Statistical Inference （统计推断）第二版 (George Casella Roger L. Berger 著) 机械工业出版社 课后答案 196页

'课后答案网：www.hackshp.cn课后答案网您最真诚的朋友www.hackshp.cn网团队竭诚为学生服务，免费提供各门课后答案，不用积分，甚至不用注册，旨在为广大学生提供自主学习的平台！课后答案网：www.hackshp.cn视频教程网：www.efanjy.comPPT课件网：www.ppthouse.com课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSolutionsManualforStatisticalInference,SecondEditionGeorgeCasellaRogerL.BergerUniversityofFloridaNorthCarolinaStateUniversityDamarisSantanaUniversityofFlorida课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn0-2SolutionsManualforStatisticalInference“WhenIhearyougiveyourreasons,”Iremarked,“thethingalwaysappearstometobesoridiculouslysimplethatIcouldeasilydoitmyself,thoughateachsuccessiveinstanceofyourreasoningIambaffleduntilyouexplainyourprocess.”Dr.WatsontoSherlockHolmesAScandalinBohemia0.1DescriptionThissolutionsmanualcontainssolutionsforalloddnumberedproblemsplusalargenumberofsolutionsforevennumberedproblems.Ofthe624exercisesinStatisticalInference,SecondEdition,thismanualgivessolutionsfor484(78%)ofthem.Thereisanobtusepatternastowhichsolutionswereincludedinthismanual.Weassembledallofthesolutionsthatwehadfromthefirstedition,andfilledinsothatallodd-numberedproblemsweredone.Inthepassagefromthefirsttothesecondedition,problemswereshuffledwithnoattentionpaidtonumbering(hencenoattentionpaidtominimizetheneweffort),butratherwetriedtoputtheproblemsinlogicalorder.Amajorchangefromthefirsteditionistheuseofthecomputer,bothsymbolicallythroughMathematicatmandnumericallyusingR.Somesolutionsaregivenascodeineitheroftheselan-guages.MathematicatmcanbepurchasedfromWolframResearch,andRisafreedownloadfromhttp://www.r-project.org/.Hereisadetailedlistingofthesolutionsincluded.ChapterNumberofExercisesNumberofSolutionsMissing1555126,30,36,422403734,38,40350424,6,10,20,30,32,34,36465528,14,22,28,36,4048,50,52,56,58,60,62569462,4,12,14,26,28allevenproblemsfrom36−68643358,16,26,28,34,36,38,42766524,14,16,28,30,32,34,36,42,54,58,60,62,648585136,40,46,48,52,56,58958412,8,10,20,22,24,26,28,30课后答案网32,38,40,42,44,50,54,56104826allevenproblemsexcept4and321141354,20,22,24,26,401231www.hackshp.cn16allevenproblems0.2AcknowledgementManypeoplecontributedtotheassemblyofthissolutionsmanual.Weagainthankallofthosewhocontributedsolutionstothefirstedition–manyproblemshavecarriedoverintothesecondedition.Moreover,throughouttheyearsanumberofpeoplehavebeeninconstanttouchwithus,contributingtoboththepresentationsandsolutions.Weapologizeinadvanceforthoseweforgettomention,andweespeciallythankJayBeder,YongSungJoo,MichaelPerlman,RobStrawderman,andTomWehrly.Thankyouallforyourhelp.And,aswesaidthefirsttimearound,althoughwehavebenefitedgreatlyfromtheassistanceand若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnACKNOWLEDGEMENT0-3commentsofothersintheassemblyofthismanual,weareresponsibleforitsultimatecorrectness.Tothisend,wehavetriedourbestbut,asawisemanoncesaid,“Youpaysyourmoneyandyoutakesyourchances.”GeorgeCasellaRogerL.BergerDamarisSantanaDecember,2001课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter1ProbabilityTheory“Ifanylittleproblemcomesyourway,Ishallbehappy,ifIcan,togiveyouahintortwoastoitssolution.”SherlockHolmesTheAdventureoftheThreeStudents1.1a.Eachsamplepointdescribestheresultofthetoss(HorT)foreachofthefourtosses.So,forexampleTHTTdenotesTon1st,Hon2nd,Ton3rdandTon4th.Thereare24=16suchsamplepoints.b.Thenumberofdamagedleavesisanonnegativeinteger.SowemightuseS={0,1,2,...}.c.Wemightobservefractionsofanhour.SowemightuseS={t:t≥0},thatis,thehalfinfiniteinterval[0,∞).d.Supposeweweightheratsinounces.TheweightmustbegreaterthanzerosowemightuseS=(0,∞).Ifweknowno10-day-oldratweighsmorethan100oz.,wecoulduseS=(0,100].e.Ifnisthenumberofitemsintheshipment,thenS={0/n,1/n,...,1}.1.2Foreachoftheseequalities,youmustshowcontainmentinbothdirections.a.x∈AB⇔x∈Aandx/∈B⇔x∈Aandx/∈A∩B⇔x∈A(A∩B).Also,x∈Aandx/∈B⇔x∈Aandx∈Bc⇔x∈A∩Bc.b.Supposex∈B.Theneitherx∈Aorx∈Ac.Ifx∈A,thenx∈B∩A,and,hencex∈(B∩A)∪(B∩Ac).ThusB⊂(B∩A)∪(B∩Ac).Nowsupposex∈(B∩A)∪(B∩Ac).Theneitherx∈(B∩A)orx∈(B∩Ac).Ifx∈(B∩A),thenx∈B.Ifx∈(B∩Ac),thenx∈B.Thus(B∩A)∪(B∩Ac)⊂B.Sincethecontainmentgoesbothways,wehaveB=(B∩A)∪(B∩Ac).(Note,amorestraightforwardargumentforthispartsimplyusestheDistributiveLawtostatethat(课后答案网B∩A)∪(B∩Ac)=B∩(A∪Ac)=B∩S=B.)c.Similartoparta).d.Frompartb).A∪B=A∪[(B∩Awww.hackshp.cn)∪(B∩Ac)]=A∪(B∩A)∪A∪(B∩Ac)=A∪[A∪(B∩Ac)]=A∪(B∩Ac).1.3a.x∈A∪B⇔x∈Aorx∈B⇔x∈B∪Ax∈A∩B⇔x∈Aandx∈B⇔x∈B∩A.b.x∈A∪(B∪C)⇔x∈Aorx∈B∪C⇔x∈A∪Borx∈C⇔x∈(A∪B)∪C.(ItcansimilarlybeshownthatA∪(B∪C)=(A∪C)∪B.)x∈A∩(B∩C)⇔x∈Aandx∈Bandx∈C⇔x∈(A∩B)∩C.c.x∈(A∪B)c⇔x/∈Aorx/∈B⇔x∈Acandx∈Bc⇔x∈Ac∩Bcx∈(A∩B)c⇔x/∈A∩B⇔x/∈Aandx/∈B⇔x∈Acorx∈Bc⇔x∈Ac∪Bc.1.4a.“AorBorboth”isA∪B.FromTheorem1.2.9bwehaveP(A∪B)=P(A)+P(B)−P(A∩B).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1-2SolutionsManualforStatisticalInferenceb.“AorBbutnotboth”is(A∩Bc)∪(B∩Ac).ThuswehaveP((A∩Bc)∪(B∩Ac))=P(A∩Bc)+P(B∩Ac)(disjointunion)=[P(A)−P(A∩B)]+[P(B)−P(A∩B)](Theorem1.2.9a)=P(A)+P(B)−2P(A∩B).c.“AtleastoneofAorB”isA∪B.Sowegetthesameanswerasina).d.“AtmostoneofAorB”is(A∩B)c,andP((A∩B)c)=1−P(A∩B).1.5a.A∩B∩C={aU.S.birthresultsinidenticaltwinsthatarefemale}b.P(A∩B∩C)=1×1×190321.6p0=(1−u)(1−w),p1=u(1−w)+w(1−u),p2=uw,p0=p2⇒u+w=1p1=p2⇒uw=1/3.Thesetwoequationsimplyu(1−u)=1/3,whichhasnosolutionintherealnumbers.Thus,theprobabilityassignmentisnotlegitimate.1.7a.(21−πrifi=0hAiP(scoringipoints)=πr2(6−i)2−(5−i)2A52ifi=1,...,5.b.P(scoringipoints∩boardishit)P(scoringipoints|boardishit)=P(boardishit)πr2P(boardishit)=Aπr2(6−i)2−(5−i)2P(scoringipoints∩boardishit)=i=1,...,5.A52Therefore,(6−i)2−(5−i)2P(scoringipoints|boardishit)=i=1,...,5课后答案网52whichisexactlytheprobabilitydistributionofExample1.2.7.1.8a.P(scoringexactlyipoints)=P(insidecirclei)−P(insidecirclei+1).Circleihasradius(6−i)r/5,sowww.hackshp.cnπ(6−i)2r2π((6−(i+1)))2r2(6−i)2−(5−i)2P(sscoringexactlyipoints)=−=.52πr252πr252b.Expandingthesquaresinparta)wefindP(scoringexactlyipoints)=11−2i,whichis25decreasingini.c.LetP(i)=11−2i.Sincei≤5,P(i)≥0foralli.P(S)=P(hittingthedartboard)=1by25definition.Lastly,P(i∪j)=areaofiring+areaofjring=P(i)+P(j).1.9a.Supposex∈(∪A)c,bythedefinitionofcomplementx6∈∪A,thatisx6∈Aforallαααααα∈Γ.Thereforex∈Acforallα∈Γ.Thusx∈∩Acand,bythedefinitionofintersectionαααx∈Acforallα∈Γ.Bythedefinitionofcomplementx6∈Aforallα∈Γ.Thereforeααx6∈∪A.Thusx∈(∪A)c.αααα若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition1-3b.Supposex∈(∩A)c,bythedefinitionofcomplementx6∈(∩A).Thereforex6∈Aforαααααsomeα∈Γ.Thereforex∈Acforsomeα∈Γ.Thusx∈∪Acand,bythedefinitionofαααunion,x∈Acforsomeα∈Γ.Thereforex6∈Aforsomeα∈Γ.Thereforex6∈∩A.Thusααααx∈(∩A)c.αα1.10ForA1,...,An!c!c[nnn[n(i)A=Ac(ii)A=Aciiiii=1i=1i=1i=1Proofof(i):Ifx∈(∪A)c,thenx/∈∪A.Thatimpliesx/∈Aforanyi,sox∈Acforeveryiiiiiandx∈∩Ai.Proofof(ii):Ifx∈(∩A)c,thenx/∈∩A.Thatimpliesx∈Acforsomei,sox∈∪Ac.iiii1.11WemustverifyeachofthethreepropertiesinDefinition1.2.1.a.(1)Theemptyset∅∈{∅,S}.Thus∅∈B.(2)∅c=S∈BandSc=∅∈B.(3)∅∪S=S∈B.b.(1)Theemptyset∅isasubsetofanyset,inparticular,∅⊂S.Thus∅∈B.(2)IfA∈B,thenA⊂S.Bythedefinitionofcomplementation,AcisalsoasubsetofS,and,hence,Ac∈B.(3)IfA,A,...∈B,then,foreachi,A⊂S.Bythedefinitionofunion,∪A⊂S.12iiHence,∪Ai∈B.c.LetB1andB2bethetwosigmaalgebras.(1)∅∈B1and∅∈B2sinceB1andB2aresigmaalgebras.Thus∅∈B1∩B2.(2)IfA∈B1∩B2,thenA∈B1andA∈B2.SinceBandBarebothsigmaalgebraAc∈BandAc∈B.ThereforeAc∈B∩B.(3)If121212A1,A2,...∈B1∩B2,thenA1,A2,...∈B1andA1,A2,...∈B2.Therefore,sinceB1andB2arebothsigmaalgebra,∪∞A∈Band∪∞A∈B.Thus∪∞A∈B∩B.i=1i1i=1i2i=1i121.12Firstwrite!![∞[n[∞PAi=PAi∪Aii=1i=1i=n+1!![n[∞=PAi+PAi(Aisaredisjoint)i=1i=n+1!Xn[∞=P(Ai)+PAi(finiteadditivity)课后答案网i=1i=n+1S∞NowdefineBk=i=kAi.NotethatBk+1⊂BkandBk→φask→∞.(Otherwisethesumoftheprobabilitieswouldbeinfinite.)Thus∞!www.hackshp.cn∞!"n#∞[[XXPAi=limPAi=limP(Ai)+P(Bn+1)=P(Ai).n→∞n→∞i=1i=1i=1i=11.13IfAandBaredisjoint,P(A∪B)=P(A)+P(B)=1+3=13,whichisimpossible.More3412generally,ifAandBaredisjoint,thenA⊂BcandP(A)≤P(Bc).ButhereP(A)>P(Bc),soAandBcannotbedisjoint.1.14IfS={s1,...,sn},thenanysubsetofScanbeconstructedbyeitherincludingorexcludings,foreachi.Thusthereare2npossiblechoices.i1.15Proofbyinduction.Theprooffork=2isgivenafterTheorem1.2.14.Assumetruefork,thatis,theentirejobcanbedoneinn1×n2×···×nkways.Fork+1,thek+1thtaskcanbedoneinnk+1ways,andforeachoneofthesewayswecancompletethejobbyperforming若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1-4SolutionsManualforStatisticalInferencetheremainingktasks.Thusforeachofthenk+1wehaven1×n2×···×nkwaysofcom-pletingthejobbytheinductionhypothesis.Thus,thenumberofwayswecandothejobis(1×(n1×n2×···×nk))+···+(1×(n1×n2×···×nk))=n1×n2×···×nk×nk+1.|{z}nk+1terms1.16a)263.b)263+262.c)264+263+262.
n1.17Thereare=n(n−1)/2piecesonwhichthetwonumbersdonotmatch.(Choose2outof2nnumberswithoutreplacement.)Therearenpiecesonwhichthetwonumbersmatch.Sothetotalnumberofdifferentpiecesisn+n(n−1)/2=n(n+1)/2.(n)2n!(n−1)(n−1)!1.18Theprobabilityisnn=2nn−2.Therearemanywaystoobtainthis.Hereisone.Thedenominatorisnnbecausethisisthenumberofwaystoplacenballsinncells.Thenumeratoristhenumberofwaysofplacingtheballssuchthatexactlyonecellisempty.Therearenwaystospecifytheemptycell.Therearen−1waysofchoosingthecellwithtwoballs.Thereare
nwaysofpickingthe2ballstogointothiscell.Andthereare(n−2)!waysofplacingthe2remainingn−2ballsintothen−2cells,oneballineachcell.Theproductoftheseisthe

nnnumeratorn(n−1)(n−2)!=n!.22
61.19a.=15.4b.Thinkofthenvariablesasnbins.Differentiatingwithrespecttooneofthevariablesisequivalenttoputtingaballinthebin.Thustherearerunlabeledballstobeplacedinn
n+r−1unlabeledbins,andtherearewaystodothis.r1.20Asamplepointspecifiesonwhichday(1through7)eachofthe12callshappens.Thusthereare712equallylikelysamplepoints.Thereareseveraldifferentwaysthatthecallsmightbeassignedsothatthereisatleastonecalleachday.Theremightbe6callsonedayand1calleachoftheotherdays.Denotethisby6111111.Thenumberofsamplepointswiththispattern

1212is76!.Thereare7waystospecifythedaywith6calls.Therearetospecifywhichof66the12callsareonthisday.Andthereare6!waysofassigningtheremaining6callstotheremaining6days.Wewillnowcountanotherpattern.Theremightbe4callsononeday,2callsoneachoftwodays,and1calloneachoftheremainingfourdays.Denotethisby4221111.



12686Thenumberofsamplepointswiththispatternis74!.(7waystopickdaywith4

4222
1268calls,topickthecallsforthatday,topicktwodayswithtwocalls,waystopick4
226twocallsforlowerednumberedday,waystopickthetwocallsforhighernumberedday,24!waystoorderremaining4calls.)Hereisalistofallthepossibilitiesandthecountsofthesamplepointsforeachone.课后答案网patternnumberofsamplepoints
12611111176!=4,656,9606

1275211111765!=83,825,2805

2

12686422111174!=523,908,000www.hackshp.cn4
2
221284311111765!=139,708,800
4
3

71296332111154!=698,544,0002
3
3
2

126975322211173!=1,397,088,000
3
33
2
2

7121086422222112!=314,344,8005222223,162,075,840Theprobabilityisthetotalnumberofsamplepointsdividedby712,whichis3,162,075,840≈712.2285.(n)22r
1.21Theprobabilityis2r.Thereare2nwaysofchoosing2rshoesfromatotalof2nshoes.(2n)2r
2r2nThusthereareequallylikelysamplepoints.Thenumeratoristhenumberofsamplepoints2r
nforwhichtherewillbenomatchingpair.Therearewaysofchoosing2rdifferentshoes2r若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition1-5styles.Therearetwowaysofchoosingwithinagivenshoestyle(leftshoeorrightshoe),which
gives22rwaysofarrangingeachoneofthenarrays.Theproductofthisisthenumerator
2rn22r.2r(31)(29)(31)(30)···(31)336335···3161.22a)1515151515b)366365336.(366)(366)180301.23XnP(samenumberofheads)=P(1sttossesx,2ndtossesx)x=0n"xn−x#2nn2Xn111Xn==.x224xx=0x=01.24a.X∞P(Awins)=P(Awinsonithtoss)i=124X∞2i+1111111=+++···==2/3.222222i=024P∞2ipb.P(Awins)=p+(1−p)p+(1−p)p+···=i=0p(1−p)=1−(1−p)2.2c.dp=p>0.Thustheprobabilityisincreasinginp,andtheminimumdp1−(1−p)2[1−(1−p)2]2pisatzero.UsingL’Hˆopital’srulewefindlimp→01−(1−p)2=1/2.1.25EnumeratingthesamplespacegivesS0={(B,B),(B,G),(G,B),(G,G)},witheachoutcomeequallylikely.ThusP(atleastoneboy)=3/4andP(bothareboys)=1/4,thereforeP(bothareboys|atleastoneboy)=1/3.Anambiguitymayariseiforderisnotacknowledged,thespaceisS0={(B,B),(B,G),(G,G)},witheachoutcomeequallylikely.1.27a.Fornoddtheproofisstraightforward.Thereareanevennumberoftermsinthesum

nn(0,1,···,n),andand,whichareequal,haveoppositesigns.Thus,allpairscancelkn−kandthesumiszero.Ifniseven,usethefollowingidentity,whichisthebasisofPascal’s


triangle:For课后答案网k>0,n=n−1+n−1.Then,fornevenkkk−1XnnX−1knnknn(−1)www.hackshp.cn=+(−1)+k0knk=0k=1nX−1nnkn−1n−1=++(−1)+0nkk−1k=1nnn−1n−1=+−−=0.0n0n−1

nn−1b.Usethefactthatfork>0,k=ntowritekk−1XnXnnX−1nn−1n−1n−1k=n=n=n2.kk−1jk=1k=1j=0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1-6SolutionsManualforStatisticalInferencePnk+1n
Pnk+1n−1
Pn−1jn−1
c.(−1)k=(−1)=n(−1)=0fromparta).k=1kk=1k−1j=0j1.28Theaverageofthetwointegralsis[(nlogn−n)+((n+1)log(n+1)−n)]/2=[nlogn+(n+1)log(n+1)]/2−n≈(n+1/2)logn−n.Letdn=logn!−[(n+1/2)logn−n],andwewanttoshowthatlimn→∞mdn=c,aconstant.Thiswouldcompletetheproblem,sincethedesiredlimitistheexponentialofthisone.Thisisaccomplishedinanindirectway,byworkingwithdifferences,whichavoidsdealingwiththefactorial.Notethat11dn−dn+1=n+log1+−1.2nDifferentiationwillshowthat((n+1))log((1+1))isincreasinginn,andhasminimum2nvalue(3/2)log2=1.04atn=1.Thusdn−dn+1>0.NextrecalltheTaylorexpansionoflog(1+x)=x−x2/2+x3/3−x4/4+···.Thefirstthreetermsprovideanupperboundonlog(1+x),astheremainingadjacentpairsarenegative.Hence11111101.12mThereforetheoutcomewithaveragex1+x2+···+xnisthemostlikely.n√b.Stirling’sapproximationisthat,asn→∞,n!≈2πnn+(1/2)e−n,andthus√!√n!2nπn!en2πnn+(1/2)e−nen=√=√=1.nnennn2nπnn2nπc.Sincewearedrawingwithreplacementfromtheset{x1,...,xn},theprobabilityofchoosinganyxis1.Thereforetheprobabilityofobtaininganorderedsampleofsizenwithoutxiniis(1−1)n.Toprovethatlim(1−1)n=e−1,calculatethelimitofthelog.Thatisnn→∞n
log1−11limnlog1−=limn.n→∞nn→∞1/nL’Hˆopital’sruleshowsthatthelimitis−1,establishingtheresult.SeealsoLemma2.3.14.1.32Thisismosteasilyseenbydoingeachpossibility.LetP(i)=probabilitythatthecandidatehiredontheithtrialisbest.Then111P(1)=,P(2)=,...,P(i)=,...,P(N)=1.NN−1N−i+11.33UsingBayesrule.05×1P(CB|M)P(M)P(M|CB)==2=.9524.P(CB|M)P(M)+P(CB|F)P(F).05×1+.0025×1221.34a.P(BrownHair)=课后答案网P(BrownHair|Litter1)P(Litter1)+P(BrownHair|Litter2)P(Litter2)213119=+=.3www.hackshp.cn25230b.UseBayesTheorem

21P(BH|L1)P(L1)10P(Litter1|BrownHair)==32=.P(BH|L1)P(L1)+P(BH|L2)P(L21919301.35ClearlyP(·|B)≥0,andP(S|B)=1.IfA1,A2,...aredisjoint,then∞!SS[P(∞A∩B)P(∞(A∩B))i=1ii=1iPAiB==P(B)P(B)i=1P∞X∞i=1P(Ai∩B)==P(Ai|B).P(B)i=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1-8SolutionsManualforStatisticalInference1.37a.UsingthesameeventsA,B,CandWasinExample1.3.4,wehaveP(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)111γ+1=γ+0+1=.3333P(A∩W)γ/3γThus,P(A|W)===where,P(W)(γ+1)/3γ+1γ=1ifγ=1γ+132γ<1ifγ<1γ+132γ>1ifγ>1.γ+132b.ByExercise1.35,P(·|W)isaprobabilityfunction.A,BandCareapartition.SoP(A|W)+P(B|W)+P(C|W)=1.But,P(B|W)=0.Thus,P(A|W)+P(C|W)=1.SinceP(A|W)=1/3,P(C|W)=2/3.(Thiscouldbecalculateddirectly,asinExample1.3.4.)SoifAcanswapfateswithC,hischanceofsurvivalbecomes2/3.1.38a.P(A)=P(A∩B)+P(A∩Bc)fromTheorem1.2.11a.But(A∩Bc)⊂BcandP(Bc)=1−P(B)=0.SoP(A∩Bc)=0,andP(A)=P(A∩B).Thus,P(A∩B)P(A)P(A|B)===P(A)P(B)1.b.A⊂BimpliesA∩B=A.Thus,P(A∩B)P(A)P(B|A)===1.P(A)P(A)Andalso,P(A∩B)P(A)P(A|B)==.P(B)P(B)c.IfAandBaremutuallyexclusive,thenP(A∪B)=P(A)+P(B)andA∩(A∪B)=A.Thus,课后答案网P(A∩(A∪B))P(A)P(A|A∪B)==.P(A∪B)P(A)+P(B)d.P(A∩B∩C)=P(Awww.hackshp.cn∩(B∩C))=P(A|B∩C)P(B∩C)=P(A|B∩C)P(B|C)P(C).1.39a.SupposeAandBaremutuallyexclusive.ThenA∩B=∅andP(A∩B)=0.IfAandBareindependent,then0=P(A∩B)=P(A)P(B).ButthiscannotbesinceP(A)>0andP(B)>0.ThusAandBcannotbeindependent.b.IfAandBareindependentandbothhavepositiveprobability,then00,soF(x)isincreasing.dx2π1+x2b.SeeExample1.5.5.−x−xd−x−xc.lime−e=0,lime−e=1,e−e=e−xe−e>0.x→−∞x→∞dxd.lim(1−e−x)=0,lim(1−e−x)=1,d(1−e−x)=e−x>0.x→−∞x→∞dx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition1-11−ye.lim1−=0,lim+1−=1,d(1−)=(1−)e>0andd(+1−)>y→−∞1+e−yy→∞1+e−ydx1+e−y(1+e−y)2dx1+e−y0,F(y)iscontinuousexceptony=0wherelim(+1−)=F(0).ThusisF(y)rightYy↓01+e−yYcontinuous.1.48IfF(·)isacdf,F(x)=P(X≤x).Hencelimx→∞P(X≤x)=0andlimx→−∞P(X≤x)=1.F(x)isnondecreasingsincetheset{x:X≤x}isnondecreasinginx.Lastly,asx↓x0,P(X≤x)→P(X≤x0),soF(·)isright-continuous.(ThisismerelyaconsequenceofdefiningF(x)with“≤”.)1.49Foreveryt,FX(t)≤FY(t).ThuswehaveP(X>t)=1−P(X≤t)=1−FX(t)≥1−FY(t)=1−P(Y≤t)=P(Y>t).Andforsomet∗,F(t∗)t∗)=1−P(X≤t∗)=1−F(t∗)>1−F(t∗)=1−P(Y≤t∗)=P(Y>t∗).XY1.50Proofbyinduction.Forn=2X21−t2tk−1=1+t=.1−tk=1Pnk−11−tnAssumetrueforn,thisist=.Thenforn+1k=11−tnX+1Xn1−tn1−tn+tn(1−t)1−tn+1tk−1=tk−1+tn=+tn==,1−t1−t1−tk=1k=1wherethesecondinequalityfollowsfromtheinductionhypothesis.1.51ThiskindofrandomvariableiscalledhypergeometricinChapter3.Theprobabilitiesareobtainedbycountingarguments,asfollows.xfX(x)=P(X=x)

.
525300≈.4616044

.
525301≈.4196134

.
525302≈.1095224

.
525303≈.0091314

.
525304≈.0002课后答案网404Thecdfisastepfunctionwithjumpsatx=0,1,2,3and4.1.52Thefunctiong(·)isclearlypositive.Also,Z∞Z∞f(x)1−F(x)www.hackshp.cng(x)dx=dx=0=1.x0x01−F(x0)1−F(x0)1.53a.limF(y)=lim0=0andlimF(y)=lim1−1=1.Fory≤1,y→−∞Yy→−∞y→∞Yy→∞y2F(y)=0isconstant.Fory>1,dF(y)=2/y3>0,soFisincreasing.Thusforally,YdyYYFYisnondecreasing.ThereforeFYisacdf.2/y3ify>1b.Thepdfisf(y)=dF(y)=YdyY0ify≤1.c.FZ(z)=P(Z≤z)=P(10(Y−1)≤z)=P(Y≤(z/10)+1)=FY((z/10)+1).Thus,(0ifz≤0FZ(z)=11−[(z/10)+1]2ifz>0.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn1-12SolutionsManualforStatisticalInferenceRπ/21.54a.sinxdx=1.Thus,c=1/1=1.0b.R∞e−|x|dx=R0exdx+R∞e−xdx=1+1=2.Thus,c=1/2.−∞−∞01.55Z31−t/1.5−2P(V≤5)=P(T<3)=edt=1−e.01.5Forv≥6,Zvv21P(V≤v)=P(2T≤v)=PT≤=e−t/1.5dt=1−e−v/3.201.5Therefore,(0−∞x)dx00Z∞Z∞=fX(y)dydx0xZ∞Zy课后答案网=dxfX(y)dy00Z∞=yfX(y)dy=EX,www.hackshp.cn0wherethelastequalityfollowsfromchangingtheorderofintegration.2.15AssumewithoutlossofgeneralitythatX≤Y.ThenX∨Y=YandX∧Y=X.ThusX+Y=(X∧Y)+(X∨Y).TakingexpectationsE[X+Y]=E[(X∧Y)+(X∨Y)]=E(X∧Y)+E(X∨Y).ThereforeE(X∨Y)=EX+EY−E(X∧Y).2.16FromExercise2.14,Z∞−λt−µt∞ET=ae−λt+(1−a)e−µtdt=−ae−(1−a)e=a+1−a.0λµλµ0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn2-6SolutionsManualforStatisticalInferenceRm23set11
1/32.17a.3xdx=m=⇒m==.794.022b.Thefunctionissymmetricaboutzero,thereforem=0aslongastheintegralisfinite.Z∞∞11dx=1tan−1(x)=1π+π=1.π−∞1+x2ππ22−∞ThisistheCauchypdf.R∞RaR∞2.18E|X−a|=|x−a|f(x)dx=−(x−a)f(x)dx+(x−a)f(x)dx.Then,−∞−∞aZaZ∞dsetE|X−a|=f(x)dx−f(x)dx=0.da−∞aThesolutiontothisequationisa=median.Thisisaminimumsinced2/da2E|X−a|=2f(a)>0.2.19Z∞Z∞d2d2d2E(X−a)=(x−a)fX(x)dx=(x−a)fX(x)dxdada−∞−∞daZ∞Z∞Z∞=−2(x−a)fX(x)dx=−2xfX(x)dx−afX(x)dx−∞−∞−∞=−2[EX−a].ThereforeifdE(X−a)2=0then−2[EX−a]=0whichimpliesthatEX=a.IfEX=athendadE(X−a)2=−2[EX−a]=−2[a−a]=0.EX=aisaminimumsinced2/da2E(X−a)2=da2>0.TheassumptionsthatareneededaretheoneslistedinTheorem2.4.3.2.20FromExample1.5.4,ifX=numberofchildrenuntilthefirstdaughter,thenP(X=k)=(1−p)k−1p,wherep=probabilityofadaughter.ThusXisageometricrandomvariable,and"#X∞X∞ddX∞EX=k(1−p)k−1p=p−(1−p)k=−p(1−p)k−1dpdpk=1k=1k=0d11=−p−1=.dpppTherefore,ifp=课后答案网1,theexpectednumberofchildrenistwo.22.21Sinceg(x)ismonotoneZ∞Z∞Z∞Eg(X)=g(xwww.hackshp.cn)f(x)dx=yf(g−1(y))dg−1(y)dy=yf(y)dy=EY,XXY−∞−∞dy−∞wherethesecondequalityfollowsfromthechangeofvariabley=g(x),x=g−1(y)anddx=dg−1(y)dy.dy222.22a.Usingintegrationbypartswithu=xanddv=xe−x/βweobtainthatZ∞2Z∞2−x2/β22β−x2/β2xedx=edx.020Theintegralcanbeevaluatedusingtheargumentonpages104-105(see3.3.14)orbytrans-22R∞−x2/β2√formingtoagammakernel(usey=−λ/β).Therefore,edx=πβ/2andhence0thefunctionintegratesto1.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition2-7√b.EX=2β/πEX2=3β2/2VarX=β23−4.2π2.23a.UseTheorem2.1.8withA={0},A=(−1,0)andA=(0,1).Theng(x)=x2onA01211andg(x)=x2onA.Then221−1/2fY(y)=y,00.Z∞Z0Z∞M(0+)=e(0+)xf(x)dx=exf(x)dx+exf(x)dxXXXX−∞−∞0Z∞Z0Z∞=e(−x)f(−x)dx+e(−x)f(−x)dx=e−xf(x)dxXXX0−∞−∞Z∞=e(0−)xf(x)dx=M(0−).XX−∞2.26a.Therearemanyexamples;herearethree.Thestandardnormalpdf(Example2.1.9)is课后答案网symmetricabouta=0because(0−)2=(0+)2.TheCauchypdf(Example2.2.4)issymmetricabouta=0because(0−)2=(0+)2.Theuniform(0,1)pdf(Example2.1.4)issymmetricaboutawww.hackshp.cn=1/2because1if0<<1f((1/2)+)=f((1/2)−)=12.0if≤<∞2b.Z∞Z∞f(x)dx=f(a+)d(changevariable,=x−a)a0Z∞=f(a−)d(f(a+)=f(a−)forall>0)0Za=f(x)dx.(changevariable,x=a−)−∞若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn2-8SolutionsManualforStatisticalInferenceSinceZZZa∞∞f(x)dx+f(x)dx=f(x)dx=1,−∞a−∞itmustbethatZZa∞f(x)dx=f(x)dx=1/2.−∞aTherefore,aisamedian.c.Z∞EX−a=E(X−a)=(x−a)f(x)dx−∞ZaZ∞=(x−a)f(x)dx+(x−a)f(x)dx−∞aZ∞Z∞=(−)f(a−)d+f(a+)d00Withachangeofvariable,=a−xinthefirstintegral,and=x−ainthesecondintegralweobtainthatEX−a=E(X−a)Z∞Z∞=−f(a−)d+f(a−)d(f(a+)=f(a−)forall>0)00=0.(twointegralsaresame)Therefore,EX=a.d.Ifa>>0,f(a−)=e−(a−)>e−(a+)=f(a+).Therefore,f(x)isnotsymmetricabouta>0.If−bfor>0.Sincebisthemodethenf(b)>f(b+)≥f(b+2)whichimpliesthatf(a−)>f(a)≥f(a+)whichcontradictthefactthef(x)issymmetric.Thusaisthemode.Forthecasewhenthemodeisnotunique,theremustexistaninterval(x1,x2)suchthatf(x)hasthesamevalueinthewholeinterval,i.e,f(x)isflatinthisintervalandforallb∈(x1,x2),bisamode.Letassumethata6∈(x1,x2),thusaisnotamode.Letalsoassumewithoutlossofgeneralitythata=(b+)>b.Sincebisamodeanda=(b+)6∈(x1,x2)thenf(b)>f(b+)≥f(b+2)whichcontradictthefactthef(x)issymmetric.Thusa∈(x1,x2)andisamode.d.f(x)isdecreasingforx≥0,withf(0)>f(x)>f(y)forall00,thereisaconstantcsuchthatZ∞Z∞1tx(logx)2/21∞eedx≥cdx=clogx|=∞.xxkkkHenceMx(t)doesnotexist.2.37a.ThegraphlooksverysimilartoFigure2.3.2exceptthatf1issymmetricaround0(sinceitisstandardnormal).b.Thefunctionslookliket2/2–itisimpossibletoseeanydifference.c.ThemgfoffiseK1(t).ThemgfoffiseK2(t).12d.Makethetransformationy=extogetthedensitiesinExample2.3.10.Rx2.39a.de−λtdt=e−λx.Verifydx0Zxx课后答案网de−λtdt=d−1e−λt=d−1e−λx+1=e−λx.dx0dxλdxλλ0b.dR∞e−λtdt=R∞de−λtdt=R∞−te−λtdt=−Γ(2)=−1.Verifydλ00dλ0λ2λ2www.hackshp.cnZ∞d−λtd11edt==−.dλ0dλλλ2R1c.d1dx=−1.Verifydttx2t2Z1!d11d1d11dx=−=−1+=−.dttx2dtxdttt2tR∞R∞R∞∞d.d1dx=d1dx=2(x−t)−3dx=−(x−t)−2=1.Verifydt1(x−t)21dt(x−t)21(1−t)21dZ∞dh∞id11−2−1(x−t)dx=−(x−t)==.dt1dt1dt1−t(1−t)2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter3CommonFamiliesofDistributions3.1ThepmfofXisf(x)=1,x=N,N+1,...,N.ThenN1−N0+1001!XN111XN1NX0−1EX=x=x−xN1−N0+1N1−N0+1x=N0x=1x=11N1(N1+1)(N0−1)(N0−1+1)=−N1−N0+122N1+N0=.2PN2Similarly,usingtheformulaforx,weobtain121N1(N1+1)(2N1+1)−N0(N0−1)(2N0−1)Ex=N1−N0+162(N1−N0)(N1−N0+2)VarX=EX−EX=.123.2LetX=numberofdefectivepartsinthesample.ThenX∼hypergeometric(N=100,M,K)whereM=numberofdefectivesinthelotandK=samplesize.a.Ifthereare6ormoredefectivesinthelot,thentheprobabilitythatthelotisaccepted(X=0)isatmost

694(100−K)·····(100−K−5)P(X=0|M=100,N=6,K)=0K
=.100100·····95课后答案网KBytrialanderrorwefindP(X=0)=.10056forK=31andP(X=0)=.09182forK=32.Sothesamplesizemustbeatleast32.www.hackshp.cnb.NowP(acceptlot)=P(X=0or1),and,for6ormoredefectives,theprobabilityisatmost



694694P(X=0or1|M=100,N=6,K)=0K
+1K−
1.100100KKBytrialanderrorwefindP(X=0or1)=.10220forK=50andP(X=0or1)=.09331forK=51.Sothesamplesizemustbeatleast51.3.3Inthesevensecondsfortheevent,nocarmustpassinthelastthreeseconds,aneventwithprobability(1−p)3.Theonlyoccurrenceinthefirstfourseconds,forwhichthepedestriandoesnotwaittheentirefourseconds,istohaveacarpassinthefirstsecondandnoothercarpass.Thishasprobabilityp(1−p)3.Thustheprobabilityofwaitingexactlyfoursecondsbeforestartingtocrossis[1−p(1−p)3](1−p)3.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn3-2SolutionsManualforStatisticalInference3.5LetX=numberofeffectivecases.Ifthenewandolddrugsareequallyeffective,thentheprobabilitythatthenewdrugiseffectiveonacaseis.8.IfthecasesareindependentthenX∼binomial(100,.8),and100X100P(X≥85)=.8x.2100−x=.1285.xx=85So,evenifthenewdrugisnobetterthantheold,thechanceof85ormoreeffectivecasesisnottoosmall.Hence,wecannotconcludethenewdrugisbetter.NotethatusinganormalapproximationtocalculatethisbinomialprobabilityyieldsP(X≥85)≈P(Z≥1.125)=.1303.3.7LetX∼Poisson(λ).WewantP(X≥2)≥.99,thatis,P(X≤1)=e−λ+λe−λ≤.01.Solvinge−λ+λe−λ=.01bytrialanderror(numericalbisectionmethod)yieldsλ=6.6384.3.8a.WewantP(X>N)<.01whereX∼binomial(1000,1/2).Sincethe1000customerschooserandomly,wetakep=1/2.Wethusrequire1000Xx1000−x100011P(X>N)=1−<.01x22x=N+1whichimpliesthat10001000X11000<.01.2xx=N+1ThislastinequalitycanbeusedtosolveforN,thatis,Nisthesmallestintegerthatsatisfies10001000X11000<.01.2xx=N+1ThesolutionisN=537.b.TousethenormalapproximationwetakeX∼n(500,250),whereweusedµ=1000(1)=5002andσ2=1000(1)(1)=250.Then课后答案网22X−500N−500P(X>N)=P√>√<.01www.hackshp.cn250250thus,N−500PZ>√<.01250whereZ∼n(0,1).FromthenormaltablewegetN−500P(Z>2.33)≈.0099<.01⇒√=2.33250⇒N≈537.Therefore,eachtheatershouldhaveatleast537seats,andtheanswerbasedontheapprox-imationequalstheexactanswer.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition3-33.9a.Wecanthinkofeachoneofthe60childrenenteringkindergartenas60independentBernoullitrialswithprobabilityofsuccess(atwinbirth)ofapproximately1.Theprobabilityofhaving905ormoresuccessesapproximatestheprobabilityofhaving5ormoresetsoftwinsenteringkindergarten.ThenX∼binomial(60,1)and90X4x60−x6011P(X≥5)=1−1−=.0006,x9090x=0whichissmallandmayberareenoughtobenewsworthy.b.LetXbethenumberofelementaryschoolsinNewYorkstatethathave5ormoresetsoftwinsenteringkindergarten.ThentheprobabilityofinterestisP(X≥1)whereX∼binomial(310,.0006).ThereforeP(X≥1)=1−P(X=0)=.1698.c.LetXbethenumberofStatesthathave5ormoresetsoftwinsenteringkindergartenduringanyofthelasttenyears.ThentheprobabilityofinterestisP(X≥1)whereX∼binomial(500,.1698).ThereforeP(X≥1)=1−P(X=0)=1−3.90×10−41≈1.3.11a.

