Nau: Game Theory 1

Introduction to Game Theory

2. Normal-Form Games Dana Nau

University of Maryland

Nau: Game Theory 2

How to reason about games?   In single-agent decision theory, look at an optimal strategy

  Maximize the agent’s expected payoff in its environment

  With multiple agents, the best strategy depends on others’ choices   Deal with this by identifying certain subsets of outcomes called solution

concepts

  Some solution concepts:   Pareto optimality   Nash equilibrium   Maximin and Minimax   Dominant strategies   Correlated equilibrium

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Pareto Optimality   Strategy profile S Pareto dominates a strategy profile Sʹ′ if

  no agent gets a worse payoff with S than with Sʹ′, i.e., ui(S) ≥ ui(Sʹ′) for all i ,

  at least one agent gets a better payoff with S than with Sʹ′, i.e., ui(S) > ui(Sʹ′) for at least one i

  Strategy profile s is Pareto optimal, or strictly Pareto efficient, if there’s no strategy s' that Pareto dominates s   Every game has at least one Pareto optimal profile   Always at least one Pareto optimal profile in which the strategies are

pure

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5, 0 1, 1

3, 3 0, 5

Examples The Prisoner’s Dilemma

  (C,C) is Pareto optimal   No profile gives both players a higher payoff

  (D,C) is Pareto optimal   No profile gives player 1 a higher payoff

  (D,C) is Pareto optimal - same argument

  (D,D) is Pareto dominated by (C,C)

Which Side of the Road   (Left,Left) and (Right,Right) are Pareto optimal

  In common-payoff games, all Pareto optimal strategy profiles have the same payoffs

  If (Left,Left) had payoffs (2,2), then (Right,Right) wouldn’t be Pareto optimal

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Best Response   Suppose agent i knows how the others are going to play

  Then agent i has the single-agent problem of choosing a utility-maximizing action

  Some notation:   S−i = (s1, …, si−1, si+1, …, sn),

•  i.e., S–i is strategy profile S without agent i’s strategy   If si' is any strategy for agent i, then

•  (si' , S−i ) = (s1, …, si−1, si', si+1, …, sn)   Hence (si , S−i ) = S

  For agent i, a best response to S−i is a mixed strategy si* such that ui (si*, S−i ) ≥ ui (si , S−i ) for every strategy si available to agent i

  A best response si* is unique if ui (si*, S−i ) > ui (si , S−i ) for every si ≠ si*

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Best Response   Unless there’s a unique best response to S–i that’s a pure strategy, the

number of best responses is infinite   Suppose a best response to S–i is a mixed strategy s* whose support

includes ≥ 2 actions •  Then every action a in support(s*) must have the same expected

utility ui(a,S–i) ›  If one of them had a higher expected utility than the others,

then it would be a better response than s* •  Thus any mixture of the actions in support(s*) is a best response

  Similarly, if there are ≥ 2 pure strategies that are best responses to S–i then any mixture of them is also a best response

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  A strategy profile s = (s1, …, sn) is a Nash equilibrium if for every i,

  si is a best response to S−i , i.e., no agent can do better by unilaterally changing his/her strategy

  A Nash equilibrium S = (s1, . . . , sn) is strict if for every i, si is the only best response to S−i ,i.e., any agent

who unilaterally changes strategy will do worse

  Otherwise the Nash equilibrium is weak

  In the Prisoner’s Dilemma, (D,D) is a strict Nash equilibrium   If either agent unilaterally switches to a different

strategy, his/her expected utility goes below 1

  In Which Side of the Road, (Left,Left) and (Right,Right) both are strict Nash equilibria

  If either agent unilaterally switches to a different strategy, his/her expected utility goes below 1

Nash Equilibrium

5, 0 1, 1

3, 3 0, 5

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Properties of Nash Equilibria   Theorem (Nash, 1951): Every game with a finite number of agents and

action profiles has at least one Nash equilibrium

  Weak Nash equilibria are less stable than strict Nash equilibria   In the weak case, at least one agent has > 1 best responses,

and only one of them is in the Nash equilibrium

  Pure-strategy Nash equilibria can be either strict or weak

  Mixed-strategy Nash equilibria are always weak   Reason: if there are ≥ 2 pure strategies that are best responses to S–i

then any mixture of them is also a best response   Thus in a strict Nash equilibrium, all of the strategies are pure

  e.g., Prisoner’s Dilemma, Which Side of the Road

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Weak and Strong Nash Equilibria

