SOLUTION MANUAL t
Game Theory Basics 1st Edition
t t t t t
By Bernhard von Stengel. Chapters 1 - 12
t t t tt t t t
1
,TABLE tOF tCONTENTS t
1 t- tNim tand tCombinatorial tGames
2 t- tCongestion tGames
3 t- tGames tin tStrategic tForm
4 t- tGame tTrees twith tPerfect tInformation
5 t- tExpected tUtility
6 t- tMixed tEquilibrium
7 t- tBrouwer’s tFixed-Point tTheorem
8 t- tZero-Sum tGames
9 t- tGeometry tof tEquilibria tin tBimatrix tGames
10 t- tGame tTrees twith tImperfect tInformation
11 t- tBargaining
12 t- tCorrelated tEquilibrium
2
,Game tTheory tBasics
Solutions t to t Exercises
© t Bernhard tvon tStengel t2022
Solution tto tExercise t1.1
(a) Let t≤ tbe tdefined tby t(1.7). t To tshow tthat t≤ tis ttransitive, tconsider tx, ty, tz twith tx t ≤ ty tand ty t≤ tz.
tIf tx t= ty tthen tx t≤ tz, tand tif ty t= tz tthen talso tx t≤ tz. tSo tthe tonly tcase tleft tis tx t< ty tand ty t<
tz, twhich timplies tx t< tz tbecause t< tis ttransitive, tand thence tx t≤ tz.
Clearly, t≤ tis treflexive tbecause tx t= tx tand ttherefore tx t≤ tx.
≤ ≤ ≤
To tshow tthat t t t t tis tantisymmetric, tconsider tx tand ty twith tx t t t t ty tand ty t t t t tx. tIf twe thad
tx t≠ ty tthen tx t< ty tand ty t< tx, tand tby ttransitivity tx t< tx twhich tcontradicts t(1.38). tHence tx t=
ty, tas trequired. t This tshows tthat t≤ tis ta tpartial torder.
≤
Finally, twe tshow t(1.6), tso twe thave tto tshow tthat tx t< ty timplies tx t t ty tand tx t≠ ty tand tvice
tversa. tLet tx t< ty, twhich timplies tx ty tby t(1.7). tIf twe thad tx t= ty tthen tx t< tx, tcontradicting
t(1.38), tso twe talso thave tx t≠ ty. tConversely, tx≤t t ty tand tx t≠ ty timply tby t(1.7) tx t< ty tor tx t= ty
twhere tthe tsecond tcase tis texcluded, thence tx t< ty, tas trequired. ≤
(b) Consider ta tpartial torder tand≤tassume t(1.6) tas ta tdefinition tof t<. tTo tshow tthat t< tis ttransitive,
tsuppose tx t< ty, tthat tis, tx ty tand tx t≠ ty, tand ty t< tz, tthat tis, ty tz tand ty t≠ tz. tBecause t t t tis
ttransitive, tx t t t tz. tIf twe thad tx t= tz ≤ ≤ tx t= ty tby
tthen tx t t t t ty tand ty t t t t tx tand thence
≤ tof t t t t, twhich ≤
tantisymmetry tcontradicts tx t≠ ty, tso twe thave ≤ tx t t t tz tand
≤ tx t≠ tz, tthat tis,tx t<
tz tby t(1.6), tas trequired.
≤ ≤
≤
Also, t< tis tirreflexive, tbecause tx t< tx twould tby tdefinition tmean tx t t tx tand tx t≠ tx, tbut tthe
tlatter tis tnot ttrue.
Finally, twe tshow t(1.7), tso twe thave tto tshow tthat tx t ≤ ty timplies tx t< ty tor tx t= ty tand tvice
tversa, tgiven tthat t< tis tdefined tby t(1.6). tLet tx t≤ ty. tThen tif tx t= ty, twe tare tdone, totherwise tx
t≠ ty tand tthen tby tdefinition tx t< ty. tHence, tx t≤ ty timplies tx t< ty tor tx t= ty. tConversely,
tsuppose tx t < t y tor tx t= ty. t If tx t < t y tthen tx t ≤ ty tby t(1.6), tand tif tx t= ty tthen tx t≤ ty
tbecause t≤ tis treflexive. t This tcompletes tthe tproof.
