Static Games of Complete Information Games and best

Static Games of Complete Information.

Games and best responses • Bob is a florist • One bunch of flowers sells for $10, but to sell 100 bunches market price should be zero • Thus, each extra bunch reduces market price by a dime • Cost is $2 a bunch

Games and best responses. $ 10 Price $2 80 Quantity 100

Games and best responses • What is Bob’s optimal quantity as a monopolist? • It turns out to be 40 units • Suppose now Ann is contemplating entry • How will the total quantity be allocated between the two?

Games and best responses • It depends on how they expect each other to act • Suppose Ann is a sophisticated thinker and responds optimally to Bob • Ann’s best response to Bob is (80 -QB)/2 • What should Bob do? • Note: Bob wants the total quantity to be the midpoint between Ann’s quantity (QA) and the break-even point (80)

Games and best responses • Demand curve facing Bob is $ 10 Price $2 QA Bob (80+QA)/2 80 100 Quantity • To maximize profits, Bob wants the total quantity to be the midpoint between Ann’s quantity (QA) and the break-even point (80)

Games and best responses • Four scenarios for Bob’s reasoning Scenario Description Bob’s conjecture QB QA Naive Ignores Ann (sticks to monopoly output) None 40 20 Primitive Reasons that Ann reacts to Bob’s previous output QA=20 30 25 Sophisticated Reasons that Ann will deduce that Bob considers Ann’s reaction to Bob’s monopoly output QA=25 Ultra sophisticated Both Bob and Ann believe that the other understands the logic of best response QA=(80 - QB)/2

What if Bob’s quantity is a given? • How many bunches should Bob bring, knowing how Ann would react to him? • Now, total quantity is (80 -QB)/2+QB= (80+QB)/2 • Thus, mkt price is $10 -$0. 10{(80+QB)/2} • Bob’s optimal quantity is 40; and so Ann’s quantity is 20 • A significant first-mover advantage! • Why does Bob produce monopoly quantity?

Strategic-Form Games • Elements of strategic-form games: - finite set of players, i I ={1, 2, …, I} - pure-strategy space Si for each player i - payoff functions ui(s) for each player i and for each strategy profile s=(s 1, s 2, …, s. I) • Each player goal is to maximize his own payoff, and NOT to ‘beat’ other players • A Zero-Sum game is where =0 • Most applications in the Social Sciences are non zero-sum games

Mixed-Strategies • A mixed-strategy σi is a probability distribution over pure-strategies • σi(si) is probability that σi assigns to si • Space of player i’s mixed-strategies is ∑i • Space of mixed-strategy profiles is ∑= with elements σ • Each player’s randomization is independent of those of other players • Player i’s payoff to profile σ is

Computing payoff from a mixedstrategy • Player 1 plays along rows; 2 along columns L M R U 4, 3 5, 1 6, 2 M 2, 1 8, 4 3, 6 D 3, 0 9, 6 2, 8 • Let σ1={⅓, ⅓, ⅓} and σ2={0, ½, ½ }. • What is player 1’s payoff?

Dominated Strategies • Is there an obvious prediction about how the previous game will be played? • Iterated Strict Dominance predicts (U, L) • Let “–i” denote all players other than i • Then strategies of other players s-i S-i and a strategy profile is (si , s-i ) • Defn: Pure strategy si is strictly dominated for player i if there exists ∑i such that, for all s-i S-i

Some implications of dominance 1. A mixed-strategy that assigns positive probability to a dominated pure strategy, is dominated 2. A mixed-strategy my be dominated even though it assigns positive probability only to undominated pure strategies. • Examples: -Prisoner’s Dilemma -Second-Price Auction

Nash Equilibrium • Defn: A mixed-strategy profile σ* is a Nash Equilibrium if, for all players i, for all s-i S-i • Common examples in economics: - Cournot quantity-setting game - Bertrand price-setting game

Assumptions • Rationality • Players aim to maximize their payoffs • Players are perfect calculators • Common Knowledge • Each player knows the rules of the game • Each player knows that each player knows the rules • Each player knows that each player knows the rules • Each player knows that each player knows the rules • Etc.

Properties and implications • Property 1: A player must be indifferent between all pure strategies in the support of a mixedstrategy • Property 2: One only needs to check for purestrategy deviations • Property 3: If a single strategy survives iterated deletion of strictly dominated strategies it is a Nash Equilibrium • Question: Why play mixed strategies when all pure strategies in the support have same payoff?

Further comments on N. E. • In many games pure strategy NE do not exist -example is game of “Matching Pennies” • Sometimes there are multiple Nash Equilibria - “Battle of the Sexes”; “Chicken” • Schelling’s theory of “focal points” • Pre-play communication

Baseball anyone? • 1986 baseball National League championship series • The New York Mets won a crucial game against he Houston Astros • Len Dykstra hit Dave Smith’s second pitch for a two-run home run • Later the two players talked about this critical play

Analysis of a home run • Dykstra said, “He threw me a fastball on the first pitch and I fouled it off. I had a gut feeling then that he’d throw me a forkball next, and he did. I got a pitch I saw real well, and I hit it real well”. • Smith said, “What it boils down to is that, it was a bad pitch selection…if I had to do it over again, it would be [another] fastball”. • Would Dykstra not have been prepared for a fastball? • Again, randomization is the only way to go

But how do you randomize? • In a game of tennis, suppose receiver’s forehand is stronger than backhand • Consider following probabilities of successfully returning serve Server’s Aim Forehand Backhand Receiver’s Move Forehand 90% 20% Backhand 30% 60%

Reducing receiver’s effectiveness by randomizing • Suppose server tosses a coin before each serve • Aims to forehand or backhand according to coin turning heads/tails • When receiver moves to forehand, his successful return rate is 55% • When receiver moves to backhand, his successful return rate is 45% • Given server’s randomization, receiver should move to forehand • The server has already an improved outcome compared to serving the same way all the time!!

