14.12 Game Theory, Fall 2001
14.12 Game Theory Fall 2001 Announcement The final exam is on Friday, Dec. 21, 9:00-12:00 in Walker. The conflict exam will be on Thursday, December 20, 9:00-12:00 in E51-085.
Faculty Professor: Muhamet Yildiz
[email protected] (Office hours: M 4:00-5:30, E52-251a, 253-5331) TA: Kenichi Amaya
[email protected] (Office hours: T 4:00-6:00, E52-303, 253-3591) TA: Astrid Dick
[email protected] Schedule Class: Room E51-372 Recitation: F10 or F3, Room E51-085
Handouts ●
syllabus
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Lecture notes 1
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Lecture notes 2
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Lecture notes 3
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Lecture notes 4
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Lecture notes 5
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Lecture notes 6
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Review notes 1
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Review notes 2
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Review notes 3 (11/30)
http://web.mit.edu/14.12/www/ (1 of 4) [2002-05-11 23:27:55]
14.12 Game Theory, Fall 2001
Lecture Slides Some of the slides for lectures 1,2, and 6 are not included, as they are made by another program. ● Slides 1 ●
Slides 2
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Slides 3
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Slides 4
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Slides 5
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Slides 6
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Slides for lecture 8
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Slides for lecture 9
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Slides for lecture 10
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Slides for lecture 12
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Slides for lecture 13
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Slides for lecture 15-18
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Slides for lecture 19-21
Homeworks ●
Homework 1 (due 9/26)
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Solutions for homework 1
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Homework 2 (due 10/3)
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Solutions for homework 2
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Homework 3 (due 10/29)
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Solutions for homework 3 Revised
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Homework 4 (due 11/7)
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Solutions for homework 4
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Homework 5 (due 12/7)
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Solutions for homework 5 Revised
Homework Grading Policies ●
You may turn in assignments during the lecture on the day they are due. After the lecture, assignments may be placed in a designated box that will be set out outside E52-303 until 5:30 p.
http://web.mit.edu/14.12/www/ (2 of 4) [2002-05-11 23:27:55]
14.12 Game Theory, Fall 2001
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m. Do not leave assignments in the professor or T. A.'s office or mailbox. You are permitted to discuss course material, including homework, with other students in the class. However, you must turn in your own individual solutions to each homework set. You need to explain how you come up with the answers.Correct answers without proper explanations will receive no credit.
Exams ●
Midterm 1 (10/10)
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Midterm 1 solutions
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Midterm 1 Grade Distribution
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Midterm 2
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Midterm 2 Solutions New!
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Midterm 2 Grade Distribution New!
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Grade distribution of the average of midterm 1 and 2 New!
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Mock Final New!
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Mock Final solutions New!
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Final exam solutions New!
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Final exam grade distribution New!
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Final grade+quiz distribution New!
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Final grade distribution New!
Past years' exams Midterm 1 ●
1995
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1995 (Solutions)
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1997
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1999
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2000
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2000 (Solutions)
Midterm 2 http://web.mit.edu/14.12/www/ (3 of 4) [2002-05-11 23:27:55]
14.12 Game Theory, Fall 2001
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1996
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1997
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1999 Mock Midterm
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2000 (11/16)
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2000 (11/17)
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Some previous midterm questions (including some questions from files above)
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Solutions for Question 7 (or 8)
Final ●
2000
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2000 solutions
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2000 make-up
http://web.mit.edu/14.12/www/ (4 of 4) [2002-05-11 23:27:55]
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3
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wkh froxpq sod|hu uhdvrqv wkdw wkh urz sod|hu zrxog qhyhu fkrrvh lw1 Lq wklv vpdoohu jdph/ 5 kdv d grplqdqw vwudwhj| zklfk lv wr frqihvv1 Wkdw lv/ li 5 lv udwlrqdo dqg nqrzv wkdw 4 lv udwlrqdo/ vkh zloo sod| frqihvv1 Lq wkh ruljlqdo jdph grq*w frqihvv glg ehwwhu djdlqvw uxq dzd|/ wkxv frqihvv zdv qrw d grplqdqw vwudwhj|1
Krzhyhu/ 4 sod|lqj uxq dzd| fdqqrw eh udwlrqdol}hg
ehfdxvh lw lv d grplqdwhg vwudwhj|1 Wklv ohdgv wr wkh Holplqdwlrq ri Vwulfwo| Grplqdwhg Vwudwhjlhv1 Zkdw kdsshqv li zh Lwhudwlyho| Holplqdwh Vwulfwo| Grplqdwhg vwudwhjlhvB Wkdw lv/ zh holplqdwh d vwulfwo| grplqdwhg vwudwhj|/ dqg wkhq orrn iru dqrwkhu vwulfwo| grplqdwhg vwudwhj| lq wkh uhgxfhg jdph1 Zh vwrs zkhq zh fdq qr orqjhu qg d vwulfwo| grplqdwhg vwudwhj|1 Fohduo|/ li lw lv frpprq nqrzohgjh wkdw sod|huv duh udwlrqdo/ wkh| zloo sod| rqo| wkh vwudwhjlhv wkdw vxuylyh wklv lwhudwlyho| holplqdwlrq ri vwulfwo| grplqdwhg vwudwhjlhv1 Wkhuhiruh/ zh fdoo vxfk vwudwhjlhv udwlrqdol}deoh1 Fdxwlrq= zh gr holplqdwh wkh vwudwhjlhv wkdw duh grplqdwhg e| vrph pl{hg vwudwhjlhv$ Lq wkh deryh h{dpsoh/ wkh vhw ri udwlrqdol}deoh vwudwhjlhv lv rqfh djdlq frqihvv/ frqihvv1 Dw wklv srlqw |rx vkrxog vwrs dqg dsso| wklv phwkrg wr wkh Frxuqrw gxrsro|$$ +Vhh Jleerqv1, Dovr/ pdnh vxuh wkdw |rx fdq jhqhudwh wkh udwlrqdolw| dv0 vxpswlrq dw hdfk holplqdwlrq1 Iru lqvwdqfh/ lq wkh jdph deryh/ sod|hu 5 nqrzv wkdw sod|hu 4 lv udwlrqdo dqg khqfh kh zloo qrw uxq dzd|> dqg vlqfh vkh lv dovr udwlrqdo/ vkh zloo sod| rqo| frqihvv/ iru wkh frqihvv lv wkh rqo| ehvw uhvsrqvh iru dq| eholhi ri sod|hu 5 wkdw dvvljqv 3 suredelolw| wr wkdw sod|hu 4 uxqv dzd|1 Wkh sureohp lv wkhuh pd| eh wrr pdq| udwlrqdol}deoh vwudwhjlhv1 Frqvlghu wkh Pdwfk0 lqj Sdqqlhv jdph= 4q5
Khdg Wdlo
Khdg
04/4
4/04
Wdlo
4/04
04/4
47
Khuh/ wkh vhw ri udwlrqdol}deoh vwudwhjlhv frqwdlqv ~Khdg/Wdlo iru erwk sod|huv1 Li 4 eholhyhv wkdw 5 zloo sod| Khdg/ kh zloo sod| Wdlo dqg li 5 eholhyhv wkdw 4 zloo sod| Wdlo/ kh zloo sod| Wdlo1 Wkxv/ wkh vwudwhj|0sdlu +Khdg/Wdlo, lv udwlrqdol}deoh1 Exw qrwh wkdw wkh eholhiv ri 4 dqg 5 duh qrw frqjuxhqw1 Wkh vhw ri udwlrqdol}deoh vwudwhjlhv lv lq jhqhudo yhu| odujh1 Lq frqwudvw/ wkh frqfhsw ri grplqdqw vwudwhj| htxloleulxp lv wrr uhvwulfwlyh= xvxdoo| lw grhv qrw h{lvw1 Wkh uhdvrq iru h{lvwhqfh ri wrr pdq| udwlrqdol}deoh vwudwhjlhv lv wkdw zh gr qrw uh0 vwulfw sod|huv* frqmhfwxuhv wr eh cfrqvlvwhqw* zlwk zkdw wkh rwkhuv duh dfwxdoo| grlqj1 Iru lqvwdqfh/ lq wkh udwlrqdol}deoh vwudwhj| +Khdg/ Wdlo,/ sod|hu 5 sod|v Wdlo e| frqmhfwxulqj wkdw Sod|hu 4 zloo sod| Wdlo/ zkloh Sod|hu 4 dfwxdoo| sod|v Khdg1 Zh frqvlghu dqrwkhu frqfhsw Qdvk Htxloleulxp +khqfhiruwk QH,/ zklfk dvvxphv pxwxdo nqrzohgjh ri frq0 mhfwxuhv/ |lhoglqj frqvlvwhqf|1
516
Qdvk Htxloleulxp
Frqvlghu wkh edwwoh ri wkh vh{hv PdqqZrpdq rshud edoohw rshud
4/7
3/3
edoohw
3/3
7/4
Lq wklv jdph/ wkhuh lv qr grplqdqw vwudwhj|1 Exw vxssrvh Z lv sod|lqj rshud1 Wkhq/ Wkxv rshud lv d ehvw0uhvsrqvh iru P
wkh ehvw wklqj P fdq gr lv wr sod| rshud/ wrr1
djdlqvw rshud1 Vlploduo|/ rshud lv d ehvw0uhvsrqvh iru Z djdlqvw rshud1 Wkxv/ dw +rshud/ rshud,/ qhlwkhu sduw| zdqwv wr wdnh d gl huhqw dfwlrq1 Wklv lv d Qdvk Htxloleulxp1 Pruh irupdoo|= Gh qlwlrq 45 Iru dq| sod|hu / d vwudwhj| r lv d ehvw uhvsrqvh wr r li dqg rqo| li 3
Er c r Er c r c ;r 5 7 3
3
Wklv gh qlwlrq lv lghqwlfdo wr wkdw ri d grplqdqw vwudwhj| h{fhsw wkdw lw lv qrw iru
doo r 5 7 exw iru d vshfl f vwudwhj| r 1 Li lw zhuh wuxh iru doo r / wkhq 7- zrxog dovr eh d grplqdqw vwudwhj|/ zklfk lv d vwurqjhu uhtxluhphqw wkdq ehlqj d ehvw uhvsrqvh 3
3
3
3
djdlqvw vrph vwudwhj| r 1 3
48
. . Gh qlwlrq 46 D vwudwhj| sur oh Er. c r lv d Qdvk Htxloleulxp li dqg rqo| li r
. . . . lv d ehvw0uhvsrqvh wr r. ' Er c r c rn c r iru hdfk 1 Wkdw lv/ iru doo / zh 3
kdyh wkdw
3
. . L Er. c r L Er c r 3
3
;r 5 7
Lq rwkhu zrugv/ qr sod|hu zrxog kdyh dq lqfhqwlyh wr ghyldwh/ li kh nqhz zklfk vwudwhjlhv wkh rwkhu sod|huv sod|1 Li d vwudwhj| sur oh lv d grplqdqw vwudwhj| htxloleulxp/ wkhq lw lv dovr d QH/ exw wkh uhyhuvh lv qrw wuxh1 Iru lqvwdqfh/ lq wkh Edwwoh ri wkh Vh{hv/ +R/R, lv d QH dqg E0E lv dq QH exw qhlwkhu duh grplqdqw vwudwhj| htxloleuld1 Ixuwkhupruh/ d grplqdqw vwudwhj| htxloleulxp lv xqltxh/ exw dv wkh Edwwoh ri wkh Vh{hv vkrzv/ QH lv qrw xqltxh lq jhqhudo1 Dw wklv srlqw |rx vkrxog vwrs/ dqg frpsxwh wkh Qdvk htxloleulxp lq Frxuqrw Gxrsro| jdph$$ Zk| grhv Qdvk htxloleulxp frlqflgh zlwk wkh udwlrqdo0 l}deoh vwudwhjlhv1 Lq jhqhudo= Duh doo udwlrqdol}deoh vwudwhjlhv Qdvk htxloleuldB Duh doo Qdvk htxloleuld udwlrqdol}deohB
\rx vkrxog dovr frpsxwh wkh Qdvk htxloleulxp lq
Frxuqrw roljrsro|/ Ehuwudqg gxrsro| dqg lq wkh frpprqv sureohp1 Wkh gh qlwlrq deryh fryhuv rqo| wkh sxuh vwudwhjlhv1 Zh fdq gh qh wkh Qdvk htxl0 oleulxp iru pl{hg vwudwhjlhv e| fkdqjlqj wkh sxuh vwudwhjlhv zlwk wkh pl{hg vwudwhjlhv1 Djdlq jlyhq wkh pl{hg vwudwhj| ri wkh rwkhuv/ hdfk djhqw pd{lpl}hv klv h{shfwhg sd|r ryhu klv rzq +pl{hg, vwudwhjlhv17 H{dpsoh Frqvlghu wkh Edwwoh ri wkh Vh{hv djdlq zkhuh zh orfdwhg wzr sxuh vwudw0 hj| htxloleuld1 Lq dgglwlrq wr wkh sxuh vwudwhj| htxloleuld/ wkhuh lv d pl{hg vwudwhj| htxloleulxp1 PdqqZrpdq rshud edoohw rshud
4/7
3/3
edoohw
3/3
7/4
Ohw*v zulwh ^ iru wkh suredelolw| wkdw P jrhv wr rshud> zlwk suredelolw| ^c kh jrhv wr edoohw1 Li zh zulwh R iru wkh suredelolw| wkdw Z jrhv wr rshud/ zh fdq frpsxwh khu 7
Lq whupv ri eholhiv/ wklv fruuhvsrqghv wr wkh uhtxluhphqw wkdw/ li l dvvljqv srvlwlyh suredelolw| wr wkh
hyhqw wkdw m pd| sod| d sduwlfxodu sxuh vwudwhj| vm / wkhq vm pxvw eh d ehvw uhvsrqvh jlyhq m*v eholhiv1
49
h{shfwhg xwlolw| iurp wklv dv L2 ER( ^ ' R^2 Ershud/rshud n R E ^ 2 Eedoohw/rshud n E R ^2 Ershud/edoohw n E R E ^ 2 Eedoohw/edoohw ' R d^2 Ershud/rshud n E ^ 2 Eedoohw/rshud o n E R d^2 Ershud/edoohw n E ^ 2 Eedoohw/edoohw o ' R d^e n E ^ fo n E R df^ n E ^ o ' Rde^o n E R d ^o Qrwh wkdw wkh whup de^o pxowlsolhg zlwk R lv khu h{shfwhg xwlolw| iurp jrlqj wr rshud/ dqg wkh whup pxowlsolhg zlwk E R lv khu h{shfwhg xwlolw| iurp jrlqj wr edoohw1 L2 ER( ^ lv vwulfwo| lqfuhdvlqj zlwk R li e^ : ^ +l1h1/ ^ : *D,> lw lv vwulfwo| ghfuhdvlqj zlwk R li e^ ^/ dqg lv frqvwdqw li e^ ' ^1 Lq wkdw fdvh/ Z*v ehvw uhvsrqvh lv R ' ri ^ : *D/ R ' f li ^ *D/ dqg R lv dq| qxpehu lq dfc o li ^ ' *D1 Lq rwkhu zrugv/ Z zrxog fkrrvh rshud li khu h{shfwhg xwlolw| iurp rshud lv kljkhu/ edoohw li khu h{shfwhg xwlolw| iurp edoohw lv kljkhu/ dqg fdq fkrrvh dq| ri rshud ru edoohw li vkh lv lqgl huhqw ehwzhhq wkhvh wzr1 Vlploduo| zh frpsxwh wkdw ^ ' lv ehvw uhvsrqvh li R : e*D> ^ ' f lv ehvw uhvsrqvh li R e*D> dqg dq| ^ fdq eh ehvw uhvsrqvh li R ' e*D1 Zh sorw wkh ehvw uhvsrqvhv lq wkh iroorzlqj judsk1 q 1 A (A, B, C) are all equilibria
B
1/5
0
C
4/5
1
p
Wkh Qdvk htxloleuld duh zkhuh wkhvh ehvw uhvsrqvhv lqwhuvhfw1 Wkhuh lv rqh dw +3/3,/ zkhq wkh| erwk jr wr edoohw/ rqh dw +4/4,/ zkhq wkh| erwk jr wr rshud/ dqg wkhuh lv rqh 4:
dw +728/428,/ zkhq Z jrhv wr rshud zlwk suredelolw| 728/ dqg P jrhv wr rshud zlwk suredelolw| 4281 Qrwh krz zh frpsxwh wkh pl{hg vwudwhj| htxloleulxp +iru 5 {5 jdphv,1 Zh fkrrvh 4*v suredelolwlhv vr wkdw 5 lv lqgl huhqw ehwzhhq klv vwudwhjlhv/ dqg zh fkrrvh 5*v suredelolwlhv vr wkdw 4 lv lqgl huhqw1
4;
47145 Jdph Wkhru| Ohfwxuh Qrwhv Ohfwxuhv 90; Pxkdphw \logl} Lq wkhvh ohfwxuhv zh dqdo|}h g|qdplf jdphv +zlwk frpsohwh lqirupdwlrq,1 Zh uvw dqdo|}h wkh shuihfw lqirupdwlrq jdphv/ zkhuh hdfk lqirupdwlrq vhw lv vlqjohwrq/ dqg ghyhors wkh qrwlrq ri edfnzdug lqgxfwlrq1 Wkhq/ frqvlghulqj pruh jhqhudo g|qdplf jdphv/ zh zloo lqwurgxfh wkh frqfhsw ri wkh vxejdph shuihfwlrq1 Zh h{sodlq wkhvh frqfhswv rq hfrqrplf sureohpv/ prvw ri zklfk fdq eh irxqg lq Jleerqv1
4
Edfnzdug lqgxfwlrq
Wkh frqfhsw ri edfnzdug lqgxfwlrq fruuhvsrqgv wr wkh dvvxpswlrq wkdw lw lv frpprq nqrzohgjh wkdw hdfk sod|hu zloo dfw udwlrqdoo| dw hdfk qrgh zkhuh kh pryhv hyhq li klv udwlrqdolw| zrxog lpso| wkdw vxfk d qrgh zloo qrw eh uhdfkhg1 Phfkdqlfdoo|/ lw lv frpsxwhg dv iroorzv1 Frqvlghu d qlwh krul}rq shuihfw lqirupdwlrq jdph1 Frqvlghu dq| qrgh wkdw frphv mxvw ehiruh whuplqdo qrghv/ wkdw lv/ diwhu hdfk pryh vwhpplqj iurp wklv qrgh/ wkh jdph hqgv1 Li wkh sod|hu zkr pryhv dw wklv qrgh dfwv udwlrqdoo|/ kh zloo fkrrvh wkh ehvw pryh iru klpvhoi1 Khqfh/ zh vhohfw rqh ri wkh pryhv wkdw jlyh wklv sod|hu wkh kljkhvw sd|r 1 Dvvljqlqj wkh sd|r yhfwru dvvrfldwhg zlwk wklv pryh wr wkh qrgh dw kdqg/ zh ghohwh doo wkh pryhv vwhpplqj iurp wklv qrgh vr wkdw zh kdyh d vkruwhu jdph/ zkhuh rxu qrgh lv d whuplqdo qrgh1 Uhshdw wklv surfhgxuh xqwlo zh uhdfk wkh ruljlq1 H{dpsoh= frqvlghu wkh iroorzlqj zhoo0nqrzq jdph/ fdoohg dv wkh fhqwlshghv jdph1 Wklv jdph looxvwudwhv wkh vlwxdwlrq zkhuh lw lv pxwxdoo| ehqh fldo iru doo sod|huv wr vwd| Wkhvh
qrwhv gr qrw lqfoxgh doo wkh wrslfv wkdw zloo eh fryhuhg lq wkh fodvv1 L zloo dgg wkrvh wrslfv
odwhu1
4
lq d uhodwlrq/ zkloh d sod|hu zrxog olnh wr h{lw wkh uhodwlrq/ li vkh nqrzv wkdw wkh rwkhu sod|hu zloo h{lw lq wkh qh{w gd|1 4
5
D
4
d
k
G
g
B
+4/4,
+3/7,
+6/6,
+5/8,
Lq wkh wklug gd|/ sod|hu 4 pryhv/ fkrrvlqj ehwzhhq jrlqj dfurvv +k, ru grzq +B,1 Li kh jrhv dfurvv/ kh zrxog jhw 5> li kh jrhv grzq/ kh zloo jhw 61 Khqfh/ zh uhfnrq wkdw kh zloo jr grzq1 Wkhuhiruh/ zh uhgxfh wkh jdph dv iroorzv= 4
D
5
G
g
+4/4,
+3/7,
d
+6/6,
Lq wkh vhfrqg gd|/ sod|hu 5 pryhv/ fkrrvlqj ehwzhhq jrlqj dfurvv +@, ru grzq +_,1 Li vkh jrhv dfurvv/ vkh zloo jhw 6> li vkh jrhv grzq/ vkh zloo jhw 71 Khqfh/ zh uhfnrq wkdw vkh zloo jr grzq1 Wkhuhiruh/ zh uhgxfh wkh jdph ixuwkhu dv iroorzv=
5
4
D
+3/7,
G
+4/4, Qrz/ sod|hu 4 jhwv 3 li kh jrhv dfurvv +,/ dqg jhwv 4 li kh jrhv grzq +(,1 Wkhuhiruh/ kh jrhv grzq1 Wkh htxloleulxp wkdw zh kdyh frqvwuxfwhg lv dv iroorzv= 4
D
5
d
4
G
g
B
+4/4,
+3/7,
+6/6,
k
+5/8,
Wkdw lv/ dw hdfk qrgh/ wkh sod|hu zkr lv wr pryh jrhv grzq/ h{lwlqj wkh uhodwlrq1 Ohw*v jr ryhu wkh dvvxpswlrqv wkdw zh kdyh pdgh lq frqvwuxfwlqj rxu htxloleulxp1 Zh dvvxphg wkdw sod|hu 4 zloo dfw udwlrqdoo| dw wkh odvw gdwh/ zkhq zh uhfnrqhg wkdw kh jrhv grzq1 Zkhq zh uhfnrqhg wkdw sod|hu 5 jrhv grzq lq wkh vhfrqg gd|/ zh dvvxphg wkdw sod|hu 5 dvvxphv wkdw sod|hu 4 zloo dfw udwlrqdoo| rq wkh wklug gd|/ dqg dovr dvvxphg wkdw vkh lv udwlrqdo/ wrr1 Rq wkh uvw gd|/ sod|hu 4 dqwlflsdwhv doo wkhvh1 Wkdw lv/ kh lv dvvxphg wr nqrz wkdw sod|hu 5 lv udwlrqdo/ dqg wkdw vkh zloo nhhs eholhylqj wkdw sod|hu 4 zloo dfw udwlrqdoo| rq wkh wklug gd|1 Wklv h{dpsoh dovr looxvwudwhv dqrwkhu qrwlrq dvvrfldwhg zlwk edfnzdug lqgxfwlrq frpplwphqw +ru wkh odfn ri frpplwphqw,1 Qrwh wkdw wkh rxwfrphv rq wkh wklug gd| +l1h1/ +6/6, dqg +5/8,, duh erwk vwulfwo| ehwwhu wkdq wkh htxloleulxp rxwfrph +4/3,1 Exw wkh| fdqqrw uhdfk wkhvh rxwfrphv/ ehfdxvh sod|hu 5 fdqqrw frpplw wr jr dfurvv/ zkhqfh sod|hu 4 h{lwv wkh uhodwlrq lq wkh uvw gd|1 Wkhuh lv dovr d ixuwkhu frpplwphqw sureohp lq wklv h{dpsoh1 Li sod|hu 4 zkhuh deoh wr frpplw wr jr dfurvv rq wkh wklug gd|/ sod|hu 6
5 zrxog gh qlwho| jr dfurvv rq wkh vhfrqg gd|/ zkhqfh sod|hu 4 zrxog jr dfurvv rq wkh uvw1 Ri frxuvh/ sod|hu 4 fdqqrw frpplw wr jr dfurvv rq wkh wklug gd|/ dqg wkh jdph hqgv lq wkh uvw gd|/ |lhoglqj wkh orz sd|r v +4/3,1 Dv dqrwkhu h{dpsoh/ ohw xv dsso| edfnzdug lqgxfwlrq wr wkh Pdwfklqj Shqqlhv zlwk Shuihfw Lqirupdwlrq= (-1, 1) Head
2
O
Tail
(1, -1)
Head
1 Tail
2
(1, -1)
Head
O Tail (-1, 1)
Li sod|hu 4 fkrrvhv Khdg/ sod|hu 5 zloo Khdg> dqg li 4 fkrrvhv Wdlo/ sod|hu 5 zloo suhihu Wdlo/ wrr1 Khqfh/ wkh jdph lv uhgxfhg wr
(-1,1) Head
Tail
(-1,1)
Lq wkdw fdvh/ Sod|hu 4 zloo eh lqgl huhqw ehwzhhq Khdg dqg Wdlo/ fkrrvlqj dq| ri wkhvh wzr rswlrq ru dq| udqgrpl}dwlrq ehwzhhq wkhvh wzr dfwv zloo jlyh xv dq htxloleulxp zlwk edfnzdug lqgxfwlrq1 Dw wklv srlqw/ |rx vkrxog vwrs dqg vwxg| wkh Vwdfnhoehuj gxrsro| lq Jle0 erqv1 \rx vkrxog dovr fkhfn wkdw wkhuh lv dovr d Qdvk htxloleulxp ri wklv jdph lq zklfk 7
wkh iroorzhu surgxfhv wkh Frxuqrw txdqwlw| luuhvshfwlyh ri zkdw wkh ohdghu surgxfhv/ dqg wkh ohdghu surgxfhv wkh Frxuqrw txdqwlw|1 Ri frxuvh/ wklv lv qrw frqvlvwhqw zlwk edfnzdug lqgxfwlrq= zkhq wkh iroorzhu nqrzv wkdw wkh ohdghu kdv surgxfhg wkh Vwdfnho0 ehuj txdqwlw|/ kh zloo fkdqjh klv plqg dqg surgxfh d orzhu txdqwlw|/ wkh txdqwlw| wkdw lv frpsxwhg gxulqj wkh edfnzdug lqgxfwlrq1 Iru wklv uhdvrq/ zh vd| wkdw wklv Qdvk htxloleulxp lv edvhg rq d qrq0fuhgleoh wkuhdw +ri wkh iroorzhu,1 Edfnzdug lqgxfwlrq lv d srzhuixo vroxwlrq frqfhsw zlwk vrph lqwxlwlyh dsshdo1 Xq0 iruwxqdwho|/ zh fdqqrw dsso| lw eh|rqg shuihfw lqirupdwlrq jdphv zlwk d qlwh krul}rq1 Lwv lqwxlwlrq/ krzhyhu/ fdq eh h{whqghg eh|rqg wkhvh jdphv wkurxjk vxejdph shuihfwlrq1
5
Vxejdph shuihfwlrq
D pdlq surshuw| ri edfnzdug lqgxfwlrq lv wkdw/ zkhq zh frq qh rxuvhoyhv wr d vxejdph ri wkh jdph/ wkh htxloleulxp frpsxwhg xvlqj edfnzdug lqgxfwlrq uhpdlqv wr eh dq htxl0 oleulxp +frpsxwhg djdlq yld edfnzdug lqgxfwlrq, ri wkh vxejdph1 Vxejdph shuihfwlrq jhqhudol}hv wklv qrwlrq wr jhqhudo g|qdplf jdphv= Gh qlwlrq 4 D Qdvk htxloleulxp lv vdlg wr eh vxejdph shuihfw li dq rqo| li lw lv d Qdvk htxloleulxp lq hyhu| vxejdph ri wkh jdph1 Zkdw lv d vxejdphB Lq dq| jlyhq jdph/ wkhuh pd| eh vrph vpdoohu jdphv hpehgghg> zh fdoo hdfk vxfk hpehgghg jdph d vxejdph1 Frqvlghu/ iru lqvwdqfh/ wkh fhqwlshghv jdph +zkhuh wkh htxloleulxp lv gudzq lq wklfn olqhv,=
4
D
5
d
4
G
g
B
+4/4,
+3/7,
+6/6,
k
Wklv jdph kdv wkuhh vxejdphv1 Khuh lv rqh vxejdph= 8
+5/8,
4
k
+5/8,
B
+6/6, Wklv lv dqrwkhu vxejdph= 5
d
4
g
B
+3/7,
+6/6,
k
+5/8,
Dqg wkh wklug vxejdph lv wkh jdph lwvhoi1 Zh fdoo wkh uvw wzr vxejdphv +h{foxglqj wkh jdph lwvhoi, surshu1 Qrwh wkdw/ lq hdfk vxejdph/ wkh htxloleulxp frpsxwhg yld edfnzdug lqgxfwlrq uhpdlqv wr eh dq htxloleulxp ri wkh vxejdph1 Qrz frqvlghu wkh pdwfklqj shqq| jdph zlwk shuihfw lqirupdwlrq1 Lq wklv jdph/ zh kdyh wkuhh vxejdphv= rqh diwhu sod|hu 4 fkrrvhv Khdg/ rqh diwhu sod|hu 4 fkrrvhv Wdlo/ dqg wkh jdph lwvhoi1 Djdlq/ wkh htxloleulxp frpsxwhg wkurxjk edfnzdug lqgxfwlrq lv d Qdvk htxloleulxp dw hdfk vxejdph1 Qrz frqvlghu wkh iroorzlqj jdph1