Solution tto tExercise t1.2
(a) In tanalysing tthe tgames tof tthree tNim theaps twhere tone theap thas tsize tone, twe tfirst tlook tat
tsome texamples, tand tthen tuse tmathematical tinduction tto tprove twhat twe tconjecture tto tbe tthe
tlosing tpositions. tA tlosing tposition tis tone twhere tevery tmove tis tto ta twinning tposition, tbecause
tthen tthe topponent twill twin. t The tpoint tof tthis texercise tis tto tformulate ta tprecise tstatement tto
tbe tproved, tand tthen tto tprove tit.
First, tif tthere tare tonly ttwo theaps trecall tthat tthey tare tlosing tif tand tonly tif tthe theaps tare tof
tequal tsize. t If tthey tare tof tunequal tsize, tthen tthe twinning tmove tis tto treduce tthe tlarger theap
tso tthat tboth theaps thave tequal tsize.
3
, ≤ ≤
Consider tthree theaps tof tsizes t1, tm, tn, twhere t1 t t t t tm t t t t tn. tWe tobserve tthe tfollowing: t1,
t1, tm tis twinning, tby tmoving tto t1, t1, t0. tSimilarly, t1, tm, tm tis twinning, tby tmoving tto t0, tm, tm.
tNext, t1, t2, t3 tis tlosing t(observed tearlier tin tthe tlecture), tand thence t1, t2, tn tfor tn t4 tis
twinning. t1, t3, tn tis twinning tfor tany tn t3 tby tmoving tto t1, t3, t2. tFor t1, t4, t5, treducing tany
theap tproduces ta twinning tposition, tso tthis tis tlosing.
≥ ≥
+
The tgeneral tpattern tfor tthe tlosing tpositions tthus tseems tto tbe: t1, tm, tm t1, tfor teven tnumbers
tm. tThis tincludes talso tthe tcase tm t= t0, twhich twe tcan ttake tas tthe tbase tcase tfor tan tinduction.
tWe tnow tproceed tto tprove tthis tformally.
≤
First twe tshow tthat tif tthe tpositions tof tthe tform t1, tm, tn twith tm t t t t t tn tare tlosing twhen tm tis
teven tand tn t= tm t1, tthen tthese tare tthe tonly tlosing tpositions tbecause tany tother tposition t1,
tm, tn t with tm t t n t is+twinning. t Namely, tif tm t = tn t then ta twinning tmove tfrom t1, tm, tm tis tto
t0, tm, tm, tso twe tcan tassume≤ tm t< tn. t If tm tis teven tthen tn t> tm t t 1 t(otherwise twe twould tbe
tin tthe tposition t1, tm, tm t t 1) tand tso tthe twinning tmove tis tto t1, tm, tm t t 1. tIf tm tis todd
+
tthen tthe twinning tmove tis tto t1, tm, tm t1, tthe tsame tas tposition t1, tm t1, tm t(this twould talso tbe ta
twinning tmove tfrom t1, tm, tm tso tthere tthe+twinning tmove tis tnot tunique). +
Second, twe tshow tthat tany tmove tfrom t1, tm, tm t+ – −
t1 twith teven tm tis tto ta twinning tposition, tusing
tas tinductive thypothesis tthat t1, tmJ, tmJ t+ t1 tfor teven tmJ tand tmJ t< tm tis ta tlosing tposition.
tThe tmove tto t0, tm, tm t+ t1 tproduces ta twinning tposition twith tcounter-move tto t0, tm, tm. tA
tmove tto t1, tmJ, tm t+ t1 tfor tmJ t< tm tis tto ta twinning tposition twith tthe tcounter-move tto t1, tmJ, tmJ
t+ t1 tif tmJ tis teven tand tto t1, tmJ, tmJ t− t1 tif tmJ tis todd. tA tmove tto t1, tm, tm tis tto ta twinning
tposition twith tcounter-move tto t0, tm, tm. tA tmove tto t1, tm, tmJ twith t mJ t< t m tis talso tto ta twinning
tposition twith tthe tcounter-move tto t1, tmJ t− t1, tmJ tif t mJ tis todd, tand tto t1, tmJ t 1, tmJ tif tmJ tis
teven t(in twhich tcase tmJ t 1 t< tm tbecause tm tis teven). tThis tconcludes tthe tinduction tproof.
