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SOLUTION MANUAL Game Theory Basics 1st Edition By Bernhard von Stengel. Chapters 1 - 12
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SOLUTION MANUAL Game Theory Basics 1st Edition By Bernhard von Stengel. Chapters 1 - 12
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SOLUTION MANUALGame Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS m m m
1 - Nim and Combinatorial Games
m m m m m
2 - Congestion Games
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3 - Games in Strategic Form
m m m m m
4 - Game Trees with Perfect Information
m m m m m m
5 - Expected Utility
m m m
6 - Mixed Equilibrium
m m m
7 - Brouwer’s Fixed-Point Theorem
m m m m
8 - Zero-Sum Games
m m m
9 - Geometry of Equilibria in Bimatrix Games
m m m m m m m
10 - Game Trees with Imperfect Information
m m m m m m
11 - Bargaining
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12 - Correlated Equilibrium
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2
,Game Theory Basics
m m
Solutions to Exercises
m m
©m BernhardmvonmStengelm2022
SolutionmtomExercisem1.1
(a) Letm≤mbemdefinedmbym(1.7).m Tomshowmthatm≤mismtransitive,mconsidermx,my,mzmwithmxm ≤mymandmym≤mz.mIf
mxm=mymthenmxm≤mz,mandmifmym=mzmthenmalsomxm≤mz.mSomthemonlymcasemleftmismxm<mymandm ym <m z,mwhic
hmimpliesmxm <m zmbecausem<mismtransitive,mandmhencemxm ≤mz.
Clearly,m≤mismreflexivembecausemxm=mxmandmthereforemxm≤mx.
Tomshowmthatmmmmm
≤ ismantisymmetric,mconsidermxmandmymwithmxmmmmmym≤andmymmmmmx.m≤Ifmwemhadmxm≠my
mthenmxm<mymandmym<mx,mandmbymtransitivitymxm<mxmwhichmcontradictsm(1.38).mHencemxm =m y,masmre
quired.m Thismshowsmthatm≤mismampartialmorder.
Finally,mwemshowm(1.6),msomwemhavemtomshowmthatmxm<mymimpliesmxmmmymand≤ mxm≠mymandmvicemversa
.mLetmxm<my,mwhichmimpliesmxmymbym(1.7).mIfmwem≤ hadmxm=mymthenmxm<mx,mcontradictingm(1.38),msomw
emalsomhavemxm≠my.m Conversely,mxmmm ymandmxm≠mymimplymbym(1.7)mxm <≤m ymorm xm =m ymwheremthemsecond
mcasemismexcluded,mhencem xm <m y,masmrequired.
(b) Considermampartialmordermand ≤massumem(1.6)masmamdefinitionmofm<.mTomshowmthatm<mismtransitiv
e,msupposemxm<my,mthatmis,mxmymandmxm≠my,≤mandmym<mz,mthatmis,mymzmandmym≠mz.mBecausemmmmismtransit
≤
ive,mxmmmmz.≤
mIfmwemhadmxm=mzmthenmxmmmmmymandmymmmmmxmandmhencemxm=mymbymantisymmetrymofmmmm,m
≤ ≤ ≤
whichmcontradictsm xm ≠m y,msomwemhavem xmmmmzmandm xm ≠m z,mthatmis,mxm <m zmbym(1.6),masmrequired.
≤ ≤
Also,m<mismirreflexive,mbecausemxm<mxmwouldmbymdefinitionmmeanmxmmmxmandm≤xm≠mx,mbutmthemlatter
mismnotmtrue.
Finally,mwemshowm(1.7),msomwemhavemtomshowmthatmxm ≤mymimpliesmxm<mymormxm=mymandmvicemversa,
mgivenmthatm<mismdefinedmbym(1.6).mLetmxm≤my.mThenmifmxm=my,mwemaremdone,motherwisemxm≠mymand
mthenmbymdefinitionmxm<my.mHence,mxm≤mymimpliesmxm<mymormxm=my.mConversely,msupposemxm <m ymor
mxm=my.m Ifmxm <m ymthenmxm ≤mymbym(1.6),mandmifmxm=mymthenmxm ≤m ymbecausem ≤mismreflexive.m Thismco
mpletesmthemproof.
SolutionmtomExercisem1.2
(a) Inm analysingm them gamesm ofm threem Nimm heapsm wherem onem heapm hasm sizem one,m wem firstm lookmatmso
memexamples,mandmthenmusemmathematicalminductionmtomprovemwhatmwemconjecturemtombemthemlosin
gmpositions.mAmlosingmpositionmismonemwheremeverymmovemismtomamwinningmposition,mbecausemth
enmthemopponentmwillmwin.m Thempointmofmthismexercisemismtomformulatemamprecisemstatementmtomb
emproved,mandmthenmtomprovemit.
First,mifmtheremaremonlymtwomheapsmrecallmthatmtheymaremlosingmifmandmonlymifmthemheapsmaremof
mequalmsize.m Ifmtheymaremofmunequalmsize,mthenmthemwinningmmovemismtomreducemthem largermheap
msomthatmbothmheapsmhavemequalmsize.
