Game Theory at Grace Baptist Church 參

(1)

Game Theory at Grace Baptist Church

Spring 2006

Professor Hsueh-I Lu （呂學一）

National Taiwan University

(2)

重要的規定（將會每週提醒）



除了白開水之外，禁止在教室內飲食，

違者立刻喪失旁聽與修課的資格。



注意路上的安全，過馬路時不要趕時間

，我們上課不會點名，所以遲到，缺席

都沒有關係，就是千萬不要「用生命趕

路」。



盡可能將腳踏車停在校園內然後走過來

，以免造成懷恩堂周遭環境的紊亂。

(3)

Outline



Mixed strategies versus pure strategies



Mixed strategy Nash equilibrium



Correlated equilibrium

(4)

Randomized Strategies in

Strategic Games

(5)

Pure v.s. mixed strategies

Rock

Scissors

Paper

Rock

0

1 -1

Scissors

-1

0

1 Paper

1 -1

0

0.4

0.3

0.3 1/3

1/3

(6)

Nash equilibrium under

mixed strategies

(7)

Definition

Consider a two-person zero-sumgamein strategic form. Let

H (x;y)

denote the expected payo®of the row player when

therow player uses mixed strategy x and thecolumn player

uses mixed strategy y.

A pair of mixed strategies (x

¤

_{; y}

¤

_{) is a}

_{saddle point}

_if

H (x;y

¤

_{) · H (x}

¤

_{; y}

¤

_{) · H (x}

¤

_{; y)}

holds for all mixed strategies x and y.

If (x

¤

_;y

¤

_{) is a saddle point of the game, then H (x}

¤

_;y

¤

_{) is}

(8)

An example

3

2

1

4 The above game has no saddle point under pure strategies.

Supposethat therow player usesa mixed strategy x = (p;1¡ p)

and the column player uses a mixed strategy y = (q;1¡ q).

The expected payo®is

H (x;y) = 3pq+ p(1¡ q) + 2q(1¡ p) + 4(1¡ p)(1¡ q)

= 4pq¡ 3p¡ 2q+ 4:

(9)

A saddle point p = 0:5 and q= 0:75.

H (x;y) = 4pq¡ 3p¡ 2q+ 4.

If p = 0:5, then H (x;y) = 2:5 no matter what q is.

If q = 0:75, then H (x;y) = 2:5 no matter what p is.

Therefore, ((0:5;0:5);(0:75;0:25)) is a saddle point of the

game under mixed strategies.

This game has a value 2:5.

3

1

(10)

Lemma

Let G bea two-person zero-sum gamewith expected payo®

H for therow player. Then, the following three statements

are equivalent.

F

1. G has a value v (and thus has a saddle point.)

F

2. max

_x

min

_y

_{H (x;y) = v = min}

_y

max

_x

H (x;y).

F

3. There exist v, x, and y such that

I

(a) H (x;

j

) ¸ v holds for each

pure strategy j

of the

column player; and

I

(b) H (i

; y) · v holds for each

pure strategy i

of the

row player.

(11)

Proof



(1)  (2).

– This can be proved exactly the same way for

the equivalence under pure strategies.



(1)  (3)

– This step is straightforward by letting (x, y) be

the saddle point. (To see this, observe that a

pure strategy is also a mixed strategy.)

(12)

Proof for (3)



_

(1)

This step can be proved by verifying the following

statements.

By 3(a) we know H (x;y) ¸ v. By 3(b), we know

H (x;y) · v. Therefore, H (x;y) = v.

By 3(a), H (x; ^

_{y) ¸ v = H (x;y) holds for any mixed}

strategy ^

y of the column player. By 3(b), H (^

x;y) ·

v = H (x;y) holds for any mixed strategy ^

x of the

row player. Therefore, (x;y) is a saddle point, and

H (x;y) = v is a value of the game.

(13)

von Neumann + Morgenstern

1944



Any

finite

two-person zero-sum game in

strategic form has a saddle point

under

mixed strategies

. (More precisely, they

show that Statement 2 of the previous

lemma holds.)

(14)

Nash gave an alternative

proof in his dissertation

using fixed-point theorem.

(15)

Fixed-Point Theorem

[Brouwer, 1900s] Let X be a compact convex subset

R

n

_{. For any continuous mapping Á : X ! X , there}

(16)

The minimax theorem

T heorem [von N eumann, 1928] For any

two-person zero-sum ¯nite game G where H is the

ex-pected payo®function for the row player, we have

that

max

(17)

Nash’s proof [1950]

For any i, j , x, and y, we de¯ne

p

_i

_{(x;y) = max(0;H (i;y) ¡ H (x;y));}

q

_j

_{(x;y) = max(0;H (x;y) ¡ H (x;j )):}

By Statement 3 of Lemma, it suf¯ces to prove that

there exist x

¤

_{and y}

¤

_{such that}

p

_i

(x

¤

_;y

¤

_{) = 0 = q}

j

(x

¤

;y

¤

)

(18)

Nash’s proof (cont.)

