Lectures 20-23 Dynamic Games with Incomplete Information ••• 14.12 Game Theory •••• ••••• Muhamet Yildiz •••• ••••• •••• •• •• • • 1
••• ••• • ••••• ••• • ••• •• ••• • •• • • Road Map • Sequential Rationality 1. Sequential Equilibrium 2. Economic Applications 3. Sequential Bargaining with incomplete 1. information Reputation 2. 2
••• ••• • ••••• What is wrong with this •••• ••••• •••• equilibrium? •• • • • x (2,6) T B L R L R (0, 1) (3,2) (-1 ,3) (1,5) 3
. _~ . _l~_ ••• ••• • ••••• •••• ••••• Beliefs •••• •• • • • x • Beliefs of an agent at a .--- (2 , 6) given information set is a probability distribution on the information set. • For each information set, we must specify the beliefs of 2 / __ ._ . ___ the agent who moves at that information set. L / R L R (0,1) (3 ,2) (-1 ,3) (1 ,5) 4
••• ••• • ••••• •••• ••••• Assessment •••• •• • • • • An assessment is a pair (0,1-1) where • a is a strategy profile and • 1.1 is a belief system: 1.1(,1/) is a probability distribution on 1 for each information set I. 5
••• ••• • ••••• •••• ••••• Sequential Rationality •••• •• • • An assessment (0,1J) is sequentially rational if 0i is a best reply to O_i given IJ(.I/) for each player i, at each information set 1 player i moves . • That is, 0i maximizes his expected payoff given IJ(.I/) and given others stick to to their strategies in the continuation game. 6
••• ••• • ••••• •••• ••••• •••• Sequential Rationality implies •• • • • x (2,6) B I-f.! L " R L ' R 2 (0,1) (3,2) (-1,3) (1 ,5) 7
••• ••• • ••••• •••• ••••• Example •••• •• • • • L R (0,10) (3,2) (-1,3) (1 ,5) 8
••• ••• • ••••• •••• ••••• Consistency •••• •• • • • (a,lJ) is consistent ifthere is a sequence m (am,lJ ) - (a, lJ) where • am is "completely mixed" and • IJm is computed from am by Bayes rule 9
••• ••• • ••••• •••• ••••• Example •••• •• • • • 1 T B 1 L R L R (0,10) (3,2) (-1,3) (1 ,5) 10
? 2 ~ ••• ••• • ••••• •••• ••••• Example •••• •• • • • 1 2 1 - / o o B L f .-.-.-~.-.-.-.-.~-.- 0.9 3 0.1 R o o 3 1 1 1 3 2 2 1 3 1 11
~(0,1J) ••• ••• • ••••• •••• ••••• Sequential Equilibrium •••• •• • • • • An assessment: (0,1J) where ° is a strategy profile and IJ is a belief system, lJ(h)E~(h) for each h. • An assessment (0,1J) is sequentially rational if at each hj , OJ is a best reply to O_j given lJ(h). m • (0,1J) is consistent if there is a sequence (om,lJ ) where om is "completely mixed" and IJm is computed from om by Bayes rule • An assessment (0,1J) is a sequential equilibrium if it is sequentially rational and consistent. 12
2~ ••• •••• •••• • •••• Example •••• • • •• • •• • • • 1 2 o T o B L R o o 3 1 1 1 3 2 2 1 3 1 13
~ ~c C.o~ ~('/ C.o~ • •• ••• • ••••• •••• ••••• •••• Beer - Quiche •• • • • I) 1 0 1 :-:... ('/ quiche beer ,\ i {. 1 } 2 3 dOl} 't , 0 tw 0 1 0 ts 0 0 OV beer {.9} quiche 2 3 ,\ dOl} 't 1 1 14
~ ~c C.o~ C.o~ ~('/ • •• ••• • ••••• •••• ••••• Beer - Quiche, An equilibrium •••• •• • • • I) 1 0 1 :-:., ('/ quiche beer 1 ,\ i .1 {. 1 } 2 3 dOl} 't . 0 tw 0 1 0 ts 0 0 OV 0 .9 beer {.9} quiche 2 3 ,\ dOl} 't 1 1 15
~c C.o~ C.o~ ••• ••• • ••••• •••• ••••• •••• • • Beer - Quiche, Another equilibriu • 1 0 1 1 :-:.. 0,,('/ quiche beer 1 {. 1 } , \ 2 3 0 tw 0 1 ts 0 o beer {.9} quiche 2 3 ,\ dOl} 't 1 1 16
~ ~c C.o~ ~('/ C.o~ • •• ••• • ••••• •••• ••••• •••• Beer - Quiche, revised •• • • • I) 1 -2 1 :-:... ('/ quiche beer o ,\ i {. 1 } 3 dOl} 't , 0 tw 0 1 0 ts 0 0 OV beer {.9} quiche 2 3 ,\ dOl} 't 1 1 17