What is server’s best mix? • Consider following graph Percentage successful returns 90 Anticipa te 60 nd eha r o e. F ipat ic Ant Backha nd 48 30 20 0 40 Percentage of times server aims serve to forehand 100

The mixing probabilities • The 40: 60 mixture of forehands to backhands is the equilibrium • This mixture is the only one that cannot be exploited by the receiver to his own advantage • With this mixture the receiver does equally well with either of his choices • Both ensure the receiver a success rate of 48%

Nash bargaining • Two players, N=1, 2, bargain over the partition of a cake • The set of Agreements, A, has elements A={(a 1, a 2)єR 2: ai≥ 0 for i=1, 2} • The event of Disagreement is D • Players have utilities ui, i=1, 2, s. t. ui: A {D}→ R • Let, S={(u 1(a), u 2(a)) for a є A}, and d=(u 1(D), u 2(D)) • A Bargaining Problem B is the set of pairs <S, d>, and the Bargaining Solution is a function f : B → R 2 that assigns to each <S, d> єB an element of S

Nash’s axioms 1. INV (invariance to equiv utility representations): If <S/, d/> is obtained from <S, d> by si , then, fi (S/, d/)= 2. SYM (symmetry) If <S, d> is symmetric, then f 1(S, d)= f 2(S, d) 3. PAR (Pareto efficiency) Suppose <S, d> is a bargaining problem, with si , ti єS , and ti >si , for i=1, 2, then f(S, d)≠s 4. IIA (independence of irrelevant alternatives) If <S, d> and <T, d> are bargaining problems with S T and f(T, d) єS, then f(S, d) = f(T, d).

Nash bargaining solution Theorem (Nash 1950) There is a unique bargaining solution f. N: B → R 2 satisfying the above axioms given by f. N(S, d)=

Application: Splitting a dollar • Set of agreements is A={(a 1, a 2)єR 2: a 1+ a 2≤ 1 and ai≥ 0 for i=1, 2} • Players are risk-averse, i. e. ui are concave • Disagreement point is d=(u 1(0), u 2(0))=(0, 0) • So <S, d> is a bargaining problem Result 1: With equal risk-aversion, players are symmetric & SYM, PAR give (u(1/2), u(1/2))

Role of risk-aversion • Let player 2 be more risk-averse than player 1 • Let player 2’s utility be v 2 =h u 2, where h is concave, and v 1 = u 1 • Let <S/, d/> be bargaining problem with utilities vi • The optimizing program for <S, d> gives solution zu where zu solves: • The optimizing program for <S/, d/> gives solution zv where zv solves:

Role of risk-aversion • The first program gives, • The second program gives, • Result 2: If player 2 becomes more risk-averse, then Player 1’s share of the dollar in the Nash solution increases.

Existence of NE Theorem (Nash 1950) Every finite strategic-form game has a mixedstrategy equilibrium. Sketch of Proof: 1. Use players’ Reaction Correspondences r(σ) 2. Realize that a NE is a Fixed Point of r(σ) 3. Show that conditions for Kakutani’s Fixed Point Theorem are satisfied in this case

Existence of NE (cont) Theorem (Shizuo Kakutani 1941) If : (i) ∑is compact and convex (ii) A correspondence r(σ) : ∑→∑ is non-empty and convex (iii) r(σ) has a closed graph Then r(σ) has a fixed point, that is, there exists σ* such that, r(σ*)= σ*

Nash equilibrium is a fixed point • For any strategy profile σ, player i’s reaction correspondence ri maps σ to the mixed strategy that maximizes his payoff, given his opponents play σ-i. Thus, ri (σ)= ri (σi , σ-i ) = ui(σ/i , σ-i ) • Each player i=1, 2, . . n, has a reaction function • Let us form the n-tuple, r(σ)=(r 1 (σ), r 2 (σ), …, rn(σ)), where r(σ)є∑ • Suppose there exists σ* є∑ such that, for all players, σ*i =ri(σ*)= ri(σ*i , σ*-i), then σ* is a Nash equilibrium • But σ*i =ri(σ*) for all i, means that σ*=r(σ*), i. e. σ* is a fixed point of the reaction correspondence r(. )

Applying Kakutani • ∑ is compact and convex: 1. Mixed strategies convexify the strategy space, and so each ∑i is a convex space of dimension (#Si-1) 2. Theorem (Heine-Borel): Closed & bounded subsets in Rn are compact • r(σ) is non-empty: 1. Theorem (Weierstrass): A continuous real-valued function defined on a compact space achieves its maximum values 2. Player’s i’s utility functions are continuous in σi

Applying Kakutani • r(σ) is convex: Suppose σ/, σ// є r(σ). ui is linear, therefore for λє(0, 1) ui(λσi/+ (1 - λ)σi//, σ-i)= λ ui(σi/, σ-i) + (1 - λ)ui(σi//, σ-i). Thus, if σi/, σi// are best responses to σ-i, then so is their weighted average. Thus, λσi/+ (1 - λ)σi// є r(σ) • r(σ) has a closed graph: If with , then