9
1 E
X
1
(2,6)
T
B
2 L
R
(0,1)
L
(3,2)
R
(-1,3)
(1,5)
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Work
(1, 2)
W Firm
Hire Shirk (0, 1)
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Nature
Low 1-p
Hire
W
Work Shirk
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(1, 1)
(-1, 2)
(0, 0)
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4
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5
S +D m E, @
S +D _ E, > S +E,
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lq dq q0shuvrq vwdwlf jdph ri lqfrpsohwh lqirupdwlrq li dqg rqo| li iru hdfk sod|hu l dqg w|sh w 5 W > vW +w , 5 duj pd{
@
[
x ^vW +w ,> ===> d > ===> vW +w ,` s +w 3 mw ,
zkhuh x lv wkh xwlolw| ri sod|hu l dqg d ghqrwhv dfwlrq1 Wkdw lv/ iru hdfk sod|hu l hdfk srvvleoh w|sh/ wkh dfwlrq fkrvhq lv rswlpdo jlyhq wkh frqglwlrqdo eholhiv ri wkdw w|sh djdlqvw wkh rswlpdo vwudwhjlhv ri doo rwkhu sod|huv1 Wkh uhpdlqlqj qrwhv duh derxw wkh dssolfdwlrqv dqg yhu| vnhwfk|> iru wkh ghwdlov vhh Jle0 erqv1 H{dpsoh= Frxuqrw zlwk Lqfrpsohwh Lqirupdwlrq1 Zh kdyh d olqhdu ghpdqg ixqfwlrq= S +T, @ d T zkhuh T @ t . t2 lv wkh wrwdo ghpdqg/ dqg frqvwdqw pdujlqdo frvwv1 Wkh pdujlqdo frvw ri Ilup 4 lv f dqg frpprq nqrzohgjh/ zkloh Ilup 5*v pdujlqdo frvw lv lwv sulydwh lqirupdwlrq1 Lwv pdujlqdo frvw +w|sh, fdq eh hlwkhu fM zlwk suredlolw| / ru fu zlwk suredelolw| 4 1 Hdfk up pd{lpl}hv lwv h{shfwhg sur w1 Krz wr qg wkh Ed|hvldq Qdvk HtxloleulxpB Ilup 5 kdv wzr srvvleoh w|shv> dqg gl huhqw dfwlrqv zloo eh fkrvhq iru wkh wzr gl huhqw w|shv1 6
it2 +fu ,> t2 +fM ,j Vxssrvh up 5 lv w|sh kljk1 Wkhq/ jlyhq wkh txdqwlw| tW fkrvhq e| sod|hu/ lwv sureohp lv pd{+S fM ,t2 @ ^d t t2 fM ` t2 =
^2
Khqfh/
t2W +fM , @
d tW fM 5
+-,
Vlploduo|/ vxssrvh up 5 lv orz w|sh=
pd{ ^d tW t2 fu ` t2 /
^2
khqfh
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d tW fu = 5
+--,
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pd{ ^d t t2W +fM , f` t . +4 , ^d t t2W +fu , f` t
^
tW @
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t2W +fM , @
d 5fM . f +4 ,+fM fu , . 6 9
t2W +fu , @
tW @
d 5fu . f +fM fu , 6 9
d 5f . fM . +4 ,fu 6 7
+---,
Kduvdu|l*v Mxvwl fdwlrq iru Pl{hg Vwudwhjlhv Frqvlghu wkh jdph R
I
R
5 . w > 4
3> 3
I
3> 3
4> 5 . w2
zkhuh w > w2 sulydwh lqirupdwlrq ri sod|huv 4 dqg 5/ uhvshfwlyho|/ dqg duh lqghshqghqw gudzq iurp xqlirup glvwulexwlrq ryhu ^> 5`1 Fkhfn wkdw + R li w 3 vW +w , @ I rwkhuzlvh/ dqg
+ v2 +w2 , @ W
I li w2 3 R rwkhuzlvh
irup d Ed|hvldq Qdvk htxloleulxp1 Qrwh wkdw sod|hu 4 sod|v 3 zlwk suredelolw| 526 dqg sod|hu 5 sod|v I zlwk suredelolw| 5261 Dv $ 3 +l1h1/ dv xqfhuwdlqw| glvdsshduv,/ wklv vwudwhj| sur oh frqyhujhv wr d pl{hg vwudwhj| htxloleulxp lq zklfk sod|hu 4 sod|v 3 zlwk suredelolw| 526 dqg sod|hu 5 sod|v I zlwk suredelolw| 5261 Vhh Jleerqv iru ghwdlov1
Dxfwlrqv Wzr elgghuv iru d xqltxh jrrg1 y = ydoxdwlrq ri elgghu l1 Ohw xv dvvxph wkdw y *v duh gudzq lqghshqghqwo| iurp d xqlirup glvwulexwlrq ryhu ^3> 4`1 y lv sod|hu l*v sulydwh lqirupdwlrq1 Wkh jdph wdnhv wkh irup ri erwk elgghuv vxeplwwlqj d elg/ wkhq wkh kljkhvw elgghu zlqv dqg sd|v khu elg1 Ohw e eh sod|hu l*v elg1 y +e > e2 > y > y2 , @ y e li e A e
K li e @ e 2 3 li e ? e 3
4 pd{+y e ,S ureie A e +y ,mjlyhq eholhiv ri sod|hu l, . +y e ,S ureie @ e +y ,m===, K 5 Ohw xv uvw frqmhfwxuh wkh irup ri wkh htxloleulxp= Frqmhfwxuh= V|pphwulf dqg olqhdu htxloleulxp
8
e @ d . fy= Wkhq/ 2 +y e ,S ureie @ e +y ,m===, @ 31 Khqfh/
pd{+y e ,S ureie d . fy j @
K
+y e ,S ureiy
e d +e d, j @ +y e , f f
IRF= y . d li y d 5 @d li y ? d
e @
+4,
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wudgh dw sulfh s @
n 1 2 K
r
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Sr +Yr ,=
Vlploduo|/ wkh vhoohu*v sureohp lv 9
Sr . SK +YK , pd{H Yr = SK +YK , Sr ` @ r 5 Sr . H^SK +YK ,mSK +YK , Sr Yr ` S ureiSK +YK , Sr j pd{ r 5
Htxloleulxp lv zkhuh Sr +Y , lv d ehvw uhvsrqvh wr SK +YK , zkloh SK +YK , lv d ehvw uhvsrqvh wr Sr +Yr ,1 Wkhuh duh pdq| Ed|hvldq Qdvk Htxloleuld1 Khuh lv rqh1 Sr @ [
li
Yr [
SK @ [
li
YK [=
Dq htxloleulxp zlwk {hg sulfh1 Zk| lv wklv dq htxloleulxpB Ehfdxvh jlyhq Sr @ [ li Yr [> wkh ex|hu grhv qrw zdqw wr wudgh zlwk YK ? [ dqg zlwk YK A [> SK @ [ lv rswlpdo1
vb v b /v s Trade Efficient
X
not to trade
0
vS Inefficient lack of trade
Qrz frqvwuxfw dq htxloleulxp zlwk olqhdu vwudwhjlhv= sK @ dK . fK yK sr @ dr . fr yr > zkhuh dK / dr / fK / dqg fr duh wr eh ghwhuplqhg1 Qrwh wkdw sK sr +yr , @ dr . fr yr l yr
sK dr = fr :
Olnhzlvh/ sr sK +yK , @ dK . fK yK l yK
sr dK = fK
Wkhq/ wkh ex|hu*v sureohp lv4 sK . sr +yr , pd{H yK = sK sr +yr , RK 5 ] RK 3@r Sr sK . sr +yr , @ pd{ yK gyr RK f 5 ] RK 3@r Sr sK . dr . fr yr gyr yK @ pd{ RK f 5 ] RK 3@r Sr sK dr sK . dr fr @ pd{ yr gyr yK RK fr 5 5 f sK dr sK . dr fr sK dr 2 @ pd{ yK RK fr 5 7 fr sK dr sK . dr sK dr yK @ pd{ RK fr 5 7 sK dr 6sK . dr = yK @ pd{ RK fr 7 I1R1F1= 4 fr
6sK . dr 6+sK dr , yK @3 7 7fr
l1h1/ 5 4 sK @ yK . dr = 6 6 Vlploduo|/ wkh vhoohu*v sureohp lv 4
Wkhuh lv vrphzkdw vlpsohu zd| lq wr jhw wkh vdph rxwfrph> vhh Jleerqv1
;
+5,
] sr . sK +yK , sr . dK . fK yK pd{H yr = sK +yK , sr @ pd{ R 3@ yr gyK Rr Rr rS K 5 5 K ] sr dK sr . dK fK @ pd{ 4 yK gyK yr . Rr fK 5 5 RrS3@K K # $ sr dK sr . dK fK sr dK 2 @ pd{ 4 yr . 4 Rr fK 5 7 fK fK sr dK sr dK sr . dK yr . . @ pd{ 4 Rr fK 5 7 7 sr dK 6sr . dK fK yr . @ pd{ 4 Rr fK 7 7 I1R1F1
4 6 sr dK 4 6sr . dK yr . . 4 @3 fK 7 7 7 fK
l1h1/ 6 fK 6sr . dK yr . . +fK +sr dK ,, @ 3> 7 7 7
l1h1/ 6sr fK 6 dK . fK dK @ . yr . +fK . dK , @ yr . 5 7 7 7 5 l1h1/ 5 dK . fK sr @ yr . = 6 6 E| +5,/ dK @ dr @6/ dqg e| +6,/ dr @ dK @ 4@451 Wkh htxloleulxp lv
+6,
@ . 2 1 Khqfh/
1 1 u(0) + u(100). 2 2
This is true if your vNM utility function is concave (u00 < 0 if u is twice differentiable). In general, if your vNM utility function is concave, you always prefer getting the avarage for sure to getting some risky lottery. In contrast, if your vNM utility function is concave, you always prefer getting some risky lottery to getting the avarage for sure. In this case, you are called risk loving. If your vNM utility function is linear, you are indifferent between getting some risky lottery and getting the avarage for sure. In this case, you are called risk neutral. 3
4
Modelling games
When we write down a game which represents some economic environment, the payoffs of the game are vNM utilities of the players, not the actual money amount received. In this way, we can assume the players maximize their expected payoffs. The only case where the payoffs of a player is equal to the actual money amount received is the case where the player is risk neutral. Actually, in many textbook modelling we find the players are assumed to be risk neutral and therefore maximize expected revenue.
4
14.12 Review Notes Dynamic games with incomplete information Kenichi Amaya∗ November 30, 2001
1
Perfect Bayesian equilibrium • s = (s1 , · · · , sn ): strategy profile. • A belief at an information set is a probability distribution over decision nodes in the information set. • A belief system µ is the collection of beliefs at all information sets.
Definition A pair of strategy profile and a belief system (s, µ) is a Perfect Bayesian (Nash) equilibrium (PBE) if 1. Given their beliefs, the players’ strategies must be sequentially rational. That is, at each information set the action taken by the player with the move (and the player’s subsequent strategy) must be optimal given the player’s belief at that information set and the other players’ subsequent strategies. 2. Beliefs are determined by Bayes’ rule and the players’ equilibrium strategies wherever possible. Note 1: When you are asked to describe a PBE, you need to show not only the strategy profile but also the belief system. (However, you usually don’t need to describe the belief at a singleton information set because it is obvious). Note 2: PBE is a stronger concept than subgame perfect Nash equilibrium. Therefore, the strayegy profile of a PBE is a subgame perfect Nash equilibrium. ∗ E52-303,
[email protected] 1
2
How do we find PBE?
Here is an abstract procedure of looking for PBE. 1. Using sequential rationality and Bayes rule, determine strategies and beliefs wherever possible. If you can determine strategies and beliefs everywhere, that’s it. 2. When you can’t do more, make an assumption about a strategy at any information set. Then, using sequential rationality and Bayes rule, determine strategies and beliefs wherever possible. 3. You may be able to determine strategies and beliefs everywhere, i.e., to find an equilibrium. Then, change your assumption and look for another equilibrium. 4. You may reach a contradiction. In this case, the assumption was wrong. Change your assumption. 5. When you can’t do more only with the assumption you made, make a further assumption. The idea we use in finding a mixed strategy Nash equilibrium applies here too: The mixing probability must be such that the other player is indifferent between the strategies he is mixing.
2
14.12 Economic Applications of Game Theory Professor: Muhamet Yildiz Lecture: MW 2:30-4:00 @4-153? Office Hours: M 4-5:30 @E52-251a TA: Kenichi Amaya F 10,3 @E51-85 Office Hours:TBA Web: http://web.mit.edu/14.12/www/
Name of the game Game Theory = Multi-person decision theory • The outcome is determined by the actions independently taken by multiple decision makers. • Strategic interaction. – Need to understand what the others will do – … what the others think that you will do – …
1
Hawk-Dove game
V − c V − c , 2 2
(V,0)
(0,V)
(V/2,V/2)
Chicken
(-1,-1)
(1,0)
(0,1)
(1/2,1/2)
2
Stag Hunt
(2,2)
(4,0)
(0,4)
(6,6)
Quiz Problem • Without discussing with anyone, each student is to write down a real number xi between 0 and 100 on a paper and submit it to the TA. • The TA will then compute the average x=
x1 + x2 + + xn . n
• The students who submit the number that is closest to x / 3 will share 100 points equally; the others will get 0.
3
14.12 Game Theory Lecture 2: Decision Theory Muhamet Yildiz
Road Map 1. Basic Concepts (Alternatives, preferences,…) 2. Ordinal representation of preferences 3. Cardinal representation – Expected utility theory 4. Applications: Risk sharing and Insurance
5. Quiz
1
Basic Concepts: Alternatives • Agent chooses between the alternatives • X = The set of all alternatives • Alternatives are – Mutually exclusive, and – Exhaustive • Example: Options = {Tea, Coffee} X = {T, C, TC, NT} where T= Tea, C = Coffee, TC = Tea and Coffee, NT = Neither Tea nor Coffee
Basic Concepts: Preferences • TeX • TeX –ordinal representation
2
Examples • Define a relation among the students in this class by – x T y iff x is at least as tall as y; – x M y iff x’s final grade in 14.04 is at least as high as y’s final grade; – x H y iff x and y went to the same high school; – X Y y iff x is strictly younger than y; – x S y iff x is as old as y;
Exercises • Imagine a group of students sitting around a round table. Define a relation R, by writing x R y iff x sits to the right of y. Can you represent R by a utility function? • Consider a relation } among positive real 2 numbers represented by u with u ( x ) = x . Can this relation be represented byu * ( x ) = x ? What about u **( x) = 1/ x ?
3
• • • •
TeX – OR Theorem TeX – Cardinal representation VNM Axioms Theorem
A Lottery
4
Two Lotteries $1000 .3 .3
$1M .00001
$10 .99999
.4 $0
$0
Exercise • Consider an agent with VNM utility 2 function u with u ( x ) = x . Can his preferences be represented by VNM utility function u * ( x ) = x ? What about u **( x) = 1/ x ?
5
Attitudes towards Risk p
x
1-p
y
• A fair gamble:
px+(1-p)y = 0.
• An agent is said to be risk neutral iff he is indifferent towards all fair gambles. He is said to be (strictly) risk averse iff he never wants to take any fair gamble, and (strictly) risk seeking iff he always wants to take fair gambles.
A utility function EU u
A u(pW1+(1- p)W2) C EU(Gamble)
B
W1
pW1+(1-p)W2
W2
6
• An agent is risk-neutral iff he has a linear utility function, i.e., u(x) = ax + b. • An agent is risk-averse iff his utility function is concave. • An agent is risk-seeking iff his utility function is convex.
Risk Sharing • Two agents, each having a utility function u with u ( x ) = x and an “asset:” .5 $100
.5
$0
• For each agent, the value of the asset is • Assume that the value of assets are independently distributed.
7
• If they form a mutual fund so that each agent owns half of each asset, each gets $100 1/4 1/2
$50
1/4 $0
Insurance • We have an agent with u(x ) = x and .5
$1M
.5
$0
• And a risk-neutral insurance company with lots of money, selling full insurance for “premium” P.
8
Lecture 3 Decision Theory/Game Theory 14.12 Game Theory Muhamet Yildiz
Road Map 1. 2.
Basic Concepts (Alternatives, preferences,…) Ordinal representation of preferences
3. Cardinal representation – Expected utility theory
4. Applications: Risk sharing and Insurance
5. Quiz 6. Representation of games in strategic and extensive forms 7. Quiz?
1
A Lottery
Two Lotteries $1000 .3 .3
.4
$1M .00001
$10 .99999 $0
$0
2
Exercise • Consider an agent with VNM utility 2 function u with u ( x ) = x . Can his preferences be represented by VNM utility function u * ( x ) = x ? What about u **( x) = 1/ x ?
Attitudes towards Risk • A fair gamble:
p
x
1-p
y
px+(1-p)y = 0.
• An agent is said to be risk neutral iff he is indifferent towards all fair gambles. He is said to be (strictly) risk averse iff he never wants to take any fair gamble, and (strictly) risk seeking iff he always wants to take fair gambles.
3
A utility function EU u
A u(pW1+(1- p)W2) C EU(Gamble)
B
W1
pW1+(1-p)W2
W2
• An agent is risk-neutral iff he has a linear utility function, i.e., u(x) = ax + b. • An agent is risk-averse iff his utility function is concave. • An agent is risk-seeking iff his utility function is convex.
4
Risk Sharing • Two agents, each having a utility function u with u ( x ) = x and an “asset:” .5 $100
.5
$0
• For each agent, the value of the asset is • Assume that the value of assets are independently distributed.
• If they form a mutual fund so that each agent owns half of each asset, each gets $100 1/4 1/2
$50
1/4 $0
5
Insurance • We have an agent with u(x ) = x and .5
$1M
.5
$0
• And a risk-neutral insurance company with lots of money, selling full insurance for “premium” P.
Quiz Problem • Without discussing with anyone, each student is to write down a real number xi between 0 and 100 on a paper and submit it to the TA. • The TA will then compute the average x=
x1 + x2 + + xn . n
• The students who submit the number that is closest to x / 3 will share 100 points equally; the others will get 0.
6
Multi-person Decision Theory • • • •
Who are the players? Who has which options? Who knows what? Who gets how much?
Knowledge 1. 2. 3. 4.
If I know something, it must be true. If I know x, then I know that I know x. If I don’t know x, then I know that I don’t know x. If I know something, I know all its logical implications.
Common Knowledge: x is common knowledge iff •Each player knows x •Each player knows that each player knows x • Each player knows that each player knows that each player knows x •Each player knows that each player knows that each player knows that each player knows x •… ad infinitum
7
Representations of games
Normal-form representation Definition (Normal form): A game is any list G = (S1 , l , S n ; u1 , l , u n )
where, for each i ∈ N = {1,2,l, n}, • Si is the set of all strategies available to i, • ui : S1 × m × Sn → ℜ is the VNM utility function of player i.
Assumption: G is common knowledge. Definition: A player i is rational iff he tries to maximize the expected value of ui given his beliefs.
8
Chicken
(-1,-1)
(1,0)
(0,1)
(1/2,1/2)
Extensive-form representation Definition: A tree is a set of nodes connected with directed arcs such that 1. For each node, there is at most one incoming arc; 2. each node can be reached through a unique path;
9
A tree?
A tree?? B
B A
A C
D C
10
A tree Non-terminal nodes
Terminal Nodes
Extensive form – definition Definition: A game consists of – a set of players – a tree – an allocation of each non-terminal node to a player – an informational partition (to be made precise) – a payoff for each player at each terminal node.
11
Information set An information set is a collection of nodes such that 1. The same player is to move at each of these nodes; 2. The same moves are available at each of these nodes. An informational partition is an allocation of each non-terminal node of the tree to an information set.
A game 1 L
R 2
(2,2)
r
l
1
1 λ (1,3)
ρ (3,1)
Λ (3,3)
u
(0,0)
Ρ (1,1)
12
Another Game x
1 T
B
2 L
R
L
R
The same game 1
x
T
B
2 L
R
L
R
13
Strategy A strategy of a player is a complete contingent-plan, determining which action he will take at each information set he is to move (including the information sets that will not be reached according to this strategy).
Matching pennies with perfect information 1 Head
Tail
2
2
head
tail
(-1,1)
(1,-1)
head (1,-1)
tail
2’s Strategies: HH = Head if 1 plays Head, Head if 1 plays Tail; HT = Head if 1 plays Head, Tail if 1 plays Tail; TH = Tail if 1 plays Head, Head if 1 plays Tail; TT = Tail if 1 plays Head, Tail if 1 plays Tail.
(-1,1)
14
Matching pennies with perfect information 2
1
HH
HT
TH
TT
Head
Tail
Matching pennies with Imperfect information 1
1
Head
2
Head
Tail
Head
(-1,1)
(1,-1)
Tail
(1,-1)
(-1,1)
Tail
2 head
tail
(-1,1)
(1,-1)
head (1,-1)
tail (-1,1)
15
A game with nature (5, 0)
Left 1 Head 1/2
Right
1/2
Left
(2, 2)
Nature 2
Tail
(3, 3)
Right (0, -5)
A centipede game 1 D (4,4)
A
2
α δ
(5,2)
1
a
(1,-5)
d (3,3)
16
Lecture 4 Representation of Games & Rationalizability 14.12 Game Theory Muhamet Yildiz
Road Map 1. Representation of games in strategic and extensive forms 2. Dominance 3. Dominant-strategy equilibrium 4. Rationalizability
1
Normal-form representation Definition (Normal form): A game is any list G = (S1 , l , S n ; u1 , l , u n )
where, for each i ∈ N = {1,2,l, n}, • Si is the set of all strategies available to i, • ui : S1 × m × Sn → ℜ is the VNM utility function of player i.
Assumption: G is common knowledge. Definition: A player i is rational iff he tries to maximize the expected value of ui given his beliefs.
Chicken
(-1,-1)
(1,0)
(0,1)
(1/2,1/2)
2
Extensive-form representation Definition: A tree is a set of nodes connected with directed arcs such that 1. For each node, there is at most one incoming arc; 2. each node can be reached through a unique path;
A tree Non-terminal nodes
Terminal Nodes
3
Extensive form – definition Definition: A game consists of – a set of players – a tree – an allocation of each non-terminal node to a player – an informational partition (to be made precise) – a payoff for each player at each terminal node.
Information set An information set is a collection of nodes such that 1. The same player is to move at each of these nodes; 2. The same moves are available at each of these nodes. An informational partition is an allocation of each non-terminal node of the tree to an information set.
4
A game 1 L
R 2
(2,2)
r
l
u
1
1 λ
ρ
(1,3)
Ρ
Λ
(3,1)
(0,0)
(1,1)
(3,3)
Another Game x
1
T
B
2 L
R
L
R
5
The same game 1
x
T
B
2 L
R
L
R
Strategy A strategy of a player is a complete contingent-plan, determining which action he will take at each information set he is to move (including the information sets that will not be reached according to this strategy).
