7/11/23,m1:42mA SolutionsmManualmformStatisticalmInferen
M ce
Solutionsm Manualm for
StatisticalmInference,mSecondmEdition
GeorgemCasellam Rogerm L.m Berger
UniversitymofmFlori NorthmCarolinamStatemUniversitym Da
da marismSantana
Universitym ofm Florida
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0-2 Solutionsm Manualm form Statisticalm Inference
“WhenmImhearmyoumgivemyourmreasons,”mImremarked,m“themthingmalwaysmappearsmtom
memtombemsom ridiculouslymsimplemthatmImcouldmeasilymdomitmmyself,mthoughmatmeach
msuccessivem instancem ofmyourm reasoningmImamm baffledmuntilm you mexplainm yourmproces
s.”
Dr.m Watsonm tom Sherlockm Holmes
AmScandalminmBohemia
0.1 Description
Thismsolutionsmmanualmcontainsmsolutionsmformallmoddmnumberedmproblemsmplusmamlargemnu
mbermofm solutionsmformevenmnumberedmproblems.mOfmthem624mexercisesminmStatisticalmInfe
rence,mSecondmEdition,m thismmanualmgivesmsolutionsmform484m(78%)mofmthem.mTheremisman
mobtusempatternmasmtomwhichmsolutionsm weremincludedminmthismmanual.mWemassembledma
llmofmthemsolutionsmthatmwemhadmfrommthemfirstmedition,m andmfilledminmsomthatmallmodd-
numberedmproblemsmweremdone.mInmthempassagemfrommthemfirstmtomthem secondmedition,m
problemsmweremshuffledmwithmnomattentionmpaidmtomnumberingm(hencemnomattentionm paid
mtomminimizemthemnewmeffort),mbutmrathermwemtriedmtomputmthemproblemsminmlogicalmord
er.
Ammajormchangemfrommthemfirstmeditionmismthemusemofmthemcomputer,mbothmsymbolicall
ymthroughm MathematicatmmandmnumericallymusingmR.mSomemsolutionsmaremgivenmasmcodem
inmeithermofmthesemlan-
m guages.mMathematicatmmcanmbempurchasedmfrommWolframmResearch,mandmRmismamfreemd
ownloadmfromm http://www.r-project.org/.
Heremism amdetailedmlistingm ofm themsolutionsmincluded.
Chapter Numberm ofm Exercises Numberm ofm Solutions Missing
1 55 51 26,m30,m36,m42
2 40 37 34,m38,m40
3 50 42 4,m6,m10,m20,m30,m32,m34,m36
4 65 52 8,m14,m22,m28,m36,m40
48,m50,m52,m56,m58, m60,m62
5 69 46 2,m4,m12,m14,m26,m28
allmevenmproblemsmfromm36m−
m68
6 43 35 8,m16,m26,m28,m34,m36,m38,m42
7 66 52 4,m14,m16,m28,m30,m32,m34,
36,m42,m54,m58,m60, m62,m64
8 58 51 36,m40,m46,m48,m52, m56,m58
9 58 41 2,m8,m10,m20,m22,m24, m26,m28,m30
32,m38,m40,m42,m44, m50,m54,m56
10 48 26 allmevenmproblemsm exceptm4m an
dm32
11 41 35 4,m20,m22,m24,m26,m40
12 31 16 allmevenmproblems
0.2 Acknowledgement
Manym peoplem contributedm tom them assemblym ofm thism solutionsm manual.m Wem againm than
km allm ofm thosem whomcontributedmsolutionsmtomthemfirstmeditionm–
mmanymproblemsmhavemcarriedmovermintomthemsecondm edition.mMoreover,mthroughoutmthe
myearsmamnumbermofmpeoplemhavembeenminmconstantmtouchmwithmus,m contributingmtombot
hmthempresentationsmandmsolutions.mWemapologizeminmadvancemformthosemwemforgetmtom m
ention,mandmwemespeciallymthankmJaymBeder,mYongmSungmJoo,mMichaelmPerlman,mRobmStr
awderman,m andmTommWehrly.mThankmyoumallmformyourmhelp.