This tresult tis tin tagreement twith tthe ttheorem ton tNim theap tsizes trepresented tas tsums tof tpowers
tof t2: t 1 t t m t t n tis tlosing tif tand tonly tif, texcept tfor t20, tthe tpowers tof t2 tmaking tuptm tand tn
∗ t +∗ +∗ 0
+ tmust tbe tthe tsame tpowers tof t2, texcept
tcome tin tpairs. tSo tthese + tfor t1 t= t2 , twhich toccurs tin
tonly tm tor tn, twhere twe thave tassumed tthat tn tis tthe tlarger tnumber, tso t1 tappears tin tthe
trepresentation tof tn: t We thave tm t = t2a t t t t t t2b t t t t t t2c for ta t > t b t > t c t > t t t t t t
t t1,tso tm tis teven, tand, twith tthe tsame t a, tb, tc, t. t. t.,+ tn t+= t2a +tt · t t· t · t 2b t t t 2c · t· t·
t≥ + + + t · t· t · t + +
1 t= tm t t t t1. t Then
∗m1 t=
t + t∗ + t∗ ≡ t∗
t t t t t tm t t t t t n t t t t t t 0. t The tfollowing tis tan texample tusing tthe tbit trepresentation twhere
t12 t(which tdetermines tthe tbit tpattern t1100, twhich tof tcourse tdepends ton tm):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) We tuse t(a). tClearly, t1, t2, t3 tis tlosing tas tshown tin t(1.2), tand tbecause tthe tNim-sum tof tthe
tbinary trepresentations t01, t10, t11 tis t00. tExamples tshow tthat tany tother tposition tis twinning.
tThe tthree tnumbers tare tn, tn t 1, tn t t 2. tIf tn tis teven tthen treducing tthe theap tof tsize tn t2 tto
t1 tcreates tthe tposition tn, tn t 1, t1 twhich+tis tlosing+ tas tshown tin t(a). tIf tn tis todd, tthen tn t 1
+ +
tis teven tand tn t t t2 t= t n t t t1 t t t1 tso tby tthe tsame targument, ta twinning tmove tis tto treduce
tthe tNim theap tof tsize tn tto t1 t(which tonly tworks tif tn t> t1).
+ + ( t + t )
t+
4
Game Theory Basics 1st Edition
t t t t t
By Bernhard von Stengel. Chapters 1 - 12
t t t tt t t t
1
,TABLE tOF tCONTENTS t
1 t- tNim tand tCombinatorial tGames
2 t- tCongestion tGames
3 t- tGames tin tStrategic tForm
4 t- tGame tTrees twith tPerfect tInformation
5 t- tExpected tUtility
6 t- tMixed tEquilibrium
7 t- tBrouwer’s tFixed-Point tTheorem
8 t- tZero-Sum tGames
9 t- tGeometry tof tEquilibria tin tBimatrix tGames
10 t- tGame tTrees twith tImperfect tInformation
11 t- tBargaining
12 t- tCorrelated tEquilibrium
2
,Game tTheory tBasics
Solutions t to t Exercises
© t Bernhard tvon tStengel t2022
Solution tto tExercise t1.1
(a) Let t≤ tbe tdefined tby t(1.7). t To tshow tthat t≤ tis ttransitive, tconsider tx, ty, tz twith tx t ≤ ty tand ty t≤ tz.
tIf tx t= ty tthen tx t≤ tz, tand tif ty t= tz tthen talso tx t≤ tz. tSo tthe tonly tcase tleft tis tx t< ty tand ty t<
tz, twhich timplies tx t< tz tbecause t< tis ttransitive, tand thence tx t≤ tz.
Clearly, t≤ tis treflexive tbecause tx t= tx tand ttherefore tx t≤ tx.
≤ ≤ ≤
To tshow tthat t t t t tis tantisymmetric, tconsider tx tand ty twith tx t t t t ty tand ty t t t t tx. tIf twe thad
tx t≠ ty tthen tx t< ty tand ty t< tx, tand tby ttransitivity tx t< tx twhich tcontradicts t(1.38). tHence tx t=
ty, tas trequired. t This tshows tthat t≤ tis ta tpartial torder.