Let X (respectively, Y ) bethespaceconsisting of all mixed

strategies of the row (respectively, column) player.

De¯ne Á : X £ Y ! X £ Y by Á(x;y) = (®;¯), where

®

_i

₌

x

i

+ p

i

(x;y)

1+

X

^

_i

p

_^

_i

(x;y)

;

¯

_j

₌

y

j

+ q

j

(x;y)

1+

X

q

_^

_j

(x;y)

:

(19)

Nash’s proof (cont.)

One can verify that

F

®is a mixed strategy of the row player;

F

¯ is a mixed strategy of the column player;

F

Á is continuous; and

F

X £ Y is a compact convex set.

By Fixed-Point Theorem of Brouwer, there exists

(x

¤

_;y

¤

_{) such that}

(20)

Nash’s proof (cont.)

That is,

x

¤

i

=

x

¤

i

+p

i

(x

¤

;y

¤

)

1+

X

^

_i

p

_^

_i

(x

¤

_;y

¤

₎

;

y

¤

j

=

y

¤

j

+q

j

(x

¤

;y

¤

)

1+

X

^

j

q

_^

_j

(x

¤

_;y

¤

₎

hold for all purestrategies i;j .

Therefore,

x

¤

i

X

^

_i

p

_^

_i

(x

¤

_;y

¤

_{) = p}

i

(x

¤

;y

¤

);

y

¤

j

X

^

j

q

_^

_j

(x

¤

_;y

¤

_{) = q}

j

(x

¤

;y

¤

)

(21)

Claim

F

Thereexists a purestrategy i of therow player

such that x

¤

i

> 0 and p

i

(x

¤

;y

¤

) = 0.

F

There exists a pure strategy j of the column

player such that y

¤

j

> 0 and q

j

(x

¤

;y

¤

) = 0.

Clearly, the above claim immediately implies that

P

i

p

i

(x

¤

;y

¤

) =

P

j

q

j

(x

¤

;y

¤

) = 0, proving the

(22)

Proof of the claim

Let us prove the ¯rst statement. T he second statement can be proved

in the same way.

Since x

¤

_{is a mixed strategy, there exists some i with x}

¤

i

> 0. Assume

for contradiction that for all pure strategies i with x

¤

i

> 0, we have

p

_i

(x

¤

_{; y}

¤

_{) > 0. It follows that}

H (x

¤

_{; y}

¤

_{) =}

X

i

x

¤

i

¢H (i;y

¤

)

>

X

i

x

¤

i

¢H (x

¤

; y

¤

)

= H (x

¤

; y

¤

);

(23)

The minimax theorem

T heorem [von N eumann, 1928] For any

two-person zero-sum ¯nite game G where H is the

ex-pected payo®function for the row player, we have

that

max

(24)

Remarks



The proof of von Neumann’s minimax theorem

is non-constructive.



It turns out that computing a saddle point (i.e.,

Nash equilibrium) under mixed strategies is

likely to be hard, even for a 2-player game.

(PPAD-complete)



Computing a mixed strategy Nash equilibrium

(25)

von Neumann and

Morgenstern’s assumption

Preference of a strategic game with respect

to mixed strategies can be realized by the

expected pay-off of some utility function

for the game.

(26)

A counter example



Four kinds of lotteries:

– (1) $2,000,000 with probability 1

– (2) $10,000,000 with probability 0.1 +

$2,000,000 with probability 0.89 +

$0 with probability 0.01

– (3) $2,000,000 with probability 0.11 +

$0 with probability 0.89

– (4) $10,000,000 with probability 0.1 +

$0 with probability 0.9

$2M

$2.78M

$0.22M

$1M

(27)

Conlisk & Camerer

and many others



Four kinds of lotteries:

– (1) $2,000,000 with probability 1

– (2) $10,000,000 with probability 0.1 +

$2,000,000 with probability 0.89 +

$0 with probability 0.01

– (3) $2,000,000 with probability 0.11 +

$0 with probability 0.89

– (4) $10,000,000 with probability 0.1 +

$0 with probability 0.9

$2M

$2.78M

$0.22M

$1M

>>

(28)

Let u() be any arbitrary

utility function

T he ¯rst preference relation says that

u(2) > 0:1¤u(10) + 0:89¤u(2) + 0:01¤u(0):

Substracting 0:89¤u(2) from both sides, we have

0:11¤u(2) > 0:1¤u(10) + 0:01¤u(0):

Adding 0:89¤u(0) to both sides, we have

0:11¤u(2) + 0:89¤u(0) > 0:1¤u(10) + 0:9¤u(0);

contradicting the second preference relation which says that

(29)

More general setting

(30)

Mixed extension

The mixed extension of the ¯nite strategic game (N;(A

_i

);(u

_i

)) is

the strategic game (N;(¢ (A

_i

));(U

_i

)), where

F

¢ (A

_i

) is the set of probability distributions over A

_i

, and

F

U

_i

maps the cross product of ¢ (A

_j

) to real numbers by

U

_i

_{(®) =}

X

a2 A

0 @

Y

j 2 N

®

_j

(a

_j

)

1 A u

_i

(a):

(31)

For example

a1 (3,1) (0,0) (0,0)

b1

b2

b3

a2 (0,0) (2,2) (0,0)

a3 (0,0) (0,0) (1,3)

.. .. (0,1,0

)

....