  Weak Nash equilibria are less stable than strict Nash equilibria   In the weak case, at least one agent has > 1 best responses,

and only one of them is in the Nash equilibrium

  Pure-strategy Nash equilibria can be either strict (e.g., the Prisoner’s Dilemma) or weak (e.g., which side of the road)

  From the corollary on the previous slide, we know that mixed-strategy Nash equilibria are always weak   Thus in a strict Nash equilibrium, all of the strategies are pure

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The Battle of the Sexes has two pure-strategy Nash equilibria

  Generally it’s tricky to compute mixed-strategy equilibria   But easy if we identify the support of the equilibrium strategies

  Assume both agents randomize, and that the husband’s mixed strategy sh is

sh(Opera) = p; sh(Football) = 1 – p

  Expected utilities of the wife’s actions: uw(Football, sh) = 0p + 1(1 − p)

uw(Opera, sh) = 2p   If the wife mixes between her two actions, they must have the same expected utility

  If one of the actions had a better expected utility, she’d do better with a pure strategy that always used that action

  Thus 0p + 1(1 – p) = 2p, so p = 1/3

  So the husband’s mixed strategy is sh(Opera) = 1/3; sh(Football) = 2/3

Husband Wife

Opera Football

Opera 2, 1 0, 0

Football 0, 0 1, 2

Finding Nash Equilibria

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Finding Nash Equilibria

  A similar calculation shows that the wife’s mixed strategy sw is sw(Opera) = 2/3, sw(Football) = 1/3

  Like all mixed-strategy equilibria, this is a weak Nash equilibrium   Each agent has infinitely many

other best-response strategies   In this equilibrium,

  Probability 5/9 that payoff is 0 for both agents   Probability 2/9 that wife gets 2 and husband gets 1   Probability 2/9 that husband gets 2 and wife gets 1   Expected utility for each agent is 2/3

  Pareto-dominated by both of the pure-strategy equilibria   In them, one agent gets 1 and the other gets 2

Husband Wife

Opera Football

Opera 2, 1 0, 0

Football 0, 0 1, 2

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Finding Nash Equilibria Matching Pennies   Easy to see that in this game, no pure strategy

could be part of a Nash equilibrium in this game   For each combination of pure strategies,

one of the agents can do better by changing his/her strategy •  e.g., for (Heads,Heads),

agent 2 can do better by switching to Tails   Thus there isn’t a strict Nash equilibrium   But again there’s a mixed-strategy equilibrium:

  (s,s), where s(Heads) = s(Tails) = ½

–1, 1 1,–1

1,–1 –1, 1

Heads Tails

Heads

Tails

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A Real-World Example   Penalty kicks in soccer

  A kicker and a goalie in a penalty kick   Kicker can kick left or right   Goalie can jump to left or right   Kicker scores iff he/she kicks to one

side and goalie jumps to the other   Analogy to Matching Pennies

•  If you use a pure strategy and the other agent uses his/her best response, the other agent will win

•  If you kick or jump in either direction with equal probability, the opponent can’t exploit your strategy

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Another Interpretation of Mixed Strategies   Another interpretation of mixed strategies is that

  Each agent’s strategy is deterministic   But each agent has uncertainty regarding the other’s strategy

  Agent i’s mixed strategy is everyone else’s assessment of how likely i is to play each pure strategy

  Example:   In a series of soccer penalty kicks, the kicker could kick left or right in

a deterministic pattern that the goalie thinks is random

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Another Real-World Example   Road Networks   Suppose that:

  1,000 drivers wish to travel from S (start) to D (destination)   Two possible paths:

•  S→A→D and S→B→D   The road from S to A is long: t = 50 minutes

•  But it’s also very wide: t = 50 minutes, no matter how many drivers

  Same for road from B to D   Road from A to E is shorter but is narrow

•  Time = (number of cars)/25   Nash equilibrium:

•  500 cars go through A, 500 cars through B •  Everyone’s time is 50 + 500/25 = 70 minutes •  If a single driver changes to the other route

