Nash Equilibrium
Jeffrey Ely
April 13, 2015
Jeffrey Ely
Nash Equilibrium
Common Knowledge Theories
We said that best-reply sets represent theories of play.
Definition
A product set ×N
i =1 Zi = Z ⊂ A has the best-reply property if for each i,
and for all ai ∈ Zi , there is an α−i ∈ ∆Z−i such that ai ∈ BRi (α−i ). ie
Zi ⊂
[
BRi (α−i ).
α−i ∈∆Z−i
Jeffrey Ely
Nash Equilibrium
Common Knowledge Theories
Implied by the assumptions of
Rationality
Common knowledge of rationality
Common knowledge that the outcome will be in Z
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Nash Equilibrium
Experience
Reasons common knowledge of Z may have emerged
Past experience.
Knowing the theory.
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Nash Equilibrium
Equilibrium
Enough past experience and the players will come to know more:
Not just that the outcome will be some element of Z
A distribution over Z .
Theories based on common knowledge of a distribution over outcomes are
called equilibrium theories.
Jeffrey Ely
Nash Equilibrium
An Equilibrium Theory
Suppose we have
Rational players,
Common knowledge of rationality, and
Common knowledge that the distribution over outcomes is some
α ∈ ∏i ∆Ai .
(i.e. independence)
Jeffrey Ely
Nash Equilibrium
Nash equilibrium
Definition
A mixed action profile α ∈ ∏N
i =1 ∆Ai is a Nash equilibrium if for every
player i and for each pure action ai ∈ Ai which is played with positive
probability, i.e. αi (ai ) > 0, we have
ai ∈ BRi (α−i ).
Jeffrey Ely
Nash Equilibrium
Hawk-Dove
D
H
D
2, 2
3, 0
H
0, 3
−1, −1
Figure: The Hawk-Dove game
Three Nash equilibria:
1
H, D (pure-strategy Nash equilbirium, asymmetric)
2
D, H (pure-strategy Nash equilibrium, asymmetric)
3
(1/2, 1/2), (1/2, 1/2) (mixed-strategy Nash equilibrium, symmetric)
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Nash Equilibrium
Mediator for Peace
Suppose the two countries hired a mediator to help them achieve peace.
What’s the best the mediator could do?
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Nash Equilibrium
Correlated Equilibrium
Definition
An outcome distribution α ∈ ∆A is a Correlated Equilibrium if for every
player i and for each pure action ai ∈ Ai which has positive probability, i.e.
αi (ai ) > 0, we have
ai ∈ BRi (α−i |ai )
(the definition is identical to Nash equilibrium except that α need not be
independent.)
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Nash Equilibrium
Existence
Proposition
Every finite game has a Nash equilibrium.
Jeffrey Ely
Nash Equilibrium
Proof of Existence
Denote K = ∏N
j =1 ∆Aj . Consider the best response correspondence
BR : K ⇒ K
defined by
BR (α) =
∏ BRj (α−j ).
j
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Nash Equilibrium
Kakutani’s Fixed Point Theorem
Theorem
Let K ⊂ RN be a non-empty, compact and convex set and let f : K ⇒ K
be a correspondence which
has non-empty values (f (x ) 6= ∅ for all x ∈ K .)
has convex values (f (x ) is convex for all x.)
has closed graph:
{(x, y ) : y ∈ f (x )}
is a closed subset of R2N .
Then f has a fixed point, i.e., there is an x ∗ ∈ A such that
x ∗ ∈ f (x ∗ ).
Jeffrey Ely
Nash Equilibrium
Proof of Existence
BR is non-empty valued because in a finite game BRi (α−i ) is non-empty
for each i and for all α−i .
Jeffrey Ely
Nash Equilibrium
Proof of Existence
BR is convex-valued because BRi (α−i ) is the set of maximizers of a linear
payoff function.
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Nash Equilibrium
Closed Graph
The graph of BR is the set
{(α, α̂) : α̂ ∈ BR (α)} .
To show that it is closed, take any convergent sequence (αk , α̂k ) of points
in the graph and let (α, α̂) be its limit point. We will show that (α, α̂)
belongs to the graph of BR.