~ ~c C.o~ C.o~ ~('/ • •• ••• • ••••• •••• ••••• Beer - Quiche, Revised •••• •• • • • I) 1 -2 1 :-:., ('/ quiche beer o ,\ i 0 1 {. 1 } 3 dOl} 't , 0 tw 0 1 0 ts 0 0 OV 1 beer {.9} 0 quiche 2 3 ,\ dOl} 't 1 1 18
~ ~c C.o~ ~('/ C.o~ • •• ••• • ••••• •••• ••••• •••• Beer - Quiche in weakland •• • • • I) 1 0 1 :-:... ('/ quiche beer ,\ i {.8} 2 3 dOl} 't , 0 tw 0 1 0 ts 0 0 OV beer {. 2} quiche 2 3 ,\ dOl} 't 1 1 19
. 5~('/ ~c I~ ~ C.o~ C.o~ • •• ••• • •• ••• Beer - Quiche in weakland •• •• •• ••• •• •• Unique PBE •• • • • 1 0 1 :-:... .5 ('/ beer quiche 3/4 2 .5 ,\ i .5 114 {.8} 3 dOl} 't , 0 tw 0 1 0 ts 0 0 OV .5 beer {. 2} quiche 0 .5 2 3 ,\ dOl} 't 1 1 20
• •• ••• • ••••• Example •••• ••••• •••• 2 •• • 1 • (1 ,-5) \ \ .9 \ \ \ \ \ \ \ (3,3) (4,4) (5,2) '. .1 \ 2 '. 1 r----+------,----- (0,-5) (-1,3 ) (-1,4 ) (0,2) 21
;or-=-:~-.-.-~ ••• ••• • ••••• Example - solved •••• ••••• •••• •• • 1 a=7 /9 2 P=1/2 • (1 ,-5) \ '. f.l= 7/8 .9 \ \ \ \ \ \ (3,3) (4,4) (5,2) '. .1 \ 1 2 '. (0,-5) (-1,3 ) (-1,4 ) (0,2) 22
••• ••• • ••••• •••• ••••• Sequential Bargaining •••• •• • • • 1-period bargaining - 2 types 1. 2-period bargaining - 2 types 2. 1-period bargaining - continuum 3. 2-period bargaining - continuum 4. 23
••• ••• • ••••• •••• ••••• Sequential bargaining 1-p •••• •• • • • • A seller S with valuation 0 H L • A buyer B with valuation v; s • B knows v, S does not • v = 2 with probability TC • = 1 with probability 1-TC • S sets a price p ~ 0; • B either • buys, yielding (p,v-p) o o • or does not, yielding (0,0). p P o o 2-p I-p 24
••• ••• • ••••• •••• ••••• •••• Solution •• • • • B buys iff v > p; 1. If P ::;; 1, both types buy: 1. 1 ------------------ ---------- ------+-----. S gets p. If 1 < P ::;; 2, only H-type 2. buys: S gets np. If P > 2, no one buys. 3. p 2 1 S offers 2. • 1 if 1C < ~, • 2 if 1C > ~ . 25
~ ••• ••• • ••••• •••• Sequential bargaining 2-perio( ••••• ••• •• 1. At t = 0, S sets a ~ rice Po 0; • A seller S with valuation B either 0 2. • buys, yielding (P o, v-P o) • A buyer B with valuation • or does not, then v , ' At t = 1, S sets another • B knows v, S does not 3. • v = price P1 ~ 0; 2 with probability 1l • = B either 1 with probability 1- 1l 4. • buys, yielding ( i:i pj,i:i( V-P 1 )) • or does not, yielding (0,0) 26
••• ••• • ••••• •••• ••••• •••• •• • Solution, 2-period • • Let J.l = Pr(v = 21history at t=1). 1. At t = 1, buy iff v ;::: P; 2. If J.l > Yz, P1 = 2 3. If J.l < Yz, P1 = 1. 4. If J.l = Yz , mix between 1 and 2. 5. B with v=1 buys at t=O if Po < 1. 6. If Po > 1, J.l = Pr(v = 21 Po,t= 1) ::; n. 7. 27
••• ••• • ••••• •••• ••••• Solution, cont. 1t <1/2 •••• •• • • • f..t = Pr(v = 2IPo,t=1) < n <1/2. I. Att = 1, buy iff v ~ P; 2. P1 = 1. 3. B with v=2 buys at t=O if 4. (2-po) > 0(2-1) = 0 ~ Po ::; 2-0. Po = 1: 5. n(2-0) + (l-n)o = 2n(1-0) + 0 < 1-0+0 = 1. 28
••• ••• • ••••• •••• ••••• Solution, cont. 1t >1/2 •••• •• • • • • If v=2 is buying at Po > 2-0, then Il = Pr(v = 21Po > 2-8,t=1) = 0; • • P1 = 1; • v = 2 should not buy at Po > 2-8 . • If v=2 is not buying at 2> Po > 2-0, then Il = Pr(v = 21Po > 2-8,t=1) = 11: > 112; • • P1 = 2; • v = 2 should buy at 2 > Po > 2-8. • No pure-strategy equilibrium. 29
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