6
Matching pennies with perfect information 2’s Strategies: HH = Head if 1 plays Head, Head if 1 plays Tail; HT = Head if 1 plays Head, Tail if 1 plays Tail; TH = Tail if 1 plays Head, Head if 1 plays Tail; TT = Tail if 1 plays Head, Tail if 1 plays Tail.
1 Head
Tail
2
2
head
tail
(-1,1)
(1,-1)
head (1,-1)
tail (-1,1)
Matching pennies with perfect information 2 1
HH
HT
TH
TT
Head
Tail
7
Matching pennies with Imperfect information 1 1 Head
(-1,1)
Head
Tail
Head
(-1,1)
(1,-1)
Tail
(1,-1)
(-1,1)
Tail
2 head
2
tail (1,-1)
head
tail
(1,-1)
(-1,1)
A game with nature Left
(5, 0)
1 Head 1/2
Right
1/2
Left
(2, 2)
Nature Tail
(3, 3)
2 Right (0, -5)
8
Mixed Strategy Definition: A mixed strategy of a player is a probability distribution over the set of his strategies. Pure strategies: Si = {si1,si2,…,sik} A mixed strategy: σi: S → [0,1] s.t. σi(si1) + σi(si2) + … + σi(sik) = 1. If the other players play s-i =(s1,…, si-1,si+1,…,sn), then the expected utility of playing σi is σi(si1)ui(si1,s-i) + σi(si2) ui(si2,s-i) + … + σi(sik) ui(sik,s-i).
How to play
9
Dominance s-i =(s1,…, si-1,si+1,…,sn) Definition: A pure strategy si* strictly dominates si if and only if
u i ( si* , s − i ) > u i ( si , s − i )
∀ s−i .
A mixed strategy σi* strictly dominates si iff σ i ( si1 )ui ( si1 , s−i ) + m + σ i ( sik )ui ( sik , s−i ) > ui ( si , s−i ) ∀si
A rational player never plays a strictly dominated strategy.
Prisoners’ Dilemma 2
Cooperate
Defect
Cooperate
(5,5)
(0,6)
Defect
(6,0)
(1,1)
1
10
A game 1
2
L
m
R
T
(3,0)
(1,1)
(0,3)
M
(1,0)
(0,10)
(1,0)
B
(0,3)
(1,1)
(3,0)
Weak Dominance Definition: A pure strategy si* weakly dominates si if and only if
u i ( si* , s − i ) ≥ u i ( si , s − i )
∀ s−i .
and at least one of the inequalities is strict. A mixed strategy σi* weakly dominates si iff
σ i ( si1 )ui ( si1 , s−i ) + + σ i ( sik )ui ( sik , s−i ) > ui ( si , s−i ) ∀si and at least one of the inequalities is strict. If a player is rational and cautious (i.e., he assigns positive probability to each of his opponents’ strategies), then he will not play a weakly dominated strategy.
11
Dominant-strategy equilibrium Definition: A strategy si* is a dominant strategy iff si* weakly dominates every other strategy si. Definition: A strategy profile s* is a dominant-strategy equilibrium iff si* is a dominant strategy for each player i. If there is a dominant strategy, then it will be played, so long as the players are …
Prisoners’ Dilemma 2
Cooperate
Defect
Cooperate
(5,5)
(0,6)
Defect
(6,0)
(1,1)
1
12
Second-price auction • N = {1,2} buyers; • The value of the house for buyer i is vi; • Each buyer i simultaneously bids bi; • i* with bi* = max bi gets the house and pays the second highest bid p = maxj≠ibj.
Question What is the probability that an nxn game has a dominant strategy equilibrium given that the payoffs are independently drawn from the same (continuous) distribution on [0,1]?
13
(2n-2)
(1/n) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
1
2
3
4
5
6
7
8
9
10
Assume: Players are rational and player 2 knows that 1 is rational.
A game
1 is rational and 2 knows this: 2 1
L
m
L
R
T
(3,0)
(1,1)
(0,3)
M
(1,0)
(0,10)
(1,0)
B
(0,3)
(1,1)
(3,0)
m
R
T
(3,0) (1,1) (0,3)
B
(0,3) (1,1) (3,0)
And 2 is rational:
L
R
T
(3,0) (0,3)
B
(0,3) (3,0)
14
Rationalizability Eliminate all the strictly dominated strategies.
Yes
Any dominated strategy In the new game? No Rationalizable strategies
The play is rationalizable, provided that …
Simplified price-competition Firm 2 Firm 1
High
Medium
Low
High
6,6
0,10
0,8
Medium
10,0
5,5
0,8
Low
8,0
8,0
4,4
Dutta
15
A strategy profile is rationalizable when … • Each player’s strategy is consistent with his rationality, i.e., maximizes his payoff with respect to a conjecture about other players’ strategies; • These conjectures are consistent with the other players’ rationality, i.e., if i conjectures that j will play sj with positive probability, then sj maximizes j’s payoff with respect to a conjecture of j about other players’ strategies; • These conjectures are also consistent with the other players’ rationality, i.e., … • Ad infinitum
16
Lecture 5 Nash equilibrium & Applications 14.12 Game Theory Muhamet Yildiz
Road Map 1. Rationalizability – summary 2. Nash Equilibrium 3. Cournot Competition 1. Rationalizability in Cournot Duopoly
4. 5. 6. 7.
Bertrand Competition Commons Problem Quiz Mixed-strategy Nash equilibrium
1
Dominant-strategy equilibrium s-i =(s1,…, si-1,si+1,…,sn) Definition: si* strictly dominates si iff u i ( si* , s − i ) > u i ( si , s − i )
∀ s−i ;
si* weakly dominates si iff u i ( s , s − i ) ≥ u i ( si , s − i ) ∀s−i and at least one of the inequalities is strict. Definition: A strategy si* is a dominant strategy iff si* weakly dominates every other strategy si. Definition: A strategy profile s* is a dominantstrategy equilibrium iff si* is a dominant strategy for each player i. Examples: Prisoners’ Dilemma; Second-Price auction. * i
Question What is the probability that an nxn game has a dominant strategy equilibrium given that the payoffs are independently drawn from the same (continuous) distribution on [0,1]?
2
(2n-2)
(1/n) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
1
2
3
4
5
6
7
8
9
10
Rationalizability Eliminate all the strictly dominated strategies.
Yes
Any dominated strategy In the new game? No Rationalizable strategies
The play is rationalizable, provided that …
3
Simplified price-competition Firm 2 Firm 1
High
Medium
Low
High
6,6
0,10
0,8
Medium
10,0
5,5
0,8
Low
8,0
8,0
4,4
Dutta
A strategy profile is rationalizable when … • Each player’s strategy is consistent with his rationality, i.e., maximizes his payoff with respect to a conjecture about other players’ strategies; • These conjectures are consistent with the other players’ rationality, i.e., if i conjectures that j will play sj with positive probability, then sj maximizes j’s payoff with respect to a conjecture of j about other players’ strategies; • These conjectures are also consistent with the other players’ rationality, i.e., … • Ad infinitum
4
Stag Hunt
(2,2)
(4,0)
(0,4)
(6,6)
A summary • If players are rational (and cautious), then they play the dominant-strategy equilibrium whenever it exists – But, typically, it does not exist
• If it is common knowledge that players are rational, then they will play a rationalizable strategy-profile – Typically, there are too many rationalizable strategies
• Now, a stronger assumption: The players are rational and their conjectures are mutually known.
5
Nash Equilibrium Definition: A strategy-profile s* =(s1*,…,sn*) is a Nash Equilibrium iff, for each player i, and for each strategy si, we have
u i ( s1* , l , si*−1 , si* , si*+1 , l , s n* ) ≥ u i ( s1* , l , si*−1 , si , si*+1 , l , s n* ), i.e., no player has any incentive to deviate if he knows what the others play. ??If players are rational, and their conjectures about what the others play are mutually known, then they must be playing a Nash equilibrium.
Stag Hunt
(2,2)
(4,0)
(0,4)
(6,6)
6
Economic Applications 1. Cournot (quantity) Competition 1. Nash Equilibrium in Cournot duopoly 2. Nash Equilibrium in Cournot oligopoly 3. Rationalizability in Cournot duopoly
2. Bertrand (price) Competition 3. Commons Problem
Cournot Oligopoly • N = {1,2,…,n} firms; • Simultaneously, each firm i produces qi units of a good at marginal cost c, • and sells the good at price P = max{0,1-Q} where Q = q1+…+qn. • Game = (S1,…,Sn; π1,…,πn) where Si = [0,∞ ∞),
P 1
πi(q1,…,qn) = qi[1-(q1+…+qn)-c] if q1+…+qn < 1, -qic otherwise.
Q 1
7
Cournot Duopoly -- profit qj=0.2 Profit
c=0.2
0 qi(1-qj-c) -cqi
-0.2
0
(1-qj-c)/2
1
1-qj-c
C-D – best responses • qiB(qj) = max{(1-qj-c)/2,0};
• Nash Equilibrium q*: q1* = (1-q2*-c)/2; q2* = (1-q1*-c)/2; • q1* = q2* = (1-c)/3
q2 q1=q1B(q2)
q* 1− c 2
q2=q2B(q1) q1
1-c
8
Cournot Oligopoly --Equilibrium • q>1-c is strictly dominated, so q ≤ 1-c. • πi(q1,…,qn) = qi[1-(q1+…+qn)-c] for each i. • FOC: ∂π ( q , , q ) ∂[ qi (1 − q1 − − qn − c )] 1 i n
=
∂qi
q=q
∂qi
*
q = q*
= (1 − q1* − − qn* − c ) − qi* = 0.
2q1* + q2* + + qn* = 1 − c
• That is,
q1* + 2q2* + + qn* = 1 − c q1* + q2* + + nqn* = 1 − c • Therefore, q1*=…=qn*=(1-c)/(n+1).
Cournot oligopoly – comparative statics P
1
n=1 n=2 n=3 n=4 Q c 1
9
Rationalizability in Cournot Duopoly q2 1-c Assume that players are rational.
1− c 2
q1 1− c 2
1-c
Players are rational: q2 1-c Assume that players know this.
1− c 2
q1 1− c 2
1-c
10
Players are rational and know that players are rational q2 1-c Assume that players know this.
1− c 2
q1 1− c 2
1-c
Players are rational; players know that players are rational; players know that players know that players are rational q 2
1-c Assume that players know this.
1− c 2
q1 1− c 2
1-c
11
Rationalizability in Cournot duopoly If i knows that qj ≤ q, then qi ≥ (1-c-q)/2. If i knows that qj ≥ q, then qi ≤ (1-c-q)/2. We know that qj ≥ q0 = 0. Then, qi ≤ q1 = (1-c-q0)/2 = (1-c)/2 for each i; Then, qi ≥ q2 = (1-c-q1)/2 = (1-c)(1-1/2)/2 for each i; … Then, qn ≤ qi ≤ qn+1 or qn+1 ≤ qi ≤ qn where qn+1 = (1-c-qn)/2 = (1-c)(1-1/2+1/4-…+(-1/2)n)/2. • As n→∞, qn → (1-c)/3. • • • • • • •
Bertrand (price) competition • N = {1,2} firms. • Simultaneously, each firm i sets a price pi; • If pi < pj, firm i sells Q = max{1 – pi,0} unit at price pi; the other firm gets 0. • If p1 = p2, each firm sells Q/2 units at price p1, where Q = max{1 – p1,0}. • The marginal cost is 0. if p1 < p2 p1 (1 − p1 ) π 1 ( p1 , p2 ) = p1 (1 − p1 ) / 2 if p1 = p2 0 otherwise.
12
Bertrand duopoly -- Equilibrium Theorem: The only Nash equilibrium in the “Bertrand game” is p* = (0,0). Proof: 1. p*=(0,0) is an equilibrium. 2. If p = (p1,p2) is an equilibrium, then p = p*. 1. If p = (p1,p2) is an equilibrium, then p1 = p2... •
If pi > pj= 0, for sufficiently small ε>0, pj’ = ε is a better response to pi for j. If pi > pj> 0, pi’ = pj is a better response for i.
2. Given any equilibrium p = (p1,p2) with p1 = p2, p = p*. •
If p1 = p2>0, for sufficiently small ε>0, pj’ = pj - ε is a better response to pj for i.
Commons Problem • N = {1,2,…,n} players, each with unlimited money; • Simultaneously, each player i contributes xi ≥ 0 to produce y = x1+…xn unit of some public good, yielding payoff Ui(xi,y) = y1/2 – xi.
13
Quiz Each student i is to submit a real number xi. We will pair the students randomly. For each pair (i,j), if xi ≠ xj, the student who submits the number that is closer to (xi+xj)/4 gets 100; the other student gets 20. If xi = xj, then each of i and j gets 50.
14
Lectures 6-7 Nash Equilibrium & Backward Induction 14.12 Game Theory Muhamet Yildiz
Road Map 1. 2. 3. 4. 5. 6. 7.
Bertrand Competition Commons Problem Mixed-strategy Nash equilibrium Bertrand competition with costly search Backward Induction Stackelberg Competition Sequential Bargaining
1
Bertrand (price) competition • N = {1,2} firms. • Simultaneously, each firm i sets a price pi; • If pi < pj, firm i sells Q = max{1 – pi,0} unit at price pi; the other firm gets 0. • If p1 = p2, each firm sells Q/2 units at price p1, where Q = max{1 – p1,0}. • The marginal cost is 0. if p1 < p2 p1 (1 − p1 ) π 1 ( p1 , p2 ) = p1 (1 − p1 ) / 2 if p1 = p2 0 otherwise.
Bertrand duopoly -- Equilibrium Theorem: The only Nash equilibrium in the “Bertrand game” is p* = (0,0). Proof: 1. p*=(0,0) is an equilibrium. 2. If p = (p1,p2) is an equilibrium, then p = p*. 1. If p = (p1,p2) is an equilibrium, then p1 = p2... •
If pi > pj= 0, for sufficiently small ε>0, pj’ = ε is a better response to pi for j. If pi > pj> 0, pi’ = pj is a better response for i.
2. Given any equilibrium p = (p1,p2) with p1 = p2, p = p*. •
If p1 = p2>0, for sufficiently small ε>0, pj’ = pj - ε is a better response to pj for i.
2
Commons Problem • N = {1,2,…,n} players, each with unlimited money; • Simultaneously, each player i contributes xi ≥ 0 to produce y = x1+…xn unit of some public good, yielding payoff Ui(xi,y) = y1/2 – xi.
Stag Hunt
(2,2)
(4,0)
(0,4)
(5,5)
3
Equilibrium in Mixed Strategies What is a strategy? – A complete contingent-plan of a player. – What the others think the player might do under various contingency.
What do we mean by a mixed strategy? – The player is randomly choosing his pure strategies. – The other players are not certain about what he will do.
Stag Hunt
(2,2)
(4,0)
(0,4)
(5,5)
4
Mixed-strategy equilibrium in Stag-Hunt game • Assume: Player 2 thinks that, with probability p, Player 1 targets for Rabbit. What is the best probability q she wants to play Rabbit? • His payoff from targeting Rabbit: U2(R;p) = 2p + 4(1-p) = 4-2p. • From Stag: U2(R;p) = 5(1-p) • She is indifferent iff 4-2p = 5(1-p) iff p = 1/3.
5 4.5 4 3.5 3 2.5
4 - 2p
2 1.5 1
5(1-p)
0.5 0
0
0.2
0.4
0.6
0.8
1
if p < 1/3 0 q ( p ) = q ∈ [0,1] if p = 1/3 1 if p > 1/3 BR
Best responses in Stag-Hunt game q
1/3 p 1/3
5
Bertrand Competition with costly search •
N = {F1,F2,B}; F1, F2 are firms; B is buyer B needs 1 unit of good, worth 6; Firms sell the good; Marginal cost = 0. Possible prices P = {1,5}. Buyer can check the prices with a small cost c > 0.
• • • •
Game: 1. Each firm i chooses price p i; 2. B decides whether to check the prices; 3. (Given) If he checks the prices, and p1≠p2, he buys the cheaper one; otherwise, he buys from any of the firm with probability ½.
Bertrand Competition with costly search F2 F1
F2 High
Low
F1
High
High
Low
Low
Check
High
Low
Don’t Check
6
Mixed-strategy equilibrium • Symmetric equilibrium: Each firm charges “High” with probability q; • Buyer Checks with probability r. • U(check;q) = q21 + (1-q2)5 – c = 5 - 4 q2 – c; • U(Don’t;q) = q1 + (1-q)5 = 5 - 4 q; • Indifference: 4q(1-q) = c; i.e., • U(high;q,r) = 0.5(1-r(1-q))5; • U(low;q,r) = qr1 + 0.5(1-qr) • Indifference = r = 4/(5-4q).
Dynamic Games of Perfect Information & Backward Induction
7
Definitions Perfect-Information game is a game in which all the information sets are singleton. Sequential Rationality: A player is sequentially rational iff, at each node he is to move, he maximizes his expected utility conditional on that he is at the node – even if this node is precluded by his own strategy. In a finite game of perfect information, the common knowledge of sequential rationality gives “Backward Induction” outcome.
A centipede game 1 D (4,4)
A
2
α δ
(5,2)
1
a
(1,-5)
d (3,3)
8
Backward Induction Take any pen-terminal node Pick one of the payoff vectors (moves) that gives ‘the mover’ at the node the highest payoff Assign this payoff to the node at the hand; Eliminate all the moves and the terminal nodes following the node Yes
Any non-terminal node No The picked moves
Battle of The Sexes with perfect information 1
T
B
2 L
(2,1)
2 R
(0,0)
L
(0,0)
R
(1,2)
9
Note • There are Nash equilibria that are different from the Backward Induction outcome. • Backward Induction always yields a Nash Equilibrium. • That is, Sequential rationality is stronger than rationality.
Matching Pennies (wpi) 1 Head
Tail
2
2
head
tail
(-1,1)
(1,-1)
head (1,-1)
tail (-1,1)
10
Stackelberg Duopoly Game: P N = {1,2} firms w MC = 0; 1. Firm 1 produces q1 units 1 2. Observing q1, Firm 2 produces q2 units 3. Each sells the good at price P = max{0,1-(q1+q2)}.
Q 1
πi(q1, q2) = qi[1-(q1+q2)] if q1+ q2 < 1, 0 otherwise.
“Stackelberg equilibrium” • If q1 > 1, q2*(q1) = 0. • If q1 ≤ 1, q2*(q1) = (1-q1)/2. • Given the function q2*, if q1 ≤ 1
P 1
π1(q1;q2*(q1)) = q1[1-(q1+ (1-q1)/2)] = q1 (1-q1)/2; 0 otherwise. • q1* = ½. • q2*(q1*) = ¼.
1
11
Sequential Bargaining • N = {1,2} • X = feasible expected-utility pairs (x,y ∈X ) • Ui(x,t) = δitxi • d = (0,0) ∈ D disagreement payoffs
D
1
1
Timeline – 2 period At t = 2,
At t = 1, – – – –
– Player 2 offers some (x2,y2), – Player 1 Accept or Rejects the offer – If the offer is Accepted, the game ends yielding payoff δ(x2,y2). – Otherwise, the game end yielding d = (0,0).
Player 1 offers some (x1,y1), Player 2 Accept or Rejects the offer If the offer is Accepted, the game ends yielding (x1,y1), Otherwise, we proceed to date 2.
1
(x1,y1)
2
(x2,y2)
2 Reject
Accept
1
(0,0) Reject
Accept
(δx2,δy2)
(x1,y1)
12
1
(x1,y1)
2
(x2,y2)
2 Reject
Accept
1
(0,0) Reject
Accept
(δx2,δy2)
(x1,y1)
At t = 2, •Accept iff y2 ≥ 0. •Offer (0,1). At t = 1, •Accept iff x2 ≥ δ. •Offer (1−δ,δ).
Timeline – 2n period T = {1,2,…,2n-1,2n} If t is odd, – – –
–
Player 1 offers some (xt,yt), Player 2 Accept or Rejects the offer If the offer is Accepted, the game ends yielding δt(xt,yt), Otherwise, we proceed to date t+1.
If t is even – Player 2 offers some (xt,yt), – Player 1 Accept or Rejects the offer – If the offer is Accepted, the game ends yielding payoff (xt,yt), – Otherwise, we proceed to date t+1, except at t = 2n, when the game end yielding d = (0,0).
13
Equilibrium • Scientific Word
14
Lectures 8 Subgame-perfect Equilibrium & Applications 14.12 Game Theory Muhamet Yildiz
Road Map 1. Subgame-perfect Equilibrium 1. 2. 3. 4.
Motivation What is a subgame? Definition Example
2. Applications 1. Bank Runs 2. Tariffs & Intra-industry trade
3. Quiz
1
A game 1 E
X
1
(2,6)
T
B
2 L
R
(0,1)
L
(3,2)
(-1,3)
R (1,5)
Sequential Bargaining 1
• N = {1,2} • X = feasible expected-utility pairs (x,y ∈X ) • Ui(x,t) = δitxi • d = (0,0) ∈ D disagreement payoffs
D
1
2
Timeline – ∞ period T = {1,2,…, n-1,n,…} If t is odd, – – –
–
Player 1 offers some (xt,yt), Player 2 Accept or Rejects the offer If the offer is Accepted, the game ends yielding δt(xt,yt), Otherwise, we proceed to date t+1.
If t is even – Player 2 offers some (xt,yt), – Player 1 Accept or Rejects the offer – If the offer is Accepted, the game ends yielding payoff (xt,yt), – Otherwise, we proceed to date t+1.
Backward induction • Can be applied only in perfect information games of finite horizon. How can we extend this notion to infinite horizon games, or to games with imperfect information?
3
A subgame A subgame is part of a game that can be considered as a game itself. • It must have a unique starting point; • It must contain all the nodes that follow the starting node; • If a node is in a subgame, the entire information set that contains the node must be in the subgame.
A game 1 D (4,4)
A
2
α δ
(5,2)
1
a
(1,-5)
d (3,3)
4
And its subgames 1 a
2
(1,-5)
d
α δ
(3,3)
1 a
(1,-5)
d
(5,2)
(3,3)
A game 1 E
X
1
(2,6)
T
B
2 L (0,1)
R (3,2)
L (-1,3)
R (1,5)
5
Definitions A substrategy is the restriction of a strategy to a subgame. A subgame-perfect Nash equilibrium is a Nash equilibrium whose substrategy profile is a Nash equilibrium at each subgame.
Example 1 E
X
1
(2,6)
T
B
2 L (0,1)
R (3,2)
L (-1,3)
R (1,5)
6
A “Backward-Induction-like” method Take any subgame with no proper subgame Compute a Nash equilibrium for this subgame Assign the payoff of the Nash equilibrium to the starting node of the subgame Eliminate the subgame Yes
Any non-terminal node No The moves computed as a part of any (subgame) Nash equilibrium
Theorem In a finite, perfect-information game, the set of subgame-perfect equilibria is the set of strategy profiles that are computed via backward induction.
7
A subgame-perfect equilibrium? X 1
(2,6)
T
B
2 L
R
(0,1)
L
(3,2)
R
(-1,3)
(1,5)
Bank Run 1
R > D > r > D/2
DW
W 2 W (r,r)
DW
DW
W
(D,2r-D) (2r-D,D)
1 W
DW
2 W (R,R)
DW
W
(2R-D,D) (D,2R-D)
DW (R,R)
8
Lectures 9 Applications of Subgame-perfect Equilibrium & Forward Induction 14.12 Game Theory Muhamet Yildiz
Road Map 1. Applications 1. Tariffs & Intra-industry trade 2. Infinite horizon bargaining – Singledeviation principle
2. Forward Induction – Examples 3. Finitely Repeated Games 4. Quiz
1
Single-Deviation principle Definition: An extensive-form game is continuous at infinity iff, given any ε > 0, there exists some t such that, for any two path whose first t arcs are the same, the payoff difference of each player is less than ε. Theorem: Let G be a game that is continuous at infinity. A strategy profile s = (s1,s2,…,sn) is a subgame-perfect equilibrium of G iff, at any information set, where a player i moves, given the other players strategies and given i’s moves at the other information sets, player i cannot increase his conditional payoff at the information set by deviating from his strategy at the information set.
Sequential Bargaining 1
• N = {1,2} • D = feasible expected-utility pairs (x,y ∈D ) • Ui(x,t) = δitxi • d = (0,0) ∈ D disagreement payoffs
D
1
2
Timeline – ∞ period T = {1,2,…, n-1,n,…} If t is odd, – – –
–
Player 1 offers some (xt,yt), Player 2 Accept or Rejects the offer If the offer is Accepted, the game ends yielding δt(xt,yt), Otherwise, we proceed to date t+1.
If t is even – Player 2 offers some (xt,yt), – Player 1 Accept or Rejects the offer – If the offer is Accepted, the game ends yielding payoff δt(xt,yt), – Otherwise, we proceed to date t+1.
SPE of ∞-period bargaining Theorem: At any t, proposer offers the other player δ/(1+δ), keeping himself 1/(1+δ), while the other player accept an offer iff he gets δ/(1+δ). “Proof:” Single-deviation principle: Take any date t, at which i offers, j accepts/rejects. According to the strategies in the continuation game, at t+1, j will get 1/(1+δ). Hence, j accepts an offer iff she gets at least δ/(1+δ). i must offer δ/(1+δ).
3
Forward Induction Strong belief in rationality: At any history of the game, each agent is assumed to be rational if possible. (That is, if there are two strategies s and s’ of a player i that are consistent with a history of play, and if s is strictly dominated but s’ is not, at this history no player j believes that i plays s.)