›  There now are 501 cars on that route, so his/her time goes up

S D

t = cars/25

t = cars/25

t = 50

t = 50

B

A

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Braess’s Paradox   Suppose we add a new road from B to A

  The road is so wide and so short that it takes 0 minutes to traverse it   Nash equilibrium:

•  All 1000 cars go S→B→A→D •  Time for S→B is 1000/25 = 40 minutes •  Total time is 80 minutes

  To see that this is an equilibrium: •  If driver goes S→A→D, his/her cost is 50 + 40 = 90 minutes •  If driver goes S→B→D, his/her cost is 40 + 50 = 90 minutes •  Both are dominated by S→B→A→D

  To see that it’s the only Nash equilibrium: •  For every traffic pattern, S→B→A→D dominates S→A→D and S→B→D

•  Choose any traffic pattern, and compute the times a driver would get on all three routes

  Carelessly adding capacity can actually be hurtful!

S D

t = cars/25

t = cars/25

t = 50

t = 50

B

A

t = 0

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Braess’s Paradox in practice   From an article about Seoul, South Korea:

  “The idea was sown in 1999,” Hwang says. “We had experienced a strange thing. We had three tunnels in the city and one needed to be shut down. Bizarrely, we found that that car volumes dropped. I thought this was odd. We discovered it was a case of ‘Braess paradox’, which says that by taking away space in an urban area you can actually increase the flow of traffic, and, by implication, by adding extra capacity to a road network you can reduce overall performance.”

  John Vidal, “Heart and soul of the city”, The Guardian, Nov. 1, 2006http://www.guardian.co.uk/environment/2006/nov/01/society.travelsenvironmentalimpact

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Maximin and Minimax Strategies   Let s be a strategy for agent i

  Then s’s worst-case payoff is the payoff if the other agents play a combination of strategies that is worst for agent i

  Agent i’s maximin value (or security level) is the best worst-case payoff of any combination of the other agents’ strategies, i.e.,

  A maximin strategy for agent i is any strategy whose worst-case payoff is i’s maximin value, i.e.,

  The maximin strategy not necessarily unique, and it often is mixed

€

maxsi

minS− i

ui si,S− i( )

€

argmaxsi

minS− i

ui si,S− i( )

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Example   Battle of the Sexes

  Opera’s worst-case payoff is 0   Football’s worst-case payoff is 0

  Consider the Nash equilibrium strategies we discussed earlier:   The wife’s strategy: sw(Opera) = 2/3, sw(Football) = 1/3

•  Recall that this strategy’s expected payoff is 2/3, regardless of what the husband does

  The husband’s strategy: sh(Opera) = 1/3, sh(Football) = 2/3 •  Recall that this strategy’s expected payoff is 2/3, regardless of what

the wife does   These are also the wife’s and husband’s maximin strategies

Husband Wife

Opera Football

Opera 2, 1 0, 0

Football 0, 0 1, 2

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Comments   If agent i plays a maximin strategy and the other agents play arbitrarily,

  i still receives an expected payoff of at least his/her maximin value   So the maximin strategy is sensible for a conservative agent

  Maximize his/her expected utility without any assumptions about the others

  E.g., don’t assume— •  that they’re rational •  that they draw their action choices from known distributions

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Minimax Strategies   The minimax strategy and minimax value are duals of their maximin

counterparts

  In a 2-player game, a minimax strategy for agent 1 against agent 2 is a strategy s1* that minimizes agent 2’s maximum payoff

  i.e., suppose player 1 first chooses a strategy s1 and then player 2 can choose a best response. Then s1* is a strategy that minimizes the payoff of 2’s best response

  Useful if you want to punish an opponent without regard for your own payoff (e.g., see repeated games)

  Agent 2’s minimax value is agent 2’s maximum payoff against s1*

€

s1* = argmin

s1maxs2

u2 s1, s2( )

€

maxs2

u2 s1*, s2( ) = min

s1maxs2

u2 s1, s2( )

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Minimax Strategies in n-Agent Games   In n-agent games (n > 2), agent i usually can’t minimize agent j’s payoff by

acting unilaterally   But suppose all the agents “gang up” on agent j

  Let S−j* be a mixed-strategy profile that minimizes j’s maximum payoff, i.e.,

  Then a minimax strategy for agent i ≠ j is i’s component of S-j*   Agent j’s minimax value is j’s maximum payoff against S–j*