And,masmwemsaidmthemfirstmtimemaround,malthoughmwemhavembenefitedmgreatlymfrommthemassi
stancemand
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ACKNOWLEDGEMENT 0-3
commentsmofmothersminmthemassemblymofmthismmanual,mwemaremresponsiblemformitsmultima
temcorrectness.m Tomthismend,mwemhavemtriedmourmbestmbut,masmamwisemmanmoncemsaid,m
“Youmpaysmyourmmoneymandmyoum takesmyourmchances.”
GeorgemCasellam R
ogermL.mBergerm
DamarismSantana
December,m 2001
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Chapterm1
Probabilitym Theory
“Ifmanymlittlemproblemmcomesmyourmway,mImshallmbemhappy,mifmImcan,mtomgivemyoum
amhintmormtwomasm tomitsmsolution.”
Sherlockm Holmes
ThemAdventurem ofmthemThreem Students
1.1 a.m Eachm samplempointm describesm them resultm ofm them tossm (HmormT)m form eachmofm th
em fourm tosses.m So,m formexamplemTHTTmdenotesmTmonm1st,mHmonm2nd,mTmonm3rd
mandmTmonm4th.mTheremarem24m=m16m suchmsamplempoints.
b. Themnumberm ofm damagedm leavesm ism am nonnegativem integer.m Som wem mightm usem Sm =m{0
,m1,m2,m.m.m.}.
c. Wemmightmobservemfractions mofmanmhour.mSomwemmightmusemSm=m{tm:mtm≥m0
},mthatmis,mthemhalfm infinitemintervalm[0,m∞).
d. Supposemwemweighmthemratsminmounces.mThemweightmmustmbemgreatermthanmzeromsomwe
mmightmuse
Sm=m(0,m∞).mIfmwemknowmnom10-day-
oldmratmweighsmmoremthanm100moz.,mwemcouldmusemSm =m(0,m100].
e. Ifmnmismthemnumbermofmitemsminmthemshipment,mthenmSm=m{0/n,m1/n,m.m.m.m,m1}.
1.2 Formeachmofmthesemequalities,myoummustmshowmcontainmentminmbothmdirections.
a. xm∈mA\Bm ⇔mxm∈mAm andm xm∈/m Bm ⇔mxm∈mAm andm xm∈/m Am∩mBm ⇔mxm∈mA\(Am
∩mB).m Also,m xm∈mAm and
xm∈/mBm⇔mxm∈mAmandmxm∈mBcm⇔mxm∈mAm∩mBc.
b. Supposem xm ∈m B.m Thenm eitherm xm ∈m Am orm xm ∈m Ac.m Ifm xm ∈m A,m thenm xm
∈m Bm ∩mA,m and,m hencem xm∈m(Bm∩mA)m∪m(Bm∩mAc).mThusmBm ⊂m(Bm∩mA)m∪m
(Bm∩mAc).m Nowm supposem xm ∈m(Bm∩mA)m∪m(Bm∩mAc).m Thenm eitherm xm ∈m(Bm
∩mA)m orm xm ∈m(Bm∩mAc).m Ifm xm ∈m(Bm∩mA),m thenm xm ∈mB.m Ifm xm ∈m(Bm∩
mAc),m then mxm∈mB.mThus m(Bm∩mA) m∪m(Bm∩mAc) m⊂mB.mSincemthemcontainmen
tmgoesmbothmways,mwemhavem Bm=m(Bm∩mA)m∪m(Bm∩mAc).m(Note,mammoremstra
ightforwardmargumentmformthismpartmsimplymusesm themDistributivemLaw mtomstat
emthatm(Bm∩mA)m∪m(Bm∩mAc)m=mBm∩ m(Am∪mAc)m=mBm∩mSm =mB.)
c. Similarm tom partm a).
d. Fromm partm b).