MN−MxK−xlim
NM/N→p,M→∞,N→∞KK!M!(N−M)!(N−K)!=limx!(K−x)!M/N→p,M→∞,N→∞N!(M−x)!(N−M−(K−x))!Inthelimit,eachofthefactorialtermscanbereplacedbytheapproximationfromStirling’sformulabecause,forexample,√√M!=(M!/(2πMM+1/2e−M))2πMM+1/2e−M√√andM!/(2πMM+1/2e−M)→1.Whenthisreplacementismade,allthe2πandexpo-nentialtermscancel.Thus,

MN−MxK−xlim
NM/N→p,M→∞,N→∞KM+1/2N−M+1/2N−K+1/2KM(N−M)(N−K)=lim.x课后答案网M/N→p,M→∞,N→∞NN+1/2(M−x)M−x+1/2(N−M−K+x)N−M−(K−x)+1/2Wecanevaluatethelimitbybreakingtheratiointoseventerms,eachofwhichhasafinitelimitwecanevaluate.InsomelimitsweusethefactthatM→∞,N→∞andM/N→pimplyN−M→∞.Thefirstterm(oftheseventerms)iswww.hackshp.cnMM111xMlim→∞M−x=limM→∞M−x
M=limM→∞−x
M=e−x=e.1+MMLemma2.3.14isusedtogetthepenultimateequality.Similarlywegettwomoreterms,N−MN−MK−xlim=eN−M→∞N−M−(K−x)andNN−K−Klim=e.N→∞N若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn3-4SolutionsManualforStatisticalInferenceNote,theproductofthesethreelimitsisone.Threeothertermsare1/2MlimM→∞=1M−x1/2N−Mlim=1N−M→∞N−M−(K−x)and1/2N−Klim=1.N→∞NTheonlytermleftisxK−x(M−x)(N−M−(K−x))limKM/N→p,M→∞,N→∞(N−K)xK−xM−xN−M−(K−x)=limM/N→p,M→∞,N→∞N−KN−K=px(1−p)K−x.b.Ifin(a)weinadditionhaveK→∞,p→0,MK/N→pK→λ,bythePoissonapproxi-mationtothebinomial,weheuristicallygetM
N−M
−λxxK−xKxK−xeλ
→p(1−p)→.Nxx!Kc.UsingStirling’sformulaasin(a),weget

MN−MxK−xlim
MKMNN,M,K→∞,→0,→λNNKe−xKxexMxex(N−M)K−xeK−x=limN,M,K→∞,M→0,KM→λx!NKeKNNxK−x1KMN−M=limx!N,M,K→∞,M→0,KM→λNNNN!KMK课后答案网1xN=λlim1−x!N,M,K→∞,M→0,KM→λKNNe−λλx=.www.hackshp.cnx!3.12ConsiderasequenceofBernoullitrialswithsuccessprobabilityp.DefineX=numberofsuccessesinfirstntrialsandY=numberoffailuresbeforetherthsuccess.ThenXandYhavethespecifiedbinomialandhypergeometricdistributions,respectively.AndwehaveFx(r−1)=P(X≤r−1)=P(rthsuccesson(n+1)storlatertrial)=P(atleastn+1−rfailuresbeforetherthsuccess)=P(Y≥n−r+1)=1−P(Y≤n−r)=1−FY(n−r).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition3-53.13ForanyXwithsupport0,1,...,wehavethemeanandvarianceofthe0−truncatedXTaregivenbyX∞X∞P(X=x)EXT=xP(XT=x)=xP(X>0)x=1x=11X∞1X∞EX=xP(X=x)=xP(X=x)=.P(X>0)P(X>0)P(X>0)x=1x=02InasimilarwaywegetEX2=EX.Thus,TP(X>0)22EXEXVarXT=−.P(X>0)P(X>0)−λ0a.ForPoisson(λ),P(X>0)=1−P(X=0)=1−eλ=1−e−λ,therefore0!e−λλxP(XT=x)=x=1,2,...x!(1−e−λ)EX=λ/(1−e−λ)TVarX=(λ2+λ)/(1−e−λ)−(λ/(1−e−λ))2.T
b.Fornegativebinomial(r,p),P(X>0)=1−P(X=0)=1−r−1pr(1−p)0=1−pr.Then0
r+x−1pr(1−p)xP(X=x)=x,x=1,2,...T1−prr(1−p)EXT=p(1−pr)r(1−p)+r2(1−p)2r(1−p)VarXT=−.p2(1−pr)p(1−pr)2P∞−(1−p)xP∞−(1−p)x3.14a.=1=1,sincethesumistheTaylorseriesforlogp.x=1xlogplogpx=1xb."∞#"∞#−1Xx−1Xx−11−11−pEX=(1−p)=(1−p)−1==−1=.logplogplogpplogppx=1x=0Sincethegeometricseriesconvergesuniformly,课后答案网−1X∞(1−p)X∞dEX2=x(1−p)x=(1−p)xlogplogpdpwww.hackshp.cnx=1x=1∞(1−p)dXx(1−p)d1−p−(1−p)=(1−p)==.logpdplogpdppp2logpx=1Thus−(1−p)(1−p)VarX=1+.p2logplogpAlternatively,themgfcanbecalculated,−1X∞hixlog(1+pet−et)tMx(t)=(1−p)e=logplogpx=1andcanbedifferentiatedtoobtainthemoments.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn3-6SolutionsManualforStatisticalInference3.15Themomentgeneratingfunctionforthenegativebinomialisrt!rp1r(1−p)(e−1)M(t)==1+,trt1−(1−p)e1−(1−p)ethetermttr(1−p)(e−1)λ(e−1)t→=λ(e−1)asr→∞,p→1andr(p−1)→λ.t11−(1−p)eThusbyLemma2.3.14,thenegativebinomialmomentgeneratingfunctionconvergestoteλ(e−1),thePoissonmomentgeneratingfunction.3.16a.Usingintegrationbypartswith,u=tαanddv=e−tdt,weobtainZ∞∞Z∞(α+1)−1−tα−tα−1−tΓ(α+1)=tedt=t(−e)−αt(−e)dt=0+αΓ(α)=αΓ(α).000√b.Makingthechangeofvariablez=2t,i.e.,t=z2/2,weobtainZ∞Z∞√√Z∞√√−1/2−t2−z2/2−z2/2π√Γ(1/2)=tedt=ezdz=2edz=2√=π.00z02wherethepenultimateequalityuses(3.3.14).3.17Z∞Z∞νν1α−1−x/β1(ν+α)−1−x/βEX=xxedx=xedx0Γ(α)βαΓ(α)βα0Γ(ν+α)βν+αβνΓ(ν+α)==.Γ(α)βαΓ(α)Note,thisformulaisvalidforallν>−α.Theexpectationdoesnotexistforν≤−α.rp3.18IfY∼negativebinomial(r,p),itsmomentgeneratingfunctionisMY(t)=t,and,1−(1−p)erpfromTheorem2.3.15,MpY(t)=pt.NowuseL’Hˆopital’sruletocalculate1−(1−p)ep11lim=lim=,课后答案网p→01−(1−p)eptp→0(p−1)tept+ept1−tsothemomentgeneratingfunctionconvergesto(1−t)−r,themomentgeneratingfunctionofagamma(r,1).www.hackshp.cn3.19Repeatedlyapplytheintegration-by-partsformula1Z∞xn−1e−x1Z∞zn−1z−zdz=+zn−2z−zdz,Γ(n)x(n−1)!Γ(n−1)xuntiltheexponentonthesecondintegraliszero.Thiswillestablishtheformula.IfX∼gamma(α,1)andY∼Poisson(x).TheprobabilisticrelationshipisP(X≥x)=P(Y≤α−1).1R∞etxtx3.21Themomentgeneratingfunctionwouldbedefinedbyπ−∞1+x2dx.On(0,∞),e>x,henceZ∞txZ∞exdx>dx=∞,01+x201+x2thusthemomentgeneratingfunctiondoesnotexist.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition3-73.22a.X∞e−λλxE(X(X−1))=x(x−1)x!x=0X∞λx−2=e−λλ2(lety=x−2)(x−2)!x=2X∞λy=e−λλ2=e−λλ2eλ=λ2y!y=0EX2=λ2+EX=λ2+λVarX=EX2−(EX)2=λ2+λ−λ2=λ.b.∞Xr+x−1xE(X(X−1))=x(x−1)pr(1−p)xx=0∞Xr+x−1x=r(r+1)pr(1−p)x−2x=22X∞(1−p)r+2+y−1y=r(r+1)pr+2(1−p)p2yy=02(1−p)=r(r−1),p2whereinthesecondequalitywesubstitutedy=x−2,andinthethirdequalityweusethefactthatwearesummingoveranegativebinomial(r+2,p)pmf.Thus,VarX=EX(X−1)+EX−(EX)2(1−p)2r(1−p)r2(1−p)2=r(r+1)+−p2pp2r(1−p)=.p2c.课后答案网Z∞Z∞221α−1−x/β1α+1−x/βEX=xxedx=xedx0Γ(α)βαΓ(α)βα0www.hackshp.cn1α+22=Γ(α+2)β=α(α+1)β.Γ(α)βαVarX=EX2−(EX)2=α(α+1)β2−α2β2=αβ2.d.(Use3.3.18)Γ(α+1)Γ(α+β)αΓ(α)Γ(α+β)αEX===.Γ(α+β+1)Γ(α)(α+β)Γ(α+β)Γ(α)α+β2Γ(α+2)Γ(α+β)(α+1)αΓ(α)Γ(α+β)α(α+1)EX===.Γ(α+β+2)Γ(α)(α+β+1)(α+β)Γ(α+β)Γ(α)(α+β)(α+β+1)α(α+1)α2αβVarX=EX2−(EX)2=−=.(α+β)(α+β+1)(α+β)2(α+β)2(α+β+1)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn3-8SolutionsManualforStatisticalInferencee.Thedoubleexponential(µ,σ)pdfissymmetricaboutµ.Thus,byExercise2.26,EX=µ.Z∞Z∞21−|x−µ|/σ21−|z|VarX=(x−µ)edx=σzeσdz−∞2σ−∞2Z∞=σ2z2e−zdz=σ2Γ(3)=2σ2.03.23a.Z∞∞x−β−1dx=−1x−β=1,ββαβααthusf(x)integratesto1.nb.EXn=βα,therefore(n−β)αβEX=(1−β)αβ2EX2=(2−β)αβ2(αβ)2VarX=−2−β(1−β)2c.Ifβ<2theintegralofthesecondmomentisinfinite.1−x/β1/γγ−yγ/βγ−13.24a.fx(x)=βe,x>0.ForY=X,fY(y)=βey,y>0.Usingthetransforma-tionz=yγ/β,wecalculateZ∞Z∞nγγ+n−1−yγ/βn/γn/γ−zn/γnEY=yedy=βzedz=βΓ+1.β00γhiThusEY=β1/γΓ(1+1)andVarY=β2/γΓ2+1−Γ21+1.γγγ12b.f(x)=e−x/β,x>0.ForY=(2X/β)1/2,f(y)=ye−y/2,y>0.WenownoticethatxβYZ√∞2−y2/22πEY=yedy=课后答案网02since√1R∞y2e−y2/2=1,thevarianceofastandardnormal,andtheintegrandissym-2π−∞metric.Useintegration-by-partstocalculatethesecondmomentwww.hackshp.cnZ∞Z∞22EY2=y3e−y/2dy=2ye−y/2dy=2,002wherewetakeu=y2,dv=ye−y/2.ThusVarY=2(1−π/4).c.Thegamma(a,b)densityis1a−1−x/bfX(x)=xe.Γ(a)baMakethetransformationy=1/xwithdx=−dy/y2togeta+1211−1/byfY(y)=fX(1/y)|1/y|=e.Γ(a)bay若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition3-9ThefirsttwomomentsareZ∞aa−111−1/byΓ(a−1)b1EY=e==Γ(a)ba0yΓ(a)ba(a−1)bΓ(a−2)ba−21EY2==,Γ(a)ba(a−1)(a−2)b2andsoVarY=1.(a−1)2(a−2)b2122d.f(x)=x3/2−1e−x/β,x>0.ForY=(X/β)1/2,f(y)=y2e−y,y>0.ToxΓ(3/2)β3/2YΓ(3/2)2calculatethemomentsweuseintegration-by-partswithu=y2,dv=ye−ytoobtainZ∞Z∞23−y22−y21EY=yedy=yedy=Γ(3/2)0Γ(3/2)0Γ(3/2)2andwithu=y3,dv=ye−ytoobtainZ∞Z∞224−y232−y23√EY=yedy=yedy=π.Γ(3/2)0Γ(3/2)0Γ(3/2)√1R∞2−y2Usingthefactthatye=1,sinceitisthevarianceofan(0,2),symmetryyields2π−∞R∞2−y2√1√yedy=π.Thus,VarY=6−4/π,usingΓ(3/2)=π.02α−yα−ye.f(x)=e−x,x>0.ForY=α−γlogX,f(y)=e−eγeγ1,−∞0,c(β)=1,w(β)=1,t(x)=−x.Γ(α)βα1β1(ii)βknown,www.hackshp.cn−x/β1f(x|α)=eexp((α−1)logx),Γ(α)βαh(x)=e−x/β,x>0,c(α)=1w(α)=α−1,t(x)=logx.Γ(α)βα11(iii)α,βunknown,1xf(x|α,β)=exp((α−1)logx−),Γ(α)βαβh(x)=I(x),c(α,β)=1,w(α)=α−1,t(x)=logx,{x>0}Γ(α)βα11w2(α,β)=−1/β,t2(x)=x.c.(i)αknown,h(x)=xα−1I(x),c(β)=1,w(β)=β−1,t(x)=log(1−x).[0,1]B(α,β)11(ii)βknown,h(x)=(1−x)β−1I(x),c(α)=1,w(x)=α−1,t(x)=logx.[0,1]B(α,β)11若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition3-11(iii)α,βunknown,h(x)=I(x),c(α,β)=1,w(α)=α−1,t(x)=logx,[0,1]B(α,β)11w2(β)=β−1,t2(x)=log(1−x).d.h(x)=1I(x),c(θ)=e−θ,w(θ)=logθ,t(x)=x.x!{0,1,2,...}11x−1rpe.h(x)=r−1I{r,r+1,...}(x),c(p)=1−p,w1(p)=log(1−p),t1(x)=x.3.29a.Forthen(µ,σ2)22!−µ/2σ1e−x2/2σ2+xµ/σ2f(x)=√e,2πσsothenaturalparameteris(η,η)=(−1/2σ2,µ/σ2)withnaturalparameterspace12{(η1,η2):η1<0,−∞<η2<∞}.b.Forthegamma(α,β),1(α−1)logx−x/βf(x)=e,Γ(α)βαsothenaturalparameteris(η1,η2)=(α−1,−1/β)withnaturalparameterspace{(η1,η2):η1>−1,η2<0}.c.Forthebeta(α,β),Γ(α+β)(α−1)logx+(β−1)log(1−x)f(x)=e,Γ(α)Γ(β)sothenaturalparameteris(η1,η2)=(α−1,β−1)andthenaturalparameterspaceis{(η1,η2):η1>−1,η2>−1}.d.ForthePoisson1
f(x)=e−θexlogθx!sothenaturalparameterisη=logθandthenaturalparameterspaceis{η:−∞<η<∞}.e.Forthenegativebinomial(r,p),rknown,r+x−1rxlog(1−p)P(X=x)=x(p)e,sothenaturalparameterisη=log(1−p)withnaturalparameterspace{η:η<0}.3.31a.Z!Xk课后答案网∂0=h(x)c(θ)expwi(θ)ti(x)dx∂θi=1Z!Xk=h(x)www.hackshp.cnc0(θ)expw(θ)t(x)dxiii=1Z!!XkXk∂w(θ)i+h(x)c(θ)expwi(θ)ti(x)ti(x)dx∂θji=1i=1Zk!"k#∂XX∂wi(θ)=h(x)logc(θ)c(θ)expwi(θ)ti(x)dx+Eti(x)∂θj∂θji=1i=1"#Xk∂∂wi(θ)=logc(θ)+Eti(x)∂θj∂θji=1hPiThereforeEk∂wi(θ)t(x)=−∂logc(θ).i=1∂θji∂θj若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn3-12SolutionsManualforStatisticalInferenceb.Z!∂2Xk0=h(x)c(θ)expwi(θ)ti(x)dx∂θ2i=1Z!Xk=h(x)c00(θ)expw(θ)t(x)dxiii=1Z!!XkXk∂w(θ)0i+h(x)c(θ)expwi(θ)ti(x)ti(x)dx∂θji=1i=1Z!!XkXk∂w(θ)0i+h(x)c(θ)expwi(θ)ti(x)ti(x)dx∂θji=1i=1Z!!2XkXk∂w(θ)i+h(x)c(θ)expwi(θ)ti(x)ti(x)dx∂θji=1i=1Z!!XkXk∂2w(θ)i+h(x)c(θ)expwi(θ)ti(x)2ti(x)dx∂θji=1i=1Z"#!∂2Xk=h(x)2logc(θ)c(θ)expwi(θ)ti(x)dx∂θji=1Z2k!c0(θ)X+h(x)c(θ)expwi(θ)ti(x)dxc(θ)i=1"k#∂X∂wi(θ)+2logc(θ)Eti(x)∂θj∂θji=1"#"#Xk∂w(θ)Xk∂2w(θ)i2i+E(ti(x))+E2ti(x)∂θj∂θji=1i=122∂∂=logc(θ)+logc(θ)∂θ2∂θjj"#"#Xk∂w(θ)Xk∂w(θ)ii−2Eti(x)Eti(x)课后答案网i=1∂θji=1∂θj"#"#Xk∂w(θ)Xk∂2w(θ)i2i+E(ti(x))+E2ti(x)∂θj∂θjwww.hackshp.cni=1i=1!"#∂2Xk∂w(θ)Xk∂2w(θ)ii=2logc(θ)+Varti(x)+E2ti(x).∂θji=1∂θji=1∂θjhiPk∂w(θ)∂2Pk∂2w(θ)ThereforeVarit(x)=−logc(θ)−Eit(x).i=1∂θji∂θ2i=1∂θ2ijj3.33a.(i)h(x)=exI(x),c(θ)=√1exp(−θ)θ>0,w(θ)=1,t(x)=−x2.{−∞0,{−∞0,w(α)=α,w(α)=α,x{0{1,-2,2.5}].3.35a.InExercise3.34(a)w(λ)=1andforan(eθ,eθ),w(θ)=1.12λ12eθb.EX=µ=αβ,thenβ=µ.Thereforeh(x)=1I(x),αx{00,w(α)=α,w(α)=α,t(x)=log(x),t(x)=−x.Γ(α)(µ)α12µ12αc.From(b)then(α,...,α,β,...,β)=(α,...,α,α1,...,αn)1n1n1nµµ3.37Thepdf(1)f((x−µ))issymmetricaboutµbecause,forany>0,σσ1(µ+)−µ111(µ−)−µf=f=f−=f.σσσσσσσσThus,byExercise2.26b,µisthemedian.3.38P(X>xα)=P(σZ+µ>σzα+µ)=P(Z>zα)byTheorem3.5.6.3.39Firsttakeµ=0andσ=1.a.Thepdfissymmetricabout0,so0mustbethemedian.Verifyingthis,writeZ∞∞P(Z≥0)=11dz=1tan−1(z)=1π−0=1.0π1+z2ππ220∞
b.P(Z≥1)=1tan−1(z)=1π−π=1.BysymmetrythisisalsoequaltoP(Z≤−1).π1π244Writingz=(x−µ)/σestablishesP(X≥µ)=1andP(X≥µ+σ)=1.243.40LetX∼f(x)havemeanµandvarianceσ2.LetZ=X−µ.Thenσ1EZ=E(X−µ)=0课后答案网σandX−µ11σ2VarZ=Var=Var(X−µ)=VarX==1.www.hackshp.cnσσ2σ2σ2ThencomputethepdfofZ,fZ(z)=fx(σz+µ)·σ=σfx(σz+µ)andusefZ(z)asthestandardpdf.3.41a.ThisisaspecialcaseofExercise3.42a.b.ThisisaspecialcaseofExercise3.42b.3.42a.Letθ1>θ2.LetX1∼f(x−θ1)andX2∼f(x−θ2).LetF(z)bethecdfcorrespondingtof(z)andletZ∼f(z).ThenF(x|θ1)=P(X1≤x)=P(Z+θ1≤x)=P(Z≤x−θ1)=F(x−θ1)≤F(x−θ2)=P(Z≤x−θ2)=P(Z+θ2≤x)=P(X2≤x)=F(x|θ2).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn3-14SolutionsManualforStatisticalInferenceTheinequalityisbecausex−θ2>x−θ1,andFisnondecreasing.Togetstrictinequalityforsomex,let(a,b]beanintervaloflengthθ1−θ2withP(a0.Letx=a+θ1.ThenF(x|θ1)=F(x−θ1)=F(a+θ1−θ1)=F(a)σ2.LetX1∼f(x/σ1)andX2∼f(x/σ2).LetF(z)bethecdfcorrespondingtof(z)andletZ∼f(z).Then,forx>0,F(x|σ1)=P(X1≤x)=P(σ1Z≤x)=P(Z≤x/σ1)=F(x/σ1)≤F(x/σ2)=P(Z≤x/σ2)=P(σ2Z≤x)=P(X2≤x)=F(x|σ2).Theinequalityisbecausex/σ2>x/σ1(becausex>0andσ1>σ2>0),andFisnondecreasing.Forx≤0,F(x|σ1)=P(X1≤x)=0=P(X2≤x)=F(x|σ2).Togetstrictinequalityforsomex,let(a,b]beanintervalsuchthata>0,b/a=σ1/σ2andP(a0.Letx=aσ1.ThenF(x|σ1)=F(x/σ1)=F(aσ1/σ1)=F(a)0,byTheorem2.1.3.Forθ>θ,YXy1211FY(y|θ1)=1−FXθ1≤1−FXθ2=FY(y|θ2)yyforally,sinceFX(x|θ)isstochasticallyincreasingandifθ1>θ2,FX(x|θ2)≤FX(x|θ1)forallx.Similarly,F(y|θ)=1−F(1|θ)<1−F(1|θ)=F(y|θ)forsomey,sinceifY1Xy1Xy2Y2θ1>θ2,FX(x|θ2)θandθ,θ>0then1>1.ThereforeX1212θ2θ1F(x|1)≤F(x|1)forallxandF(x|1)2/9=EX2/b2.√ThusEX2/b2isabetterbound.Butforb=2,√E|X|/b=1/2<1=EX2/b2.ThusE|X|/bisabetterbound.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition3-153.45a.Z∞Z∞M(t)=etxf(x)dx≥etxf(x)dxXXX−∞aZ∞≥etaf(x)dx=etaP(X≥a),Xawhereweusethefactthatetxisincreasinginxfort>0.b.Z∞ZaM(t)=etxf(x)dx≥etxf(x)dxXXX−∞−∞Za≥etaf(x)dx=etaP(X≤a),X−∞whereweusethefactthatetxisdecreasinginxfort<0.c.h(t,x)mustbenonnegative.3.46ForX∼uniform(0,1),µ=1andσ2=1,thus212√1k1k1−√2kk<3,P(|X−µ|>kσ)=1−P−√≤X≤+√=12√2122120k≥3,ForX∼exponential(λ),µ=λandσ2=λ2,thus1+e−(k+1)−ek−1k≤1P(|X−µ|>kσ)=1−P(λ−kλ≤X≤λ+kλ)=e−(k+1)k>1.FromExample3.6.2,Chebychev’sInequalitygivestheboundP(|X−µ|>kσ)≤1/k2.Comparisonofprobabilitiesku(0,1)exp(λ)Chebychevexactexact.1.942.926100.5.711.6174课后答案网1.423.13511.5.134.0821.44√300.0651.332www.hackshp.cn00.0498.25400.00674.06251000.0000167.01SoweseethatChebychev’sInequalityisquiteconservative.3.47Z∞1−x2/2P(|Z|>t)=2P(Z>t)=2√edx2πtrZ∞221+x−x2/2=edxπt1+x2rZZ∞∞221−x2/2x−x2/2=edx+edx.πt1+x2t1+x2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn3-16SolutionsManualforStatisticalInferencex−x2/2−x2/21−x2Toevaluatethesecondterm,letu=1+x2,dv=xedx,v=−e,du=(1+x2)2,toobtainZ∞2∞Z∞2x−x2/2x−x2/21−x−x2/2edx=(−e)−(−e)dxt1+x21+x2t(1+x2)2tZ∞2t−t2/21−x−x2/2=e+edx.1+t2t(1+x2)2Therefore,rrZ∞22t−t2/2211−x−x2/2P(Z≥t)=e++edxπ1+t2πt1+x2(1+x2)2rrZ∞2t−t2/222−x2/2=e+edxπ1+t2πt(1+x2)2r2t−t2/2≥e.π1+t23.48Forthenegativebinomialr+x+1−1rx+1r+xP(X=x+1)=p(1−p)=(1−p)P(X=x).x+1x+1Forthehypergeometric(M−x)(k−x+x+1)(x+1)ifx0)=2.4c.ClearlyP(|X+Y|<2)=1.4.2Theseareallfundamentalpropertiesofintegrals.TheproofisthesameasforTheorem2.2.5withbivariateintegralsreplacingunivariateintegrals.4.3Fortheexperimentoftossingtwofairdice,eachofthepointsinthe36-pointsamplespaceareequallylikely.Sotheprobabilityofaneventis(numberofpointsintheevent)/36.Thegivenprobabilitiesareobtainedbynotingthefollowingequivalencesofevents.61P({X=0,Y=0})=P({(1,1),(2,1),(1,3),(2,3),(1,5),(2,5)})==36661P({X=0,Y=1})=P({(1,2),(2,2),(1,4),(2,4),(1,6),(2,6)})==366P({X=1,Y=0})=P({(3,1),(4,1),(5,1),(6,1),(3,3),(4,3),(5,3),(6,3),(3,5),(4,5),(5,5),(6,5)})121==363P({X=1,Y=1})=P(课后答案网{(3,2),(4,2),(5,2),(6,2),(3,4),(4,4),(5,4),(6,4),(3,6),(4,6),(5,6),(6,6)})121==363R1R2www.hackshp.cn14.4a.C(x+2y)dxdy=4C=1,thusC=.00R4111(x+2y)dy=(x+1)0Y)=√(x+y)dxdy=.0y20RR√21y1b.P(Xx)=1−P(B−A>x)=1−1dbda=+x−.1a+x224.7Wewillmeasuretimeinminutespast8A.M.SoX∼uniform(0,30),Y∼uniform(40,50)andthejointpdfis1/300ontherectangle(0,30)×(40,50).Z50Z60−y11P(arrivebefore9A.M.)=P(X+Y<60)=dxdy=.40030024.9P(a≤X≤b,c≤Y≤d)=P(X≤b,c≤Y≤d)−P(X≤a,c≤Y≤d)=P课后答案网(X≤b,Y≤d)−P(X≤b,Y≤c)−P(X≤a,Y≤d)+P(X≤a,Y≤c)=F(b,d)−F(b,c)−F(a,d)−F(a,c)=FX(b)FY(d)−FX(b)FY(c)−FX(a)FY(d)−FX(a)FY(c)=P(X≤b)[www.hackshp.cnP(Y≤d)−P(Y≤c)]−P(X≤a)[P(Y≤d)−P(Y≤c)]=P(X≤b)P(c≤Y≤d)−P(X≤a)P(c≤Y≤d)=P(a≤X≤b)P(c≤Y≤d).4.10a.ThemarginaldistributionofXisP(X=1)=P(X=3)=1andP(X=2)=1.The42marginaldistributionofYisP(Y=2)=P(Y=3)=P(Y=4)=1.But311P(X=2,Y=3)=06=()()=P(X=2)P(Y=3).23Thereforetherandomvariablesarenotindependent.b.ThedistributionthatsatisfiesP(U=x,V=y)=P(U=x)P(V=y)whereU∼XandV∼Yis若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition4-3U123211112612V3111126124111126124.11Thesupportofthedistributionof(U,V)is{(u,v):u=1,2,...;v=u+1,u+2,...}.Thisisnotacross-productset.Therefore,UandVarenotindependent.Moresimply,ifweknowU=u,thenweknowV>u.4.12Oneinterpretationof“astickisbrokenatrandomintothreepieces”isthis.Supposethelengthofthestickis1.LetXandYdenotethetwopointswherethestickisbroken.LetXandYbothhaveuniform(0,1)distributions,andassumeXandYareindependent.ThenthejointdistributionofXandYisuniformontheunitsquare.Inorderforthethreepiecestoformatriangle,thesumofthelengthsofanytwopiecesmustbegreaterthanthelengthofthethird.Thiswillbetrueifandonlyifthelengthofeachpieceislessthan1/2.Tocalculatetheprobabilityofthis,weneedtoidentifythesamplepoints(x,y)suchthatthelengthofeachpieceislessthan1/2.Ify>x,thiswillbetrueifx<1/2,y−x<1/2and1−y<1/2.Thesethreeinequalitiesdefinethetrianglewithvertices(0,1/2),(1/2,1/2)and(1/2,1).(Drawagraphofthisset.)Becauseoftheuniformdistribution,theprobabilitythat(X,Y)fallsinthetriangleistheareaofthetriangle,whichis1/8.Similarly,ifx>y,eachpiecewillhavelengthlessthan1/2ify<1/2,x−y<1/2and1−x<1/2.Thesethreeinequalitiesdefinethetrianglewithvertices(1/2,0),(1/2,1/2)and(1,1/2).Theprobabilitythat(X,Y)isinthistriangleisalso1/8.Sotheprobabilitythatthepiecesformatriangleis1/8+1/8=1/4.4.13a.E(Y−g(X))22=E((Y−E(Y|X))+(E(Y|X)−g(X)))=E(Y−E(Y|X))2+E(E(Y|X)−g(X))2+2E[(Y−E(Y|X))(E(Y|X)−g(X))].Thecrosstermcanbeshowntobezerobyiteratingtheexpectation.ThusE(Y−g(X))2=E(Y−E(Y|X))2+E(E(Y|X)−g(X))2≥E(Y−E(Y|X))2,forallg(·).Thechoiceg(X)=E(Y|X)willgiveequality.b.Equation(2.2.3)isthespecialcaseofa)wherewetaketherandomvariableXtobeaconstant.Then,课后答案网g(X)isaconstant,sayb,andE(Y|X)=EY.4.15WewillfindtheconditionaldistributionofY|X+Y.Thederivationoftheconditionaldistri-butionofX|X+Yissimilar.LetU=X+YandV=Y.InExample4.3.1,wefoundthejointpmfof(U,V).Notethatforfixedwww.hackshp.cnu,f(u,v)ispositiveforv=0,...,u.Thereforetheconditionalpmfisθu−ve−θλve−λvu−vf(u,v)(u−v)!v!uλθf(v|u)==(θ+λ)ue−(θ+λ)=,v=0,...,u.f(u)vθ+λθ+λu!ThatisV|U∼binomial(U,λ/(θ+λ)).4.16a.Thesupportofthedistributionof(U,V)is{(u,v):u=1,2,...;v=0,±1,±2,...}.IfV>0,thenX>Y.Soforv=1,2,...,thejointpmfisfU,V(u,v)=P(U=u,V=v)=P(Y=u,X=u+v)=p(1−p)u+v−1p(1−p)u−1=p2(1−p)2u+v−2.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn4-4SolutionsManualforStatisticalInferenceIfV<0,thenX0,y>0.Changingtopolarcoordinates,xwww.hackshp.cn=rcosθandy=rsinθ,weobtainZ∞Z∞Zπ/2Z∞Zπ/2Z∞Zπ/22g(r)22f(x,y)dxdy=rdrdθ=g(r)drdθ=1dθ=1.0000πrπ00π0.√.4.19a.Since(X−X)2∼n(0,1),(X−X)22∼χ2(seeExample2.1.9).12121b.Makethetransformationy=x1,y=x+xthenx=yy,x=y(1−y)and1x1+x2212112221|J|=y2.Thenf(y,y)=Γ(α1+α2)yα1−1(1−y)α2−11yα1+α2−1e−y2,12Γ(α)Γ(α)11Γ(α+α)21212thusY1∼beta(α1,α2),Y2∼gamma(α1+α1,1)andareindependent.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition4-54.20a.Thistransformationisnotone-to-onebecauseyoucannotdeterminethesignofX2fromY1andY2.Sopartitionthesupportof(X1,X2)intoA0={−∞0}andA2={−∞0,y>0}ismappedontotheset{z1>0,00.SoZandWareindependent.4.27FromTheorem4.2.14weknowU∼n(µ+γ,2σ2)andV∼n(µ−γ,2σ2).Itremainstoshowthattheyareindependent.ProceedasinExercise4.24.课后答案网1−1[(x−µ)2+(y−γ)2]fXY(x,y)=e2σ2(byindependence,sofXY=fXfY)2πσ2Letu=x+y,v=x−y,thenx=1(u+v),y=1(u−v)and22www.hackshp.cn1/21/21|J|=1/2−1/2=.2Theset{−∞0dxdxΓ(r)λrxλforsomex,ifr>1.But,Z∞−x/νR∞1−x/νde−0ν2eqλ(ν)dνlogqλ(ν)dν=R∞1−x/ν<0∀x.dx0ν0νeqλ(ν)dν4.39a.Withoutlossofgeneralityletsassumethati0)+P(−X≤zandXY<0)=P(X≤zandY<0)+P(X≥−zandY<0)(sincez<0)=P(X≤z)P(Y<0)+P(X≥−z)P(Y<0)(independence)=P(X≤z)P(Y<0)+P(X≤z)P(Y>0)(symmetryofXandY)=P(X≤z)(P(Y<0)+P(Y>0))=P(X≤z).Byasimilarargument,forz>0,wegetP(Z>z)=P(X>z),andhence,P(Z≤z)=P(X≤z).Thus,Z∼X∼n(0,1).b.BydefinitionofZ,Z>0⇔either(i)X<0andY>0or(ii)X>0andY>0.SoZandYalwayshavethesamesign,hencetheycannotbebivariatenormal.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn4-18SolutionsManualforStatisticalInference4.49a.ZfX(x)=(af1(x)g1(y)+(1−a)f2(x)g2(y))dyZZ=af1(x)g1(y)dy+(1−a)f2(x)g2(y)dy=af1(x)+(1−a)f2(x).ZfY(y)=(af1(x)g1(y)+(1−a)f2(x)g2(y))dxZZ=ag1(y)f1(x)dx+(1−a)g2(y)f2(x)dx=ag1(y)+(1−a)g2(y).b.(⇒)IfXandYareindependentthenf(x,y)=fX(x)fY(y).Then,f(x,y)−fX(x)fY(y)=af1(x)g1(y)+(1−a)f2(x)g2(y)−[af1(x)+(1−a)f2(x)][ag1(y)+(1−a)g2(y)]=a(1−a)[f1(x)g1(y)−f1(x)g2(y)−f2(x)g1(y)+f2(x)g2(y)]=a(1−a)[f1(x)−f2(x)][g1(y)−g2(y)]=0.Thus[f1(x)−f2(x)][g1(y)−g2(y)]=0since01P(X/Y≤t)=21+(1−t)t≤12P(XY≤t)=t−tlogt014t2t4.53P(RealRoots)=P(B2>4AC)=P(2logB>log4+logA+logC)=P(−2logB≤−log4−logA−logC)=P(−2logB≤−log4+(−logA−logC)).LetX=−2logB,Y=−logA−logC.ThenX∼exponential(2),Y∼gamma(2,1),indepen-dent,andP(RealRoots)=P(X<−log4+Y)Z∞=P(X<−log4+y)fY(y)dylog4Z∞Z−log4+y1−x/2−y=edxyedylog402Z∞−1log4−y/2−y=1−e2eyedy.log4R∞Integration-by-partswillshowthatye−y/b=b(a+b)e−a/bandhencea课后答案网112P(RealRoots)=(1+log4)−+log4=.511.4243QnQnPn4.54LetY=i=1Xi.ThenP(Y≤y)=P(i=1Xi≤y)=P(Pi=1−logXi≥−logy).Now,www.hackshp.cnn−logXi∼exponential(1)=gamma(1,1).ByExample4.6.8,i=1−logXi∼gamma(n,1).Therefore,Z∞1n−1−zP(Y≤y)=zedz,−logyΓ(n)andZ∞d1n−1−zfY(y)=zedzdy−logyΓ(n)1n−1−(−logy)d=−(−logy)e(−logy)Γ(n)dy1n−1=(−logy),00fY(y)=0y≤0.4.57a.XnXn1111A1=[xi]1=xi,thearithmeticmean.nnx=1x=1Xn1−1−11A−1=[nxi]=111,theharmonicmean.(+···+)x=1nx1xnXnXn1Pnr−11r111rni=1rxilimlogAr=limlog[xi]r=limlog[xi]=limPnr→0r→0nr→0rnr→01xrx=1x=1ni=1i1PnrXnYnni=1xilogxi11=lim1Pnr=logxi=log(xi).r→0xnnni=1ii=1i=11QnQn1ThusA0=limr→0Ar=exp(log(i=1xi))=(i=1xi)n,thegeometricmean.Thetermnrxr−1=xrlogxsincerxr−1=dxr=dexp(rlogx)=exp(rlogx)logx=xrlogx.iiiidridriiiiib.(i)iflogArisnondecreasingthenforr≤r0logAr≤logAr0,thenelogAr≤elogAr0.ThereforeAr≤Ar0.ThusArisnondecreasinginPr.PP1nr−1rnxrlogxPd−11nr1nPi=1rxi1Pi=1ii1nr(ii)drlogAr=r2log(nx=1xi)+r1nxr=r2nxr−log(nx=1xi),ni=1ix=1ir−1wherewe课后答案网usetheidentityforrxishowedina).(iii)PnrXnrPi=1xilogxi1rnwww.hackshp.cnr−log(xi)xnx=1ix=1PnrXnrPi=1xilogxir=log(n)+nr−log(xi)x=1xix=1"#XnxrxrXn=log(n)+Pirlogx−Pilog(xr)nxrinxrii=1i=1ii=1ix=1"#XnxrXn=log(n)+Pi(rlogx−log(xr))nxriii=1i=1ix=1XnrPnrXnPxix=1xi1=log(n)−nrlog(r)=log(n)−ailog().i=1i=1xixii=1ai若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition4-21Pn1Weneedtoprovethatlog(Pn)≥i=1ailog(aPi).UsingJenseninequalitywehavethatElog(1)=nalog(1)≤log(E1)=log(na1)=log(n)whichestablishtheai=1iaiai=1iairesult.4.59AssumethatEX=0,EY=0,andEZ=0.ThiscanbedonewithoutlossofgeneralitybecausewecouldworkwiththequantitiesX−EX,etc.ByiteratingtheexpectationwehaveCov(X,Y)=EXY=E[E(XY|Z)].AddingandsubtractingE(X|Z)E(Y|Z)givesCov(X,Y)=E[E(XY|Z)−E(X|Z)E(Y|Z)]+E[E(X|Z)E(Y|Z)].SinceE[E(X|Z)]=EX=0,thesecondtermaboveisCov[E(X|Z)E(Y|Z)].ForthefirsttermwriteE[E(XY|Z)−E(X|Z)E(Y|Z)]=E[E{XY−E(X|Z)E(Y|Z)|Z}]wherewehavebroughtE(X|Z)andE(Y|Z)insidetheconditionalexpectation.ThiscannowberecognizedasECov(X,Y|Z),establishingtheidentity.4.61a.Tofindthedistributionoff(X|Z),letU=X2−1andV=X.Thenx=h(u,v)=uv+1,1X1121x1=h2(u,v)=v.Thereforef(u,v)=f(h(u,v),h(u,v))|J|=e−(uv+1)e−vv,U,VX,Y12andZ∞e−1f(u)=ve−(uv+1)e−vdv=.U(u+1)20ThusV|U=0hasdistributionve−v.ThedistributionofX|Xise−x1sinceXandX1212areindependent.b.ThefollowingMathematicacodewilldrawthepicture;thesolidlinesareB1andthedashedlinesareB2.Notethatthesolidlinesincreasewithx1,whilethedashedlinesareconstant.ThusB1isinformative,astherangeofX2changes.e=1/10;Plot[{-e*x1+1,e*x1+1,1-e,1+e},{x1,0,5},PlotStyle->{Dashing[{}],Dashing[{}],Dashing[{0.15,0.05}],Dashing[{0.15,0.05}]}]课后答案网c.Rv∗Rve−(uv+1)e−vdudv∗R0R−P(X1≤x|B1www.hackshp.cn)=P(V≤v|−1,unlessP(Z=EZ=0)=1.4.64a.Letaandbberealnumbers.Then,|a+b|2=(a+b)(a+b)=a2+2ab+b2≤|a|2+2|ab|+|b|2=(|a|+|b|)2.Takethesquarerootofbothsidestoget|a+b|≤|a|+|b|.b.|X+Y|≤|X|+|Y|⇒E|X+Y|≤E(|X|+|Y|)=E|X|+E|Y|.4.65WithoutlossofgeneralityletusassumethatEg(X)=Eh(X)=0.Forpart(a)Z∞E(g(X)h(X))=g(x)h(x)fX(x)dx−∞ZZ=g(x)h(x)fX(x)dx+g(x)h(x)fX(x)dx{x:h(x)≤0}{x:h(x)≥0}ZZ≤g(x0)h(x)fX(x)dx+g(x0)h(x)fX(x)dx{x:h(x)≤0}{x:h(x)≥0}Z∞=h(x)fX(x)dx−∞=g(x0)Eh(X)=0.wherex0isthenumbersuchthath(x0)=0.Notethatg(x0)isamaximumin{x:h(x)≤0}andaminimumin{x:h(x)≥0}sinceg(x)isnondecreasing.Forpart(b)whereg(x)andh(x)arebothnondecreasingZ∞E(g(X)h(X))=g(x)h(x)fX(x)dx−∞ZZ=g(x)h(x)fX(x)dx+g(x)h(x)fX(x)dx{x:h(x)≤0}{x:h(x)≥0}ZZ≥g(x0)h(x)fX(x)dx+g(x0)h(x)fX(x)dx{x:h(x)≤0}{x:h(x)≥0}Z∞=h(x)fX(x)dx课后答案网−∞=g(x0)Eh(X)=0.Thecasewheng(x)andwww.hackshp.cnh(x)arebothnonincreasingcanbeprovedsimilarly.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter5PropertiesofaRandomSample5.1LetX=#colorblindpeopleinasampleofsizen.ThenX∼binomial(n,p),wherep=.01.TheprobabilitythatasamplecontainsacolorblindpersonisP(X>0)=1−P(X=0),
whereP(X=0)=n(.01)0(.99)n=.99n.Thus,0P(X>0)=1−.99n>.95⇔n>log(.05)/log(.99)≈299.5.3NotethatPYi∼Bernoulliwithpi=P(Xi≥µ)=1−F(µ)foreachi.SincetheYi’sareiidnBernoulli,i=1Yi∼binomial(n,p=1−F(µ)).5.5LetY=X1+···+Xn.ThenX¯=(1/n)Y,ascaletransformation.ThereforethepdfofX¯isf(x)=1fx=nf(nx).X¯1/nY1/nY015.6a.ForZ=X−Y,setW=X.ThenY=W−Z,X=W,and|J|==1.Then−11R∞fZ,W(z,w)=fX(w)fY(w−z)·1,thusfZ(z)=−∞fX(w)fY(w−z)dw.01b.ForZ=XY,setW=X.ThenY=Z/Wand|J|=2=−1/w.Then1/w−z/wR∞fZ,W(z,w)=fX(w)fY(z/w)·|−1/w|,thusfZ(z)=−∞|−1/w|fX(w)fY(z/w)dw.01c.ForZ=X/Y,setW=X.ThenY=W/Zand|J|==w/z2.Then−w/z21/zR∞f(z,w)=f(w)f(w/z)·|w/z2|,thusf(z)=|w/z2|f(w)f(w/z)dw.Z,WXYZ−∞XY5.7Itis,perhaps,easiesttorecovertheconstantsbydoingtheintegrations.WehaveZ∞Z∞BDω
2dω=σπB,ω−z
2dω=τπD−∞1+−∞1+στand课后答案网Z"#∞AωCωω
2−ω−z
2dω−∞1+1+στZwww.hackshp.cn"#Z∞∞AωC(ω−z)1=ω
2−ω−z
2dω−Czω−z
2dω−∞1+1+−∞1+σττ"2#∞σ2ω2Cτ2ω−z=Alog1+−log1+−τπCz.2σ2τ−∞TheintegralisfiniteandequaltozeroifA=M2,C=M2forsomeconstantM.Henceσ2τ212πMz11fZ(z)=2σπB−τπD−=2,πσττπ(σ+τ)1+(z/(σ+τ))2ifB=τ,D=σ,M=−στ1.σ+τσ+τ)2z(σ+τ)1+(z)2σ+τ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-2SolutionsManualforStatisticalInference5.8a.1XnXn2(Xi−Xj)2n(n−1)i=1j=11XnXn=(X−X¯+X¯−X)2ij2n(n−1)i=1j=11XnXnhi22=(X−X¯)−2(X−X¯)(X−X¯)+(X−X¯)2n(n−1)iijji=1j=1XnXnXnXn1=n(X−X¯)2−2(X−X¯)(X−X¯)+n(X−X¯)22n(n−1)iijji=1i=1j=1j=1|{z}=0nXnnXn=(X−X¯)2+(X−X¯)2ij2n(n−1)2n(n−1)i=1j=1Xn122=(Xi−X¯)=S.n−1i=1b.Althoughallofthecalculationsherearestraightforward,thereisatediousamountofbook-keepingneeded.Itseemsthatinductionistheeasiestroute.(Note:Withoutlossofgeneralitywecanassumeθ1=0,soEXi=0.)21P4P42(i)Provetheequationforn=4.WehaveS=24i=1j=1(Xi−Xj),andtocalculateVar(S2)weneedtocalculateE(S2)2andE(S2).ThelatterexpectationisstraightforwardandwegetE(S2)=24θ.TheexpectedvalueE(S2)2=E(S4)contains256(=44)terms2ofwhich112(=4×16+4×16−42)arezero,wheneveri=j.Oftheremainingterms,•24areoftheformE(X−X)4=2(θ+3θ2)ij42•96areoftheformE(X−X)2(X−X)2=θ+3θ2ijik42•24areoftheformE(X−X)2(X−X)2=4θ2ijk`2Thus,hi21222112Var(S)=224×2(θ4+3θ2)+96(θ4+3θ2)+24×4θ4−(24θ2)=θ4−θ2.课后答案网2443(ii)Assumethattheformulaholdsforn,andestablishitforn+1.(LetSndenotethevariancebasedonnobservations.)Straightforwardalgebrawillestablishwww.hackshp.cn1XnXnXnS2=(X−X)2+2(X−X)2n+12n(n+1)ijkn+1i=1j=1k=1def’n1=[A+2B]2n(n+1)where2n−32Var(A)=4n(n−1)θ4−θ2(inductionhypothesis)n−1Var(B)=n(n+1)θ−n(n−3)θ2(XandXareindependent)42kn+1Cov(A,B)=2n(n−1)θ−θ2(someminorbookkeepingneeded)42若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition5-3Hence,211n−22Var(Sn+1)=22[Var(A)+4Var(B)+4Cov(A,B)]=θ4−θ2,4n(n+1)n+1nestablishingtheinductionandverifyingtheresult.c.Againassumethatθ1=0.Then1XnXnXn22Cov(X,S¯)=2EXk(Xi−Xj).2n(n−1)k=1i=1j=1Thedoublesumoveriandjhasn(n−1)nonzeroterms.Foreachofthese,theentireexpectationisnonzeroforonlytwovaluesofk(whenkmatcheseitheriorj).Thus22n(n−1)21Cov(X,S¯)=2EXi(Xi−Xj)=θ3,2n(n−1)nandX¯andS2areuncorrelatedifθ=0.35.9ToestablishtheLagrangeIdentityconsiderthecasewhenn=2,(ab−ab)2=a2b2+a2b2−2abab122112211221=a2b2+a2b2−2abab+a2b2+a2b2−a2b2−a2b21221122111221122=(a2+a2)(b2+b2)−(ab+ab)2.12121122Assumethatistrueforn,then!!!2nX+1nX+1nX+1a2b2−abiiiii=1i=1i=1!!!2XnXnXn=a2+a2b2+b2−ab+abin+1in+1iin+1n+1i=1i=1i=1!!!2XnXnXn=a2b2−abiiiii=1i=1i=1课后答案网n!n!n!XXX+a2b2+a2b2−2ababin+1n+1iiin+1n+1i=1i=1i=1nXwww.hackshp.cn−1XnXn=(ab−ab)2+(ab−ab)2ijjiin+1n+1ii=1j=i+1i=1XnnX+1=(ab−ab)2.ijjii=1j=i+1IfallthepointslieonastraightlinethenY−µy=c(X−µx),forsomeconstantc6=0.LetPnPn+12bi=Y−µyandai=(X−µx),thenbi=cai.Thereforei=1j=i+1(aibj−ajbi)=0.Thusthecorrelationcoefficientisequalto1.5.10a.θ1=EXi=µ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-4SolutionsManualforStatisticalInferenceθ=E(X−µ)2=σ22iθ=E(X−µ)33i=E(X−µ)2(X−µ)(Stein’slemma:Eg(X)(X−θ)=σ2Eg0(X))ii=2σ2E(X−µ)=0iθ=E(X−µ)4=E(X−µ)3(X−µ)=3σ2E(X−µ)2=3σ4.4iiii4b.VarS2=1(θ−n−3θ2)=1(3σ4−n−3σ4)=2σ.n4n−12nn−1n−1c.Usethefactthat(n−1)S2/σ2∼χ2andVarχ2=2(n−1)togetn−1n−1!2(n−1)SVar=2(n−1)σ22whichimplies((n−1))VarS2=2(n−1)andhenceσ42(n−1)2σ4VarS2==.(n−1)2/σ4n−1Remark:Anotherapproachtob),notusingtheχ2distribution,istouselinearmodeltheory.ForanymatrixAVar(X0AX)=2µ2trA2+4µθ0Aθ,whereµisσ2,θ=EX=µ1.WriteP222S2=1n(X−X¯)=1X0(I−J¯)X.Wheren−1i=1in−1n1−1−1···−1nnn.−11−1..I−J¯n=nn..........−1······1−1nnNoticethattrA2=trA=n−1,Aθ=0.So11
2σ4VarS2=Var(X0AX)=2σ4(n−1)+0=.(n−1)2(n−1)2n−15.11Letg(s)=s2.Sinceg(·)isaconvexfunction,weknowfromJensen’sinequalitythatEg(S)≥g(ES),whichimpliesσ2=ES2≥(ES)2.Takingsquareroots,σ≥ES.FromtheproofofJensen’sInequality,itisclearthat,infact,theinequalitywillbestrictunlessthereisanintervalIsuchthatgislinearonIandP(X∈I)=1.Sinces2is“linear”onlyonsinglepoints,wehaveET2课后答案网>(ET)2foranyrandomvariableT,unlessP(T=ET)=1.5.13rr!√σ2S2(n−1)EcS2=cEwww.hackshp.cnn−1σ2rZ2∞σ√1(n−1)−1−q/2=cq
q2edq,n−10Γn−12(n−1)/22pSinceS2(n−1)/σ2isthesquarerootofaχ2randomvariable.Nowadjusttheintegrandtobeanotherχ2pdfandgetrZ√σ2Γ(n/2)2n/2∞11EcS2=c·q(n−1)/2−e−q/2dq.n−1Γ((n−1)/2)2((n−1)/20Γ(n/2)2n/22|{z}=1sinceχ2pdfn√n−1Γ()n−1Soc=√2givesE(cS)=σ.2Γ(n)2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition5-55.15a.Pn+1PnX¯=i=1Xi=Xn+1+i=1Xi=Xn+1+nX¯n.n+1n+1n+1n+1b.nX+1
2n2nSn+1=Xi−X¯n+1(n+1)−1i=1nX+12Xn+1+nX¯n=Xi−(use(a))n+1i=1nX+12Xn+1nX¯n=Xi−−n+1n+1i=1nX+1X¯2
Xn+1n
=Xi−X¯n−−±X¯nn+1n+1i=1n+1"#X
2
Xn+1−X¯n1
2=Xi−X¯n−2Xi−X¯n+2Xn+1−X¯nn+1(n+1)i=1Xn