≤
Finally, twe tshow t(1.6), tso twe thave tto tshow tthat tx t< ty timplies tx t t ty tand tx t≠ ty tand tvice
tversa. tLet tx t< ty, twhich timplies tx ty tby t(1.7). tIf twe thad tx t= ty tthen tx t< tx, tcontradicting
t(1.38), tso twe talso thave tx t≠ ty. tConversely, tx≤t t ty tand tx t≠ ty timply tby t(1.7) tx t< ty tor tx t= ty
twhere tthe tsecond tcase tis texcluded, thence tx t< ty, tas trequired. ≤
(b) Consider ta tpartial torder tand≤tassume t(1.6) tas ta tdefinition tof t<. tTo tshow tthat t< tis ttransitive,
tsuppose tx t< ty, tthat tis, tx ty tand tx t≠ ty, tand ty t< tz, tthat tis, ty tz tand ty t≠ tz. tBecause t t t tis
ttransitive, tx t t t tz. tIf twe thad tx t= tz ≤ ≤ tx t= ty tby
tthen tx t t t t ty tand ty t t t t tx tand thence
≤ tof t t t t, twhich ≤
tantisymmetry tcontradicts tx t≠ ty, tso twe thave ≤ tx t t t tz tand
≤ tx t≠ tz, tthat tis,tx t<
tz tby t(1.6), tas trequired.
≤ ≤
≤
Also, t< tis tirreflexive, tbecause tx t< tx twould tby tdefinition tmean tx t t tx tand tx t≠ tx, tbut tthe
tlatter tis tnot ttrue.
Finally, twe tshow t(1.7), tso twe thave tto tshow tthat tx t ≤ ty timplies tx t< ty tor tx t= ty tand tvice
tversa, tgiven tthat t< tis tdefined tby t(1.6). tLet tx t≤ ty. tThen tif tx t= ty, twe tare tdone, totherwise tx
t≠ ty tand tthen tby tdefinition tx t< ty. tHence, tx t≤ ty timplies tx t< ty tor tx t= ty. tConversely,
tsuppose tx t < t y tor tx t= ty. t If tx t < t y tthen tx t ≤ ty tby t(1.6), tand tif tx t= ty tthen tx t≤ ty
tbecause t≤ tis treflexive. t This tcompletes tthe tproof.
Solution tto tExercise t1.2
(a) In tanalysing tthe tgames tof tthree tNim theaps twhere tone theap thas tsize tone, twe tfirst tlook tat
tsome texamples, tand tthen tuse tmathematical tinduction tto tprove twhat twe tconjecture tto tbe tthe
tlosing tpositions. tA tlosing tposition tis tone twhere tevery tmove tis tto ta twinning tposition, tbecause
tthen tthe topponent twill twin. t The tpoint tof tthis texercise tis tto tformulate ta tprecise tstatement tto
tbe tproved, tand tthen tto tprove tit.
First, tif tthere tare tonly ttwo theaps trecall tthat tthey tare tlosing tif tand tonly tif tthe theaps tare tof
tequal tsize. t If tthey tare tof tunequal tsize, tthen tthe twinning tmove tis tto treduce tthe tlarger theap
tso tthat tboth theaps thave tequal tsize.
3
, ≤ ≤
Consider tthree theaps tof tsizes t1, tm, tn, twhere t1 t t t t tm t t t t tn. tWe tobserve tthe tfollowing: t1,
t1, tm tis twinning, tby tmoving tto t1, t1, t0. tSimilarly, t1, tm, tm tis twinning, tby tmoving tto t0, tm, tm.
tNext, t1, t2, t3 tis tlosing t(observed tearlier tin tthe tlecture), tand thence t1, t2, tn tfor tn t4 tis
twinning. t1, t3, tn tis twinning tfor tany tn t3 tby tmoving tto t1, t3, t2. tFor t1, t4, t5, treducing tany
theap tproduces ta twinning tposition, tso tthis tis tlosing.
≥ ≥
+
The tgeneral tpattern tfor tthe tlosing tpositions tthus tseems tto tbe: t1, tm, tm t1, tfor teven tnumbers
tm. tThis tincludes talso tthe tcase tm t= t0, twhich twe tcan ttake tas tthe tbase tcase tfor tan tinduction.
tWe tnow tproceed tto tprove tthis tformally.
