1 / 34

# Static Games of Complete Information - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Static Games of Complete Information' - cricket

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Static Games of Complete Information

.

• 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

.

\$ 10

Price

\$ 2

80

100

Quantity

• 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?

• 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)

• Demand curve facing Bob is

• To maximize profits, Bob wants the total quantity to be the midpoint between Ann’s quantity (QA) and the break-even point (80)

\$ 10

Price

\$ 2

QA

(80+QA)/2

80

100

Quantity

Bob

• Four scenarios for Bob’s reasoning

• 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?

• 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=(s1, s2,…, sI)

• 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

• A mixed-strategy σiis 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

• Player 1 plays along rows; 2 along columns

• Let σ1={⅓, ⅓, ⅓} andσ2={0, ½, ½ }.

• What is player 1’s payoff?

• 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

• A mixed-strategy that assigns positive probability to a dominated pure strategy, is dominated

• A mixed-strategy my be dominated even though it assigns positive probability only to undominated pure strategies.

• Examples: -Prisoner’s Dilemma

-Second-Price Auction

• Defn: A mixed-strategy profile σ* is a Nash Equilibrium if, for all players i,

for all for all s-i S-i

• Common examples in economics:

- Cournot quantity-setting game

- Bertrand price-setting game

• 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 that each player knows the rules

• Each player knows that each player knows that each player knows that each player knows the rules

• Each player knows that each player knows that each player knows that each player knows that each player knows the rules

• Etc. Etc. Etc.

• Property 1: A player must be indifferent between all pure strategies in the support of a mixed-strategy

• Property 2: One only needs to check for pure-strategy 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?

• 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

• 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

• 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

• In a game of tennis, suppose receiver’s forehand is stronger than backhand

• Consider following probabilities of successfully returning serve

Server’s Aim

• 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!!

• Consider following graph

90

Anticipate Forehand

60

Anticipate Backhand

48

Percentage successful returns

30

20

0

40

100

Percentage of times server aims serve to forehand

• 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%

• Two players, N=1, 2, bargain over the partition of a cake

• The set of Agreements, A, has elements

A={(a1, a2)єR2: 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={(u1(a), u2(a)) for a є A}, and d=(u1(D), u2(D))

• A Bargaining ProblemB is the set of pairs <S, d>, and the Bargaining Solution is a function f :B → R2 that assigns to each <S, d> єB an element of S

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 f1(S, d)= f2(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).

Theorem (Nash 1950)

There is a unique bargaining solution fN:B → R2 satisfying the above axioms given by

fN(S, d)=

• Set of agreements is

A={(a1, a2)єR2: a1+ a2≤1 and ai≥0 for i=1, 2}

• Players are risk-averse, i.e. uiare concave

• Disagreement point is d=(u1(0), u2(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))

• Let player 2 be more risk-averse than player 1

• Let player 2’s utility be v2 =hu2,where h is concave, and v1 = u1

• 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:

• 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.

Theorem (Nash 1950)

Every finite strategic-form game has a mixed-strategy 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

Theorem (Shizuo Kakutani 1941)

If :

• ∑ is compact and convex

• A correspondence r(σ) :∑→∑ is non-empty and convex

• r(σ) has a closed graph

Then r(σ) has a fixed point, that is, there exists σ*

such that, r(σ*)= σ*

• 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)=(r1 (σ), r2 (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(.)

• ∑ 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’sutility functions are continuous in σi

• r(σ) is convex:

Suppose σ/,σ//є r(σ) . uiis 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 responsestoσ-i, then so is their weighted average. Thus, λσi/+ (1- λ)σi// є r(σ)

• r(σ) has a closed graph:

If with , then