(X−X¯)2
22n+1nn+12=Xi−X¯n+Xn+1−X¯n−2+2Xn+1−X¯nn+1(n+1)i=1!Xnsince(Xi−X¯n)=012n
2=(n−1)Sn+Xn+1−X¯n.n+1P3Xi−i
225.16a.i=1i∼χ3,vu,Xi−1
utP3Xi−i
2b.ii=2i2∼t2c.Squaretherandomvariableinpartb).5.17a.LetU∼χ2andV∼χ2,independent.Theirjointpdfisp课后答案网q1p−1q−1−(u+v)p
q
u2v2e2.ΓΓ2(p+q)/222FromDefinition5.3.6,therandomvariablewww.hackshp.cnX=(U/p)/(V/q)hasanFdistribution,sowemakethetransformationx=(u/p)/(v/q)andy=u+v.(Ofcourse,manychoicesofywilldo,butthisonemakescalculationseasy.Thechoiceispromptedbytheexponentialterminthepdf.)Solvingforuandvyieldspqqxyypyu=q,v=q,and|J|=2.1+x1+xqpp1+xpWethensubstituteintofU,V(u,v)toobtain!p−1!q−1pxy22qy1qy−ypfX,Y(x,y)=p
q
qqe2.ΓΓ2(p+q)/21+x1+xq222pp1+xp若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-6SolutionsManualforStatisticalInferenceNotethatthepdffactors,showingthatXandYareindependent,andwecanreadoffthepdfsofeach:XhastheFdistributionandYisχ2.Ifweintegrateoutytorecoverthep+qproperconstant,wegettheFpdfp+q
p/2p/2−1ΓqxfX(x)=
2
.pqpp+qΓ2Γ2q21+xp2χ/pb.SinceF=p,letU∼χ2,V∼χ2andUandVareindependent.Thenwehavep,qχ2/qpqqU/pUqEFp,q=E=EE(byindependence)V/qpVp1=qE(EU=p).pVThenZ∞Z∞111q−1−v1q−2−1−vE=q
v2e2dv=q
v2e2dvV0vΓ2q/2Γ2q/2022Γq−2
2(q−2)/21
q−2(q−2)/221=Γ2=

=.Γq2q/22Γq−2q−22q/2q−2222pqqHence,EFp,q=pq−2=q−2,ifq>2.Tocalculatethevariance,firstcalculateU2q2q21E(F2)=E=E(U2)E.p,qp2V2p2V2NowE(U2)=Var(U)+(EU)2=2p+p2andZ∞111(q/2)−1−v/21E=vedv=.V20v2Γ(q/2)2q/2(q−2)(q−4)Therefore,q21q2(p+2)EF2=p(2+p)=,课后答案网p,qp2(q−2)(q−4)p(q−2)(q−4)and,hence222q(p+2)qqq+p−2Var(Fp,q)=www.hackshp.cn−2=2,q>4.p(q−2)(q−4)(q−2)q−2p(q−4)c.WriteX=U/pthen1=V/q∼F,sinceU∼χ2,V∼χ2andUandVareindependent.V/pXU/pq,ppq(p/q)XpXqYdxq−2d.LetY==,soX=and=(1−y).Thus,Yhaspdf1+(p/q)Xq+pXp(1−Y)dypp−2
pqy2q+pΓp2p(1−y)qfY(y)=
2
pqqp+q2Γ2Γ2pqy2p(1−y)1+qp(1−y)hi−1pqp−1q−1pq=B,y2(1−y)2∼beta,.2222若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition5-7p5.18IfX∼t,thenX=Z/V/pwhereZ∼n(0,1),V∼χ2andZandVareindependent.ppppa.EX=EZ/V/p=(EZ)(E1/V/p)=0,sinceEZ=0,aslongastheotherexpectationisfinite.Thisissoifp>1.Frompartb),X2∼F.ThusVarX=EX2=p/(p−2),ifp>21,p(fromExercise5.17b).b.X2=Z2/(V/p).Z2∼χ2,sotheratioisdistributedF.11,pc.ThepdfofXis"#p+1Γ()1f(x)=2.X√Γ(p/2)pπ(1+x2/p)(p+1)/2DenotethequantityinsquarebracketsbyCp.FromanextensionofStirling’sformula(Exercise1.28)wehave√
p−1+1p−1p−122−2πe21limC=lim2p→∞pp→∞√
p−2+1p−2√pπp−222−2πe22
p−1+1e−1/2p−122e−1/2e1/2=√lim2=√√,πp→∞
p−2+1√πp−222p22byanapplicationofLemma2.3.14.Applyingthelemmaagainshowsthatforeachx2
(p+1)/2x2/2lim1+x/p=e,p→∞establishingtheresult.d.AstherandomvariableF1,pisthesquareofatp,weconjecturethatitwouldconvergetothesquareofan(0,1)randomvariable,aχ2.1e.TherandomvariableqFq,pcanbethoughtofasthesumofqrandomvariables,eachatpsquared.Thus,byalloftheabove,weexpectittoconvergetoaχ2randomvariableasqp→∞.5.19a.χ2∼χ2+χ2whereχ2andχ2areindependentχ2randomvariableswithqandd=p−qpqdqddegreesoffreedom.Sinceχ2isapositiverandomvariable,foranya>0,dP(χ>a)=P(χ2+χ2>a)>P(χ2>a).课后答案网pqdqb.Fork>k,kF∼(U+V)/(W/ν),whereU,VandWareindependentandU∼χ2,121k1,νk2V∼χ2andW∼χ2.Foranya>0,becauseV/(W/ν)isapositiverandomvariable,k1−k2νwehavewww.hackshp.cnP(k1Fk1,ν>a)=P((U+V)/(W/ν)>a)>P(U/(W/ν)>a)=P(k2Fk2,ν>a).c.α=P(Fk,ν>Fα,k,ν)=P(kFk,ν>kFα,k,ν).So,kFα,k,νistheαcutoffpointfortherandomvariablekFk,ν.BecausekFk,νisstochasticallylargerthat(k−1)Fk−1,ν,theαcutoffforkFk,νislargerthantheαcutofffor(k−1)Fk−1,ν,thatiskFα,k,ν>(k−1)Fα,k−1,ν.5.20a.ThegivenintegralisZ∞1−t2x/2√1(ν/2)−1−νx/2√eνx(νx)edx2πν/20Γ(ν/2)2ν/2Z∞1ν−t2x/2((ν+1)/2)−1−νx/2=√exedx2πν/2Γ(ν/2)20若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-8SolutionsManualforStatisticalInferenceν/2Z∞1ν((ν+1)/2)−1−(ν+t2)x/2integrandiskernelof=√xedx2πν/2gamma((ν+1)/2,2/(ν+t2)Γ(ν/2)20ν/2(ν+1)/21ν2=√Γ((ν+1)/2)2πν/2ν+t2Γ(ν/2)21Γ((ν+1)/2)1=√,νπΓ(ν/2)(1+t2/ν)(ν+1)/2thepdfofatνdistribution.b.DifferentiatebothsideswithrespecttottoobtainZ∞νfF(νt)=yf1(ty)fν(y)dy,0wherefFistheFpdf.Nowwriteoutthetwochi-squaredpdfsandcollecttermstoget−1/2Z∞t(ν−1)/2−(1+t)y/2νfF(νt)=yedy(ν+1)/2Γ(1/2)Γ(ν/2)20t−1/2Γ(ν+1)2(ν+1)/2=2.(ν+1)/2(ν+1)/2Γ(1/2)Γ(ν/2)2(1+t)Nowdefiney=νttogetΓ(ν+1)−1/2(y/ν)f(y)=2,FνΓ(1/2)Γ(ν/2)(ν+1)/2(1+y/ν)thepdfofanF1,ν.c.Againdifferentiatebothsideswithrespecttot,writeoutthechi-squaredpdfs,andcollecttermstoobtain−m/2Z∞t(m+ν−2)/2−(1+t)y/2(ν/m)fF((ν/m)t)=yedy.(ν+m)/2Γ(m/2)Γ(ν/2)20Now,asbefore,integratethegammakernel,collectterms,anddefiney=(ν/m)ttogetΓ(ν+m)mm/2ym/2−1f(y)=2,FΓ(m/2)Γ(ν/2)ν(ν+m)/2课后答案网(1+(m/ν)y)thepdfofanFm,ν.5.21Letmdenotethemedian.Then,forgeneralwww.hackshp.cnnwehaveP(max(X1,...,Xn)>m)=1−P(Xi≤mfori=1,2,...,n)nn1=1−[P(X1≤m)]=1−.25.22CalculatingthecdfofZ2,weobtain√2FZ2(z)=P((min(X,Y))≤z)=P(−z≤min(X,Y)≤z)√√=P(min(X,Y)≤z)−P(min(X,Y)≤−z)√√=[1−P(min(X,Y)>z)]−[1−P(min(X,Y)>−z)]√√=P(min(X,Y)>−z)−P(min(X,Y)>z)√√√√=P(X>−z)P(Y>−z)−P(X>z)P(Y>z),若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition5-9whereweusetheindependenceofXandY.SinceXandYareidenticallydistributed,P(X>a)=P(Y>a)=1−FX(a),so√√√22FZ2(z)=(1−FX(−z))−(1−FX(z))=1−2FX(−z),√√since1−FX(z)=FX(−z).Differentiatingandsubstitutinggivesd√11−z/2−1/2fZ2(z)=FZ2(z)=fX(−z)√=√ez,dzz2πthepdfofaχ2randomvariable.Alternatively,122P(Z≤z)=P[min(X,Y)]≤z√√=P(−z≤min(X,Y)≤z)√√√√=P(−z≤X≤z,X≤Y)+P(−z≤Y≤z,Y≤X)√√=P(−z≤X≤z|X≤Y)P(X≤Y)√√+P(−z≤Y≤z|Y≤X)P(Y≤X)1√√1√√=P(−z≤X≤z)+P(−z≤Y≤z),22usingthefactsthatXandYareindependent,andP(Y≤X)=P(X≤Y)=1.Moreover,2sinceXandYareidenticallydistributed2√√P(Z≤z)=P(−z≤X≤z)andd√√1−z/21−1/2−z/21−1/2fZ2(z)=P(−z≤X≤z)=√(ez+ez)dz2π221−1/2−z/2=√ze,2πthepdfofaχ2.15.23X∞X∞P(Z>z)课后答案网=P(Z>z|x)P(X=x)=P(U1>z,...,Ux>z|x)P(X=x)x=1x=1X∞Yx=www.hackshp.cnP(Ui>z)P(X=x)(byindependenceoftheUi’s)x=1i=1X∞X∞1=P(U>z)xP(X=x)=(1−z)xi(e−1)x!x=1x=11X∞(1−z)xe1−z−1==0v,fX(i)|X(j)(u|v)www.hackshp.cn(n−j)!n−ii−1−jj−n=fX(u)[1−FX(u)][FX(u)−FX(v)][1−FX(v)](n−1)!(i−1−j)!i−j−1n−i(n−j)!fX(u)FX(u)−FX(v)FX(u)−FX(v)=1−.(i−j−1)!(n−i)!1−FX(v)1−FX(v)1−FX(v)Thisisthepdfofthe(i−j)thorderstatisticfromasampleofsizen−j,fromapopulationwithpdfgivenbythetruncateddistribution,f(u)=fX(u)/(1−FX(v)),u>v.b.FromExample5.4.7,n−2nn(n−1)r/a1fV|R(v|r)=n−2n=a−r,r/2100.4=Pi=1Xi/100>1.004=P(X>¯1.004).BytheCLT,P(X>¯1.004)≈P(Z>(1.004−1)/(.05/10))=P(Z>.8)=.2119.5.30FromtheCLTwehave,approximately,X¯∼n(µ,σ2/n),X¯∼n(µ,σ2/n).SinceX¯andX¯1212areindependent,X¯−X¯∼n(0,2σ2/n).Thus,wewant12
.99≈PX¯1−X¯2<σ/5!−σ/5X¯1−X¯2σ/5=Pp0,p√p√p√p√PXn−a>=PXn−aXn+a>Xn+ap√课后答案网=P|Xn−a|>Xn+a√
≤P|Xn−a|>a→0,√√asn→∞,sinceXnwww.hackshp.cn→ainprobability.ThusXn→ainprobability.b.Forany>0,aaaP−1≤=P≤Xn≤Xn1+1−aa=Pa−≤Xn≤a+1+1−aaaa≥Pa−≤Xn≤a+a+0thereexistNsuchthatifn>N,thenP(Xn+Yn>c)>1−.ChooseN1suchthatP(Xn>−m)>1−/2andN2suchthatP(Yn>c+m)>1−/2.ThenP(Xn+Yn>c)≥P(Xn>−m,+Yn>c+m)≥P(Xn>−m)+P(Yn>c+m)−1=1−.5.34UsingEX¯=µandVarX¯=σ2/n,weobtainnn√√√n(X¯n−µ)nnE=E(X¯n−µ)=(µ−µ)=0.σσσ√n(X¯−µ)nnnσ2nVar=Var(X¯n−µ)=VarX¯==1.σσ2σ2σ2n5.35a.Xi∼exponential(1).µX=1,VarX=1.FromtheCLT,X¯nisapproximatelyn(1,1/n).SoX¯n−1X¯n−1√→Z∼n(0,1)andP√≤x→P(Z≤x).1/n1/nb.dd1−x2/2P(Z≤x)=FZ(x)=fZ(x)=√e.dxdx2πdX¯n−1P√≤xdx1/n!!dXn√Xn=Xi≤xn+nW=Xi∼gamma(n,1)dxi=1i=1d√√√1√√√=F(xn+n)=f(xn+n)·n=(xn+n)n−1e−(xn+n)n.WWdxΓ(n)√√√2Therefore,(1/Γ(n))(xn+n)n−1e−(xn+n)n≈√1e−x/2asn→∞.Substitutingx=0√2πyieldsn!≈nn+1/2e−n2π.5.37a.Fortheexactcalculations,usethefactthatVnisitselfdistributednegativebinomial(10r,p).Theresultsaresummarizedinthefollowingtable.Notethattherecursionrelationofproblem课后答案网3.48canbeusedtosimplifycalculations.P(Vn=v)www.hackshp.cn(a)(b)(c)vExactNormalApp.Normalw/cont.0.0008.0071.00561.0048.0083.01132.0151.0147.02013.0332.0258.02634.0572.0392.05495.0824.0588.06646.1030.0788.08827.1148.0937.10078.1162.1100.11379.1085.1114.114410.0944.1113.1024若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition5-13b.Usingthenormalapproximation,wehaveµv=r(1−p)/p=20(.3)/.7=8.57andqp2σv=r(1−p)/p=(20)(.3)/.49=3.5.Then,Vn−8.571−8.57P(Vn=0)=1−P(Vn≥1)=1−P≥=1−P(Z≥−2.16)=.0154.3.53.5AnotherwaytoapproximatethisprobabilityisV−8.570−8.57P(Vn=0)=P(Vn≤0)=P≤=P(Z≤−2.45)=.0071.3.53.5ContinuinginthiswaywehaveP(V=1)=P(V≤1)−P(V≤0)=.0154−.0071=.0083,etc.(k−.5)−8.57(k+.5)−8.57c.Withthecontinuitycorrection,computeP(V=k)byP≤Z≤,so3.53.5P(V=0)=P(−9.07/3.5≤Z≤−8.07/3.5)=.0104−.0048=.0056,etc.Noticethatthecontinuitycorrectiongivessomeimprovementovertheuncorrectednormalapproximation.5.39a.Ifhiscontinuousgiven>0thereexitsδsuchthat|h(xn)−h(x)|<for|xn−x|<δ.SinceX1,...,XnconvergesinprobabilitytotherandomvariableX,thenlimn→∞P(|Xn−X|<δ)=1.Thuslimn→∞P(|h(Xn)−h(X)|<)=1.b.DefinethesubsequenceXj(s)=s+I[a,b](s)suchthatinI[a,b],aisalways0,i.e,thesubse-quenceX1,X2,X4,X7,....Forthissubsequencensifs>0Xj(s)→s+1ifs=0.5.41a.Let=|x−µ|.(i)Forx−µ≥0P(|Xn−µ|>)=P(|Xn−µ|>x−µ)=P(Xn−µ<−(x−µ))+P(Xn−µ>x−µ)≥P(Xn−µ>x−µ)=P(Xn>x)=1−P(Xn≤x).Therefore,0=lim课后答案网n→∞P(|Xn−µ|>)≥limn→∞1−P(Xn≤x).Thuslimn→∞P(Xn≤x)≥1.(ii)Forx−µ<0www.hackshp.cnP(|Xn−µ|>)=P(|Xn−µ|>−(x−µ))=P(Xn−µ−(x−µ))≥P(Xn−µ)≥limn→∞P(Xn≤x).By(i)and(ii)theresultsfollows.b.Forevery>0,P(|Xn−µ|>)≤P(Xn−µ<−)+P(Xn−µ>)=P(Xn<µ−)+1−P(Xn≤µ+)→0asn→∞.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-14SolutionsManualforStatisticalInferencepp5.43a.P(|Yn−θ|<)=P(n)(Yn−θ)<(n).Therefore,pplimP(|Yn−θ|<)=limP(n)(Yn−θ)<(n)=P(|Z|<∞)=1,n→∞n→∞whereZ∼n(0,σ2).ThusY→θinprobability.n√b.BySlutsky’sTheorem(a),g0(θ)n(Y−θ)→g0(θ)XwhereX∼n(0,σ2).Therefore√√nn[g(Y)−g(θ)]=g0(θ)n(Y−θ)→n(0,σ2[g0(θ)]2).nn5.45Wedopart(a),theotherpartsaresimilar.UsingMathematica,theexactcalculationisIn[120]:=f1[x_]=PDF[GammaDistribution[4,25],x]p1=Integrate[f1[x],{x,100,[Infinity]}]//N1-CDF[BinomialDistribution[300,p1],149]Out[120]=e^(-x/25)x^3/2343750Out[121]=0.43347Out[122]=0.0119389.TheanswercanalsobesimulatedinMathematicaorinR.HereistheRcodeforsimulatingthesameprobabilityp1<-mean(rgamma(10000,4,scale=25)>100)mean(rbinom(10000,300,p1)>149)Ineachcase10,000randomvariablesweresimulated.Weobtainedp1=0.438andabinomialprobabilityof0.0108.5.47a.−2log(U)∼exponential(2)∼χ2.ThusYisthesumofνindependentχ2randomvariables.j22ByLemma5.3.2(b),Y∼χ2.2νb.βlog(Uj)∼exponential(2)∼gamma(1,β).ThusYisthesumofindependentgammarandomvariables.ByExample4.6.8,Y∼gamma(a,β)PaPbc.LetV=j=1log(Uj)∼gamma(a,1).SimilarlyW=j=1log(Uj)∼gamma(b,1).ByExercise4.课后答案网24,V∼beta(a,b).V+W5.49a.SeeExample2.1.4.b.X=g(U)=−log1−U.Theng−1(x)=1.ThusUwww.hackshp.cn1+e−ye−ye−yfX(x)=1×=−∞meanobs[1]5.231>variance[1]1.707346b.obs<-rhyper(1000,8,2,4)meanobs<-mean(obs)variance<-var(obs)课后答案网hist(obs)Output:>meanobswww.hackshp.cn[1]3.169>variance[1]0.4488879c.obs<-rnbinom(1000,5,1/3)meanobs<-mean(obs)variance<-var(obs)hist(obs)Output:>meanobs[1]10.308>variance[1]29.51665若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-16SolutionsManualforStatisticalInference5.55LetXdenotethenumberofcomparisons.ThenX∞X∞EX=P(X>k)=1+P(U>Fy(yk−1))k=0k=1X∞X∞=1+(1−Fy(yk−1))=1+(1−Fy(yi))=1+EYk=1k=05.57a.Cov(Y1,Y2)=Cov(X1+X3,X2+X3)=Cov(X3,X3)=λ3sinceX1,X2andX3areindependent.b.n1ifXi=X3=0Zi=0otherwisep=P(Z=0)=P(Y=0)=P(X=0,X=0)=e−(λi+λ3).ThereforeZareiiii3iBernoulli(pi)withE[Zi]=pi,Var(Zi)=pi(1−pi)andE[Z1Z2]=P(Z1=1,Z2=1)=P(Y1=0,Y2=0)=P(X1+X3=0,X2+X3=0)=P(X1=0)P(X2=0)P(X3=0)=e−λ1e−λ2e−λ3.Therefore,Cov(Z1,Z2)=E[Z1Z2]−E[Z1]E[Z2]=e−λ1e−λ2e−λ3−e−(λi+λ3)e−(λ2+λ3)=e−(λi+λ3)e−(λ2+λ3)(eλ3−1)=pp(eλ3−1).12λ3√p1p2(e√−1)ThusCorr(Z1,Z2)=.p1(1−p1)p2(1−p2)c.E[Z1Z2]≤pi,thereforeCov(Z1,Z2)=E[Z1Z2]−E[Z1]E[Z2]≤p1−p1p2=p1(1−p2),andCov(Z1,Z2)≤p2(1−p1).Therefore,pp1(1−p2)p1(1−p2)课后答案网Corr(Z1,Z2)≤pp=pp1(1−p1)p2(1−p2)p2(1−p1)andpp2(1−p1)p2(1−p1)Corr(www.hackshp.cnZ1,Z2)≤pp=pp1(1−p1)p2(1−p2)p1(1−p2)whichimpliestheresult.5.59P(V≤y,U<1f(V))1cYP(Y≤y)=P(V≤y|U0andb−[b]>0andy∈(0,1).Γ([a])Γ([b])若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition5-17Γ(a+b)ya−1(1−y)b−1Γ(a)Γ(b)b.M=supyΓ([a]+b)y[a]−1(1−y)b−1<∞,sincea−[a]>0andy∈(0,1).Γ([a])Γ(b)Γ(a+b)ya−1(1−y)b−1Γ(a)Γ(b)0c.M=supyΓ([a]+1+β)y[a]+1−1(1−y)b0−1<∞,sincea−[a]−1<0andy∈(0,1).b−b>0Γ([a]+1)Γ(b0)whenb0=[b]andwillbeequaltozerowhenb0=b,thusitdoesnotaffecttheresult.d.Letf(y)=yα(1−y)β.Thendf(y)α−1βαβ−1α−1β−1=αy(1−y)−yβ(1−y)=y(1−y)[α(1−y)+βy]dywhichismaximizeaty=α.Thereforefor,α=a−a0andβ=b−b0α+βΓ(a+b)a−a0b−b0a−a0b−b0Γ(a)Γ(b)M=.Γ(a0+b0)a−a0+b−b0a−a0+b−b0Γ(a0)Γ(b0)a−a0b−b000WeneedtominimizeMina0andb0.Firstconsidera−ab−b.Leta−a0+b−b0a−a0+b−b0α
αc−α
c−αα1c=α+β,thenthistermbecomes.Thistermismaximizeat=,thisccc2Γ(a+b)0011(a−a+b−b)Γ(a)Γ(b)isatα=c.ThenM=().NotethattheminimumthatMcouldbe22Γ(a0+b0)Γ(a0)Γ(b0)isone,whichitisattainwhena=a0andb=b0.Otherwisetheminimumwilloccurwhena−a0andb−b0areminimumbutgreaterorequalthanzero,thisiswhena0=[a]andb0=[b]ora0=aandb0=[b]ora0=[a]andb0=b.2−y√1e225.63M=sup2π.Letf(y)=−y+|y|.Thenf(y)ismaximizeaty=1wheny≥0andaty−|y|2λλ1eλ2λ−1√1e2λ21−12π0y=wheny<0.ThereforeinbothcasesM=−1.TominimizeMletM=λe2λ2.λ1eλ22λ0ThendlogM=1−1,thereforeMisminimizeatλ=1orλ=−1.Thusthevalueofλthatdλλλ3willoptimizethealgorithmisλ=1.5.65XmXm1Pmf(Yi)I(Y≤x)∗∗mi=1g(Yi)iP(X≤x)=P(X≤x|qi)qi=I(Yi≤x)qi=1Pmf(Yi)i=1i=1mi=1g(Yi)f(Y)Rxf(y)Zx课后答案网−→Egg(Y)I(Y≤x)−∞g(y)g(y)dy=R=f(y)dy.m→∞f(Y)∞f(y)Egg(Y)−∞g(y)g(y)dy−∞5.67AnRcodetogeneratethesampleofsize100fromthespecifieddistributionisshownforpartwww.hackshp.cnc).TheMetropolisAlgorithmisusedtogenerate2000variables.Amongotheroptionsonecanchoosethe100variablesinpositions1001to1100ortheonesinpositions1010,1020,...,2000.a.WewanttogenerateX=σZ+µwhereZ∼Student’stwithνdegreesoffreedom.Thereforewefirstcangenerateasampleofsize100fromaStudent’stdistributionwithνdegreesoffreedomandthenmakethetransformationtoobtaintheX’s.ThusfZ(z)=ν+1Γ(2)√1

