FRM一級強(qiáng)化段定量分析（標(biāo)準(zhǔn)版）

AnalysisFRM一級培訓(xùn)講義-強(qiáng)化班FrameworkFundamentalsRandomandstatisticsDistributionHypothesisLinearTimeMeasuringandSimulationandBootstrapping2-150Fundamentals3-150AxiomsofProbabilityAnyeventAintheeventspacehasPA≥0PrA≥0).1.IfeventsAAPA∪A=PA+PA.121212Extensions購課后務(wù)必加唯一售后微：xuebajun888sTheprobabilityofaneventoritsmustbe1CCPA∪A=PA+PA=1Thetheunionanytwosetsbedecomposedinto:PA∪B=PA+PB?PA∩B4-150ConditionalUnconditionalProbability(MarginalProbability)Thewithoutany(orlackinganypriorinformation),commonlyknownP(A).ConditionalProbabilityTheonconditionthatanotheroccursfirst.TheconditionalB,conditionalonA,byP(A∩B)PBA=;PA>0P(A)Jointprobability∩B)thejointprobability,meansthethattwooccur5-150ConditionalIndependentevents(orunconditionalindependentevents):Ifthe(B)notinfluencedbywhethertheotherevent(A)occurs,thensaythoseindependent,otherwisetheydependent.PA∩B=P(A)×P(B)Conditionalindependence,independencecanholdconditionalonanotherevent(C),twoAandBconditionallyindependent:PA∩BC=PAC×PBCthattwotypesindependence—unconditionalandconditional—donotimplyeachEventscanbothunconditionallydependentandconditionallyindependent.Eventscanindependent,conditionalonanothertheymaybedependent.6-150ProbabilityFormulaIfAmustonethemutually?,?,12?3,……,??,then??=????|?+????|?+...+?????1122??…A11?∩?=?(?≠?)A2??????=Ω…?=1A37-150rulethehappeningonlyanotherhappensfirst.Bayes’ruleamethodconstructconditionalusingotherItbothasimplethedefinitionconditionalandimportanttool.??∩???????==×??????posteriorprobabilitypriorprobability8-150ruleExample:Supposeusingdatathemanagersormanagersoutperformthe70%thetime.managersoutperformtheonly50%thetime.Assumethatthemanagersoutperformingtheforanytheirperformanceyears.ABCInsurancehasfoundthatonly20%fundmanagersmanagersandthe80%managers.Anewfundtheportfoliostartedyearsandoutperformedtheyears.9-150ruleQuestions:1.Whatistheprobabilitythatthenewmanagerwasanexcellent23.Whataretheprobabilitiesthatthenewmanagerisanexcellentor.Whatistheprobabilitythatthenewmanagerwillbeatthemarketnextgiventhatthenewmanageroutperformedthemarketthelastthreeyears?TIPS:P(O):ThethatthefundoutperformtheP(E):ThethatthefundP(A):Thethatthefundaverage10-150ruleAnswer1:P(E)=20%Answer2:Theoutperformingthe70%[P(O|E)=70%].Theoutperformingthe0%[P(O|A)=50%].5Next,useBayes’thethatthenewthattheoutperformedtheyearsa???=0.73=0.343??=0.233??=???×??+???×??=0.7×0.2+0.5×0.8=0.1686???×?(?)0.343×0.2???===0.4069=40.7%??)0.168611-150ruleAnswer3:Thisquestionfindingtheunconditionalthenewoutperformingthenowuse40.7%theforthethattheand59.3%theforthethatthe??=???×??+???×??=0.7×0.407+0.5×0.593=0.581412-150Question1followingamatrixforX={1,2,3}andY={1,2,3}.X1231236%12%2%15%30%5%9%18%3%YEachthefollowingtrueexcept:A.XandYindependent.The(X,Y)C.Y=3conditionalonX=110%.TheunconditionalthatX=250%.CorrectAnswer:B13-150Randomstatistics14-150DefinitionarandomvariableArandomafunction?(i.e.,thesampleΩ)thatanumberx.classesrandomvariablesDiscreterandomvariableAssignsaadistinctsets,eitherorcontainacountablysetWhenthesetthesetmustbecountable.massfunction(PMF)Cumulativedistributionfunction(CDF):??=PX≤?Continuousrandomvariableuncountableset.densityfunction(PDF)usedPMF?(?=?)