A jatekelmelet. February 5, 2009. These notes were written by Yoav Zemel, and
the lecturer, Itai Arieli, should not be held responsible for the mistakes.
A jatekelmelet February 5, 2009
These notes were written by Yoav Zemel, and the lecturer, Itai Arieli, should not be held responsible for the mistakes. Comments are welcome at
[email protected].
1
Games of two players in a strategic form
A game is an ordered set ({1, 2}, S1 , S2 , U1 , U2 ), where {1, 2} are the players, and Si are non empty sets of strategies of the player i. Ui : S1 × S2 → R is the payment or utility function of player i. Let G be such game. An EQUILIBRIUM in G is the pair (s1 , s2 ) ∈ S1 × S2 for which the following conditions apply :
U1 (s1 , s2 ) ≥ U1 (t, s2 ) ∀t ∈ S1 and
U2 (s1 , s2 ) ≥ U1 (s1 , t) ∀t ∈ S2 Both players are so-called satisfied if the other player sticks with his strategy.
PRISONER’S DILEMMA 1
S1 = {T, B}, and S2 = {L, R}
L R The game can be described as T (3, 3) (0, 4) B (4, 0) (1, 1) The equilibrium is at the point (B, R), as 1 = U1 (B, R) ≥ U1 (T, R) = 0 and 1 = U2 (B, R) ≥ U2 (B, L) = 0. (T, L) is not an equilibrium because 3 = U1 (T, L) < U2 (B, L) = 4 even though it is a better result for both players! B dominates T for player 1, and R dominates L for player 2. For the maximal security level, player 1 will also choose B and player 2 will choose R. L R Now we look at the game T (2, 1) (0, 0) which has two equilibrium points. B (0, 0) (1, 2) the maximal security level of both player is 0. Let G be a game of two players and we define the functions Br1 : S2 → P (S1 ), and Br2 : S1 → P (S2 ), by
Br1 (s2 ) = {s1 ∈ S1 |U1 (s1 , s2 ) ≥ U1 (t, s2 ) ∀t ∈ S1 } = ArgMaxt U1 (t, s2 )
and
Br2 (s1 ) = {s2 ∈ S2 |U2 (s1 , s2 ) ≥ U1 (s1 , s) ∀s ∈ S2 } = ArgMaxs U2 (s1 , s)
THEOREM: (s1 , s2 ) is an equilibrium if and only if s1 ∈ Br1 (s2 ) and s2 ∈ Br2 (s1 ). The proof is trivial.
2
2
Games in the unit square
Let S1 = S2 = [0, 1], U1 (x, y) = 3xy − 2x − 2y + 2 and U2 (x, y) = −4xy + 2x + y U1 = x(3y − 2) − 2y + 2, thus {1} y < Br1 (y) = [0, 1] y = {0} y >
2 3 2 3 2 3
{1} x < Br2 (x) = [0, 1] x = {0} x >
1 4 1 4 1 4
U2 = y(1 − 4x) + 2x, thus
The equilibrium point is ( 14 , 23 ) because both are the best replies for each other (though not the only best replies). What can player 1 ensure for himself ? maxx∈[0,1] miny∈[0,1] U1 (x, y). We calculate the minimum by y. we calculate f (x) = miny U1 (x, y). U1 = y(3x − 2) − 2x + 2 thus x
x< 2 f (x) = 3 x= 2 − 2x x >
2 3 2 3 2 3
The maximum of this function is obtained when x = 2 3
2 3
thus maxx∈[0,1] miny∈[0,1] U1 (x, y) =
and is obtained when x = 23 . Thus, the strategy of maximal secuitry level is
x = 23 . Now we will do a similar calculation and find that U2 = x(2 − 4y) + y thus y < 21 y min U2 (x, y) = 12 y = 12 x 2 − 3y y > 12 The maximum of this function is obtained when y = 1 2
1 2
thus maxy∈[0,1] minx∈[0,1] U2 (x, y) =
and is obtained when y = 12 . Thus, the strategy of maximal secuitry level is
y = 21 .
3
3
Zero sum games - mixed strategies
Let G be a zero sum game, represented by the matrix A ∈ Mn×m , where S1 = {1, . . . , n} and S2 = {1, . . . , m}. The utility function which is the payment player 2 pays player 1, is H : S1 × S2 :→ R, H(i, j) = aij . THE MIXED EXPANSION of the game is S10
n
n
= ∆ = {p ∈ R |pi ≥ 0,
n X
pi = 1}
i=1
S20 = ∆m = {p ∈ Rm |pi ≥ 0,
m X
pi = 1}
i=1
h : ∆n × ∆m → R : h(p, q) =
n X m X
pi qj aij = pt Aq
i=1 j=1
Note that h is multilinear, so h(p, q) =
n X
pi h(εi , q) =
i=1
m X
qi h(p, εi )
i=1
What can player 1 ensure for himself ? maxp minq h(p, q). What can player 2 ensure for himself not to pay more than ? maxq minp h(p, q).
MINIMAX THEOREM: max min h(p, q) = max min h(p, q) = v p
q
q
p
where v is called the VALUE of the game. A strategy p ∈ ∆n is called OPTIMAL for player 1 if it ensures v for any strategy of player 2, that is h(p, q) ≥ v ∀q ∈ ∆m Similarly, a strategy q ∈ ∆m is called OPTIMAL for player 2 if it ensures for him to pay no more than v for any strategy of player 1, that is h(p, q) ≤ v∀p ∈ ∆n 4
THEOREM: max min h(p, q) = max min h(p, εi ) = p
q
p
1≤j≤m
n X
pi aij
i=1
PROOF : For a constant p, minq h(p, q) ≡ f (p), and min1≤j≤m h(p, q) ≡ g(p). Clearly ∀p : g(p) ≤ f (p). We need to show that g(p) = f (p) for any p ∈ ∆n PROOF : min h(p, q) = min q
q
m X
qj h(p, εj ) ≥ min q
j=1
m X
qj ( min h(p, εi ))
j=1
1≤i≤m
but min q
m X
qj ( min h(p, εj )) = min h(p, εj )
j=1
1≤i≤m
i≤j≤m
as qj ’s sum to 1 and the minimum is a constant. Thus, g(p) ≥ f (p) and therefore g(p) = h(p). QED.
4
Calculating the values of 2x2 games
Consider
5 0 3 4
The strategies of the playes give
q 1−q p 5 0 1−p 3 4
and h(p, q) = 5pq+0p(1−q)+3(1−p)q+4(1−p)(1−q) = 3q+2pq+4+4pq−4p−4q = 6pq−4p−q+4 The value of the game is maxp minq h(p, q), and by the previous theorem, max min h(p, q) = max min h(p, εj ) = max min{h(p, ε1 ), h(p, ε2 )} =
0≤p≤1 0≤q≤1
0≤p≤1 j=1,2
0≤p≤1
5
( 4 − 4p p ≥ 1/6 10 = max min{5p+3−3p, 4−4p} = max min{2p+3, 4−4p} = max = 0≤p≤1 0≤p≤1 0≤p≤1 2p + 3 p ≤ 1/6 3 What is the optimal strategy of player 1 ? p∗ =
1 6
and matches the strategy
( 16 , 65 ) What is the optimal strategy of player 2 ? we want to calculate maxp minq h(p, q) : min max h(p, q) = min max h(ei , q) = min max{h(ε1 , q), h(ε2 , q)} =
0≤q≤1 0≤p≤1
0≤q≤1 i=1,2
0≤q≤1
( 4−q min0≤q≤1 max{5q, 3q+4−4q} = min max{5q, 4−q} = min 0≤q≤1 0≤q≤1 5q The optimal strategy of player 2 is q ∗ =
2 3
p≥ p≤
4 6 4 6
=
and matches the strategy ( 13 , 23 )
Let A ∈ Mn×m be a payment matrix. The following conditions are equivalent : (1) p is an optimal strategy for player 1. (2) ∀j : h(p, εj ) ≥ val(A) = v. Note : there is equality in (2) for any j for which there is an optimal q ∈ ∆s2 for player 2 in which qj > 0. Clearly (1)→(2). We need to show that (2)→(1). PROOF : Fix q ∈ ∆m and we need to show that h(p, q) ≥ v. but h(p, q) =
m X
qj h(p, εj) ≥
j=1
m X
qj v = v
j=1
QED. Proof of the note : Let q be optimal for player 2 and qk > 0. v = h(p, q) =
X qj 6=0
6
qj h(p, εj)
10 3
As all the elements of the sum are not smaller than v, had one of them been strictly bigger than v, we would have obtained that the sum is strictly bigger than v, while it equals v exactly, QED.
2 6 Consider 5 5. It is obviously easier to calculate min max as player 2 7 4 has less possibilities: 1 6 − 4q q ≤ 4 1 v = min max h(εi , q) = min 5 ≤ q ≤ 13 4 j 0≤j≤1 i=1,2,3 4 + 3q q ≥ 31 =5, which is the value of the game. The optimal strategies for player 2 are {(q, 1 − q)|1/3 ≥ q ≥ 1/4} What are the best strategies for player 1 ? {(p1 , p2 , 1 − p1 − p2 )} By the previous claim, p is optimal if and only if h(p, εj ) = v = 5 for j = 1, 2. This yields 2p1 + 5p2 + 7 − 7p1 − 7p2 = 5 and 6p1 + 5p2 + 4 − 4p1 − 4p2 = 5 or 2p2 + 5p1 = 2 2p1 + p2 = 1 This yields p2 = 1 which means that the single optimal strategy for player 1 is (0, 1, 0). 7
5
WEAK DOMINATION
Let A ∈ Mn×m be a payment matrix. A strategy i0 of 1 is WEAKLY DOMINATED if ∃t ∈ ∆n : ti0 = 0, and
n X
ti aij ≥ ai0 j ∀1 ≤ j ≤ m
i=1
That is, if player 1 will play t, in any case he will obtain a non-worse result than by playing εi0 . Similarly, we define STRICT DOMINATION with a strict inequation instead of the weak one. THEOREM : Let i0 be a dominated line, and A0 the matrix obtained after the deletion of the line, then (1) V al(A) = val(A0 ) = v (2) The set of optimal strategies of player 2 does not change. (3) If the domination is strict, then the set of optimal strategies for player 1 does not change. Example : 3 0 2 1 2 4 2 1 1 The lower line is weakly dominated as 12 R2 + 21 R1 ≥ R3 , thus the matrix has the same value as
3 0 2 1 2 4
Qw qill now prove the theorem. Let A be a payment matrix of n × m of a two players zero sum game, and i0 be such that there is t ∈ ∆S2 , ti0 = 0 and ∀1 ≤ j ≤ m
n X i=1
8
ti aij ≥ ai0 j
That is, the line i0 is dominated (if the ineqation is strict then the domination is strict). Let A0 be the matrix obtained by removing line i0 for A, then : 1 v(A) = v(A0 ). 2 The column player has the same set of optimal strategies. 3 The set of optimal strategies of player 1 in A0 is a subset of the set of his optimal strategies in A. PROOF : v(A) ≥ v(A0 ) as we minimized the number of strategies of player 1. We need to see that v(A0 ) ≥ v(A). Let b be an optimal strategy of player 1 in A. This is equivalent to hA (p, εj ) ≥ v(A) ∀1 ≤ j ≤ m. We will build a strategy pb that give i0 = 0 and ensures the value vA . If pi0 = 0 we are done (b p = p). If not, then instead of giving this value to pi0 , we will give it to the convex combination that dominates it. Define p by pi =
pi , i0 6= i, pi0 = 0 (1 − pi0 )
and pb = (1 − pi0 )p + pi0 t It is indeed a stragtegy. Now, for j, hA (b p, εj ) = (1−pi0 )hA (p, εj )+pi0 hA (t, εj ) ≥ (1−pi0 )hA (p, εj )+pi0 hA (εi0 , εj ) = = (1 − pi0 )
X i6=i0
pi aij + pi0 ai0 j = hA (p, εj ) = v(A) 1 − pi0
QED 1. Any strategy of player 2 that ensured v(A) will still ensure that value (or a better one). We have to show that we did not add more strategies for player 9
2. Let q be an optimal strategy for player 2 in A0 . For any i 6= i0 hA (εi , q) = v(A0 ) = v(A) hA (εi0 , q) ≤ hA (t, q) ≤ v(A) ∀1 ≤ i ≤ n : hA (εi , q) ≤ v(A0 ) And thus q is also an optimal strategy in A, QED. Example : Consider the game :
7 −5 2 1 3 2 4 −1 1 1 R 2 1
+ 12 R2 = (4, −1, 2) ≥ (4, −1, 1) and we can ignore the third line without
changing the value of the game. This yields the game
1 C + 21 C2 2 1
7 −5 2 1 3 2
= (1, 2) ≤ (2, 2) and we can ignore the third column without chang-
ing the value of the game.
In this game, v =
13 ,p 7
7 −5 1 3
= ( 17 , 76 ), q = ( 47 , 37 ). Now we go back to see if we
missed some strategy. In this case we could have made a strict domination and thus we did not miss any strategy in the second case. Now we we will check if we missed a strategy of player 1. Let p = (x, y, 1 − x − y). In order for p to be an optimal strategy, we must have h(p, ε1 ) = v =
13 7
= h(p, ε2 ), and h(p, ε3 ) ≥
13 . 7
equations yield that there are more optimal strategies. 10
Solving the resulting
6
Games in a strategic form
(N, (Si )i∈N , (gi )i∈N ) where N = {1, . . . , n} is the set pf players, and Si = {1, . . . , mi } the set of pure strategies of player i, and g : s1 × · · · × sn → RRR is the payment matrix. The MIXED EXPANSION of the game is defined by 1
n
Gi : X1 ×· · ·×Xn → RRR : Gi (X , . . . , X ) =
m1 X j1 =1
···
mn X
x1j1 . . . xnjm g i (j1 , . . . , jn )
jm =1
When Xi = ∆mi is the set of mixed strategies for player i. xi ∈ ∆mi . For any y i ∈ ∆mi , we denote Gi (x−i , y i ) = Gi (x1 , . . . , y i , xi+1 , . . . xn ) the payment fot player i if he plays y i , and the rest of the players play by x, when X = (x1 , . . . xn ) ∈ X1 × · · · × XN . DEFINITION : X = (x1 , . . . xn ) ∈ X1 × · · · × Xn is a NASH EQUILIBRIUM if for any 1 ≤ i ≤ n and for any yi ∈ Xi , Gi (X) ≥ Gi (X−i , y i ) That is, player i will gain nothing by changing his strategy. THOEREM : (1) X = (x1 , . . . xn ) is a nash equilibrium if and only for any player i, and for any 1 ≤ j ≤ mi , Gi (X) ≥ Gi (X−i , εij ) where εij is a strategy of player i that gives a probability of 1 to the pure strategy 1 ≤ j ≤ mi . (2) If X is a nash equilibrium and xij > 0 then Gi (X) = Gi (X−i , εij ) 11
PROOF : Gi is multi-linear and thus i
G (X) =
mi X j=1
≤
mi X j=1
X
Gi (X−i , εij )xij =
Gi (X−i , εij )xij ≤
j|xij >0
Gi (x−i , εij0 ) max Gi (X−i , εij0 )xij = max i
j0 |xij >0
j0 |xj
0
0
as the sum of xij is 1. Had there been a strategy that gives less than Gi (X) then there must be another one that gives more then Gi (X) and player i will choose that one. Suppose X is an equilibrium then Gi (X) ≥ Gi (X−i , εij ) for any 1 ≤ i ≤ n, q ≤ j ≤ mi . On the other hand, suppose that for any i and for any j there is no offset to gain more profit in the pure strategies - that is - for any 1 ≤ i ≤ n and 1 ≤ j ≤ mi Gi (X) ≥ Gi (X−i , εij ) As we know that Gi (X) ≤ max Gi (X−i , εij0 ) i j0 |xj
0
CALCULATION OF NASH EQUILIBRIUM POINTS
1, 1 0, 0 0, 0 2, 2
There are two types of equilibrium points : (1) both players play pure strategies (2) one plays a pure strategy and the othere plays a mixed one. (3) both players play mixed strategies. At this game, there are two equilibrium points in pure strategies : ((1,0),(1,0)) and both players get 1, and ((0,1),(0,1)) and both players get 2. Are the equilibria of type (2) ? There cannot be ! suppose player 1 plays a 12
pure strategy. If he plays (1,0) then in order for the existence of an equilibrium point of type ((1,0),(y,1-y)) for 0 < y < 1, the payment for player 2 when he plays (1,0) must equal the payment that he will obtain when playing (0,1) : but the former is 1 and the latter is 0. A similar calculation yields that there cannot be an equilibrium when player 1 plays (0,1) and player 2 plays a strict mixed startegy, and similarly when we switch the roles of the players. Now we need to find the equilibria of type (3). ((x, 1 − x), (y, 1 − y)) when 0 < x, y < 1. By the assumption that 0 < x < 1, we must have y = G1 ((1, 0), (y, 1 − y)) = G1 ((0, 1), (y, 1 − y)) = 2 − 2y and thus y = 32 . A similar calculation yields x = G2 ((x, 1 − x), (1, 0)) = G1 ((x, 1 − x), (0, 1)) = 2 − 2x and x = 23 . Thus the equilibrium is (( 32 , 13 ), ( 23 , 13 )). Now consider the game
2, 3 1, 3 1, 0 2, 2
If player 1 playes the player the first row, we must have G1 ((1, 0), (y, 1 − y)) ≥ G1 ((0, 1), (y, 1 − y)) Reminder : x = (x0 , . . . , xn ) is an equilibrium if and only if for any i ∈ N , 1≤j≤m Gi (x) ≥ Gi (x−i , εij ) Clearly if x is an equilibrium then the condition takes place. Now suppose that the condition takes place and we need to show that x is an equilibrium. 13
PROOF : Let y i ∈ X i and we show that Gi (x−i , y i ) ≤ Gi (x). i
i
G (x−i , y ) =
mi X
yji Gi (x−i , εij )
≤
j=1
mi X
yji Gi (x) = Gi (x)
j=1
QED. THEOREM : If x be an equilibrium then for any i ∈ N , for any 1 ≤ j ≤ mi for which xij > 0, Gi (x) = Gi (x−i , εij ) PROOF : i
G (x) =
mi X
xij Gi (x−i , εij )
j=1
As the right side contains an average of elements that are not bigger then Gi (x), and the left side equals Gi (x), all theses elements must equal Gi (x). Now we need to show that for any j, k ≤ mi , xij Gi (x−i , εij ) ≥ xij Gi (x−i , εij ) PROOF : The case xij = 0 is trivial. The case xij > 0 is also trivial by the previous theorem : xij Gi (x−i , εij ) = xij Gi (x) ≥ xij Gi (x−i , εij ) as x is an equilibrium. Now prove the converse : In this case, for any j, k for which xij > 0 and xik > 0 we can switch the the jobs of j, k and this will yield that Gi (x−i , εij ) = Gi (x−i , εij ) = A Furthermore, if xij > 0 then for any 1 ≤ j 0 ≤ mi Gi (x−i , εij ) ≥ Gi (x−i , εij 0 ) 14
Thus the pure strategy j gives the best value player i can get. As x ensures this value as well : Gi (x) =
X
xij Gi (x−i , εij ) = A
j:xij >0
x is not worse than any pure strategy and by an earlier theorem it is an equilibrium. Consider the following game :
y 1−y x 2, 1 0, 4 1 − x 2, 3 1, 2 (2,3) is an equilibrium of pure strategy by the strategies ((0,1),(1,0)). Note that if y < 1 there will be a strict domination of the second row : 2y = G1 (ε2 , (y, 1 − y)) > G1 (ε1 , (y, 1 − y)) = 2y + 1 − y and thus there are no equilibriums when both players play strict mixed startegies, as player 1 will play the pure strategy ε2 . If player 1 playes a pure strategy then there will be no equilibrium as player 2 will choose a pure startegy. Thus the only case of pure and mixed startegies that give equilibrium is when player 2 playes ε1 , and then player 1 can choose any strategy. In order for it to be an equilibrium, we need the following to take place : G2 ((x, 1 − x), (1, 0)) ≥ G2 ((x, 1 − x), (0, 1)) or x + 3 − 3x ≥ 4x + 2 − 2x 15
or x ≤ 41 . Thus ((x,1-x),(1,0)) is an equilibrium for any x ≤ 14 . Now consider the game
y 1−y x 0, 0, 0 0, 0, 0 1 − x 0, 0, 0 0, 0, 0
y 1−y x 1, 1, 1 0, 0, 0 1 − x 0, 0, 0 2, 3, 4
Player 3 choose z ∈ {0, 1}. If z = 1 then the game takes place in the right matrix. If z = 0 then the game takes place in the left matrix. There are 4 pure startegies equilibriums. Suppose player 3 playes a pure strategy. If one of the players will play a mixed strategies, player 3 will strictly favor z = 1. Thus when player 3 plays a pure strategy, there will only be an equilibrium of pure strategies.
7
security levels
For the player i, the maximal security level of player i in G is Vi = max
xi ∈∆Si X−i ∈
Qmin ∆(s ) G0 (X−i , xi ) j6=i
j
Let X be an equilibrium in G then Vi ≤ Gi (x) : By the definition of Vi there exists a strategy of player i : x bi that ensures at least Vi against any set of decisions of the other players. Thus, Gi (x) must be more than or equal Vi , as had it been smaller player i would have changed his strategy to gain Vi . Consider battle of the sexes
2, 1 0, 0 0, 0 1, 2
The mixed equilibrium is (( 32 , 13 ), ( 13 , 23 )) Gi (x) = 23 , for both players. What is the maximal security level of player 1 ? Consider 2 0 0 1 16
Where player 2 just wants to minimize the profits of player 1 regardless of this own. It can be calculated that the maximal security level of player 1 is ( 13 , 32 ). Similarly the maximal security level for player 2 is ( 23 , 13 ). Thus, in equilibrium player 1 plays ( 13 , 23 ), and player 2 plays ( 13 , 32 ) which are not the optimal security levels. Consider the following symmetric game of three players −1, −1, −1 −1, 0, −1 −1, −1, 0 −4, −3, −3 0, −1, −1 −3, −3, −4 −3, −4, −3 −3, −3, −3 There are 4 equilibriums of pure strategies : As the game is symmetric, 3 of them are permutations of themselves, and 1 is an equilibrium where all players play the same strategies. Now pure-mixed strategies. As the game is symmetric it is enough to find all the equilibriums where player 3 plays a pure strategy. There exist two possibilities : (1) player 3 plays (1,0) and the game for players 1,2 is y 1−y x −1, −1 −1, 0 1 − x 0, −1 −3, −3 If another player plays a pure startegy, the third one will also play a pure strategy (this is trivial to see just by looking at the matrix). So we look for an equilibriums when both players 1,2 play a strict mixed strategy. This is equivalent to solving this matrix. This gives the suspected equilibrium (( 23 , 31 ), ( 23 , 31 ), (1, 0)). We need to see that 2 1 2 1 2 1 2 1 G3 (( , ), ( , ), (1, 0)) ≥ G3 (( , ), ( , ), (0, 1)) 3 3 3 3 3 3 3 3 This indeed takes place and thus (( 23 , 31 ), ( 23 , 13 ), (1, 0)) and its permutations are equilibriums. 17
Similarly we check the case z = 0 which yields the matrix y 1−y x −1, −1, 0 −4, −3, −3 1 − x −3, −4, −3 −3, −3, −3 And the solution is (( 13 , 32 ), ( 13 , 23 ), (1, 0)). In order for this to be an equilibrium we need that 1 2 1 2 1 2 1 2 G3 (( , ), ( , ), (0, 1)) ≥ G3 (( , ), ( , ), (1, 0)) 3 3 3 3 3 3 3 3 Which does not take place and thus (( 13 , 32 ), ( 13 , 23 ), (1, 0)) is not an equilibrium. Now we need to consider the case where z playes a strict mixed strategy. This can happen only if 3y + 3z − 6yz = 1 3x + 3z − 6xz = 1 3y + 3x − 6yx = 1 Thus, 3y−6yz−3x+6xz = 0 or −6z(y−x)+3(y−x) = 0 or (1−2z)(y−x) = 0. Thus either z =
1 2
which cannot work, or y = x, and this will yield the solution.
—- MISSED THE TUTORIAL OF 31.12.2008 —– I will write brief notes from Ayelet’s notes. GENERAL GAMES N is the finite set of players and Si is the set of strategies for player i. Ui are real functions of the utility. BIDS Any player has a value for the item on bid, which is denoted by vi > 0. The strategies are Si = [0, ∞). vi − B−i ∀b = (b1 , . . . , bn ) Ui (b) = 0 vi −B−i Ni
bi > maxk6=j bk = B−i bi < B−i bi = B−i
Where N−i = {k|bk = bi }. Domination (weak and strict) are defined similarly to the previous games. 18
THEOREM : For any i, vi dominates any other strategy. PROOF : If B−i ≥ vi then vi ensures 0 and player i cannot gain more than 0. If B−i < vi then vi ensures vi − B−i > 0 and thus any strategy that is strictly bigger than B−i will be optimal. QED. —– END OF TUTORIAL OF 31.12.2008 —Find the equilibriums of the game: 0, 0 6, 7 7, 6 7, 6 0, 0 6, 7 6, 7 7, 6 0, 0 There are no equilibriums of pure strategies. Look for equilibriums where one player mixes two strategies. As the game is symmetric it is sufficient to check just one pair. Suppose there is an equilibrium ((x, 1−x, 0), (y1 , y2 , 1−y1 −y2 )). Then we must have y1 = 0 as G2(X, e1 ) < G2 (x, e3 ) Thus the equilibrium is of the sort ((x, 1 − x, 0), (0, y, 1 − y)). Thus player one will play x = 1 since in this case G1 (e1 , Y ) > G2 (e2 , Y ). Therefore there cannot be such an equilibriums and all equilibriums are of strict pure strategies for all three choices for both players. The equations yield that (( 13 , 13 , 31 ), ( 13 , 13 , 13 )) is the unique equilibrium of this game.
8
Matching
Given a set X, a PREFERENCE RELATION < on X is a relation which is complete, antireflexive and transitive. A matching problem is composed of n women and n men such that any woman has a preference relation of the men and any man has a preference relation of the women. A matching is a bijective from the men to the women. We have 19
defined a STABLE matching and proved that it always exists. Let A, B be two stable matchings. We say that A ≥m B if any man that is matched to different women in A and B prefers the women that is matched to him in A. THEOREM : If A ≥m B then B ≥w A. PROOF : Suppose that a Shushan is matched to shoshana in A and to rivka in B, then Shushan prefers shoshana over rivka and thus shoshana has a man that she prefers better than Shushan (otherwise B would not have been stable). Thus for shoshana the matching in B is better than the matching in A. The same idea works on each women and thus B ≥w A, QED. THEOREM : When the men go to the women they get the best stable matching for themselves and thus the worst matching for the women. PROOF : Let B be a stable matching. It is sufficient to see that no man has been rejected in the process by the women to which he is matched in B. This is done by induction on the iteration of the algorithm : If Shushan is matched to shoshana in B then if he goes to her at the first iteration then she is the best girl in his list. If she rejects him then she rejects him for Andras, so she is the first in Andras list which means that B is not stable. QED first iteration. Suppose this is true for any iteration until k. The induction step is trivial.
9
Bargaining problems
Geometrical properties. Let S ⊆ RRR2 be a convex set and d ∈ S a point in the boundary. 20
A line l is called a SUPPORTER if all the set S is on the same side of the line. THEOREM : A point y ∈ S that satisfies xi > di is the solution of the bargaining problem (d, S) if and only if there exists a supporter of S through y whose slope equals −1 times the slope of the line that connects y and d. Thus, the two lines generate a triangle with two equal sides. PROOF : Let (S, d) be a bargaining problem and a = (a1 , a2 ) = N (S, d). Consider the transformation T (x) = (
x1 − d1 x2 − d2 , ) a1 − d1 a2 − d2
T (d) = (0, 0) and N (T (S), T (d)) = T (a) = (1, 1). The line x1 + x2 = 2 is a supporter of T (S) in the point (1,1). Thus, T (S) ⊂ {(x1 , x2 ) : x1 + x2 = 2} (Exercise). Thus x1 + x2 is a supporter of T (S) in (1,1). The triangle with two equal sides will remain a triangle with two equal sides under a linear transformation.
10 EXERCISE 11 QUESTION 4 : Find the solution for S = {(x1 , y1 ) : (3x − 2)2 + (2x − 1)2 ≤ 4}, d = ( 13 , 0). SOLUTION : Use the linear transformation T (x1 , x2 ) = (3x1 , 2x2 ). Thus T (S) = {(x1 , x2 ) : f (T −1 (x1 , x2 )) ≤ 4} = {(x1 , x2 ) : (x1 − 2)2 + (x2 − 1)2 ≤ 4} and T (d) = (1, 0). Now Consider L(x1 , x2 ) = (x1 − 2, x2 − 1). It yields L(T (S)) = {(x1 , x2 ) : x21 + x22 ≤ 4} 21
√ √ L(T (d)) = (−1, −1). The solution of L(T (S)) is clearly ( 2, 2) by symmetry √ √ and efficiency. Thus the solution of the original problem is T −1 (L−1 ( 2, 2)) = √
√
( 2+3 2 , 1+2 2 ).
11
Shapley’s value
A COMMUTATIVE GAME is (N, v) where N is a set of players and v : P (N )\ → RRR such that v({φ}) = 0. Shapley’s value satisfies the following : (1) Efficiency. (2) An inefficient player will get 0. (3) Two symmetric players get the same. (4) For two games (N, v), (N, w), xi (N, v) + xi (N, w) = xi (N, v + w). There exists a unique value ϕ(v) ∈ RRRn that satisfies the four axioms and it is given by ϕi (v) =
1 X v(Piπ ∪ {i}) − v(Piπ ) n! π∈S n
Example : majority game q, wi ≥ 0, where [q, w1 , . . . , wn ]. The team must obtain q, as described by the utility function. ( P 1 i∈S wi ≥ q v(s) = 0 otherwise This is a monotone game : If U ⊂ V then v(U ) ≤ v(V ). Shapley’s value is 1
q{π :
P
i∈P1π
ϕ (v) =
< q,
P
i∈P1π
n!
Explicit example : ϕ([3, 2, 1, 1]). ϕ1 = (1 − 2α, α, α) 22
wi + w1 ≥ q}
As players 2 and 3 are symmetric and the sum must equal 1. We will now calculate α For which places in π, player 1 makes the difference ? If π(1) = 1 then he is not ! If π(1) = 2 then he is, and so if π(1) = 3. Thus ϕi (v) = 46 . This yields ϕ = ( 46 , 16 , 61 ). For a non empty coalition T ⊆ N , the UNANIMOUS game is defined by ( 1 T ⊆S UT (S) = 0 otherwise
12
Threats game
Let C be convex and compact in R2 . The PARETTO EFFICIENT boundary is C ∗ = {x ∈ C : there does not exist y ∈ C such that y ≥ x} y ≥ x ⇔ y1 ≥ x1 , y 6= x. Given a point d ∈ C, the expansion of Nash function is defined by N (d, c) = argmaxx∈C ∗ ,xi ≥di (x1 − d1 )(x2 − d2 ) Given a two players game of strategic form G, the threats game that is derived from it is a new two player game Γ where the set of strategies of player i is the set of mixed strategies of i in G. The payment function for the players is defined by U (y 1 , y 2 ) = N ((u1 (y 1 , y 2 ), u2 (y 1 , y 2 )), C) where C is the convex-hull of the payment matrix, and u1 is the payment function in the original game. THEOREM : Let Γ be the threats game derived by the strategies game G. There exist (y 1 , y 2 ) ∈ X1 × X2 and a unique b ∈ C ∗ such that (1) u1 (y 1 , y 2 ) ≥ b1 23
∀y2 ∈ X2
and (2) u2 (y1 , y 2 ) ≥ b2
∀y1 ∈ X1
Note that (1) is equivalent to ∀y2 ∈ X2
u2 (y 1 , y2 ) ≤ b2
∀y1 ∈ X1
u1 (y1 , y 2 ) ≤ b1
and (2) is equivalent to
as for any a, b ∈ C ∗ , a1 > b1 → b2 > a2 and b1 > a1 → a2 > b2 . DEFINITIONS : yi is called the OPTIMAL THREAT, and b is the FINAL AGREEMENT. Under these terms, the theorem states that the optimal threats and the final agreement exist and the final agreement is unique. Consider the game
0, 0 2, 1 −1, 2 3, −1
Find the optimal threats and the final agreement point. First, draw the convex-hull. The lines are x1 + 3x2 = 5 and 2x1 + x2 = 5. Split C to three parts, C1 , C2 , C3 where C1 is the best for player 2, and C3 is the best for player 1. The lines in C1 are of the form x1 − 3x2 = 5, and those in C3 are of the form 2x1 − x2 = l. Note that k changes linearly in C : Any point on the line determines k. If a = (a1 , a2 ) satisfies a1 − 3a2 = k1 and b = (b1 , b2 ) satisfies b1 − 3b2 = k2 then the point ta + (1 − t)b satisfies c1 − 3c2 = tk1 + (1 − t)k2 . Player 1 wants k to be as big as possible. Second part : Playing for k = x1 − 3x2 . The struggle for k in C1 corresponds to the following zero sum game
0 −1 −7 6 24
Solve this zero sum game : −7 + 7y = −y + 6 − 6y so p =
13 14
is the optimal
strategy for player 1. It ensures k ≥ − 12 . This k is outside C1 and thus player 1 can ensure himself at least 2. Now play for k in C3 . This corresponds to the game k = 2x1 − x2 :
0 3 −4 7
By domination trivially k ∗ = 0, and q ∗ = (1, 0). As in C3 k ∈ [3, 7], player 2 can ensure disagreement points in C1 or C2 . Thus, in the final agreement he will obtain at least -1 and the final agreement will be (2, −1), optimal threats ( 13 , 1 ), (1, 0) (not necessarily unique of course). 14 14
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