1.LetV∼n(0,ν)sincegivenνwecansetΓ(ν)νπ(v+1)/2ν−221+z2ννEV=EZ=0,andVar(V)=Var(Z)=.ν−2Now,followthealgorithmonpage254andgeneratethesampleZ1,Z2...,Z100andthencalculateXi=σZi+µ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-18SolutionsManualforStatisticalInference22−(logx−µ)/2σb.f(x)=√1e.LetV∼gamma(α,β)whereX2πσx222(eµ+(σ/2))2e2(µ+σ)−e2µ+σα=,andβ=,e2(µ+σ2)−e2µ+σ2eµ+(σ2/2)sincegivenµandσ2wecanset2EV=αβ=eµ+(σ/2)=EXand22Var(V)=αβ2=e2(µ+σ)−e2µ+σ=Var(X).Now,followthealgorithmonpage254.α−xc.f(x)=αeβxα−1.LetV∼exponential(β).Now,followthealgorithmonpage254whereXβ()α−1−Vα+Vi−Zi−1+ZαViii−1ρi=mineβ,1α−1Zi−1AnRcodetogenerateasamplesizeof100fromaWeibull(3,2)is:#initializeaandbb<-2a<-3Z<-rexp(1,1/b)ranvars<-matrix(c(Z),byrow=T,ncol=1)for(iinseq(2000)){U<-runif(1,min=0,max=1)V<-rexp(1,1/b)p<-pmin((V/Z)^(a-1)*exp((-V^a+V-Z+Z^a)/b),1)if(U<=p)Z<-Vranvars<-cbind(ranvars,Z)}#Oneoption:chooseelementsinposition1001,1002,...,1100tobethesamplevector.1<-ranvars[1001:1100]课后答案网mean(vector.1)var(vector.1)#Anotheroption:chooseelementsinposition1010,1020,...,2000www.hackshp.cntobethesamplevector.2<-ranvars[seq(1010,2000,10)]mean(vector.2)var(vector.2)Output:[1]1.048035[1]0.1758335[1]1.130649[1]0.17787245.69Letw(v,z)=fY(v)fV(z),andthenρ(v,z)=min{w(v,z),1}.WewillshowthatfV(v)fY(z)Zi∼fY⇒P(Zi+1≤a)=P(Y≤a).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition5-19WriteP(Zi+1≤a)=P(Vi+1≤aandUi+1≤ρi+1)+P(Zi≤aandUi+1>ρi+1).SinceZi∼fY,suppressingtheunnecessarysubscriptswecanwriteP(Zi+1≤a)=P(V≤aandU≤ρ(V,Y))+P(Y≤aandU>ρ(V,Y)).AddandsubtractP(Y≤aandU≤ρ(V,Y))togetP(Zi+1≤a)=P(Y≤a)+P(V≤aandU≤ρ(V,Y))−P(Y≤aandU≤ρ(V,Y)).ThusweneedtoshowthatP(V≤aandU≤ρ(V,Y))=P(Y≤aandU≤ρ(V,Y)).WriteouttheprobabilityasP(V≤aandU≤ρ(V,Y))ZaZ∞=ρ(v,y)fY(y)fV(v)dydv−∞−∞ZaZ∞fY(v)fV(y)=I(w(v,y)≤1)fY(y)fV(v)dydv−∞−∞fV(v)fY(y)ZaZ∞+I(w(v,y)≥1)fY(y)fV(v)dydv−∞−∞ZaZ∞=I(w(v,y)≤1)fY(v)fV(y)dydv−∞−∞ZaZ∞+I(w(v,y)≥1)fY(y)fV(v)dydv.−∞−∞Now,noticethatw(v,y)=1/w(y,v),andthusfirsttermabovecanbewrittenZaZ∞I(w(v,y)≤1)fY(v)fV(y)dydv−∞−∞ZaZ∞=I(w(y,v)>1)fY(v)fV(y)dydv课后答案网−∞−∞=P(Y≤a,ρ(V,Y)=1,U≤ρ(V,Y)).ThesecondtermisZaZwww.hackshp.cn∞I(w(v,y)≥1)fY(y)fV(v)dydv−∞−∞ZaZ∞=I(w(y,v)≤1)fY(y)fV(v)dydv−∞−∞ZaZ∞fV(y)fY(v)=I(w(y,v)≤1)fY(y)fV(v)dydv−∞−∞fV(y)fY(v)ZaZ∞fY(y)fV(v)=I(w(y,v)≤1)fV(y)fY(v)dydv−∞−∞fV(y)fY(v)ZaZ∞=I(w(y,v)≤1)w(y,v)fV(y)fY(v)dydv−∞−∞=P(Y≤a,U≤ρ(V,Y),ρ(V,Y)≤1).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn5-20SolutionsManualforStatisticalInferencePuttingitalltogetherwehaveP(V≤aandU≤ρ(V,Y))=P(Y≤a,ρ(V,Y)=1,U≤ρ(V,Y))+P(Y≤a,U≤ρ(V,Y),ρ(V,Y)≤1)=P(Y≤aandU≤ρ(V,Y)),andhenceP(Zi+1≤a)=P(Y≤a),sofYisthestationarydensity.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter6PrinciplesofDataReduction6.1BytheFactorizationTheorem,|X|issufficientbecausethepdfofXis21−x2/2σ21−|x|2/2σ22f(x|σ)=√e=√e=g(|x||σ)·1.|{z}2πσ2πσh(x)6.2BytheFactorizationTheorem,T(X)=mini(Xi/i)issufficientbecausethejointpdfisYnf(x,...,x|θ)=eiθ−xiI(x)=einθI(T(x))·e−Σixi.1n(iθ,+∞)i(θ,+∞)|{z}|{z}i=1h(x)g(T(x)|θ)Notice,weusethefactthati>0,andthefactthatallxis>iθifandonlyifmini(xi/i)>θ.6.3Letx(1)=minixi.ThenthejointpdfisnnY1eµ/σf(x,...,x|µ,σ)=e−(xi−µ)/σI(x)=e−Σixi/σI(x)·1.1nσ(µ,∞)iσ(µ,∞)(1)|{z}i=1|{z}h(x)g(x(1),Σixi|µ,σ)P
Thus,bytheFactorizationTheorem,X(1),iXiisasufficientstatisticfor(µ,σ).6.4Thejointpdfis(!)YnXkXkXnYnh(x课后答案网)c(θ)expw(θ)t(x)=c(θ)nexpw(θ)t(x)·h(x).jiijiijjj=1i=1i=1j=1j=1|{z}|{z}www.hackshp.cng(T(x)|θ)h(x)PPnnBytheFactorizationTheorem,j=1t1(Xj),...,j=1tk(Xj)isasufficientstatisticforθ.6.5ThesampledensityisgivenbyYnYn1f(xi|θ)=I(−i(θ−1)≤xi≤i(θ+1))2iθi=1i=1n!1nY1xxii=Imin≥−(θ−1)Imax≤θ+1.2θiiii=1Thus(minXi/i,maxXi/i)issufficientforθ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn6-2SolutionsManualforStatisticalInference6.6Thejointpdfisgivenbynnn!α−1Y11Yf(x,...,x|α,β)=xα−1e−xi/β=xe−Σixi/β.1nΓ(α)βαiΓ(α)βαii=1i=1QnPnBytheFactorizationTheorem,(i=1Xi,i=1Xi)issufficientfor(α,β).6.7Letx(1)=mini{x1,...,xn},x(n)=maxi{x1,...,xn},y(1)=mini{y1,...,yn}andy(n)=maxi{y1,...,yn}.Thenthejointpdfisf(x,y|θ)Yn1=I(θ1,θ3)(xi)I(θ2,θ4)(yi)(θ3−θ1)(θ4−θ2)i=1n1=I(θ1,∞)(x(1))I(−∞,θ3)(x(n))I(θ2,∞)(y(1))I(−∞,θ4)(y(n))·|{z}1.(θ3−θ1)(θ4−θ2)|{z}h(x)g(T(x)|θ)
BytheFactorizationTheorem,X(1),X(n),Y(1),Y(n)issufficientfor(θ1,θ2,θ3,θ4).6.9UseTheorem6.2.13.a.("!#)−n/2−Σi(xi−θ)2/2XnXnf(x|θ)(2π)e122f(y|θ)=−n/2−Σ(y−θ)2/2=exp−2xi−yi+2θn(y¯−x¯).(2π)eiii=1i=1Thisisconstantasafunctionofθifandonlyif¯y=¯x;thereforeX¯isaminimalsufficientstatisticforθ.b.Note,forX∼locationexponential(θ),therangedependsontheparameter.NowQne−(xi−θ)I(x)
f(x|θ)i=1(θ,∞)i=Qn
f(y|θ)e−(yi−θ)I(θ,∞)(y)i=1ienθe−ΣixiQnI(x)e−ΣixiI(minx)Qi=1(θ,∞)i(θ,∞)i=n=.enθe−ΣiyiI(θ,∞)(yi)e−ΣiyiI(θ,∞)(minyi)课后答案网i=1Tomaketheratioindependentofθweneedtheratioofindicatorfunctionsindependentofθ.Thiswillbethecaseifandonlyifmin{x1,...,xn}=min{y1,...,yn}.SoT(X)=min{X1,...,Xn}isaminimalsufficientstatistic.www.hackshp.cnc.−Σi(xi−θ)Qn1+e−(yi−θ)
2f(x|θ)ei=1=Qn
2−Σi(y−θ)f(y|θ)1+e−(xi−θ)eii=1Q
!2n1+e−(yi−θ)=e−Σi(yi−xi)Qi=1
.n1+e−(xi−θ)i=1Thisisconstantasafunctionofθifandonlyifxandyhavethesameorderstatistics.Therefore,theorderstatisticsareminimalsufficientforθ.d.Thisisadifficultproblem.Theorderstatisticsareaminimalsufficientstatistic.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition6-3e.Fixsamplepointsxandy.DefineA(θ)={i:xi≤θ},B(θ)={i:yi≤θ},a(θ)=thenumberofelementsinA(θ)andb(θ)=thenumberofelementsinB(θ).Thenthefunctionf(x|θ)/f(y|θ)dependsonθonlythroughthefunctionXnXn|xi−θ|−|yi−θ|i=1i=1XXXX=(θ−xi)+(xi−θ)−(θ−yi)−(yi−θ)i∈A(θ)i∈A(θ)ci∈B(θ)i∈B(θ)c=(a(θ)−[n−a(θ)]−b(θ)+[n−b(θ)])θXXXX+−xi+xi+yi−yii∈A(θ)i∈A(θ)ci∈B(θ)i∈B(θ)cXXXX=2(a(θ)−b(θ))θ+−xi+xi+yi−yi.i∈A(θ)i∈A(θ)ci∈B(θ)i∈B(θ)cConsideranintervalofθsthatdoesnotcontainanyxisoryis.Thesecondtermisconstantonsuchaninterval.Thefirsttermwillbeconstant,ontheintervalifandonlyifa(θ)=b(θ).Thiswillbetrueforallsuchintervalsifandonlyiftheorderstatisticsforxarethesameastheorderstatisticsfory.Therefore,theorderstatisticsareaminimalsufficientstatistic.6.10ToproveT(X)=(X(1),X(n))isnotcomplete,wewanttofindg[T(X)]suchthatEg[T(X)]=0forallθ,butg[T(X)]6≡0.AnaturalcandidateisR=X(n)−X(1),therangeofX,becausebyExample6.2.17itsdistributiondoesnotdependonθ.FromExample6.2.17,R∼beta(n−1,2).ThusER=(n−1)/(n+1)doesnotdependonθ,andE(R−ER)=0forallθ.Thusg[X(n),X(1)]=X(n)−X(1)−(n−1)/(n+1)=R−ERisanonzerofunctionwhoseexpectedvalueisalways0.So,(X(1),X(n))isnotcomplete.Thisproblemcanbegeneralizedtoshowthatifafunctionofasufficientstatisticisancillary,thenthesufficientstatisticisnotcomplete,becausetheexpectationofthatfunctiondoesnotdependonθ.Thatprovidestheopportunitytoconstructanunbiased,nonzeroestimatorofzero.6.11a.Thesearealllocationfamilies.LetZ(1),...,Z(n)betheorderstatisticsfromarandomsampleofsizenfromthestandardpdff(z|0).Then(Z(1)+θ,...,Z(n)+θ)hasthesamejointdistributionas(X(1),...,X(n)),and(Y(1),...,Y(n−1))hasthesamejointdistributionas(Z(n)+θ−(Z(1)+θ),...,Z(n)+θ−(Z(n−1)+θ))=(Z(n)−Z(1),...,Z(n)−Z(n−1)).Thelastvectordependsonlyon(Z1,...,Zn)whosedistributiondoesnotdependonθ.So,(Y(1),...,Y课后答案网(n−1))isancillary.b.Fora),Basu’slemmashowsthat(Y1,...,Yn−1)isindependentofthecompletesufficientstatistic.Forc),d),ande)theorderstatisticsaresufficient,so(Y1,...,Yn−1)isnotinde-pendentofthesufficientstatistic.Forb),www.hackshp.cnX(1)issufficient.DefineYn=X(1).Thenthejointpdfof(Y1,...,Yn)isnY−1−n(y−θ)−(n−1)yyi0µ.Y1XXNow,writeEg(Y)=R∞g(y)ne−n(y−µ)dy.Ifthisiszeroforallµ,thenR∞g(y)e−nydy=0µ1µµforallµ(becausenenµ>0forallµanddoesnotdependony).Moreover,Z∞d−ny−nµ0=g(y)edy=−g(µ)edµµforallµ.Thisimpliesg(µ)=0forallµ,soX(1)iscomplete.b.Basu’sTheoremsaysthatifX(1)isacompletesufficientstatisticforµ,thenX(1)isinde-pendentofanyancillarystatistic.Therefore,weneedtoshowonlythatS2hasdistributionindependentofµ;thatis,S2isancillary.Recognizethatf(x|µ)isalocationfamily.SowecanwriteXi=Zi+µ,whereZ1,...,Znisarandomsamplefromf(x|0).Then1X1X1XS2=(X−X¯)2=((Z+µ)−(Z¯+µ))2=(Z−Z¯)2.iiin−1n−1n−1BecauseS2isafunctionofonlyZ,...,Z,thedistributionofS2doesnotdependonµ;1nthatis,S2isancillary.Therefore,byBasu’stheorem,S2isindependentofX.(1)6.31a.(i)ByExercise3.28thisisaone-dimensionalexponentialfamilywitht(x)=x.ByTheoremPP6.2.25,iXiisacompletesufficientstatistic.X¯isaone-to-onefunctionofiXi,soX¯isalsoacompletesufficientstatistic.FromTheorem5.3.1weknowthat(n−1)S2/σ2∼χ2=gamma((n−1)/2,2).S2=[σ2/(n−1)][(n−1)S2/σ2],asimplescalen−1transformation,hasagamma((n−1)/2,2σ2/(n−1))distribution,whichdoesnotdependonµ;thatis,S2isancillary.ByBasu’sTheorem,X¯andS2areindependent.(ii)TheindependenceofX¯andS2isdeterminedbythejointdistributionof(X,S¯2)foreachvalueof(课后答案网µ,σ2).Bypart(i),foreachvalueof(µ,σ2),X¯andS2areindependent.b.(i)µisalocationparameter.ByExercise6.14,M−X¯isancillary.Asinpart(a)X¯isacompletesufficientstatistic.ByBasu’sTheorem,X¯andM−X¯areindependent.Becausetheyareindependent,byTheorem4.5.6Varwww.hackshp.cnM=Var(M−X¯+X¯)=Var(M−X¯)+VarX¯.(ii)IfS2isasamplevariancecalculatedfromanormalsampleofsizeN,(N−1)S2/σ2∼χ2.Hence,(N−1)2VarS2/(σ2)2=2(N−1)andVarS2=2(σ2)2/(N−1).BothMN−1andM−X¯areasymptoticallynormal,so,M1,...,MNandM1−X¯1,...,MN−X¯Nareeachapproximatelynormalsamplesifnisreasonablelarge.Thus,usingtheaboveexpressionwegetthetwogivenexpressionswhereinthestraightforwardcaseσ2referstoVarM,andintheswindlecaseσ2referstoVar(M−X¯).c.(i)k"k#kkXXk
indep.Xk
E(X)=EY=EY=EEY.YYY
DividebothsidesbyEYktoobtainthedesiredequality.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn6-8SolutionsManualforStatisticalInferenceP(ii)Ifαisfixed,T=iXiisacompletesufficientstatisticforβbyTheorem6.2.25.Becauseβisascaleparameter,ifZ1,...,Znisarandomsamplefromagamma(Pα,P1)distribution,thenX(i)/Thasthesamedistributionas(βZ(i))/(βiZi)=Z(i)/(iZi),andthisdistributiondoesnotdependonβ.Thus,X(i)/Tisancillary,andbyBasu’sTheorem,itisindependentofT.WehaveX(i)X(i)indep.X(i)part(i)E(X(i))E(X(i)|T)=ETT=TET=TE=T.TTTETNote,thisexpressioniscorrectforeachfixedvalueof(α,β),regardlesswhetherαis“known”ornot.6.32IntheFormalLikelihoodPrinciple,takeE1=E2=E.ThentheconclusionisEv(E,x1)=Ev(E,x2)ifL(θ|x1)/L(θ|x2)=c.Thusevidenceisequalwheneverthelikelihoodfunctionsareequal,andthisfollowsfromFormalSufficiencyandConditionality.6.33a.Forallsamplepointsexcept(2,x∗)(butincluding(1,x∗)),T(j,x)=(j,x).Hence,21jjg(T(j,x)|θ)h(j,x)=g((j,x)|θ)1=f∗((j,x)|θ).jjjjFor(2,x∗)wealsohave2∗∗∗∗∗1∗g(T(2,x2)|θ)h(2,x2)=g((1,x1)|θ)C=f((1,x1)|θ)C=Cf1(x1|θ)21∗1∗1∗∗∗=CL(θ|x1)=L(θ|x2)=f2(x2|θ)=f((2,x2)|θ).222BytheFactorizationTheorem,T(J,XJ)issufficient.b.Equations6.3.4and6.3.5followimmediatelyfromthetwoPrinciples.CombiningthemwehaveEv(E,x∗)=Ev(E,x∗),theconclusionoftheFormalLikelihoodPrinciple.1122c.ToprovetheConditionalityPrinciple.LetoneexperimentbetheE∗experimentandtheotherEj.Then∗11L(θ|(j,xj))=f((j,xj)|θ)=fj(xj|θ)=L(θ|xj).22Letting(j,x)andxplaytherolesofx∗andx∗intheFormalLikelihoodPrinciplewejj12canconcludeEv(E∗,(j,x))=Ev(E,x),theConditionalityPrinciple.NowconsiderthejjjFormalSufficiencyPrinciple.IfT(X)issufficientandT(x)=T(y),thenL(θ|x)=CL(θ|y),whereC=课后答案网h(x)/h(y)andhisthefunctionfromtheFactorizationTheorem.Hence,bytheFormalLikelihoodPrinciple,Ev(E,x)=Ev(E,y),theFormalSufficiencyPrinciple.6.35Let1=successand0=failure.Thefoursamplepointsare{0,10,110,111}.Fromthelikelihoodprinciple,inferenceaboutwww.hackshp.cnpisonlythroughL(p|x).Thevaluesofthelikelihoodare1,p,p2,andp3,andthesamplesizedoesnotdirectlyinfluencetheinference.6.37a.Foroneobservation(X,Y)wehave∂22Y2EYI(θ)=−Elogf(X,Y|θ)=−E−=.∂θ2θ3θ3But,Y∼exponential(θ),andEY=θ.Hence,I(θ)=2/θ2forasampleofsizeone,andI(θ)=2n/θ2forasampleofsizen.b.(i)ThecdfofTisPPPiYi22PiYi/θ2222P(T≤t)=P≤t=P≤t/θ=P(F2n,2n≤t/θ)iXi2iXiθ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition6-9whereF2n,2nisanFrandomvariablewith2ndegreesoffreedominthenumeratoranddenominator.Thisfollowssince2Yi/θand2Xiθareallindependentexponential(1),orχ2.Differentiating(int)andsimplifyinggivesthedensityofTas22n2nΓ(2n)2tθfT(t)=,Γ(n)2tt2+θ2t2+θ2andthesecondderivative(inθ)ofthelogdensityist4+2t2θ2−θ42n22n=1−,θ2(t2+θ2)2θ2(t2/θ2+1)2andtheinformationinTis"#!222n12n11−2E=1−2E.θ2T2/θ2+1θ2F22n,2n+1Theexpectedvalueis!2Z1Γ(2n)∞1wn−1Γ(2n)Γ(n)Γ(n+2)n+1E===.F22n,2n+1Γ(n)20(1+w)2(1+w)2nΓ(n)2Γ(2n+2)2(2n+1)SubstitutingthisabovegivestheinformationinTas2nn+1n1−2=I(θ),θ22(2n+1)2n+1whichisnottheanswerreportedbyJoshiandNabar.PP(ii)LetW=iXiandV=iYi.Ineachpair,XiandYiareindependent,soWandVareindependent.Xi∼exponential(1/θ);hence,W∼gamma(n,1/θ).Yi∼exponential(θ);hence,V∼gamma(n,θ).Usethisjointdistributionof(W,V)toderivethejointpdfof(T,U)as22n−1uθutf(t,u|θ)=uexp−−,u>0,t>0.[Γ(n)]2ttθNow,theinformationin(课后答案网T,U)is∂22UT2V2nθ2n−Elogf(T,U|θ)=−E−=E==.∂θ2θ3θ3θ3θ2www.hackshp.cnPP(iii)Thepdfofthesampleisf(x,y)=exp[−θ(ixi)−(iyi)/θ].Hence,(W,V)definedasinpart(ii)issufficient.(T,U)isaone-to-onefunctionof(W,V),hence(T,U)isalsosufficient.But,EU2=EWV=(n/θ)(nθ)=n2doesnotdependonθ.SoE(U2−n2)=0forallθ,and(T,U)isnotcomplete.6.39a.ThetransformationfromCelsiustoFahrenheitisy=9x/5+32.Hence,5∗5(T(y)−32)=((.5)(y)+(.5)(212)−32)995=((.5)(9x/5+32)+(.5)(212)−32)=(.5)x+50=T(x).9b.T(x)=(.5)x+506=(.5)x+106=T∗(x).Thus,wedonothaveequivariance.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn6-10SolutionsManualforStatisticalInference6.40a.BecauseX1,...,Xnisfromalocationscalefamily,byTheorem3.5.6,wecanwriteXi=σZi+µ,whereZ1,...,Znisarandomsamplefromthestandardpdff(z).ThenT1(X1,...,Xn)T1(σZ1+µ,...,σZn+µ)σT1(Z1,...,Zn)T1(Z1,...,Zn)===.T2(X1,...,Xn)T2(σZ1+µ,...,σZn+µ)σT2(Z1,...,Zn)T2(Z1,...,Zn)BecauseT1/T2isafunctionofonlyZ1,...,Zn,thedistributionofT1/T2doesnotdependonµorσ;thatis,T1/T2isanancillarystatistic.b.R(x1,...,xn)=x(n)−x(1).Becausea>0,max{ax1+b,...,axn+b}=ax(n)+bandmin{ax1+b,...,axn+b}=ax(1)+b.Thus,R(ax1+b,...,axn+b)=(ax(n)+b)−(ax(1)+b)=a(x(n)−x(1))=aR(x1,...,xn).Forthesamplevariancewehave1XS2(ax+b,...,ax+b)=((ax+b)−(ax¯+b))21nin−11X=a2(x−x¯)2=a2S2(x,...,x).i1nn−1Thus,S(ax1+b,...,axn+b)=aS(x1,...,xn).Therefore,RandSbothsatisfytheabovecondition,andR/Sisancillarybya).6.41a.Measurementequivariancerequiresthattheestimateofµbasedonybethesameastheestimateofµbasedonx;thatis,T∗(x+a,...,x+a)−a=T∗(y)−a=T(x).1nb.TheformalstructuresfortheprobleminvolvingXandtheprobleminvolvingYarethesame.Theybothconcernarandomsampleofsizenfromanormalpopulationandestimationofthemeanofthepopulation.Thus,formalinvariancerequiresthatT(x)=T∗(x)forallx.Combiningthiswithpart(a),theEquivariancePrinciplerequiresthatT(x1+a,...,xn+a)−a=T∗(x+a,...,x+a)−a=T(x,...,x),i.e.,T(x+a,...,x+a)=T(x,...,x)+a.1n1n1n1nPPc.W(x1+a,...,xn+a)=i(xi+a)/n=(ixi)/n+a=W(x1,...,xn)+a,soW(x)isequivariant.Thedistributionof(X1,...,Xn)isthesameasthedistributionof(Z1+θ,...,Zn+Pθ),whereZ1,...,Znarearandomsamplefromf(x−0)andEZi=0.Thus,EθW=Ei(Zi+θ)/n=θ,forallθ.6.43a.Foralocation-scalefamily,ifX∼f(x|θ,σ2),thenY=g(X)∼f(y|cθ+a,c2σ2).Soa,cforestimatingσ2,¯g(σ2)=c2σ2.Anestimatorofσ2isinvariantwithrespecttoGifa,c1W(cx+a,...,cx+a)=c2W(x,...,x).AnestimatoroftheformkS2isinvariant1n1nbecause!2kXnXn2kS课后答案网(cx1+a,...,cxn+a)=(cxi+a)−(cxi+a)/nn−1i=1i=1Xnk2=((cxi+a)−(cx¯+a))www.hackshp.cnn−1i=1Xn2k222=c(xi−x¯)=ckS(x1,...,xn).n−1i=1ToshowinvariancewithrespecttoG2,usetheaboveargumentwithc=1.ToshowinvariancewithrespecttoG3,usetheaboveargumentwitha=0.(G2andG3arebothsubgroupsofG1.SoinvariancewithrespecttoG1impliesinvariancewithrespecttoG2andG3.)b.ThetransformationsinGleavethescaleparameterunchanged.Thus,¯g(σ2)=σ2.An2aestimatorofσ2isinvariantwithrespecttothisgroupifW(x1+a,...,xn+a)=W(ga(x))=¯ga(W(x))=W(x1,...,xn).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition6-11Anestimatorofthegivenformisinvariantif,forallaand(x1,...,xn),x¯+a2x¯2W(x1+a,...,xn+a)=φs=φs=W(x1,...,xn).ssInparticular,forasamplepointwiths=1and¯x=0,thisimplieswemusthaveφ(a)=φ(0),foralla;thatis,φmustbeconstant.Ontheotherhand,ifφisconstant,thentheestimatorsareinvariantbyparta).Sowehaveinvarianceifandonlyifφisconstant.InvariancewithrespecttoG1alsorequiresφtobeconstantbecauseG2isasubgroupofG1.Finally,anestimatorofσ2isinvariantwithrespecttoGifW(cx,...,cx)=c2W(x,...,x).31n1nEstimatorsofthegivenformareinvariantbecausecx¯222x¯22W(cx1,...,cxn)=φcs=cφs=cW(x1,...,xn).css课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter7PointEstimation7.1Foreachvalueofx,theMLEθˆisthevalueofθthatmaximizesf(x|θ).Thesevaluesareinthefollowingtable.x01234θˆ112or333Atx=2,f(x|2)=f(x|3)=1/4arebothmaxima,sobothθˆ=2orθˆ=3areMLEs.7.2a.n"n#α−1Y11YL(β|x)=xα−1e−xi/β=xe−Σixi/βΓ(α)βαiΓ(α)nβnαii=1i=1"#PYnxlogL(β|x)=−logΓ(α)n−nαlogβ+(α−1)logx−iiiβi=1P∂logLnαixi=−+∂βββ2PSetthepartialderivativeequalto0andsolveforβtoobtainβˆ=ixi/(nα).Tocheckthatthisisamaximum,calculateP∂2logLnα2xi(nα)32(nα)3(nα)3=−i=P−P=−P<0.∂β2β2β3(x)2(x)2(x)2β=βˆβ=βˆiiiiiiBecauseβˆistheuniquepointwherethederivativeis0anditisalocalmaximum,itisaglobalmaximum.Thatis,βˆistheMLE.b.Nowthelikelihoodfunctionis"n#α−1课后答案网1YL(α,β|x)=xe−Σixi/β,Γ(α)nβnαii=1thesameasinpart(a)exceptwww.hackshp.cnαandβarebothvariables.ThereisnoanalyticformfortheMLEs,ThevaluesˆαandβˆthatmaximizeL.Oneapproachtofindingˆαandβˆwouldbetonumericallymaximizethefunctionoftwoarguments.Butitisusuallybesttodoasmuchaspossibleanalytically,first,andperhapsreducethecomplexityofthenumericalproblem.PFrompart(a),foreachfixedvalueofα,thevalueofβthatmaximizesLisixi/(nα).SubstitutethisintoL.Thenwejustneedtomaximizethefunctionoftheonevariableαgivenby"n#α−11YPxe−Σixi/(Σixi/(nα))Γ(α)n(x/(nα))nαiiii=1"n#α−11Y=Pxe−nα.Γ(α)n(x/(nα))nαiiii=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-2SolutionsManualforStatisticalInferencePForthegivendata,n=14andixi=323.6.Manycomputerprogramscanbeusedtomaximizethisfunction.FromPROCNLINinSASweobtainˆα=514.219and,hence,βˆ=323.6=.0450.14(514.219)7.3Thelogfunctionisastrictlymonotoneincreasingfunction.Therefore,L(θ|x)>L(θ0|x)ifandonlyiflogL(θ|x)>logL(θ0|x).SothevalueθˆthatmaximizeslogL(θ|x)isthesameasthevaluethatmaximizesL(θ|x).7.5a.ThevalueˆzsolvestheequationY(1−p)n=(1−xz),iiwhere0≤z≤(maxx)−1.Letkˆ=greatestintegerlessthanorequalto1/zˆ.ThenfromiiExample7.2.9,kˆmustsatisfyYYnn[k(1−p)]≥(k−xi)and[(k+1)(1−p)]<(k+1−xi).iiBecausetheright-handsideofthefirstequationisdecreasinginˆz,andbecausekˆ≤1/zˆ(sozˆ≤1/kˆ)andkˆ+1>1/zˆ,kˆmustsatisfythetwoinequalities.ThuskˆistheMLE.
4b.Forp=1/2,wemustsolve1=(1−20z)(1−z)(1−19z),whichcanbereducedtothe2cubicequation−380z3+419z2−40z+15/16=0.Therootsare.9998,.0646,and.0381,leadingtocandidatesof1,15,and26forkˆ.Thefirsttwoarelessthanmaxixi.Thuskˆ=26.QQ
7.6a.f(x|θ)=θx−2I(x)=x−2θnI(x).Thus,Xisasufficientstatisticforii[θ,∞)iii[θ,∞)(1)(1)θbytheFactorizationTheorem.Q
b.L(θ|x)=θnx−2I(x).θnisincreasinginθ.Thesecondtermdoesnotinvolveθ.ii[θ,∞)(1)SotomaximizeL(θ|x),wewanttomakeθaslargeaspossible.Butbecauseoftheindicatorfunction,L(θ|x)=0ifθ>x(1).Thus,θˆ=x(1).R∞−1∞c.EX=θxdx=θlogx|=∞.Thusthemethodofmomentsestimatorofθdoesnotθθexist.(ThisistheParetodistributionwithα=θ,β=1.)Q√7.7L(0|Qx)=1,0√¯1/2,L(θ|x)isanincreasingfunctionofθon[0,1/2]andobtainsitsmaximumattheupperboundofθwhichis1/2.SotheMLEisθˆ=minX,¯1/2.b.TheMSEofθ˜isMSE(θ˜)=Varθ˜+bias(θ˜)2=(θ(1−θ)/n)+02=θ(1−θ)/n.ThereisnosimpleformulaforMSE(θˆ),butanexpressionisnXnMSE(θˆ)=E(θˆ−θ)2=(θˆ−θ)2θy(1−θ)n−yyy=0[Xn/2]y2nXn12n=−θθy(1−θ)n−y+−θθy(1−θ)n−y,ny2yy=0y=[n/2]+1PwhereY=iXi∼binomial(n,θ)and[n/2]=n/2,ifniseven,and[n/2]=(n−1)/2,ifnisodd.c.Usingthenotationusedin(b),wehaveXn22ynyn−yMSE(θ˜)=E(X¯−θ)=−θθ(1−θ).nyy=0Therefore,课后答案网n"2#Xy21nMSE(θ˜)−MSE(θˆ)=−θ−−θθy(1−θ)n−yn2yy=[n/2]+1www.hackshp.cnXny1y1nyn−y=+−2θ−θ(1−θ).n2n2yy=[n/2]+1Thefactsthaty/n>1/2inthesumandθ≤1/2implythateveryterminthesumispositive.ThereforeMSE(θˆ)n/2.Ifniseven,2j−n=0ifj=n/2.Sothelikelihoodisconstantbetweenx(n/2)andx((n/2)+1),andanyvalueinthisintervalistheMLE.UsuallythemidpointofthisintervalistakenastheMLE.Ifnisodd,thelikelihoodisminimizedatθˆ=x((n+1)/2).7.15a.Thelikelihoodis()λn/2λX(x−µ)2iL(µ,λ|x)=Qexp−.(2π)nxi2µ2xiiiForfixedλ,maximizingwithrespecttoµisequivalenttominimizingthesumintheexpo-nential.dX(x−µ)2dX((x/µ)−1)2X2((x/µ)−1)xiiii==−.dµµ2xidµxixiµ2iiiSettingthisequaltozeroisequivalenttosettingXxi−1=0,µiandsolvingforµyieldsˆµn=¯x.Plugginginthisˆµnandmaximizingwithrespecttoλamountstomaximizinganexpressionoftheformλn/2e−λb.SimplecalculusyieldsnX(x−x¯)2λˆ=whereb=i.n2b2¯x2xiiFinally,XxiX1X1nX1X112b=−2+=−+=−.x¯2x¯xix¯xixix¯iiiiib.ThisisthesameasExercise6.27b.c.ThisinvolvedalgebracanbefoundinSchwarzandSamanta(1991).7.17a.ThisisaspecialcaseofthecomputationinExercise7.2a.b.Makethetransformation课后答案网z=(x2−1)/x1,w=x1⇒x1=w,x2=wz+1.TheJacobeanis|w|,andZZwww.hackshp.cn1−1/θ−w(1+z)/θfZ(z)=fX1(w)fX2(wz+1)wdw=2ewedw,θwheretherangeofintegrationis00.Thus,(R1−1/zwe−w(1+z)/θdwifz<0f(z)=e−1/θR0Zθ2∞we−w(1+z)/θdwifz≥00RUsingthefactthatwe−w/adw=−e−w/a(aw+a2),wehave((1+z)/zθzθ+e(1+z−zθ)f(z)=e−1/θθz(1+z)2ifz<0Z1(1+z)2ifz≥0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-6SolutionsManualforStatisticalInferencec.Frompart(a)wegetθˆ=1.Frompart(b),X2=1impliesZ=0which,ifweusetheseconddensity,givesusθˆ=∞.d.Theposteriordistributionsarejustthenormalizedlikelihoodtimesprior,soofcoursetheyaredifferent.7.18a.TheusualfirsttwomomentequationsforXandYare1Xx¯=EX=µ,x2=EX2=σ2+µ2,XiXXni1Xy¯=EY=µ,y2=EY2=σ2+µ2.YiYYniWealsoneedanequationinvolvingρ.1Xxiyi=EXY=Cov(X,Y)+(EX)(EY)=ρσXσY+µXµY.niSolvingthesefiveequationsyieldstheestimatorsgiven.FactssuchasXP2P2P2122ixi−(ixi)/ni(xi−x¯)xi−x¯==nnniareused.b.Twoanswersareprovided.First,usetheMiscellanea:For!XkL(θ|x)=h(x)c(θ)expwi(θ)ti(x),i=1PPnnthesolutionstothekequationsj=1ti(xj)=Eθj=1ti(Xj)=nEθti(X1),i=1,...,k,providetheuniqueMLEforθ.Multiplyingouttheexponentinthebivariatenormalpdfshowsithasthisexponentialfamilyformwithk=5andt1(x,y)=x,t2(x,y)=y,t3(x,y)=x2,t(x,y)=y2andt(x,y)=xy.Settingupthemethodofmomentequations,wehave45XXx=nµ,x2=n(µ2+σ2),iXiXXiiXXy=nµ,y2=n(µ2+σ2),iYiYYii课后答案网XXxiyi=[Cov(X,Y)+µXµY]=n(ρσXσY+µXµY).iiThesearethesameequationsasinpart(a)ifyoudivideeachonebywww.hackshp.cnn.SotheMLEsarethesameasthemethodofmomentestimatorsinpart(a).Forthesecondanswer,usethehintinthebooktowriteL(θ|x,y)=L(θ|x)L(θ,x|y)()X2−n12=(2πσX)2exp−(x−µX)2σ2iXi|{z}A"2#22
−n−1XσY×2πσ(1−ρ)2expy−µ+ρ(x−µ)Y2σ2(1−ρ2)iYσiXYiX|{z}B若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition7-7PWeknowthat¯xandˆσ2=(x−x¯)2/nmaximizesA;thequestioniswhethergivenσ,XiiYµ,andρ,does¯x,ˆσ2maximizeB?Letusfirstfixσ2andlookforˆµ,thatmaximizesB.YXXXWehave!∂logBXρσYρσYset∝−2(yi−µY)−(xi−µX)=0∂µXσXσXiXρσY⇒(yi−µY)=Σ(xi−µˆX).σXiPPSimilarlydothesameprocedureforL(θ|y)L(θ,y|x)Thisimplies(x−µ)=ρσX(y−iiPXσYiiµPˆY).ThesolutionsˆµXandˆµYthereforemustsatisfybothequations.IfPPi(yi−µˆY)6=0ori(xi−µˆX)6=0,wewillgetρ=1/ρ,soweneedi(yi−µˆY)=0andi(xi−µˆX)=0.2Thisimpliesˆµ=¯xandˆµ=¯y.(∂logB<0.Thereforeitismaximum).Togetˆσ2takeXY∂µ2XX∂logBXρσYρσYset∂σ2∝σ2(xi−µˆX)(yi−µY)−σ(xi−µX)=0.XiXXXρσXY2⇒(xi−µˆX)(yi−µˆY)=(xi−µˆX).σˆXiPPSimilarly,(x−µˆ)(y−µˆ)=ρσX(y−µˆ)2.Thusσˆ2andˆσ2mustsatisfytheiiXiYσˆYiiYXYabovetwoequationswithˆµX=X¯,ˆµY=Y¯.ThisimpliesPPσˆXσˆX(x−x¯)2(y−y¯)2Y(x−x¯)2=X(y−y¯)2⇒ii=ii.σˆiσˆiσˆ2σˆ2XiYiXYPPTherefore,σˆ2=a(x−x¯)2,ˆσ2=a(y−y¯)2whereaisaconstant.CombiningtheXPii
