统计推断(Statistical Inference)第二版 练习题 答案

SolutionsManualforStatisticalInference,SecondEditionGeorgeCasellaUniversityofFloridaRogerL.BergerNorthCarolinaStateUniversityDamarisSantanaUniversityofFlorida0-2SolutionsManualforStatisticalInference“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,andRisafreedownloadfrom−68643358,16,26,28,34,36,38,42766524,14,16,28,30,32,34,36,42,54,58,60,62,648585136,40,46,48,52,56,58958412,8,10,20,22,24,26,28,3032,38,40,42,44,50,54,56104826allevenproblemsexcept4and321141354,20,22,24,26,40123116allevenproblems0.2AcknowledgementManypeoplecontributedtotheassemblyofthissolutionsmanual.Weagainthankallofthosewhocontributedsolutionstothefirstedition–manyproblemshavecarriedoverintothesecondedition.Moreover,throughouttheyearsanumberofpeoplehavebeeninconstanttouchwithus,contributingtoboththepresentationsandsolutions.Weapologizeinadvanceforthoseweforgettomention,andweespeciallythankJayBeder,YongSungJoo,MichaelPerlman,RobStrawderman,andTomWehrly.Thankyouallforyourhelp.And,aswesaidthefirsttimearound,althoughwehavebenefitedgreatlyfromtheassistanceandACKNOWLEDGEMENT0-3commentsofothersintheassemblyofthismanual,weareresponsibleforitsultimatecorrectness.Tothisend,wehavetriedourbestbut,asawisemanoncesaid,“Youpaysyourmoneyandyoutakesyourchances.”GeorgeCasellaRogerL.BergerDamarisSantanaDecember,2001Chapter1ProbabilityTheory“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∈A\B⇔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(B∩A)∪(B∩Ac)=B∩(A∪Ac)=B∩S=B.)c.Similartoparta).d.Frompartb).A∪B=A∪[(B∩A)∪(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).1-2SolutionsManualforStatisticalInferenceb.“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)=190×13×121.6p0=(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.P(scoringipoints)=￿1−πr2Aifi=0πr2A￿(6−i)2−(5−i)252￿ifi=1,...,5.b.P(scoringipoints|boardishit)=P(scoringipoints∩boardishit)P(boardishit)P(boardishit)=πr2AP(scoringipoints∩boardishit)=πr2A￿(6−i)2−(5−i)252￿i=1,...,5.Therefore,P(scoringipoints|boardishit)=(6−i)2−(5−i)252i=1,...,5whichisexactlytheprobabilitydistribution