Table for the bidding game Ui = 20(2+2minjbidj – bidi)
min
1
2
3
1
60
-
-
2
40
80
-
3
20
60
100
bid
4
Nash equilibria of bidding game • 3 equilibria: s1 = everybody plays 1; s2 = everybody plays 2; s3 = everybody plays 3. • Assume each player trembles with probability ε < 1/2, and plays each unintended strategy w.p. ε/2, e.g., w.p. ε/2, he thinks that such other equilibrium is to be played. – s3 is an equilibrium iff – s2 is an equilibrium iff – s1 is an equilibrium iff
1 0.8 0.6 0.4 0.2 0.047
0 -0.2
n
(1-ε/2) -1/2 -0.4 -0.6 n
-0.8 -1
n
(1-ε) +(1-ε/2) -1 0
0.032 0.05
0.1
0.15
0.2
5
Bidding game with entry fee Each player is first to decide whether to play the bidding game (E or X); if he plays, he is to pay a fee p > 60.
1
2
3
60 40 20
80 60
100
min Bid
1 2 3
For each m =1,2,3, ∃SPE: (m,m,m) is played in the bidding game, and players play the game iff 20(2+m) ≥ p. Forward induction: when 20(2+m) < p, (Em) is strictly dominated by (Xk). After E, no player will assign positive probability to min bid ≤ m. FI-Equilibria: (Em,Em,Em) where 20(2+m) ≥ p. What if an auction before the bidding game?
Burning Money 0
B
B 3,1
S 0,0
S
0,0
1,3
1
D
BB
B
B 2,1
S -1,0
S
-1,0
0,3
BS
SB
SS
0B 0S DB DS
O T
H E
R
6
Repeated Games
Entry deterrence 1 X
Enter
2
Acc.
(1,1)
Fight (0,2)
(-1,-1)
7
Entry deterrence, repeated twice 1
(1,3)
2 Acc.
Enter
X 2 Enter 1
Acc. Fight
(-1,1)
X (0,4)
Fight
1 X (-1,1)
1
Enter
2 Acc.
(2,2)
Fight
X
(1,3) (0,0) Enter 2 Acc.
(0,0)
Fight (-2,-2)
8
Lectures 10 -11 Repeated Games
14.12 Game Theory Muhamet Yildiz
Road Map 1.
Forward Induction – Examples
2. Finitely Repeated Games with observable actions 1. Entry-Deterrence/Chain-store paradox 2. Repeated Prisoners’ Dilemma 3. A general result 4. When there are multiple equilibria
3. Infinitely repeated games with observable actions 1. 2. 3. 4. 5.
Discounting / Present value Examples The Folk Theorem Repeated Prisoners’ Dilemma, revisited –tit for tat Repeated Cournot oligopoly
4. Infinitely repeated games with unobservable actions
Forward Induction Strong belief in rationality: At any history of the game, each agent is assumed to be rational if possible. (That is, if there are two strategies s and s’ of a player i that are consistent with a history of play, and if s is strictly dominated but s’ is not, at this history no player j believes that i plays s.)
Bidding game with entry fee Each player is first to decide whether to play the bidding game (E or X); if he plays, he is to pay a fee p > 60.
min
1
2
3
1 2 3
60 40 20
80 60
100
Bid
For each m =1,2,3, ∃SPE: (m,m,m) is played in the bidding game, and players play the game iff 20(2+m) ≥ p. Forward induction: when 20(2+m) < p, (Em) is strictly dominated by (Xk). After E, no player will assign positive probability to min bid ≤ m. FI-Equilibria: (Em,Em,Em) where 20(2+m) ≥ p. What if an auction before the bidding game?
Burning Money 0
B
B 3,1
S 0,0
S
0,0
1,3
1
D
BB
B
B 2,1
S -1,0
S
-1,0
0,3
BS
SB
SS
0B 0S DB DS
O T
H E
R
Repeated Games
Entry deterrence 1 X
Enter
2 Acc. Fight
(0,2)
(-1,-1)
(1,1)
Entry deterrence, repeated twice, many times 1
(1,3)
Enter
2 Acc.
X 2 Enter 1
Acc. Fight
(-1,1)
X (0,4)
Fight
1 X (-1,1)
1
Enter
2 Acc.
X
Fight
(1,3) (0,0) Enter 2 Acc.
Fight (-2,-2)
What would happen if repeated n times?
(0,0)
(2,2)
Prisoners’ Dilemma, repeated twice, many times • Two dates T = {0,1}; • At each date the prisoners’ dilemma is played:
C C 5,5
D 0,6
D 6,0
1,1
• At the beginning of 1 players observe the strategies at 0. Payoffs= sum of stage payoffs.
Twice-repeated PD 1
C
D 2
D
C 1
C 10 10
1 D 2
C
D C
5 11
6 6
11 5
5 11
1 C
D
C
D C
1 D
2
6 6
1 7
C
D 2
2 D
D C 0 12
D
C
C 11 5
D
C
D
C
D C
D
6 6
12 0
7 1
6 6
1 7
2 2
What would happen if T = {0,1,2,…,n}?
7 1
A general result • G = “stage game” = a finite game • T = {0,1,…,n} • At each t in T, G is played, and players remember which actions taken before t; • Payoffs = Sum of payoffs in the stage game. • Call this game G(T). Theorem: If G has a unique subgame-perfect equilibrium s*, G(T) has a unique subgameperfect equilibrium, in which s* is played at each stage.
With multiple equilibria T = {0,1} 2
L
M2
R
T
1,1
5,0
0,0
M1
0,5
4,4
0,0
B
0,0
0,0
3,3
1
s* = •At t = 0, each i play Mi; •At t = 1, play (B,R) if (M1,M2) at t = 0, play (T,L) otherwise.
L
M2
R
T
2,2
6,1
1,1
M1
1,6
7,7
1,1
B
1,1
1,1
4,4
Infinitely repeated Games with observable actions • T = {0,1,2,…,t,…} • G = “stage game” = a finite game • At each t in T, G is played, and players remember which actions taken before t; • Payoffs = Discounted sum of payoffs in the stage game. • Call this game G(T).
Definitions The Present Value of a given payoff stream π = (π0,π1,…,πt,…) is PV(π;δ) = Σ∞t=1 δtπt = π0 + δπ1 + … + δtπt +… The Average Value of a given payoff stream π is (1−δ)PV(π;δ) = (1−δ)Σ∞t=1 δtπt The Present Value of a given payoff stream π at t is PVt(π;δ) = Σ∞s=t δs-t πs = πt + δπt+1 + … + δsπt+s +…
Infinite-period entry deterrence 1 X (0,2)
Enter
2 Acc. Fight (-1,-1)
(1,1)
Strategy of Entrant: Enter iff Accomodated before. Strategy of Incumbent: Accommodate iff accomodated before.
Incumbent: • V(Acc.) = VA = 1/(1−δ); • V(Fight) = VF = 2/(1−δ); • Case 1: Accommodated before. – Fight => -1 + δVA – Acc. => 1 + δVA.
• Case 2: Not Accommodated – Fight => -1 + δVF – Acc. => 1 + δVA – Fight Ù -1 + δVF ≥ 1 + δVA ÙVF − VA = 1/(1−δ) ≥ 2/δ Ùδ ≥ 2/3.
Entrant: • Accommodated – Enter => 1+VAE – X => 0 +VAE
• Not Acc. – Enter =>-1+VFE – X => 0 +VFE
Infinitely-repeated PD C C 5,5
D 0,6
D 6,0
1,1
A Grimm Strategy: Defect iff someone defected before.
• • • •
VD = 1/(1−δ); VC = 5/(1−δ) = 5VD; Defected before (easy) Not defected – D => – C => –CÙ
Tit for Tat • Start with C; thereafter, play what the other player played in the previous round. • Is (Tit-for-tat,Tit-for-tat) a SPE? • Modified: Start with C; if any player plays D when the previous play is (C,C), play D in the next period, then switch back to C.
Folk Theorem Definition: A payoff vector v = (v1,v2,…,vn) is feasible iff v is a convex combination of some pure-strategy payoff-vectors, i.e., v = p1u(a1) + p2u(a2) +…+ pku(ak), where p1 + p2 +…+ pk = 1, and u(aj) is the payoff vector at strategy profile aj of the stage game. Theorem: Let x = (x1,x2,…,xn) be s feasible payoff vector, and e = (e1,e2,…,en) be a payoff vector at some equilibrium of the stage game such that xi > ei for each i. Then, there exist δ < 1 and a strategy profile s such that s yields x as the expected average-payoff vector and is a SPE whenever δ > δ.
Folk Theorem in PD C C 5,5
D 0,6
D 6,0
1,1
• A SPE with PV (1.1,1.1)? – With PV (1.1,5)? – With PV (6,0)? – With PV (5.9,0.1)?
Infinitely-repeated Cournot oligopoly • N firms, MC = 0; P = max{1-Q,0}; • Strategy: Each is to produce q = 1/(2n); if any firm defects produce q = 1/(1+n) forever. • VC = • VD = • V(D|C) = • Equilibrium Ù
1 0.95 0.9 0.85 0.8 0.75 0.7 0.65 0.6 0.55 0.5
0
20
40
60
80
100
1 0.95 0.9 0.85 0.8 0.75 0.7 0.65 0.6 0.55 0.5
0
200
400
600
800
1000
IRCD (n=2) • Strategy: Each firm is to produce q*; if any one deviates, each produce 1/(n+1) thereafter. • VC = q*(1-2q*)/(1-δ); • VD = 1/(9(1-δ)); δ 2 • VD|C = max q(1-q*-q) +δVD = (1 − q *) / 4 + 9(1 − δ ) • Equilibrium iff 2 q * (1 − 2q *) ≥ (1 − δ )(1 − q *) / 4 + δ / 9 • Ù
9 − 5δ q* ≥ 3(9 − δ )
x = δ, y = (3-5/3 δ)/(9-δ )
0.4 0.3
y
0.2 0.1 0 0
0.2
0.4
0.6 x
0.8
1
Carrot and Stick Produce ¼ at the beginning; at ant t > 0, produce ¼ if both produced ¼ or both produced x at t-1; otherwise, produce x. Two Phase: Cartel & Punishment VC = 1/8(1-δ). Vx = x(1-2x) + δVC. VD|C = max q(1-1/4-q) + δVX = (3/8)2 + δVX VD|x = max q(1-x-q) + δVX = (1-x)2/4 + δVX VC ≥ VD|C Ù VC ≥ (3/8)2 + δ2VC + δ x(1-2x) Ù (1-δ2) VC - (3/8)2 ≥ δ x(1-2x) Ù(1+δ)/8 - (3/8)2 ≥ δ x(1-2x) VX ≥ VD|C Ù (1-δ)Vx ≥ (1-x)2/4 Ù (1-δ)(x(1-2x) + δ/8(1-δ)) ≥ (1-x)2/4 Ù (1-δ)x(1-2x) + δ/8 ≥ (1-x)2/4 2x2 – x + 1/8 – 9/64δ ≥ 0 (9/4-2δ)x2 – (3-2δ)x +δ/8(1-δ) ≤ 0
Lectures 12-13 Incomplete Information Static Case 14.12 Game Theory Muhamet Yildiz
Road Map 1. Examples 2. Bayes’ rule 3. Definitions 1. Bayesian Game 2. Bayesian Nash Equilibrium
4. Mixed strategies, revisited 5. Economic Applications 1. Cournot Duopoly 2. Auctions 3. Double Auction
1
Incomplete information We have incomplete (or asymmetric) information if one player knows something (relevant) that some other player does not know.
An Example W ork
(1, 2)
W Firm
H ire Shirk (0, 1)
H igh p D o not hire
N ature
Low 1-p
H ire
(0, 0) W
W ork Shirk
D o not hire
(1, 1)
(-1, 2)
(0, 0)
2
The same example Work
(1, 2)
Shirk
(0, 1)
W Nature
High p
Hire
hire
Firm
Work
Low 1-p
(1, 1)
W Shirk
Do not
(-1, 2) (0, 0)
Another Example Buy
(p, 2-p)
B Seller
p
Don’t
p’
Buy
High 0.5
B
Nature
Low .5
What would you ask if you were to choose p from [0,4]?
B
Don’t
Buy
(0, 0) (p’,2-p’)
(0, 0) (p, 1-p)
p Don’t p’
B
(0,0)
Buy
(p’, 1-p’)
Don’t
(0, 0)
3
Same “Another Example” Buy
(p, 2-p)
B High 0.5 Don’t
Nature
Buy p
Low .5
B
Don’t
Seller p’
Nature High 0.5
What would you ask if you were to choose p from [0,4]?
B
Buy Don’t
Low .5
B
(0, 0) (p’,2-p’)
(0, 0) (p, 1-p)
(0,0)
Buy
(p’, 1-p’)
Don’t
(0, 0)
Bayes’ Rule Prob(A and B) • Prob(A|B) = Prob(B) • Prob(A and B) = Prob(A|B)Prob(B) = Prob(B|A)Prob(A)
Prob(B|A)Prob(A) • Prob(A|B) = Prob(B)
4
Example p Work µ
Success 1-p
• Prob(Work|Success) = µp/[µp + (1−µ)(1-p)] • Prob(Work|Failure) = (1-µ)p/[µ(1−p) + (1−µ)p]
1-p Failure
Shirk 1−µ
p
1 0.9
P(W|S)
0.8
P (w|S ),P (W|F)
0.7 0.6 0.5
P(W|F)
0.4 0.3 0.2 0.1 0
0
0.2
0.4
0.6
0.8
1
µ
5
Bayesian Game (Normal Form) A Bayesian game is a list G = {A1,…,An;T1,…,Tn;p1,…,pn;u1,…,un} where • Ai is the action space of i (ai in Ai) • Ti is the type space of i (ti) • pi(t-i|ti) is i’s belief about the other players • ui(a1,…,an;t1,…,tn) is i’s payoff.
An Example Work
(1, 2)
W Firm
Hire Shirk (0, 1)
High p Do not (0, 0) hire
Nature
Low 1-p
Hire
W
Work Shirk
Do not hire
TFirm={tf}; TW = {High,Low} AFirm = {Hire, Don’t} AW = {Work,Shirk} pF(High) = p pF(Low) = 1-p
(1, 1)
(-1, 2)
(0, 0)
6
Bayesian Nash equilibrium A Bayesian Nash equilibrium is a Nash equilibrium of a Bayesian game. Given any Bayesian game G = {A1,…,An;T1,…,Tn;p1,…,pn;u1,…,un} a strategy of a player i in a is any function si:Ti → Ai; A strategy profile s* = (s1*,…, s1*) is a Bayesian Nash equilibrium iff si*(ti) solves
max ∑ u (s (t ),..., s (t ), a , s (t ),..., s (t ); t )p (t ai ∈Ai
t− i ∈T− i
i
* 1
* i −1
1
i −1
i
* i +1
i +1
* n
n
i
−i
| ti )
i.e., si* is a best response to s-i*.
An Example Work
(1, 2)
W Firm
Hire Shirk (0, 1)
High p Do not (0, 0) hire
Nature
Low 1-p
Hire
W
Work Shirk
Do not hire
(0, 0)
TFirm={tf}; TW = {High,Low} AFirm = {Hire, Don’t} AW = {Work,Shirk} pF(High) = p >1/2 pF(Low) = 1-p
(1, 1)
(-1, 2)
sF* = Hire, sF* (High) = Work sF* (Low) = Shirk Another equilibrium?
7
Stag Hunt, Mixed Strategy
(2,2)
(4,0)
(0,4)
(6,6)
Mixed Strategies • t and v are iid with uniform distribution on [−ε,ε]. • t and v are privately known by 1 and 2, respectively, i.e., are types of 1 and 2, respectively. • Pure strategy: 2+t,2+v 4+t,0 – s1(t) = Rabbit iff t > 0; – s2(v) = Rabbit iff t > 0. 0,4+v 6,6 • p = Prob(s1(t)=Rabbit|v) = Prob(t > 0) = 1/2. U1(R|t) = t +2q+4(1-q) = t + 4 – 2q • q = Prob(s2(v)=Rabbit|t) = U1(S|t) = 6(1-q); 1/2. U1(R|t) > U1(S|t) t+4–2q > 6(1-q) t > 6-6q+2q-4 = 2 – 4q = 0.
8
Economic Applications with Incomplete Information 14.12 Game Theory Muhamet Yildiz
Road Map 1. Cournot duopoly with Incomplete Information 2. A ¿rst price auction 3. A double auction 4. Quiz
1
1 Cournot duopoly with Incomplete Information Demand:
E' ' @ ' where ' ' ^ n ^2. The marginal cost of Firm 1 = S common knowledge. Firm 2’s marginal cost: SM with probaility w, Su with probaility w its private information. Each ¿rm maximizes its expected pro¿t.
2
1.1 Bayesian Nash Equilibrium Firm 2 of high type: 4@ E SM ^2 ' 4@ d@ ^ ^2 SM o ^2 ^5
^5
@ ^ SM ' 2 Firm 2 of low type: ^2ESM
(*)
4@ d@ ^ ^2 Suo ^2, ^5
^2ESu Firm 1:
@ ^ Su ' 2
(**)
4@ w d@ ^ ^2ESM So ^ ^4
^
n E w d@ ^ ^2ESu So ^
w d@ ^2ESM So n E w d@ ^2ESu So ' 2 (***)
3
Solve *, **, and *** for ^c ^2ESu c ^2ESM .
3
^
4
6 3
5
2 w w C ^2ESM D ' 7 2 f 8 f 2 ^2ESu
4
@S C @ SM D ( @ Su
@ 2SM n S E w ESM Su n ' S @ 2Su n S wESM Su ^2 ESu ' S @ 2S n wSM n E w Su ^ '
^2ESM
4
2 A First-price Auction One object, two bidders ' Valuation of bidder i iid with uniform distribution over dfc o. Simultaneously, each bidder submits a bid K, then the highest bidder wins the object and pays her bid. The payoffs: ; ? K if K : K c l EKc K2c c 2 ' lK if K ' K c 2 = f if K K Objective of : 4@ . dEKc K2c c 2 o Kl
where . do ' E K hiK : K E j n E K hiK ' K E j 2
5
2.1 Symmetric, linear equilibrium K ' @ n S Then,
hiK ' K E j ' f Equilibrium condition: @ K . Hence, . do ' E K hiK @ n S j K @ ' E K hi j S K @ ' E K S FOC: ln@ if @ 2 K ' (1) @ if @ Therefore, K ' 2 6
2.2 Any symmetric equilibrium K ' K E Hence,
. do ' E K hiK KE j ' E K hi K EK j ' E K K EK FOC (Y*YK ' f): EK _K K EK n E K ' f _K ' f n E K E 3 K E K3 E n K E ' _ dK E o ' _ K E ' 2*2 n SJ?r| K E ' *2 n SJ?r|* K Ef ' f, =: cons = 0. K E ' *2 7
3 Double Auction 1. Simultaneously, Seller names Rr and Buyer names RK. a. If RK Rr, then no trade b. if RK Rr, trade at price R '
Re nRv . 2
2. Valuations are private information: K, r iid w/ uniform on dfc o. 3. Payoffs: v if RK Rr K RenR 2 K ' f otherwise RenRv r if RK Rr 2 r ' f otherwise 4. The buyer’s problem: RK n RrEr G RK RrEr 4@ . K Re 2 5. The seller’s problem: Rr n RKEK r G RKEK Rr 4@ . Rv 2
8
An Equilibrium: f RK ' f f Rr '
if K f c otherwise if r f otherwise
vb v b /v s Trade Efficient
X
not to trade
vS
0 Inefficient lack of trade
9
Equilibrium with linear strategies: RK ' @K n SKK Rr ' @r n Srr
RK @r RK Rr Er ' @r n Srr +, r Sr Rr RK EK ' @K n SKK +, K
10
Rr @K SK
RK n RrEr G RK RrEr . dKo ' . K 2 s d ] e v fv RK n RrEr K _r ' 2 f s d ] e vk fv RK n @r n Srr l K _r ' 2 f ] sedv fv Sr RK @r RK n @r K ' r_r Sr 2 2 f 2 RK @r RK n @r Sr RK @r K ' Sr 2 e Sr RK @r RK n @r RK @r K ' Sr 2 e RK @r RK n @r ' K Sr e F.O.C. (4@ Re . dKo): ERK @r RK n @r K 'f Sr e eSr i.e., 2 RK ' K n @r (2) 11
Similarly, Rr n RKEK r G RKEK Rr . dro ' . 2 ] k l Rr n @K n SKK ' r _K sv de 2 k fe ] l SK Rr n @K Rr @K r n ' K_K s d v SK 2 2 f e e k l Rr n @K Rr @K r ' 2 $ # SK 2 Rr @K SK n e SK Rr n @K Rr @K SK Rr @K r n n ' SK 2 e e Rr n @K SK Rr @K r n ' SK e e
12
F.O.C. (4@ Rv . dro): Rr @K Rr n @K 'f r n n SK e e e SK
Rr n @K SK r n n ESK ERr @K ' fc e e e Rr @K SK @K n SK ' n r n ESK n @K ' r n 2 e e e 2 i.e.,
2 @K n SK Rr ' r n
@K ' @r* @r ' @K* n 2*b Hence, b@r ' @r n 2, @r ' *e @K ' *2. 2 RK ' K n 2 2 Rr ' r n e 13
(3)
(4) (5)
We have trade iff RK Rr iff
2 2 K n r n 2 e iff ' ' K r 2 e 2 2S e
3/4
ps
1/4 1/12
pb
14
Lectures 15-18 Dynamic Games with Incomplete Information 14.12 Game Theory Muhamet Yildiz
Road Map 1. 2. 3. 4.
Examples Sequential Rationality Perfect Bayesian Nash Equilibrium Economic Applications 1. Sequential Bargaining with incomplete information 2. Reputation
1
An Example Firm
W
Work
Hire Shirk
High 0.7 Do not hire
Nature
Hire
Low 0.3
(1, 2)
(0, 1)
(0, 0) W
Work Shirk
Do not hire
(1, 1)
(-1, 2)
(0, 0)
What is wrong with this equilibrium? Firm
W
Work
Hire Shirk
High .7 Do not hire
Nature
Low .3
(1, 2)
Hire
(0, 0) W
Work Shirk
Do not hire
(0, 1)
(1, 1)
(-1, 2)
(0, 0)
2
What is wrong with this equilibrium? X 1
(2,6)
T
B
2 L (0,1)
R (3,2)
L (-1,3)
R (1,5)
Beliefs X • Beliefs of an agent at a given 1 (2,6) information set is a probability distribution on T B the information set. • For each information set, we µ 1−µ must specify the beliefs of 2 the agent who moves at that L R L R information set. (0,1) (3,2) (-1,3) (1,5)
3
Sequential Rationality A player is said to be sequentially rational iff, at each information set he is to move, he maximizes his expected utility given his beliefs at the information set (and given that he is at the information set) – even if this information set is precluded by his own strategy.
An Example Firm
W
Work
Hire Shirk
High 0.7 Do not hire
Nature
Low 0.3
(1, 2)
Hire
(0, 0) W
Work Shirk
Do not hire
(0, 1)
(1, 1)
(-1, 2)
(0, 0)
4
Another example X 1
(2,6)
T
B
2 L
R
(0,1)
L
(3,2)
R
(-1,3)
(1,5)
Example 1 T
2 L (0,10)
B .9
.1
R (3,2)
L (-1,3)
R (1,5)
5
“Consistency” Definition: Given any (possibly mixed) strategy profile s, an information set is said to be on the path of play iff the information set is reached with positive probability if players stick to s. Definition: Given any strategy profile s and any information set I on the path of play of s, a player’s beliefs at I is said to be consistent with s iff the beliefs are derived using the Bayes’ rule and s.
Example 1 T
B
2 L (0,10)
R (3,2)
L (-1,3)
R (1,5)
6
Example 1 E
X
2 T
B
2 0 0
3 L
R 1 2 1
3 3 3
L
R 0 1 2
0 1 1
“Consistency” • Given s and an information set I, even if I is off the path of play, the beliefs must be derived using the Bayes’ rule and s “whenever possible,” e.g., if players tremble with very small probability so that I is on the path, the beliefs must be very close to the ones derived using the Bayes’ rule and s.
7
Example 1 E
X
2 T
B
2 0 0
3 L
R 1 2 1
3 3 3
L
R 0 1 2
0 1 1
Perfect Bayesian Nash Equilibrium A Perfect Bayesian Nash Equilibrium is a pair (s,b) of strategy profile and a set of beliefs such that 1. Strategy profile s is sequentially rational given beliefs b, and 2. Beliefs b are consistent with s. Nash Bayesian Nash
Subgame-perfect Perfect Bayesian
8
Example 1 E
X
2 T
B
2 0 0
3 L
R
L
R
3 3 3
1 2 1
0 1 2
0 1 1
Beer – Quiche
2 0
3 1
’ don
’t don
quiche
beer
{.1}
t
el du
1 0
du el
tw ts beer {.9}
quiche
e du
1 1
l
don
3 0
’t
0 0
du el
0 1
don ’
t
2 1
9
Example 2
1
1
(1,-5)
.9
.1
(4,4)
(3,3)
(5,2) 1
(-1,4)
2
(0,2)
1
(0,-5)
(-1,3)
Sequential Bargaining 1. 2. 3. 4.
1-period bargaining – 2 types 2-period bargaining – 2 types 1-period bargaining – continuum 2-period bargaining – continuum
10
Sequential bargaining 1-p • A seller S with valuation 0 • A buyer B with valuation v;
H S
– B knows v, S does not – v = 2 with probability π – = 1 with probability 1-π
p
• S sets a price p ≥ 0; • B either
Y
– buys, yielding (p,v-p) – or does not, yielding (0,0).
L
p 2-p
p N 0 0
N
Y
0 0
p 1-p
Solution 1. B buys iff v ≥ p; 1. If p ≤ 1, both types buy: 1 S gets p. 2. If 1 < p ≤ 2, only H-type buys: S gets πp. 3. If p > 2, no one buys.
1
2
2. S offers • 1 if π < ½, • 2 if π > ½.
11
Sequential bargaining 2-period • A seller S with valuation 0 • A buyer B with valuation v; – B knows v, S does not – v = 2 with probability π – = 1 with probability 1-π
1. At t = 0, S sets a price p0 ≥ 0; 2. B either – –
buys, yielding (p0,v-p0) or does not, then
3. At t = 1, S sets a price p0 ≥ 0; 4. B either – –
buys, yielding (δp0,δ(v-p0)) or does not, yielding (0,0)
Solution, 2-period 1. 2. 3. 4. 5. 6. 7.