Yiiknowledgethatx,¯1(x−x¯)2=(ˆµ,σˆ2)maximizesA,weconcludethata=1/n.niiXXLastly,wefindˆρ,theMLEofρ.WritelogL(¯x,y,¯σˆ2,σˆ2,ρ|x,y)XYn1X(x−x¯)22ρ(x−x¯)(y−y¯)(y−y¯)2=−log(1−ρ2)−i−ii+i22(1−ρ2)σˆ2σˆX,σˆYσˆ2iXYn21X(x−x¯)(y−y¯)=−log(1−ρ)−2n−2ρii22(1−ρ2)σˆXσˆY课后答案网|i{z}APPbecauseσˆ2=1(x−x¯)2andˆσ2=1(y−y¯)2.NowXniwww.hackshp.cniYniin2nρlogL=−log(1−ρ)−+A21−ρ21−ρ2and∂logLnnρA(1−ρ2)+2Aρ2set=−+=0.∂ρ1−ρ2(1−ρ2)2(1−ρ2)2ThisimpliesA+Aρ2−nρˆ−nρˆ3=0⇒A(1+ˆρ2)=nρˆ(1+ˆρ2)(1−ρ2)2A1X(x−x¯)(y−y¯)⇒ρˆ==ii.nnσˆXσˆYi若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-8SolutionsManualforStatisticalInference7.19a.Y11L(θ|y)=√exp−(y−βx)222σ2ii2πσi!1X=(2πσ2)−n/2exp−(y2−2βxy+β2x2)2σ2iiiiiP!β2x21XβX=(2πσ2)−n/2exp−iiexp−y2+xy.2σ22σ2iσ2iiiiPPByTheorem6.1.2,(Y2,xY)isasufficientstatisticfor(β,σ2).iiiiib.nn1XβXβ2XlogL(β,σ2|y)=−log(2π)−logσ2−y2+xy−x2.222σ2iσ2ii2σ2iiiForafixedvalueofσ2,P∂logL1XβX2setixiyi=2xiyi−2xi=0⇒βˆ=P2.∂βσiσiixiAlso,∂2logL1X=x2<0,∂β2σ2iisoitisamaximum.Becauseβˆdoesnotdependonσ2,itistheMLE.AndβˆisunbiasedbecausePPPixiEYiiPxi·βxiEβˆ===β.x2x2iiiiPPc.βˆ=aY,wherea=x/x2areconstants.ByCorollary4.6.10,βˆisnormallydis-iiiiijjtributedwithmeanβ,and!2PXXxx2σ2Varβˆ=a2VarY=Piσ2=Piiσ2=P.iix2(x2)2x2课后答案网iijjjjii7.20a.PiYi1X1XEP=PEYi=Pβxi=β.www.hackshp.cnixiixiiixiib.PPY1Xσ2nσ2σ2VarPii=PVarY=Pi==.x(x)2i(x)2n2x¯2nx¯2iiiiiiiPPPBecausex2−nx¯2=(x−x¯)2≥0,x2≥nx¯2.Hence,iiiiiiPσ2σ2YPPiiVarβˆ=≤=Var.x2nx¯2xiiii(Infact,βˆisBLUE(BestLinearUnbiasedEstimatorofβ),asdiscussedinSection11.3.2.)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition7-97.21a.1XYi1XEYi1XβxiE===β.nxinxinxiiiib.1XY1XVarYσ2X1iiVar==.nxin2x2n2x2iiiiiUsingExample4.7.8witha=1/x2weobtainii1X1n≥P.nx2x2iiiiThus,σ2σ2X11XYiVarβˆ=P≤=Var.x2n2x2nxiiiiiiBecauseg(u)=1/u2isconvex,usingJensen’sInequalitywehave11X1≤.x¯2nx2iiThus,PYσ2σ2X11XYPiiiVar=≤=Var.xinx¯2n2x2inxiiii7.22a.√n−n(¯x−θ)2/(2σ2)1−(θ−µ)2/2τ2f(¯x,θ)=f(¯x|θ)π(θ)=√e√e.2πσ2πτb.Factortheexponentinpart(a)as−n2121212(¯x−θ)−(θ−µ)=−(θ−δ(x))−(¯x−µ),2σ22τ22v2τ2+σ2/n.whereδ(x)=(τ2x¯+(σ2/n)µ)/(τ2+σ2/n)andv=(σ2τ2/n)(τ+σ2/n).Letn(a,b)denotethepdfofanormaldistributionwithmean课后答案网aandvarianceb.Theabovefactorizationshowsthatf(x,θ)=n(θ,σ2/n)×n(µ,τ2)=n(δ(x),v2)×n(µ,τ2+σ2/n),wherethemarginaldistributionofX¯isn(µ,τ2+σ2/n)andtheposteriordistributionofθ|xisn(δ(x),v2).Thisalsocompletespart(c).www.hackshp.cn7.23Lett=s2andθ=σ2.Because(n−1)S2/σ2∼χ2,wehaven−1[(n−1)/2]−11n−1−(n−1)t/2θn−1f(t|θ)=te.Γ((n−1)/2)2(n−1)/2θθWithπ(θ)asgiven,wehave(ignoringtermsthatdonotdependonθ)"#((n−1)/2)−11−(n−1)t/2θ11−1/βθπ(θ|t)∝eeθθθα+1((n−1)/2)+α+111(n−1)t1∝exp−+,θθ2β若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-10SolutionsManualforStatisticalInferencewhichwerecognizeasthekernelofaninvertedgammapdf,IG(a,b),with−1n−1(n−1)t1a=+αandb=+.22βDirectcalculationshowsthatthemeanofanIG(a,b)is1/((a−1)b),son−1t+1n−1s2+12β2βE(θ|t)==.n−1+α−1n−1+α−122ThisisaBayesestimatorofσ2.P7.24Fornobservations,Y=iXi∼Poisson(nλ).a.ThemarginalpmfofYisZ∞y−nλ(nλ)e1α−1−λ/βm(y)=λedλ0y!Γ(α)βαyZ∞yy+αn(y+α)−1−λnβ=λeβ/(nβ+1)dλ=Γ(y+α).y!Γ(α)βα0y!Γ(α)βαnβ+1Thus,−λf(y|λ)π(λ)λ(y+α)−1eβ/(nβ+1)βπ(λ|y)==y+α∼gammay+α,.m(y)βnβ+1Γ(y+α)nβ+1b.ββ1E(λ|y)=(y+α)=y+(αβ).nβ+1nβ+1nβ+1β2Var(λ|y)=(y+α).2(nβ+1)7.25a.Wewillusetheresultsandnotationfrompart(b)todothisspecialcase.Frompart(b),theXisareindependentandeachXihasmarginalpdfZ∞Z∞22221−(x−θ)2/2σ2−(θ−µ)2/2τ2m(x|µ,σ,τ)=f(x|θ,σ)π(θ|µ,τ)dθ=eedθ.−∞−∞2πστCompletethesquarein课后答案网θtowritethesumofthetwoexponentsashi22µσ2θ−xτ+σ2+τ2σ2+τ2(x−µ)2−−.2σ2τ22(σ2+τ2)www.hackshp.cnσ2+τ2Onlythefirstterminvolvesθ;callit−A(θ).Also,e−A(θ)isthekernelofanormalpdf.Thus,Z∞√−A(θ)στedθ=2π√,−∞σ2+τ2andthemarginalpdfis1√στ(x−µ)2m(x|µ,σ2,τ2)=2π√exp−2πστσ2+τ22(σ2+τ2)1(x−µ)2=√√exp−,2πσ2+τ22(σ2+τ2)an(µ,σ2+τ2)pdf.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition7-11b.ForoneobservationofXandθthejointpdfish(x,θ|τ)=f(x|θ)π(θ|τ),andthemarginalpdfofXisZ∞m(x|τ)=h(x,θ|τ)dθ.−∞Thus,thejointpdfofX=(X1,...,Xn)andθ=(θ1,...,θn)isYh(x,θ|τ)=h(xi,θi|τ),iandthemarginalpdfofXisZ∞Z∞Ym(x|τ)=···h(xi,θi|τ)dθ1...dθn−∞−∞iZ∞Z∞Yn=···h(x1,θ1|τ)dθ1h(xi,θi|τ)dθ2...dθn.−∞−∞i=2Thedθ1integralisjustm(x1|τ),andthisisnotafunctionofθ2,...,θn.So,m(x1|τ)canbepulledoutoftheintegrals.DoingeachintegralinturnyieldsthemarginalpdfYm(x|τ)=m(xi|τ).iBecausethismarginalpdffactors,thisshowsthatmarginallyX1,...,Xnareindependent,andtheyeachhavethesamemarginaldistribution,m(x|τ).7.26Firstwrite−n(¯x−θ)2−|θ|/af(x1,...,xn|θ)π(θ)∝e2σ2wheretheexponentcanbewrittenn2|θ|nn22
2σ2(¯x−θ)−a=2σ2(θ−δ±(x))+2σ2x¯−δ±(x)2withδ(x)=¯x±σ,whereweusethe“+”ifθ>0andthe“−”ifθ<0.Thus,theposterior±nameanisR∞−n(θ−δ±(x))2θe2σ2dθ课后答案网−∞E(θ|x)=R∞−n(θ−δ(x))2.±e2σ2dθ−∞Nowusethefactsthatforconstantsaandb,Z∞www.hackshp.cnZ0r−a(t−b)2−a(t−b)2πe2dt=e2dt=,0−∞2aZ∞Z∞Z∞r−a(t−b)2−a(t−b)2−a(t−b)21−ab2πte2dt=(t−b)e2dt+be2dt=e2+b,000a2aZr0−a(t−b)21−ab2πte2dt=−e2+b,−∞a2atogetqπσ2σ2−nδ2(x)−nδ2(x)(δ−(x)+δ+(x))+e2σ2+−e2σ2−2nnE(θ|x)=q.πσ222n若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-12SolutionsManualforStatisticalInference7.27a.TheloglikelihoodisXnlogL=−βτi+yilog(βτi)−τi+xilog(τi)−logyi!−logxi!i=1anddifferentiationgivesXnPn∂yiτii=1yilogL=−τi+⇒β=Pn∂βi=1βτii=1τi∂yjβxjxj+yjlogL=−β+−i+⇒τj=∂τjβτjτj1+βnPnPnXj=1xj+j=1yj⇒τj=.1+βj=1PnPnxj+yjCombiningtheseexpressionsyieldsβˆ=j=1yj/j=1xjandˆτj=1+βˆ.b.ThestationarypointoftheEMalgorithmwillsatisfyPnβˆ=Pi=1yinτˆ1+i=2xiτˆ1+y1τˆ1=βˆ+1xj+yjτˆj=.βˆ+1ThesecPondequationyieldsPτ1=y1/β,andsubstitutingthisintothefirstequationyieldsnnPβ=j=2Pyj/j=2xj.SummingoverPjinthethirdequation,andsubstitutingPβ=nnnnj=2yj/j=2xjshowsusthatj=2τˆj=j=2xj,andpluggingthisintothefirstequa-tiongivesthedesiredexpressionforβˆ.Theothertwoequationsin(7.2.16)areobviouslysatisfied.c.Theexpressionforβˆwasderivedinpart(b),asweretheexpressionsforˆτi.7.29a.Thejointdensityistheproductoftheindividualdensities.b.Theloglikelihoodis课后答案网XnlogL=−mβτi+yilog(mβτi)+xilog(τi)+logm!−logyi!−logxi!i=1www.hackshp.cnandPn∂i=1yilogL=0⇒β=Pn∂βi=1mτi∂xj+yjlogL=0⇒τj=.∂τjmβPPnPnPnPPSinceτj=1,βˆ=Pi=1yi/m=i=1yi/i=1xi.Also,Pjτj=j(yj+xj)=1,whichimpliesthatmβ=j(yj+xj)andˆτj=(xj+yj)/i(yi+xi).c.Inthelikelihoodfunctionwecanignorethefactorialterms,andtheexpectedcomplete-datalikelihoodisobtainedbyontherthiterationbyreplacingxwithE(X|τˆ(r))=mτˆ(r).1111SubstitutingthisintotheMLEsofpart(b)givestheEMsequence.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition7-13TheMLEsfromthefulldatasetareβˆ=0.0008413892andτˆ=(0.06337310,0.06374873,0.06689681,0.04981487,0.04604075,0.04883109,0.07072460,0.01776164,0.03416388,0.01695673,0.02098127,0.01878119,0.05621836,0.09818091,0.09945087,0.05267677,0.08896918,0.08642925).PTheMLEsfortheincompletedatawerecomputedusingR,wherewetakem=xi.TheRcodeis#mlesontheincompletedata#xdatam<-c(3560,3739,2784,2571,2729,3952,993,1908,948,1172,1047,3138,5485,5554,2943,4969,4828)ydata<-c(3,4,1,1,3,1,2,0,2,0,1,3,5,4,6,2,5,4)xdata<-c(mean(xdatam),xdatam);for(jin1:500){xdata<-c(sum(xdata)*tau[1],xdatam)beta<-sum(ydata)/sum(xdata)tau<-c((xdata+ydata)/(sum(xdata)+sum(ydata)))}betatauTheMLEsfromtheincompletedatasetareβˆ=0.0008415534andτˆ=(0.06319044,0.06376116,0.06690986,0.04982459,0.04604973,0.04884062,0.07073839,0.01776510,0.03417054,0.01696004,0.02098536,0.01878485,0.05622933,0.09820005,0.09947027,0.05268704,0.08898653,0.08644610).7.31a.BydirectsubstitutionwecanwritehihilogL(θ|y)=ElogL(θ|y,X)|θˆ(r),y−Elogk(X|θ,y)|θˆ(r),y.Thenextiterate,θˆ(r+1)isobtainedbymaximizingtheexpectedcomplete-dataloglikelihood,hihisoforanyθ,ElogL(θˆ(r+1)y,X)(r),y≥ElogL(θ|y,X)|θˆ(r),yθˆb.WriteZZE[logk(X|θ,y)|θ0,y]=logk(x|θ,y)logk(x|θ0,y)dx≤logk(x|θ0,y)logk(x|θ0,y)dx,hihifromthehint.HenceElogk(X|θˆ(r+1),y)(r),y≤Elogk(X|θˆ(r),y)(r),y,andsothe课后答案网θˆθˆentirerighthandsideinpart(a)isdecreasing.p2np(1−p)np+α7.33Substituteα=β=n/www.hackshp.cn4intoMSE(ˆpB)=(α+β+n)2+α+β+n−pandsimplifytoobtainnMSE(ˆpB)=√,4(n+n)2independentofp,asdesired.7.35a.δp(g(x))=δp(x1+a,...,xn+a)R∞QR∞Qtf(x+a−t)dt(y+a)f(x−y)dyR−∞ii−∞Rii=∞Q=∞Q(y=t−a)f(x+a−t)dtf(x−y)dy−∞ii−∞ii=a+δp(x)=¯g(δp(x)).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-14SolutionsManualforStatisticalInferenceb.Y1−1Σ21−1212i(x−t)n(¯x−t)−(n−1)sf(x−t)=e2i=e2e2,i(2π)n/2(2π)n/2iso√√R∞−1n(¯x−t)2(n/2π)te2dtx¯−∞δp(x)=√√R∞−1n(¯x−t)2==¯x.(n/2π)e2dt1−∞c.YY1111f(xi−t)=It−≤xi≤t+=Ix(n)−≤t≤x(1)+,2222iisoRx(1)+1/2x(n)+1/2tdtx(1)+x(n)δp(x)=Rx(1)+1/2=2.1dtx(n)+1/27.37Tofindabestunbiasedestimatorofθ,firstfindacompletesufficientstatistic.ThejointpdfisnYn11f(x|θ)=I(−θ,θ)(xi)=I[0,θ)(max|xi|).2θ2θiiBytheFactorizationTheorem,maxi|Xi|isasufficientstatistic.Tocheckthatitisacompletesufficientstatistic,letY=max|X|.NotethatthepdfofYisf(y)=nyn−1/θn,0.nn2nn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-18SolutionsManualforStatisticalInferenceTocalculatetheCram´er-Raolowerbound,wehave22∂logf(X|θ)∂1−(X−θ)2/2Eθ=Eθlog√e∂θ2∂θ22π∂21∂=Elog(2π)−1/2−(X−θ)2=E(X−θ)=−1,θ∂θ22θ∂θandτ(θ)=θ2,[τ0(θ)]2=(2θ)2=4θ2sotheCram´er-RaoLowerBoundforestimatingθ2is[τ0(θ)]24θ2
=.∂2−nEθ∂θ2logf(X|θ)nThus,theUMVUEofθ2doesnotattaintheCram´er-Raobound.(However,theratioofthevarianceandthelowerbound→1asn→∞.)7.45a.BecauseES2=σ2,bias(aS2)=E(aS2)−σ2=(a−1)σ2.Hence,MSE(aS2)=Var(aS2)+bias(aS2)2=a2Var(S2)+(a−1)2σ4.b.Thereweretwotyposinearlyprintings;κ=E[X−µ]4/σ4and21n−34Var(S)=κ−σ.nn−1SeeExercise5.8bfortheproof.c.Therewasatypoinearlyprintings;undernormalityκ=3.Undernormalitywehave44E[X−µ]X−µ4κ==E=EZ,σ4σwhereZ∼n(0,1).Now,usingLemma3.6.5withg(z)=z3wehaveκ=EZ4=Eg(Z)Z=1E(3Z2)=3EZ2=3.TominimizeMSE(S2)ingeneral,writeVar(S2)=Bσ4.ThenminimizingMSE(S2)isequivalenttominimizing课后答案网a2B+(a−1)2.Setthederivativeofthisequalto0(Bisnotafunctionofa)toobtaintheminimizingvalueofais1/(B+1).Usingtheexpressioninpart(b),undernormalitytheminimizingvalueofaiswww.hackshp.cn11n−1==.B+113−n−3+1n+1nn−1d.Therewasatypoinearlyprintings;theminimizingaisn−1a=.(κ−3)(n−1)(n+1)+nToobtainthissimplycalculate1/(B+1)with(frompart(b))1n−3B=κ−.nn−1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition7-19e.Usingtheexpressionforainpart(d),ifκ=3thesecondterminthedenominatoriszeroanda=(n−1)/(n+1),thenormalresultfrompart(c).Ifκ<3,thesecondterminthedenominatorisnegative.Becausewearedividingbyasmallervalue,wehavea>(n−1)/(n+1).BecauseVar(S2)=Bσ4,B>0,and,hence,a=1/(B+1)<1.Similarly,ifκ>3,thesecondterminthedenominatorispositive.Becausewearedividingbyalargervalue,wehavea<(n−1)/(n+1).7.46a.Fortheuniform(θ,2θ)distributionwehaveEX=(2θ+θ)/2=3θ/2.Sowesolve3θ/2=X¯forθtoobtainthemethodofmomentsestimatorθ˜=2X/¯3.b.Letx(1),...,x(n)denotetheobservedorderstatistics.Then,thelikelihoodfunctionis1L(θ|x)=θnI[x(n)/2,x(1)](θ).Because1/θnisdecreasing,thisismaximizedatθˆ=x/2.Soθˆ=X/2istheMLE.Use(n)(n)thepdfofXtocalculateEX=2n+1θ.SoEθˆ=2n+1θ,andifk=(2n+2)/(2n+1),(n)(n)n+12n+2Ekθˆ=θ.c.FromExercise6.23,aminimalsufficientstatisticforθis(X(1),X(n)).θ˜isnotafunctionofthisminimalsufficientstatistic.SobytheRao-BlackwellTheorem,E(θ˜|X(1),X(n))isanunbiasedestimatorofθ(θ˜isunbiased)withsmallervariancethanθ˜.TheMLEisafunctionof(X(1),X(n)),soitcannotbeimprovedwiththeRao-BlackwellTheorem.d.θ˜=2(1.16)/3=.7733andθˆ=1.33/2=.6650.7.47X∼n(r,σ2),soX¯∼n(r,σ2/n)andEX¯2=r2+σ2/n.ThusE[(πX¯2−πσ2/n)]=πr2isibestunbiasedbecauseX¯isacompletesufficientstatistic.Ifσ2isunknownreplaceitwiths2andtheconclusionstillholds.7.48a.TheCram´er-RaoLowerBoundforunbiasedestimatesofpishi2dpdp11p(1−p)=no=no=,d2d2X1−XX(1−X)n−nEdp2logL(p|X)−nEdp2log[p(1−p)]−nE−p2−(1−p)2PbecauseEX=p.TheMLEofpisˆp=iXi/n,withEˆp=pandVarˆp=p(1−p)/n.ThuspˆattainstheCRLBandisthebestunbiasedestimatorofp.Qb.Byindependence,E(XXXX)=EX=p4,sotheestimatorisunbiased.BecauseP1234iiPiXiisacompletesufficientstatistic,Theorems7.3.17and7.3.23implythatE(X1X2X3X4|X)isthebestunbiasedestimatorof课后答案网p4.Evaluatingthisyieldsii!PXP(X=X=X=X=1,nX=t−4)123P4i=5iEX1X2X3X4Xi=t=www.hackshp.cnP(iXi=t)i4n−4
t−4n−t.pt−4p(1−p)n−4n=
=,npt(1−p)n−tt−4ttPfort≥4.Fort<4oneoftheXismustbezero,sotheestimatorisE(X1X2X3X4|iXi=t)=0.7.49a.FromTheorem5.5.9,Y=X(1)haspdfn!1−y/λh−y/λin−1n−ny/λfY(y)=e1−(1−e)=e.(n−1)!λλThusY∼exponential(λ/n)soEY=λ/nandnYisanunbiasedestimatorofλ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-20SolutionsManualforStatisticalInferencePb.BecausefPX(x)isintheexponentialfamily,iXiisacompletesufficientstatisticandPE(nX(1)|iXPi)isthebestunbiasedestimatorofPλ.BecauseE(iXi)=nλP,wemusthaveE(nX(1)|iXi)=iXi/nbycompleteness.Ofcourse,anyfunctionofiXithatisanunbiasedestimatorofλisthebestunbiasedestimatorofλ.Thus,weknowdirectlyPPthatbecauseE(iXi)=nλ,iXi/nisthebestunbiasedestimatorofλ.c.Frompart(a),λˆ=601.2andfrompart(b)λˆ=128.8.Maybetheexponentialmodelisnotagoodassumption.7.50a.E(aX¯+(1−a)cS)=aEX¯+(1−a)E(cS)=aθ+(1−a)θ=θ.SoaX¯+(1−a)cSisanunbiasedestimatorofθ.b.BecauseX¯andS2areindependentforthisnormalmodel,Var(aX¯+(1−a)cS)=a2V+(1−1a)2V,whereV=VarX¯=θ2/nandV=Var(cS)=c2ES2−θ2=c2θ2−θ2=(c2−1)θ2.212UsecalculustoshowthatthisquadraticfunctionofaisminimizedatV(c2−1)θ2(c2−1)2a===.V1+V2((1/n)+c2−1)θ2((1/n)+c2−1)c.UsethefactorizationinExample6.2.9,withthespecialvaluesµ=θandσ2=θ2,toshowthat(X,S¯2)issufficient.E(X¯−cS)=θ−θ=0,forallθ.SoX¯−cSisanonzerofunctionof(X,S¯2)whoseexpectedvalueisalwayszero.Thus(X,S¯2)isnotcomplete.7.51a.Straightforwardcalculationgives:2Eθ−(aX¯+acS)=a2VarX¯+a2c2VarS+θ2(a+a−1)2.1212122BecauseVarX¯=θ2/nandVarS=ES2−(ES)2=θ2c−1,wehavec2h.iX¯+a222222Eθ−(a12cS)=θa1n+a2(c−1)+(a1+a2−1),andweonlyneedminimizetheexpressioninsquarebrackets,whichisindependentofθ.2−12−1Differentiatingyieldsa2=(n+1)c−nanda1=1−(n+1)c−n.b.TheestimatorT∗hasminimumMSEoveraclassofestimatorsthatcontainthoseinExercise7.50.c.Becauseθ>0,restrictingT∗≥0willimprovetheMSE.d.No.Itdoesnotfitthedefinitionofeitherone.课后答案网P7.52a.BecausethePoissonfamilyisanexponentialfamilywithPt(x)=x,iXiisacompletesufficientstatistic.AnyfunctionofiXithatisanunbiasedestimatorofPλistheuniquebestunbiasedestimatorofwww.hackshp.cnλ.BecauseX¯isafunctionofiXiandEX¯=λ,X¯isthebestunbiasedestimatorofλ.b.S2isanunbiasedestimatorofthepopulationvariance,thatis,ES2=λ.X¯isaone-to-onePfunctionofX.SoX¯isalsoacompletesufficientstatistic.Thus,E(S2|X¯)isanunbiasediiestimatorofλand,byTheorem7.3.23,itisalsotheuniquebestunbiasedestimatorofλ.ThereforeE(S2|X¯)=X¯.Thenwehave
VarS2=VarE(S2|X¯)+EVar(S2|X¯)=VarX¯+EVar(S2|X¯),soVarS2>VarX¯.c.Weformulateageneraltheorem.LetT(X)beacompletesufficientstatistic,andletT0(X)beanystatisticotherthanT(X)suchthatET(X)=ET0(X).ThenE[T0(X)|T(X)]=T(X)andVarT0(X)>VarT(X).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition7-217.53LetabeaconstantandsupposeCovθ0(W,U)>0.ThenVar(W+aU)=VarW+a2VarU+2aCov(W,U).θ0θ0θ0θ0.Choosea∈−2Covθ0(W,U)Varθ0U,0.ThenVarθ0(W+aU)Xn+1=h(p).i=1Pn+1Pn+1b.i=1Xiisacompletesufficientstatisticforθ,soETi=1Xiisthebestunbiasedestimatorofh(p).Wehave!!nX+1XnnX+1ETXi=y=PXi>Xn+1Xi=yi=1i=1i=1!!XnnX+1.nX+1=PXi>Xn+1,Xi=yPXi=y.i=1i=1i=1
Thedenominatorequalsn+1py(1−p)n+1−y.Ify=0thenumeratorisy!XnnX+1PXi>Xn+1,Xi=0=0.i=1i=1Ify>0thenumeratoris!!XnnX+1XnnX+1PXi>Xn+1,Xi=y,Xn+1=0+PXi>Xn+1,Xi=y,Xn+1=1i=1i=1i=1i=1whichequals!!XnXnXnXnPXi课后答案网>0,Xi=yP(Xn+1=0)+PXi>1,Xi=y−1P(Xn+1=1).i=1i=1i=1i=1Forally>0,www.hackshp.cnnn!n!XXXnPX>0,X=y=PX=y=py(1−p)n−y.iiiyi=1i=1i=1Ify=1or2,then!XnXnPXi>1,Xi=y−1=0.i=1i=1Andify>2,thennn!n!XXXnPX>1,X=y−1=PX=y−1=py−1(1−p)n−y+1.iiiy−1i=1i=1i=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition7-23Therefore,theUMVUEis0ify=0n+1!(n)py(1−p)n−y(1−p)(n)Xy=y=1ify=1or2ETX=y=(n+1)py(1−p)n−y+1(n+1)(n+1)(n+1−y)iyy((n)ny(1−p)n−y+1nni=1y+(y−1))p=(y)+(y−1)=1ify>2.(n+1)py(1−p)n−y+1(n+1)yy7.59WeknowT=(n−1)S2/σ2∼χ2.Thenn−1Z∞pp+n−1
p/21p+n−1−1−t22Γ2ET=
t2e2dt=
=Cp,n.n−1n−1n−1Γ2220Γ2Thus!p/22(n−1)SE=Cp,n,σ2.so(n−1)p/2SpCisanunbiasedestimatorofσp.FromTheorem6.2.25,(X,S¯2)isap,n.complete,sufficientstatistic.Theunbiasedestimator(n−1)p/2SpCisafunctionof(X,S¯2).p,nHence,itisthebestunbiasedestimator.7.61ThepdfforY∼χ2isν1ν/2−1−y/2f(y)=ye.Γ(ν/2)2ν/2ThusthepdfforS2=σ2Y/νis2ν/2−12ν1sν−s2ν/(2σ2)g(s)=e.σ2Γ(ν/2)2ν/2σ2Thus,thelog-likelihoodhastheform(gatheringtogetherconstantsthatdonotdependons2orσ2)1s2s2logL(σ2|s2)=log+Klog−K0+K00,σ2σ2σ2whereK>0andK0>0.ThelossfunctioninExample7.3.27is课后答案网2aaL(σ,a)=−log−1,www.hackshp.cnσ2σ2sothelossofanestimatoristhenegativeofitslikelihood.7.63Leta=τ2/(τ2+1),sotheBayesestimatorisδπ(x)=ax.ThenR(µ,δπ)=(a−1)2µ2+a2.Asτ2increases,R(µ,δπ)becomesflatter.7.65a.Figureomitted.b.TheposteriorexpectedlossisE(L(θ,a)|x)=ecaEe−cθ−cE(a−θ)−1,wheretheexpectationiswithrespecttoπ(θ|x).Thendca−cθsetE(L(θ,a)|x)=ceEe−c=0,daanda=−1logEe−cθisthesolution.Thesecondderivativeispositive,sothisisthemini-cmum.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn7-24SolutionsManualforStatisticalInferencec.π(θ|x)=n(x,σ¯2/n).So,substitutingintotheformulaforanormalmgf,wefindEe−cθ=22e−cx¯+σc/2n,andtheLINEXposteriorlossis22E(L(θ,a)|x)=ec(a−x¯)+σc/2n−c(a−x¯)−1.22SubstituteEe−cθ=e−cx¯+σc/2nintotheformulainpart(b)tofindtheBayesruleisx¯−cσ2/2n.d.ForanestimatorX¯+b,theLINEXposteriorloss(frompart(c))is22E(L(θ,x¯+b)|x)=ecbecσ/2n−cb−1.22ForX¯theexpectedlossisecσ/2n−1,andfortheBayesestimator(b=−cσ2/2n)theexpectedlossisc2σ2/2n.ThemarginaldistributionofX¯ism(¯x)=1,sotheBayesriskisinfiniteforanyestimatoroftheformX¯+b.2e.ForX¯+b,thesquarederrorriskisE(X¯+b)−θ=σ2/n+b2,soX¯isbetterthantheBayesestimator.TheBayesriskisinfiniteforbothestimators.P7.66LetS=iXi∼binomial(n,θ).2a.Eθˆ2=ES=1ES2=1(nθ(1−θ)+(nθ)2)=θ+n−1θ2.n2n2n2nnP2.b.T(i)X(n−1)2.ForSvaluesofi,T(i)2/(n−1)2becausetheXn=j6=ijn=(S−1)ithatisdroppedoutequals1.Fortheothern−Svaluesofi,T(i)2/(n−1)2becausen=StheXithatisdroppedoutequals0.Thuswecanwritetheestimatoras!S2n−1(S−1)2S2S2−SJK(Tn)=nn2−nS2+(n−S)2=n(n−1).(n−1)(n−1)222c.EJK(T)=1(nθ(1−θ)+(nθ)2−nθ)=nθ−nθ=θ2.nn(n−1)n(n−1)d.Forthisbinomialmodel,Sisacompletesufficientstatistic.BecauseJK(Tn)isafunctionofSthatisanunbiasedestimatorofθ2,itisthebestunbiasedestimatorofθ2.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter8HypothesisTesting8.1LetX=#ofheadsoutof1000.Ifthecoinisfair,thenX∼binomial(1000,1/2).So1000Xxn−x100011P(X≥560)=≈.0000825,x22x=560whereacomputerwasusedtodothecalculation.Forthisbinomial,EX=1000p=500andVarX=1000p(1−p)=250.Anormalapproximationisalsoverygoodforthiscalculation.X−500559.5−500P{X≥560}=P√≥√≈P{Z≥3.763}≈.0000839.250250Thus,ifthecoinisfair,theprobabilityofobserving560ormoreheadsoutof1000isverysmall.Wemighttendtobelievethatthecoinisnotfair,andp>1/2.8.2LetX∼Poisson(λ),andweobservedX=10.Toassessiftheaccidentratehasdropped,wecouldcalculate10Xe−1515i1521510P(X≤10|λ=15)==e−151+15++···+≈.11846.i!2!10!i=0Thisisafairlylargevalue,notoverwhelmingevidencethattheaccidentratehasdropped.(Anormalapproximationwithcontinuitycorrectiongivesavalueof.12264.)8.3TheLRTstatisticissupθ≤θ0L(θ|y1,...,ym)λ(y)=.课后答案网supΘL(θ|y1,...,ym)PmLety=i=1yi,andnotethattheMLEinthenumeratorismin{y/m,θ0}(seeExercise7.12)whilethedenominatorhasy/mastheMLE(seeExample7.2.7).Thuswww.hackshp.cn(1ify/m≤θ0λ(y)=(θ0)y(1−θ0)m−y(y/m)y(1−y/m)m−yify/m>θ0,andwerejectH0ifym−y(θ0)(1−θ0)b,wecouldshowλ(y)isdecreasinginysothatλ(y)b>mθ0.Itiseasiertoworkwithlogλ(y),andwehaveym−ylogλ(y)=ylogθ0+(m−y)log(1−θ0)−ylog−(m−y)log,mm若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-2SolutionsManualforStatisticalInferenceanddy1m−y1logλ(y)=logθ0−log(1−θ0)−log−y+log+(m−y)dymymm−y
!m−yθ0m=log.y/m1−θ0Fory/m>θ0,1−y/m=(m−y)/m<1−θ0,soeachfractionaboveislessthan1,andthelogislessthan0.Thusdlogλ<0whichshowsthatλisdecreasinginyandλ(y)b.8.4Fordiscreterandomvariables,L(θ|x)=f(x|θ)=P(X=x|θ).Sothenumeratoranddenomi-natorofλ(x)arethesupremumofthisprobabilityovertheindicatedsets.8.5a.Thelog-likelihoodis!YlogL(θ,ν|x)=nlogθ+nθlogν−(θ+1)logxi,ν≤x(1),iwherex(1)=minixi.Foranyvalueofθ,thisisanincreasingfunctionofνforν≤x(1).SoboththerestrictedandunrestrictedMLEsofνareˆν=x(1).TofindtheMLEofθ,set!∂nYlogL(θ,x(1)|x)=+nlogx(1)−logxi=0,∂θθiandsolveforθyieldingnnθˆ=Qn=.log(ixi/x(1))T(∂2/∂θ2)logL(θ,x|x)=−n/θ2<0,forallθ.Soθˆisamaximum.(1)b.UnderH0,theMLEofθisθˆ0=1,andtheMLEofνisstillˆν=x(1).Sothelikelihoodratiostatisticisxn/(Qx)2n−Tn(1).iiTeT−T+nλ(x)=nn2/TQn/T+1=n−Tn/T=ne.(n/T)x(1)(ixi)(e)(∂/∂T)log课后答案网λ(x)=(n/T)−1.Hence,λ(x)isincreasingifT≤nanddecreasingifT≥n.Thus,T≤cisequivalenttoT≤c1orT≥c2,forappropriatelychosenconstantsc1andc2.c.Wewillnotusethehint,althoughtheproblemcanbesolvedthatway.Instead,makethefollowingthreetransformations.First,letYi=logXi,i=1,...,n.Next,makethen-to-1transformationthatsetswww.hackshp.cnZ1=miniYiandsetsZ2,...,ZnequaltotheremainingYis,withtheirorderunchanged.Finally,letW1=Z1andWi=Zi−Z1,i=2,...,n.ThenyoufindthattheWsareindependentwithW∼f(w)=nνne−nw,w>logν,iP1W1nandWi∼exponential(1),i=2,...,n.NowT=i=2Wi∼gamma(n−1,1),and,hence,2T∼gamma(n−1,2)=χ2.2(n−1)8.6a.supQn1e−xi/θQm1e−yj/θsupΘ0L(θ|x,y)θi=1θj=1θλ(x,y)==Qn1−x/θQm1−yj/µsupΘL(θ|x,y)supθ,µi=1eij=1eθµnPP.osup1exp−nx+myθθθm+ni=1ij=1j=PnPo.sup1exp{−nx/θ}1exp−my/µθ,µθni=1iµmj=1j若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-3PPDifferentiationwillshowthatinthenumeratorθˆ0=(ixi+jyj)/(n+m),whileinthedenominatorθˆ=¯xandˆµ=¯y.Therefore,n+mPPPn+mPexp−Pn+mPx+yxi+yjxi+yjiijjijijλ(x,y)=nmPPPnexp−PnxPmexp−PmyxixiiiyjyjjjiijjPPmn(n+m)n+m(ixi)jyj=nnmmPPn+m.ixi+jyjAndtheLRTistorejectH0ifλ(x,y)≤c.b.P!nP!mn+myn+m(n+m)PixPiPjPj(n+m)nmλ==T(1−T).nnmmxi+yjxi+yjnnmmijijThereforeλisafunctionofT.λisaunimodalfunctionofTwhichismaximizedwhenT=n.Rejectionforλ≤cisequivalenttorejectionforT≤aorT≥b,whereaandbm+nareconstantsthatsatisfyan(1−a)m=bn(1−b)m.PPc.WhenH0istrue,iXi∼gamma(n,θ)andjYj∼gamma(m,θ)andtheyareindepen-dent.SobyanextensionofExercise4.19b,T∼beta(n,m).8.7a.Ynn1−(x−θ)/λ1−(Σixi−nθ)/λL(θ,λ|x)=eiI[θ,∞)(xi)=eI[θ,∞)(x(1)),λλi=1whichisincreasinginθifx(1)≥θ(regardlessofλ).SotheMLEofθisθˆ=x(1).ThenP∂logLnixi−nθˆsetX=−+2=0⇒nλˆ=xi−nθˆ⇒λˆ=x¯−x(1).∂λλλiBecauseP∂2logLnxi−nθˆn2n(¯x−x(1))−n=−2i=−=<0,∂λ2λ2λ3(¯x−x)2(¯x−x)3(¯x−x)2x¯−x(1)(1)(1)(1)wehaveθˆ=课后答案网x(1)andλˆ=¯x−x(1)astheunrestrictedMLEsofθandλ.Undertherestrictionθ≤0,theMLEofθ(regardlessofλ)isθˆ=0ifx(1)>00xifx≤0.www.hackshp.cn(1)(1)Forx(1)>0,substitutingθˆ0=0andmaximizingwithrespecttoλ,asabove,yieldsλˆ0=¯x.Therefore,(supΘ0L(θ,λ|x)sup{(λ,θ):θ≤0}L(λ,θ|x)1ifx(1)≤0λ(x)===L(¯x,0|x)supΘL(θ,λ|x)L(θ,ˆλˆ|x)L(λ,ˆθˆ|x)ifx(1)>0,wheren!nnL(x,¯0|x)(1/x¯)e−nx/¯x¯λˆx¯−x(1)x(1)n=n===1−.L(λ,ˆθˆ|x)1/λˆe−n(¯x−x(1))/(¯x−x(1))x¯x¯x¯Sorejectingifλ(x)≤cisequivalenttorejectingifx/x¯≥c∗,wherec∗issomeconstant.(1)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-4SolutionsManualforStatisticalInferenceb.TheLRTstatisticissup(1/βn)e−Σixi/ββλ(x)=nnQγ−1−Σixγ/β.supβ,γ(γ/β)(ixi)eiThenumeratorismaximizedatβˆ0=¯x.Forfixedγ,thedenominatorismaximizedatβˆPγγ=ixi/n.Thusx¯−ne−nx¯−nλ(x)=nQγ−1−Σixγ/βˆ=nnQγ−1.supγ(γn/βˆ)(xi)eiγsupγ(γ/βˆγ)(ixi)γiThedenominatorcannotbemaximizedinclosedform.Numericmaximizationcouldbeusedtocomputethestatisticforobserveddatax.8.8a.WewillfirstfindtheMLEsofaandθ.WehaveYn1−(x−θ)2/(2aθ)L(a,θ|x)=√ei,2πaθi=1Xn112logL(a,θ|x)=−log(2πaθ)−(xi−θ).22aθi=1ThusXnXn∂logL112n12set∂a=−2a+2θa2(xi−θ)=−2a+2θa2(xi−θ)=0i=1i=1n∂logLX1121=−+(x−θ)+(x−θ)∂θ2θ2aθ2iaθii=1Xnn12nx¯−nθset=−+(xi−θ)+=0.2θ2aθ2aθi=1WehavetosolvethesetwoequationssimultaneouslytogetMLEsofaandθ,sayˆaandθˆ.SolvethefirstequationforaintermsofθtogetXn12a=(xi−θ).课后答案网nθi=1Substitutethisintothesecondequationtogetwww.hackshp.cnnnn(¯x−θ)−++=0.2θ2θaθSowegetθˆ=¯x,and1Xnσˆ2aˆ=(x−x¯)2=,inx¯x¯i=1theratiooftheusualMLEsofthemeanandvariance.(Verificationthatthisisamaximumislengthy.Weomitit.)Fora=1,wejustsolvethesecondequation,whichgivesaquadraticinθthatleadstotherestrictedMLEq−1+1+4(ˆσ2+¯x2)θˆR=.2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-5Notingthataˆθˆ=ˆσ2,weobtainQn1−(x−θˆR)2/(2θˆ)√eiRL(θˆ|x)i=12πθˆRRλ(x)==Qn√1−(x−θˆ)2/(2ˆaθˆ)L(ˆa,θˆ|x)i=1ei2πaˆθˆn/221/(2πθˆ)e−Σi(xi−θˆR)/(2θˆR)R=2n/2−Σ(x−x¯)2/(2ˆσ2)(1/(2πσˆ))eiin/22=σˆ2/θˆe(n/2)−Σi(xi−θˆR)/(2θˆR).Rb.InthiscasewehavenX11logL(a,θ|x)=−log(2πaθ2)−(x−θ)2.22aθ2ii=1ThusXnXn∂logL112n12set∂a=−2a+2a2θ2(xi−θ)=−2a+2a2θ2(xi−θ)=0.i=1i=1n∂logLX1121=−+(x−θ)+(x−θ)∂θθaθ3iaθ2ii=1n1Xn1Xn2set=−+(xi−θ)+(xi−θ)=0.θaθ3aθ2i=1i=1SolvingthefirstequationforaintermsofθyieldsXn12a=(xi−θ).nθ2i=1Substitutingthisintothesecondequation,wegetPnn(x−θ)−++nii=0.P2θθ(x−θ)iiSoagain,θˆ课后答案网=¯xand1Xnσˆ2aˆ=(x−x¯)2=nx¯2ix¯2www.hackshp.cni=1intheunrestrictedcase.Intherestrictedcase,seta=1inthesecondequationtoobtain∂logLn1Xn1Xn2set=−+(xi−θ)+(xi−θ)=0.∂θθθ3θ2i=1i=1Multiplythroughbyθ3/ntoget1XnθXn−θ2+(x−θ)2−(x−θ)=0.iinni=1i=1Add±x¯insidethesquareandcompleteallsumstogettheequation−θ2+ˆσ2+(¯x−θ)2+θ(¯x−θ)=0.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-6SolutionsManualforStatisticalInferenceThisisaquadraticinθwithsolutionfortheMLEqθˆ=¯x+x¯+4(ˆσ2+¯x2)2.RwhichyieldstheLRTstatisticQn122pe−(xi−θˆR)/(2θˆR)i=12nL(θˆ|x)2πθˆσˆ2λ(x)=R=R=e(n/2)−Σi(xi−θˆR)/(2θˆR).Qn122θˆL(ˆa,θˆ|x)√e−(xi−θˆ)/(2ˆaθˆ)Ri=12πaˆθˆ2Y¯