  In two-player games, an agent’s minimax value = his/her maximin value   For n-player games, an agent’s maximin value ≤ his/her minimax value

€

S− j* = argmin

S− jmaxs j

u j s j ,S− j( )

€

maxs j

uj s j ,S− j*( ) = min

S− jmaxs j

uj s j ,S− j( )

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Minimax Theorem (von Neumann, 1928)   Theorem. Let G be any two-person finite zero-sum game. Then there are a

number VG called the value of G, and strategies s1* and s2* for agents 1 and 2, such that   If agent 2 uses s2*, agent 1’s expected utility is ≤ VG, i.e.,

•  maxs U(s,s2*) = VG   If agent 1 uses s1*, agent 1’s expected utility is ≥ VG, i.e.,

•  mint U(s1*,t) = VG   How this relates to the last few slides:

  agent 1’s maximin value = agent 1’s minimax value = VG   agent 2’s maximin value = agent 2’s minimax value = –VG   si* = any minimax strategy for agent i

= any maximin strategy for agent i   (s1*, s2*) = a Nash equilibrium

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  Agent i’s mixed strategy is to display heads with some probability pi

  The graph shows u1(p1, p2) for every possible p1, p2

  u2(p1, p2) = –u1(p1, p2)

  The Nash equilibrium is the saddle point, p1 = p2 = 0.5

  The value of the game is 0

  If agent i uses a strategy with pi ≠ 0.5,   The other agent has a

“best response” strategy that will lower i’s expected utility below 0

Example: Matching Pennies

–1, 1 1,–1

1,–1 –1, 1

Heads Tails

Heads

Tails

p1 p2

u1(p1, p2)

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Two-Finger Morra   There are several versions of this game

  Here’s the one I’ll use:

  Each agent holds up 1 or 2 fingers   If the total number of fingers is odd

  Agent 1 gets that many points   If the total number of fingers is even

  Agent 2 gets that many points

  Agent 1 has no dominant strategy

  Agent 2 plays 1 => agent 1’s best response is 2   Agent 2 plays 2 => agent 1’s best response is 1

  Similarly, agent 2 has no dominant strategy   Thus there’s no pure-strategy Nash equilibrium

  Look for a mixed-strategy equilibrium

3, –3 –4, 4

–2, 2 3, –3

2

1

1 2

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3, –3 –4, 4

–2, 2 3, –3

2

1

1 2

  Suppose agent 2 plays 1 with probability p2

  If agent 1 mixes between 1 and 2, they must have the same expected utility   Agent 1 plays 1 => expected utility is –2p2 + 3(1−p2) = 3 – 5p2

  Agent 1 plays 2 => expected utility is 3p2 – 4(1−p2) = 7p2 – 4   Thus 3 – 5p2 = 7p2 – 4, so p2 = 7/12

  Agent 1’s expected utility is 3–5(7/12) = 1/12

  Suppose agent 1 plays 1 with probability p1

  If agent 2 mixes between 1 and 2, they must have the same expected utility   Agent 2 plays 1 => agent 2’s expected utility is 2p1 – 3(1−p1) = 5p1 – 3

  Agent 2 plays 2 => agent 2’s expected utility is –3p1 + 4(1−p1) = 4 – 7p1

  Thus 5p1 – 3 = 4 – 7p1 , so p1 = 7/12   Agent 2’s expected utility is 5(7/12) – 3 = –1/12

Two-Finger Morra

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Dominant Strategies  Let si and si' be two strategies for agent i

  Intuitively, si dominates si' if s gives agent i a greater payoff than si’ for every strategy profile s−i of the remaining agents

 Mathematically, there are three gradations of dominance: 1.  si strictly dominates sʹ′i if ui (si, s−i) > ui (sʹ′i, s−i) for all s−i ∈ S−i 2.  si weakly dominates sʹ′i if

ui (si, s−i) ≥ ui (sʹ′i, s−i) for all s−i ∈ S−i

and ui (si, s−i ) > ui (sʹ′i, s−i ) for at least one s−i ∈ S−i

3.  si very weakly dominates sʹ′i if ui (si, s−i ) ≥ ui (sʹ′i, s−i) for all s−i ∈ S−i