(0.5,0,0.5)

.

..

(0,1,0)

(2,2)

(0,0)

..

(0.5,0,0.5)

(0,0)

(1,1)

(32)

Mixed strategy Nash equilibrium

A mixed strategy Nash equilibrium of a ¯nite strategic game is a

Nash equilibrium of its mixed extension.

(33)

Theorem 1

(34)

Proof

Let G = (N;(A

i

);(u

i

)) be a strategic game, and for each player

i let m

_i

be the number of members of the set A

_i

. Then, we

can identify the set ¢ (A

_i

) of player is mixed strategies with the

set of vectors (p

₁

;:::;p

_m

_i

) for which p

_k

¸ 0 holds for all k and

P

_m

_i

k=1

p

k

= 1. This set is nonempty, convex, and compact. Since

expected payo®is linear in the probabilities, each players payo®

function in the mixed extension of G is both quasi-concave in his

own strategy and continuous. Since the mixed extension of G

sat-is¯es all therequirements of thenext lemma (i.e., Proposition 20.3

of our textbook), the theorem is proved.

(35)

Lemma

The strategic game (N;(A

_i

);(%

_i

)) has a Nash equilibrium if for each

i in N we have that

F

the set A

_i

of actions of player i is a non-empty compact convex

subset of a Euclidian space, and

F

the preference relation %

_i

is continuous and quasi-concave on

A

_i

.

C omment

F

A preference relation %

_i

over A

_i

is quasi-concave on A

_i

if for

every a

¤

_{2 A, the set f a}

i

2 A

i

j (a

¤

_{¡ i}

;a

i

) %

i

a

¤

g is convex.

(36)

Theorem 2

Terminology Let G be a ¯nite game. Let ® be a mixed-strategy

action of G, i.e., ®

_i

is a mixed strategy of the i-th player in G.

F

A pure strategy a

_i

of the i-th player is in the

support

of ®

_i

if ®

_i

assigns positive (i.e., non-zero) probability ®

_i

(a

_i

) to a

_i

.

F

A mixed strategy x of the i-th player is one of her

best response

to ®

_{¡ i}

if she does not prefer (y;®

_{¡ i}

) to (x; ®

_{¡ i}

) for any mixed

strategy y of the i-th player.

T heorem 2 ®is a mixed strategy Nash equilibrium of G

if and only

if

for each i, each pure strategy of the i-th player in the support of ®

_i

is

(37)

Illustration

F

If ®

_i

_{= (}

0:7

; 0;

0:2

; 0; 0;

0:1

), then the 1-st, 3-rd, and 6-th pure

strategies of the i-th player is in the support of ®

_i

.

F

If

u

₁

_{((p; 1¡ p); (q; 1¡ q)) = 10¡ 7¤(q¡ p¡ 0:4)}

2 is the payo®of the ¯rst player in a two-player zero-sum game,

then

I

(0:4; 0:6) is the best response of player 1 to the mixed strategy

(0:8; 0:2) of player 2.

I

(0; 1) and (1; 0) are the best responses of player 2 to the mixed

strategy (0:1; 0:9) of player 1.

F

Comment: ®is mixed strategy Nash equilibrium if, for each

(38)

The if direction

Observe that ®

_i

is a convex combination of the pure

strategies of the i-th players that are in the support

of ®

_i

. Since each of those pure strategies in the

sup-por ®

_i

is a best response to ®

_{¡ i}

, so is ®

_i

.

(39)

Proof for the only-if

direction

Suppose that ®is a mixed strategy Nash equilibrium.

That is, each ®

_i

is a best response to ®

_{¡ i}

. Let a

_i

be an

arbitrary pure strategy of the i-th player in the support

of ®

_i

. (That is, ®

_i

assigns positive probability p

_i

to a

_i

.)

Assumefor a contradiction that a

_i

isnot a best response

to ®

_{¡ i}

for the i-th player. If we move p

_i

of a

_i

in ®

_i

to

a best response x

_i

to ®

_{¡ i}

, the resulting mixed strategy

has to havean improved expected pay-o®, contradicting

the assumption that ®

_i

is a best action to ®

_{¡ i}

.

