

40= 0.975 = 97.5%

193