−1−18.9a.TheMLEofλunderH0isλˆ0=,andtheMLEofλiunderH1isλˆi=Yi.TheLRTstatisticisboundedaboveby1andisgivenby
−nY¯e−n1≥Q.(Y)−1e−niiQ1/nRearrangementofthisinequalityyieldsY¯≥(iYi),thearithmetic-geometricmeaninequality.b.ThepdfofXisf(x|λ)=(λ/x2)e−λi/xi,x>0.TheMLEofλunderHisλˆ=Piiiiii00n/[i(1/Xi)],andtheMLEofλiunderH1isλˆi=Xi.Now,theargumentproceedsasinpart(a).P8.10LetY=iXi.Theposteriordistributionofλ|yisgamma(y+α,β/(β+1)).a.y+αZλ0(β+1)y+α−1−t(β+1)/βP(λ≤λ0|y)=tedt.Γ(y+α)βy+α0P(λ>λ0|y)=1−P(λ≤λ0|y).b.Becauseβ/(β+1)isascaleparameterintheposteriordistribution,(2(β+1)λ/β)|yhasagamma(y+α,2)distribution.If2αisaninteger,thisisaχ2distribution.So,for2y+2αα=5/2andβ=2,2(β+1)λ2(β+1)λ02P(λ≤λ0|y)=P≤y=P(χ2y+5≤3λ0).课后答案网ββ8.11a.FromExercise7.23,theposteriordistributionofσ2givenS2isIG(γ,δ),whereγ=α+(n−1)/2andδ=[(n−1)S2/2+1/β]−1.LetY=2/(σ2δ).ThenY|S2∼gamma(γ,2).(Note:If2αisaninteger,thisisaχ2distribution.)LetMdenotethemedianofagamma(γ,2)www.hackshp.cn2γdistribution.NotethatMdependsononlyαandn,notonS2orβ.ThenwehaveP(Y≥2/δ|S2)=P(σ2≤1|S2)>1/2ifandonlyif2222M−2/βM>=(n−1)S+,thatis,S<.δβn−1b.FromExample7.2.11,theunrestrictedMLEsareˆµ=X¯andˆσ2=(n−1)S2/n.UnderH,0µˆisstillX¯,becausethiswasthemaximizingvalueofµ,regardlessofσ2.ThenbecauseL(¯x,σ2|x)isaunimodalfunctionofσ2,therestrictedMLEofσ2isˆσ2,ifˆσ2≤1,andis1,ifˆσ2>1.SotheLRTstatisticis1ifˆσ2≤1λ(x)=2n/2−n(ˆσ2−1)/22(ˆσ)eifˆσ>1.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-7Wehavethat,forˆσ2>1,∂n1logλ(x)=−1<0.∂(ˆσ2)2σˆ2Soλ(x)isdecreasinginˆσ2,andrejectingHforsmallvaluesofλ(x)isequivalenttorejecting0forlargevaluesofˆσ2,thatis,largevaluesofS2.TheLRTacceptsHifandonlyifS2k.Then,asβvariesbetween0and∞,(M−2/β)/(n−1)variesbetween−∞andM/(n−1).So,forsomechoiceofβ,(M−2/β)/(n−1)=kandtheacceptanceregionsmatch.√8.12a.ForH0:µ≤0vs.H1:µ>0theLRTistorejectH0if¯x>cσ/n(Example8.3.3).Forα=.05takec=1.645.Thepowerfunctionis√X¯−µµnµβ(µ)=P√>1.645−√=PZ>1.645−.σ/nσ/nσ√Notethatthepowerwillequal.5whenµ=1.645σ/n.√b.ForH0:µ=0vs.HA:µ6=0theLRTistorejectH0if|x¯|>cσ/n(Example8.2.2).Forα=.05takec=1.96.Thepowerfunctionis√√
β(µ)=P−1.96−nµ/σ≤Z≤1.96+nµ/σ.√Inthiscase,µ=±1.96σ/ngivespowerofapproximately.5.8.13a.Thesizeofφ1isα1=P(X1>.95|θ=0)=.05.Thesizeofφ2isα2=P(X1+X2>C|θ=0).If1≤C≤2,thisisZ1Z12(2−C)α2=P(X1+X2>C|θ=0)=1dx2dx1=.1−CC−x12√SettingthisequaltoαandsolvingforCgivesC=2−2α,andforα=.05,weget√C=2−.1≈1.68.b.Forthefirsttestwehavethepowerfunction(0ifθ≤−.05课后答案网β1(θ)=Pθ(X1>.95)=θ+.05if−.05<θ≤.951if.95<θ.UsingthedistributionofY=X1+X2,givenbywww.hackshp.cn(y−2θif2θ≤y<2θ+1fY(y|θ)=2θ+2−yif2θ+1≤y<2θ+20otherwise,weobtainthepowerfunctionforthesecondtestas0ifθ≤(C/2)−1(2θ+2−C)2/2if(C/2)−1<θ≤(C−1)/2β2(θ)=Pθ(Y>C)=21−(C−2θ)/2if(C−1)/2<θ≤C/21ifC/2<θ.c.Fromthegraphitisclearthatφ1ismorepowerfulforθnear0,butφ2ismorepowerfulforlargerθs.φ2isnotuniformlymorepowerfulthanφ1.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-8SolutionsManualforStatisticalInferenced.IfeitherX1≥1orX2≥1,weshouldrejectH0,becauseifθ=0,P(Xi<1)=1.Thus,considertherejectionregiongivenby[[{(x1,x2):x1+x2>C}{(x1,x2):x1>1}{(x1,x2):x2>1}.Thefirstsetistherejectionregionforφ2.Thetestwiththisrejectionregionhasthesamesizeasφ2becausethelasttwosetsbothhaveprobability0ifθ=0.Butfor0<θc,weneedtofindcandntosatisfy!!c−n(.49)c−n(.51)PZ>p=.01andPZ>p=.99.n(.49)(.51)n(.51)(.49)Wethuswantc−n(.49)c−n(.51)p=2.33andp=−2.33.n(.49)(.51)n(.51)(.49)Solvingtheseequationsgivesn=13,567andc=6,783.5.8.15FromtheNeyman-PearsonlemmatheUMPtestrejectsH0if−n/222n()f(x|σ)(2πσ2)e−Σixi/(2σ1)σ1X111=1=0expx2−>kf(x|σ)2−n/2−Σix2/(2σ2)σ2iσ2σ20(2πσ0)ei01i01forsomek≥0.Aftersomealgebra,thisisequivalenttorejectingifXn22log(k(σ1/σ0))11xi>=cbecause2−2>0.1−1σ0σ1iσ2σ201PThisistheUMPtestofsizeα,whereα=P(X2>c).TodeterminectoobtainaspecifiedPσ0iiα,usethefactthatX2/σ2∼χ2.Thusii0n!X
α=PX2/σ2>c/σ2=Pχ2>c/σ2,σ0i00n0isowemusthave课后答案网c/σ2=χ2,whichmeansc=σ2χ2.0n,α0n,α8.16a.Size=www.hackshp.cnP(rejectH0|H0istrue)=1⇒TypeIerror=1.Power=P(rejectH0|HAistrue)=1⇒TypeIIerror=0.b.Size=P(rejectH0|H0istrue)=0⇒TypeIerror=0.Power=P(rejectH0|HAistrue)=0⇒TypeIIerror=1.8.17a.Thelikelihoodfunctionis!µ−1θ−1YYL(µ,θ|x,y)=µnxθny.ijij若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-9Maximizing,bydifferentiatingthelog-likelihood,yieldstheMLEsnmµˆ=−Pandθˆ=−P.ilogxijlogyjUnderH0,thelikelihoodisθ−1YYL(θ|x,y)=θn+mxy,ijijandmaximizingasaboveyieldstherestrictedMLE,n+mθˆ0=−PP.ilogxi+jlogyjTheLRTstatisticis!θˆθˆ0−θˆ0−µˆθˆm+nYYλ(x,y)=0xy.ijµˆnθˆmijQθˆ−µˆQθˆ0−θˆb.Substitutingintheformulasforθˆ,ˆµandθˆyields(x)0y=1and0iijjθˆm+nθˆnθˆmmn000m+nm+nmnλ(x,y)===(1−T)T.µˆnθˆmµˆnθˆmmnThisisaunimodalfunctionofT.Sorejectingifλ(x,y)≤cisequivalenttorejectingifT≤c1orT≥c2,wherec1andc2areappropriatelychosenconstants.c.Simpletransformationsyield−logXi∼exponential(1/µ)and−logYi∼exponential(1/θ).Therefore,T=W/(W+V)whereWandVareindependent,W∼gamma(n,1/µ)andV∼gamma(m,1/θ).UnderH0,thescaleparametersofWandVareequal.Then,asimplegeneralizationofExercise4.19byieldsT∼beta(n,m).Theconstantsc1andc2aredeterminedbythetwoequationsP(T≤c)+P(T≥c)=αand(1−c)mcn=(1−c)mcn.课后答案网1211228.18a.|X¯−θ0||X¯−θ0|β(θ)=www.hackshp.cnPθ√>c=1−Pθ√≤cσ/nσ/ncσcσ=1−Pθ−√≤X¯−θ0≤√nn√√−cσ/n+θ0−θX¯−θcσ/n+θ0−θ=1−Pθ√≤√≤√σ/nσ/nσ/nθ0−θθ0−θ=1−P−c+√≤Z≤c+√σ/nσ/nθ0−θθ0−θ=1+Φ−c+√−Φc+√,σ/nσ/nwhereZ∼n(0,1)andΦisthestandardnormalcdf.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-10SolutionsManualforStatisticalInferenceb.Thesizeis.05=β(θ0)=1+Φ(−c)−Φ(c)whichimpliesc=1.96.Thepower(1−typeIIerror)is√√√√.75≤β(θ0+σ)=1+Φ(−c−n)−Φ(c−n)=1+Φ(−1.96−n)−Φ(1.96−n).|{z}≈0√Φ(−.675)≈.25implies1.96−n=−.675impliesn=6.943≈7.8.19ThepdfofYis1(1/θ)−1−y1/θf(y|θ)=ye,y>0.θBytheNeyman-PearsonLemma,theUMPtestwillrejectif1−1/2y−y1/2f(y|2)ye=>k.2f(y|1)Toseetheformofthisrejectionregion,wecompute1/2d1−1/2y−y1/21−3/2y−y1/2y1ye=yey−−dy2222whichisnegativefory<1andpositivefory>1.Thusf(y|2)/f(y|1)isdecreasingfory≤1andincreasingfory≥1.Hence,rejectingforf(y|2)/f(y|1)>kisequivalenttorejectingfory≤c0ory≥c1.Toobtainasizeαtest,theconstantsc0andc1mustsatisfy−c0−c1f(c0|2)f(c1|2)α=P(Y≤c0|θ=1)+P(Y≥c1|θ=1)=1−e+eand=.f(c|1)f(c|1)01Solvingthesetwoequationsnumerically,forα=.10,yieldsc0=.076546andc1=3.637798.TheTypeIIerrorprobabilityisZc1c1−1/2−y1/2−y1/21P(c0kisUMPofitssize.ByExercise8.25c,theratiof(y|1/2)/f(y|1/4)isincreasinginy.Sotheratiof(y|1/4)/f(y|1/2)isdecreasinginy,andrejectingforlargevalueoftheratioisequivalenttorejectingforsmallvaluesofy.Togetα=.0547,wemustfindcsuchthatP(Y≤c|p=1/2)=.0547.Tryingvaluesc=0,1,...,wefindthatforc=2,P(Y≤2|p=1/2)=.0547.SothetestthatrejectsifY≤2istheUMPsizeα=.0547test.ThepowerofthetestisP(Y≤2|p=1/4)≈.526.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-11P1010
1
k1
10−kb.ThesizeofthetestisP(Y≥6|p=1/2)=≈.377.ThepowerP
k=6k22functionisβ(θ)=1010θk(1−θ)10−kk=6kc.ThereisanonrandomizedUMPtestforallαlevelscorrespondingtotheprobabilitiesP(Y≤i|p=1/2),whereiisaninteger.Forn=10,αcanhaveanyofthevalues0,1,11,56,176,386,638,848,968,1013,1023,and1.10241024102410241024102410241024102410248.23a.ThetestisRejectH0ifX>1/2.SothepowerfunctionisZ11β(θ)=P(X>1/2)=Γ(θ+1)xθ−1(1−x)1−1dx=θ1xθ=1−1.θΓ(θ)Γ(1)θ2θ1/21/2Thesizeissupβ(θ)=sup(1−1/2θ)=1−1/2=1/2.θ∈H0θ≤1b.BytheNeyman-PearsonLemma,themostpowerfultestofH0:θ=1vs.H1:θ=2isgivenbyRejectH0iff(x|2)/f(x|1)>kforsomek≥0.Substitutingthebetapdfgives1x2−1(1−x)1−1f(x|2)β(2,1)Γ(3)==x=2x.f(x|1)1x1−1(1−x)1−1Γ(2)Γ(1)β(1,1)Thus,theMPtestisRejectH0ifX>k/2.Wenowusetheαleveltodeterminek.WehaveZ1Z111−11−1kα=supβ(θ)=β(1)=fX(x|1)dx=x(1−x)dx=1−.θ∈Θ0k/2k/2β(1,1)2Thus1−k/2=α,sothemostpowerfulαleveltestisrejectH0ifX>1−α.c.Forθ>θ,f(x|θ)/f(x|θ)=(θ/θ)xθ2−θ1,anincreasingfunctionofxbecauseθ>θ.21212121SothisfamilyhasMLR.BytheKarlin-RubinTheorem,thetestthatrejectsH0ifX>tistheUMPtestofitssize.Bytheargumentinpart(b),uset=1−αtogetsizeα.8.24ForH0:θ=θ0vs.H1:θ=θ1,theLRTstatisticisL(θ0|x)1ifL(θ0|x)≥L(θ1|x)λ(x)==max{L(θ0|x),L(θ1|x)}L(θ0|x)/L(θ1|x)ifL(θ0|x)k.Ifk=1/c>1,thisisequivalenttoL(θ0|x)/L(θ1|x)θ1,22−(x−θ2)/2σg(x|θ2)ex(θ2−θ1)/σ2(θ2−θ2)/2σ2==ee12.课后答案网−(x−θ22g(x|θ1)e1)/2σBecauseθ−θ>0,theratioisincreasinginx.Sothefamiliesofn(θ,σ2)haveMLR.21b.Forθ2>θ1,www.hackshp.cng(x|θ)e−θ2θx/x!θx2=2=2eθ1−θ2,g(x|θ1)e−θ1θ1x/x!θ1whichisincreasinginxbecauseθ2/θ1>1.ThusthePoisson(θ)familyhasanMLR.c.Forθ2>θ1,
nθx(1−θ)n−xxng(x|θ2)x22θ2(1−θ1)1−θ2=
=.g(x|θ)nθx(1−θ)n−xθ(1−θ)1−θ1x11121Bothθ2/θ1>1and(1−θ1)/(1−θ2)>1.Thustheratioisincreasinginx,andthefamilyhasMLR.(Note:Youcanalsousethefactthatanexponentialfamilyh(x)c(θ)exp(w(θ)x)hasMLRifw(θ)isincreasinginθ(Exercise8.27).Forexample,thePoisson(θ)pmfise−θexp(xlogθ)/x!,andthefamilyhasMLRbecauselogθisincreasinginθ.)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-12SolutionsManualforStatisticalInference8.26a.Wewillprovetheresultforcontinuousdistributions.ButitisalsotruefordiscreteMLRfamilies.Forθ1>θ2,wemustshowF(x|θ1)≤F(x|θ2).Nowdf(x|θ1)[F(x|θ1)−F(x|θ2)]=f(x|θ1)−f(x|θ2)=f(x|θ2)−1.dxf(x|θ2)BecausefhasMLR,theratioontheright-handsideisincreasing,sothederivativecanonlychangesignfromnegativetopositiveshowingthatanyinteriorextremumisaminimum.Thusthefunctioninsquarebracketsismaximizedbyitsvalueat∞or−∞,whichiszero.b.FromExercise3.42,locationfamiliesarestochasticallyincreasingintheirlocationparam-eter,sothelocationCauchyfamilywithpdff(x|θ)=(π[1+(x−θ)2])−1isstochasticallyincreasing.ThefamilydoesnothaveMLR.8.27Forθ2>θ1,g(t|θ2)c(θ2)[w(θ2)−w(θ1)]t=eg(t|θ1)c(θ1)whichisincreasingintbecausew(θ2)−w(θ1)>0.Examplesincluden(θ,1),beta(θ,1),andBernoulli(θ).8.28a.Forθ2>θ1,thelikelihoodratiois2f(x|θ)1+ex−θ12=eθ1−θ2.f(x|θ1)1+ex−θ2Thederivativeofthequantityinbracketsisd1+ex−θ1ex−θ1−ex−θ2=.dx1+ex−θ2(1+ex−θ2)2Becauseθ>θ,ex−θ1>ex−θ2,and,hence,theratioisincreasing.ThisfamilyhasMLR.21b.ThebesttestistorejectH0iff(x|1)/f(x|0)>k.Frompart(a),thisratioisincreasinginx.Thusthisinequalityisequivalenttorejectingifx>k0.Thecdfofthislogisticis.F(x|θ)=ex−θ(1+ex−θ).Thus01ek−1α=1−F(k0|0)=andβ=F(k0|1)=.1+ek01+ek0−1Foraspecifiedα,k0=log(1−α)/α.Soforα=.2,k0≈1.386andβ≈.595.c.TheKarlin-RubinTheoremissatisfied,sothetestisUMPofitssize.课后答案网8.29a.Letθ2>θ1.Then222f(x|θ2)1+(x−θ1)1+(1+θ1)/x−2θ1/x==.www.hackshp.cnf(x|θ1)1+(x−θ)21+(1+θ)2/x2−2θ/x222Thelimitofthisratioasx→∞orasx→−∞is1.Sotheratiocannotbemonotoneincreasing(ordecreasing)between−∞and∞.Thus,thefamilydoesnothaveMLR.b.BytheNeyman-PearsonLemma,atestwillbeUMPifitrejectswhenf(x|1)/f(x|0)>k,forsomeconstantk.Examinationofthederivativeshowsthatf(x|1)/f(x|0)isdecreasing√√√forx≤(1−5)/2=−.618,isincreasingfor(1−5)/2≤x≤(1+5)/2=1.618,andis√decreasingfor(1+5)/2≤x.Furthermore,f(1|1)/f(1|0)=f(3|1)/f(3|0)=2.Sorejectingiff(x|1)/f(x|0)>2isequivalenttorejectingif10.8.30a.Forθ2>θ1>0,thelikelihoodratioanditsderivativearef(x|θ)θθ2+x2df(x|θ)θθ2−θ22=21and2=221x.f(x|θ1)θ1θ22+x2dxf(x|θ1)θ1(θ2+x2)22Thesignofthederivativeisthesameasthesignofx(recall,θ2−θ2>0),whichchanges21sign.Hencetheratioisnotmonotone.b.Becausef(x|θ)=(θ/π)(θ2+|x|2)−1,Y=|X|issufficient.Itspdfis2θ1f(y|θ)=,y>0.πθ2+y2Differentiatingasabove,thesignofthederivativeisthesameasthesignofy,whichispositive.HencethefamilyhasMLR.PP8.31a.BytheKarlin-RubinTheorem,theUMPtestistorejectPH0ifiXi>k,becauseiXiissufficientandPiXi∼Poisson(nλ)whichhasMLR.ChoosetheconstantktosatisfyP(iXi>k|λ=λ0)=α.b.!X√
setPXi>kλ=1≈PZ>(k−n)/n=.05,i!X√setPXi>kλ=2≈PZ>(k−2n)/2n=.90.iThus,solveforkandnink−nk−2n√=1.645and√=−1.28,课后答案网n2nyieldingn=12andk=17.70.8.32a.ThisisExample8.3.15.www.hackshp.cnb.ThisisExample8.3.19.8.33a.FromTheorems5.4.4and5.4.6,themarginalpdfofY1andthejointpdfof(Y1,Yn)aref(y|θ)=n(1−(y−θ))n−1,θ1and1n1nalwaysacceptsiff(y,y|θ)/f(y,y|0)<1.Forθ≥k,usek0=0.Thegiventestalways1n1nrejectsiff(y1,yn|θ)/f(y1,yn|0)>0andalwaysacceptsiff(y1,yn|θ)/f(y1,yn|0)<0.ThusthegiventestisUMPbyCorollary8.3.13.d.Accordingtothepowerfunctioninpart(b),β(θ)=1forallθ≥k=1−α1/n.Sotheseconditionsaresatisfiedforanyn.8.34a.ThisisExercise3.42a.b.ThisisExercise8.26a.8.35a.WewillusetheequalityinExercise3.17whichremainstruesolongasν>−α.RecallthatY∼χ2=gamma(ν/2,2).Thus,usingtheindependenceofXandYwehaveν0X√−1/2√Γ((ν−1)/2)课后答案网ET=Ep=(EX)νEY=µν√Y/νΓ(ν/2)2ifν>1.Tocalculatethevariance,computewww.hackshp.cnX2Γ((ν−2)/2)(µ2+1)νE(T0)2=E=(EX2)νEY−1=(µ2+1)ν=Y/νΓ(ν/2)2ν−2ifν>2.Thus,ifν>2,220(µ+1)ν√Γ((ν−1)/2)VarT=−µν√ν−2Γ(ν/2)2b.Ifδ=0,allthetermsinthesumfork=1,2,...arezerobecauseoftheδkterm.Theexpressionwithjustthek=0termandδ=0simplifiestothecentraltpdf.c.TheargumentthatthenoncentralthasanMLRisfairlyinvolved.ItmaybefoundinLehmann(1986,p.295).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-15√√
8.37a.P(X>¯θ0+zασ/n|θ0)=P(X¯−θ0)/(σ/n)>zα|θ0=P(Z>zα)=α,whereZ∼n(0,1).Because¯xistheunrestrictedMLE,andtherestrictedMLEisθ0if¯x>θ0,theLRTstatisticis,for¯x≥θ0.2222−n/2−Σ(x−θ)2/2σ2−[n(¯x−θ0)+(n−1)s]]2σ(2πσ)eii0e22λ(x)===e−n(¯x−θ0)/2σ.2−n/2−Σi(x−x¯)2/2σ2e−(n−1)s2/2σ2(2πσ)eiandtheLRTstatisticis1for¯x<θ0.Thus,rejectingifλc0(aslongasc<1–seeExercise8.24).0b.ThetestisUMPbytheKarlin-RubinTheorem.√c.P(X>θ¯0+tn−1,αS/n|θ=θ0)=P(Tn−1>tn−1,α)=α,whenTn−P1isaStudent’strandomvariablewithn−1degreesoffreedom.Ifwedefineˆσ2=1(x−x¯)2andPniσˆ2=1(x−θ)2,thenfor¯x≥θtheLRTstatisticisλ=(ˆσ2/σˆ2)n/2,andfor¯x<θthe0ni0000LRTstatisticisλ=1.Writingˆσ2=n−1s2andˆσ2=(x¯−θ)2+n−1s2,itisclearthatthen00nLRTisequivalenttothet-testbecauseλtn−1,α/2S2/nnppo=1−Pθ0−tn−1,α/2S2/n≤X¯−θ0≤tn−1,α/2S2/n()!X¯−θX¯−θ00=1−Pθ0−tn−1,α/2≤p≤tn−1,α/2p∼tn−1underH0S2/nS2/n=1−(1−α)=α.Pb.TheunrestrictedMLEsareθˆ=X¯andˆσ2=(X−X¯)2/n.TherestrictedMLEsarePiiθˆ=θandˆσ2=(X−θ)2/n.SotheLRTstatisticis00课后答案网0ii0−n/222(2πσˆ0)exp{−nσˆ0/(2ˆσ0)}λ(x)=−n/222www.hackshp.cn(2πσˆ)exp{−nσˆ/(2ˆσ)}"P#n/2"P#n/222(x−x¯)(x−x¯)=ii=ii.P2P22i(xi−θ0)i(xi−x¯)+n(¯x−θ0)Foraconstantc,theLRTis"P#2i(xi−x¯)12/nrejectH0ifP22=2P2c−1(n−1).n若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-16SolutionsManualforStatisticalInferenceWenowchoosetheconstantctoachievesizeα,andweprejectif|x¯−θ0|>tn−1,α/2s2/n.c.Again,seeChapter5ofLehmann(1986).8.39a.FromExercise4.45c,W=X−Y∼n(µ,σ2),whereµ−µ=µandσ2+σ2−iiiWWXYWXYρσσ=σ2.TheWsareindependentbecausethepairs(X,Y)are.XYWiiib.ThehypothesesareequivalenttoHp0:µW=0vsH1:µW6=0,and,fromExercise8.38,ifwerejectH0when|W¯|>tn−1,α/2S2/n,thisistheLRT(basedonW1,...,Wn)ofsizeWα.(Notethatifρ>0,VarWicanbesmallandthetestwillhavegoodpower.)8.41a.supL(µ,µ,σ2|x,y)L(ˆµ,σˆ2|x,y)H0XY0λ(x,y)==.supL(µX,µY,σ2|x,y)L(ˆµ,µˆY,σˆ12|x,y)XUnderH,theXsandYsareonesampleofsizem+nfroman(µ,σ2)population,where0iiµ=µX=µY.SotherestrictedMLEsarePPP2P2iXi+iYinx¯+ny¯2i(Xi−µˆ)+i(Yi−µˆ)µˆ==andσˆ0=.n+mn+mn+mToobtaintheunrestrictedMLEs,ˆµ,ˆµ,ˆσ2,usexy222L(µ,µ,σ2|x,y)=(2πσ2)−(n+m)/2e−[Σi(xi−µX)+Σi(yi−µY)]/2σ.XYFirstly,notethatˆµX=¯xandˆµY=¯y,becausemaximizingoverµXdoesnotinvolveµYandviceversa.Then"#∂logLn+m11X2X21set∂σ2=−2σ2+2(xi−µˆX)+(yi−µˆY)22=0(σ)iiimplies"#XnXm1222σˆ=(x−x¯)+(y−y¯).iin+mi=1i=1Tocheckthatthisisamaximum,"#∂2logLn+m1XX1课后答案网2222=222−(xi−µˆX)+(yi−µˆY)23∂(σ)(σ)(σ)σˆ2iiσˆ2n+m11n+m1=−(n+m)=−<0.www.hackshp.cn2(ˆσ2)2(ˆσ2)22(ˆσ2)2Thus,itisamaximum.Wethenhavenhion+mPP(2πσˆ2)−2exp−1n(x−µˆ)2+m(y−µˆ)2−n+m02ˆσ2i=1ii=1iσˆ22n0hio=0λ(x,y)=−n+m1Pn2Pm22(2πσˆ2)2exp−2(x−x¯)+(y−y¯)σˆ12ˆσi=1ii=1iandtheLRTisrejectsHifˆσ2/σˆ2>k.Inthenumerator,firstsubstituteˆµ=(nx¯+00my¯)/(n+m)andwriteXnnx¯+my¯2Xnnx¯+my¯2Xnnm2x−=(x−x¯)+x¯−=(x−x¯)2+(¯x−y¯)2,in+min+mi2(n+m)i=1i=1i=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-17becausethecrosstermiszero.PerformingasimilaroperationontheYsumyields2P2P2nm22σˆ0(xi−x¯)+(yi−y¯)+n+m(¯x−y¯)nm(¯x−y¯)==n+m+.σˆ2σˆ2n+mσˆ2..Becauseˆσ2=n+m−2S2,largevaluesofˆσ2σˆ2areequivalenttolargevaluesof(¯x−y¯)2S2n+mp0pandlargevaluesof|T|.Hence,theLRTisthetwo-samplet-test.b..pX¯−Y¯(X¯−Y¯)σ2(1/n+1/m)T=q=q.S2(1/n+1/m)[(n+m−2)S2/σ2]/(n+m−2)ppUnderH,(X¯−Y¯)∼n(0,σ2(1/n+1/m)).Underthemodel,(n−1)S2/σ2and(m−1)S2/σ20XYareindependentχ2randomvariableswith(n−1)and(m−1)degreesoffreedom.Thus,(n+m−2)S2/σ2=(n−1)S2/σ2+(m−1)S2/σ2∼χ2.Furthermore,X¯−Y¯ispXYn+m−2independentofS2andS2,and,hence,S2.SoT∼t.XYpn+m−2c.Thetwo-samplettestisUMPunbiased,buttheproofisratherinvolved.SeeChapter5ofLehmann(1986).d.Forthesedatawehaven=14,X¯=1249.86,S2=591.36,m=9,Y¯=1261.33,S2=176.00XYandS2=433.13.Therefore,T=−1.29andcomparingthistoatdistributiongivesap21p-valueof.21.Sothereisnoevidencethatthemeanagediffersbetweenthecoreandperiphery.8.42a.TheSatterthwaiteapproximationstatesthatifY∼χ2,wheretheY’sareindependent,iriithenPXχ2(aY)2approxνˆPiiiaiYi∼whereνˆ=22.iνˆiaiYi/riWehaveY=(n−1)S2/σ2∼χ2andY=(m−1)S2/σ2∼χ2.Nowdefine1XXn−12YYm−1σ2σ2a=Xanda=Y.1n(n−1)[(σ2/n)+(σ2/m)]2m(m−1)[(σ2/n)+(σ2/m)]XYXYThen,Xσ2(n−1)S2aY=XXiin(n−1)[(σ2/n)+(σ2/m)]σ2XYXσ2(m−1)S2+YY课后答案网m(m−1)[(σ2/n)+(σ2/m)]σ2XYYS2/n+S2/mχ2=XY∼νˆσ2/n+σ2/mνˆwww.hackshp.cnXYwhere2222SX/n+SY/mS2/n+S2/mσ2/n+σ2/mXYνˆ=XY=.1S41S4S4S4X+YX+Y(n−1)n2(σ2/n+σ2/m)2(m−1)m2(σ2/n+σ2/m)2n2(n−1)m2(m−1)XYXY
22b.BecauseX¯−Y¯∼nµ−µ,σ2/n+σ2/mandSX/n+SY/mapprox∼χ2/νˆ,underH:XYXYσ2/n+σ2/mνˆ0XYµX−µY=0wehave.pX¯−Y¯(X¯−Y¯)σ2/n+σ2/m0XYapproxT=q=r∼tνˆ.22(S2/n+S2/m)S/n+S/mXYXY22(σ/n+σ/m)XY若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-18SolutionsManualforStatisticalInferencec.UsingthevaluesinExercise8.41d,weobtainT0=−1.46andˆν=20.64.Sothep-valueis.16.Thereisnoevidencethatthemeanagediffersbetweenthecoreandperiphery.d.F=S2/S2=3.36.ComparingthiswithanFdistributionyieldsap-valueof2P(F≥XY13,83.36)=.09.Sothereissomeslightevidencethatthevariancediffersbetweenthecoreandperiphery.8.43Thereweretyposinearlyprintings.Thetstatisticshouldbe(X¯−Y¯)−(µ1−µ2)qq,1ρ2(n1−1)s2+(n2−1)s2/ρ2+XYn1n2n1+n2−2andtheFstatisticshouldbes2/(ρ2s2).MultiplyanddividethedenominatorofthetstatisticYXbyσtoexpressitas(X¯−Y¯)−(µ1−µ2)qσ2ρ2σ2+n1n2dividedbys(n1−1)s2/σ2+(n2−1)s2/(ρ2σ2)XY.n1+n2−2Thenumeratorhasan(0,1)distribution.Inthedenominator,(n−1)s2/σ2∼χ2and1Xn1−1(n−1)s2/(ρ2σ2)∼χ2andtheyareindependent,sotheirsumhasaχ2distribution.2Yn2−1pn1+n2−2Thus,thestatistichastheformofn(0,1)/χ2ν/νwhereν=n1+n2−2,andthenumeratoranddenominatorareindependentbecauseoftheindependenceofsamplemeansandvariancesinnormalsampling.Thusthestatistichasatn1+n2−2distribution.TheFstatisticcanbewrittenass2s2/(ρ2σ2)[(n−1)s2/(ρ2σ2)]/(n−1)Y=Y=2Y2ρ2s2s2/σ2[(n1−1)s2/(σ2)]/(n1−1)XXXwhichhastheform[χ2/(n−1)]/[χ2/(n−1)]whichhasanFdistribution.n2−12n1−11n2−1,n1−1(Note,earlyprintingshadatypowiththenumeratoranddenominatordegreesoffreedomswitched.)√√8.44Test3rejectsH0:θ=θ0infavorofH1:θ6=θ0ifX>θ¯0+zα/2σ/norX<θ¯0−zα/2σ/n.LetΦandφdenotethestandardnormalcdfandpdf,respectively.BecauseX¯∼n(θ,σ2/n),thepowerfunctionofTest3is√√课后答案网β(θ)=Pθ(X<θ¯0−zα/2σ/n)+Pθ(X>θ¯0+zα/2σ/n)θ0−θθ0−θ=Φ√−zα/2+1−Φ√+zα/2,www.hackshp.cnσ/nσ/nanditsderivativeis√√dβ(θ)nθ0−θnθ0−θ=−φ√−zα/2+φ√+zα/2.dθσσ/nσσ/nBecauseφissymmetricandunimodalaboutzero,thisderivativewillbezeroonlyifθ0−θθ0−θ−√−zα/2=√+zα/2,σ/nσ/nthatis,onlyifθ=θ0.So,θ=θ0istheonlypossiblelocalmaximumorminimumofthepowerfunction.β(θ0)=αandlimθ→±∞β(θ)=1.Thus,θ=θ0istheglobalminimumofβ(θ),and,foranyθ06=θ,β(θ0)>β(θ).Thatis,Test3isunbiased.00若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-198.45TheverificationofsizeαisthesamecomputationasinExercise8.37a.Example8.3.3showsthatthepowerfunctionβm(θ)foreachofthesetestsisanincreasingfunction.Soforθ>θ0,βm(θ)>βm(θ0)=α.Hence,thetestsareallunbiased.8.47a.ThisisverysimilartotheargumentforExercise8.41.+b.Byanargumentsimilartopart(a),thisLRTrejectsH0ifX¯−Y¯−δT+=q.
≤−tn+m−2,αS21+1pnm+−c.BecauseH0istheunionofH0andH0,bytheIUTmethodofTheorem8.3.23thetestthatrejectsH0ifthetestsinparts(a)and(b)bothrejectisalevelαtestofH0.Thatis,thetestrejectsHifT+≤−tandT−≥t.0n+m−2,αn+m−2,αd.UseTheorem8.3.24.ConsiderparameterpointswithµX−µY=δandσ→0.Foranyσ,P(T+≤−t)=α.ThepoweroftheT−testiscomputedfromthenoncentraltn+m−2,αdistributionwithnoncentralityparameter|µx−µY−(−δ)|/[σ(1/n+1/m)]=2δ/[σ(1/n+1/m)]whichconvergesto∞asσ→0.Thus,P(T−≥t)→1asσ→0.ByTheoremn+m−2,α8.3.24,thisIUTisasizeαtestofH0.8.49a.Thep-valueis7ormoresuccesses1Pθ=outof10Bernoullitrials27382911001011101110111011=+++7228229221022=.171875.b.P-value=P{X≥3|λ=1}=1−P(X<3|λ=1)e−112e−111e−110=1−++≈.0803.2!1!0!c.XP-value=P{Xi≥9|3λ=3}=1−P(Y<9|3λ=3)课后答案网i383736353130=1−e−3++++···++≈.0038,8!7!6!5!1!0!P3www.hackshp.cnwhereY=i=1Xi∼Poisson(3λ).8.50FromExercise7.26,rn−n(θ−δ22π(θ|x)=e±(x))/(2σ),2πσ22whereδ(x)=¯x±σandweusethe“+”ifθ>0andthe“−”ifθ<0.±naa.ForK>0,rZ∞√n−n(θ−δ2/(2σ2)nP(θ>K|x,a)=e+(x))dθ=PZ>[K−δ(x)],2πσ2σ+KwhereZ∼n(0,1).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn8-20SolutionsManualforStatisticalInference√nb.Asa→∞,δ+(x)→x¯soP(θ>K)→PZ>σ(K−x¯).c.ForK=0,theanswerinpart(b)is1−(p-value)forH0:θ≤0.8.51Ifαx¯θ=0=1−P|X¯|≤x¯θ=0
√
课后答案网=1−P−x¯≤X¯≤x¯θ=0=21−Φx/¯(σ/n),whereΦisthestandardnormalcdf.d.Forσ2=τ2=1andn=9wehaveap-valueof2(1−Φ(3¯x))andwww.hackshp.cnr!−1181¯x2/20P(θ=0|x¯)=1+e.10Thep-valueof¯xisusuallysmallerthantheBayesposteriorprobabilityexceptwhen¯xisveryclosetotheθvaluespecifiedbyH0.Thefollowingtableillustratesthis.Somep-valuesandposteriorprobabilities(n=9)x¯0±.1±.15±.2±.5±.6533±.7±1±2p-valueof¯x1.7642.6528.5486.1336.05.0358.0026≈0posteriorP(θ=0|x¯).7597.7523.7427.7290.5347.3595.3030.0522≈0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition8-218.54a.FromExercise7.22,theposteriordistributionofθ|xisnormalwithmean[τ2/(τ2+σ2/n)]¯xandvarianceτ2/(1+nτ2/σ2).So!0−[τ2/(τ2+σ2/n)]¯xP(θ≤0|x)=PZ≤pτ2/(1+nτ2/σ2)!!ττ=PZ≤−px¯=PZ≥px¯.(σ2/n)(τ2+σ2/n)(σ2/n)(τ2+σ2/n)b.Usingthefactthatifθ=0,X¯∼n(0,σ2/n),thep-valueisx¯−01P(X¯≥x¯)=PZ≥√=PZ≥√x¯σ/nσ/nc.Forσ2=τ2=1,!!11P(θ≤0|x)=PZ≥px¯andP(X¯≥x¯)=PZ≥px¯.(1/n)(1+1/n)1/nBecause11p0.Letα=P(1zαisalsosizeαandisuniformlymorepowerfulthanδ,thatis,βδz(µ)>βδ(µ)forallµ>0.Hence,αR(µ,δzα)=1−βδz(µ)<1−βδ(µ)=R(µ,δ),forallµ>0.www.hackshp.cnαNowreversetherolesofHandHandconsidertestingH∗:µ>0versusH∗:µ≤0.Consider0101thetestδ∗thatrejectsH∗ifX≤1orX≥2,andthetestδ∗thatrejectsH∗ifX≤z.Itis0zα0αeasilyverifiedthatfor0–1lossδandδ∗havethesameriskfunctions,andδ∗andδhavethezαzαsameriskfunctions.Furthermore,usingtheKarlin-RubinTheoremasbefore,wecanconcludethatδ∗isuniformlymorepowerfulthanδ∗.ThuswehavezαR(µ,δ)=R(µ,δ∗)≥R(µ,δ∗)=R(µ,δ),forallµ≤0,zαzαwithstrictinequalityifµ<0.Thus,δzαisbetterthanδ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter9IntervalEstimation9.1DenoteA={x:L(x)≤θ}andB={x:U(x)≥θ}.ThenA∩B={x:L(x)≤θ≤U(x)}and1≥P{A∪B}=P{L(X)≤θorθ≤U(X)}≥P{L(X)≤θorθ≤L(X)}=1,sinceL(x)≤U(x).Therefore,P(A∩B)=P(A)+P(B)−P(A∪B)=1−α1+1−α2−1=1−α1−α2.9.3a.TheMLEofβisX(n)=maxXi.Sinceβisascaleparameter,X(n)/βisapivot,andα0ncβα0n.05=Pβ(X(n)/β≤c)=Pβ(allXi≤cβ)==cβimpliesc=.051/α0n.Thus,.95=P(X/β>c)=P(X/c>β),and{β:β<β(n)β(n)X/(.051/α0n)}isa95%upperconfidencelimitforβ.(n)b.From7.10,ˆα=12.59andX=25.Sotheconfidenceintervalis(0,25/[.051/(12.59·14)])=(n)(0,25.43).9.4a.
supLσ2,σ2x,yλ(x,y)=λ=λ0XYsupλ∈(0,+∞)L(σ2,σ2|x,y)XY22TheunrestrictedMLEsofσ2andσ2areˆσ2=ΣXiandˆσ2=ΣYi,asusual.UndertheXYXnYmrestriction,λ=λ,σ2=λσ2,and0Y0X22x,y
2
−n/22
−m/2−Σx2/(2σ2)−Σy2/(2λ0σ2)LσX,λ0σX=2πσX2πλ0σXeiX·eiX2
−(m+n)/2−m/2−(λ0Σx2i+Σyi2)/(2λ0σX2)=2πσXλ0eDifferentiatingtheloglikelihoodgives课后答案网dlogLdm+nm+nmλΣx2+Σy2=−logσ2−log(2π)−logλ−0ii22dσ22X2202λσ2d(σX)www.hackshp.cnX0Xm+n
−1λΣx2+Σy2
−2set=−σ2+0iiσ2=0XX22λ0whichimpliesλΣx2+Σy2σˆ2=0ii.0λ0(m+n)Toseethisisamaximum,checkthesecondderivative:d2logLm+n
−21