(40)

So, verifying a Nash

is relatively easy…

3 1

2 4

We can verify as follows that ((0:5;0:5);(0:75;0:25)) is a

mixed strategy Nash equilibrium of the two-person

zero-sum game on the right.

F

If (0:75;0:25) is the mixed strategy of player 2, then

the expected payo®of player 1 with mixed strategy

(p;1¡ p) is 3¤0:75¤p+ 0:25¤p+ 2¤0:75¤(1¡ p) +

4¤0:25¤(1¡ p) = 2:5. Clearly, (1;0) and (0;1) are

best response to (0:75;0:25).

F

If (0:5;0:5) is the mixed strategy of player 1, then

the expected payo®of player 2 with mixed strategy

(q;1¡ q) is 0:5¤(5¤q+ 5¤(1¡ q)) = 2:5. Clearly,

(1;0) and (0;1) are best responses to (0:5;0:5).

(41)

Guessing 2/3

of the average

Each of n people announces a number in the set

f 1;2;::: ;100g. A prize of one million dollars is

split equality between all the people whose

num-ber is closest to 2=3 of the average numnum-ber.

Question: What is a mixed strategy Nash

equi-librium of the game?

(42)

A Nash equilibrium

All n players announcing 1 is a Nash equilibrium.

To see this, suppose that the i-th player announces j > 1 and all the other

n ¡ 1 players announce 1. One can verify that n ¸ 2 and j > 1 together

imply

j ¡

n ¡ 1+ j

n

¢

2

3 >

n ¡ 1+ j

n

¢

2

3 ¡ 1:

That is, announcing 1 yields payo®

1 n

and announcing j > 1 yields payo®

zero for player i. Therefore, announcing 1 is the best response of player i

to the condition that everybody else also announces 1.

Comment: Observe that announcing 1 is the only pure strategy of player i

in the support of the mixed strategy that announces 1 with probability 1.

(43)

Exercise



Prove that “all players announcing 1” is

the unique mixed strategy Nash

equilibrium of “guessing 2/3 of the

average”.

(44)

斬首行動

(45)

The game



The commander-in-chief of the enemy is

supposed to be on one of those 9 tanks.



You have only one anti-tank missile to fire at

one of those 9 tanks.



In this two-person zero-sum game, if your

missile hits the right tank, you gain 1 unit of

payoff. Otherwise, you gain 0 unit of payoff.

(46)

Missileman’s payoff

t1 t2 t3 t4 t5 t6 t7 t8 t9

m1

1 0 0 0 0 0 0 0 0

m2 0

1 0 0 0 0 0 0 0

m3 0 0

1 0 0 0 0 0 0

m4 0 0 0

1 0 0 0 0 0

m5 0 0 0 0

1 0 0 0 0

m6 0 0 0 0 0

1 0 0 0

m7 0 0 0 0 0 0

1 0 0

m8 0 0 0 0 0 0 0

1

0 m9 0 0 0 0 0 0 0 0

1

(47)

A Nash equilibrium

((1=9;1=9;:::;1=9);(1=9;1=9;:::;1=9)) is a mixed strategy

Nash equilibrium.

It suf¯ces to verify the following statements.

F

If player 1 assigns probability 1=9 to each pure

strat-egy, then each of those 9 pure strategies is a best

response for player 2.

F

If player 2 assigns probability 1=9 to each pure

strat-egy, then each of those 9 pure strategies is a best

response for player 1.

(48)

No other Nash equilibrium

Suppose that (m;t) is a Nash equilibrium,

where m

_i

is the probability that the

mis-sileman assigns to the i-th pure strategy

and t

_j

is the probability of that the enemy

(49)

m = (1/9, 1/9, …, 1/9).

Given that t is themixed strategy of theenemy, theexpected

pay-o®of the missileman using the i-th pure strategy is t

_i

. T herefore,

if m

_i

> 0 (i.e., the i-th pure strategy is in the support of m), then

t

_i

_{= max}

_{1· j · 9}

t

_j

, thereby t

_i

> 0.

Given that m is the mixed strategy of the missleman, the

ex-pected payo®of the enemy using the j -th pure strategy is ¡ m

_j

.

T herefore, if t

_j

> 0, then m

_j

_{= min}

_{1· i · 9}

m

_i

, thereby m

_j

· 1=9.

It follows that if m

_i

> 0, then m

_i

· 1=9. In order to satisfy

P

₉

(50)

t = (1/9, 1/9, …, 1/9).

Since m = (1=9;1=9;:::;1=9) is the mixed strategy

of themissileman in any Nash equilibrium, each pure

strategy of the missileman has to be a best response

to the mixed strategy t of the enemy.

Therefore, all purestrategies of themissleman must

have the same expected payo®, i.e., t

₁

_{= t}

₂

_{= ¢¢¢=}

t

₉

. By t

₁

+t

₂

+¢¢¢+t

₉

_{= 1, we know that t}

₁

_{= t}

₂

₌

¢¢¢= t

9 = 1=9.