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Dominant Strategy Equilibria   A strategy is strictly (resp., weakly, very weakly) dominant for an agent

if it strictly (weakly, very weakly) dominates any other strategy for that agent   It’s not hard to show that a strictly dominant strategy must be pure

  Obviously, a strategy profile (s1, . . . , sn) where every si is dominant for agent i (strictly, weakly, or very weakly) is a Nash equilibrium

  Such a strategy profile forms an equilibrium in strictly (weakly, very weakly) dominant strategies

  An equilibrium in strictly dominant strategies is necessarily the unique Nash equilibrium

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5, 0 1, 1

3, 3 0, 5

Examples   Example: the Prisoner’s Dilemma

  D is strongly dominant for agent 1   it strongly dominates C if agent 2 uses C   it strongly dominates C if agent 2 uses D

  Similarly, D is strongly dominant for agent 2

  So (D,D) is a Nash equilibrium in dominant strategies   Ironically, of the pure strategy profiles,

(D,D) is the only one that’s not Pareto optimal

C D

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Example: Matching Pennies   Matching Pennies

  Heads isn’t strongly dominant for agent 1, because u1(Heads,Tails) < u1(Tails,Tails)

  Tails isn’t strongly dominant for agent 1, because u1(Heads,Heads) > u1(Tails,Heads)

  Agent 1 doesn’t have a dominant strategy   No Nash equilibrium in dominant strategies

  Which Side of the Road   No Nash equilibrium in dominant strategies   Same kind of argument as above

–1, 1 1,–1

1,–1 –1, 1

Heads Tails

Heads

Tails

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Elimination of Dominated Strategies  A strategy si is strictly (weakly, very weakly) dominated for an agent i if

some other strategy sʹ′i strictly (weakly, very weakly) dominates si

  A strictly dominated strategy can’t be a best response to any move

  So we can eliminate it (remove it from the payoff mtrix)

  Once a pure strategy is eliminated, another strategy may become dominated

  This elimination can be repeated

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Elimination of Dominated Strategies   A pure strategy s may be dominated by a mixed strategy t

without being dominated by any of the strategies in t’s support

  Example: the three games shown below

  In the first game, agent 2’s strategy R is dominated by L (and also by C) •  Eliminate it, giving the second game

  In the second game, M is dominated by neither U nor D •  But it’s dominated by the mixed strategy P(U) = 0.5, P(D) = 0.5

›  (It wasn’t dominated before we removed R) •  Removing it gives the third game (maximally reduced)

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Elimination of Dominated Strategies   This gives a solution concept:

  The set of all strategy profiles that assign 0 probability to playing any action that would be removed through iterated removal of strictly dominated strategies

  A much weaker solution concept than Nash equilibrium   The set of strategy profiles includes all the Nash equilibria   But it includes many other mixed strategies

•  In some games, it’s equal to S (the set of all mixed strategies)   Use this techniques to simplify finding Nash equilibria

  In the current example, go from a 3 × 3 to a 2 × 2 game   Sometimes (e.g., Prisoner’s dilemma), end with a single cell

•  In this case, the single cell must be a Nash equilibrium

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Elimination of Dominated Strategies   Each flavor of domination has advantages and disadvantages

  Strict domination yields a reduced game   Get the same game, independent of the elimination order

  Weak domination can yield a smaller reduced game   But which reduced game may depend on the elimination order

  Very weak domination can yield even smaller reduced games   But again, which one might depend on elimination order   And it doesn’t impose a strict order on strategies:

•  Can have two strategies that each very weakly dominates the other •  For this reason, very weak domination usually is considered the

least important

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Correlated Equilibrium   Motivating example: the mixed-strategy equilibrium

for the Battle of the Sexes   P(both agents choose Opera) = (2/3)(1/3) = 2/9

  Wife’s payoff is 2, husband’s payoff is 1   P(both agents choose Football) = (1/3)(2/3) = 2/9

  Wife’s payoff is 1, husband’s payoff is 2

  P(the agents choose different activities) = (2/3)(2/3) + (1/3)(1/3) = 5/9   In this case, payoff is 0 for both agents