−3=σ2−λΣx2+Σy2σ2222Xλ0iiXd(σX)0σ2=ˆσ2X0m+n2−2=−(ˆσ0)<0,2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn9-2SolutionsManualforStatisticalInferencethereforeˆσ2istheMLE.TheLRTstatisticis02
n/22
m/2σˆXσˆY,m/22(m+n)/2λ0(ˆσ0)andthetestis:RejectH0ifλ(x,y)0.Furthermore,ifa>0,thentherealwaysexistsatleastonerealroot.Thisfollowsfromthefactthatatθ=¯y/x¯,thevalueofthefunctionisnegative.Forθ¯=¯y/x¯wehave
y¯2y¯
x¯2−ts2−2(¯xy¯−ts)+y¯2−as2XXYYx¯x¯y¯2y¯=−ts2−2s+s2x¯2Xx¯XYY"n#Xy¯2y¯=−t(x−x¯)2−2(x−x¯)(y−y¯)+(y−y¯)2x¯2ix¯iiii=1"#Xny¯2=−t(x−x¯)−(y−y¯)x¯iii=1whichisnegative.b.Theparabolaopensdownwardifa<0,thatis,if¯x2zα/2σ/n},acceptanceregionA(θ0)=√{x:−zα/2σ/n≤x¯−√θ0≤zα/2σ/n},and1−αconfidenceintervalC(θ)={θ:¯x−zα/2σ/n≤θ≤x¯+zα/2σ/n}.√b.WehaveaUMPtestwithrejectionregion√{x:¯x−θ0<−zασ/n},acceptanceregion√A(θ0)={x:¯x−θ0≥−zασ/n},and1−αconfidenceintervalC(θ)={θ:¯x+zασ/n≥θ}.√c.Similartob),theUMPtesthasrejectionregion√{x:¯x−θ0>zασ/n},acceptanceregion√A(θ0)={x:¯x−θ0≤zασ/n},and1−αconfidenceintervalC(θ)={θ:¯x−zασ/n≤θ}.9.17a.SinceX−θ∼uniform(−1/2,1/2),P(a≤X−θ≤b)=b−a.Anyaandbsatisfyingb=a+1−αwilldo.Onechoiceisa=−1+α,b=1−α.2222b.SinceT=课后答案网X/θhaspdff(t)=2t,0≤t≤1,ZbP(a≤X/θ≤b)=2tdt=b2−a2.www.hackshp.cnappAnyaandbsatisfyingb2=a2+1−αwilldo.Onechoiceisa=α/2,b=1−α/2.
9.18a.P(X=1)=3p1(1−p)3−1=3p(1−p)2,maximumatp=1/3.p1
P(X=2)=3p2(1−p)3−2=3p2(1−p),maximumatp=2/3.p2
b.P(X=0)=3p0(1−p)3−0=(1−p)3,andthisisgreaterthanP(X=2)if(1−p)2>3p2,0or2p2+2p−1<0.Atp=1/3,2p2+2p−1=−1/9.c.Toshowthatthisisa1−α=.442interval,comparewiththeintervalinExample9.2.11.Thereareonlytwodiscrepancies.Forexample,P(p∈interval|.362.442bycomparisonwithSterne’sprocedure,whichisgivenby若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition9-7xinterval0[.000,.305)1[.305,.634)2[.362,.762)3[.695,1].9.19ForF(t|θ)increasinginθ,thereareuniquevaluesθ(t)andθ(t)suchthatF(t|θ)<1−αTULT2ifandonlyifθ<θ(t)andF(t|θ)>αifandonlyifθ>θ(t).Hence,UT2LP(θL(T)≤θ≤θU(T))=P(θ≤θU(T))−P(θ≤θL(T))αα=PFT(T)≤1−−PFT(T)≤22=1−α.9.21Toconstructa1−αconfidenceintervalforpoftheform{p:`≤p≤u}withP(`≤p≤u)=1−α,weusethemethodofTheorem9.2.12.Wemustsolvefor`anduintheequationsXxXnαnkn−kαnkn−k(1)=u(1−u)and(2)=`(1−`).2k2kk=0k=xInequation(1)α/2=P(K≤x)=P(Y≤1−u),whereY∼beta(n−x,x+1)andK∼binomial(n,u).ThisisExercise2.40.LetZ∼F2(n−x),2(x+1)andc=(n−x)/(x+1).ByTheorem5.3.8c,cZ/(1+cZ)∼beta(n−x,x+1)∼Y.SowewantcZ1cuα/2=P≤1−u=P≥.(1+cZ)Z1−uFromTheorem5.3.8a,1/Z∼F2(x+1),2(n−x).Soweneedcu/(1−u)=F2(x+1),2(n−x),α/2.Solvingforuyieldsx+1Fn−x2(x+1),2(n−x),α/2u=.1+x+1Fn−x2(x+1),2(n−x),α/2Asimilarmanipulationonequation(2)yieldsthevaluefor`.9.23a.TheLRTstatisticforH0:λ=λ0versusH1:λ6=λ0isg(y)=e−nλ0(nλ)y/e−nλˆ(nλˆ)y,课后答案网0PwhereY=Xi∼Poisson(nλ)andλˆ=y/n.TheacceptanceregionforthistestisA(λ0)={y:g(y)>c(λ0))wherec(λ0)ischosensothatP(Y∈A(λ0))≥1−α.g(y)isaunimodalfunctionofwww.hackshp.cnysoA(λ0)isanintervalofyvalues.ConsiderconstructingA(λ0)foreachλ0>0.Then,forafixedy,therewillbeasmallestλ0,callita(y),andalargestλ0,callitb(y),suchthaty∈A(λ0).TheconfidenceintervalforλisthenC(y)=(a(y),b(y)).Thevaluesa(y)andb(y)arenotexpressibleinclosedform.Theycanbedeterminedbyanumericalsearch,constructingA(λ0)fordifferentvaluesofλ0anddeterminingthosevaluesforwhichy∈A(λ0).(JayBederoftheUniversityofWisconsin,Milwaukee,remindsusthatsincecisafunctionofλ,theresultingconfidencesetneednotbeahighestdensityregionofalikelihoodfunction.Thisisanexampleoftheeffectoftheimpositionofonetypeofinference(frequentist)onanothertheory(likelihood).)b.Theprocedureinparta)wascarriedoutfory=558andtheconfidenceintervalwasfoundtobe(57.78,66.45).FortheconfidenceintervalinExample9.2.15,weneedthevaluesχ2=1116,.951039.444andχ2=1196.899.Thisconfidenceintervalis(1039.444/18,1196.899/18)=1118,.05(57.75,66.49).Thetwoconfidenceintervalsarevirtuallythesame.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn9-8SolutionsManualforStatisticalInference9.25TheconfidenceintervalderivedbythemethodofSection9.2.3is1α1αC(y)=µ:y+log≤µ≤y+log1−n2n2wherey=minixi.TheLRTmethodderivesitsintervalfromthetestofH0:µ=µ0versusH1:µ6=µ0.SinceYissufficientforµ,wecanusefY(y|µ).WehavesupL(µ|y)ne−n(y−µ0)I[µµ=µ00,∞)(y)λ(y)==supµ∈(−∞,∞)L(µ|y)ne−(y−y)I[µ,∞)(y)−n(y−µ0)0ify<µ0=eI[µ0,∞)(y)=e−n(y−µ0)ify≥µ.0WerejectHifλ(y)=e−n(y−µ0)µ0−orY<µ0µ=µ0nZ∞logcα−n(y−µ0)=PY>µ0−µ=µ0=nedynµ−logcα0n∞−n(y−µ0)logcα=−e=e=cα.logcαµ0−nTherefore,cα=αandthe1−αconfidenceintervalislogα1C(y)=µ:µ≤y≤µ−=µ:y+logα≤µ≤y.nnTousethepivotalmethod,notethatsinceµisalocationparameter,anaturalpivotalquantityisZ=Y−µ.Then,f(z)=ne−nzI(z).LetP{a≤Z≤b}=1−α,whereaandbsatisfyZ(0,∞)Zaα−nz−nza−na−naα=nedz=−e=1−e⇒e=1−2020
−log1−α⇒a=2nZ∞α−nz−nz∞−nbα=nedz=−e=e⇒−nb=log课后答案网2b2b1α⇒b=−logwww.hackshp.cnn2Thus,thepivotalintervalisY+log(α/2)/n≤µ≤Y+log(1−α/2),thesameintervalasfromExample9.2.13.Tocomparetheintervalswecomparetheirlengths.Wehave11LengthofLRTinterval=y−(y+logα)=−logαnn1111−α/2LengthofPivotalinterval=y+log(1−α/2)−(y+logα/2)=lognnnα/2Thus,theLRTintervalisshorterif−logα0,P(µ∈Ca(X))=P(µ≤max{0,X+a})=P(µ≤X+a)(sinceµ>0)=P(Z≥−a)(Z∼n(0,1))=.95(a=1.645.)Asimilarcalculationholdsforµ<0.b.ThecredibleprobabilityisZmax(0,x+a)Zmax(−x,a)1−1(µ−x)21−1t2√e2dµ=√e2dtmin(0,x−a)2πmin(−x,−a)2π=P(min(−x,−a)≤Z≤max(−x,a)).Toevaluatethisprobabilitywehavetwocases:课后答案网(i)|x|≤a⇒credibleprobability=P(|Z|≤a)(ii)|x|>a⇒credibleprobability=P(−a≤Z≤|x|)Thusweseethatfora=1.645,thecredibleprobabilityisequalto.90if|x|≤1.645andincreasesto.95as|xwww.hackshp.cn|→∞.√√9.34a.A1−αconfidenceintervalforµis{µ:¯x−1.96σ/n≤µ≤x¯+1.96σ/n}.Weneed√√2(1.96)σ/n≤σ/4orn≥4(2)(1.96).Thusweneedn≥64(1.96)2≈245.9.Son=246suffices.√b.Thelengthofa95%confidenceintervalis2tn−1,.025S/n.ThusweneedSσS2σ2P2t√≤≥.9⇒P4t2≤≥.9n−1,.0254n−1,.025n16n2(n−1)S(n−1)n⇒P≤≥.9.σ2t2n−1,.025·64|{z}∼χ2n−1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition9-11Weneedtosolvethisnumericallyforthesmallestnthatsatisfiestheinequality(n−1)n22≥χn−1,.1.tn−1,.025·64Tryingdifferentvaluesofnwefindthatthesmallestsuchnisn=276forwhich(n−1)n22=306.0≥305.5=χn−1,.1.tn−1,.025·64Astobeexpected,thisissomewhatlargerthanthevaluefoundina).√√9.35length=2zα/2σ/n,andifitisunknown,E(length)=2tα/2,n−1cσ/n,where√n−1Γ(n−1)c=√22Γ(n/2)√andEcS=σ(Exercise7.50).Thusthedifferenceinlengthsis(2σ/n)(zα/2−ctα/2).Alittleworkwillshowthat,asn→∞,c→constant.(ThiscanbedoneusingStirling’sformulaalongwithLemma2.3.14.Infact,somecarefulalgebrawillshowthatc→1asn→∞.)Also,weknow√that,asn→∞,tα/2,n−1→zα/2.Thus,thedifferenceinlengths(2σ/n)(zα/2−ctα/2)→0asn→∞.9.36ThesamplepdfisYnf(x,...,x|θ)=eiθ−xiI(x)=eΣ(iθ−xi)I[min(x/i)].1n(iθ,∞)i(θ,∞)ii=1ThusT=min(Xi/i)issufficientbytheFactorizationTheorem,andYnYnZ∞Ynn(n+1)iθ−xi(θ−t)−(t−θ)P(T>t)=P(Xi>it)=edx=e=e2,i=1i=1iti=1andn(n+1)−n(n+1)(t−θ)fT(t)=e2,t≥θ.2Clearly,θisalocationparameterandY=T−θisapivot.Tofindtheshortestconfidenceintervaloftheform[课后答案网T+a,T+b],wemustminimizeb−asubjecttotheconstraintP(−b≤Y≤−a)=1−α.NowthepdfofYisstrictlydecreasing,sotheintervallengthisshortestif−b=0andasatisfiesn(n+1)−awww.hackshp.cnP(0≤Y≤−a)=e2=1−α.Soa=2log(1−α)/(n(n+1)).9.37a.ThedensityofY=Xisf(y)=nyn−1/θn,00.Letµ−∞R∞bethepointofsymmetryandletb=2µ−a.Thenf(b)=f(a)andf(x)dx=α/2.a≤µRRbaµsincef(x)dx=α/2≤1/2=f(x)dx.Similarly,b≥µ.And,f(b)=f(a)>0since−∞R−∞af(a)≥f(x)forallx≤aandf(x)dx=α/2>0⇒f(x)>0forsomex0.−∞SotheconditionsofTheorem9.3.2aresatisfied.9.41a.Weshowthatforanyinterval[a,b]and>0,theprobabilitycontentof[a−,b−]isgreater(aslongasb−>a).WriteZaZb−ZbZaf(x)dx−f(x)dx=f(x)dx−f(x)dxba−b−a−≤f(b−)[b−(b−)]−f(a)[a−(a−)]≤[f(b−)−f(a)]≤0,wherealloftheinequalitiesfollowbecausef(x)isdecreasing.Somovingtheintervaltowardzeroincreasestheprobability,anditisthereforemaximizedbymovingaallthewaytozero.b.T=Y−µisapivotwithdecreasingpdff(t)=ne−ntI(t).Theshortest1−αintervalT[0,∞]onTis[0,−1logα],sincenZb−nt1nedt=1−α⇒b=−logα.0nSincea≤T≤bimpliesY−b≤µ≤Y−a,thebest1−αintervalonµisY+1logα≤µ≤Y.n9.43a.UsingTheorem8.3.12,identifyg(t)withf(x|θ1)andf(t)withf(x|θ0).DefineRφ(t)=1ift∈Cand0otherwise,andletφ0betheindicatorofanyothersetC0satisfyingf(t)dt≥C01−α.Then(φ(t)−φ0(t))(g(t)−λf(t))≤0andZZZZZZZ0≥(φ−φ0)(g−λf)=g−g−λf−f≥g−g,CC0CC0CC0showingthatCisthebestset.b.ForExercise9.37,thepivotT=Y/θhasdensityntn−1,andthepivotalintervala≤T≤bresultsintheθintervalY/b≤θ≤Y/a.Thelengthisproportionalto1/a−1/b,andthusg(t)=1/t2.Thebestsetis{t:1/t2≤λntn−1},whichisasetoftheform{t:a≤t≤1}.Thishasprobabilitycontent1−αifa=α1/n.ForExercise9.24(orExample9.3.4),thegfunctionisthesameandthedensityofthepivotisfk,thedensityofagamma(k,1).TheRbset{t:1/t2≤λf(t)}={t:f(t)≥λ0},sothebestaandbsatisfyf(t)dt=1−α课后答案网kk+2akandfk+2(a)=fk+2(b).P9.45a.SinceY=Xi∼gamma(n,λ)hasMLR,theKarlin-RubinTheorem(Theorem8.3.2)showsthattheUMPtestistorejectH0ifYθ¯0+zασ/n|σ)=√α,hencethetesthasunconditionalsizeα.Theconfidencesetis{(θ,σ):θ≥x¯−zασ/n},whichhasconfidencecoefficient1−αconditionallyand,hence,unconditionally.b.FromtheKarlin-RubinTheorem,theUMPtestistorejectH0ifX>c.Tomakethissizeα,Pθ0(X>c)=Pθ0(X>c|σ=10)P(σ=10)+P(X>c|σ=1)P(σ=1)X−θ0c−θ0=pP>+(1−p)P(X−θ0>c−θ0)1010c−θ0=pPZ>+(1−p)P(Z>c−θ0),10whereZ∼n(0,1).Withoutlossofgeneralitytakeθ0=0.Forc=z(α−p)/(1−p)wehavefortheproposedtest
课后答案网Pθ0(reject)=p+(1−p)PZ>z(α−p)/(1−p)(α−p)=p+(1−p)=p+α−p=α.www.hackshp.cn(1−p)ThisisnotUMP,butmorepowerfulthanparta.TogetUMP,solveforcinpP(Z>c/10)+(1−p)P(Z>c)=α,andtheUMPtestistorejectifX>c.Forp=1/2,α=.05,wegetc=12.81.Ifα=.1andp=.05,c=1.392andz.1−.05=.0526=1.62..959.51
Pθ(θ∈C(X1,...,Xn))=PθX¯−k1≤θ≤X¯+k2
=Pθ−k2≤X¯−θ≤k1X=Pθ−k2≤Zi/n≤k1,whereZi=Xi−θ,i=1,...,n.Sincethisisalocationfamily,foranyθ,Z1,...,Znareiidwithpdff(z),i.e.,theZisarepivots.Sothelastprobabilitydoesnotdependonθ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn9-14SolutionsManualforStatisticalInference9.52a.TheLRTofH0:σ=σ0versusH1:σ6=σ0isbasedonthestatisticsupµ,σ=σ0L(µ,σ0|x)λ(x)=.supµ,σ∈(0,∞)L(µ,σ2|x)PInthedenominator,ˆσ2=(x−x¯)2/nandˆµ=¯xaretheMLEs,whileinthenumerator,iσ2andˆµaretheMLEs.Thus0Σ(xi−x¯)2Σ(x−x¯)2
−n/2−2−i22σ2−n/22σ22πσ0e0σ0e0λ(x)==,−n/2−Σ(xi−x¯)2σˆ2e−n/2(2πσˆ2)e2σ2and,writingˆσ2=[(n−1)/n]s2,theLRTrejectsHif02−n/2(n−1)s2σ0−22σe01/2πthederivativeisalwayspositivesincee−c/2<1.9.55E[L((µ,σ),C)]=E[L((µ,σ),C)|SK]P(S>K)0=EL((µ,σ),C)|SK]P(S>K)0=RL((µ,σ),C)+E[L((µ,σ),C)|S>K]P(S>K),wherethelastequalityfollowsbecauseC0=∅ifS>K.TheconditionalexpectationinthesecondtermisboundedbyE[L((µ,σ),C)|S>K]=E[blength(C)−IC(µ)|S>K]=E[2bcS−IC(µ)|S>K]>E[2bcK−1|S>K](sinceS>KandIC≤1)=2bcK−1,whichispositiveifK>1/2bc.ForthosevaluesofK,C0dominatesC.9.57a.ThedistributionofX−X¯isn[0,σ2(1+1/n)],son+1pPXn+1∈X¯±zα/2σ1+1/n=P(|Z|≤zα/2)=1−α.b.ppercentofthenormalpopulationisintheintervalµ±zp/2σ,so¯x±kσisa1−αtoleranceintervalifP(µ±zp/2⊆σX¯±kσ)=P(X¯−kσ≤µ−zp/2σandX¯+kσ≥µ+zp/2σ)≥1−α.Thiscanbeattainedbyrequiring课后答案网P(X¯−kσ≥µ−zp/2σ)=α/2andP(X¯+kσ≤µ+zp/2σ)=α/2,√whichisattainedfork=zp/2+zα/2/n.www.hackshp.cnpc.Frompart(a),(pXn+1−X¯)/(S1+1/n)∼tn−1,soa1−αpredictionintervalisX¯±tn−1,α/2S1+1/n.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter10AsymptoticEvaluations10.1Firstcalculatesomemomentsforthisdistribution.1θ2EX=θ/3,EX2=1/3,VarX=−.39So3X¯nisanunbiasedestimatorofθwithvarianceVar(3X¯)=9(VarX)/n=(3−θ2)/n→0asn→∞.nSobyTheorem10.1.3,3X¯nisaconsistentestimatorofθ.10.3a.Theloglikelihoodisn1X−log(2πθ)−(xi−θ)/θ.22Differentiateandsetequaltozero,andalittlealgebrawillshowthattheMLEistheroot√ofθ2+θ−W=0.Therootsofthisequationare(−1±1+4W)/2,andtheMLEistherootwiththeplussign,asithastobenonnegative.Pb.Thesecondderivativeoftheloglikelihoodis(−2x2+nθ)/(2θ3),yieldinganexpectediFisherinformationofP−2X2+nθ2nθ+nI(θ)=−Ei=,θ2θ32θ2andbyTheorem10.1.12thevarianceoftheMLEis1/I(θ).10.4a.WritePPPXiYiXi(Xi+i)XiiP=P=1+P.课后答案网X2X2X2iiiFromnormalityandindependenceEX=0,Varwww.hackshp.cnX=σ2(µ2+τ2),EX2=µ2+τ2,VarX2=2τ2(2µ2+τ2),iiiiiiandCov(Xi,Xii)=0.ApplyingtheformulasofExample5.5.27,theasymptoticmeanandvariancearePPXYXYnσ2(µ2+τ2)σ2iiiiEP≈1andVarP≈=Xi2Xi2[n(µ2+τ2)]2n(µ2+τ2)b.PPYiiP=β+PXiXiwithapproximatemeanβandvarianceσ2/(nµ2).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn10-2SolutionsManualforStatisticalInferencec.1XYi1Xi=β+nXinXiwithapproximatemeanβandvarianceσ2/(nµ2).10.5a.TheintegralofET2isunboundednearzero.WehavenrZ1rZ12n1−(x−µ)2/2σ2n1ETn>2πσ2x2edx>2πσ2Kx2dx=∞,0022whereK=maxe−(x−µ)/2σ0≤x≤1b.Ifwedeletetheinterval(−δ,δ),thentheintegrandisbounded,thatis,overtherangeofintegration1/x2<1/δ2.c.Assumeµ>0.Asimilarargumentworksforµ<0.Then√√√√P(−δj12n(n−1)2=nσ+2ρσn22σ2n−1=+ρσ2nnb.InthiscasewehaveXXnXi−1E(X−µ)(X−µ)=σ2ρi−j.iji>ji=2j=1Inthedoublesumρappearsn−1times,ρ2appearsn−2times,etc..soXnXi−1nX−1ni−jiρ1−ρρ=(n−i)ρ=n−,1−ρ1−ρi=2j=1i=1wheretheseriescanbesummedusing(1.5.4),thepartialsumofthegeometricseries,orusingMathematica.c.ThemeanandvarianceofXiareEX=E[E(X|X)]=EρX=···=ρi−1EXiii−1i−11andVarX=VarE(X|X)+EVar(X|X)=ρ2σ2+1=σ2iii−1ii−1forσ2=1/(1−ρ2).Also,byiteratingtheexpectationEX1Xi=E[E(X1Xi|Xi−1)]=E[E(X1|Xi−1)E(Xi|Xi−1)]=ρE[X1Xi−1],whereweusedthefactsthatX1andXiareindependentconditionalonXi−1.ContinuingwiththeargumentwegetthatE课后答案网XX=ρi−1EX2.Thus,1i1ρi−1EX2−ρi−1(EX)2ρi−1σ2Corr(X,X)=√11=√=ρi−1.1iwww.hackshp.cnVarX1VarXiσ2σ210.21a.Ifanyx→∞,s2→∞,soithasbreakdownvalue0.Toseethis,supposethatx→∞.i1Write!1Xn11Xns2=(x−x¯)2=[(1−)x−x¯]2+(x−x¯)2,i1−1in−1n−1ni=1i=2where¯x−1=(x2+...+xn)/n.Itiseasytoseethatasx1→∞,eachterminthesum→∞.b.Iflessthan50%ofthesample→∞,themedianremainsthesame,andthemedianof|xi−M|remainsthesame.Ifmorethan50%ofthesample→∞,M→∞andsodoestheMAD.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn10-8SolutionsManualforStatisticalInference10.23a.TheAREis[2σf(µ)]2.WehaveDistributionParametersvariancef(µ)AREnormalµ=0,σ=11.3989.64logisticµ=0,β=1π2/3.25.82doubleexp.µ=0,σ=12.52b.IfX,X,...,XareiidfwithEX=µandVarX=σ2,theAREisσ2[2∗f(µ)]2.12nX11XIfwetransformtoYi=(Xi−µ)/σ,thepdfofYiisfY(y)=σfX(σy+µ)withARE[2∗f(0)]2=σ2[2∗f(µ)]2YXc.Themedianismoreefficientforsmallerν,thedistributionswithheaviertails.νVarXf(0)ARE33.3671.6255/3.379.960105/4.389.7572525/23.395.6785025/24.397.657∞1.399.637d.Againtheheaviertailsfavorthemedian.δσARE.012.649.12.747.52.895.015.777.151.83.552.9810.25Bytransformingy=x−θ,Z∞Z∞ψ(x−θ)f(x−θ)dx=ψ(y)f(y)dy.−∞−∞Sinceψisanoddfunction,ψ(y)=−ψ(−y),andZ∞Z0Z∞课后答案网ψ(y)f(y)dy=ψ(y)f(y)dy+ψ(y)f(y)dy−∞−∞0Z0Z∞=−ψ(−y)f(y)dy+ψ(y)f(y)dy−∞0www.hackshp.cnZ∞Z∞=−ψ(y)f(y)dy+ψ(y)f(y)dy=0,00whereinthelastlinewemadethetransformationy→−yandusedthefactthefissymmetric,sof(y)=f(−y).FromthediscussionprecedingExample10.2.6,θˆMisasymptoticallynormalwithmeanequaltothetrueθ.10.27a.1δ(x−µ)lim[(1−δ)µ+δx−µ]=lim=x−µ.δ→0δδ→0δb.P(X≤a)=P(X≤a|X∼F)(1−δ)+P(x≤a|X=x)δ=(1−δ)F(a)+δI(x≤a)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition10-9and1−11(1−δ)F(a)=⇒a=F22(1−δ)1−δ1−12(1−δ)F(a)+δ=⇒a=F22(1−δ)c.Thelimitisaδ−a00lim=aδ|δ=0δ→0δbythedefinitionofderivative.SinceF(a)=1,δ2(1−δ)dd1F(aδ)=dδdδ2(1−δ)or0101f(aδ)aδ=2(1−δ)2⇒aδ=2(1−δ)2f(a).δSincea0=m,theresultfollows.Theotherlimitcanbecalculatedinasimilarmanner.10.29a.Substitutingcl0forψmakestheAREequalto1.b.Foreachdistributionisthecasethatthegivenψfunctionisequaltocl0,hencetheresultingM-estimatorisasymptoticallyefficientby(10.2.9).10.31a.BytheCLT,√pˆ1−p1√pˆ2−p2n1p→n(0,1)andn2p→n(0,1),p1(1−p1)p2(1−p2)soifˆp1andˆp2areindependent,underH0:p1=p2=p,pˆ1−pˆ2r→n(0,1)1+1pˆ(1−pˆ)n1n2whereweuseSlutsky’sTheoremandthefactthatˆp=(S1+S2)/(n1+n2)istheMLEofpunderHandconvergestopinprobability.Therefore,T→χ2.课后答案网01b.SubstituteˆpisforSiandFistogetn2(ˆp−pˆ)2n2(ˆp−pˆ)2T∗=www.hackshp.cn11+22n1pˆn2pˆn2[(1−pˆ)−(1−pˆ)]2n2[(1−pˆ)−(1−pˆ)]2+11+22n1(1−pˆ)n2pˆn(ˆp−pˆ)2n(ˆp−pˆ)21122=+pˆ(1−pˆ)pˆ(1−pˆ)Writeˆp=(n1pˆ1+n2pˆ2)/(n1+n2).Substitutethisintothenumerator,andsomealgebrawillget(ˆp−pˆ)2n(ˆp−pˆ)2+n(ˆp−pˆ)2=12,112211+n1n2soT∗=T.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn10-10SolutionsManualforStatisticalInferencec.UnderH0,pˆ1−pˆ2r→n(0,1)1+1p(1−p)n1n2andbothˆp1andˆp2areconsistent,soˆp1(1−pˆ1)→p(1−p)andˆp2(1−pˆ2)→p(1−p)inprobability.Therefore,bySlutsky’sTheorem,pˆ1−pˆ2q→n(0,1),pˆ1(1−pˆ)pˆ2(1−pˆ)1+2n1n2and(T∗∗)2→χ2.ItiseasytoseethatT∗∗6=Tingeneral.1d.Theestimator(1/n1+1/n2)ˆp(1−pˆ)istheMLEofVar(ˆp1−pˆ2)underH0,whiletheestimatorˆp1(1−pˆ1)/n1+ˆp2(1−pˆ2)/n1istheMLEofVar(ˆp1−pˆ2)underH1.Onemightarguethatinhypothesistesting,thefirstoneshouldbeused,sinceunderH0,itprovidesabetterestimatorofvariance.Ifinterestisinfindingtheconfidenceinterval,however,wearemakinginferenceunderbothH0andH1,andthesecondoneispreferred.e.Wehaveˆp1=34/40,ˆp2=19/35,ˆp=(34+19)/(40+35)=53/75,andT=8.495.Sinceχ2=3.84,wecanrejectHatα=.05.1,.05010.32a.FirstcalculatetheMLEsunderp1=p2=p.Wehaven−1!m−x1−x2−···−xn−1XL(p|x)=px1px2px3···pxn−11−2p−p.n−1ii=3TakinglogsanddifferentiatingyieldthefollowingequationsfortheMLEs:Pn−1∂logLx1+x22m−i=1xi=−P=0∂pp1−2p−n−1pi=3i∂logLxixn=−P=0,i=3,...,n−1,∂pipi1−2p−n−1pi=3iPwithsolutionsˆp=x1+x2,ˆp=xi,i=3,...,n−1,andˆp=m−n−1x/m.Except2mimni=1iforthefirstandsecondcells,wehaveexpected=observed,sincebothareequaltoxi.Forthefirsttwoterms,expected=课后答案网mpˆ=(x1+x2)/2andwegetX(observed−expected)2x1+x2
2x1+x2
22x1−2x2−2(x1−x2)=+=.expectedx1+x2x1+x2x+xwww.hackshp.cn2212b.NowthehypothesisisaboutconditionalprobabilitiesisgivenbyH0:P(change—initialp1p2agree)=P(change—initialdisagree)or,intermsoftheparametersH0:p1+p3=p2+p4.Thisisthesameasp1p4=p2p3,whichisnotthesameasp1=p2.10.33Theorem10.1.12andSlutsky’sTheoremimplythatθˆ−θq→n(0,1)1I(θˆ)nnandtheresultfollows.√10.35a.Sinceσ/nistheestimatedstandarddeviationofX¯inthiscase,thestatisticisaWaldstatistic若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition10-11Pb.TheMLEofσ2isˆσ2=(x−µ)2/n.Theinformationnumberisµii!d2n1σˆ2n2µ−−logσ−=.d(σ2)222σ22ˆσµ2σ2=ˆσ2µqUsingtheDeltamethod,thevarianceofˆσ=σˆ2isˆσ2/8n,andaWaldstatisticisµµµσˆµ−σ0q.σµ2/8n10.37a.Theloglikelihoodisn1XlogL=−logσ2−(x−µ)2/σ2i22iwithd1Xn=(xi−µ)=(¯x−µ)dµσ2σ2id2n=−,dµ2σ2sotheteststatisticforthescoretestisn(¯x−µ)√σ2x¯−µp=nσ2/nσb.WetesttheequivalenthypothesisH:σ2=σ2.ThelikelihoodisthesameasExercise0010.35(b),withfirstderivativen(ˆσ2−σ2)dµ−=课后答案网dσ22σ4andexpectedinformationnumber!n(2ˆσ2−σ2)n(2σ2−σ2)nµwww.hackshp.cnE==.2σ62σ62σ4Thescoreteststatisticisrσˆ2−σ2nµ02σ2010.39Wesummarizetheresultsfor(a)−(c)inthefollowingtable.Weassumethattheunderlyingdistributionisnormal,andusethatforallscorecalculations.Theactualdataisgeneratedfromnormal,logistic,anddoubleexponential.Thesamplesizeis15,weuse1000simulationsanddraw20bootstrapsamples.Hereθ0=0,andthepoweristabulatedforanominalα=.1test.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn10-12SolutionsManualforStatisticalInferenceUnderlyingpdfTestθ0θ0+.25σθ0+.5σθ0+.75σθ0+1σθ0+2σLaplaceNaive0.1010.3660.7740.9570.9931.Boot0.0970.3640.7490.9320.9861.Median0.0650.2450.7060.9620.9951.LogisticNaive0.1370.3410.6830.8960.971.Boot0.1330.3120.6410.8710.9671.Median0.2970.4480.7720.9440.9931.NormalNaive0.1680.3160.6280.8780.9671.Boot0.1480.3060.580.8360.9571.Median0.0960.1910.4790.7610.9351.HereisMathematicacode:ThisprogramcalculatessizeandpowerforExercise10.39,SecondEditionWedoourcalculationsassumingnormality,butsimulatepowerandsizeunderotherdistri-butions.WetestH0:θ=0.theta_0=0;Needs["Statistics‘Master‘"]Clear[x]f1[x_]=PDF[NormalDistribution[0,1],x];F1[x_]=CDF[NormalDistribution[0,1],x];f2[x_]=PDF[LogisticDistribution[0,1],x];f3[x_]=PDF[LaplaceDistribution[0,1],x];v1=Variance[NormalDistribution[0,1]];v2=Variance[LogisticDistribution[0,1]];v3=Variance[LaplaceDistribution[0,1]];Calculatem-estimateClear[k,k1,k2,t,x,y,d,n,nsim,a,w1]ind[x_,k_]:=If[Abs[x]k,1,0];Psi1[x_,k_]=If[Abs[x]<=k,1,0];num=Table[Psi[w1[[j]][[i]],k1],{j,1,nsim},{i,1,n}];den=Table[Psi1[w1[[j]][[i]],k1],{j,1,nsim},{i,1,n}];varnaive=Map[Mean,num^2]/Map[Mean,den]^2;naivestat=Table[Table[m1[[i]][[j]]-theta_0/Sqrt[varnaive[[j]]/n],{j,1,nsim}],{i,1,ntheta}];absnaive=Map[Abs,naivestat];N[Table[Mean[Table[If[absnaive[[i]][[j]]>1.645,1,0],{j,1,nsim}]],{i,1,ntheta}]]Calculationofbootstrapvarianceandteststatisticnboot=20;u:=Random[DiscreteUniformDistribution[n]]databoot=Table[data[[jj]][[u]],{jj,1,nsim},{j,1,nboot},{i,1,n}];m1boot=Table[Table[a/.mest[k1,databoot[[j]][[jj]]],{jj,1,nboot}],{j,1,nsim}];varboot=Map[Variance,m1boot];bootstat=Table[Table[m1[[i]][[j]]-theta_0/Sqrt[varboot[[j]]],{j,1,nsim}],{i,1,ntheta}];absboot=Map[Abs,bootstat];N[Table[Mean[Table[If[absboot[[i]][[j]]>1.645,1,0],{j,1,nsim}]],{i,1,ntheta}]])Calculationofmediantest-usethescorevarianceattherootdensity(normal)med=Map[Median,data];medsd=1/(n*2*f1[theta_0]);medstat=Table[Table[med[[j]]+theta[[i]]-theta_0/medsd,{j,1,nsim}],{i,1,ntheta}];absmed=Map[Abs,medstat];N[Table[Mean[Table[If[(absmed[[i]][[j]]>1.645,1,0],{j,1,nsim}]],{i,1,ntheta}]]10.41a.Theloglikelihoodis课后答案网logL=nrlogp+nx¯log(1−p)withdnrnx¯d2nrnx¯logL=−andlogL=−−,dpwww.hackshp.cnp1−pdp2p2(1−p)2expectedinformationnrand(Wilks)scoreteststatisticp2(1−p)rnx¯r√p−1−pn(1−p)r+px¯nq=√.rr1−pp2(1−p)Sincethisisapproximatelyn(0,1),a1−αconfidencesetisrn(1−p)r−px¯p:√≤zα/2.r1−p若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn10-14SolutionsManualforStatisticalInferenceb.Themeanisµ=r(1−p)/p,andalittlealgebrawillverifythatthevariance,r(1−p)/p2canbewrittenr(1−p)/p2=µ+µ2/r.Thusrn(1−p)r−px¯√µ−x¯√=np.r1−pµ+µ2/rTheconfidenceintervalisfoundbysettingthisequaltozα/2,squaringbothsides,andsolvingthequadraticforµ.Theendpointsoftheintervalareqqr(8¯x+z2)±rz216rx¯+16¯x2+rz2α/2α/2α/2.8r−2z2α/2Forthecontinuitycorrection,replace¯xwith¯x+1/(2n)whensolvingfortheupperendpoint,andwith¯x−1/(2n)whensolvingforthelowerendpoint.c.Wetabletheendpointsforα=.1andarangeofvaluesofr.Notethatr=∞isthePoisson,andsmallervaluesofrgiveawidertailtothenegativebinomialdistribution.rlowerboundupperbound122.1796364.42536.2315107.991038.456595.285040.680785.7110041.001584.53100041.300883.46∞41.334883.3410.43a.Since!XPX=0=(1−p)n=α/2⇒p=1−α1/niiand!XPX=n=pn=α/2⇒p=α1/n,iitheseendpointsareexact,andaretheshortestpossible.b.Sincep∈[0,1],anyvalueoutsidehaszeroprobability,sotruncatingtheintervalshortensitatnocost.课后答案网10.45Thecontinuitycorrectedrootsarerz22ˆp+z2/n±www.hackshp.cn1±α/2[±2n(1−2ˆp)−1]+(2ˆp+z2/n)2−4ˆp2(1+z2/n)α/2nn3α/2α/22(1+z2/n)α/2whereweusetheuppersignfortheupperrootandthelowersignforthelowerroot.Notethattheonlydifferencesbetweenthecontinuity-correctedintervalsandtheordinaryscoreintervalsarethetermswith±infront.Butitisstilldifficulttoanalyticallycomparelengthswiththenon-correctedinterval-wewilldoanumericalcomparison.Forn=10andα=.1wehavethefollowingtableoflengthratios,withthecontinuity-correctedlengthinthedenominatorn012345678910Ratio0.790.820.840.850.860.860.860.850.840.820.79Thecoverageprobabilitiesare若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition10-15p0.1.2.3.4.5.6.7.8.91score.99.93.97.92.90.89.90.92.97.93.99cc.99.99.97.92.98.98.98.92.97.99.99Mathematicacodetodothecalculationsis:Needs["Statistics‘Master‘"]Clear[p,x]pbino[p_,x_]=PDF[BinomialDistribution[n,p],x];cut=1.645^2;n=10;Thequadraticscoreintervalwithandwithoutcontinuitycorrectionslowcc[x_]:=p/.FindRoot[(x/n-1/(2*n)-p)^2==cut*(p*((1-p))/n,{p,.001}]supcc[x_]:=p/.FindRoot[(x/n+1/(2*n)-p)^2==cut*(p*((1-p)/n,{p,.999}]slow[x_]:=p/.FindRoot[(x/n-p))^2==cut*(p*(1-p))/n,{p,.001}]sup[x_]:=p/.FindRoot[(x/n-p)^2==cut*(p*(1-p)/n,{p,.999}]scoreintcc=Partition[Flatten[{{0,sup[0]},Table[{slowcc[i],supcc[i]},{i,1,n-1}],{slowcc[n],1}},2],2];scoreint=Partition[Flatten[{{0,sup[0]},Table[{slow[i],sup[i]},{i,1,n-1}],{slowcc[n],1}},2],2];LengthComparisonTable[(sup[i]-slow[i])/(supcc[i]-slowcc[i]),{i,0,n}]Nowwe’llcalculatecoverageprobabilitiesscoreindcc[p_,x_]:=If[scoreintcc[[x+1]][[1]]<=p<=scoreintcc[[x+1]][[2]],1,0]scorecovcc[p_]:=scorecovcc[p]=Sum[pbino[p,x]*scoreindcc[p,x],{x,0,n}]scoreind[p_,x_]:=If[scoreint[[x+1]][[1]]<=p<=scoreint[[x+1]][[2]],1,0]scorecov[p_]:=scorecov[p]=Sum[pbino[p,x]*scoreind[p,x],{x,0,n}]{scorecovcc[.0001],Table[scorecovcc[i/10],{i,1,9}],scorecovcc[.9999]}//N{scorecov[.0001],Table[scorecov[i/10],{i,1,9}],scorecov[.9999]}//N10.47a.Since2pY∼χ2(approximately)课后答案网nrP(χ2≤2pY≤χ2)=1−α,www.hackshp.cnnr,1−α/2nr,α/2andrearrangmentgivestheinterval.b.TheintervalisoftheformP(a/2Y≤p≤b/2Y),sothelengthisproportionaltob−a.RbThismustbeminimizedsubjecttotheconstraintf(y)dy=1−α,wheref(y)isthepdfaofaχ2.Treatingbasafunctionofa,differentiatinggivesnrb0−1=0andf(b)b0−f(a)=0whichimpliesthatweneedf(b)=f(a).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter11AnalysisofVarianceandRegression11.1a.ThefirstorderTaylor’sseriesapproximationisVar[g(Y)]≈[g0(θ)]2·VarY=[g0(θ)]2·v(θ).Ryb.Ifwechooseg(y)=g∗(y)=√1dx,thenav(x)∗Zθdg(θ)d11=pdx=p,dθdθav(x)v(θ)bytheFundamentalTheoremofCalculus.Then,foranyθ,!2∗1Var[g(Y)]≈pv(θ)=1.v(θ)2∗√dg∗(λ)1∗dg∗(λ)11.2a.v(λ)=λ,g(y)=y,=√,Varg(Y)≈·v(λ)=1/4,independentofλ.dλ2λdλb.TousetheTaylor’sseriesapproximation,weneedtoexpresseverythingintermsofθ=EY=np.Thenv(θ)=θ(1−θ/n)and2∗2dg(θ)1111=q·q·=.dθθθn4nθ(1−θ/n)1−2nnTherefore∗2∗dg(θ)1Var[g(Y)]≈v(θ)=,课后答案网dθ4nindependentofθ,thatis,independentofp.2dg∗(θ)1∗1
22c.v(θ)=Kθ,=andVar[g(Y)]≈·Kθ=K,independentofθ.dθθθ11.3a.g∗(y)isclearlycontinuouswiththepossibleexceptionofwww.hackshp.cnλ=0.Forthatvalueuseλl’Hˆopital’sruletogetyλ−1(logy)yλlim=lim=logy.λ→0λλ→01b.FromExercise11.1,wewanttofindv(λ)thatsatisfiesλZyy−11=pdx.λav(x)TakingderivativesλZydy−1λ−1d11=y=pdx=p.dyλdyav(x)v(y)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-2SolutionsManualforStatisticalInferenceThusv(y)=y−2(λ−1).FromExercise11.1,2yλ−1dθλ−1Var≈v(θ)=θ2(λ−1)θ−2(λ−1)=1.λdyλNote:Ifλ=1/2,v(θ)=θ,whichagreeswithExercise11.2(a).Ifλ=1thenv(θ)=θ2,whichagreeswithExercise11.2(c).11.5ForthemodelYij=µ+τi+εij,i=1,...,k,j=1,...,ni,takek=2.Thetwoparameterconfigurations(µ,τ1,τ2)=(10,5,2)(µ,τ1,τ2)=(7,8,5),havethesamevaluesforµ+τ1andµ+τ2,sotheygivethesamedistributionsforY1andY2.11.6a.UndertheANOVAassumptionsY=θ+,where∼independentn(0,σ2),soY∼ijiijijijindependentn(θ,σ2).ThereforethesamplepdfisiYkYni(yij−θi)21XkXni2−1/2−2−Σni/22(2πσ)e2σ2=(2πσ)exp−(yij−θi)2σ2i=1j=1i=1j=1()Xk2−Σni/212=(2πσ)exp−2niθi2σi=11XX2Xk×exp−y2+θnY¯.2σ2ij2σ2iii·iji=1Therefore,bytheFactorizationTheorem,XXY¯,Y¯,...,Y¯,Y21·2·k·ijij
isjointlysufficientforθ,...,θ,σ2.Since(Y¯,...,Y¯,S2)isa1-to-1functionofthis1k1·k·pvector,(Y¯,...,Y¯,S2)isalsojointlysufficient.课后答案网1·k·pb.Wecanwrite1XkXni(2πσ2)−Σni/2exp−(y−θ)2www.hackshp.cn2σ2ijii=1j=11XkXni=(2πσ2)−Σni/2exp−([y−y¯]+[y¯−θ])22σ2iji·i·ii=1j=1()1XkXni1Xk=(2πσ2)−Σni/2exp−[y−y¯]2exp−n[¯y−θ]2,2σ2iji·2σ2ii·ii=1j=1i=1so,bytheFactorizationTheorem,Y¯i·,i=1,...,n,isindependentofYij−Y¯i·,j=1,...,ni,soS2isindependentofeachY¯.pi·c.JustidentifyniY¯i·withXiandredefineθiasniθi.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-311.7LetUi=Y¯i·−θi.ThenXkXkn[(Y¯−Y¯¯)−(θ−θ¯)]2=n(U−U¯)2.ii·iiii=1i=1TheUareclearlyn(0,σ2/n).ForK=2wehaveiiS2=n(U−U¯)2+n(U−U¯)221122U¯U¯2U¯U¯2n11+n22n11+n22=n1U1−+n2U2−n1+n2n1+n2"22#2n2n1=(U1−U2)n1+n2n1+n2n1+n2(U−U)212=.1+1n1n2SinceU−U∼n(0,σ2(1/n+1/n)),S2/σ2∼χ2.LetU¯betheweightedmeanofkUs,121221kiandnotethatU¯=U¯+nk+1(U−U¯),k+1kk+1kNk+1PkwhereNk=j=1nj.ThenkX+1kX+1222nk+1Sk+1=ni(Ui−U¯k+1)=ni(Ui−U¯k)−(Uk+1−U¯k)Nk+1i=1i=12nk+1Nk2=Sk+(Uk+1−U¯k),Nk+1wherewehaveexpandedthesquare,notedthatthecross-term(summeduptok)iszero,anddidaboat-loadofalgebra.NowsinceU−U¯∼n(0,σ2(1/n+1/N))=n(0,σ2(N/nN)),k+1kk+1kk+1k+1kindependentofS2,therestoftheargumentisthesameasintheproofofTheorem5.3.1(c).k11.8UndertheonewayANOVAassumptions,Y∼independentn(θ,σ2).Therefore课后答案网iji
Y¯∼nθ,σ2/n(Y’sareindependentwithcommonσ2.)i·iiij
aY¯∼naθ,a2σ2/nii·iiiiwww.hackshp.cn!XkXXkaY¯∼naθ,σ2a2/n.ii·iiiii=1i=1AllthesedistributionsfollowfromCorollary4.6.10.11.9a.FromExercise11.8,XXXT=aY¯∼naθ,σ2a2,iiiiiandunderH0,ET=δ.Thus,underH0,PaiY¯i−δqP∼tN−k,S2a2pi若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-4SolutionsManualforStatisticalInferencePwhereN=ni.Therefore,thetestistorejectH0ifPaiY¯i−δqP>tN−k,α.2S2a2/npiiPPb.SimilarlyforH0:aiθi≤δvs.H1:aiθi>δ,werejectH0ifPaiY¯i−δqP>tN−k,α.S2a2/npii11.10a.LetHi,i=1,...,4denotethenullhypothesisusingcontrasta,oftheform0iXHi:aθ≥0.0ijjjIfH1isrejected,itindicatesthattheaverageofθ,θ,θ,andθisbiggerthanθwhich023451isthecontrolmean.IfallHi’sarerejected,itindicatesthatθ>θfori=1,2,3,4.Tosee05ithis,supposeH4andH5arerejected.Thismeansθ>θ5+θ4>θ;thefirstinequalityis00523impliedbytherejectionofH5andthesecondinequalityistherejectionofH4.Asimilar00argumentimpliesθ5>θ2andθ5>θ1.But,forexample,itdoesnotmeanthatθ4>θ3orθ3>θ2.Italsoindicatesthat111(θ5+θ4)>θ3,(θ5+θ4+θ3)>θ2,(θ5+θ4+θ3+θ2)>θ1.234b.Inparta)allofthecontrastsareorthogonal.Forexample,0X50111111a2ia3i=0,1,−,−,−1=−++=0,i=13331366212andthisholdsforallpairsofcontrasts.Now,fromLemma5.4.2,课后答案网!XXσ2XY¯Y¯Covajii·,aj0ii·=ajiaj0i,nwww.hackshp.cniiiwhichiszerobecausethecontrastsareorthogonal.Notethattheequalnumberofobser-0vationspertreatmentisimportant,sinceifni6=ni0forsomei,i,then!XkXkXk2XkY¯Y¯σ2ajiaj0iCovajii,aj0ii=ajiaj0i=σ6=0.ninii=1i=1i=1i=1c.Thisisnotasetoforthogonalcontrastsbecause,forexample,a1×a2=−1.However,eachcontrastcanbeinterpretedmeaningfullyinthecontextoftheexperiment.Forexample,a1teststheeffectofpotassiumalone,whilea5looksattheeffectofaddingzinctopotassium.11.11ThisisadirectconsequenceofLemma5.3.3.11.12a.Thisisaspecialcaseof(11.2.6)and(11.2.7).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-5b.FromExercise5.8(a)Weknowthat1Xk1X222s=(¯yi·−y¯¯)=(¯yi·−y¯i0·).k−12k(k−1)i=1i,i0ThenXX2Xk2121(¯yi·−y¯i0·)(¯yi·−y¯¯)tii0=2=2k(k−1)i,i02k(k−1)i,i0sp/ni=1(k−1)sp/nP2in(¯yi·−y¯¯)/(k−1)=,s2pwhichisdistributedasFk−1,N−kunderH0:θ1=···=θk.NotethatXXkXkt2t2ii0=ii0,i,i0i=1i0=1
thereforet22arebothincluded,whichiswhythedivisorisk(k−1),notk(k−1)=k.ii0andti0i22Also,tousetheresultofExample5.9(a),wetreatedeachmeanY¯i·asanobservation,with¯¯overallmeanY.Thisistrueforequalsamplesizes.11.13a.Nk/2PP1−1kni(y22ij−θi)/σL(θ|y)=e2i=1j=1.2πσ2NotethatXkXniXkXniXk(y−θ)2=(y−y¯)2+n(¯y−θ)2ijiiji·ii·ii=1j=1i=1j=1i=1Xk=SSW+n(¯y−θ)2,ii·ii=1andtheLRTstatisticis课后答案网λ=(ˆτ2/τˆ2)Nk/20whereXτˆ2=SSWwww.hackshp.cnandτˆ2=SSW+n(¯y−y¯)2=SSW+SSB.0ii···iThusλk|η)=P>k≥P>k=α,χ2/(N−k)χ2/(N−k)N−kN−kwheretheinequalityfollowsfromthefactthatthenoncentralchisquaredisstochasticallyincreasinginthenoncentralityparameter.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-6SolutionsManualforStatisticalInference11.14LetX∼n(θ,σ2).ThenfromExercise11.11iiPP√PCov√aiX,cvX=σ2aviciiiiiiiiPaPa2P√
PVar√iX=σ2i,VarcvX=σ2cv2,iciiciiiiiiiandtheCauchy-SchwarzinequalitygivesX.XXava2/c≤cv2.iiiiiiIfai=civithisisanequality,hencetheLHSismaximized.ThesimultaneousstatementisequivalenttoP2ki=1ai(¯yi·−θi)P≤Mforalla1,...,ak,s2ka2/npi=1iandtheLHSismaximizedbyai=ni(¯yi·−θi).ThisproducestheFstatistic.11.15a.Sincet2=F,itfollowsfromExercise5.19(b)thatfork≥2ν1,νP[(k−1)F≥a]≥P(t2≥a).k−1,ννSoifa=t2,theFprobabilityisgreaterthanα,andthustheα-levelcutofffortheFν,α/2mustbegreaterthant2.ν,α/2b.Theonlydifferenceintheintervalsisthecutoffpoint,sotheScheff´eintervalsarewider.c.Bothsetsofintervalshavenominallevel1−α,butsincetheScheff´eintervalsarewider,testsbasedonthemhaveasmallerrejectionregion.Infact,therejectionregioniscontainedinthetrejectionregion.Sothetismorepowerful.11.16a.Ifθi=θjforalli,j,thenθi−θj=0foralli,j,andtheconverseisalsotrue.b.H:θ∈∩ΘandH:θ∈∪(Θ)c.0ijij1ijij11.17a.Ifallofthemeansareequal,theScheff´etestwillonlyrejectαofthetime,sothettestswillbedoneonlyαofthetime.Theexperimentwiseerrorrateispreserved.b.Thisfollowsfromthefactthatthettestsuseasmallercutoffpoint,sotherecanberejectionusingthettestbutnorejectionusingScheff´e.SinceScheff´ehasexperimentwiselevelα,thettesthasexperimentwiseerrorgreaterthanα.c.Thepooledstandarddeviationis2.358,andthemeansand课后答案网tstatisticsareMeantstatisticLowMediumHighMed-LowHigh-MedHigh-Low3.51.www.hackshp.cn9.2724.933.8610.4914.36Thetstatisticsallhave12degreesoffreedomand,forexample,t12,.01=2.68,soallofthetestsrejectandweconcludethatthemeansareallsignificantlydifferent.11.18a.P(Y>a|Y>b)=P(Y>a,Y>b)/P(Y>b)=P(Y>a)/P(Y>b)(a>b)>P(Y>a).(P(Y>b)<1)b.IfaisacutoffpointthenwewoulddeclaresignificanceifY>a.ButifweonlycheckifYissignificantbecauseweseeabigY(Y>b),thepropersignificancelevelisP(Y>a|Y>b),whichwillshowlesssignificancethanP(Y>a).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-711.19a.ThemarginaldistributionsoftheYiParesomewhatstraightforwardtoderive.AsPXi+1∼iigamma(λi+1,1)and,independently,j=1Xj∼gamma(j=1λj,1)(Example4.6.8),weonlyneedtoderivethedistributionoftheratiooftwoindependentgammas.LetX∼gamma(λ1,1)andY∼gamma(λ2,1).Makethetransformationu=x/y,v=y⇒x=uv,y=v,withJacobianv.Thedensityof(U,V)is1uλ1−1f(u,v)=(uv)λ1−1vλ2−1ve−uve−v=vλ1+λ2−1e−v(1+u).Γ(λ1)Γ(λ2)Γ(λ1)Γ(λ2)TogetthedensityofU,integratewithrespecttov.Notethatwehavethekernelofagamma(λ1+λ2,1/(1+u)),whichyieldsΓ(λ+λ)uλ1−112f(u)=.Γ(λ1)Γ(λ2)(1+u)λ1+λ2−1Thejointdistributionisanightmare.Wehavetomakeamultivariatechangeofvariable.Thisismadeabitmorepalatableifwedoitintwosteps.FirsttransformW1=X1,W2=X1+X2,W3=X1+X2+X3,...,Wn=X1+X2+···+Xn,withX1=W1,X2=W2−W1,X3=W3−W2,...Xn=Wn−Wn−1,andJacobian1.ThejointdensityoftheWiisYnf(w,w,...,w)=1(w−w)λi−1e−wn,w≤w≤···≤w,12nii−112nΓ(λi)i=1wherewesetw0=0andnotethattheexponenttelescopes.Nextnotethatw2−w1w3−w2wn−wn−1y1=,y2=,...yn−1=,yn=wn,w1w2wn−1withynwi=Qn−1,i=1,...,n−1,wn=yn.课后答案网j=i(1+yj)Sinceeachwionlyinvolvesyjwithj≥i,theJacobianmatrixistriangularandthedeterminantistheproductofthediagonalelements.Wehavedwiwww.hackshp.cnyndwn=−Q,i=1,...,n−1,=1,dyi(1+y)n−1(1+y)dynij=ijand!λ1−11ynf(y1,y2,...,yn)=Qn−1Γ(λ1)j=1(1+yj)n−1!λi−1Y1ynyn−yn×Γ(λ)Qn−1−Qn−1ei=2ij=i(1+yj)j=i−1(1+yj)nY−1yn×Q.n−1i=1(1+yi)j=i(1+yj)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-8SolutionsManualforStatisticalInferenceFactoroutthetermswithynanddosomealgebraonthemiddletermtoget!λ1−1Σiλi−1−yn11f(y1,y2,...,yn)=yneΓ(λ)Qn−11j=1(1+yj)n−1!λi−1Y1yi−11×QΓ(λi)1+yi−1n−1(1+y)i=2j=ijnY−11×Q.n−1i=1(1+yi)j=i(1+yj)WeseethatYnisindependentoftheotherYi(andhasagammadistribution),buttheredoesnotseemtobeanyotherobviousconclusiontodrawfromthisdensity.b.TheYiarerelatedtotheFdistributionintheANOVA.Forexample,aslongasthesumoftheλiareintegers,χ2Xi+12Xi+1λi+1PYi=Pi=Pi=2∼Fλi.χPi+1,λjj=1Xj2j=1Xjij=1λjj=1NotethattheFdensitymakessenseeveniftheλiarenotintegers.11.21a.188.54Grandmean¯y··==12.5715X3X52Totalsumofsquares=(yij−y¯··)=1295.01.i=1j=1X3X52WithinSS=(yij−y¯i·)11X5X5X5222=(y1j−3.508)+(y2j−9.274)+(y3j−24.926)111=1.089+2.189+63.459=66.74!X32课后答案网BetweenSS=5(yij−y¯i·)1www.hackshp.cn=5(82.120+10.864+152.671)=245.65·5=1228.25.ANOVAtable:SourcedfSSMSFTreatment21228.25614.125110.42Within1266.745.562Total141294.99NotethatthetotalSShereisdifferentfromabove–roundofferroristoblame.Also,F2,12=110.42ishighlysignificant.b.Completingtheproofof(11.2.4),wehaveXkXniXkXni22(yij−y¯¯)=((yij−y¯i·)+(¯yi−y¯¯))i=1j=1i=1j=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-9XkXniXkXni22=(yij−y¯i·)+(¯yi·−y¯¯)i=1j=1i=1j=1XkXni+(yij−y¯i·)(¯yi·−y¯¯),i=1j=1wherethecrossterm(thesumoverj)iszero,sothesumofsquaresispartitionedasXkXniXk22(yij−y¯i·)+ni(¯yi−y¯¯)i=1j=1i=1c.Froma),theFstatisticfortheANOVAis110.42.Theindividualtwo-samplet’s,usings2=1(66.74)=5.5617,arep15−3(3.508−9.274)233.247t2===14.945,12(5.5617)(2/5)2.2247(3.508−24.926)2t2==206.201,132.2247(9.274−24.926)2t2==110.122,232.2247and2(14.945)+2(206.201)+(110.122)=110.42=F.611.23a.EYij=E(µ+τi+bj+ij)=µ+τi+Ebj+Eij=µ+τiVarY=Varb+Var=σ2+σ2,ijjijBbyindependenceofbjandij.b.!XnXnXY¯2Varaii·=aiVarY¯i·+2Cov(aiYi·,ai0Yi0·).i=1i=1i>i0Thefirsttermis课后答案网XnXn1Xr1Xna2VarY¯=a2Varµ+τ+b+=a2(rσ2+rσ2)ii·irijijr2iBi=1www.hackshp.cni=1j=1i=1frompart(a).ForthecovarianceEY¯i·=µ+τi,and1X1XE(Y¯i·Y¯i0·)=Eµ+τi+(bj+ij)µ+τi0+(bj+i0j)rrjj1XX=(µ+τi)(µ+τi0)+2E(bj+ij)(bj+i0j)rjj若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-10SolutionsManualforStatisticalInferencesincethecrosstermshaveexpectationzero.Next,expandingtheproductinthesecondtermagaingivesallzerocrossterms,andwehaveY¯12E(Y¯i·i0·)=(µ+τi)(µ+τi0)+2(rσB),rand2Cov(Y¯i·,Y¯i0·)=σB/r.Finally,thisgives!Xn1XnXY¯2222Varaii·=2ai(rσB+rσ)+2aiai0σB/rri=1i=1i>i0"#1XnXn=a2σ2+σ2(a)2iBiri=1i=1Xn122=σairi=1Xn1222=(σ+σB)(1−ρ)ai,ri=1Pwhere,inthethirdequalityweusedthefactthatiai=0.11.25Differentiationyields∂PsetPPa.∂cRSS=2[yi−(c+dxi)](−1)=0⇒nc+dxi=yi∂PsetPP2P∂dRSS=2[yi−(ci+dxi)](−xi)=0⇒cxi+dxi=xiyi.PPb.Notethatnc+dxi=yi⇒c=¯y−dx¯.ThenXXXXXX(¯y−dx¯)x+dx2=xyanddx2−nx¯2=xy−xy¯iiiiiiiiPPwhichsimplifiestod=x(y−y¯)/(x−x¯)2.Thuscanddaretheleastsquaresiiiestimates.c.Thesecondderivativesare∂2∂2X∂2X课后答案网RSS=n,RSS=x,RSS=x2.∂c2∂c∂di∂d2iThustheJacobianofthesecond-orderpartialsiswww.hackshp.cnPXX2XPnPxi=nx2−x=n(x−x¯)2>0.xx2iiiiiP11.27ForthelinearestimatoriaiYitobeunbiasedforαwehave!XXXXEaiYi=ai(α+βxi)=α⇒ai=1andaixi=0.iiiiPPSinceVaraY=σ2a2,weneedtosolve:iiiiiXXXminimizea2subjecttoa=1andax=0.iiiiiii若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-11AsolutioncanbefoundwithLagrangemultipliers,butverifyingthatitisaminimumisexcruciating.SoinsteadwenotethatX1ai=1⇒ai=+k(bi−¯b),niforsomeconstantsk,b1,b2,...,bn,andX−x¯1x¯(bi−¯b)aixi=0⇒k=Pandai=−P.ii(bi−¯b)(xi−x¯)ni(bi−¯b)(xi−x¯)Now2PXX1x¯(b−¯b)1x¯2(b−¯b)2a2=−Pi=+Pii,in(b−¯b)(x−x¯)n[(b−¯b)(x−x¯)]2iiiiiiiisincethecrosstermiszero.Soweneedtominimizethelastterm.FromCauchy-SchwarzweknowthatP(b−¯b)21PiiP≥,[(bi−¯b)(xi−x¯)]2i(xi−x¯)]2iandtheminimumisattainedatbi=xi.Substitutingbackwegetthattheminimizingaiis1Px¯(xi−x¯)Pn−(x−x¯)2,whichresultsiniaiYi=Y¯−βˆx¯,theleastsquaresestimator.ii11.28Tocalculaten/211Σ222−i[yi−(ˆα+βˆxi)]/σmaxL(σ|y,αˆβˆ)=maxe2σ2σ22πσ2takelogsanddifferentiatewithrespecttoσ2togetPdn1[y−(ˆα+βxˆ)]2logL(σ2|y,α,ˆβˆ)=−+iii.dσ22σ22(σ2)2Setthisequaltozeroandsolveforσ2.Thesolutionisˆσ2.11.29a.课后答案网Eˆi=E(Yi−αˆ−βxˆi)=(α+βxi)−α−βxi=0.b.Varˆ=E[Ywww.hackshp.cn−αˆ−βxˆ]2iii=E[(Y−α−βx)−(ˆα−α)−x(βˆ−β)]2iii=VarY+Varˆα+x2Varβˆ−2Cov(Y,αˆ)−2xCov(Y,βˆ)+2xCov(ˆα,βˆ).iiiiii11.30a.Straightforwardalgebrashowsαˆ=y¯−βˆx¯PX1x¯(xi−x¯)yi=yi−Pn(xi−x¯)2X1x¯(xi−x¯)=−Pyi.n(xi−x¯)2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-12SolutionsManualforStatisticalInference1Px¯(xi−x¯)PPb.Notethatforci=n−(x−x¯)2,ci=1andcixi=0.TheniXXEˆα=EciYi=ci(α+βxi=α,XXVarˆα=c2VarY=σ2c2,iiiandXX2X1P2221Px¯(xi−x¯)x¯(xi−x¯)ci=n−(x−x¯)2=n2+P22(crossterm=0)i((xi−x¯))P1x¯2x2=+P=i.n(x−x¯)2nSxxiPc.Writeβˆ=diyi,wherexi−x¯di=P.(xi−x¯)2FromExercise11.11,XXXCov(ˆα,βˆ)=CovcY,dY=σ2cdiiiiiiX1x¯(x−x¯)(x−x¯)−σ2x¯2PiPiP=σ−=.n(xi−x¯)2(xi−x¯)2(xi−x¯)211.31ThefactthatXˆi=[δij−(cj+djxi)]YjiPfollowsdirectlyfrom(11.3.27)andthedefinitionofcjanddj.Sinceˆα=iciYi,fromLemma11.3.2XCov(ˆ,αˆ)=σ2c[δ−(c+dx)]ijijjjijX=σ2c−c(c+dx)ijjjijXX=σ2c−c2−xcd.课后答案网ijijjjjSubstitutingforcjanddjgiveswww.hackshp.cn1(xi−x¯)¯xci=−nSxxX1x¯2c2=+jnSxxjXxix¯xicjdj=−,SxxjandsubstitutingthesevaluesshowsCov(ˆi,αˆ)=0.Similarly,forβˆ,XXCov(ˆ,βˆ)=σ2d−cd−xd2iijjijjj若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-13with(xi−x¯)di=SxxXx¯cjdj=−SxxjX1xd2=,ijSxxjandsubstitutingthesevaluesshowsCov(ˆi,βˆ)=0.11.32Writethemodelsas3yi=α+βxi+iy=α0+β0(x−x¯)+iii=α0+β0z+.iia.Since¯z=0,PPβˆ=(xPi−x¯)(yi−y¯)=zPi(yi−y¯)=βˆ0.(xi−x¯)2z2ib.αˆ=y¯−βˆx¯,αˆ0=y¯−βˆ0z¯=¯ysince¯z=0.αˆ0∼n(α+βz,σ¯2/n)=n(α,σ2/n).c.WriteX1Xzαˆ0=yβˆ0=Piy.niz2iiThenX1z2PiCov(α,ˆβˆ)=−σ=0,nz2iPsincezi课后答案网=0.11.33a.From(11.23.25),β=ρ(σY/σX),soβ=0ifandonlyifρ=0(sinceweassumethatthevariancesarepositive).b.Startfromthedisplayfollowing(11.3.35).Wehavewww.hackshp.cnβˆ2S2/Sxyxx=S2/SxxRSS/(n−2)S2xy=(n−2)
Syy−Sxy2/SxxSxxS2xy=(n−2)
,SyySxx−Sxy2anddividingtopandbottombySyySxxfinishestheproof.√√√c.From(11.3.33)ifρ=0(equivalentlyβ=0),thenβ/ˆ(S/Sxx)=n−2r/1−r2hasatn−2distribution.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-14SolutionsManualforStatisticalInference11.34a.ANOVAtableforheightdataSourcedfSSMSFRegression160.3660.3650.7Residual67.141.19Total767.50Theleastsquareslineisˆy=35.18+.93x.b.Sinceyi−y¯=(yi−yˆi)+(ˆyi−y¯),wejustneedtoshowthatthecrosstermiszero.XnXnhihi(yi−yˆi)(ˆyi−y¯)=yi−(ˆα+βxˆi)(ˆα+βxˆi)−y¯i=1i=1Xnhihi=(ˆyi−y¯)−βˆ(xi−x¯)βˆ(xi−x¯)(ˆα=¯y−βˆx¯)i=1XnXn=βˆ(x−x¯)(y−y¯)−βˆ2(x−x¯)2=0,iiii=1i=1fromthedefinitionofβˆ.c.XXS2(ˆy−y¯)2=βˆ2(x−x¯)2=xy.iiSxx11.35a.Fortheleastsquaresestimate:dXX(y−θx2)2=2(y−θx2)x2=0iiiiidθiiwhichimpliesPyx2θˆ=Piii.x4iib.Theloglikelihoodisn1XlogL=−log(2πσ2)−(y−θx2)2,22σ2ii课后答案网iandmaximizingthisisthesameastheminimizationinpart(a).c.Thederivativesoftheloglikelihoodarewww.hackshp.cnd1XlogL=(y−θx2)x2dθσ2iiiid2−1XlogL=x4,dθ2σ2iiPsotheCRLBisσ2/x4.ThevarianceofθˆisiiP!yx2Xx2XVarθˆ=VarPiii=Piσ2=σ2/x4,x4x4iiiijjisoθˆisthebestunbiasedestimator.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-1511.36a.hiEˆα=E(Y¯−βˆX¯)=EE(Y¯−βˆX¯|X¯)=Eα+βX¯−βX¯=Eα=α.Eβˆ=E[E(βˆ|X¯)]=Eβ=β.b.RecallVarY=Var[E(Y|X)]+E[Var(Y|X)]Cov(Y,Z)=Cov[E(Y|X),E(Z|X)]+E[Cov(Y,Z|X)].ThushX.iVarˆα=E[Var(ˆα|X)]=σ2EX2SiXXVarβˆ=σ2E[1/S]XXCov(ˆα,βˆ)=E[Cov(ˆα,βˆ|Xˆ)]=−σ2E[X/S¯].XX11.37ThisisalmostthesameproblemasExercise11.35.Theloglikelihoodisn1XlogL=−log(2πσ2)−(y−βx)2.22σ2iiiPPPTheMLEisxy/x2,withmeanβandvarianceσ2/x2,theCRLB.iiiiiiiPP11.38a.Themodelisy=θx+,sotheleastsquaresestimateofθisxy/x2(regressioniiiiiithroughtheorigin).PPxiYixi(xiθ)EP=P=θx2x2iiPPPxYx2(xθ)x3VarPii=ii=θi.x2P22P22i(xi)(xi)Theestimatorisunbiased.b.ThelikelihoodfunctionisnQYe−θxi(θx)yie−θΣxi(θx)yiiiL(θ|x)==Q(yi)!yi!i=1课后答案网∂∂hXXYilogL=−θxi+yilog(θxi)−logyi!∂θ∂θXXxiyiset=−xi+=0www.hackshp.cnθxiwhichimpliesPθˆ=PyixiPPPθxiyiθxiθEθˆ=P=θandVarθˆ=VarP=P=P.xixi(x)2xiic.PPP∂2∂Xy−y∂2xiiilogL=−xi+=andE−logL=.∂θ2∂θθθ2∂θ2θPThus,theCRLBisθ/xi,andtheMLEisthebestunbiasedestimator.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-16SolutionsManualforStatisticalInference11.39LetAibethesetshi.1(x−x¯)2A=α,ˆβˆ:(ˆα+βxˆ)−(α+βx)S+0i≤t.i0i0in−2,α/2mnSxxThenP(∩mA)istheprobabilityofsimultaneouscoverage,andusingtheBonferroniIn-i=1iequality(1.2.10)wehaveXmXmαP(∩mA)≥P(A)−(m−1)=1−−(m−1)=1−α.i=1iimi=1i=111.41Assumethatwehaveobserveddata(y1,x1),(y2,x2),...,(yn−1,xn−1)andwehavexnbutnoty.Letφ(y|x)denotethedensityofY,an(a+bx,σ2).niiiia.Theexpectedcomplete-dataloglikelihoodis!XnnX−1Elogφ(Yi|xi)=logφ(yi|xi)+Elogφ(Y|xn),i=1i=1wheretheexpectationisrespecttothedistributionφ(y|xn)withthecurrentvaluesoftheparameterestimates.Thusweneedtoevaluate1212Elogφ(Y|xn)=E−log(2πσ1)−2(Y−µ1),22σ1whereY∼n(µ,σ2).Wehave00E(Y−µ)2=E([Y−µ]+[µ−µ])2=σ2+[µ−µ]2,1001001sincethecrosstermiszero.Puttingthisalltogether,theexpectedcomplete-dataloglikelihoodisn1nX−1σ2+[(a+bx)−(a+bx)]2−log(2πσ2)−[y−(a+bx)]2−000n11n212σ2i11i2σ21i=11n1Xnσ2=−log(2πσ2)−[y−(a+bx)]2−0课后答案网212σ2i11i2σ21i=11ifwedefineyn=a0+b0xn.b.Forfixeda0andb0www.hackshp.cn,maximizingthislikelihoodgivestheleastsquaresestimates,whilethemaximumwithrespecttoσ2is1Pn222i=1[yi−(a1+b1xi)]+σ0σˆ1=.nSotheEMalgorithmisthefollowing:Atiterationt,wehaveestimatesˆa(t),ˆb(t),andˆσ2(t).Wethensety(t)(t)+ˆb(t)x(whichisessentiallytheE-step)andthentheM-stepisn=ˆantocalculateˆa(t+1)andˆb(t+1)astheleastsquaresestimatorsusing(y,x),(y,x),...1122(t)(yn−1,xn−1),(yn,xn),andPn[y−(a(t+1)+b(t+1)x)]2+σ2(t)σˆ2(t+1)=i=1ii0.1n若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition11-17c.TheEMcalculationsaresimplehere.Sincey(t)(t)+ˆb(t)x,theestimatesofaandbn=ˆanmustconvergetotheleastsquaresestimates(sincetheyminimizethesumofsquaresoftheobserveddata,andthelasttermaddsnothing.Forˆσ2wehave(substitutingtheleastsquaresestimates)thestationarypointPn222i=1[yi−(ˆa+ˆbxi)]+ˆσ22σˆ=⇒σˆ=σobs,nwhereσ2istheMLEfromthen−1observeddatapoints.SotheMLEsarethesameasobsthosewithouttheextraxn.d.Nowweusethebivariatenormaldensity(seeDefinition4.5.10andExercise4.45).Denotethedensitybyφ(x,y).Thentheexpectedcomplete-dataloglikelihoodisnX−1logφ(xi,yi)+Elogφ(X,yn),i=1whereafteriterationtthemissingdatadensityistheconditionaldensityofXgivenY=yn,X|Y=y∼nµ(t)+ρ(t)(σ(t)/σ(t))(y−µ(t)),(1−ρ2(t))σ2(t).nXXYnYXDenotingthemeanbyµandthevariancebyσ2,theexpectedvalueofthelastpiecein00thelikelihoodisElogφ(X,yn)1222=−log(2πσXσY(1−ρ))2"22#1X−µX(X−µX)(yn−µY)yn−µY−E−2ρE+2(1−ρ2)σXσXσYσY1222=−log(2πσXσY(1−ρ))2"22#1σ2µ−µ(µ−µ)(y−µ)y−µ00X0XnYnY−+−2ρ+.2(1−ρ2)σ2σXσXσYσYXSotheexpectedcomplete-dataloglikelihoodis课后答案网nX−1σ20logφ(xi,yi)+logφ(µ0,yn)−2(1−ρ2)σ2.www.hackshp.cni=1XTheEMalgorithmissimilartothepreviousone.FirstnotethattheMLEsofµandσ2YYaretheusualones,¯yandˆσ2,anddon’tchangewiththeiterations.WeupdatetheotherY(t)estimatesasfollows.Atiterationt,theE-stepconsistsofreplacingxnby(t)(t+1)(t)(t)σXxn=ˆµX+ρ(t)(yn−y¯).σY(t+1)Thenµ=¯xandwecanwritethelikelihoodasX11S+σ2SS−log(2πσ2σˆ2(1−ρ2))−xx0−2ρxy+yy.2XY2(1−ρ2)σ2σσˆσˆ2XXYY若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn11-18SolutionsManualforStatisticalInferencewhichistheusualbivariatenormallikelihoodexceptthatwereplaceSwithS+σ2.xxxx0SotheMLEsaretheusualones,andtheEMiterationsare(t)(t+1)(t)(t)σXxn=µˆX+ρ(t)(yn−y¯)σYµˆ(t+1)=x¯(t)XS(t)2(t))ˆσ2(t)2(t+1)xx+(1−ρˆXσˆ=Xn(t)(t+1)Sxyρˆ=q.(S(t)2(t))σˆ2(t+1))Sxx+(1−ρˆXyyHereisRcodefortheEMalgorithm:nsim<-20;xdata0<-c(20,19.6,19.6,19.4,18.4,19,19,18.3,18.2,18.6,19.2,18.2,18.7,18.5,18,17.4,16.5,17.2,17.3,17.8,17.3,18.4,16.9)ydata0<-(1,1.2,1.1,1.4,2.3,1.7,1.7,2.4,2.1,2.1,1.2,2.3,1.9,2.4,2.6,2.9,4,3.3,3,3.4,2.9,1.9,3.9,4.2)nx<-length(xdata0);ny<-length(ydata0);#initialvaluesfrommlesontheobserveddata#xmean<-18.24167;xvar<-0.9597797;ymean<-2.370833;yvar<-0.8312327;rho<--0.9700159;for(jin1:nsim){#Thisistheaugmentedx(O2)data#xdata<-c(xdata0,xmean+rho*(4.2-ymean)/(sqrt(xvar*yvar)))xmean<-mean(xdata);Sxx<-(ny-1)*var(xdata)+(1-rho^2)*xvarxvar<-Sxx/nyrho<-cor(xdata,ydata0)*sqrt((ny-1)*var(xdata)/Sxx)}Thealgorithmconvergesveryquickly.TheMLEsareµˆ=18.24,µˆ=2.37,σˆ2=.969,σˆ2=.831,ρˆ=−0.969.课后答案网XYXYwww.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnChapter12RegressionModels12.1Thepoint(ˆx0,yˆ0)istheclosestifitliesonthevertexoftherighttrianglewithvertices(x0,y0)and(x0,a+bx0).BythePythagoreantheorem,wemusthaveh
2ihi(ˆx0−x0)2+yˆ0−(a+bx0)+(ˆx0−x0)2+(ˆy0−y0)2=(x0−x0)2+(y0−(a+bx0))2.Substitutingthevaluesofˆx0andˆy0from(12.2.7)weobtainfortheLHSabove"0202#"022#b(y−bx0−a)b2(y−bx0−a)b(y−bx0−a)y0−bx−a)+++1+b21+b21+b21+b2"#b2+b4+b2+1002002=(y−(a+bx))=(y−(a+bx)).(1+b2)2set212.3a.Differentiationyields∂f/∂ξi=−2(xi−ξi)−2λβ[yi−(α+βξi)]=0⇒ξi(1+λβ)=x−λβ(y−α),whichistherequiredsolution.Also,∂2f/∂ξ2=2(1+λβ2)>0,sothisisaiiminimum.b.Partsi),ii),andiii)areimmediate.Foriv)justnotethat√√DisEuclideandistancebetween(x1,λy1)and(x2,λy2),hencesatisfiesthetriangleinequality.12.5DifferentiatelogL,forLin(12.2.17),toget∂−n1λXnhi2∂σ2logL=σ2+222yi−(ˆα+βxˆi).δδ2(σδ)1+βˆi=1Setthisequaltozeroandsolveforσ2.Theansweris(12.2.18).δ12.7a.Suppressingthesubscriptiandtheminussign,theexponentis课后答案网222222(x−ξ)[y−(α+βξ)]σ+βσδ2[y−(α+βx)]+=(ξ−k)+,σ2σ2σ2σ2σ2+β2σ2δδδ22wherek=σx+σδβ(y−α).Thus,integratingwithrespecttoξeliminatesthefirstterm.σ2+β2www.hackshp.cnσ2δb.TheresultingfunctionmustbethejointpdfofXandY.Thedoubleintegralisinfinite,however.12.9a.Fromthelasttwoequationsin(12.2.19),21211Sxyσˆδ=Sxx−σˆξ=Sxx−,nnnβˆwhichispositiveonlyifSxx>Sxy/βˆ.Similarly,2122121Sxyσˆ=Syy−βˆσˆξ=Syy−βˆ,nnnβˆwhichispositiveonlyifSyy>βSˆxy.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn12-2SolutionsManualforStatisticalInferenceb.Wehavefromparta),ˆσ2>0⇒S>S/βˆandˆσ2>0⇒S>βSˆ.Furthermore,δxxxyyyxyσˆ2>0impliesthatSandβˆhavethesamesign.ThusS>|S|/|βˆ|andS>|βˆ||S|.ξxyxxxyyyxyCombiningyields|Sxy|Syy<βˆ<.Sxx|Sxy|12.11a.Cov(aY+bX,cY+dX)=E(aY+bX)(cY+dX)−E(aY+bX)E(cY+dX)
22=EacY+(bc+ad)XY+bdX−E(aY+bX)E(cY+dX)=acVarY+ac(EY)2+(bc+ad)Cov(X,Y)+(bc+ad)EXEY+bdVarX+bd(EX)2−E(aY+bX)E(cY+dX)=acVarY+(bc+ad)Cov(X,Y)+bdVarX.b.Identifya=βλ,b=1,c=1,d=−β,andusing(12.3.19)Cov(βλY+X,Y−βX)=βλVarY+(1−λβ2)Cov(X,Y)−βVarXiiii