(51)

Question

Does the mixed extension of a strictly

competitive game have to be strictly

(52)

Not necessarily

b1

a1 (5,1)

a2 (2,2)

a3 (1,5)

1 ..

..

(0,1,0)

(2,2)

..

(0.5,0,0.5

)

(3,3)

..

(53)

Correlated equilibrium

(54)

Warm-up

(55)

Battle of Sexes



The Da Vinci Code



Mission Impossible III

DVC MI:3

DVC

2,1

0,0

(56)

Pure Nash



The Da Vinci Code



Mission Impossible III

DVC MI:3

DVC

2,1

0,0

(57)

Mixed Nash



The Da Vinci Code



Mission Impossible III

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

There are three mixed strategy Nash equilibria:

F

((1;0);(1;0)).

F

((0;1);(0;1)).

(58)

Theorem (recall)

Terminology Let G be a ¯nite game. Let ® be a mixed-strategy

action of G, i.e., ®

_i

is a mixed strategy of the i-th player in G.

F

A pure strategy a

_i

of the i-th player is in the

support

of ®

_i

if ®

_i

assigns positive (i.e., non-zero) probability ®

_i

(a

_i

) to a

_i

.

F

A mixed strategy x of the i-th player is one of her

best response

to ®

_{¡ i}

if she does not prefer (y;®

_{¡ i}

) to (x; ®

_{¡ i}

) for any mixed

strategy y of the i-th player.

T heorem 2 ®is a mixed strategy Nash equilibrium of G

if and only

if

for each i, each pure strategy of the i-th player in the support of ®

_i

is

(59)

Verification

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

T here are three mixed strategy Nash equilibria:

F

((1;0);(1; 0)): trivial.

F

((0;1);(0; 1)): trivial.

F

((

2

3 ;

1

3 ); (

1

3 ;

2

3 )):

I

If (

2

3 ;

1

3 ) is the strategy of the row player, then

the expected payo®of the column player is

al-ways

2q

3 +

2(1¡ q)

3 =

2

3 .

I

If (

1

3 ;

2

3 ) is thestrategy of thecolumn player, then

the expected payo®of the row player is always

2p

₊

2(1¡ p)

(60)

No more Nash

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

Suppose that ((p;1 ¡ p);(q;1 ¡ q)) is a Nash equilibrium. T he

expected payo®s of both players are 2pq + (1 ¡ p)(1 ¡ q) and

pq+ 2¢(1¡ p)(1¡ q).

F

If 0 < p < 1, then both pure strategies of the row player

are in the support of (p;1 ¡ p). It follows that both pure

strategies have to be best responses to (q;1¡ q). Therefore,

we have 2q = 1¡ q, i.e., q=

1 ₃

.

F

If 0 < q < 1, then both pure strategies of the column player

are in the support of (q;1 ¡ q). It follows that both pure

strategies have to be best responses to (p;1¡ p). Therefore,

we have 2(1¡ p) = p, i.e., p =

2 ₃

.

(61)

What if there is a coin toss,

observable by both players?

And, the strategy of a player takes the following form:

F

If the outcome of the coin toss is head, then use pure

strategy a

₁

.

(62)

DVC or MI:3 with an unbiased

public coin

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

(a

₁

;b

₁

)

(a

₂

;b

₂

)

(63)

An example

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

(DVC; MI:3)

An equilibrium

(64)

?

Verification for player 1

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

(DVC; MI:3)

When (DVC;MI:3) is thestrategy of player 2, theexpected

payo®s of four possible strategies of the ¯rst player are as

follows.

F

(DVC;DVC): 1.

F

(DVC;MI:3): 3=2.

F

(MI:3;DVC): 0.

(65)

(DVC; MI:3)

Verification for player 2

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

?

When (DVC;MI:3) is thestrategy of player 1, theexpected

payo®s of four possible strategies of the second player are

as follows.

F

(DVC;DVC): 1=2.

F

(DVC;MI:3): 3=2.

F

(MI:3;DVC): 0.

(66)

Expected payoff = (3/2, 3/2)

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

(DVC; MI:3)

(67)

The public information for both players

can be less than perfectly correlated



The first player (who is red-color-blind but not

yellow-color-blind) can only tell the difference

between {

r

} and {

o

,

y

} balls.



The second player (who is yellow-color-blind

but not red-color-blind) can only tell the

difference between {

y

} and {

o

,

r

} balls.

(68)

Strategies



Player 1:

– If the ball is

red

, then use a

₁

;

– Otherwise, use b

1 .



Player 2:

– If the ball is

yellow

, then use a

₂

;

– Otherwise, use b

2 .