Jeffrey Ely
Nash Equilibrium
Closed Graph
That is we want to show that α̂i ∈ BRi (α−i ) for all i, i.e.
ui (α̂i , α−i ) ≥ ui (ai , α−i )
for any action ai ∈ Ai . By construction, α̂ki was at least as good as ai for
all k:
ui (α̂ki , αk−i ) ≥ ui (ai , αk−i )
and by continuity the inequality is preserved in the limit.
Jeffrey Ely
Nash Equilibrium
Some Basic Properties
A Nash equilibrium attaches probability 1 to rationalizable strategies.
The addition of a strictly dominated strategy does not affect the set of
Nash equilibria.
A Nash equilibrium can involve the play of a weakly dominated
strategy.
L
R
A 1, 3 2, 0
B
1, 1
0, 0
A strict equilibrium need not exist.
A pure-strategy equilibrium need not exist.
(Properly) mixed equilibria necessarily involve indifferences.
H
T
H
1, −1
0, 0
T
0, 0
−1, 1
Figure: Matching Pennies
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Nash Equilibrium
Comparative Statics Can Be Subtle
L
C
R
L
0, 0
1, −1
1, −1
C
1, −1
0, 0
1, −1
R
1, −1
1, −1
0, 0
Figure: Penalty Kick
In the unique Nash equilibrium,
both players randomize uniformly over their three strategies.
a shot in any direction is equally likely to score.
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Nash Equilibrium
But There’s This
Humiliation
Jeffrey Ely
Nash Equilibrium
Comparative Statics Can Be Subtle
L
C
R
L
0, 0
1, −1
1, −1
C
1, −1
−5, 0
1, −1
R
1, −1
1, −1
0, 0
Figure: Penalty Kick with public shaming
Jeffrey Ely
Nash Equilibrium
Comparative Statics Can Be Subtle
In the unique Nash equilibrium:
The penalty-taker’s strategy is unaffected by the threat of humiliation.
The goalkeeper’s strategy is affected.
In fact the goalkeeper is now less likely to choose C .
The probability of scoring is highest when shot down the middle.
Jeffrey Ely
Nash Equilibrium
Games With Infinitely Many Strategies
To construct a mixed-strategy equilibrium in a two-player game
We guess the support of each players mixed-strategy, say S1 ⊂ A1 and
S2 ⊂ A2 .
We find a mixed-strategy for player 2 that makes player 1 indifferent
among all the actions in S1 .
And we find a mixed strategy for player 1 that makes 2 indifferent
among all actions in S2 .
Then we make check that no actions outside of S1 or S2 give higher
payoffs.
The same method applies when there are infinitely many strategies.
Jeffrey Ely
Nash Equilibrium
First-Price Auction
MWG has a simple version (ex. 8.D.3) where bids are integers, we will
consider a version where any real number is a feasible bid.
2 bidders with maximum willingness to pay:
I v = 1
1
I v = 2
2
The winning bidder’s payoff is vi − bi . The loser’s payoff is zero.
Ties broken at random.
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Nash Equilibrium
No Pure-Strategy Nash Equilibrium
This game does not have a pure-strategy Nash equilibrium.
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Nash Equilibrium
Mixed Equilibrium
Let’s construct a mixed-strategy equilbrium.
It seems intuitive that 2 should win. Let’s construct an equilibrium in
which he wins with probability 1.
The lowest 2 can bid and win with probability 1 is b2 = 1. Let’s take
that to be 2’s strategy.
So it will be 1 that mixes.
Let’s construct a strategy that has no atoms at any b1 > 0.
What will be the probability F1 (t ) that b1 ≤ t?
I
I
I
I
I
This probability must be chosen in a way that makes 2 unwilling to bid
t.
If no atom at t, then if 2 bid would get payoff F1 (t )(2 − t ) from
bidding t.
So if F1 (t ) = 1/(2 − t ), then bidding t is no better than bidding 1.
This leaves an atom with mass 1/2 at b1 = 0.
So we check separately that b2 = 0 is no better than b2 = 1.
Jeffrey Ely
Nash Equilibrium