Am∪mBm=mAm∪m[(Bm∩mA)m∪m(Bm∩mAc)]m =m Am∪m(Bm ∩mA)m∪mAm∪m(Bm ∩mAc)m =
m Am∪m[Am∪m(Bm∩mAc)] m =
Am∪m(Bm∩mAc).
1.3 a.m xm∈mAm∪mBm ⇔m xm∈mAm orm xm∈mBm ⇔m xm∈mBm∪mA
xm∈mAm∩mBm ⇔m xm∈mAm andm xm∈mBm ⇔m xm∈mBm∩mA.
b.m xm∈mAm∪m(Bm∪mC)m⇔m xm∈mAm orm xm∈mBm∪mCm ⇔m xm∈mAm∪mBm orm x
m∈mCm ⇔m xm∈m(Am∪mB)m∪mC.m (Itm canm similarlym bem shownm thatm Am∪m
(Bm∪mC)m=m(Am∪mC)m∪mB.)
xm∈mAm∩m(Bm∩mC)m⇔m xm∈mAm andm xm∈mBm andm xm∈mCm ⇔m xm ∈m(Am∩mB)m∩mC.
c.m xm∈m(Am∪mB)cm⇔m xm∈/mAmormxm∈/mBm ⇔m xm∈mAcm andmxm∈mBcm ⇔m xm∈mAcm
∩mBc
xm∈m(Am∩mB)cm⇔m xm∈/mAm∩mBm⇔m xm∈/mAmandmxm∈/mBm ⇔m xm∈mAcm orm xm∈
mBcm ⇔m xm∈mAcm∪mBc.
1.4 a.m “AmormBmormboth”mismA∪B.mFrommTheoremm1.2.9bmwemhavemPm(A∪B)m=mPm(A)+Pm
(B)−Pm(A∩B).
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M ce
Solutionsm Manualm for
StatisticalmInference,mSecondmEdition
GeorgemCasellam Rogerm L.m Berger
UniversitymofmFlori NorthmCarolinamStatemUniversitym Da
da marismSantana
Universitym ofm Florida
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0-2 Solutionsm Manualm form Statisticalm Inference
“WhenmImhearmyoumgivemyourmreasons,”mImremarked,m“themthingmalwaysmappearsmtom
memtombemsom ridiculouslymsimplemthatmImcouldmeasilymdomitmmyself,mthoughmatmeach
msuccessivem instancem ofmyourm reasoningmImamm baffledmuntilm you mexplainm yourmproces
s.”
Dr.m Watsonm tom Sherlockm Holmes
AmScandalminmBohemia
0.1 Description
Thismsolutionsmmanualmcontainsmsolutionsmformallmoddmnumberedmproblemsmplusmamlargemnu
mbermofm solutionsmformevenmnumberedmproblems.mOfmthem624mexercisesminmStatisticalmInfe
rence,mSecondmEdition,m thismmanualmgivesmsolutionsmform484m(78%)mofmthem.mTheremisman
mobtusempatternmasmtomwhichmsolutionsm weremincludedminmthismmanual.mWemassembledma
llmofmthemsolutionsmthatmwemhadmfrommthemfirstmedition,m andmfilledminmsomthatmallmodd-
numberedmproblemsmweremdone.mInmthempassagemfrommthemfirstmtomthem secondmedition,m
problemsmweremshuffledmwithmnomattentionmpaidmtomnumberingm(hencemnomattentionm paid
mtomminimizemthemnewmeffort),mbutmrathermwemtriedmtomputmthemproblemsminmlogicalmord
er.
Ammajormchangemfrommthemfirstmeditionmismthemusemofmthemcomputer,mbothmsymbolicall
ymthroughm MathematicatmmandmnumericallymusingmR.mSomemsolutionsmaremgivenmasmcodem
inmeithermofmthesemlan-
m guages.mMathematicatmmcanmbempurchasedmfrommWolframmResearch,mandmRmismamfreemd
ownloadmfromm http://www.r-project.org/.