Let µ = Pr(v = 2|history at t=1). At t = 1, buy iff v ≥ p; If µ > ½, p1 = 2 If µ < ½, p1 = 1. If µ = ½, mix between 1 and 2. B with v=1 buys at t=0 if p0 ≤ 1. If p0 > 1, µ = Pr(v = 2|p0,t=1) ≤ π.
12
Solution, cont. π 2−δ, then – µ = Pr(v = 2|p0 > 2−δ,t=1) = 0; – p1 = 1; – v = 2 should not buy at p0 > 2−δ.
• If v=2 is not buying at 2> p0 > 2−δ, then – µ = Pr(v = 2|p0 > 2−δ,t=1) = π > 1/2; – p1 = 2; – v = 2 should buy at 2 > p0 > 2−δ.
• No pure-strategy equilibrium.
13
Mixed-strategy equilibrium, π >1/2 1. For p0 > 2−δ, µ(p0) = ½; 2. β(p0) = 1- Pr(v=2 buys at p0) µ=
β ( p0 )π 1 1−π = ⇔ β ( p0 )π = 1 − π ⇔ β ( p0 ) = . π β ( p0 )π + (1 − π ) 2
3. v = 2 is indifferent towards buying at p0: 2- p0 = δγ(p0) Ù γ(p0) = (2- p0)/δ where γ(p0) = Pr(p1=1|p0).
Sequential bargaining, v in [0,1] • 1 period: – – – –
B buys at p iff v ≥ p; S gets U(p) = p Pr(v ≥ p); v in [0,a] => U(p) = p(a-p)/a; p = a/2.
14
Sequential bargaining, v in [0,1] • 2 periods: (p0,p1) – At t = 0, B buys at p0 iff v ≥ a(p0); – p1 = a(p0)/2; – Type a(p0) is indifferent: a(p0) – p0 = δ(a(p0) – p1 ) = δa(p0)/2 Ùa(p0) = p0/(1-δ/2) 2 • S gets p p 0 p0 + 0 1 − 1 δ / 2 − 2 −δ
• FOC:
1−
(1 − δ / 2 ) 2 p0 2 p0 + = 0 ⇒ p0 = 1−δ / 2 2 −δ 2(1 − 3δ / 4 ) 2
Reputation
15
Centipede Game 1
2
1
1
2
1
2
98 98
97 100
99 99
98 101
100 100
… 1 1
0 3
2 2
Centipede Game – with doubt
197
{.999} 1
5 2
2
4 1
µ5
3
2
2
1
=n
1 2
100 100
µ1
µ3
… 1 1 {.001}
96 99
0 3 1
2
98 98
97 100
99 99
98 101
2
1
2
1
2
0 99
-1 98
0 100
-1 99
0 101
0 100
… -1 1
0 3
16
Facts about the Centipede • Every information set of 2 is reached with positive probability. • 2 always goes across with positive probability. • If 2 strictly prefers to go across at n, then – she must strictly prefer to go across at n+1, – her posterior at n is her prior.
• For any n > 2, 1 goes across with positive probability. If 1 goes across w/p 1 at n, then 2’s posterior at n-1 is her prior.
If 2’s payoff at any n is x and 2 is mixing, then x = µn(x+1) + (1- µn)[(x-1)pn +(1-pn)(x+1)] = µn(x+1) + (1- µn)[ (x+1) -2pn] = x+1 - 2pn(1- µn) Ù(1- µn) pn= 1/2 µn −1 =
µn µn = = 2 µn µn + (1 − µn )(1 − pn ) µn + (1 − µn ) − pn (1 − µn ) µn =
µn −1 2
17
0.5 0.45 0.4 0.35 0.3 0.25
Go Across
0.2 0.15 0.1 0.05 0
Mix 0
20
40
60
80
100
18
Lecture 19-21 14.12 Game Theory Muhamet Yildiz
Road Map • • • •
Market for Lemons (Adverse Selection) Insurance Market (Screening) Signaling – theory Job Market Signaling
1
Example for Adverse selection • A buyer and a seller, who owns an object. • The value of the object is – v for the seller, – v+b for the buyer.
• v is uniformly distributed on [0,1]. Seller knows v. b > 0 is a known constant. • Buyers offers a price p; seller decides on whether to sell.
Solution • Seller sells at p iff p ≥ v; • Buyer’s payoff: p 1 p U ( p ) = ∫ (v + b − p )dv = p 2 + p (b − p ) = p b − 2 2 0 • Buyer offers p = 2b.
2
Example for Adverse selection • A buyer and a seller, who owns an object. • The value of the object is – v for the seller, – 3v/2 for the buyer.
• v is uniformly distributed on [0,1]. Seller knows v. • Buyers offers a price p; seller decides on whether to sell.
Solution • Seller sells at p iff p ≥ v; • Buyer’s payoff: p 3 3 p U ( p ) = ∫ v − p dv = p − p = − p2 / 4 2 2 2 0 • Buyer offers p = 0.
3
Market for Lemons • Two types of cars: Lemons and Peaches. – A Peach is worth $2500 to seller, $3000 to buyer; – A lemon is worth $1000 to seller, $2000 to buyer;
• Each seller knows whether his car is a Peach or Lemon; buyers cannot tell. • There are 200 Lemons and 100 Peaches. • Equilibrium: a market clearing price p.
Demand/Supply – Market for Lemons Supply
P
2000
1000
200
300
Q
4
Quiz • The students are grouped in pairs; • In each pair, one is assigned to be seller, the other one is buyer. The buyer is given a valuation v, an integer between 20 and 40 with uniform distribution. – Buyer offers a price p; – seller accepts or rejects; – If seller accepts, seller gets p, buyer gets 20+v-p. – If seller rejects, w/probability 0.1 each gets 20; w/p 0.95 we proceed to next date, and – Seller offers a price p; – If buyer accepts, seller gets p, buyer gets 20+v-p; otherwise each gets 20.
Insurance with adverse selection • Two states: s1, s2. • A risk averse agent with – risky endowment (Y1,Y2) where Y1 > Y2, and – Utility function u.
• Two types: – High risk: U(y1,y2) = (1−πH)u(y1) + πHu(y2) – Low risk: U(y1,y2) = (1−πL)u(y1) + πLu(y2), where πH > πL.
• A risk neutral insurance company offers a menu ((y1H,y2H), (y1L,y2L)) of insurance policies, and the risk averse agent either chooses one of the policies in the menu or rejects the offer. • Agent knows his type, the company does not.
5
High-risk Agent
Y2
Y1
Low-risk Agent
Y2
Y1
6
High- and Low-risk Agent
Y2
Y1
Optimal menu
Y2
Y1
7
Signaling Games
Beer – Quiche
2 0
3 1
’ don
’t don
quiche
beer
{.1}
t
el du
1 0
du el
tw ts beer {.9}
quiche
e du
1 1
l
don
3 0
’t
0 0
du el
0 1
don ’
t
2 1
8
Signaling Game -- Definition • Two Players: (S)ender, (R)eceiver 1. Nature selects a type ti from T = {t1,…,tI} with probability p(ti); 2. Sender observes ti, and then chooses a message mj from M = {m1,…,mI}; 3. Receiver observes mj (but not ti), and then chooses an action ak from A = {a1,…,aK}; 4. Payoffs are US(ti,mj,ak) and US(ti,mj,ak).
Beer – Quiche
2 0
3 1
’ don
’t don
quiche
beer
{.1}
t
el du
1 0
du el
tw ts beer {.9}
quiche
e du
1 1
l
don
3 0
’t
0 0
du el
0 1
don ’
t
2 1
9
Types of Equilibria • A pooling equilibrium is an equilibrium in which all types of sender send the same message. • A separating equilibrium is an equilibrium in which all types of sender send different messages. • A partially separating/pooling equilibrium is an equilibrium in which some types of sender send the same message, while some others sends some other messages.
A Pooling equilibrium
2 0
3 1
’ don
’t don
quiche
beer
{.1}
t
el du
1 0
du el
tw ts beer {.9}
quiche
e du
1 1
l
don
3 0
’t
0 0
du el
0 1
don ’
t
2 1
10
A Separating equilibrium
0.5 0
3 1
’ don
quiche
beer
{.1}
t
tw
el du
1 0
du el
ts beer {.9}
quiche
e du
1 1
l
don
don ’
’t don
3 0
’t
0 0
du el
0 1
2 1
t
A Mixed equilibrium du el
’ don
2 0
3 1
{.1}
t
el du
1 0
’t don
quiche
beer tw ts beer {.9}
quiche
e du
1 1
l
don
3 0
’t
0 0
du el
0 1
don ’
t
2 1
11
Job Market Signaling
Model • A worker – with ability t = H or t = L (his private information) – Pr(t = H) = q, – obtains an observable education level e, – incurring cost c(t,e) where c(H,e) < c(L,e), and – finds a job with wage w(e), where he – produces y(t,e).
• Firms compete for the worker: in equilibrium, w(e) = µ(H|e)y(H,e) + (1– µ(H|e))y(L,e).
12
Equilibrium (eH, eL, w(e), µ(H|e)) where • et = argmaxe w(e) – c(t,e) for each t; • w(e) = µ(H|e)y(H,e) + (1– µ(H|e))y(L,e); qPr(eH = e)
• µ(H|e) =
qPr(eH = e) + (1-q)Pr(eL = e) whenever well-defined.
If t were common knowledge w
w (e
e*(t)
)
) (t,e y =
e
13
No need to imitate e) , y (H
w
,e) L ( y
e*(L)
e*(H)
No need to imitate
e
w(.)
, e) H ( y
w
,e) L ( y
e*(L)
e*(H)
e
14
want to imitate w
e) , y (H
,e) L ( y
e
e*(L) e*(H)
A pooling equilibrium
, e) H y(
w qy
e)+ (H ,
(1-
(L q) y
,e)
,e) L ( y
e*(L) e*(H)
e
15
A separating equilibrium w
e) , y (H
,e) L ( y
eL=e*(L)
e*(H)
e
eH
An intuitive separating equilibrium w
, e) H ( y
,e) L ( y
eL=e*(L)
eH
e
16
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Krphzrun 4
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4
14.12 Solutions for Homework 2 Kenichi Amaya1 October 5, 2001 Question 1 (a) L 1,1 0,1
T B
R 1,0 0,10000
Fact: The set of rationalizable strategies is same as the set of strategies which survives iterated elimination of strictly dominated strategies. Step 1: For player 1, B is strictly dominated by T. Now we have T
L 1,1
R 1,0
Step 2: For player 2, R is strictly dominated by L. Now we have L 1,1
T
Therefore, the rationalizable strategies are T for player 1 and L for player 2. (b) There was a typo in the problem. If player 2 intends to play L, L is played with probability 1 − , and R is played with probability . Of course, if you interpreted the problem differently, you will get credit as long as you are solving the problem consistently with your interpretation. We can write a new game which represents this situation as following: We look for the players’ expected payoffs as functions of the intended play. If player 1 intends to play T and player intends to play L, what are the expected payoffs? (T,L) is realized with probability (1 − )2 , (T,R) and (B,L) are realized each with probability (1 − ), and (B,R) is realized with probability 2 . Therefore, the expected payoff of player 1 is (1 − )2 · 1 + (1 − ) · 1 + (1 − ) · 0 + 2 · 0 = 1 − = .999. In the same way, player 2’s expected payoff is (1 − )2 · 1 + (1 − ) · 0 + (1 − ) · 1 + 2 · 10000 = (1 − ) + 2 · 10000 = 1.009 Therefore we must have 1 E52-303,
[email protected], 253-3591
1
L .999, 1.009
T B
R
In the same way, we can calculate expected payoffs for the remaining of the payoff matrix and we have T B
L .999, 1.009 .001, 10.999
R .999, 9.991 .001, 9980.01
For player 1, B is strictly dominated by T, and for player 2, L is strictly dominated by R. Now we have T
R .999, 9.991
Therefore, the rationalizable strategies are T for player 1 and R for player 2. (c) For player 1, the possibility that opponent trembles doesn’t affect her optimal choice because T gives higher payoff than B no matter what the opponent is doing. For player 2, in the case of (a), the large payoff differnce between (B,L) and (B,R) didn’t matter because he was sure B is never played. However, when we introduce trembles, this large payoff difference matters with positive probability even though player 2 knows that player 1 never intends to play B. In this situation, the payoff difference between (T,L) and (T,R) is so small that it is outweighed by the payoff differnce between (B,L) and (B,R). Question 2 Fact All the Nash equilibrium strategies are rationalizable, and therefore survives iterated elimination of strictly dominated strategies. When we look for Nash equilibria, we can first eliminate strictly dominated strategies and then find Nash equilibria of the game left. A B C
L 3,1 0,0 1,1
M 0,0 1,3 0,1
R 1,0 1,1 0,10
Step 1 For player 1, C is strictly dominated by 12 A + 12 B. Now we have A B
L 3,1 0,0
M 0,0 1,3 2
R 1,0 1,1
Step 2 For player 2, R is stricly dominated by 12 L + 12 M . Now we have A B
L 3,1 0,0
M 0,0 1,3
Step 3 In the game left, (A,L) and (B,M) are pure strategy Nash equilibria, because players are taking best response to each other. Step 4 We want to find the mixed strategy equilibrium. Let (pA + (1 − p)B, qL + (1 − q)M ) be the equilibrium. First, player 1 must be indifferent between playing A and B, which implies 3q = 1(1 − q), where the LHS is the expected payoff of playing A and the RHS is that of B, when player 2 is playing the equilibrium strategy. This implies q=
1 . 4
Next, player 2 must be indifferent between playing L and M, implying 1p = (1 − p)3 3 p= . 4 To conclude, the Nash equilibria are; (A,L), (B,M), and ( 34 A+ 14 B, 14 L+ 34 M ). Question 3 This question is hard! Notice that in the Cournot model with fixed cost, there can be equilibria where not every firm is producing. For example, if there are two firms, there can be an equilibrium where one firm is producing the monopoly quiantity and the other firm is producing nothing. (Of course, whether this is actually a Nash equilibrium or not depends on the cost and demand structure.) To see why this can be an equilibium, consider each firm’s best response. If the opponent firm is producing the monopoly quantity, the revenue from selling some amount can not cover the fixed cost, if the fixed cost is large enough, and thus producing nothing is the best response. On the other hand, if the opponent is producing nothing, you can get a revenue equal to the monopoly profit, which is higher than the fixed cost. It is hard to characterize all the equilibria, so we limit attention to pure strategy equilibria which are symmetric in the sense that all the producing firms are choosing the same quantity. (Keep in mind that there might be other equilibria! Actually, it is not very hard to show that all pure strategy equilibria are symmetric. This would be a good exercise.) 3
First, consider the equilibrium where all firms are producing q ∗ . What we need is that q ∗ is a best response to all other firms choosing q ∗ . The profit of firm i when it is producing qi and all other firms are producing q ∗ is πi (qi , q ∗ ) = (max{1 − (qi + (1 − n)q ∗ ), 0} − c)qi − F.
(1)
In the equilibrium all firms must be getting positive revenue (because there is fixed cost), so it must be 1 − (qi + (n − 1)q ∗ ) > 0. Therefore πi (qi , q ∗ ) = (1 − (qi + (n − 1)q ∗ ) − c)qi − F.
(2)
Taking first order condition, qi =
1 − (n − 1)q ∗ − c . 2
(3)
In the symmetric equilibium, we must have qi = q ∗ , thus q∗ =
1−c . n+1
(4)
For this quantity to be actually construct equilibrium, each firm must be getting nonnegative profit. Plugging this into (2), πi =
1 − c 2 n+1
− F.
(5)
1−c Therefore there exists an equilibrium where all firms are producing q ∗ = n+1 if 1−c 2 and only if ( n+1 ) ≥ F and c ≤ 1. Next consider an equilibrium where only k < n firms are producing q ∗ . We need to check incentives of firms which are producing and incentives of firms which are not producing. For a firm which is producing, qi = q ∗ must be a best response to k − 1 of other firms producing q ∗ and the rest producing zero. The firm’s profit when producing qi is
πi (qi , q ∗ ) = (max{1 − (qi + (1 − k)q ∗ ), 0} − c)qi − F.
(6)
Notice this is same as equation (1), n being replaced by k. Therefore the same argument as above holds. 1−c q∗ = , (7) k+1 1−c 2 For this to be an equilibrium, we need ( k+1 ) ≥ F and c ≤ 1. However, this is not a sufficient condition. It must also be true that for nonproducing firms, not producing is actually the best response. From the point
4
of view of a non-producing firm i, k other firms are producing q ∗ . By the same argument above, if this firm does choose to produce, the optimal amount is qi∗ = and the payoff is
1 − kq ∗ − c , 2
(8)
1 − kq ∗ − c 2
− F. (9) 2 ∗ If this profit is positive, the firm wants to produce qi rather than producing 2 ∗ nothing. Therefore, we need 1−kq2 −c ≤ F , i.e., πi =
1 − c 2 ≤ F. 2(k + 1)
(10)
Finally, consider an equilibrium where no firm produces. This is the case where c ≥ 1 or where even the monopoly profit is less than the fixed cost. That is, 1 − c 2 ≤ F. (11) 2 Question 4 Each player i = 1, · · · , n chooses his price of meal pi . The payoff of player i is ui = =
√
√
pi −
pi pi − − n
Pn
j=1
P n
j6=i
n
pj
pj
.
Taking first order condition, 1 1 √ = 2 pi n pi =
n2 . 4
This gives the best response, but actually this optimum does not depend on what other players are doing, which means that it is the dominant strategy. To conclude, the Nash equilibrium is pi =
n2 ∀i. 4
Notice if n = 1 then p = 12 . This is the efficient choice, because there is no externality. If n → ∞, then pi → ∞. This is because every marginal increase in the price of the meal will splitted by all the people and the marginal increase of the payment of the player herself converges to 0. 5
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5
14.12 Solutions for Homework 3 Kenichi Amaya1 November 2, 2001 Question 1 We solve this problem by backward induction. First, we solve for the quantity choice (stage 2) given the entrance decision, and then we solve for the entrance decision (stage 1). Stage 2 Suppose m firms has entered in stage 1. Assuming 1 − Q ≥ 0 in equilibrium (we check this is actually true later), each firm i maximizes πi = qi (1 − Q) X = qi (1 − qj − qi ). j6=i
Notice, since the fixed cost F was payed in stage 1 already, i.e., it is already sunk, the firm doesn’t care about it any more. Taking first order condition, P 1 − j6=i qj . qi = 2 Assuming the equilibrium is symmetric (qi is same for all i), qi =
1 − (m − 1)qi , 2
implying 1 . m+1 (It is not hard to prove the equilibrium is actually symmetric.) Now we can confirm that our assumption 1 − Q ≥ 0 is actually satisfied because m < 1. Q = mqi = m+1 qi =
The equilibrium payoff of each firm is π=
1 . (m + 1)2
Stage 1 Let π(m) be the profit (gained in stage 2) of each firm when m firms are entering. As we solved above, π(m) =
1 . (m + 1)2
There are three possible classes of equilibria: 1 E52-303,
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1
1. All n firms enter. 2. No firm enters 3. k firms enter, where 0 < k < n. The first class of equilibrium exists if it is better to enter and gain π(n)−F than to exit and gain 0. That is, π(n) − F ≥ 0, 1 ≥ F. (n + 1)2 The second class of equilibrium exists if it is better to exit and get 0 than to enter to be the only entering firm and gain π(1) − F . That is, π(1) − F < 0, 1 ≤ F. 4 To see when the third class of equilibrium exists, we need to check the incentives of the firms which are entering and the firms which are not entering. Let’s consider the incentive of an entering firm first. It gets π(k) − F if it enters, and gets 0 if it exits. Therefore, there is no incentive to deviate if π(k) − F ≥ 0, 1 ≥ F. (k + 1)2 Next consider the incentive of an exiting firm. It gets 0 if it exits. If it enters, it gets π(k + 1) − F because k other firms are also entering. Therefore, there is no incentive to deviate if π(k + 1) − F ≤ 0, 1 ≤ F. (k + 2)2 Notice, the argument above says nothing about which k firms enter. If the condition above is satisfied, any selection of k firms constitutes an equilibrium. To summarize, let’s formally describe the equilibrium strategies. Remember, we need to describe all the contingent plans for stage 2, even for the information sets which are not achieved on the equilibrium path. The equilibrium strategies are as the following. 2
1.
• All firms enter in stage 1. • If m firms enter in stage 1, each entering firm produces 2.
1 m+1
in stage
1 The equilibrium payoff is (n+1) 2 −F for every firm. This equilibrium exists 1 if and only if (n+1)2 ≥ F .
2.
• No firm enter in stage 1. • If m firms enter in stage 1, each entering firm produces 2.
1 m+1
in stage
The equilibrium payoff is 0 for every firm. This equilibrium exists if and only if 14 ≤ F . 3.
• k firms enter in stage 1, where 0 < k < n. • If m firms enter in stage 1, each entering firm produces 2.
1 m+1
in stage
1 The equilibrium payoff is (k+1) 2 − F for every entering firm, and 0 for 1 every exiting firm. This equilibrium exists if and only if (k+1) 2 ≥ F and 1 ≤ F . (k+2)2
Question 2 (a) 1 A X A N A U2 A (2,2) @ x @ n @ R1 @ (2,2) @ L @ R @ R2 @ A A l A r l A r A A U A AU (3,3) (0,0) (0,0) (1,1)
We can solve for subgame perfect equilibria by backward induction. This game has three subgames: the whole game itself, a subgame starting from player 3
2’s first decision node, and a subgame starting from player 1’s second decision node. First we look at the subgame starting from player 1’s second decision node. This subgame is equivalent to the following nomal form game: L R
l 3,3 0,0
r 0,0 1,1
Obviously, pure strategy equilibria are (L,l) and (R,r). If (L,l) is played in this subgame, player 2 chooses n at his first decision node, because he gets payoff of 2 by playing x and 3 by playing n. Given this, player 1 chooses N at her first decision node because she gets payoff of 2 by playing X and 3 by playing N. Therefore, player 1’s strategy (N,L) and player 2’s strategy (n,l) constitute a subgame perfect Nash equilibrium. If (R,r) is played in the subgame starting from player 1’s second decision node, player 2 chooses x at his first decision node, because he gets payoff of 2 by playing x and 1 by playing n. Given this, player 1 is indifferent between X and N at her first decision node because she gets payoff of 2. Therefore, player 1’s strategy (X,R) and player 2’s strategy (x,r) constitute a subgame perfect Nash equilibrium, and player 1’s strategy (N,r) and player 2’s strategy (x,r) constitute another subgame perfect Nash equilibrium. To conclude, there are three subgame perfect Nash equilibria: 1. Player 1: (N,L), Player 2: (n,l) 2. Player 1: (X,R), Player 2: (x,r) 3. Player 1: (N,R), Player 2: (x,r) (b) Let’s apply the forward induction argument here. Suppose player 1’s second decision node is reached. This means that player 2 has chosen N at his first decision node, i.e., his strategy is either (n,l) or (n,r) (or any mixture of them). Should player 1 think that player 2 is choosing (n,r)? Player 2 can get payoff of 0 or 1 by playing (n,r), whereas he can get 2 for sure by playing (x,l) or (x,r). In other words, (n,r) is a strictly dominated strategy. If player 1 knows player 2 is rational, she must know that there is no reason player 2 play (n,r). Therefore, given player 2 has played N at his first decision node, player 1 must conclude player 2 is choosing (X,l), and choose the best response to it, namely L. To conclude, in any subgame perfect equilibrium which is consistent with common knowlegde of rationality, player 1’s choice at his second decision node must be L. Therefore, only the first equilibrium (N,L), (n,l) is consistent.
4
Question 3 (a) The monopoly price is pm = 21 . Construct the following “trigger strategy”. • Start with playing pm , and play pm if everybody has played pm all the time. • Play 0 if anybody has played something different from pm at least once. By the single deviation principle, it is sufficient to check incentives to deviate only once, in order to check when this is actually a subgabe perfect Nash equilibrium. Firstly, check the incentive when nobody has deviated from pm before. If a firm plays pm , as suggested by the strategy, then everyone will be playing pm all the time in future, and a firm’s profit per period is pm (1 − pm ) 1 = , 2 8 and the present discounted sum of the profit sequence from today is 1 1 1 1 1 + δ + δ2 + · · · = . 8 8 8 1−δ8 If the firm is to deviate, the best it can do today is to set a price slightly less than pm and get the whole monopoly profit pm (1 − pm ) =
1 . 4
If the firm does deviate, everyone will be playing 0 from the next period, yielding per period payoff of zero. Therefore, the present discounted sum of the profit sequence from today is 1 1 + δ0 + δ 2 0 + · · · = . 4 4 The firm has no incentive to deviate if and only if 1 1 1 ≥ 1−δ8 4 1 δ≥ . 2 Secondly, check the incentive when somebody has ever deviated from pm before. In this case, a firm’s action today doesn’t affect it’s future payoff because everyone will be playing 0 in future no matter what happens today. Therefore it is sufficient to check if the firm can increase the present period profit by deviating. If it plays p = 0, as suggested by the strategy, it gets a payoff of 0. 5
If it deviates and plays p > 0, it gets a payoff of 0 again because all consumers buy from the other firm which is choosing p = 0. Therefore it can’t benefit from deviating. To conclude, the trigger strategy constitutes a subgame perfect Nash equilibrium if δ ≥ 21 . (b) When there are n firms, we can use the trigger strategy constructed in part (a) again. As we did in part (a), we check incentives to deviate before and after any deviation. Firstly, check the incentive when nobody has deviated from pm before. If a firm plays pm , as suggested by the strategy, then everyone will be playing pm all the time in future, and a firm’s profit per period is 1 pm (1 − pm ) = , n 4n and the present discounted sum of the profit sequence from today is 1 1 1 1 1 +δ + δ2 + ··· = . 4n 4n 4n 1 − δ 4n If the firm is to deviate, the best it can do today is to set a price slightly less than pm and get the whole monopoly profit pm (1 − pm ) =
1 . 4
If the firm does deviate, everyone will be playing 0 from the next period, yielding per period payoff of zero. Therefore, the present discounted sum of the profit sequence from today is 1 1 + δ0 + δ 2 0 + · · · = . 4 4 The firm has no incentive to deviate if and only if 1 1 1 ≥ 1 − δ 4n 4 n−1 δ≥ . n To check the incentive when somebody has ever deviated from pm before, the same argument as in part (a) holds. To conclude, the trigger strategy constitutes a subgame perfect Nash equilibrium if δ ≥ n−1 n . Question 4 (a) 6
To begin with, let’s find the subgame perfect Nash equilibrium of the stage game. We do this by backward induction. The long run firm, observing the short run firm’s quantity xt , chooses its quantity yt to maximize its profit πtL = yt (1 − (xt + yt )). Solving first order condition, the optimum is yt∗ (xt ) =
1 − xt . 2
The short run firm chooses its quantity xt to maximize its profit πtS = xt (1 − (xt + yt )), knowing that if it choose xt , the long run firm reacts with yt∗ (xt ) =
1 − xt . 2
Therefore, the short run firm’s objective function can be rewritten as a function of xt ; 1 − xt πtS = xt (1 − (xt + )). 2 Solving first order condition, the optimum is x∗t =
1 . 2
Therefore, the subgame perfect Nash equilibrium of the stage game is x∗t =
1 1 − xt , yt∗ (xt ) = . 2 2
Now let’s solve for the subgame perfect Nash equilibrium of the finitely repeated game. Since it is finite, we can use backward induction. At the last period, t = T , the players don’t care about future and they concern only about the payoff of that period. Therefore they must play the subgame perfect Nash equilibrium of the stage game, regardless of what happened in the past. At time t = T − 1, the players know that the actions today doesn’t affect tomorrow’s outcome, so they concern only about the payoff of that period. Therefore they must play the subgame perfect Nash equilibrium of the stage game, regardless of what happened in the past. We can repeat the same argument until we reach the very first period. Therefore, the subgame perfect Nash equilibrium of the repeated game is x∗t =
1 1 − xt , yt∗ (xt ) = for all t, 2 2
regardless of the history. 7
(b) Construct the following trigger strategy. Long rum firm
• Start with playing the following strategy: 1/2 if xt ≤ 1/2 (∗) yt (xt ) = 1 − xt if xt ≥ 1/2
Keep playing this strategy as long it has not deviated from it. • Play yt∗ (xt ) =
1−xt 2
if it has deviated from (*) at least once.