  Each agent’s expected utility is 2(2/9) + 1(2/9) + 0(5/9) = 2/3   If the agents could coordinate their choices, they could

eliminate the case where they choose different activities

  Each agent’s payoff would always be 1 or 2   Higher expected utility for both agents

  This leads to the notion of a correlated equilibrium   Generalization of a Nash equilibrium

Husband Wife

Opera Football

Opera 2, 1 0, 0

Football 0, 0 1, 2

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General Setting   Start out with an n-agent game G

  Let v1, . . . , vn be random variables called signals, one for each agent   For each i, let Di be the domain (the set of possible values) of vi

  Let π be a joint distribution over v1, . . . , vn   π(d1, …, dn) = P(v1=d1, …, vn=dn)

  Nature uses π to choose values d1, …, dn for v1, …, vi

  Nature tells each agent i the value of vi   An agent can condition his/her action on the value of vi

  An agent’s strategy is a mapping σi : Di → Ai

•  σi is deterministic (i.e., a pure strategy)

•  Mixed strategies wouldn’t give any greater generality

  A strategy profile is (σ1, …, σn)   The games we’ve been considering before now are a degenerate case in which the

random variables v1, . . . , vn are independent

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Correlated Equilibrium   Given an n-player game G

  Let v, π, and Σ be as defined on the previous slide   Given a strategy profile (σ1, . . . , σn), the expected utility for agent i is

  (v, π, Σ) is a correlated equilibrium if for every agent i and every mapping σiʹ′ from Di to Ai ,   ui(σ1, …, σi–1, σi, σi+1, …, σn) ≥ ui(σ1, …, σi–1, σiʹ′, σi+1, …, σn)

€

ui(σ1,…,σn ) ≥ π d1,…,dn( )ui σ1 d1( ),…,σn dn( )( )d1 ,…,dn

∑

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Correlated Equilibrium Theorem. For every Nash equilibrium (s1, …, sn), there’s a corresponding correlated

equilibrium (σ1, . . . , σn)   “Corresponding” means they produce the same distribution on outcomes

Sketch of Proof   Let each vi = (v1, …, vn) be independently distributed a random variables, where

•  each vi has domain Ai and probability distribution si(ai ), for ai in Ai

  Hence the joint distribution is π(a1, …, an) = Πi si(ai )   Let each σi be the identity function

  When the agents play the strategy profile (σ1, …, σn), the distribution over outcomes is identical to that under (s1, …, sn)

  The vi’s are uncorrelated and no agent i can benefit by deviating from σi, so (σ1, . . . , σn) is a correlated equilibrium

  But not every correlated equilibrium is equivalent to a Nash equilibrium

  e.g., the correlated equilibrium for the Battle of the Sexes

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Convex Combinations   Correlated equilibria can be combined to form new correlated equilibria

  First we need a definition

  Let x1, x2, …, xn be a finite number of points (vectors, scalars, …)   A convex combination of x1, x2, …, xn is a linear combination of them

in which all coefficients are non-negative and sum up to 1, i.e., a1x1 + a2x2 + … + anxn

where every ai ≥ 0 and a1 + a2 + … + an = 1   Any convex combination of two points x1 and x2 lies on the straight line

segment between x1 and x2   Any convex combination of x1, x2, …, xn is within the convex hull of

x1, x2, …, xn

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Convex Combinations   Theorem. Any convex combination of correlated equilibrium payoffs can

be realized as the payoff profile of some correlated equilibrium

  To understand this, imagine a public random device that randomly selects which correlated equilibrium to play   Each agent’s overall expected payoff is the weighted sum of the

payoffs from the correlated equilibria that were combined   No agent has an incentive to deviate regardless of the probabilities

governing the first random device

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Summary   I’ve discussed several solution concepts, and ways of finding them:

  Pareto optimality •  Prisoner’s Dilemma, Which Side of the Road

  best responses and Nash equilibria •  Battle of the Sexes, Matching Pennies

  real-world examples •  soccer penalty kicks, road networks (Braess’s Paradox)

  maximin and minimax strategies, and the Minimax Theorem •  Matching Pennies, Two-Finger Morra

  dominant strategies •  Prisoner’s Dilemma, Which Side of the Road, Matching Pennies

  correlated equilibrium •  Battle of the Sexes