Heremism amdetailedmlistingm ofm themsolutionsmincluded.
Chapter Numberm ofm Exercises Numberm ofm Solutions Missing
1 55 51 26,m30,m36,m42
2 40 37 34,m38,m40
3 50 42 4,m6,m10,m20,m30,m32,m34,m36
4 65 52 8,m14,m22,m28,m36,m40
48,m50,m52,m56,m58, m60,m62
5 69 46 2,m4,m12,m14,m26,m28
allmevenmproblemsmfromm36m−
m68
6 43 35 8,m16,m26,m28,m34,m36,m38,m42
7 66 52 4,m14,m16,m28,m30,m32,m34,
36,m42,m54,m58,m60, m62,m64
8 58 51 36,m40,m46,m48,m52, m56,m58
9 58 41 2,m8,m10,m20,m22,m24, m26,m28,m30
32,m38,m40,m42,m44, m50,m54,m56
10 48 26 allmevenmproblemsm exceptm4m an
dm32
11 41 35 4,m20,m22,m24,m26,m40
12 31 16 allmevenmproblems
0.2 Acknowledgement
Manym peoplem contributedm tom them assemblym ofm thism solutionsm manual.m Wem againm than
km allm ofm thosem whomcontributedmsolutionsmtomthemfirstmeditionm–
mmanymproblemsmhavemcarriedmovermintomthemsecondm edition.mMoreover,mthroughoutmthe
myearsmamnumbermofmpeoplemhavembeenminmconstantmtouchmwithmus,m contributingmtombot
hmthempresentationsmandmsolutions.mWemapologizeminmadvancemformthosemwemforgetmtom m
ention,mandmwemespeciallymthankmJaymBeder,mYongmSungmJoo,mMichaelmPerlman,mRobmStr
awderman,m andmTommWehrly.mThankmyoumallmformyourmhelp.
And,masmwemsaidmthemfirstmtimemaround,malthoughmwemhavembenefitedmgreatlymfrommthemassi
stancemand
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ACKNOWLEDGEMENT 0-3
commentsmofmothersminmthemassemblymofmthismmanual,mwemaremresponsiblemformitsmultima
temcorrectness.m Tomthismend,mwemhavemtriedmourmbestmbut,masmamwisemmanmoncemsaid,m
“Youmpaysmyourmmoneymandmyoum takesmyourmchances.”
GeorgemCasellam R
ogermL.mBergerm
DamarismSantana
December,m 2001
about:blank 3/197
, 7/11/23,m1:42mA SolutionsmManualmformStatisticalmInferen
M ce
Chapterm1
Probabilitym Theory
“Ifmanymlittlemproblemmcomesmyourmway,mImshallmbemhappy,mifmImcan,mtomgivemyoum
amhintmormtwomasm tomitsmsolution.”
Sherlockm Holmes
ThemAdventurem ofmthemThreem Students
1.1 a.m Eachm samplempointm describesm them resultm ofm them tossm (HmormT)m form eachmofm th
em fourm tosses.m So,m formexamplemTHTTmdenotesmTmonm1st,mHmonm2nd,mTmonm3rd
mandmTmonm4th.mTheremarem24m=m16m suchmsamplempoints.
b. Themnumberm ofm damagedm leavesm ism am nonnegativem integer.m Som wem mightm usem Sm =m{0
,m1,m2,m.m.m.}.
c. Wemmightmobservemfractions mofmanmhour.mSomwemmightmusemSm=m{tm:mtm≥m0
},mthatmis,mthemhalfm infinitemintervalm[0,m∞).
d. Supposemwemweighmthemratsminmounces.mThemweightmmustmbemgreatermthanmzeromsomwe
mmightmuse
Sm=m(0,m∞).mIfmwemknowmnom10-day-
oldmratmweighsmmoremthanm100moz.,mwemcouldmusemSm =m(0,m100].
e. Ifmnmismthemnumbermofmitemsminmthemshipment,mthenmSm=m{0/n,m1/n,m.m.m.m,m1}.
