- 111 Views
- Uploaded on
- Presentation posted in: General

Static Games of Complete Information

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

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

- 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

Receiver’s Move

- 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