(69)

Information Partitions



The information partition for player 1

– {{

red

}, {

orange

,

yellow

}}



The information partition for player 2

(70)

Correlated Equilibrium



A

correlated equilibrium

of a strategic game

consists of

– a probability space of a public random variable

observable by the players;

– an information partition (for the above probability

space) for each player; and

– a strategy (i.e., vector of pure strategies) for each

player

such that given the strategies of other players,

each player has no incentive to change her own

strategy.

(71)

Formally,…

A correlated equilibrium of a strategic game (N ;(A

_i

);(u

_i

)) consists of

F

a ¯nite probability space ( ;¼), where

is a set of states and ¼is

a probability measure on ;

F

for each player i 2 N , a partition P

_i

of

(i.e., the information

partition of the i-th player); and

F

for each player i 2 N , a function ¾

_i

: P

_i

_{¡ ! A}

_i

(i.e., the strategy

of the i-th player)

such that

X

! 2

¼(! ) ¢u

_i

(¾

_{¡ i}

(! );¾

_i

(! )) ¸

X

! 2

¼(! ) ¢u

_i

(¾

_{¡ i}

(! );¿

_i

(! ))

holds for each i 2 N and each strategy ¿

_i

of the i-th player.

(72)

Comment



The state space and

the information

partition is part of

the correlated

(73)

Question:

What is the relation between mixed

strategy Nash equilibrium and

correlated equilibrium?

Correlated

equilibrium

Mixed

Nash

equilibrium

(74)

The set of correlated equilibria “contains”

the set of mixed Nash equilibria.

T heorem 1

For each mixed strategy Nash equilibrium

®

of a ¯nite

strategic game (N;(A

_i

);(u

_i

)), thereis a correlated

equi-librium

(( ;¼);(P

_i

);(¾

_i

))

in which for each player i 2 N , the distribution on A

_i

induced by ¾

_i

is ®

_i

.

(75)

Proof

Let

_{= A. For each action a 2 A, de¯ne}

¼(a) =

Y

j 2 N

®

_j

(a

_j

):

For each i 2 N and b

_i

2 A

_i

, let

P

_i

(b

_i

_{) = f a 2 A j a}

_i

_{= b}

_i

g

and let P

_i

consist of the jA

_i

j sets P

_i

(b

_i

). For each a

_i

2 A

_i

, de¯ne

¾

_i

(P

_i

(a

_i

_{)) = a}

_i

:

(76)

Proof

C laim 1 The distribution on A

_i

induced by ¾

_i

is ®

_i

.

Observe that the probability that ¾

_i

outputs a

_i

is

X

c2 P

_i

(a

_i

)

(77)

Proof

C laim 2 (( ;¼); (P

_i

);(¾

_i

)) a correlated equilibrium.

F

T heexpected payo®of thei-th player with respect

to ¾ is the same as that in the mixed strategy

equilibrium ®.

F

T he distribution on A

_i

induced by any other ¿

_i

is

a mixed strategy on A

_i

for thei-th player, thereby

cannot yield a better expected payo®(because ®

is a mixed strategy Nash equilibrium.)

(78)

The set of correlated equilibria “contains”

the set of mixed Nash equilibria.

T heorem 1

For each mixed strategy Nash equilibrium

®

of a ¯nite

strategic game (N;(A

_i

);(u

_i

)), thereis a correlated

equi-librium

(( ;¼);(P

_i

);(¾

_i

))

in which for each player i 2 N , the distribution on A

_i

induced by ¾

_i

is ®

_i

.

(79)

Convex combination

T heorem 2

Let G = (N;(A

i

);(u

i

)) be a strategic game. T hen, any

convex combination of correlated equilibrium payo®s of

G has to be a correlated equilibrium payo®of G.

(80)

Mixed Nash



The Da Vinci Code



Mission Impossible III

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

There are three mixed strategy Nash equilibria:

F

_{((1;0);(1;0)): payo®= (2;1).}

F

_{((0;1);(0;1)): payo®= (1;2).}

F

((

2 _;

1 _);(

1 _;

2 _{)): payo®= (}

2 _;

2 _).

(81)

Convex combination

(2;1)

(1;2)

(

2

3 ;

2

3 )

(82)

Expected payoff = (3/2, 3/2)

DVC MI:3

DVC

2,1

0,0

MI:3

0,0

1,2

(DVC; MI:3)

Correlated

equilibrium

(83)

Convex combination

(2;1)

(1;2)

(

2

3 ;

2

3 )

(

3

2 ;

3

2 )

(84)

Proof

For each k, let ((

k

_;¼

k

_);(P

k

i

);(¾

i

k

)) be a correlated

equilibrium with expected payo®u

k

_{. Without loss of}

generality, we may assume that all

k

_{are disjoint.}

Let u be a convex combination of those u

k

_{. That is,}

there are non-negative numbers ¸

k

_with

P

k

¸

k

= 1

(85)

Proof

We de¯ne (( ; ¼); (P

_i

);(u

_i

)) as follows.