=βλσ2+β2σ2+(1−λβ2)βσ2−βσ2+σ2ξξδξ=βλσ2−βσ2=0δifλσ2=σ2.(Notethatwedidnotneedthenormalityassumption,justthemoments.)δc.LetW√i=βλY√i+Xi,Vi=Yi+βXi.Exercise11.33showsthatifCov(√pWi,Vi)=0,thenn−2r/1−r2hasatn−2distribution.Thusn−2rλ(β)/1−r2(β)hasatn−2λdistributionforallvaluesofβ,bypart(b).Also(2)!(n−2)r(n−2)r2(β)λ(β)λPβ:2≤F1,n−2,α=P(X,Y):2≤F1,n−2,α=1−α.1−r1−r(β)λ(β)λ12.13a.Rewrite(12.2.22)toget2tσˆβtσˆβ(βˆ−β)β:βˆ−√≤β≤βˆ+√=β:.≤F.n−2n−2σ2(n−2)课后答案网βb.Forβˆof(12.2.16),thenumeratorofrλ(β)in(12.2.22)canbewritten221βλSyy+(1−βλ)Sxy−βSxy=β(λSxy)+β(Sxx−λSyy)+Sxy=λSxy(β−βˆ)β+.www.hackshp.cnλβˆAgainfrom(12.2.22),wehaver2(β)λ1−r2(β)λ
2βλS+(1−β2λ)S−βSyyxyxy=,(β2λ2S+2βλS+S)(S−2βS+β2S)−(βλS+(1−β2λ)S−βS)2yyxyxxyyxyxxyyxyxxandagreatdealofstraightforward(buttedious)algebrawillshowthatthedenominatorofthisexpressionisequalto
(1+λβ2)2SS−S2.yyxxxy若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition12-3Thus22λ2S2β−βˆβ+1r2(β)xyλλβˆ=y
1−r2(β)(1−λβ2)2SS−S2λyyxxy2!2β−βˆ1+λββˆ(1+λβˆ2)2S2xy=hi,σˆ21+λβ2βˆ222β(Sxx−λSyy)+4λSxyaftersubstitutingˆσ2frompage588.Nowusingthefactthatβˆand−1/λβˆarebothrootsβofthesamequadraticequation,wehave222(S−λS)2+4λS2(1+λβˆ)1xxyyxy=+λβˆ=.βˆ2βˆS2xyThustheexpressioninsquarebracketsisequalto1.12.15a.eα+β(−α/β)e01π(−α/β)===.1+eα+β(−α/β)1+e02b.eα+β((−α/β)+c)eβcπ((−α/β)+c)==,1+eα+β((−α/β)+c)1+eβcande−βceβc1−π((−α/β)−c)=1−=.1+e−βc1+eβcc.deα+βxπ(x)=β=βπ(x)(1−π(x)).dx[1+eα+βx]2d.Becauseπ(x)α+βx=e,课后答案网1−π(x)theresultfollowsfromdirectsubstitution.e.Followsdirectlyfrom(d).f.Followsdirectlyfromwww.hackshp.cn∂∂F(α+βx)=f(α+βx)andF(α+βx)=xf(α+βx).∂α∂βg.ForF(x)=ex/(1+ex),f(x)=F(x)(1−F(x))andtheresultfollows.ForF(x)=π(x)off(12.3.2),frompart(c)iffollowsthat=β.F(1−F)12.17a.ThelikelihoodequationsandsolutionarethesameasinExample12.3.1withtheexceptionthathereπ(xj)=Φ(α+βxj),whereΦisthecdfofastandardnormal.b.Ifthe0−1failureresponseindenoted“oring”andthetemperaturedatais“temp”,thefollowingRcodewillgeneratethelogitandprobitregression:summary(glm(oring~temp,family=binomial(link=logit)))summary(glm(oring~temp,family=binomial(link=probit)))若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cn12-4SolutionsManualforStatisticalInferenceForthelogitmodelwehaveEstimateStd.ErrorzvaluePr(>|z|)Intercept15.04297.37192.0410.0413temp−0.23220.1081−2.1470.0318andfortheprobitmodelwehaveEstimateStd.ErrorzvaluePr(>|z|)Intercept8.770843.862222.2710.0232temp−0.135040.05632−2.3980.0165Althoughthecoefficientsaredifferent,thefitisqualitativelythesame,andtheprobabilityoffailureat31◦,usingtheprobitmodel,is.9999.12.19a.UsingthenotationofExample12.3.1,thelikelihood(jointdensity)isJy∗n∗JPPYeα+βxjj1j−yjY1nj∗∗αyj+βxjyj=ejj.1+eα+βxj1+eα+βxj1+eα+βxjj=1j=1PPBytheFactorizationTheorem,y∗andxy∗aresufficient.jjjjjb.Straightforwardsubstitution.12.21Sincedlog(π/(1−π))=1/(π(1−π)),dπ2πˆ1π(1−π)1Varlog≈=1−πˆπ(1−π)nnπ(1−π)P12.23a.Ifai=0,XXXEaiYi=ai[α+βxi+µ(1−δ)]=βaixi=βiiiforai=xi−x¯.b.1XE(Y¯−βx¯)=[α+βxi+µ(1−δ)]−βx¯=α+µ(1−δ),nisotheleastsquaresestimateaisunbiasedinthemodelY=α0+βx+,whereα0=iiiα+µ(1−课后答案网δ).12.25a.Theleastabsolutedeviationlineminimizes|www.hackshp.cny1−(c+dx1)|+|y2−(c+dx1)|+|y3−(c+dx3)|.Anylinethatliesbetween(x1,y1)and(x1,y2)hasthesamevalueforthesumofthefirsttwoterms,andthisvalueissmallerthanthatofanylineoutsideof(x1,y1)and(x2,y2).Ofallthelinesthatlieinside,theonesthatgothrough(x3,y3)minimizetheentiresum.b.Fortheleastsquaresline,a=−53.88andb=.53.Anylinewithbbetween(17.9−14.4)/9=.39and(17.9−11.9)/9=.67anda=17.9−136bisaleastabsolutedeviationline.12.27IntheterminologyofM-estimators(seetheargumentonpages485−486),βˆLisconsistentPfortheβ0thatsatisfiesEβ0iψ(Yi−β0xi)=0,sowemusttakethe“true”βtobethisvalue.WethenseethatXψ(Yi−βˆLxi)→0iaslongasthederivativetermisbounded,whichweassumeisso.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn 课后答案网：www.hackshp.cnSecondEdition12-512.29TheargumentforthemedianisaspecialcaseofExample12.4.3,wherewetakexi=1soσ2=1.Theasymptoticdistributionisgivenin(12.4.5)which,forσ2=1,agreeswithxxExample10.2.3.12.31TheLADestimates,fromExample12.4.2are˜α=18.59andβ˜=−.89.HereisMathematicacodetobootstrapthestandarddeviations.(Mathematicaisprobablynotthebestchoicehere,asitissomewhatslow.Also,theminimizationseemedabitdelicate,andworkedbetterwhendoneiteratively.)Sadisthesumoftheabsolutedeviations,whichisminimizediterativelyinbminandamin.Theresidualsarebootstrappedbygeneratingrandomindicesufromthediscreteuniformdistributionontheintegers1to23.1.FirstenterdataandinitializeNeeds["Statistics‘Master‘"]Clear[a,b,r,u]a0=18.59;b0=-.89;aboot=a0;bboot=b0;y0={1,1.2,1.1,1.4,2.3,1.7,1.7,2.4,2.1,2.1,1.2,2.3,1.9,2.4,2.6,2.9,4,3.3,3,3.4,2.9,1.9,3.9};x0={20,19.6,19.6,19.4,18.4,19,19,18.3,18.2,18.6,19.2,18.2,18.7,18.5,18,17.4,16.5,17.2,17.3,17.8,17.3,18.4,16.9};model=a0+b0*x0;r=y0-model;u:=Random[DiscreteUniformDistribution[23]]Sad[a_,b_]:=Mean[Abs[model+rstar-(a+b*x0)]]bmin[a_]:=FindMinimum[Sad[a,b],{b,{.5,1.5}}]amin:=FindMinimum[Sad[a,b/.bmin[a][[2]]],{a,{16,19}}]2.Hereistheactualbootstrap.Thevectorsabootandbbootcontainthebootstrappedvalues.B=500;Do[rstar=Table[r[[u]],{i,1,23}];astar=a/.amin[[2]];bstar=b/.bmin[astar][[2]];aboot=Flatten[{aboot,astar}];bboot=Flatten[{bboot,bstar}],{i,1,B}]3.SummaryStatistics课后答案网Mean[aboot]StandardDeviation[aboot]Mean[bboot]StandardDeviation[bboot]www.hackshp.cn4.TheresultsareIntercept:Mean18.66,SD.923Slope:Mean−.893,SD.050.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn'