NORMAL FORM GAMES - euler.fd.cvut.czeuler.fd.cvut.cz/predmety/game_theory/lecture... · 2005. 11....

NORMAL FORM GAMES

☛ Example. Advertising Strategies

Two mountain hotels in the same district, Krakonoš and Trau-tenberg, compete for tourists from three different countries: Ger-many, Czech Republic and Poland. The capacity of both hotels issufficient for the accommodation of all tourists in only one of them.Both hotels have financial resources for the advertising campaignin only one country, the effectiveness of their campaigns is thesame. If only one hotel runs the campaign in a given country, itgains all tourists from this country and hence the following profit:in the case of Germany 150 thousand EUR, in the case of theCzech Republic 90 thousand EUR and in the case of Poland 72thousand EUR. If both firms run the campaign in the same coun-try, receives each of them the half of the profit from this country’scustomers, similarly in the case that no firm runs the campaign ina specified country.What are the optimal strategies for both hotels?

1

Germany . . . . . . . . . 150Czech Republic . . . 90Poland . . . . . . . . . . . . 72

312

Trautenberg

Strategy Germany Czech Rep. Poland

Germany

Krakonoš Czech Rep.

Poland

2


312

Trautenberg


Germany (156, 156)

Krakonoš Czech Rep. (156, 156)

Poland (156, 156)

3


312

Trautenberg


Germany (156, 156) (186, 126)

Krakonoš Czech Rep. (156, 156)

Poland (156, 156)

4


312

Trautenberg


Germany (156, 156) (186, 126)

Krakonoš Czech Rep. (126, 186) (156, 156)

Poland (156, 156)

5


312

Trautenberg


Germany (156, 156) (186, 126) (195, 117)

Krakonoš Czech Rep. (126, 186) (156, 156)

Poland (117, 195) (156, 156)

6


312

Trautenberg


Germany (156, 156) (186, 126) (195, 117)

Krakonoš Czech Rep. (126, 186) (156, 156) (165, 147)

Poland (117, 195) (147, 165) (156, 156)

7


312

Trautenberg


Germany (156, 156) (186, 126) (195, 117)

Krakonoš Czech Rep. (126, 186) (156, 156) (165, 147)

Poland (117, 195) (147, 165) (156, 156)

8

Definition. n-player normal form game (NFG) is defined as the(2n+ 1)-tuple

(Q; S1, . . . , Sn; u1(s1, . . . , sn), . . . , un(s1, . . . , sn)) ,

where n ≥ 2 is a natural number; Q = {1, 2, . . . , n} is a givenfinite set, so-called set of players, its elements are called players;for every i ∈ {1, 2, . . . , n}, Si is an arbitrary set, so-called setof strategies of the player i, and

ui : S1 × S2 × · · · × Sn → R

is a real function called payoff function of the player i.

9

☛ Example

Player 2

Strategy t1 t2

s1 (2, 0) ← (2,−1)Player 1 ↑ ↓

s2 (1, 1) ← (3,−2)

10

☛ Example

Player 2

Strategy t1 t2

s1 (2, 0) ← (2,−1)Player 1 ↑ ↓

s2 (1, 1) ← (3,−2)

Definition. An n-tuple of strategies s∗ = (s∗1, . . . , s∗n) is called

an equilibrium point or Nash equilibrium of the game (HNT),if and only if for every i ∈ {1, 2, . . . , n} end every si ∈ Si thefollowing condition holds:

ui(s∗1, . . . , s∗i−1, si, s

∗i+1, . . . , s

∗n) ≤

≤ ui(s∗1, . . . , s∗i−1, s

∗i , s

∗i+1, . . . , s

∗n). (1.1)

11

1979 B. A. Baldwin, G. B. Meese: Skinner sty

12

1979 B. A. Baldwin, G. B. Meese: Skinner sty

13

Strategy Press the lever Sit by the trough

Press the lever (8,−2) → (5, 3)

Sit by the trough (10,−2) → (0, 0)

26

☛ ExamplePlayer 2

Strategy t1 t2

s1 (1,−1) → (−1, 1) pPlayer 1 ↑ ↓

s2 (−1, 1) ← (1,−1) 1− p

27

☛ ExamplePlayer 2

Strategy t1 t2

s1 (1,−1) → (−1, 1) pPlayer 1 ↑ ↓

s2 (−1, 1) ← (1,−1) 1− p

q 1− q

28

☛ ExamplePlayer 2

Strategy t1 t2

s1 (1,−1) → (−1, 1) pPlayer 1 ↑ ↓

s2 (−1, 1) ← (1,−1) 1− p

q 1− qMixed strategies:

{s1, s2} {(p, 1− p), p ∈ 〈0, 1〉}

{t1, t2} {(q, 1− q), q ∈ 〈0, 1〉}

Payoff functions u1, u2 expected payoffs π1, π2 :

π1(p, q) = −1pq + 1(1− p)q + 1p(1− q)− 1(1− p)(1− q)

π2(p, q) = 1pq − 1(1− p)q − 1p(1− q) + 1(1− p)(1− q)

29

Finite Normal Form GameFinite normal form game is a normal form game where all the setsS1, S2, . . . , Sn are finite.

Definition. Consider a finite n-player normal form game. Denotethe number of elements of the strategy set Si of an arbitrary playeri by the symbol mi. Mixed strategy of the player i is defined asthe vector of probabilities

pi = (pi1, pi2, . . . , p

imi), with pij ≥ 0 for all 1 ≤ j ≤ mi,

mi∑j=1

pij = 1,

(1.2)where pij, j = 1, 2, . . . , mi, expresses the probability of choo-sing the j-th strategy from the strategy space Si.

30

A mixed strategy is therefore again a certain strategy that can be cha-racterized in the following way:

”Use the strategy si1 ∈ Si with the probability pi1,. . . ,use the strategy simi ∈ Si with the probability p

imi

.”

For the case of distinction, the elements of the strategy set Si are calledpure strategies.

Theorem (J. Nash). In mixed strategies, every finite normal form gamehas at least one equilibrium point.

31

MATHEMATIZATION OF GAME THEORY

JOHN VON NEUMANN (1903 – 1957)

1926 proof of minimax theorem (Gött.Math.Soc.)

1928 Sur la théorie des jeux (Comptes Rendus)Zur Theorie der Gesellschaftsspiele (Math. Annalen)

• Mathematization of strategy games

• The proof of ”minimax theorem”

Formulation: finite n-player zero sum game

More results: n = 2

({1, 2}; {s1, . . . , sk}, {t1, . . . , tl};u1, u2)

u1(si, tj) + u2(si, tj) = 0

32

Player 1

s1

s2

...

sk

Player 2

t1 t2 . . . tl

u1(s1, t1) u1(s1, t2) . . . u1(s1, tl)

u1(s2, t1) u1(s2, t2) . . . u1(s2, tl)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

u1(sk, t1) u1(sk, t2) . . . u1(sk, tl)

Player 1: mintj u1(si, tj) MAX

Player 2: maxsi u1(si, tj) MIN

It is:max

simin

tju1(si, tj) ≤min

tjmax

siu1(si, tj)

33

Player 1

s1

s2

sk

Player 2

t1 t2 t3 t4

5 4 4 5

−4 5 3 9

7 8 −1 8

4

−4

−1

min

max: 7 8 4 9

maxsmin

tu1(si, tj) = 4 = min

tmax

su1(si, tj)

34

Player 1

s1

s2

sk

Player 2

t1 t2 t3

0 1 −1

−1 0 1

1 −1 0

−1

−1

−1

min

max: 1 1 1

maxsmin

tu1(si, tj) = −1 < min

tmax

su1(si, tj) = 1

35

Mixed strategies – expected payoff for player 1:

π1(p, q) =k∑

i=1

l∑j=1

u1(si, tj)piqj

Theorem. There always exist mixed strategies (p∗, q∗),such that

π1(p∗, q∗) = maxpmin

qπ1(si, tj) = min

qmax

pπ1(si, tj)

36

GAME THEORY = MATHEMATICAL DISCIPLINEJohn von Neumann (1903 – 1957), Oskar Morgenstern (1902 – 1976)

1944 Theory of Games and Economic Behavior

• Application possibilities of game theory – detailed formulationof economical problem

• Axiomatic utility theory

• General formal description of a game of strat.

• 2-player antagonistic finite games

• n-player cooperative games (with transferable payoffs) von Neumann-Morgenstern’s solution

(it is not unique, does not necessarily exist). . .

Massive development of game theory and its applications

The next step: Non-constant sum noncooperative games, cooperativegames without transferable payoff

37

(3,−3) (2,−2)(0, 0) (1,−1)

(3, 3) (2, 4)(0, 6) (1, 5)

(3, 3) → (2, 4)

↑ ↓(0, 2) → (4, 5)

(4, 5) . . . mutually best replies – equilibrium point

38

JOHN FORBES NASH (*1928)

Equilibrium Concept

1949 Non-Cooperative Games (thesis; Ph.D. 1950)

• Nash equilibrium – introduction of the concept

• Nash theorem – proof of its existence

Motivations:

– Rational considerations of payoffs

– Interactive adjustment processes, in which boudedlyrational agents observe the strategies played by theirlikely opponents over time, and gradually learn to ad-just their own strategies to earn higher payoffs, willeventually converge to a Nash equilibrium – if theirconverge to anything at all (only in the thesis)

confirmed by later experimental works.39

1950 Equilibrium Points in n-Person GamesProceedings of Nat. Acad. of Sciences of USA

1951 Non-Cooperative Games, Annals of Math.

Axiomatic Bargaining Theory

1950 The Bargaining Problem, Econometrica 18

1953 Two-Person Cooperative Games, ibid. 21

• Non-transferable payoffs

• The approach to cooperative games by theirreduction to non-cooperative

• Axioms that a solution shall satisfy

• The proof of the existence of the unique solution: Nashbargaining solution

40

BEGINNINGS OF UTILITY THEORY

Daniel Bernoulli (1700 – 1782)Exposition of a New Theory of Risk Evaluation

(Petersburg 1725 – 1733, publ. 1838)

A gamble should be evaluated not in terms of the value of itsalternative pay-offs but rather in terms of the value of its utilities .

Illustration example: Somehow a very poor fellow obtains a lottery ticket

that will yield with equal probability either nothing ofr twenty thousand

ducats. Will this man evaluate his chance of winning at ten thousand

ducats? Would he not be ill-advised to sell this lottery ticket for nine

thousand ducats? To me it seems that the answer is negative. On the

other hand I am inclined to believe that a rich man would be ill-advised

to refuse to buy the lottery ticket for nine thousand ducats. If I am not

wrong then it seems clear that all men cannot use the same rule to

evaluate the gamble . . .

41

Utility function u(x). . . amount of utility units the number of utility units for the

financial amount x;

Assumption:

the increase of utility du(x) is proportionate to the increase of theamount dx and inversely proportionate to the quantity previouslypossessed (a poor man generally obtains more utility than doesa rich man from an equal gain):

du(x) =bdx

xb > 0 (constant of proportion)

u(x) = b lnx+ c c ∈ R

= b lnx− b lnα α ∈ (0,+∞)

u(x) = b lnx

αα – quantity originally possessed

Application: elucidation of St. Petersburg Paradox:42

St. Petersburg ParadoxPeter tosses a coin and continues to do so until it should land”heads” when it comes to the ground. He agrees to give Paul oneducat if he gets ”heads” on the very first throw, two ducats if hegets it on the second, four if on the third, eight if on the fourth, andso on, so that with each additional throw the number of ducats hemust pay is doubled. Suppose we seek to determine the value ofPaul’s expectation.

The mean value of the win:

1

2+2·

(1

2

)2+· · ·+2n−1·

(1

2

)n+· · · =

1

2+1

2+· · ·

1

2+· · · =∞

Paradox:although an expected value of the win is infinite, a reasonableperson sells – with a great pleasure – the engagement in the playfor 20 ducats.

43

Bernoulli: the mean value of the utility brought by the win:

∞∑n=1

1

2nb ln

α+ 2n−1

α=

= b ln[(α+ 1)12 (α+ 2)

14 · · · (α+ 2n−1) 12n · · · ]− b lnα

Amount D, whose addition to the initial possession brings thesame utility:

b lnα+D

α= b ln[(α+1)

12 (α+2)

14 · · · (α+2n−1) 12n · · · ]−b lnα

D = [(α+ 1)12 (α+ 2)

14 · · · (α+ 2n−1) 12n · · · ]− α

For a zero initial possession:

D = 2√1 · 4√2 · 8√4 · 16√8 · · · = 2

44

Imperfections of Bernoulli’s utility function

• Defined for positive values of q only, while in the real worldthe losses are important, too

• Utility function is different for different people, depends onother than property conditions, too

Important stimulus for further development.

Similar – but independent – considerations(Bernoulli cites at the end of his treatise):

45

Gabriel Cramer (1704 – 1752)

Letter to Nicholas Bernoulli, 1728

Idea: people evaluate money in proportion to the utility they canobtain from it.

Assumption:

Any amount above 10 millions, or (for the sake of simplicity) above224 ducats be deemed by me equal in value to 224 ducats or, betteryet, that I can never win more than that amount, no matter howlong it takes before the coin falls with its cross upward [withoutany further discussion]. In this case, my expectation is

46

1

2·1+1

4·2+1

8·4+· · ·+

1

224·224+

1

225·224+

1

226·224+· · · =

=1

2+1

2+ · · ·+

1

2+1

4+1

8+ · · · = 12+1 = 13 .

Thus, my moral expectation is reduced in value to 13 ducats andthe equivalent to be paid for it is similarly reduced – a result whichseems much more reasonable than does rendering it infinite.

47

Daniel Bernoulli(1700 – 1782)

48

Gabriel Cramer(1704 – 1752)

49

Normal Form Games whose Strategy Sets are OpenIntervals

Theorem – EQUILIBRIUM TEST.Let G be a normal form game whose strategy sets are open in-tervals and payoff functions are twice differentiable. Assume thata strategy profile (s∗1, . . . , s

∗n) satisfies the following conditions:

1)∂ui(s∗1, . . . , s

∗n)

∂si= 0 for each i = 1, 2, . . . , n.

2) Each s∗i is the only stationary point of the function

ui(s∗1, . . . , s∗i−1, si, s

∗i+1, . . . , s

∗n), si ∈ Si.

3)∂2ui(s∗1, . . . , s

∗n)

∂s2i< 0 for each i = 1, 2, . . . , n.

Then (s∗1, . . . , s∗n) is an equilibrium point of the game G.

50

Remark. In practice, we usually find the solution of the system of equati-ons

∂ui(s∗1, . . . , s∗n)

∂si= 0 for each i = 1, 2, . . . , n,

and then use other economic considerations to verify that the solutionis the Nash equilibrium of the game.

51

A typical example of the normal form game whose strategy sets are openintervals is an oligopoly model where several firms produce the sameproduct, each of them contributing significantly to the total production.The price is given by the demand equation which describes the marketbehavior and gives the relation between the price and the total numberof products that should be sold. In other words, it gives the highest pricefor which it is possible to sell the given amount of products. In Cournotmodels discussed below the demand equation has the simplest possibleform:

p+ q =M, M >> c, (1.3)

where p is the price of the product, q is the demand for this productin the market, c expresses the production costs of one piece, M is aconstant much greater than c.

52

☛ Example. Cournot Monopoly Model

If only one firm – a monopolist – produces the given product, thesituation is clear. The monopolist knows that when he producesq products, the highest price for which he can sell one piece tosell all the production out is given by the demand equation (1.3):

p =M − q. (1.4)

Since nobody else influences the total amount of products, themonopolist is faced with the problem of a mere profit maximi-zation, i.e. finding the maximum of one variable function

u(q) = p · q− c · q =Mq− q2− cq = (M − q− c)q. (1.5)

Maximum can easily be found with the help of a derivative:

u′(q) =M − c− 2q = 0

q∗mon =12(M − c)

(1.6)

53

Since for q < q∗mon the derivative is positive and the function u in-creasing, for q > q∗mon is the derivative negative and the functionu decreasing, it is really a maximum. Hence the monopolist rea-ches the maximal profit by the production of q∗mon =

12(M − c)

pieces, namely

u∗mon = u(q∗mon) =

(M − 12(M − c)− c

)12(M − c) =

=[12(M − c)

]2(1.7)

The corresponding price is then

p∗mon =12(M + c) (1.8)

54

☛ Example. Cournot Duopoly Model (1938).

Now consider two firms producing the same product, each ofthem contributing significantly to the total production. Monopo-list’s problem lied in finding the maximum of a simple quadraticfunction Duopolists are faced with a game, since each of themcontrols only a part of the total production; the price he receivesfor his products therefore depends not only on his own decision,but on the decision of the opponent, too. Now we assume thatthe duopolists decide simultaneously and independently of eachother.

Let us find an equilibrium point in this game, i.e. an optimalamounts that should be produced by particular duopolists, whenit is not advantageous for none of them to deviate.

55

Denote q1, q2 the production of the first and the second duopolistrespectively. Maximal price for which all the products can be soldis again given by the demand equation (1.3):

p =M − q1 − q2 (1.9)

The situation can be modelled by the normal form game whereplayers are duopolists, each of them choosing in general thenumber of an interval (0, M); strategy sets are therefore theintervals

S1 = S2 = (0, M),

payoff functions are the profits:

u1(q1, q2) = (p− c)q1 = (M − c− q1 − q2)q1

u2(q1, q2) = (p− c)q2 = (M − c− q1 − q2)q2(1.10)

56

Although the strategy sets are infinite, the equilibrium point canbe found very simply. The first duopolist looks for a function whichto each opponent’s strategy, i.e. each number q2 assigns such anumber q1 = R1(q2), which is the best reply to q2 in the sencethat the value of the function u1(q1, q2) is maximal. In other words,for every fixed (but arbitrary) q2 ∈ S2 the first duopolist searchesthe maximum of the function u1(q1, q2) which is now a functionof one variable q1 :

57

∂u1

∂q1=M − c− q2 − 2q1 = 0

R1(q2) = q1 = 12(M − c− q2) (1.11)

(verify this is really a maximum). Similarly, for every strategy q1of the first duopolist, the second duopolist searches his best replyq2 = R2(q1), i.e. such number which for a given q1 maximizesthe profit u2 :

∂u2

∂q2=M − c− q1 − 2q2 = 0

R2(q1) = q2 = 12(M − c− q1) (1.12)

Functions R1(q2) and R2(q1) are called reaction curves; wecan represent them in the following way:

58

From the definition it can be easily deduced that (q∗1, q∗2) is an

equilibrium point if and only if it is an intersection of reactioncurves:

q∗1 = R1(q∗2) end q

∗2 = R2(q

∗1).

In our example:

(q∗1, q∗2) =

(13(M − c),

13(M − c)

). (1.13)

The price for which duopolists will sell the products is then

p =M − 23(M − c) =13M +

23c. (1.14)

The corresponding profit for each duopolist is

u1(q∗1, q∗2) = u2(q

∗1, q

∗2) =

[13(M − c)

]2. (1.15)

In equilibrium the total production of duopolists is

q∗D = q∗1 + q

∗2 =

23(M − c) >

12(M − c) = q

∗mon , 60

hence they sell the greater total production for the lower pricethan a monopolist.

Comparing the results for the monopoly and duopoly, it is clearthat it would be the best for duopolists to get together in a cosycartel and collude to avoid the inroads into their profits that com-petition brings. Then they could together produce only

q1 + q2 = q∗mon =12(M − c) (1.16)

(these points form the green line segment) and with respect tocircumstances divide the profit that emerged in this way – insymmetrical situation fifty-fifty:(

12q

∗mon,

12q

∗mon

)=

(14(M − c),

14(M − c)

).

Nevertheless, this outcome is unstable, because for each duo-polist it is profitable to deviate to his best reply to the opponent’schoice and receive more for himself.

61

The problem lies in the fact that such deals are collusive and, withrespect to anti-monopoly laws, usually lawless – and a collusivedeal made in smoke-filled hotel room is cheap and certainly notbinding.

In the end – fortunately for the customer (and for this reason thereare those anti-monopoly laws) – the only deal when no duopolisthas a temptation to deviate, is the above mentioned equilibriumpoint

(q∗1, q∗2) =

(23q

∗mon,

23q

∗mon

)=

(13(M − c),

13(M − c)

).

But the situation changes radically when the game is repeated,i.e. when the same duopolists find themselves in the same si-tuation repeatedly: if the probability that there will be one moreround is high in each time the game is played, it can be advan-tageous for each duopolist to keep the deal.

63

☛ Example. Cournot Oligopoly Model

Consider n firms producing the same product, each of them con-tributing significantly to the total production. Now we have ann-player game where each player searches the optimal amountof products qi he shall produce. Let us find the equilibrium pointof this game.

Likewise in the case of duopoly, profits of particular oligopolistsare the following:

u1(q1, . . . , qn) = (p− c) q1 = (M − c− q1 − q2 − · · · − qn) q1

u2(q1, . . . , qn) = (p− c) q2 = (M − c− q1 − q2 − · · · − qn) q2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

un(q1, . . . , qn) = (p− c) qn = (M − c− q1 − q2 − · · · − qn) qn

(1.17)64

Again, it is easy to find the equilibrium point. From the conditions

∂u1

∂q1=M − c− 2q1 − q2 − · · · − qn = 0

∂u2

∂q2=M − c− q1 − 2q2 − · · · − qn = 0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∂un

∂qn=M − c− q1 − q2 − · · · − 2qn = 0

we get the system of equations:

2q1 + q2 + · · · + qn = M − c

q1 + 2q2 + · · · + qn = M − c

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

q1 + q2 + · · · + 2qn = M − c65

Its solution is:

q∗1 = q∗2 = · · · = q

∗n =

M − cn+ 1

(1.18)

Oligopolists therefore produce the total production

q∗O = q∗1+q

∗2+ · · ·+q

∗n = n

M − cn+ 1

=n

n+ 1(M − c) (1.19)

From the result it is clear that the greater the number of firms is,the greater is the total production q∗O and the lower is the price p

∗

and the total profit u∗ of the firms:

p∗ =1

n+ 1M +

n

n+ 1c (1.20)

u∗ =n

(n+ 1)2(M − c)2 (1.21)

A limit case of oligopoly where n→∞ is a perfect competition:here the great number of small firms contribute to the total pro-duction and no firm can significantly influence the total outcome.

66

The total production is now given by

limn→∞

n

n+ 1(M − c) = (M − c), (1.22)

the price of a product in the market is equal to production costsc,

p∗ =M − (M − c) = c, (1.23)

and the profit of particular firms is equal to zero,

u∗ = 0. (1.24)

The following table summarizes the above results:

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Total Production q∗ Price p∗ Total Profit u∗

Monopoly 12(M − c)12M +

12c

14(M − c)

2

Duopoly 23(M − c)13M +

23c

29(M − c)

2

Oligopoly nn+1(M − c)

1n+1M +

nn+1c

n(n+1)2 (M − c)

2

Perfect Compet. (M − c) c 0

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NORMAL FORM GAMES - euler.fd.cvut.czeuler.fd.cvut.cz/predmety/game_theory/lecture... · 2005. 11....

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