Mixed Strategy Equilibrium
Advanced Microeconomics II
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Zero‐Sum Game
• Matching Pennies Player 2 Player 1
Heads Tails
Player 1
Heads 1
-1
-1 1
Tails -1
1
1 -1
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No Nash Equilibrium?
• The distinguishing feature of zero‐sum game is that each player would like to outguess the othersince there is No“win‐win” situation.
• Examples are:
P k bl ff t
–Porker: bluff or not
–Battle: attack (/defend) by land or by sea –Tennis: left or right to serve (/receive)
• When each player would always like outguess the other(s), there is no Nash equilibrium.
Is there no solution or stable outcome?
Mixed Strategy
• A mixed strategy for a player is a probability distributionover some (or all) of her strategies.
• The strategy we have studied so far, i.e., taking some action for sure, is called a pure‐strategy.
• When the outcome of the game is uncertain, we assume that each player maximizes expected value of her payoff.
Expected utility theory(von Neumann and Morgenstern, 1944)
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Matching Pennies Again
• Introducing mixed strategies Player 2
Player 1
Heads (p) Tails (1-p) y
Heads (q)
1 -1
-1 1
Tails (1-q)
-1 1
1 -1
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How to Find Equilibrium?
•
If a player takes both “Heads” and “Tails” with
positive probability, she must be indifferent
between these two pure strategies, i.e., the
expected payoff derived by choosing Heads
expected payoff derived by choosing Heads
must be equal to that by choosing Tails.
•
‐p+(1‐p)=p‐(1‐p), hence p=0.5.
•
q‐(1‐q)=‐q+(1‐q), hence q=0.5.
How to Verify Equilibrium?
• Note that if p=0.5, Player 1 does not have a strict incentive to change her strategy from q=0.5.
• Similarly, Player 2 does not have a strict incentive to change his strategy from p=0.5, if q=0.5.
Therefore, p=q=0.5 constitutes a mixed‐strategy
Therefore, p q 0.5 constitutes a mixed strategy equilibrium.
Q: Is this equilibrium reasonable/stable?
A: Yes (each player ends up randomizing two strategies equally if the rival is smart enough).
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Modified Matching Pennies
• Suppose the payoffs in the up‐left cell changes as the following:
Player 2 Heads (p) Tails (1-p) Player 1
Heads (q)
2 -2
-1 1
Tails (1-q)
-1 1
1 -1
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Indifference Property
•
Under mixed‐strategy NE, Player 1 must be
indifferentbetween choosing H and T:
‐2p+(1‐p)=p‐(1‐p), hence p=0.4.
Si il l l 2 b
i diffb
•
Similarly, Player 2 must be indifferent between
choosing H and T:
2q‐(1‐q)=‐q+(1‐q), hence q=0.4.
•
You can easily verify that (p,q)=(0.4,0.4)
indeed becomes a mixed‐strategy NE.
Existence of NE
Theorem (Nash, 1950)
• If a game has finitenumber of players and actions, then there exists at least one Nash equilibrium, possibly involving mixed strategies.
The proof uses the Kakutani’s fixed‐point theorem.
Best response mappings satisfy the condition of the fixed‐point theorem, and hence have a fixed point, which is equivalent to a NE!
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Infinite Games
• The finite game assumption is not necessary but gives a sufficientcondition for the existence of NE.
• There are many games that do not satisfy the conditions of the Theorem but nonetheless have one or more NE (e g Bertrand model Cournot model) or more NE. (e.g., Bertrand model, Cournot model) Q: Are there any games which do not even have a
mixed‐strategy equilibrium? A: Yes, e.g., Integer game.
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Remarks
• Indifference property in mixed strategy NE. –If a player chooses more than one pure strategy with
strictly positive probability, then she must be indifferentbetween all of these pure strategies.
P t t i i l f i d t t
• Pure strategy is a special case of mixed strategy assigning a strategy probability one.
• Any finite game has a Nash equilibrium, possibly in mixed strategies.
• Interesting example: Reporting a Crime
Check the following slides.
Reporting a Crime
• A crime is observed by a group of n people. Each person would like the police to be informed but prefers that someone else make the phone call.
• Players: The n people.
S i “C ll” “D ’ ll”
• Strategies: “Call” or “Don’t call”
• Payoffs: 0if no one calls.
vif someone else calls but she does not. v‐cif the player calls.
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Pure‐Strategy Equilibrium
• This game has n pure‐strategy NE, in each of which exactly one person calls.
–If that person switches to not calling, her payoff falls from v‐c to 0; if any other person switches to calling, his payoff falls from v to v‐c.
f h b f h d ff
• If the members of the group differ in some respect, then these asymmetricequilibria may be compelling as steady states.
• For example, the social norm in which the oldest person in the group makes the phone call is stable.
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Mixed‐Strategy Equilibrium
• There is a symmetricmixed strategy equilibrium in which each person calls with positive probability p<1.
• In any such equilibrium, each person’s expected payoff to calling is equal toher expected payoff to not calling
not calling.
v‐c=v(1‐Pr{no one else calls})
c/v=Pr{no one else calls}
• Notice that the probability that no one else calls is independent of n, but constant!