1.2 Formeachmofmthesemequalities,myoummustmshowmcontainmentminmbothmdirections.
a. xm∈mA\Bm ⇔mxm∈mAm andm xm∈/m Bm ⇔mxm∈mAm andm xm∈/m Am∩mBm ⇔mxm∈mA\(Am
∩mB).m Also,m xm∈mAm and
xm∈/mBm⇔mxm∈mAmandmxm∈mBcm⇔mxm∈mAm∩mBc.
b. Supposem xm ∈m B.m Thenm eitherm xm ∈m Am orm xm ∈m Ac.m Ifm xm ∈m A,m thenm xm
∈m Bm ∩mA,m and,m hencem xm∈m(Bm∩mA)m∪m(Bm∩mAc).mThusmBm ⊂m(Bm∩mA)m∪m
(Bm∩mAc).m Nowm supposem xm ∈m(Bm∩mA)m∪m(Bm∩mAc).m Thenm eitherm xm ∈m(Bm
∩mA)m orm xm ∈m(Bm∩mAc).m Ifm xm ∈m(Bm∩mA),m thenm xm ∈mB.m Ifm xm ∈m(Bm∩
mAc),m then mxm∈mB.mThus m(Bm∩mA) m∪m(Bm∩mAc) m⊂mB.mSincemthemcontainmen
tmgoesmbothmways,mwemhavem Bm=m(Bm∩mA)m∪m(Bm∩mAc).m(Note,mammoremstra
ightforwardmargumentmformthismpartmsimplymusesm themDistributivemLaw mtomstat
emthatm(Bm∩mA)m∪m(Bm∩mAc)m=mBm∩ m(Am∪mAc)m=mBm∩mSm =mB.)
c. Similarm tom partm a).
d. Fromm partm b).
Am∪mBm=mAm∪m[(Bm∩mA)m∪m(Bm∩mAc)]m =m Am∪m(Bm ∩mA)m∪mAm∪m(Bm ∩mAc)m =
m Am∪m[Am∪m(Bm∩mAc)] m =
Am∪m(Bm∩mAc).
1.3 a.m xm∈mAm∪mBm ⇔m xm∈mAm orm xm∈mBm ⇔m xm∈mBm∪mA
xm∈mAm∩mBm ⇔m xm∈mAm andm xm∈mBm ⇔m xm∈mBm∩mA.
b.m xm∈mAm∪m(Bm∪mC)m⇔m xm∈mAm orm xm∈mBm∪mCm ⇔m xm∈mAm∪mBm orm x
m∈mCm ⇔m xm∈m(Am∪mB)m∪mC.m (Itm canm similarlym bem shownm thatm Am∪m
(Bm∪mC)m=m(Am∪mC)m∪mB.)
xm∈mAm∩m(Bm∩mC)m⇔m xm∈mAm andm xm∈mBm andm xm∈mCm ⇔m xm ∈m(Am∩mB)m∩mC.
c.m xm∈m(Am∪mB)cm⇔m xm∈/mAmormxm∈/mBm ⇔m xm∈mAcm andmxm∈mBcm ⇔m xm∈mAcm
∩mBc
xm∈m(Am∩mB)cm⇔m xm∈/mAm∩mBm⇔m xm∈/mAmandmxm∈/mBm ⇔m xm∈mAcm orm xm∈
mBcm ⇔m xm∈mAcm∪mBc.
1.4 a.m “AmormBmormboth”mismA∪B.mFrommTheoremm1.2.9bmwemhavemPm(A∪B)m=mPm(A)+Pm
(B)−Pm(A∩B).
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