1、 在NE中加入mixed strategymixed strategy for player i is a probability distribution over ( some or all of ) the strategies in SiMixed 2Pure strategies : 在Si中的strategies都是pure strategies故在上例中:Si = T , H 有2個pure strategies : H , T而mixed strategy為 ( q , 1-q ) q = Prob ( H ) 1-q = Prob ( T ) 0 q 1= ( pure st
2、rategy is limiting case of mixed strategy )MixedPure(0 , 1)表示THstrategy notation :mixed : PiPure : SiDef : In the normal form game G = S , , S = Si Suppose Si =si1 , , sik Then a mixed strategy for player i is a probability distributionPi = (Pi1 , , Pik), where 0 Pik 1 , for k = 1 , , K And Pi1 + +
3、Pik = 1Mixed 3按定義若Si為strictly dominated strategy 表示不存在任何belief讓i認為選Si是optimal choice若Si為strictly dominated strategy,則沒有belief可以讓i不選Si作為optimal choice若無任何belief使得i認為必須以Si為optimal choice,則必存在一個是strictly dominates Si看下例:( 2 )LRPlayer ( 1 )3 , -0 , -MB1 , -對player 1而言,T , M或B無一個是被其他choice strictly domin
4、ated(注意)但在某一belief下 ( q , 1 - q ) , q = Prob ( L )則player 1s best response is T if q player 1s best response is M if q B is strictly dominated by mixed strategies , one example : Mixed 41 22 , -B is not a best response to the other pure strategies 但B can be a best response to a mixed strategy若player
5、 2 play L , player 1s best response is T player 2 play R , player 1s best response is M若player 2 play ( q , 1 - q ) , 則B為best response to ( q , 1 - q ) c.f player 1s mixed strategy ( P , 1 P , 0 ),注意不是若表示dominance故:E ( 1 ) = P3q + P0( 1 q ) + ( 1 P )0q + ( 1 P )3( 1 q ) = 3 P q + 3 ( 1 P ) ( 1 q ) =
6、 6 P q 3 - 3 P 3 q若E 1 ( P , 1-P , 0 ) 1 ( B ) = 2 B : ( 0 , 0 , 1 )?mixed 5 Matching Pennies Game1 believes 2s strategy : ( q , 1-q ) , q = Prob ( H )Give this belief , player 1s E ( ) are令1的mixed strategy為( r , 1-r )NE : 求best response for player 2 s ( q , 1-q ) : r* ( q )Mixed 6一般化:若player 1相信2的m
7、ixed strategy為( P21 , , P2K )若1 play Sij ( pure strategy )若1 plays ( P11 , , P1J ) : mixed strategiesMixed 7P = ( P11 , , P1J )為player 1 對player 2s mixed strategy P2的best response它必須符合下述條件P1j 0 only if 亦即: 0 只有在其所對應的pure strategy S1j是P2的best response在best response mixed strategy中只有可能的pure strategy才能
8、放入正的機率(若該alternation strategy並非best response則其機率 = 0),如第32頁中的strategy B,其機率 = 0此外,若player 1有許多best response to P2 那麼:任何mixed strategy放P 0於全部或部分位於B R ( S1 ) 中的pure strategies , P = 0 於非BR者,皆為player 1對P2的best responser* ( q ) 中包含r = 0 ( 0 , 1 ),r = 1 ( 1 , 0 ),Mixed 8Nash equilibrium ( include mixed s
9、trategy ) In a two player normal form gameG =S1, S2 ; 1 , 2, the mixed strategies are a NE if each players mixed strategy is a best response to the other players mixed strategy , that is for to be NE.Mixed 9找NE解:回到Matching Penniesr H( 1-r ) T找player 1s mixed strategy ( r , 1-r ) 的best response q* (
10、r )Best response correspondence : r* ( q ) , q* ( r )注意:兩個players 都不是隨機選擇其pure strategy(亦即,並非在選擇前擲銅板來決定方向)例如:打擊者若不清楚投手在練習時丟什麼球最順手,那麼他必認為丟直球、曲球的機率相同,根據此,打擊手選即使投手所依據的選擇標準為private information ( unavailable / unobservable ) to打擊手Mixed 10妻OperaFight夫Opera ( o )2 , 10 , 0Fight ( F )1 , 2另一例:Battle of the
11、Sexes夫:(r , 1-r )妻:( q , 1-q )此處有三個交點,另外兩個NE為(q = 0 , r = 0),(q = 1 , r = 1)Mixed 11用圖形來討論NE(1) 找best response correspondence for each player (BRC)(2) NE為BRC之交點例:q1-qLeftRightrUpX , -Y , -1-rDownZ , -W , -主要cases :( i ) X Z , Y W Up r* = 1 ( ii ) X Z , Y W( iv ) X Z Y q q/ = Up : r* = 1 q Down : r* = 0 q = q/ Mix 12此外:( v ) X=Z q=1 YW( vi )Y=W q=0 XZ同理,也可以找r 以及q*( r ) for player 2 必可以找到r* ( q ) 與q*( r ) 之交點:其NE可能為單一,可能為多數主要case有4種,加上a=6又有四種若只討論主要case的game 4*4種可能圖形否則8*8=64種可能圖形
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