≤
First twe tshow tthat tif tthe tpositions tof tthe tform t1, tm, tn twith tm t t t t t tn tare tlosing twhen tm tis
teven tand tn t= tm t1, tthen tthese tare tthe tonly tlosing tpositions tbecause tany tother tposition t1,
tm, tn t with tm t t n t is+twinning. t Namely, tif tm t = tn t then ta twinning tmove tfrom t1, tm, tm tis tto
t0, tm, tm, tso twe tcan tassume≤ tm t< tn. t If tm tis teven tthen tn t> tm t t 1 t(otherwise twe twould tbe
tin tthe tposition t1, tm, tm t t 1) tand tso tthe twinning tmove tis tto t1, tm, tm t t 1. tIf tm tis todd
+
tthen tthe twinning tmove tis tto t1, tm, tm t1, tthe tsame tas tposition t1, tm t1, tm t(this twould talso tbe ta
twinning tmove tfrom t1, tm, tm tso tthere tthe+twinning tmove tis tnot tunique). +
Second, twe tshow tthat tany tmove tfrom t1, tm, tm t+ – −
t1 twith teven tm tis tto ta twinning tposition, tusing
tas tinductive thypothesis tthat t1, tmJ, tmJ t+ t1 tfor teven tmJ tand tmJ t< tm tis ta tlosing tposition.
tThe tmove tto t0, tm, tm t+ t1 tproduces ta twinning tposition twith tcounter-move tto t0, tm, tm. tA
tmove tto t1, tmJ, tm t+ t1 tfor tmJ t< tm tis tto ta twinning tposition twith tthe tcounter-move tto t1, tmJ, tmJ
t+ t1 tif tmJ tis teven tand tto t1, tmJ, tmJ t− t1 tif tmJ tis todd. tA tmove tto t1, tm, tm tis tto ta twinning
tposition twith tcounter-move tto t0, tm, tm. tA tmove tto t1, tm, tmJ twith t mJ t< t m tis talso tto ta twinning
tposition twith tthe tcounter-move tto t1, tmJ t− t1, tmJ tif t mJ tis todd, tand tto t1, tmJ t 1, tmJ tif tmJ tis
teven t(in twhich tcase tmJ t 1 t< tm tbecause tm tis teven). tThis tconcludes tthe tinduction tproof.
This tresult tis tin tagreement twith tthe ttheorem ton tNim theap tsizes trepresented tas tsums tof tpowers
tof t2: t 1 t t m t t n tis tlosing tif tand tonly tif, texcept tfor t20, tthe tpowers tof t2 tmaking tuptm tand tn
∗ t +∗ +∗ 0
+ tmust tbe tthe tsame tpowers tof t2, texcept
tcome tin tpairs. tSo tthese + tfor t1 t= t2 , twhich toccurs tin
tonly tm tor tn, twhere twe thave tassumed tthat tn tis tthe tlarger tnumber, tso t1 tappears tin tthe
trepresentation tof tn: t We thave tm t = t2a t t t t t t2b t t t t t t2c for ta t > t b t > t c t > t t t t t t
t t1,tso tm tis teven, tand, twith tthe tsame t a, tb, tc, t. t. t.,+ tn t+= t2a +tt · t t· t · t 2b t t t 2c · t· t·
t≥ + + + t · t· t · t + +
1 t= tm t t t t1. t Then
∗m1 t=
t + t∗ + t∗ ≡ t∗
t t t t t tm t t t t t n t t t t t t 0. t The tfollowing tis tan texample tusing tthe tbit trepresentation twhere
t12 t(which tdetermines tthe tbit tpattern t1100, twhich tof tcourse tdepends ton tm):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) We tuse t(a). tClearly, t1, t2, t3 tis tlosing tas tshown tin t(1.2), tand tbecause tthe tNim-sum tof tthe
tbinary trepresentations t01, t10, t11 tis t00. tExamples tshow tthat tany tother tposition tis twinning.
tThe tthree tnumbers tare tn, tn t 1, tn t t 2. tIf tn tis teven tthen treducing tthe theap tof tsize tn t2 tto
t1 tcreates tthe tposition tn, tn t 1, t1 twhich+tis tlosing+ tas tshown tin t(a). tIf tn tis todd, tthen tn t 1
+ +
tis teven tand tn t t t2 t= t n t t t1 t t t1 tso tby tthe tsame targument, ta twinning tmove tis tto treduce
tthe tNim theap tof tsize tn tto t1 t(which tonly tworks tif tn t> t1).
+ + ( t + t )
t+
4