Short run firms • Play xt = 1/4 if the long run firm has not deviated from (*) before. • Play xt =
1 2
if the long run firm has deviated from (*) at least once.
To see this is actually a subgame perfect Nash equilibrium, let’s check incentives to deviate. Firstly, consider the long run firm’s incentive when it has never deviated from (*) before. Case 1: xt ≤ 1/2 If it follows the strategy and choose yt = 1/2, the present period profit of the long run firm is 1 1 (1 − ( + xt )). 2 2 Starting from the next period, the outcome will be xt = 1/4 and yt = 1/2 every period, and the long run firm’s per period profit is 1/8 and therefore the present discounted value of the profit sequence is 1 1 δ (1 − ( + xt )) + . 2 2 8(1 − δ) If it is to deviate, the best it can do today is to play yt =
1 − xt 2
and get payoff of (1 − xt )2 . 4 However, starting from next period, the outcome will be xt = 1/2 and yt = 1/4 and the long run firm’s per period profit is 1/16. Therefore the present discounted value of the profit sequence is (1 − xt )2 δ + . 4 16(1 − δ) 8
If δ = 0.99, 1 1 δ (1 − xt )2 δ (1 − ( + xt )) + > + , 2 2 8(1 − δ) 4 16(1 − δ) (check it!), and therefore it is better to follow the equilibrium strategy than to deviate. Case 2: xt ≥ 1/2 If it follows the strategy, pt = 0 and today’s payoff is 0, and the outcome will be xt = 1/4 and yt = 1/2 every period, starting the next period. The long run firm’s per period profit is 1/8 and therefore the present discounted value of the profit sequence is δ . 8(1 − δ) 1−xt and get 2 (1−xt )2 . However, starting from next period, the outcome will be 4 and yt = 1/4 and the long run firm’s per period profit is 1/16.
If it is to deviate, the best it can do today is to play yt =
payoff of xt = 1/2 Therefore the present discounted value of the profit sequence is δ (1 − xt )2 + . 4 16(1 − δ) This value is the largest when xt = 1/2 and is equal to 1 δ + . 16 16(1 − δ)
If δ = 0.99, it is better to follow the equilibrium strategy than to deviate (check it!). Secondly, consider the long run firm’s incentive when it has deviated from (*) before. According to the strategy strategy profile, future outcomes don’t depend on today’s behavior. Therefore the long run firm cares only about its payoff today. Actually, by following the strategy, it is taking best response to the short run firm. Finally, consider the short run firm’s incentives. Since they never care about future payoff, it must be playing a best response to the long run firm’s strategy, which is actually true. (Check it!) (c) In the equilibrium we saw in part (b), the per period profit on the equilibrium path were 1/8 for the long run firm and 1/16 for the short run firms.
9
If there is a subgame perfect Nash equilibrium where xt = yt − 1/4 on th eequilibrium path, then the per period profit on the equilibrium path are 1/8 for the long run firm and 1/8 for the short run firms. Construct the following trigger strategy, which is different from (b) only in (*) where xt = 1/4: Long rum firm
• Start with playing the if 1/4 1/2 if (∗) yt (xt ) = 1 − xt if
following strategy: xt = 1/4 xt ≤ 1/2 and xt 6= 1/4 xt ≥ 1/2
Keep playing this strategy as long it has not deviated from it. • Play yt∗ (xt ) =
1−xt 2
if it has deviated from (*) at least once.
Short run firms • Play xt = 1/4 if the long run firm has not deviated from (*) before. • Play xt =
1 2
if the long run firm has deviated from (*) at least once.
The incentive of the long run firm when it is supposed to play (*) and xt = 1/4 is satisfied because it gets current payoff of 1/8, which is same as part (b), and what it can get by deviating is same as in part (b). The incentive problem of the long run firm is same as in (b). The incentives of the short run firms are also satisfied because the payoff from following the strategy is larger than part (b) and payoff when deviating is same as in (b).
10
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4
14.12 Solutions for Homework 4 Kenichi Amaya1 November 9, 2001 Question 1 (a) Action spaces A1 = A2 = {B, S}. Type spaces T1 = {α}, T2 = {β1 , β2 }. Since player 1 has no private information, we can model it so that her type can take only one value. Player 2 knows that the game above is played when his type is β1 and he knows that the game below is played when his type is β2 . Belief Player i’s belief µi (tj |ti ) is the probability that player j’s type is tj conditional on that player i’s type is ti . In this model, since it is assumed that the types are independent, 1 , 2 µ2 (α|β1 ) = µ2 (α|β2 ) = 1.
µ1 (β1 |α) = µ1 (β2 |α) =
vNM utility function ui (a1 , a2 , t1 , t2 ) is the vNM utility when player 1’s action is a1 , player 2’s action is a2 , player 1’s type is t1 and player 2’s type is t2 . u1 (B, B, α, β1 ) = 2, u2 (B, B, α, β1 ) = 1, u1 (B, S, α, β1 ) = 0, u2 (B, S, α, β1 ) = 0, u1 (S, B, α, β1 ) = 0, u2 (S, B, α, β1 ) = 0, u1 (S, S, α, β1 ) = 1, u2 (S, S, α, β1 ) = 2, u1 (B, B, α, β2 ) = 2, u2 (B, B, α, β2 ) = 0, u1 (B, S, α, β2 ) = 0, u2 (B, S, α, β2 ) = 2, u1 (S, B, α, β2 ) = 0, u2 (S, B, α, β2 ) = 1, u1 (S, S, α, β2 ) = 1, u2 (S, S, α, β2 ) = 0. (b) First consider player 1’s incentive. Since she doesn’t know the game which is being played, she maximizes her expected payoff. If she plays B, with probability 1/2 the top game is being played and player 2 chooses B and thus she gets a payoff of 2, and with probability 1/2 the bottom game is being played and player 2 chooses S and thus she gets a payoff of 0. 1 E52-303,
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1
Therefore her expected payoff is 1. If she plays S, with probability 1/2 the top game is being played and player 2 chooses B and thus she gets a payoff of 0, and with probability 1/2 the bottom game is being played and player 2 chooses S and thus she gets a payoff of 1. Therefore her expected payoff is 1/2. Therefore B is actually player 1’s best response against player 2’s strategy. Next consider player 2’s incentive. When he knows that the top game is being played, B is the best response given player 1 is choosing B. When he knows that the bottom game is being played, S is the best response given player 1 is choosing B. Therefore choosing B when the top game is being played and choosing S when the bottom game is being played is actually player 2’s best response against player 1’s strategy. Since both players are taking a best response to each other, the strategy profile constitutes a Bayesian Nash equilibrium. Question 2 Each player’s action is the choice of price. A price can take any nonnegative real number. Therefore the action space is R+ for both players. Player i’s type is her private information. In this model, bi is player i’s type. bi is either bH or bL . Therefore, the type space of each player is {bH , bL }. Player i’s belief µi (bj |bi ) is the probability that player j’s type is bj conditional on that player i’s type is bi . In this model, since it is assumed that the types are independent, θ if bj = bH µi (bj |bi ) = . 1 − θ if bj = bL (von Neuman Morgenstern) utility in this model is the profit of each player (assuming the firms are risk neutral) as a function of the actions and types of both players: ui (pi , pj , bi , bj ) = pi (a − pi − bi pj ). Player i’s strategy specifies what action to take for any realization of her type. In this model, it is a two dimensional vector (pi (bH ), pi (bL )), where pi (bH ) is the price when its type is bH and pi (bL ) is the price when its type is bL . The 2 for each i. strategy space is R+ A strategy profile {(p∗1 (bH ), p∗1 (bL )), (p∗2 (bH ), p∗2 (bL ))} constitutes a Bayesian Nash equilibrium if each p∗i (bi ) is a best response, i.e., a maximizer of player i’s expected payoff, conditional on that her type is bi and the opponent is choosing strategy (p∗j (bH ), p∗j (bL )). That is: p∗1 (bH ) = argmaxp1 θp1 (a − p1 − bH p∗2 (bH )) + (1 − θ)p1 (a − p1 − bH p∗2 (bL )), p∗1 (bL ) = argmaxp1 θp1 (a − p1 − bL p∗2 (bH )) + (1 − θ)p1 (a − p1 − bL p∗2 (bL )), p∗2 (bH ) = argmaxp2 θp2 (a − p2 − bH p∗1 (bH )) + (1 − θ)p2 (a − p2 − bH p∗1 (bL )), p∗2 (bL ) = argmaxp2 θp2 (a − p2 − bL p∗1 (bH )) + (1 − θ)p2 (a − p2 − bL p∗1 (bL )). 2
Taking first order conditions, a − bH (θp∗2 (bH ) + (1 − θ)p∗2 (bL )) , 2 a − bL (θp∗2 (bH ) + (1 − θ)p∗2 (bL )) p∗1 (bL ) = , 2 a − bH (θp∗1 (bH ) + (1 − θ)p∗1 (bL )) p∗2 (bH ) = , 2 ∗ ∗ a − bL (θp1 (bH ) + (1 − θ)p1 (bL )) p∗2 (bL ) = . 2 p∗1 (bH ) =
Since the game is symmetric, let’s look for a symmetric equilibrium where p∗1 (bH ) = p∗2 (bH ) = p∗H and p∗1 (bL ) = p∗2 (bL ) = p∗L . Then the conditions are reduced to a − bH (θp∗H + (1 − θ)p∗L ) , 2 a − bL (θp∗H + (1 − θ)p∗L ) . p∗L = 2
p∗H =
Solving these, we get a bH (1 − ), 2 2 + θbH + (1 − θ)bL a bL ). p∗L = (1 − 2 2 + θbH + (1 − θ)bL
p∗H =
Question 3 Consider the following incomplete information game. H T
H 1 + α, −1 + β −1, 1 + β
T −1 + α, 1 1, −1
Here, α is player 1’s type, β is player 2’s type, α and β are independent draw from uniform distribution in the interval [0, 1], and > 0 is a constant. The interpretation is like this: Player 1 (2) can get an extra payoff of α (β) by playing head, in addition to the payoff generated from the original game. This extra payoff is her private information. Notice, as we converge to 0, the game converges to the original game. From the construction, player 1 (2) is more tempted to play H if she has a higher value of α (β). Observing this, try to find the pure strategy Bayesian Nash equilibrium of the following form. • Player 1 chooses H if α ≥ α ¯ and chooses L if α ≤ α ¯ , where α ¯ ∈ [0, 1] is a constant. 3
¯ where β¯ ∈ [0, 1] is a • Player 2 chooses H if β ≥ β¯ and chooses L if β ≤ β, constant. This strategy profile is actually a Bayesian Nash equilibrium if player 1 with cutoff type α ¯ and player 2 with cutoff type β¯ are indifferent between choosing H and T. ¯ and According to the strategy, the probability that player 2 plays H is 1 − β, ¯ the probability that player 2 plays L is β. Therefore, if player 1 of type α ¯ plays H, her expected payoff is ¯ + (−1)β¯ + ¯ 1(1 − β) α, and if she plays T, she gets ¯ + 1β. ¯ (−1)(1 − β) Therefore, we need ¯ + (−1)β¯ + ¯ ¯ + 1β. ¯ 1(1 − β) α = (−1)(1 − β)
(1)
¯ According to the strategy, We do the same thing for player 2 with type β. the probability that player 1 plays H is 1 − α ¯ , and the probability that player 1 plays L is α ¯ . Therefore, if player 2 of type β¯ plays H, his expected payoff is ¯ (−1)(1 − α ¯ ) + 1α ¯ + β, and if she plays T, she gets 1(1 − α ¯ ) + (−1)¯ α. Therefore, we need (−1)(1 − α ¯ ) + 1¯ α + β¯ = 1(1 − α ¯ ) + (−1)¯ α.
(2)
Solving (1) and (2), we get 8 − 2 , 16 + 2 8 + 2 β¯ = . 16 + 2
α ¯=
(3) (4)
Now, as we converge to 0, i.e., as we converge the game to the complete information game, both α ¯ and β¯ converge to 1/2. Therefore, in the limit, both players are choosing H and T with probability 1/2, which is equivalent to the mixed strategy equilibrium of the complete information game. Question 4 4
Let i = 1, · · · , n be the index of bidders, vi be bidder i’s valuation of the good, and bi be player i’s bid. We denote player i’s strategy by a function xi (vi ), meaning that player i bids bi = xi (vi ) when her valuation is vi . We want to show that the strategy profile xi (vi ) =
(n − 1)vi for all i n
constitutes a Bayesian Nash equilibrium. Since the game is symmetric and strategy profile is symmetric, it is sufficient to check one player’s incentive because every player is facing the same incentive problem. We are going to show that if player i’s valuation is vi and all other players are taking the strategy (n − 1)vj xj (vj ) = , n then the bid which maximizes her expected payoff is bi = xi (vi ) =
(n − 1)vi . n
First consider the probability of winning the auction by bidding bi . She wins if and only if all other players’ bids are less than bi , i.e., xj (vj ) =
(n − 1)vj ≤ bi for all j 6= i. n
This is equivalent to vj ≤
nbi for all j 6= i. n−1
Since P rob(vj ≤
nbi nbi )= n−1 n−1
for all j because vj is uniformly distributed in [0.1], nbi for all j 6= i) n−1 nbi n−1 = P rob(vj ≤ ) n−1 nbi n−1 = . n−1
P rob(win) = P rob(vj ≤
Therefore, the expected payoff from bidding bi is Ui = (vi − bi )P rob(win) = (vi − bi )
5
nbi n−1 . n−1
Taking first order condition, nbi n−2 n nbi n−1 ∂Ui = (vi − bi )(n − 1) − = 0, ∂bi n−1 n−1 n−1 nbi (vi − bi )n = , n−1 (n − 1)vi bi = . n Therefore, the strategy of player i bi = xi (vi ) =
(n − 1)vi n
is actually the best response to other player’s playing xj (vj ) =
(n − 1)vj . n
Q.E.D.
6
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5
14.12 Game Theory Fall 2001 Problem Set 5 Solutions
December 14, 2001 1. The possibilities are: Pooling on A, pooling on D, and hybrid equilibria (since there are only two possible actions for player 1, but three possible types of player 1). Let us write the strategies in the form (t11 , t21 , t31 ; t12 , t22 , t32 ) for player 1, where ti j indicates the action taken by type i at the jth move, and (a) for the strategy of player 2, which gets to move only once and has only one information set. The only equilibrium here is (A, A, D; d, a, d) and (δ). Player 2 moves δ since u2 (α) = 5/6(3) + 1/6(−5) = 5/3 < 2 = u2 (δ) – note how beliefs are updated. Finally, each type of player 1 does not have an incentive to deviate: top type is getting 5 as opposed to 4 (if deviates and plays D); middle type is getting 0 as opposed to -1 (if deviates and plays D); bottom type is getting 0 as opposed to -1 (if deviates and plays A). 2. In the second stage of the game, it is clear that each type will always play the static best response: a = 2 type will fight and a = −1 type will accomodate. The possibilities for the first stage are: (a) Pooling on Fight: If π = 0.4, then at t = 2, given no updating of beliefs (so that µ(a = 2|F ) = π), the entrant stays out since EU (Enter) = 2π − 1 = −0.2 < 0 = EU (Out). Then, in t = 1, incumbents pool on fight. The PBE equilibrium is [(F, F ; F, A), (O, O, E)], where we write the strategy for player 1 as (t11 , t21 ; t12 , t22 ) where ti j is the action taken by type i in stage j and for player 2 as (a1 , a21 , a22 where a1 refers to the first information set in t = 1, and the other two refer to the information sets in t = 2: a21 being the information set after the incumbent fights in the first period, and a22 being the information set after the incumbent fights in the first period. Beliefs are not updated on a21 and off the equilibrium belief on a22 is µ(a = −1|accomodate) = 1. (b) Separating equilibrium: If π = 0.9, then entrant always enters since EU (Enter) = 2π − 1 = 0.8 > 0 = EU (Out), a=2 type fights and a=1 type accomodates. The PBE is [(F, A; F, A), (E, O, E)]. Beliefs are updated such that at a22 we have µ(a = −1|accomodate) = 1 and at a21 we have µ(a = 2|f ight) = 1. 1
If the game is repeated 100 times a similar structure applies. 3. In t = 1, buyer of type 1 will accept if and only if: 0.9(1 − p1 ) ≥ 0, that is p1 ≤ 1 0.9(2 − p1 ) ≥ 0, that is p1 ≤ 2 and seller will offer p1 ≥ 0. If seller offers 1 ≤ p1 ≤ 2, then type 1 rejects the offer and type 2 accepts it: EU (seller) = (1 − π)(0) + p1 π = p1 π or simply 2π, which is the value that maximizes the seller’s utility. If seller offers 0 ≤ p1 ≤ 1 then both types accept the offer: EU (seller) = p1 (1 − π) + p1 π = p1 or simply 1, which is the value that maximizes the seller’s utility. If π > 1/2, then seller offers p1 = 2 in t = 1 and she accepts offers in t = 0 p0 ≥ 0.9 since if she waits until t = 1 she can get 0.9(1)=0.9 . If π < 1/2, then seller offers p1 = 1 and she accepts offers in t = 0 p0 ≥ 1.8 since in t = 1 she can get 0.9(2)=1.8. 4.
• Proposed pooling strategy of Sender: Suppose there is an equilibrium in which the Sender’s strategy is (R, R), where (m0 , m00 ) means that type t1 chooses m0 and type t2 chooses m00 . • Receiver’s belief at R: Then the Receiver’s information set corresponding corresponding to R is on the equilibrium path, so the Receiver’s belief (p, 1 − p) at this information set is determined by Bayes’rule and the Sender’s strategy: p = .5, the prior distribution (since there is no updating of belief in pooling). • Receiver’s best response at R: Given this belief, the Receiver’s best response following R is to play d since u2 (u|R) = 1(.5) + 0(.5) = .5 < u2 (d|R) = 0(.5) + 2(.5) = 1 • Receiver’s belief and best response at information set for L: To determine whether both Sender types are willing to choose R, we need to specify how the Receiver would react to L. In particular, we need to pin down the beliefs such that u2 (u|L) > u2 (d|L) in order for the Sender’s willingness to choose R. The Receiver’s belief at the information set corresponding to L needs to be, in order for u to be the best response, q ≥ 1/3. Thus, if there is an equilibrium in which the Sender’s strategy is (R, R) then the Receiver’s strategy must be (d, u) where (a0 , a00 ) means that the Receiver plays a0 following R and a00 following L. • Pooling perfect Bayesian equilibrium: [(R, R), (d, u), p = .5, q] for q ≥ 1/3.
5. Part a.
2
• Pooling on R: Beliefs on the equilibrium path: µ(t1 |R) = .5 Receiver’s best response: u2 (u|R) = .5(2) = 1 > u2 (d|R) = .5(1) = .5 Sender’s payoff: ut1 (R) = 2,ut2 (R) = 1. Any play by Receiver after observing L sustains (R, R). PBE: [(R, R), (u, u), q > 1/2], [(R, R), (u, d), q < 1/2], [(R, R), (u, αu + (1 − α)d), q = 1/2] • Pooling on L: Not an equilibrium, since t2 will always want to play R. • t1 plays R, t2 plays L: Then both of the Receiver’s information sets are on the equilibrium path, so both beliefs are determined by Bayes’ rule and the Sender’s strategy: µ(t1 |R) = 1, µ(t2 |L) = 1. Receiver’s best response: (u, d) t2 , however, will always play R. • t1 plays L, t2 plays R: µ(t1 |L) = 1, µ(t2 |R) = 1. Receiver’s best response: (u, d) Sender’s payoff: ut1 (L) = 1,ut2 (R) = 1. PBE: [(L, R), (u, d)] Part b. • Pooling on R: Beliefs on the equilibrium path: µ(t1 |R) = .5 Receiver’s best response: u2 (u|R) = .5(2) = 1 > u2 (d|R) = .5(1) = .5 Sender’s payoff: ut1 (R) = 0,ut2 (R) = 1. However, this is not an equilibrium since t1 would always play L. • Pooling on L: Beliefs on the equilibrium path: µ(t1 |R) = .5 Receiver’s best response: u2 (u|L) = .5(3) = 1.5 > u2 (d|L) = .5(1) + .5(1) = 1 Sender’s payoff: ut1 (L) = 3,ut2 (L) = 3. Receiver’s belief and best response at information set for R: u2 (u|R) = 2(1 − p) > u2 (d|R) = p when p < 2/3. PBE: [(L, L), (u, u), p < 2/3] • t1 plays R, t2 plays L: Then both of the Receiver’s information sets are on the equilibrium path, so both beliefs are determined by Bayes’ rule and the Sender’s strategy: µ(t1 |R) = 1, µ(t2 |L) = 1. Receiver’s best response: (d, u) No deviations are profitable. PBE: [(R, L), (d, u)] • t1 plays L, t2 plays R: µ(t1 |L) = 1, µ(t2 |R) = 1. Receiver’s best response: (d, u) No deviations are profitable. PBE: [(L, R), (d, u)]
3
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61 Vlqfh Sod|hu 4 nqrzv wkdw Sod|hu 5 lv udwlrqdo dqg wkdw Sod|hu 5 nqrzv wkdw Sod|hu 4 lv udwlrqdo/ kh zloo nqrz wkdw 5 zloo qrw sod| U1 Jlyhq wklv/ P vwulfwo| grplqdwhv W1 Vlqfh Sod|hu 4 lv udwlrqdo/ kh zloo qrw sod| W/ hlwkhu1 Zh duh ohiw zlwk 4q5 O P P 5/7 5/4 71 Vlqfh Sod|hu 5 nqrzv wkdw Sod|hu 4 lv udwlrqdo/ dqg wkdw Sod|hu 4 nqrzv wkdw Sod|hu 5 lv udwlrqdo/ dqg wkdw Sod|hu 4 nqrzv wkdw Sod|hu 5 nqrzv wkdw Sod|hu 4 lv udwlrqdo/ kh zloo nqrz wkdw Sod|hu 4 zloo qrw sod| W ru E1 Jlyhq wklv/ O vwulfwo| grplqdwhv P1 Vlqfh Sod|hu 5 lv udwlrqdo/ kh zloo qrw sod| P/ hlwkhu1 Kh zloo sod| O1 4q5 O P 5/7 Wkxv/ wkh rqo| vwudwhjlhv wkdw duh frqvlvwhqw zlwk wkh frpprq nqrzohgjh ri udwlrqdolw| duh P iru sod|hu 4 dqg O iru sod|hu 51 61 Frqvlghu wkh iroorzlqj h{whqvlyh irup jdph1 1 X
E 2
2,1)
L
R M
(1,2)
1
1 l (3,1)
r
λ
(1,3)
(1,3)
ρ (3,1)
+d, Xvlqj Edfnzdug Lqgxfwlrq/ frpsxwh dq htxloleulxp ri wklv jdph1
5
1 X
E 2
(2,1)
L
R M
(1,2)
1
1 r
λ
(1,3)
(1,3)
l (3,1)
ρ (3,1)
+e, Ilqg wkh qrupdo irup uhsuhvhqwdwlrq ri wklv jdph1 4q5 [ob [o4 [ub [u4 Hob Ho4 Hub Hu4
O 5/4 5/4 5/4 5/4 4/5 4/5 4/5 4/5
P 5/4 5/4 5/4 5/4 6/4 6/4 4/6 4/6
U 5/4 5/4 5/4 5/4 4/6 6/4 4/6 6/4
Wkh srlqwv zloo eh wdnhq r iurp wkh shrsoh zkr glg qrw glvwlqjxlvk wkh vwudwhjlhv wkdw vwduw zlwk [ iurp hdfk rwkhu1 +f, Ilqg doo sxuh vwudwhj| Qdvk htxloleuld1 4q5 [ob [o4 [ub [u4 Hob Ho4 Hub Hu4
O 5/4 5/4 5/4 5/4 4/5 4/5 4/5 4/5
P 5/4 5/4 5/4 5/4 6/4 6/4 4/6 4/6
U 5/4 5/4 5/4 5/4 4/6 6/4 4/6 6/4
Wkh Qdvk htxloleuld duh +[ob/O,/ +[o4/O,/ +[ub/O,/ +[u4/O,1
6
71 Lq wklv txhvwlrq |rx duh dvnhg wr frpsxwh wkh udwlrqdol}deoh vwudwhjlhv lq olqhdu Ehuwudqg0 gxrsro| zlwk glvfuhwh sulfhv1 Zh frqvlghu d zruog zkhuh wkh sulfhv pxvw eh wkh srvlwlyh pxowlsohv ri fhqwv/ l1h1/ ' iffc ff2c c ff?c j lv wkh vhw ri doo ihdvleoh sulfhv1 Iru hdfk sulfh R 5 / wkh ghpdqg lv ' ER ' 4@ i Rc fj Zh kdyh wzr upv ' ic 2j/ hdfk zlwk }hur pdujlqdo frvw1 Vlpxowdqhrxvo|/ hdfk up vhwv d sulfh R 5 1 Revhuylqj wkh sulfhv R dqg R / frqvxphuv ex| iurp wkh up zlwk wkh orzhvw sulfh> zkhq wkh sulfhv duh htxdo/ wkh| glylgh wkhlu ghpdqg htxdoo| ehwzhhq wkh upv1 Hdfk up pd{lpl}hv lwv rzq sur w ; li R R ? R ' ER R ' ER *2 li R ' R Z ER c R ' = f rwkhuzlvh/
2
2
zkhuh 9' 1 +d, Vkrz wkdw dq| sulfh R juhdwhu wkdq wkh prqrsro| sulfh R6J? ' fD lv vwulfwo| grplqdwhg e| vrph vwudwhj| wkdw dvvljqv vrph suredelolw| " : f wr wkh sulfh R ' ff dqg suredelolw| " wr wkh sulfh R6J? ' fD1 4?
