Durrent习题答案.pdf
- 文档编号:18861566
- 上传时间:2024-02-02
- 格式:PDF
- 页数:122
- 大小:6.09MB
Durrent习题答案.pdf
《Durrent习题答案.pdf》由会员分享,可在线阅读,更多相关《Durrent习题答案.pdf(122页珍藏版)》请在冰点文库上搜索。
Probability:
TheoryandExamplesSolutionsManualThecreationofthissolutionmanualwasoneofthemostimportantim-provementsinthesecondeditionofProbability:
TheoryandExamples.Thesolutionsarenotintendedtobeaspolishedastheproofsinthebook,butaresupposedtogiveenoughofthedetailssothatlittleislefttothereadersimag-ination.Itisinevitablethatsomeofthemanysolutionswillcontainerrors.Ifyoundmistakesorbettersolutionssendthemviae-mailtortd1cornell.eduorviaposttoRickDurrett,Dept.ofMath.,523MalottHall,CornellU.,IthacaNY14853.RickDurrettContents1LawsofLargeNumbers11.BasicDenitions12.RandomVariables33.ExpectedValue44.Independence75.WeakLawsofLargeNumbers126.Borel-CantelliLemmas157.StrongLawofLargeNumbers198.ConvergenceofRandomSeries209.LargeDeviations242CentralLimitTheorems261.TheDeMoivre-LaplaceTheorem262.WeakConvergence273.CharacteristicFunctions314.CentralLimitTheorems356.PoissonConvergence397.StableLaws438.InnitelyDivisibleDistributions459.LimittheoremsinRd463Randomwalks481.StoppingTimes484.Renewaltheory51Contentsiii4Martingales541.ConditionalExpectation542.Martingales,AlmostSureConvergence573.Examples434.DoobsInequality,LpConvergence645.UniformIntegrability,ConvergenceinL1666.BackwardsMartingales687.OptionalStoppingTheorems695MarkovChains741.DenitionsandExamples742.ExtensionsoftheMarkovProperty753.RecurrenceandTransience794.StationaryMeasures825.AsymptoticBehavior846.GeneralStateSpace886ErgodicTheorems911.DenitionsandExamples912.BirkhosErgodicTheorem933.Recurrence956.ASubadditiveErgodicTheorem967.Applications977BrownianMotion981.DenitionandConstruction982.MarkovProperty,Blumenthals0-1Law993.StoppingTimes,StrongMarkovProperty1004.MaximaandZeros1015.Martingales1026.DonskersTheorem1057.CLTsforDependentVariables1068.EmpiricalDistributions,BrownianBridge1079.LawsoftheIteratedLogarithm107ivContentsAppendix:
MeasureTheory1081.Lebesgue-StieltjesMeasures1082.CaratheodarysExtensionTheorem1093.Completion,etc.1094.Integration1095.PropertiesoftheIntegral1126.ProductMeasures,FubinisTheorem1148.Radon-NikodymTheorem1161LawsofLargeNumbers1.1.BasicDenitions1.1.(i)AandBAaredisjointwithB=A(BA)soP(A)+P(BA)=P(B)andrearranginggivesthedesiredresult.(ii)LetAn=AnA,B1=A1andforn1,Bn=Ann1m=1Am.SincetheBnaredisjointandhaveunionAwehaveusing(i)andBmAmP(A)=m=1P(Bm)m=1P(Am)(iii)LetBn=AnAn1.ThentheBnaredisjointandhavem=1Bm=A,nm=1Bm=AnsoP(A)=m=1P(Bm)=limnnm=1P(Bm)=limnP(An)(iv)AcnAcso(iii)impliesP(Acn)P(Ac).SinceP(Bc)=1P(B)itfollowsthatP(An)P(A).1.2.(i)SupposeAFiforalli.ThensinceeachFiisa-eld,AcFiforeachi.SupposeA1,A2,.isacountablesequenceofdisjointsetsthatareinFiforalli.ThensinceeachFiisa-eld,A=mAmFiforeachi.(ii)Wetaketheinteresectionofallthe-eldscontainingA.Thecollectionofallsubsetsofisa-eldsothecollectionisnotempty.1.3.ItsucestoshowthatifFisthe-eldgeneratedby(a1,b1)(an,bn),thenFcontains(i)theopensetsand(ii)allsetsoftheformA1AnwhereAiR.For(i)notethatifGisopenandxGthenthereisasetoftheform(a1,b1)(an,bn)withai,biQthatcontainsxandliesinG,soanyopensetisacountableunionofourbasicsets.For(ii)xA2,.,Anand2Chapter1LawsofLargeNumbersletG=A:
AA2AnF.SinceFisa-elditiseasytoseethatifGthenGisa-eldsoifGAthenG(A).FromthelastresultitfollowsthatifA1R,A1(a2,b2)(an,bn)F.Repeatingthelastargumentn1moretimesproves(ii).1.4.ItisclearthatifAFthenAcF.NowletAibeacountablecollectionofsets.IfAciiscountableforsomeithen(iAi)ciscountable.OntheotherhandifAiiscountableforeachitheniAiiscountable.TocheckadditivityofPnow,supposetheAiaredisjoint.IfAciiscountableforsomeithenAjiscountableforallj=isokP(Ak)=1=P(kAk).OntheotherhandifAiiscountableforeachitheniAiisandkP(Ak)=0=P(kAk).1.5.Thesetsoftheform(a1,b1)(ad,bd)whereai,biQisacountablecollectionthatgeneratesRd.1.6.IfBRthenZB=(XBA)(YBAc)F1.7.P(4)
(2)1/241e8=3.3345105Thelowerboundis15/16softheupperbound,i.e.,3.1261051.8.Theintervals(F(x),F(x),xRaredisjointandeachonethatisnonemptycontainsarationalnumber.1.9.LetF1(x)=supy:
F(y)xandnotethatF(F1(x)=xwhenFiscontinuous.Thisinversewearsahatsinceitisdierentfromtheonedenedintheproofof(1.2).ToprovetheresultnownotethatP(F(X)x)=P(XF1(x)=F(F1(x)=x1.10.Ify(g(),g()thenP(g(X)y)=P(Xg1(y)=F(g1(y).Dierentiatingwithrespecttoygivesthedesiredresult.1.11.Ifg(x)=extheng1(x)=logxandg(g1(x)=xsousingtheformulainthepreviousexercisegives
(2)1/2e(logx)2/2/x.1.12.(i)LetF(x)=P(Xx).P(X2y)=F(y)F(y)fory0.DierentiatingweseethatX2hasdensityfunction(f(y)+f(y)/2y(ii)Inthecaseofthenormalthisreducesto(2y)1/2ey/2.Section1.2RandomVariables31.2.RandomVariables2.1.LetGbethesmallest-eldcontainingX1(A).Since(X)isa-eldcontainingX1(A),wemusthaveG(X)andhenceG=XB:
BFforsomeSFA.However,ifGisa-eldthenwecanassumeFis.SinceAgeneratesS,itfollowsthatF=S.2.2.IfX1+X2xthentherearerationalnumbersriwithr1+r2xandXirisoX1+X2x=r1,r2Q:
r1+r2xX1r1X20=G,soweneedalltheopensetstomakeallthecontinuousfunctionsmeasurable.2.5.Iffisl.s.c.andxnisasequenceofpointsthatconvergetoxandhavef(xn)athenf(x)a,i.e.,x:
f(x)aisclosed.Toarguetheconversenotethatify:
f(y)aisopenforeachaRandf(x)athenitisimpos-sibletohaveasequenceofpointsxnxwithf(xn)asoliminfyxf(y)aandsinceaaisopenforeachaR.Todothiswenotethatiff(x)athenthereisan0andazwith|zx|abutthenif|yx|a.Asimilarargumentshowsthatx:
f(x)0.3.2.(3.1c)istrivialifEX=orEY=.WhenEX+andEY,wehaveE|X|,E|Y|sinceEXEYandEX+EY+.Toprove(3.1a)wecanwithoutlossofgeneralitysupposeEX,EYandalsothatEX+=(forifE|X|,E|Y|theresultfollowsfromthetheorem).Inthiscase,E(X+Y)EX+EYandE(X+Y)+EX+EY=soE(X+Y)=EX+EY.Toprove(3.1b)wenotethatitiseasytoseethatifa=0E(aX)=aEX.TocompletetheproofnowitsucestoshowthatifEY=thenE(Y+b)=,whichisobviousifb0andeasytoprovebycontradictionifb(x)forx=EXsowemusthaveX=EXa.s.3.4.Thereisalinearfunction(x)=(EX1,.,EXn)+ni=1ai(xiEXi)sothat(x)(x)forallx.Takingexpectedvaluesnowandusing(3.1c)nowgivesthedesiredresult.3.5.(i)LetP(X=a)=P(X=a)=b2/2a2,P(X=0)=1(b2/a2).Section1.3ExpectedValue5(ii)Asawehavea21(|X|a)0a.s.SincealltheserandomvariablesaresmallerthanX2,thedesiredresultfollowsfromthedominatedconvergencetheorem.3.6.(i)FirstnotethatEY=EXandvar(Y)=var(X)impliesthatEY2=EX2andsince(x)=(x+b)2isaquadraticthatE(Y)=E(X).Applying(3.4)wehaveP(Ya)E(Y)/(a+b)2=E(X)/(a+b)2=p(ii)By(i)wewanttondp,b0sothatapb(1p)=0anda2p+b2(1p)=2.Lookingattheanswerwecanguessp=2/(2+a2),pickb=2/asothatEX=0andthencheckthatEX2=2.3.7.(i)LetP(X=n)=P(X=n)=1/2n2,P(X=0)=11/n2forn1.(ii)LetP(X=1)=11/nandP(X=1+b)=1/nforn2.TohaveEX=1,var(X)=2weneed(11/n)+b(1/n)=02(11/n)+b2(1/n)=2Therstequationimplies=b/(n1).Usingthisinthesecondweget2=b21n(n1)+b21n=b2n13.8.Cauchy-SchwarzimpliesEY1(Ya)2EY2P(Ya)Thelefthandsideislargerthan(EYa)2sorearranginggivesthedesiredresult.3.9.EX2/n=n2(1/n1/(n+1)=n/(n+1)1.IfYXnforallnthenYnon(1/(n+1),1/n)butthenEYn=1n1/(n+1)=since1.3.10.Ifg=1Athisfollowsfromthedenition.Linearityofintegrationextendstheresulttosimplefunctions,andthenmonotoneconvergencegivestheresultfornonnegativefunctions.Finallybytakingpositiveandnegativepartswegettheresultforintegrablefunctions.3.11.Toseethat1A=1ni=1(11Ai)notethattheproductiszeroifandonlyifAisomei.Expandingouttheproductgives1ni=1(11Ai)=ni=11Aiij1Ai1Aj+
(1)nnj=11Aj6Chapter1LawsofLargeNumbers3.12.Therstinequalityshouldbeclear.Toprovetheseconditsucestoshow1Ani=11Aiij1Ai1AjTodothisweobservethatifisinexactlymofthesetsAithentherighthandsideismm2whichis1forallm1.Forthethirdinequalityitsucestoshow1Ani=11Aiij1Ai1Aj+ijk1Ai1Aj1AkThistimeifisinexactlymofthesetsAithentherighthandsideismm(m1)2+m(m1)(m2)6Wewanttoshowthistobe1whenm1.Whenm5thethirdtermisthesecondandthisistrue.Computingthevaluewhenm=1,2,3,4gives1,1,1,2andcompletestheproof.3.13.If0jkthen|x|j1+|x|ksoE|X|kimpliesE|X|j.Toprovetheinequalitynotethat(x)=|x|k/jisconvexandapply(3.2)to|X|j.3.14.Jensensinequalityimplies(EX)E(X)sothedesiredresultfollowsbynotingE(X)=nm=1p(m)ymand(EX)=expnm=1p(m)logym=nm=1yp(m)m3.15.LetYn=Xn+X1.ThenYn0andYnX+X1sothemonotoneconvergencetheoremimpliesE(Xn+X1)E(X+X1).Using(3.1a)nowitfollowsthatEXn+EX1EX+EX1.TheassumptionthatEX1y)1andconvergesto0a.s.asysotherstresultfollowsfromtheboundedconvergencetheorem.Toprovethesecondresult,weuseourrstobservationtoseethatif0yy)P(0Xy)P(0X)Section1.4Independence7andthedesiredresultfollowssinceisarbitrary.3.17.LetYN=Nn=0Xn.Usingthemonotoneconvergencetheorem,thelin-earityofexpectation,andthedenitionoftheinnitesumofasequenceofnonnegativenumbersEn=0Xn=ElimNYN=limNEYN=limNNn=0EXn=n=0EXn3.18.LetYn=|X|1An.Jensensinequalityandthepreviousexerciseimplyn=0|E(X;An)|n=0EYn=En=0YnE|X|LetBn=nm=0Am,andXn=X1Bn.Asn,X1BnX1AandE|X|1.IfI1,.,ndoesnotcontain1itisclearthatP(iIBi)=iIP(Bi).Supposenowthat1IandletJ=I1.SubtractingP(iIAi)=iIP(Ai)fromP(iJAi)=iJP(Ai)givesP(Ac1iJAi)=P(Ac1)iJP(Ai).(ii)Iterating(i)weseethatifBiAi,AcithenB1,.,Bnareindependent.ThusifCiAi,Aci,P(ni=1Ci)=ni=1P(Ci).ThelastequalityholdstriviallyifsomeCi=,sonoting1Ai,Ai,Aci,thedesiredresultfollows.4.4.Letcm=g(xm)dxm.Ifsomecm=0thengm=0andhencef=0a.e.,acontradiction.Integratingoverthewholespacewehave1=nm=1cmsoeachcm.Letfm(x)=gm(x)/cmandFm(y)=yfm(x)dxforx.Integratingoverx:
xmym,1mnwehaveP(Xmym,1mn)=nm=1Fm(ym)Takingyk
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- Durrent 习题 答案