The Second Anti-Mathima on Game Theory
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1 The Second Ant-Mathma on Game Theory Ath. Kehagas December Introducton In ths note we wll examne the noton of game equlbrum for three types of games 1. 2-player 2-acton zero-sum games 2. 2-player 2-acton non-zero-sum noncooperatve games 3. 2-player 2-acton non-zero-sum cooperatve games. In all cases t wll be easy to see how to generalze for the case of M-player N-acton games (except for the case of cooperatve games where our anlyss s good only for two players M = 2!!!.. 2 Zero-Sum Games [ a11 a 1. Ths game s determned by a 2 2 payoff matrx A = a 21 a 22 player when row player plays and column player plays. where a s payoff to row 2. Mnmax nequalty. The best worst payoff for the row player s less than or equal to the best worst for the column player. In other words max mn mn a a max a a a mn max a. We wll call max mn a the securty level of the row player; he s guaranteed to get t no matter what the column player does. Smlarly mn max a s the securty level of the column player. 3. Suppose there s a saddle pont a such that u = max mn a = a = mn max a = v. When there s a saddle pont each player should go for the saddle pont f he assumes the opponent wll do the same. These assumptons (by both players are self-renforcng; the game s n equlbrum. 1
2 [ 1 4 (a Example. A = 2 3 max mn where a = max ( 1 2 = 2 = mn (2 4 = mn max a so there s a saddle pont a 21 = 2. Suppose the row player decdes to go for the saddle pont.e. play 2; he mght hope to mprove hs payoff by playng 1 only f he expected column player to play 2; but he also knows that the column player s securty level s 2 acheved by playng 1; f he expects the column player to go for the securty level the row player has no reason to change from 2 to 1 (because then he would lose 1 unt. Smlar remarks hold for the column player. Hence f each player expects the other to go for hs securty level these expectatons are self-renforcng. (b Example. The saddle [ pont argument holds a fortor when there s a domnatng strategy. 1 2 For example take A = where the row player has move 2 as a domnatng strategy. 2 3 Also max mn a = max ( 2 2 = 2 = mn (2 3 = mn max a.e. the saddle pont s acheved at a domnatng strategy. 4. However a game does not necessarly have a domnatng strategy or a saddle pont. I ths case each player has an ncentve to change hs play f he assumes the other player wll go for hs securty level; such a game s not n equlbrum. [ 2 4 (a Example. The game does not have a saddle pont 1 3 max mn a = max ( 2 3 = 2 < 1 = mn (4 1 = mn max a. Now f the row player goes for hs securty level wll play 1. But f n addton he expects the column player to go for hs securty level he wll expect hm to play 1 n whch case row player has an ncentve to change and play 2 (to get a payoff of 1. But f the column player takes ths reasonng nto account (and so expects row to play 2 then he has an ncentve to play 2 (to get a payoff of 3. Now f the row player takes ths nto account he has an ncentve to play 1... and so the game s not n equlbrum. [ 5 1 (b Example. The game does not have a saddle pont 3 4 max mn a = max (1 3 = 3 < 4 = mn (5 4 = mn max a. Now f the column player goes for hs securty level he wll play 2. But f n addton he expects the row player to go for hs securty level he wll expect hm to play 2 n whch case column player has an ncentve to change and play 1 (to get a payoff of 3. But f the row player takes ths reasonng nto account (and so expects column to play 1 then he has an ncentve to play 1 (to get a payoff of 5. Now f the column player takes ths nto account he has an ncentve to play 2... and so the game s not n equlbrum. 5. To obtan an extenson of the saddle pont concept (an extenson whch exsts for every zerosum game we wll extend the set of strateges a player may use. A mxed strategy s a par of probablty vectors p = [p 1 p 2 q = [q 1 q 2 whch s used as follows: 2
3 (a row plays 1 wth prob p 1 = p and 2 wth prob. p 2 = 1 p (b column plays 1 wth prob q 1 = q and 2 wth prob. q 2 = 1 q. 6. Usng mxed stateges p = [p 1 p and q = [q 1 q the expected payoff for player 1 s u = pqa 11 + p (1 q a + (1 p qa 21 + (1 p (1 q a 22 = p T Aq; the expected payoff for player 2 s v = u. 7. Theorem (Mnmax: For all zero-sum games A there exst mxed stateges p q such that u = max p mn q The u s called the value of the game. pt Aq = (p T Aq = mn q max p pt Aq. 8. Mxed strateges and the mnmax theorem also work for pure strateges. [ 1 2 (a Example. Wth A = we use p 2 3 = [0 1 q = [1 0 and get u = max mn u = [ 0 1 [ a = max [ 1 0 ( 2 2 = 2 = mn = 2. (2 3 = mn max a So the Mnmax Theorem works on pure strateges too. [ 1 4 (b Example. Take A = then we use p 2 3 = [0 1 q = [1 0 and get u = max mn u = [ 0 1 [ a = max [ 1 0 ( 2 2 = 2 = mn = 2. (2 3 = mn max a [ a11 a 9. Let us solve the general 2 2 zero sum game. We have A =. To not have a saddle a 21 a 22 pont we must have ether mn (a 11 a 22 > max(a a 21 or mn (a a 21 > max(a 11 a 22. We want to have [ [ [ a11 a u = max mn p 1 p q p q a 21 a 22 1 q = [ p 1 p [ [ a 11 a q a 21 a 22 1 q = mn q [ [ a max p 1 p 11 a p a 21 a 22 Usng q = [1 0 and q = [0 1 we get (for the optmal p [ a 11 p + a 21 (1 p = u (p q u a p + a 22 (1 p = u (p q u q 1 q. 3
4 usng p = [1 0 and p = [0 1 we get (for the optmal q a 11 q + a (1 q = u ( p q u a 21 q + a 22 (1 q = u ( p q u We want to fnd p and q such that the above hold as equaltes. Assumng the equaltes hold we can elmnate u from the above and get the equatons These can be solved to get p = a 11 p + a 21 (1 p = a p + a 22 (1 p a 11 q + a (1 q = a 21 q + a 22 (1 q a 22 a 21 (a 11 + a 22 (a + a 21 q = a 22 a (a 11 + a 22 (a + a 21. We have (why? 0 p 1 and 0 q 1. And the value of the game s [ u = = a 22 a 21 (a 11 +a 22 (a +a 21 1 a 11 a 22 a 21 a 22 (a 11 + a 22 (a + a 21 (a Example. Wth A = [ p = a 22 a 21 (a 11 +a 22 (a +a 21 we have [ a 11 a a 21 a 22 [ a 22 a (5 + 4 (1 + 3 = 1 [ 1 5 p = q 4 1 = (5 + 4 (1 + 3 = 3 5 q = u = [ [ [ = (a 11 +a 22 (a +a 21 a 22 a (a 11 +a 22 (a +a [ Note that f row player uses hs optmal mxed strategy and column player uses an arbtrary strategy then row player gets payoff [ [ [ a 22 a 21 a (a 11 +a 22 (a +a a 21 a 11 a q (a 11 +a 22 (a +a 21 a 21 a 22 1 q a 11 a 22 a 21 a 22 = (a 11 + a 22 (a + a 21 = u whch s not only optmal but also ndependent of q. The correspondng remark also holds for column player. Hence each player has no ncentve to change hs mxed strategy and the game s n equlbrum! 4
5 3 Non-Zero-Sum Noncooperatve Games [ a11 b 1. The game s determned by a 2 2 payoff b-matrx 11 a b where a a 21 b 21 a 22 b s payoff to 22 row player and b s payoff to column player (when row player plays and column player plays. Or equvalently by two matrces: [ [ a11 a A = b11 b B =. a 21 a 22 b 21 b We need a concept of stable soluton of the game (lke mnmax. 3. Defnton. The mxed strateges p q are a Nash Equlbrum ff or n other words (p q : u (p q u (p q v (p q v (p q (p q : (p T Aq p T Aq (p T Bq (p T Bq 4. Example. Prsoner s dlemma: A = q = [ 0 1 then we have Smlarly [ B = [ If we use p = [ 0 1 and u (p q = ( ( (1 0 (1 0 = 1 u (p q = 3 p p ( (1 p (1 p (1 0 = 1 p v (p q v (p q. Hence p = [ 0 1 and q = [ 0 1 (all-d and all-d s a Nash equlbrum (the unque one for PD. [ [ Example. Battle of the Sexes: A = B =. If we use p = [ 1 0 and q = [ 0 1 then we have Smlarly u ([ 1 0 [ 0 1 = = 2 u ( p [ 0 1 = 0 p p (1 p (1 p 1 = 2p v ([ 1 0 [ 0 1 v ([ 0 1 q. But also f we use [ 0 1 and [ 1 0 then we have Smlarly u ([ 1 0 [ 0 1 = = 5 u ( p [ 0 So we have two Nash equlbra! 1 = 0 p p (1 p (1 p 1 = 5 5p v ([ 1 0 [ 0 1 v ([ 1 0 q. 5
6 6. Theorem (Nash. Every 2 2 (nonzero sum game has at least one Nash equlbrum p q. 7. Defnton. The mxed strateges p q are Pareto optmal ff (p q (p q : u (p q u (p q v (p q v (p q. 8. Compare wth the Nash equlbrum (p q : u (p q u (p q v (p q v (p q 9. Not every Nash equlbrum s Pareto optmal (and conversely. For example n Prsoner s dlemma the Nash equlbrum [ 0 1 [ 0 1 s not Pareto optmal: u ([ 0 1 [ 0 1 u ([ 1 0 [ 1 0 v ([ 0 1 [ 0 1 v ([ 1 0 [ 1 0 So [ 0 1 [ 0 1 s a Nash equlbrum but not Pareto optmal. And [ 1 0 [ 1 0 s Pareto optmal but not a Nash equlbrum. 10. The computaton of Nash equlbra s hard we wll not dscuss t. 6
7 3.1 Appendx: Computng Nash Equlbrqa for Contnuous Games The problem of the commons. Two players A and B. Player A uses x goats and has proft functon c (x y = x 1 (x + y 2 Player B uses y goats and has proft functon d (x y = y 1 (x + y 2 Each of A B wants to selfshly optmze hs proft. We wll compute the (selfsh Nash strategy for each one of them and also the globally optmal strategy Computng the Selfsh Optmum For player A to Nash optmze x we must have D x c (x y = 0. We have D x (x 1 (x + y 2 = 1 + 2x2 + 3xy + y 2 1 x 2 2xy y 2 and for 1 + 2x2 + 3xy + y 2 1 x 2 2xy y = 0 2 the soluton s: x = 3 4 y y x = 3 4 y 4 1 y (nadmssble. Player B uses y goats and has proft functon d (x y = y 1 (x + y 2 We have D y (y 1 (x + y 2 = 1 + x2 + 3xy + 2y 2 1 x 2 2xy y 2 and for 1 + x2 + 3xy + 2y 2 1 x 2 2xy y = 0 2 the soluton s: y = 3 4 x x y = 3 4 x 1 4 x (nadmssble. Now we must solve y = 3 4 x x x = 3 4 y y The soluton s: x = y = The payofffor each player s ( = and the total payoff s =
8 3.1.2 Computng the Global Optmum The global proft s Or settng z = x + y The central planner sets c (x y + d (x y = C (x + y = (x + y 1 (x + y 2 C (z = z 1 z 2. ( D z z 1 z z2 = = 0 1 z 2 and the soluton s: z = x = y = 2/4 = The payofffor each player s ( > and the total payoff s = 0.50 > Concluson If the players play selfshly the overutlze the Commons.e. they rase more goats than would maxmze the common good. 8
9 4 Non-Zero-Sum Cooperatve Games [ a11 b 1. Agan the game s determned by a 2 2 payoff b-matrx 11 a b where a a 21 b 21 a 22 b s payoff 22 to row player and b s payoff to column player (when row player plays and column player plays. Or equvalently by two matrces: [ [ a11 a A = b11 b B =. a 21 a 22 b 21 b Agan we wll look for a soluton concept. Now however we assume that the players can negotate before the game and reach a bndng agreement on how to coordnate ther moves. 3. The assumpton of coordnated moves can be descrbed mathematcally as follows. Movng drectly to mxed strateges we assume that the players can decde on ont acton probabltes (ont mxed strateges. I.e. we defne {1 2} : p = Pr ( Player 1 plays and player 2 plays p 0 and 2 p = 1. =1 4. (Why play wth ont probabltes? To acheve the maxmum value for both players. We can also thnk of the p s as proportons. Note that ths s more general than the noncooperatve case where no coordnaton was possble and we had p = p q. 5. For every value p the payoffs to the two players are (p a p b. All of these ponts taken together form a set of ponts whch can be plotted on a plane wth a set of axes: the horzontal axs corresponds to player 1 payoff and the vertcal axs to player 2 payoff. In fact ths set s a polygon the payoff polygon M = 2 =1 p [ a b : 0 p 1 and more specfcally M s the convex hull of the vectors [ 10 0 (a For the game 0 1 [ 0 1 (b For the game 1 5 [ 1 10 (c For the game 0 0 [ p = 1 =1 [ a11 b 11 [ a b see Fgure 1 (9.3G6 / F9.6. [ 5 1 see Fgure 2 / (9.2P2 / F [ 10 1 see Fgure 3 (9.2G4 / F [ a21 b 21 [ a22 b Some of the payoffs are preferrable / better (to both players than others. In Example a (5 6 s preferrable to (5 5 whch s preferrable to (4 4. But (5 6 s not (globally preferrable to ( To formalze ths dea of preferrablty we ntroduce a new order. We wrte { } u1 u (u 1 v 1 (u 2 v 2 ff 2. v 1 v 2. 9
10 (a Ths s a partal order. In other words we have the basc order propertes (u 1 v 1 (u 1 v 1 { } (u1 v 1 (u 2 v 2 (u (u 2 v 2 (u 1 v 1 1 v 1 = (u 2 v 2 { } (u1 v 1 (u 2 v 2 (u (u 2 v 2 (u 3 v 3 1 v 1 (u 3 v 3 But we do not have the property that every par of elements s comparable.e. we may have (u 1 v 1 (u 2 v 2 and (u 2 v 2 (u1 v Then the negotaton set N M s the set of maxmal elements of M. 9. The negotaton set can be further restrcted f there s a status quo pont (u 0 v 0 where u 0 s the mnmum payoff acceptable to player 1 and v 0 s the mnmum payoff acceptable to player 2. (For the tme beng we take the status quo pont as gven; later we wll see how t s determned. 10. Hence fnally the negotaton set s defned to be N = {(u v : (u v M and (x y (u v : (u v (x y and (u v (u 0 v 0 }. Intutvely N s the set of payoff pars whch the two players wll consder as canddate outcomes of the game (they have no reason to consder any (u v M N because they can both do better than (u v. (a Example. Fgure 1. (b Example. Fgure 2. (c Example. Fgure We are now lookng for an equlbrum of the game.e. a par of payoffs (u v whch s somehow optmal or reasonable or somethng lke that. (u v wll depend on M (and N and also on the status quo (u 0 v 0. In other words (u v = F [M (u 0 v 0. Nash has obtaned one such equlbrum usng the axomatc method. He postulated that (u v must satsfy the followng. (a (Pareto Optmalty (u v N. (b (Invarance If L s a lnear transformaton of the payoffs of the form L (u v = (au + b cv + d then F [L (M L (u 0 v 0 = L (u v. (c (Symmetry If (u v M (v u M and u 0 = v 0 then u = v. (d (Independence of rrelevant alternatves If M M and (u v M then F [ L ( M L (u 0 v 0 = F [L (M L (u 0 v 0. 10
11 13. Theorem (Nash. For every 2 2 cooperatve game there s exactly one pont (u v whch satsfes the above four axoms; t s (u v = arg max uv N (u u 0 (v v We have left open the choce of status quo pont (u 0 v 0. One way to choose t s f player 1 consders hs game A as zero sum game and takes u 0 to be hs mnmax value; and smlarly for player 2. (There are many other possbltes whch we wll not examne here. 15. Examples. (a (9.3G6 / F9.6 [ 10 0 Player 1: max 0 1 mn A = 0. [ [ Player 2: max mn B T = 0. Securty levels (0 0 Negotaton set: StrLnSeg from (10 1 to (1 10 so we solve 1 = 10a + b 10 = a + b Soluton s: {b = 11 a = 1}.e. v = u + 11 Maxmzaton uv= u ( u + 11 = u u hence (u v = (11/2 11/2 (b (9.2P2 / F9.7 [ 0-1 Player 1: max 1 5 mn A = 1 (acton 2 domnates 1. [ [ Player 2: max mn B T = 0 Securty levels (1 0 Negotaton set: StrLnSeg from (0 5 to (5 0 so we solve 5 = b 0 = 5a + b Soluton s: {a = 1 b = 5}.e. v = u + 5 Maxmzaton (u 1 v= (u 1 ( u + 5 = u 2 + 6u 5 hence (u v = (3 2 (c (9.2P2 / F9.7 - Varaton [ 0-1 Player 1: max 1 2 mn A = 1 (acton 2 domnates 1. [ [ Player 2: max mn B T = 0 Securty levels (1 0 Negotaton set: StrLnSeg from (0 7 to (2 0 so we solve 7 = b 0 = 2a + b 11
12 Soluton s: { b = 7 a = 7 2}.e. v = 7 2u + 7 ; but we must restrct for u 1 so we take the part of the segment from (1 7/2 to (2 0. Maxmzaton (u 1 v= (u 1 ( ( 7 2 u + 7 = 7 2 u u 7 whch s maxmzed at = (u v (d (9.2G4 / F9.8 [ 1 10 Player 1: acton 1 domnates 2; he has securty level of [ [ Player 2: max mn B T = 9. Securty levels (1 9 Negotaton set: StrLnSeg from (1 10 to (10 1 so we solve 10 = a + b 1 = 10a + b Soluton s: {b = 11 a = 1} and so v = u + 11 Maxmzaton (u 1 (v + 9= (u 1 ( u = u 2 +21u 20. Ths s maxmzed at u = 21 2 outsde of N. So the maxmum on N s at (u v = (10 1 (e (Mne / Fgure 999 wth 0 < a < b < c < d < 1. [ 1 a Player 1: max c 0 mn A = max ( a c = a. [ [ 0 b 0 d Player 2: max d 1 b 1 mn B T = max ( d b = b. Securty levels ( a b Negotaton set: StrLnSeg from (1 0 to (0 1 so we solve 1 = b 0 = a + b Soluton s: {b = 1 a = 1} and so v = u + 1 Maxmzaton (u + a (v + b= (u + a ( u b = (u + a ( u b = u 2 + u + ub au + a + ab whch s maxmzed at u = (b a then v = (a b. (f Alternatve soluton (usng dfferent status quo The row player s game soluton s The column player s game soluton s c p = 1 + (a + c u ac 0 = 1 + (a + c. 1 + b q = 1 + (b + d v db 0 = 1 + (b + d. ( Hence the status quo pont s ac 1+(a+c db 1+(b+d. To fnd the soluton we must maxmze G (u v = ( u + ( ac v (a + c db 1 + (b + d
13 subect to the constrant (negotaton set v = 1 u. So we maxmze ( ( ac bd G (v = u + 1 u (a + c 1 + (b + d ( = u 2 + u bd 1 + b + d ac 1 + a + c Takng the dervatve and settng equal to zero we get u = ( bd b + d ac 1 + a + c v = ( bd b + d ac 1 + a + c whch s the soluton of the game. abcd (1 + a + c (1 + b + d. 13
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