Dqvzhu= Wdnh dq| sod|hu dqg dq| sulfh R : R6J? 1 Zh zdqw wr vkrz wkdw wkh pl{hg vwudwhj| j " zlwk j" ER6J? ' " dqg j" R ' " vwulfwo| grplqdwhv R iru vrph " : f1 Wdnh dq| vwudwhj| R : R6J? ri wkh rwkhu sod|hu 1 Zh kdyh 4?
Z ER c R R ' ER ' R E R fD feb ' f2ebbc zkhuh wkh uvw lqhtxdolw| lv e| gh qlwlrq dqg wkh odvw lqhtxdolw| lv gxh wr wkh idfw wkdw R fD1 Rq wkh rwkhu kdqg/ Z Ej " c R ' E " R6J? E R6J? n "R R : E " R6J? E R6J? ' f2D E " 4?
4?
Wkxv/ Z Ej" c R : f2ebb Z ER c R zkhqhyhu f " ffffe1 Fkrrvh " ' ffffe1 Qrz/ slfn dq| R R6J? 1 Vlqfh R : R6J? / zh qrz kdyh Z ER c R ' f1 Exw Z Ej" c R ' E " R6J? E R6J? n "R R "R R : f 4?
4?
4?
Wkdw lv/ Z Ej " c R : Z ER c R 1 Wkhuhiruh/ j " vwulfwo| grplqdwhv R 1
7
4?
+e, Lwhudwlyho| holplqdwlqj doo wkh vwulfwo| grplqdwhg vwudwhjlhv/ vkrz wkdw wkh rqo| udwlrqdol}deoh vwudwhj| iru d up lv R ' ff 4?
Dqvzhu= Zh kdyh douhdg| holplqdwhg wkh vwudwhjlhv wkdw duh odujhu wkdq R6J? 1 Dw dq| lwhudwlrq | dvvxph wkdw/ iru hdfk sod|hu/ wkh vhw ri doo uhpdlqlqj vwudwhjlhv duh | ' iffc ff2c c R7j zkhuh R R7 R6J? 1 Zh zdqw wr vkrz wkdw R7 lv vwulfwo| R ff ' " dqg j"R R grplqdwhg e| wkh pl{hg vwudwhj| j "R zlwk j"R E7 ' "/ r dqg holplqdwh wkh vwudwhj| R71 Wklv surfhvv zloo hqg zkhq ' iffj/ frpsohwlqj wkh surri1 Qrz/ iru sod|hu / R7 E R7 *2 li R ' R7c Rc R ' Z E7 f rwkhuzlvh1 4?
4?
7
7
7
Rq wkh rwkhu kdqg/ R ff E R7 n ff n "R Z j"R c R7 ' E " E7 : E " E7 R ff E R7 n ff ' E " d7 R E R7 ff E 27 R o Wkhq/ Z j"R c R7 : Z E7 Rc R zkhqhyhu
4?
7
R
4?
7
"
R7 E R7 *2 R7 E R7 ff E 27 R
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Z j"R c R ' E " E7 R ff E R7 n ff n "R Rc R c "R R : f ' Z E7
4?
7
4?
R
4?
4?
vkrzlqj wkdw j "R vwulfwo| grplqdwhv R7/ dqg frpsohwlqj wkh surri1 7
+f, Zkdw duh wkh Qdvk htxloleuld ri wklv jdphB Dqvzhu= Vlqfh dq| Qdvk htxloleulxp lv udwlrqdol}deoh/ dqg vlqfh wkh rqo| udwlr0 qdol}deoh vwudwhj| sur oh lv R c R / wkh rqo| Qdvk htxloleulxp lv R c R 1 +Vlqfh wklv lv d qlwh jdph/ wkhuh lv dozd|v d Qdvk htxloleulxp srvvleo| lq pl{hg vwudwhjlhv1, 4?
4?
8
4?
4?
Grade Distribution For Midterm 1 Average = 67.96 Std.Dev.= 18.05 100
Assuming that you will exhibit similar performance in the future, if you got above 70, you will probably get A(+/-); if you got above 50, you will probably get B(+/-) or above; if you got below 40, you should reconsider ...
90
80
70
grade
60
50
40
30
20
10
0 1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 rank
14.12 Game Theory — Midterm II 10/10/2001 Prof. Muhamet Yildiz Instructions. This is an open book exam; you can use any written material. You have one hour and 20 minutes. Each question is 25 points. Good luck! 1. Consider the following game: 2 X
E 1
5/2 5/2
L
2 l
3 3
R
r
l
0 2
2 0
r 2 2
Compute all the pure-strategy subgame-perfect equilibria. Use a forward induction argument to eliminate one of these equilibria. 2. Below, there are pairs of stage games and strategy profiles. For each pair, check whether the strategy profile is a subgame-perfect equilibrium of the game in which the stage game is repeated infinitely many times. Each agent tries to maximize the discounted sum of his expected payoffs in the stage game, and the discount rate is δ = 0.99. (a) Stage Game: 1\2 L M R T 2,-1 0,0 -1,2 M 0,0 0,0 0,0 B -1,2 0,0 2,-1 Strategy profile: Until some player deviates, player 1 plays T and player 2 alternates between L and R. If anyone devites, then each play M thereafter. (b) Stage Game: 1\2 A B A 2,2 1,3 B 3,1 0,0 Strategy profile: The play depends on three states. In state S0 , each player plays A; in states S1 and S2 , each player plays B. The game start at state S0 . In state S0 , if each player plays A or if each player plays B, we stay at S0 , but if a player i plays B while the other is playing A, then we switch to state Si . At any Si , if player i plays B, we switch to state S0 ; otherwise we state at state Si .
1
3. Consider the following first-price, sealed-bid auction where an indivisible good is sold.There are n ≥ 2 buyers indexed by i = 1, 2, . . . , n. Simultaneously, each buyer i submits a bid bi ≥ 0. The agent who submits the highest bid wins. If there are k > 1 players submitting the highest bid, then the winner is determined randomly among these players – each has probability 1/k of winning. The winner i gets the object and pays his bid bi , obtaining payoff vi − bi , while the other buyers get 0, where v1 , . . . , vn are independently and identically distributed with probability density function f where ½ 2 3x x ∈ [0, 1] f (x) = 0 otherwise. (a) Compute the symmetric, linear Bayesian Nash equilibrium. (b) What happens as n → ∞? [Hint: Since v1 , v2 , . . . , vn is independently distributed, for any w1 , w2 , . . . , wk , we have Pr(v1 ≤ w1 , v2 ≤ w2 , . . . , vk ≤ wk ) = Pr(v1 ≤ w1 ) Pr(v2 ≤ w2 ) . . . Pr(vk ≤ wk ).] 4. This question is about a thief and a policeman. The thief has stolen an object. He can either hide the object INSIDE his car on in the TRUNK. The policeman stops the thief. He can either check INSIDE the car or the TRUNK, but not both. (He cannot let the thief go without checking, either.) If the policeman checks the place where the thief hides the object, he catches the thief, in which case the thief gets -1 and the police gets 1; otherwise, he cannot catch the thief, and the thief gets 1, the police gets -1. (a) Compute all the Nash equilibria. (b) Now imagine that we have 100 thieves and 100 policemen, indexed by i = 1, . . . , 100, and j = 1, . . . , 100. In addition to their payoffs above, each thief i gets extra payoff bi form hiding the object in the TRUNK, and each policeman j gets extra payoff di from checking the TRUNK. We have b1 < b2 < · · · < b50 < 0 < b51 < · · · < b100 , d1 < d2 < · · · < d50 < 0 < d51 < · · · < d100 . Policemen cannot distinguish the thieves from each other, nor can the thieves distinguish the policemen from each other. Each thief has stolen an object, hiding it either in the TRUNK or INSIDE the car. Then, each of them is randomly matched to a policeman. Each matching is equally likely. Again, a policeman can either check INSIDE the car or the TRUNK, but not both. Compute a pure-strategy Bayesian Nash equilibrium of this game.
2
The game for problem 1.
2 X
E 1
5/2 5/2
L
R 2
l 3 3
r
l
0 2
2 0
r 2 2
2 X
E 1
5/2 5/2
L
2 l
3 3
R
r
l
0 2
2 0
3
r 2 2
14.12 Game Theory — Midterm II 10/10/2001 Prof. Muhamet Yildiz Instructions. This is an open book exam; you can use any written material. You have one hour and 20 minutes. Each question is 25 points. Good luck! 1. Consider the following game: 2 X
E 1
5/2 5/2
L
R 2
l 3 3
r
l
0 2
2 0
r 2 2
Compute all the pure-strategy subgame-perfect equilibria. Use a forward induction argument to eliminate one of these equilibria. Answer: There are two pure strategy Nash equilibria in the proper subgame, yielding subgame-perfect equilibria:
2
2
X
E
X
E
1 5/2 5/2
L
5/2 5/2
R 2
l 3 3
1
r
l
0 2
2 0
r
L
2 l
2 2
3 3
R
r
l
0 2
2 0
For player 2, Er is strictly dominated by Xr, while El is not dominated. Hence, if player 1 keeps believing that 2 is rational whenever it is possible, then when he sees that 2 played E, he ought to believe that 2 plays strategy El – not the dominated strategy Ex. In that case, 1 would play L, and 2 would play E. Therefore, the equilibrium on the left is eliminated. 1
r 2 2
2. Below, there are pairs of stage games and strategy profiles. For each pair, check whether the strategy profile is a subgame-perfect equilibrium of the game in which the stage game is repeated infinitely many times. Each agent tries to maximize the discounted sum of his expected payoffs in the stage game, and the discount rate is δ = 0.99. (a) Stage Game: 1\2 L M R T 2,-1 0,0 -1,2 M 0,0 0,0 0,0 B -1,2 0,0 2,-1 Strategy profile: Until some player deviates, player 1 plays T and player 2 alternates between L and R. If anyone deviates, then each play M thereafter. Answer: It is subgame perfect. Since (M,M) is a Nash equilibrium of the stage game, we only need to check if any player wants to deviate when player 1 plays T and player 2 alternates between L and R. In this regime, the present value of player 1’s payoffs is V1L =
2 δ 2−δ − = >0 1−δ 1−δ 1−δ
when 2 is to play L and V1R =
2δ 1 2δ − 1 − = = 98 1−δ 1−δ 1−δ
when 2 is to play R. When 2 plays L, 1 cannot gain by deviating. When 2 plays R, the best 1 gets by deviating is 2 + 0 < 98 (when he plays B). The only possible profitable deviation for player 2 is to play R when he is supposed to play left. In that contingency, if he follows the strategy he gets V1R = 98; if he deviates, he gets 2 + 0 < V1R . (b) Stage Game: 1\2 A B A 2,2 1,3 B 3,1 0,0 Strategy profile: The play depends on three states. In state S0 , each player plays A; in states S1 and S2 , each player plays B. The game start at state S0 . In state S0 , if each player plays A or if each player plays B, we stay at S0 , but if a player i plays B while the other is playing A, then we switch to state Si . At any Si , if player i plays B, we switch to state S0 ; otherwise we state at state Si . Answer: It is not subgame-perfect. At state S2 , player 2 is to play B, and we will switch to state S0 no matter what 1 plays. In that case, 1 would gain by deviating and playing A (in state S2 ).
2
3. Consider the following first-price, sealed-bid auction where an indivisible good is sold. There are n ≥ 2 buyers indexed by i = 1, 2, . . . , n. Simultaneously, each buyer i submits a bid bi ≥ 0. The agent who submits the highest bid wins. If there are k > 1 players submitting the highest bid, then the winner is determined randomly among these players – each has probability 1/k of winning. The winner i gets the object and pays his bid bi , obtaining payoff vi − bi , while the other buyers get 0, where v1 , . . . , vn are independently and identically distributed with probability density function f where ½ 2 3x x ∈ [0, 1] f (x) = 0 otherwise. (a) Compute the symmetric, linear Bayesian Nash equilibrium. Answer: We look for an equilibrium of the form bi = a + cvi where c > 0. Then, the expected payoff from bidding bi with type vi is U (bi ; vi ) = (vi − bi ) Pr (bi > a + cvj ∀j 6= i) Y Pr (bi > a + cvj ) = (vi − bi ) j6=i
¶ µ bi − a Pr vj < = (vi − bi ) c j6=i Y µ bi − a ¶3 = (vi − bi ) c j6=i µ ¶3(n−1) bi − a = (vi − bi ) c Y
for bi ∈ [a, a + c]. The first order condition is ¶3(n−1) ¶3(n−1)−1 µ µ bi − a 1 bi − a ∂U (bi ; vi ) =− + 3 (n − 1) (vi − bi ) = 0; ∂bi c c c i.e.,
µ
bi − a − c i.e.,
¶
+ 3 (n − 1)
1 (vi − bi ) = 0; c
a + 3 (n − 1) vi . 3 (n − 1) + 1 Since this is an identity, we must have bi =
a=
a =⇒ a = 0, 3 (n − 1) + 1
and c= 3
3 (n − 1) . 3 (n − 1) + 1
(b) What happens as n → ∞? Answer: As n → ∞,
bi → vi .
In the limit, each bidder bids his valuation, and the seller extracts all the gains from trade. [Hint: Since v1 , v2 , . . . , vn is independently distributed, for any w1 , w2 , . . . , wk , we have Pr(v1 ≤ w1 , v2 ≤ w2 , . . . , vk ≤ wk ) = Pr(v1 ≤ w1 ) Pr(v2 ≤ w2 ) . . . Pr(vk ≤ wk ).] 4. This question is about a thief and a policeman. The thief has stolen an object. He can either hide the object INSIDE his car on in the TRUNK. The policeman stops the thief. He can either check INSIDE the car or the TRUNK, but not both. (He cannot let the thief go without checking, either.) If the policeman checks the place where the thief hides the object, he catches the thief, in which case the thief gets -1 and the police gets 1; otherwise, he cannot catch the thief, and the thief gets 1, the police gets -1. (a) Compute all the Nash equilibria. Answer: This is a matching-pennies game. There is a unique Nash equilibrium, in which Thief hides the object INSIDE or the TRUNK with equal probabilities, and the Policeman checks INSIDE or the TRUNK with equal probabilities. (b) Now imagine that we have 100 thieves and 100 policemen, indexed by i = 1, . . . , 100, and j = 1, . . . , 100. In addition to their payoffs above, each thief i gets extra payoff bi form hiding the object in the TRUNK, and each policeman j gets extra payoff dj from checking the TRUNK. We have b1 < b2 < · · · < b50 < 0 < b51 < · · · < b100 , d1 < d2 < · · · < d50 < 0 < d51 < · · · < d100 . Policemen cannot distinguish the thieves from each other, nor can the thieves distinguish the policemen from each other. Each thief has stolen an object, hiding it either in the TRUNK or INSIDE the car. Then, each of them is randomly matched to a policeman. Each matching is equally likely. Again, a policeman can either check INSIDE the car or the TRUNK, but not both. Compute a pure-strategy Bayesian Nash equilibrium of this game. Answer: A Bayesian Nash equilibrium: A thief i hides the object in INSIDE if bi < 0 TRUNK if bi > 0; a policeman j checks INSIDE if dj < 0 TRUNK if dj > 0. This is a Bayesian Nash equilibrium, because, from the thief’s point of view the policeman is equally likely to to check TRUNK or INSIDE the car, hence it is the best response for him to hide in the trunk iff the extra benefit from hiding in the trunk is positive. Similar for the policemen.
4
14.12 Midterm 2 Grade Distribution 100 90 80 70
50 40 30 20 10
Rank
45
43
41
39
37
35
33
31
29
27
25
23
21
19
17
15
13
11
9
7
5
3
0 1
grade
60
14.12 Grade Distribution (Midterm 1 + Midterm 2)/2 100
90
80
60
50
40
30
rank
45
43
41
39
37
35
33
31
29
27
25
23
21
19
17
15
13
11
9
7
5
3
20 1
grade
70
14.12 Game Theory — Final 12/10/2001 Prof. Muhamet Yildiz Instructions. This is an open book exam; you can use any written material. You have 2 hours 50 minutes. Each question is 20 points. Good luck! 1. Consider the following extensive form game. 1 O
I
2
2 2
L
R 1
L
3 1
R
L
0 0
R
0 0
1 3
(a) Find the normal form representation of this game. (b) Find all rationalizable pure strategies. (c) Find all pure strategy Nash equilibria. (d) Which strategies are consistent with all of the following assumptions? (i) (ii) (iii) (iv)
1 is rational. 2 is sequentially rational. at the node she moves, 2 knows (i). 1 knows (ii) and (iii).
2. This question is about a milkman and a customer. At any day, with the given order, • Milkman puts m ∈ [0, 1] liter of milk and 1 − m liter of water in a container and closes the container, incurring cost cm for some c > 0; • Customer, without knowing m, decides on whether or not to buy the liquid at some price p. If she buys, her payoff is vm − p and the milkman’s payoff is p − cm. If she does not buy, she gets 0, and the milkman gets −cm. If she buys, then she learns m.
1
(a) Assume that this is repeated for 100 days, and each player tries to maximize the sum of his or her stage payoffs. Find all subgame-perfect equilibria of this game. (b) Now assume that this is repeated infinitely many times and each player tries to maximize the discounted sum of his or her stage payoffs, where discount rate is δ ∈ (0, 1). What is the range of prices p for which there exists a subgame perfect equilibrium such that, everyday, the milkman chooses m = 1, and the customer buys on the path of equilibrium play? 3. For the game in question 3.a, assume that with probability 0.001, milkman strongly believes that there is some entity who knows what the milkman does and will punish him severely on the day 101 for each day the milkman dilutes the milk (by choosing m < 1). Call this type irrational. Assume that this is common knowledge. For v > p > c, find a perfect Bayesian equilibrium of this game. Bonus: [10 points] Discuss what would happen if the irrational type were known to dilute the milk by accident with some small but positive probability. 4. Find a perfect Bayesian equilibrium of the following game.
0 0
3 1
quiche
beer
{.1}
’t don
el du
1 0
du el
tw ts beer {.9}
quiche
el du
don ’t
don ’
’t don
1 1 3 0 0 0
du el
-1 1
t
2 1
5. A risk-neutral entrepreneur has a project that requires $100,000 as an investment, and will yield $300,000 with probability 1/2, $0 with probability 1/2. There are two types of entrepreneurs: rich who has a wealth of $1,000,000, and poor who has $0. For some reason, the wealthy entrepreneur cannot use his wealth as an investment towards this project. There is also a bank that can lend money with interest rate π. That is, if the entrepreneur borrows $100,000 to invest, after the project is completed he will pay back $100, 000 (1 + π) – if he has that much money. If his wealth is less than this amount at the end of the project, he will pay all he has. The order of the events is as follows: • First, bank posts π. 2
• Then, entrepreneur decides whether to borrow ($100,000) and invest. • Then, uncertainty is resolved.
(a) Compute the subgame perfect equilibrium for the case when the wealth is common knowledge. (b) Now assume that the bank does not know the wealth of the entrepreneur. The probability that the entrepreneur is rich is 1/4. Compute the perfect Bayesian equilibrium.
3
14.12 Game Theory — Final 12/10/2001 Prof. Muhamet Yildiz Instructions. This is an open book exam; you can use any written material. You have 2 hours 50 minutes. Each question is 20 points. Good luck! 1. Consider the following extensive form game. 1 O
I
2
2 2
L
R 1
L
3 1
R
L
0 0
R
0 0
(a) Find the normal form representation of this game. A: L R OL 2,2 2,2 OR 2,2 2,2 IL 3,1 0,0 IR 0,0 1,3 (b) Find all rationalizable pure strategies. L R OL 2,2 2,2 OR 2,2 2,2 IL 3,1 0,0 (c) Find all pure strategy Nash equilibria. L R OL 2,2 2,2 OR 2,2 2,2 IL 3,1 0,0 1
1 3
(d) Which strategies are consistent with all of the following assumptions? (i) (ii) (iii) (iv)
1 is rational. 2 is sequentially rational. at the node she moves, 2 knows (i). 1 knows (ii) and (iii).
ANSWER: By (i) 1 does not play IR. Hence, by (iii), at the node she moves, 2 knows that 1 does not play IR, hence he knows IL. Then, by (ii), 2 must play L. Therefore, by (i) and (iv), 1 must play IL. The answer is (IL,L). 2. This question is about a milkman and a customer. At any day, with the given order, • Milkman puts m ∈ [0, 1] liter of milk and 1 − m liter of water in a container and closes the container, incurring cost cm for some c > 0; • Customer, without knowing m, decides on whether or not to buy the liquid at some price p. If she buys, her payoff is vm − p and the milkman’s payoff is p − cm. If she does not buy, she gets 0, and the milkman gets −cm. If she buys, then she learns m. (a) Assume that this is repeated for 100 days, and each player tries to maximize the sum of his or her stage payoffs. Find all subgame-perfect equilibria of this game. ANSWER: The stage game has a unique Nash equilibrium, in which m = 0 and the customer does not buy. Therefore, this finitely repeated game has a unique subgame-perfect equilibrium, in which the stage equilibrium is repeated. (b) Now assume that this is repeated infinitely many times and each player tries to maximize the discounted sum of his or her stage payoffs, where discount rate is δ ∈ (0, 1). What is the range of prices p for which there exists a subgame perfect equilibrium such that, everyday, the milkman chooses m = 1, and the customer buys on the path of equilibrium play? ANSWER: The milkman can guarantee himself 0 by always choosing m = 0. Hence, his continuation value at any history must be at least 0. Hence, in the worst equilibrium, if he deviates customer should not buy milk forever, giving the milkman exactly 0 as the continuation value. Hence, the SPE we are looking for is the milkman always chooses m=1 and the customer buys until anyone deviates, and the milkman chooses m=0 and the customer does not buy thereafter. If the milkman does not deviate, his continuation value will be V =
p−c . 1−δ
The best deviation for him (at any history on the path of equilibrium play) is to choose m = 0 (and not being able to sell thereafter). In that case, he will get Vd = p + δ0 = p. In order this to be an equilibrium, we must have V ≥ Vd ; i.e., p−c ≥ p, 1−δ
2
i.e.,
c p≥ . δ In order that the customer buy on the equilibrium path, we must also have p ≤ v. Therefore, c v≥p≥ . δ
3. For the game in question 2.a, assume that with probability 0.001, milkman strongly believes that there is some entity who knows what the milkman does and will punish him severely on the day 101 for each day the milkman dilutes the milk (by choosing m < 1). Call this type irrational. Assume that this is common knowledge. For some v > p > c, find a perfect Bayesian equilibrium of this game. [If you find it easier, take the customers at different dates different, but assume that each customer knows whatever the previous customers knew.] ANSWER: [It is very difficult to give a rigorous answer to this question, so you would get a big partial grade for an informal answer that shows that you understand the reputation from an incomplete-information point of view.] Irrational type always sets m = 1. Since he will be detected whenever he sets m < 1 and the customer buys, the rational type will set either m = 1 or m = 0. We are looking for an equilibrium in which early in the relation the rational milkman will always set m = 1 and the customer will always buy, but near the end of the relation the rational milkman will mix between m = 1 and m = 0, and the customer will mix between buy and not buy. In this equilibrium, if the milkman sets m < 1 or the customer does not buy at any t, then the rational milkman sets m = 0 at each s > t. In that case, if in addition the costumer buys at some dates in the period {t + 1, t + 2, . . . , s − 1} and if the milkman chooses m = 1 at each of those days, then the costumer will assign probability 1 to that the milkman is irrational and buy the milk at s; otherwise, he will not buy the milk. On the path of such play, if the milkman sets m < 1 or the customer does not buy at any t, then the rational milkman sets m = 0 and the costumer does not buy at each s > t. In order this to be an equilibrium, the probability µt that the milkman is irrational at such history must satisfy µt (100 − t) (v − p) − (1 − µt ) p ≤ 0, where the first term is the expected benefit from experimenting (if the milkman happens to be irrational) and the second term is the cost (if he is rational). That is, µt ≤
p . p + (100 − t) (v − p)
Now we determine what happens if the milkman has always been setting m = 1, and the customer has been buying. In the last date, the rational type will set m = 0, and the rational type will set m = 1; hence, the buyer will buy iff µ100 (v − p) − (1 − µ100 ) p ≥ 0, 3
i.e.,
p µ100 ≥ . v
Since we want him to mix, we set
p µ100 = . v We derive µt for previous dates using the Bayes’ rule and the indifference condition necessary for the customer’s mixing. Let’s write αt for the probability that the rational milkman sets m = 1 at t, and at = µt + (1 − µt ) αt for the total probability that m = 1 at date t. Since the customer will be indifferent between buying and not buying at t + 1, his expected payoff at t + 1 will be 0. Hence, his expected payoff from buying at t is at (v − p) + (1 − at ) (−p) . For indifference, this must be equal to zero, thus p at = . v On the other hand, by Bayes’ rule, µt+1 = Therefore,
µt . at
p µt = at µt+1 = µt+1 . v
That is, µ100 = µ99 = µ98 =
p v ³ p ´2 v ³ p ´3 v
.. . Note that
p − µt p v . at = = µt + (1 − µt ) αt ⇒ αt = v 1 − µt
Assume that (p/v)100 < 0.001. Then, we will have a date t∗ such that ³ p ´101−t∗ v
< 0.001
t∗ , we will have µt = (p/v)101−t and the players will mix so that at = vp . At each date t < t∗ , the milkman will set m = 1 and the customer will buy. At date t∗ , the rational milkman will mix so that ³ p ´100−t∗ µ∗ 0.001 = µt∗ +1 = t = , v at∗ at∗ 4
hence
0.001 at∗ = ¡ p ¢100−t∗ . v
Note that at∗ > p/v, hence the customer will certainly buy at t∗ . Let’s write β t for the probability that the customer will buy at day t. In the day 99, if the rational milkman sets m = 1, he will get U = β 99 (p − c) + β 99 β 100 p + (1 − β 99 ) (−c) , where the first term is the profit from selling at day 99, the second term is the profit from day 100 (when he will set m = 0), and the last term is the loss if the customer does not buy at day 99. If he sets m = 0, he will get β 99 p (from the sale at 99, and will get zero thereafter). Hence, he will set m = 1 iff β 99 (p − c) + β 99 β 100 p + (1 − β 99 ) (−c) ≥ β 99 p i.e.,
c β 99 β 100 ≥ . p We are looking for an indifference, hence we set c β 99 β 100 = . p Similarly, at day 98 the rational milkman will set m = 1 iff β 98 (p − c) + β 98 β 99 p + (1 − β 98 ) (−c) ≥ β 98 p, where the second term is due to the fact that at date 99 he will be indifferent between choosing m = 0 and m = 1. For indifference, we set c β 98 β 99 = . p We will continue on like this as long as we need the milkman to mix. That is, we will have c , p c , = p .. . c = , p c = . p
β t∗ β t∗ +1 = β t∗ +1 β t∗ +2
β 98 β 99 β 99 β 100
5
As we noted before, β t∗ = 1. Hence, β t∗ +1 = pc . Hence, β t∗ +2 = 1, ... That is, β t∗ = 1, c β t∗ +1 = , p β t∗ +2 = 1, c β t∗ +3 = , p .. . Bonus: [10 points] Discuss what would happen if the irrational type were known to dilute the milk by accident with some small but positive probability. 4. Find a perfect Bayesian equilibrium of the following game.
0 0
3 1
quiche
beer
{.1}
’t don
el du
1 0
du el
tw ts beer {.9}
quiche
el u d
don ’t
don ’
’t don
ANSWER:
6
1 1 3 0 0 0
du el
-1 1
t
2 1
0 0
3 1
beer t {0} ’ n do
el du
1 0
du el
quiche {.1}
tw
el u d
{1} d on’ t
ts quiche {0}
{1}beer {.9}
don ’
’t don
1 1 3 0 0 0
du el
-1 1
t
2 1
5. A risk-neutral entrepreneur has a project that requires $100,000 as an investment, and will yield $300,000 with probability 1/2, $0 with probability 1/2. There are two types of entrepreneurs: rich who has a wealth of $1,000,000, and poor who has $0. For some reason, the wealthy entrepreneur cannot use his wealth as an investment towards this project. There is also a bank that can lend money with interest rate π. That is, if the entrepreneur borrows $100,000 to invest, after the project is completed he will pay back $100, 000 (1 + π) – if he has that much money. If his wealth is less than this amount at the end of the project, he will pay all he has. The order of the events is as follows: • First, bank posts π.
• Then, entrepreneur decides whether to borrow ($100,000) and invest. • Then, uncertainty is resolved.
(a) Compute the subgame perfect equilibrium for the case when the wealth is common knowledge. ANSWER: The rich entrepreneur is always going to pay back the loan in full amount, hence his expected payoff from investing (as a change from not investing) is (0.5)(300, 000) − 100, 000 (1 + π) . Hence, he will invest iff this amount is non-negative, i.e., π ≤ 1/2. Thus, the bank will set the interest rate at π R = 1/2. 7
The poor entrepreneur is going to pay back the loan only when the project succeeds. Hence, his expected payoff from investing is (0.5)(300, 000 − 100, 000 (1 + π)). He will invest iff this amount is non-negative, i.e., π ≤ 2. Thus, the bank will set the interest rate at π P = 2. (b) Now assume that the bank does not know the wealth of the entrepreneur. The probability that the entrepreneur is rich is 1/4. Compute the perfect Bayesian equilibrium. ANSWER: As in part (a), the rich type will invest iff π ≤ π R = .5, and the poor type will invest iff π ≤ π P = 2. Now, if π ≤ πR , the bank’s payoff is · ¸ 3 1 1 1 100, 000 (1 + π) + 100, 000 (1 + π) + 0 − 100, 000 U (π) = 4 4 2 2 5 100, 000 (1 + π) − 100, 000 = 8 5 ≤ 100, 000 (1 + π R ) − 100, 000 8 5 1 = 100, 000 (1 + 1/2) − 100, 000 = − 100, 000 < 0. 8 16 If π R < π ≤ π P , the bank’s payoff is · ¸ 1 3 1 100, 000 (1 + π) + 0 − 100, 000 U (π) = 4 2 2 3 = 100, 000 (π − 1) , 8 which is maximized at πP , yielding 38 100, 000. If π > π P , U (π) = 0. Hence, the bank will choose π = πP .
8
14.12 Game Theory — Final 12/21/2001 Prof. Muhamet Yildiz Instructions. This is an open book exam; you can use any written material. You have 2 hours 50 minutes. Each question is 20 points. Good luck! 1. Consider the following extensive form game.
1
A
2
D
α
1
δ (1,1)
a
(1,4)
d
(0,3)
(2,2)
(a) Find the normal form representation of this game. 1\2 Aa Ad Da Dd
α 1,4 2,2 1,1 1,1
δ 0,3 0,3 1,1 1,1
(b) Find all rationalizable pure strategies. 1\2 α δ Ad 2,2 0,3 Da 1,1 1,1 Dd 1,1 1,1 (c) Find all pure strategy Nash equilibria. 1\2 α δ Ad 2,2 0,3 Da 1,1 1,1 Dd 1,1 1,1 (d) Which strategies are consistent with all of the following assumptions? (i) (ii) (iii) (iv)
1 is rational. 2 is sequentially rational. at the node she moves, 2 knows (i). 1 knows (ii) and (iii).
1
ANSWER: By (i) 1 does not play Aa. Hence, by (iii), at the node she moves, 2 knows that 1 does not play Aa, hence he knows that 1 plays Ad. Then, by (ii), 2 must play δ. Therefore, by (i) and (iv), 1 must play Ad or Aa. The answer is 1 plays A, given chance 2 would play δ. 2. This question is about a game between a possible applicant (henceforth student) to a Ph.D. program in Economics and the Admission Committee. Ex-ante, Admission Committee believes that with probability .9 the student hates economics and with probability .1 he loves economics. After Nature decides whether student loves or hates economics with the above probabilities and reveals it to the student, the student decides whether or not to apply to the Ph.D. program. If the student does not apply, both the student and the committee get 0. If student applies, then the committee is to decide whether to accept or reject the student. If the committee rejects, then committee gets 0, and student gets -1. If the committee accepts the student, the payoffs depend on whether the student loves or hates economics. If the student loves economics, he will be successful and the payoffs will be 20 for each player. If he hates economics, the payoffs for both the committee and the student will be -10. Find a separating equilibrium and a pooling equilibrium of this game. ANSWER: A separating equilibrium:
(-10,-10) Accept
Apply {0}
Reject (-1,0)
Don’t
.9
Hate (0,0) .1
Love
Apply
{1}
(20,20) Accept Reject (-1,0)
Don’t (0,0)
A pooling equilibrium:
2
(-10,-10) Accept
Apply {.9}
Reject (-1,0)
Don’t
.9
Hate (0,0) .1
Love
Apply
{.1}
(20,20) Accept
Reject (-1,0)
Don’t (0,0)
3. Consider a bargaining problem where two risk-neutral players are trying to divide a dollar they own, which they cannot use until they reach an agreement. The players do not discount the future, but at the end of each rejection of an offer the bargaining breaks down with probability 1 − δ ∈ (0, 1) and each player gets 0. (a) Consider the following bargaining procedure. Player 1 makes an offer (x, 1 − x), where x is player 1’s share. Then, player 2 decides whether or not to accept the offer. If she accepts, they implement the offer, yielding division (x, 1 − x). If she rejects the offer, then with probability 1 − δ, the bargaining breaks down an each gets 0; with probability δ, player 1 makes another offer, which will be accepted or rejected by player 2 as above. (If player 2 rejects the offer, bargaining will break down with probability 1 − δ again.) If the offer is rejected and the bargaining did not break down, now player 2 makes a counter offer, and player 1 accepts or rejects this counter offer as above. If the offer is rejected, this time the game will end, and each will get 0. Find the subgame-perfect Nash equilibrium of this game. Compute the expected payoff of each player at the beginning of the game in this equilibrium. ANSWER: On the last day, 1 accepts any offer, so 2 offers (0,1). Hence, on the previous day, 2 accepts an offer iff she gets at least δ. Hence, 1 offers (1 − δ, δ) – accepted. in¢ the first day, 2 accepts an offer iff she gets at least¢ δ 2 . Hence, ¡ Thus, ¡ 1 offers 1 − δ 2 , δ 2 – accepted. The expected payoffs are 1 − δ 2 , δ 2 .
(b) Compute the subgame-perfect equilibrium of the game in which the procedure in part (a) is repeated 2 times. (The probability of bargaining breakdown after each rejection is 1 − δ, except for the end of the game.) ANSWER: The last period as above. Let’s look at the first period. ¡ ¢On the last 2 day of the first period, 1 accepts an offer iff he gets at least δ 1 − δ , so 2 offers 3
¡ ¡ ¢ ¡ ¢¢ 2 δ 1 − δ2 , 1 − δ 1 − δ . ¢¢ Hence, on the previous day, 2 accepts an offer iff she ¡ ¡ 2 gets at least δ 1 − δ 1 − δ . Hence, 1 offers ¢¢ ¡ ¡ ¢¢¢ ¡ ¡ ¢ ¡ ¡ ¢¢¢ ¡ ¡ ¡ = 1 − δ + δ2 1 − δ2 , δ 1 − δ 1 − δ2 1 − δ 1 − δ 1 − δ2 , δ 1 − δ 1 − δ2
–¡accepted. Thus, in the first day, 2 accepts an offer iff she gets at least ¡ ¢¢ 2 2 δ 1 − δ 1 − δ . Hence, 1 offers ¡ ¡ ¢¢ ¡ ¡ ¢¢¢ ¡ ¡ ¢ ¡ ¡ ¢¢¢ ¡ = 1 − δ2 + δ3 1 − δ2 , δ2 1 − δ 1 − δ2 1 − δ2 1 − δ 1 − δ2 , δ 1 − δ 1 − δ2 – accepted.
(c) Find the subgame-perfect equilibrium of the game in which this procedure is repeated until they reach an agreement. Note that player 1 makes two offers, then 2 makes one offer, then 1 makes two offers, and so on. You need to show that the proposed strategy profile is in fact a subgame-perfect equilibrium. (The probability of bargaining breakdown after each rejection is 1 − δ.) [Hint: One way is to compute the SPE for the game in which the procedure is repeated n times and let n → ∞. A somewhat easier way is to consider an alternating offer bargaining procedure with some effective discount rates – different for a different player.] ANSWER: If you compare the calculations above with the calculations with the alternating offer case with asymmetric discount rates, you should realize that the first offer player 1 makes and the offer player 2 makes are identical to the offers players 1 and 2 make, respectively, if the discount rates were δ1 = δ and δ 2 = δ 2 . Intuitively, in his second offer player 1 makes player 2 indifferent between accepting 1’s second offer and making an offer next day, and in his first offer he makes her indifferent between accepting the offer and waiting for the second offer. Therefore, 2 is indifferent between accepting 1’s first offer and waiting two days two make an offer, as in the the alternating offer case when her discount rate is δ 2 . Now conjecture that the subgame-perfect equilibrium would be as in the alternating offer game with above discount rates. That is, • in his first offer, player 1 offers ¶ µ ¶ µ ¶ µ 1 − δ 2 δ 2 (1 − δ 1 ) 1 − δ 2 δ2 (1 − δ) 1 − δ2 1 − δ2 ≡ ≡ ,1 − , , 1 − δ1δ2 1 − δ1δ2 1 − δ1 δ2 1 − δ1 δ2 1 − δ3 1 − δ3 µ ¶ 1+δ δ2 ≡ ; , 1 + δ + δ2 1 + δ + δ2 • in his secon offer, he will offer µ ¶ µ ¶ δ (1 − δ 1 ) δ (1 − δ1 ) δ 1 + δ2 1− ≡ ; , , 1 − δ1δ2 1 − δ1δ2 1 + δ + δ2 1 + δ + δ2 • player 2 will offer µ ¶ µ ¶ µ ¶ 1 − δ1 δ 1 (1 − δ 2 ) 1 − δ 1 δ + δ2 1 − δ1 1 1− ≡ ≡ . , , , 1 − δ1 δ2 1 − δ1 δ2 1 − δ1δ2 1 − δ1δ2 1 + δ + δ2 1 + δ + δ2 4
Player 1’s first offer and player 2’s offer are by formula for alternating offer, player 1’s second offer is calculated by backward induction using the player 2’s offer in the next period. Using single deviation property, you need to check that this is an equilibrium. 4. We have an employer and a worker, who will work as a salesman. The worker may be a good salesman or a bad one. In expectation, if he is a good salesman, he will make $200,000 worth of sales, and if he is bad, he will make only $100,000. The employer gets 10% of the sales as profit. The employer offers a wage w. Then, the worker accepts or rejects the offer. If he accepts, he will be hired at wage w. If he rejects the offer, he will not be hired. In that case, the employer will get 0, the worker will get his outside option, which will pay $15,000 if he is good, $8,000 if he is bad. Assume that all players are risk-neutral. (a) Assume that the worker’s type is common knowledge, and compute the subgameperfect equilibrium. ANSWER: A worker will accept a wage iff it is at least as high as his outside option, and the employer will offer the outside option – as he still makes profit. That is, 15,000 for the good worker 8,000 for the bad. (b) Assume that the worker knows his type, but the employer does not. Employer believes that the worker is good with probability 1/4. Find the perfect Bayesian Nash equilibrium. ANSWER: Again a worker will accepts an offer iff his wage at least as high as his outside option. Hence if w ≥ 15, 000 the offer will be accepted by both types, yielding U (w) = (1/4) (.1) 200, 000 + (3/4) (.1) 100, 000 − w = 12, 500 − w < 0 as the profit for the employer. If 8, 000 ≤ w < 15, 000, then only the bad worker will accept the offer, yielding U (w) = (3/4) [(.1) 100, 000 − w] = (3/4) [10, 000 − w] as profit. If w < 0, no worker will accept the offer, and the employer will get 0. In that case, the employer will offer w = 8, 000, hiring the bad worker at his outside option. (c) Under the information structure in part (b), now consider the case that the employer offers a share s in the sales rather than the fixed wage w. Compute the perfect Bayesian Nash equilibrium. ANSWER: Again a worker will accept the share s iff his income is at least as high as his outside option. That is, a bad worker will accept s iff 100, 000s ≥ 8, 000 5
i.e., s ≥ sB =
8, 000 = 8%. 100, 000
A good worker will accept s iff s ≥ sG =
15, 000 = 7.5%. 200, 000
In that case, if s < sG no one will accept the offer, and the employer will get 0; if sG ≤ s < sB , the good worker will accept the offer and the employer will get (1/4) (10% − s) 200, 000 = 50, 000 (10% − s) , and if s ≥ sB , each type will accept the offer and the employer will get (10% − s) [(1/4) 200, 000 + (3/4) 100, 000] = 125, 000 (10% − s) . Since 125, 000 (10% − sB ) = 2%125, 000 = 2, 500 is larger than 50, 000 (10% − sG ) = 2.5%50, 000 = 1, 250, he will offer s = sB , hiring both types. 5. As in question 4, We have an employer and a worker, who will work as a salesman. Now the market might be good or bad. In expectation, if the market is good, the worker will make $200,000 worth of sales, and if the market is bad, he will make only $100,000 worth of sales. The employer gets 10% of the sales as profit. The employer offers a wage w. Then, the worker accepts or rejects the offer. If he accepts, he will be hired at wage w. If he rejects the offer, he will not be hired. In that case, the employer will get 0, the worker will get his outside option, which will pay $12,000. Assume that all players are risk-neutral. (a) Assume that whether the market is good or bad is common knowledge, and compute the subgame-perfect equilibrium. ANSWER: A worker will accept a wage iff it is at least as high as his outside option 12,000. If the market is good, the employer will offer the outside option w = 12, 000, and make 20, 000 − 12, 000 = 8, 000 profit. If the market is bad, the return 10,000 is lower than the worker’s outside option, and the worker will not be hired. (b) Assume that the employer knows whether the market is good or bad, but the worker does not. The worker believes that the market is good with probability 1/4. Find the perfect Bayesian Nash equilibrium. ANSWER: As in part (a). [We will have a separating equilibrium.] (c) Under the information structure in part (b), now consider the case that the employer offers a share s in the sales rather than the fixed wage w. Compute a perfect Bayesian Nash equilibrium. ANSWER: Note that, since the return is 10% independent of whether the market is good or bad, the employer will make positive profit iff s < 10%. Hence, except 6
for s = 10%, we must have a pooling equilibrium. Hence, at any s, the worker’s income is [(1/4) 200, 000 + (3/4) 100, 000] s = 125, 000s. This will be at least as high as his outside option iff s ≥ s∗ =
12, 000 = 9.6% < 10%. 125, 000
Hence an equilibrium: the worker will accept an offer s iff s ≥ s∗ , and the employer will offer s∗ . The worker’s beliefs at any offer s is that the market is good with probability 1/4. [Note that this is an inefficient equilibrium. When the market is bad, the gains from trade is less than the outside option.] There are other inefficient equilibria where there is no trade (i.e., worker is never hired). In any such equilibrium, worker take any high offer as a sign that the market is bad, and does not accept an offer s unless s ≥ 12, 000/100, 000 = 12%, and the employer offers less than 12%. When the market is good, in any such pure strategy equilibrium, he must in fact be offering less than s∗ . (why?) For instance, employer offers s = 0 independent of the market, and the worker accept s iff s > 12%.
7
14.12 Final Exam Grade Distribution Average = 73.82
Std. Dev = 20.74
120
100
60
40
20
rank
45
43
41
39
37
35
33
31
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25
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0 1
grade
80
Final Grade, Final Grade + Quiz 100
A+ 90 88
A
80 76 70
A-
60 B+ 52 50
B
45
B-
40 C 30 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 Series1
Series2
Final Grade (.25M1+.25M2+.4F+.1PS) 100
A+ 90 88
A
80 76 70
A-
60 B+ 52 50
B
45
B-
40 C 30 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 rank
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6
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5
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5
47145 Jdph Wkhru| Plgwhup LL 4424925333 Suri1 Pxkdphw \logl} Lqvwuxfwlrqv1 Wklv lv dq rshq errn h{dp> |rx fdq xvh dq| zulwwhq pdwhuldo1 \rx kdyh rqh krxu dqg 53 plqxwhv1 Hdfk txhvwlrq lv 66 srlqwv1 Jrrg oxfn$ 41 Frpsxwh wkh shuihfw Ed|hvldq htxloleulxp ri wkh iroorzlqj jdph1 1 {0.5} Nature
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4
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14.12 Solutions for Question 7 (or 8) Kenichi Amaya1 November 9, 2001 (a) Let’s denote firm 2’s by c2 , i.e., c2 ∈ {cL , cH }. Let p1 and p2 be each firm’s price (action). Then, each firm’s payoff functions can be written as p1 − c if p1 < p2 p1 −c π1 (p1 , p2 ) = if p1 = p2 , 2 0 if p1 > p2 p2 − c2 if p2 < p1 p2 −c2 π2 (p1 , p2 ; c2 ) = if p2 = p1 2 0 if p2 > p1 (b) A pure strategy profile {p∗1 , (p∗2L , p∗2H )} is a Bayesian Nash equilibrium if p∗1 = argmaxp1 θπ1 (p1 , p∗2L ) + (1 − θ)π1 (p1 , p∗2H ), p∗2L = argmaxp2 π2 (p∗1 , p2 ; cL ), p∗2H = argmaxp2 π2 (p∗1 , p2 ; cH ), (c) If there exists a pure strategy Bayesia Nash equilibrium, it must be true that p∗1 = min{p∗2L , p∗2H }. To see this, if p∗1 < min{p∗2L , p∗2H }, firm 1 is always selling to the whole market. If it increases the price slightly, it still sells the whole market. So it can increase the payoff. If p∗2k < p∗1 , where k is either L or H, firm 2 is selling to the whole market when its type is k. If it increases the price slightly, it still sells the whole market. So it can increase the payoff. Next, if there exists a pure strategy Bayesia Nash equilibrium, it must be true that p∗1 = min{p∗2L , p∗2H } = c. To see this, if p∗1 < c, firm 1 is selling positive amount and losing money. Therefore it’s better to charge a price higher than R and not to sell to get zero profit. On the other hand, if p∗1 < c, firm 1 can increase it’s profit by slightly reducing its price. Finally, p∗1 = min{p∗2L , p∗2H } = c can not be an equilibrium because firm 2 of type L can increase its profit by choosing a price slightly smaller than c. 1 E52-303,
[email protected], 253-3591
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5
14.12 Game Theory Fall 2000 Final Exam Solutions
December 14, 2001
1. (a) Given perfect competition among firms, wages will be w1 = 1,w2 = 2 and w3 = 3 for types t = 1, 2, 3 respectively. The incentive compatibility constraints give us the conditions that the education levels must fulfill in equilibrium: e∗2 ≥ 1 + e∗1 and e∗3 ≥ 2 + e∗1 for t = 1, e∗2 ≤ 2 + e∗1 and e∗3 ≥ 2 + e∗2 for t = 2, and e∗3 ≤ 3 + e∗2 and e∗3 ≤ 6 + e∗1 for t = 3. If we let, for example, e∗1 = 0, then e∗2 = 1 and e∗3 = 3. Beliefs are µ(t = 1|e0 ) = 1, where e0 is in [0, inf), µ(t = 2|e = 1) = 1 and µ(t = 3|e = 3) = 1. (b) Let w1 = 1 as before, since type 1 is separating. Given that types 2 and 3 are pooling, their wage is the expected marginal productivity or 1/2(2) + 1/2(3) = 5/2. Letting e∗1 = 0, the compatibility constraints provide the conditions that the education levels must fulfill in equilibrium: e∗ ≥ 3/2 and e∗ ≤ 3. For instance, we choose e∗ = 2. Beliefs are µ(t = 1|e 6= e∗ ) = 1, µ(t = 2|e∗ ) = 1/2 and µ(t = 3|e∗ ) = 1/2. 2. (a) Let the separating equilibrium be one where top plays R, bottom plays L. The beliefs are µ(top|R) = 1 and µ(bottom|R) = 1. Player 2, the receiver, plays up in the information set on the right-hand side, and down, on the information set on the left-hand side. Now it is easy to check that player 1’s types do not want to deviate: if top type plays L instead, she gets 0 as opposed to 3, while bottom type, if she plays R instead, gets 1 as opposed to 2. (b) Pooling on R: player 2 plays down when sees R since EU (up|R) = 0.4(1) + 0.6(0) = 0.4 < EU (down|R) = 0.4(0) + 0.6(1) = 0.6. Any beliefs about types given L actually support this equilibrium. (c) Mixed strategies equilibrium: Bottom mixes αR + (1 − α)L. Player 2 mixes 1/2up+1/2down when sees R. By applying Bayes’ rule and forcing it to be equal to 1/2, because this is the value that would make the bottom type want to mix, we have µ(top|R) = 0.4(1)/(0.4+0.6(α)) = 1/2. In particular, for player 2, EU (up|R) = P (top|R) and EU (down|R) = 1 − P (top|R). Setting these two equal, because we need player 2 to mix 1
after observing R in order for bottom type to want to mix, we obtain that P (top|R) = 1/2. From the Bayes’ rule equation we solve for α above, such that bottom type’s strategy is 2/3R + 1/3L. Top type always plays R. Player 2 plays 1/2up + 1/2down when sees R and plays down when sees L. Beliefs are µ(bottom|L) = 1 and µ(top|R) = 1/2. 3. Given no discounting, and the fact that the plaintiff gets to make the first offer, the game ends in the first day, with the plaintiff making an offer w and the defendant accepting the offer. In particular, taking into account the legal fees both parties must pay until court day, and the amount the defendant must pay then, which is certain and common knowledge, the plaintiff will offer 1, 110K and the defendant will accept all offers less than or equal to this amount, and reject all others. 4. One pooling equilibrium of this game would be for the firm to reject all offers w0 > 5 and accept all others, for both types of workers to offer w0 = 5, and to offer w1 = 5 in the following period if reached. Beliefs: µ(type = 0|w 6= 5) = 1. Workers do not want to deviate because that would lead to a rejection and a wage of 5 in the following period, which is what they can get now (so they are indifferent).
2
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