3
, Considermthreemheapsmofmsizesm1,mm,mn,mwherem1mmmmmm≤mmmmmn. ≤mWemobservemthemfollowing:m1,m1,m
mmismwinning,mbymmovingmtom1,m1,m0.mSimilarly,m1,mm,mmmismwinning,mbymmovingmtom0,mm,mm.mN
ext,m1,m2,m3mismlosingm(observedmearlierminmthemlecture),mandmhencem1,m2,mnmformnm4mismwinning.
m1,m3,mnmismwinningmformanymnm3mbymmovingmtom1,m3,m2.mForm1,m4,m5,mreducingmanymheapmprodu
≥ ≥
cesmamwinningmposition,msomthismismlosing.
Themgeneralmpatternmformthemlosingmpositionsmthusmseemsmtombe:m1,mm,mmm1,mform+evenmnumber
smm.m Thismincludesmalsomthemcasemmm=m0,mwhichmwemcanmtakemasmthembasemcasemformanminduction
.m Wemnowmproceedmtomprovemthismformally.
Firstmwemshowmthatmifmthempositionsmofmthemformm1,mm,mnmwithmmmmmmmmnm≤aremlosingmwhenmmmisme
venmandmnm=mmm1,mthenm+ thesemaremthemonlymlosingmpositionsmbecausemanymothermpositionm1,mm,m
nm withmmm m nm ismwinning.m ≤ Namely,mifmmm =mnm thenmamwinningmmovemfromm1,mm,mmmismtom0,mm,mm,m
somwemcanmassumemmm<mn.m Ifmmmismevenmthenmnm>mmm m 1m(otherwisemwemwouldmbeminmthempositionm
+
1,mm,mmm m 1)mandmsomthemwinningmmovemismtom1,mm,mmm m 1.mIfmmmismoddmthenmthemwinningmmovemi
+ +
smtom1,mm,mmm1,mthemsamemasmpositionm1,mmm1,mmm(thismwouldm alsom bem am winningm movem fromm 1,mm,m
mm som therem them winningm movem ism notm unique). – −
Second,mwemshowmthatmanymmovemfromm1,mm,mmm+m1mwithmevenmmmismtomamwinningmposition,musingm
asminductivemhypothesismthatm1,mmJ,mmJm+m1mformevenmmJmandmmJm<mmmismamlosingmposition.mThem
movemtom0,mm,mmm+m1mproducesmamwinningmpositionmwithmcounter-
movemtom0,mm,mm.mAmmovemtom1,mmJ,mmm+m1mformmJm<mmmismtomamwinningmpositionmwithmthemcounter-
movemtom1,mmJ,mmJm+m1mifmmJmismevenmandmtom1,mmJ,mmJm−m1mifmmJmismodd.mAmmovemtom1,mm,mmmismt
omamwinningmpositionmwithmcounter-
movemtom0,mm,mm.mAmmovemtom1,mm,mmJmwithm mJm<m mmismalsomtomamwinningmpositionmwithmthemcount
er-
movemtom1,mmJm−m1,mmJmifm mJmismodd,mandmtom1,mmJm 1,mmJmifmmJmismevenm(inmwhichmcasemmJm 1m<mmm
+ mconcludesmtheminductionmproof.
becausemmmismeven).mThis +
ThismresultmisminmagreementmwithmthemtheoremmonmNimmheapmsizesmrepresentedmasmsumsmofmpowers
0
mofm2:m 1m m mm m nmismlosingmifmandmonlymif,mexceptmform2 ,mthempowersmofm2mmakingmupmmmandmnmco
∗m +∗ +∗
meminmpairs.mSomthesemmustmbemthemsamempowersmofm2,mexceptmform1m=m20,mwhichmoccursminmonlym
mmormn,mwheremwemhavemassumedmthatmnmismthemlargermnumber,msom1mappearsminm them representatio
nm ofm n:m Wem havem mm =m 2ammmmmm2bmmmmmm2c
+ + +m ·m ·m ·
form am >m bm >m cm >mmmmmmmm 1,mso
a b·mm ·+
mm ·mm ≥c + +m ·m ·m ·m + +
m mm ism even,m and,m withm them samem a,mb, mc, m.m. m.,m nm =m 2 2 2 1m =m mmmmm 1.m Then
m m m
∗1mmmmmmm
+m ∗ m+mmmmm
m∗ nmmmmmm 0.m Them followingm ism anmexamplem usingm them bitmrepresentationmwhere
mm=m12m(which
≡m∗ mdeterminesmthembitmpatternm1100,mwhichmofmcoursemdependsmonmm):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) Wemusem(a).mClearly,m1,m2,m3mismlosingmasmshownminm(1.2),mandmbecausemthemNim-
summofmthembinarymrepresentationsm01,m10,m11mism00.mExamplesmshowmthatmanymothermpositio
nmismwinning.mThemthreemnumbersmaremn,mnm+1,mnm m+2.mIfmnmismevenmthenmreducingmthemheapmofmsize
mnm2mtom1mcreatesmthempositionmn,mnm 1,m1mwhichmismlosingmasmshownminm(a).mIfmnmismodd,mthenm
+ +
nm 1mismevenmandmnmmm2m=m nmmm1mmm1msombymthemsamemargument,mamwinningmmovemismtomreduce
+ +
mthemNimmheapmofmsizemnmtom1m(whichmonlymworksmifmnm >m1).
(m +m )m+
4
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