F

Let

₌

S

_k

k

.

F

_{For each ! 2 , de¯ne¼(! ) = ¸}

k

_¢¼

k

_{(! ), where}

k is the index with k 2

k

_.

F

For each i 2 N , let P

_i

₌

S

_k

P

_i

k

.

F

For each i 2 N , let ¾

_i

(P

_i

_{) = ¾}

k

i

(P

i

), where k is

the index with P

_i

2 P

k

i

.

E xercise Prove that (( ;¼); (P

_i

); (u

_i

)) is indeed a

correlated equilibrium of G.

(86)

Convex combination

T heorem 2

Let G = (N;(A

i

);(u

i

)) be a strategic game. T hen, any

convex combination of correlated equilibrium payo®s of

G has to be a correlated equilibrium payo®of G.

(87)

Question:

(1; 2)

(2;1)

(

2 3

;

23

)

?

Can there be correlated equilibria whose

pay-o®s lie outside the convex hull of the paypay-o®s

of mixed strategy Nash equilibria?

(88)

Yes!

Left Right

Top

6,6

2,7

Bottom

7,2

0,0

(7; 2)

(2; 7)

(4

2 3

; 4

23

)

Exercise: Verify that (2;7), (7;2), and (4

2

3 ;4

2

3 ) are

the only payo®s of mixed strategy equilibria.

Now we de¯ne (( ;¼); (P

_i

); (¾

_i

)) as follows.

F

Let

_{= f x;y;zg.}

F

_{Let ¼(x) = ¼(y) = ¼(z) =}

1

3 .

F

Let P

₁

_{= f f xg;f y; zgg and P}

₂

_{= f f x;yg;f zgg.}

F

Let ¾

₁

_{(f xg) = Bottom and ¾}

₁

_{(f y;zg) = Top.}

(89)

Left Right

Top

6,6

2,7

Bottom

7,2

0,0

(7; 2)

(2; 7)

(4

2 3

; 4

23

)

Is this a correlated equilibrium? (Checking the column player.)

Given ¾

₁

_{(f xg) = Bottom and ¾}

₁

_{(f y; zg) = Top, the expected payo®s of}

four possible ¾

₂

are

F

If ¾

₂

_{(f x; yg) = Left and ¾}

₂

_{(f zg) = Left, payo®is (19=3;14=3).}

F

If ¾

₂

_{(f x; yg) = Left and ¾}

₂

_{(f zg) = Right, payo®is (15=3;15=3).}

F

If ¾

₂

_{(f x; yg) = Right and ¾}

₂

_{(f zg) = Left, payo®is (8=3;13=3).}

(90)

Left Right

Top

6,6

2,7

Bottom

7,2

0,0

(7; 2)

(2; 7)

(4

2 3

; 4

23

)

Is this a correlated equilibrium? (Checking the row player.)

Given ¾

₂

_{(f x;yg) = Left and ¾}

₂

_{(f zg) = Right, the expected payo®s of four}

possible ¾

₂

are

F

If ¾

₁

_{(f xg) = Top and ¾}

₁

_{(f y;zg) = Top, payo®is (14=3;19=3).}

F

If ¾

₁

_{(f xg) = Top and ¾}

₁

_{(f y;zg) = Bottom, payo®is (13=3;8=3).}

F

If ¾

₁

_{(f xg) = Bottom and ¾}

₁

_{(f y;zg) = Top, payo®is (15=3;15=3).}

(91)

Left Right

Top

6,6

2,7

Bottom

7,2

0,0

(7; 2)

(2; 7)

(4

2 3

; 4

23

)

The expected payo®of this correlated equilibrium is

(5;5), which is outside the above triangle.

Conclusion: The set of correlated equilibria can be

a proper superset of the set consisting of the convex

combinations of mixed strategy Nash equilibria.

(92)

So, finding all correlated equilibria

is an impossible mission?

(2;1)

(1;2)

(

2

3 ;

2

3 )

?

(93)

Fortunately, …

T heorem 3

Let G = (N;(A

i

);(u

i

)) be a ¯nite strategic game. Any expected

payo®that can be obtained in a correlated equilibrium of G can

also be obtained in a correlated equilibrium of G in which

F

the set of states is A and

F

for each i 2 N , the information partition of the i-th player

consists of all sets of the form

f a 2 A j a

_i

_{= b}

_i

g

for some action b

_i

2 A

_i

.

(94)

Proof sketch

Let (( ;¼);(P

_i

);(¾

_i

)) be a correlated equilibrium of G. Then,

onecan verify that ((A; ^

¼);( ^

P

_i

);(^

¾

_i

)) is also a correlated

equi-librium of G, where

F

P

^

_i

consists of the sets f a 2 A j a

_i

_{= b}

_i

g for each b

_i

2 A

_i

;

F

¾

^

_i

(f a 2 A j a

_i

_{= b}

_i

_{g) = b}

_i

for each b

_i

2 A

_i

; and

F

for each a 2 A, let ^

¼(a) bethesumof ¼(! ) over all ! 2

such that ¾maps the set in P

_i

containing ! to a.

It is left as an exercise to show that (1) the above

((A; ^

¼);( ^

P

_i

);(^

¾

_i

)) is indeed a correlated equilibrium of G and

(2) both correlated equilibria have the same expected payo®.

(95)

Evolutionary equilibrium

(96)

Evolutionary equilibrium



John Maynard Smith (1972) & George R. Price (1973)



A steady state in which all organism take this action and

(97)

Two types of bees

Aggressive Passive

Aggressive

1,1

4,0

Passive

0,4

2,2

(98)

Normal bees

are aggressive

Suppose that normal bees are aggressive. Passive bees are

mu-tants. More precisely,

F

let ² be the probability that a bee (normal or mutant)

en-counters a mutant (i.e., passive bee), and

F

let 1¡ ² be the probability that a bee encounters a normal

(i.e., aggressive) bee,

where 0 < ² < 0:5.

The expected payo®s:

F

_{Normal (i.e., aggressive) bee: (1¡ ²) + 4¢² = 1+ 3¢².}

F

Mutant (i.e., passive) bee: 2¢².

Since ² > 0, we have 1+ 3¢² > 2². Therefore, being aggressive is

(99)

Interpretation



Each bee regularly produces offspring (the

reproduction is asexual).



A bee always follows its parent’s behavior.

– An aggressive bee produces aggressive bees.

– A passive bee produces passive bees.



The two types of bees multiply at a rate

(100)

Normal bees

are passive

Suppose that normal bees are passive. Aggressive bees are

mu-tant. More precisely,

F

let ² be the probability that a bee (normal or mutant)

en-counters a mutant (i.e., aggressive bee), and

F

let 1¡ ² be the probability that a bee encounters a normal

(i.e., passive) bee,

where 0 < ² < 0:5.

The expected payo®s:

F

_{Normal (i.e., passive) bee: (1¡ ²) ¢2 = 2¡ 2².}

F

_{Mutant (i.e., aggressive) bee: 4¢(1¡ ²) + ² = 4¡ 3².}

(101)

Definition

Let G = (f 1;2g;(B;B);(u

i

)) be a symmetric strategic game,

where u

₁

_{(a; b) = u}

₂

_{(b; a) = u(a;b) for some function u. An action}

b

¤

_{2 B of G is an}

_{evolutionarily stable strategy}

_{(ESS) of G if there}

exists an ¹² > 0 such that

(1¡ ²) ¢u(b

¤

; b

¤

) + ² ¢u(b

¤

; b)

>

(1¡ ²) ¢u(b;b

¤

) + ² ¢u(b;b)

holds for all b6

_{= b}

¤

_{and all ² with 0 < ² < ¹².}

Comments:

T he red

>

cannot be relaxed to

¸

.

(102)

An equivalent definition

Let G = (f 1;2g;(B ;B );(u

i

)) be a symmetric

strate-gic game, whereu

₁

_{(a;b) = u}

₂

_{(b;a) = u(a;b) for some}

function u. An action b

¤

_{2 B of G is an}

evolutionar-ily stable strategy

(ESS) of G if

F

(b

¤

_;b

¤

_{) is a Nash equilibrium of G}

_and

F

u(b;b) < u(b

¤

_{;b) holds for every best response}

b2 B to b

¤

_{with b6}

_{= b}

¤

_.

(103)

Example

Aggressive Passive

Aggressive

1,1

4,0

Passive

0,4

2,2



There are two symmetric Nash equilibria:

(A, A) and (P, P).



(A, A) is an evolutionary equilibrium.

– A is the only best response to A. (There are no

other best response to A.)



(P, P) is NOT an evolutionary

equilibrium.

– A is the only best response to P.

– u(A, A) = 1 > 0 = u(P, A).

(104)

Another example

X 2,2 0,0

X

Y

Y 0,0 1,1



There are two symmetric Nash equilibria:



(X, X) is an evolutionary equilibrium.

– X is the only best response to X.



(Y, Y) is also an evolutionary equilibrium.

– Y is the only best response to Y.



That is, both X and Y are evolutionarily

stable strategies.

(105)

Question

Are there games having no

evolutionary equilibrium?

(106)

Yes

X

Y

X 1,1 1,1

Y 1,1 1,1



There are two symmetric Nash equilibria:



(X, X) is NOT an evolutionary equilibrium.

– Y is the only best response to X other than X.

– u(Y, Y) ≥ u(X, Y)



(Y, Y) is NOT an evolutionary equilibrium.

– X is the only best response to Y other than Y.

– u(X, X) ≥ u(Y, X)



That is, neither X nor Y is an evolutionarily