=0thoughx15-150randomtheasinglerandomincludetwoorrandom.theircounterparts,theybeorcontinuous.Probabilitymatrix(probabilitytable):Whenwiththejointtwoconvenientsummarizetheamatrix.Example:X1or2withthesameY[1,X]withthesameXY12120.500.000.250.25Theblanksjointprobabilities16-150Marginaldistributions:Thedistributionasinglecomponentarandomadistribution.WhenaPMFamatrix,thetwodistributionseithersummingcolumnsorsumming.E.g.TheblanksmarginalprobabilitiesThecolumnandthedistributionYandX.X2Y1f??120.500.0000.750.250.50f??17-150ConditionalDistributions:Theconditionaldistributionsummarizesthetheforonerandomconditionalontheothertakingaspecific.E.g.WhattheconditionaldistributionYconditiononX=2X,Y2,yf2X,Y2,y0.5Y|Xyx=2==XTheconditionaldistributionthenY1250%50%18-150IndependenceThecomponentsarandom?,?x,?=f?×f???E.g.XandYnotthef?=1×f?=1=0.375≠f1,1=0.5???,?X2Y1f??120.500.0000.750.250.50f??19-1502.ExpectationsMoment:Asstatedmomentsasetcommonlyusedthatimportantrandom.typesmomentsCentralmoment:?=?????,?≥2?()Non-centralmoment:?=??,?≥1)???(?20-1502.ExpectationconditionalexpectationsTheexpectationafunctiong(X,Y)aweightedaveragethefunctiontheg(X,Y).Theexpectationdefinedas:???,?=????,???(?,?)Example,considerthefollowingPMF:X123415%60%10%15%Yconsiderthefunction:??,?=??.Theexpectation??,????,?=????,??(?,?)?∈}?∈}3434???,?=1×0.15+1×0.60+2×0.10+2×0.15=3.9521-1502.ExpectationconditionalexpectationsAconditionalexpectationisanexpectationwhenonerandomvariableaspecificorfallsadefinedrangeE.g.InthethedistributionYconditiononX=2X,?2,?X,?2,??|??x=2==f2?0.5TheconditionaldistributionthenY1250%50%TheconditionalexpectationYthenE?X=2=1×50%+2×50%=1.522-1502.ExpectationsAcentral–thefirstmomentE(X)=?P(x)x=P(x+P(x+...+P(xnii1122nPropertiesIfbaconstant,E(b)=b.Ifaaconstant,E(aX)=aE(X).Ifaandbconstants,thenEaX+b=X+E(b).E(X2)≠[E(X)]2.EX+Y=E(X)+E(Y).InEXY≠E(X)E(Y);IfXandYindependentrandomthenEXY=E(X)E(Y).23-1502.ExpectationsThemedianthe50%quantilethedistribution.First,thedatasortedsmallestWhenthesamplesizeodd,???????=(?+1)?2Whenthesamplesize???????=1?2??2+??2+1)othercommonlyreportedquantilestheandquantiles.First,thedatasortedsmallestSecond,theunitinterval=100%/(??1)the25%quantileand75%quantilewithThesetwoquantile(?and?)usedtogether2575theinter-quartile????=???257524-150Adispersion–thesecondmomentn(Xi?μ)2∑i=122σ==EX?μxnformulathedefinitiontheusethefollowingformula:222σX=EX?[E(X)]hownoisyorthatrandomis.Thepositive??2,,knownthedeviation,?also25-150PropertiesTheaconstantIfaconstant,then:σ(aX)=aσ(X)222.Ifbaconstant,then:σ+b)=σ(X)22.Ifaandbconstant,then:σ+b)=aσ(X)222.IfXandYrandomandaandbconstants,thenσ2+bY=a2σ2X+b2σ2.IfXandYtwoindependentrandomthen:222222σX+Y=σX+σ(Y)σX?Y=σX+σ(Y)Theexpectationand222σX=EX?EX26-150Covariance????,?=?????????=?????????1?=???（???）（???）????1?=1howonerandomwithanotherrandominfinitypositive27-150PropertiesCovarianceIfXandYrandomtheirTheand222σ=σ+σ±2CovX,YX±YxyIfbandcconstant,thenCov(a+bX,cY)=Cov(a,cY)+Cov(bX,cY)=b×c×Cov(X,Y)TheXanditselftheXCovX,X=EX?EXX?EX=σ2X28-150CorrelationcoefficientCovX,Y???????[(???)(??????===??????????PropertiesCorrelationcoefficienthasnounits,ranges-1+1.thetworandomIftwoindependent,theirthecoefficientbeTheconverse,nottrue.ForY=X2.222?=?+?±2????±???????29-150AasymmetryaPDF–themoment.3thirdcentralmomentcubeofdeviation??????==3??andnonsymmetricaldistributionsandNegative-skewedMeanPositive-skewedMean==MedianMeanMode<<havingafattailMode>>havingaleftfattail30-150KurtosisKurtosisAtallnessorflatnessaPDF–thefourthmoment.fourthcentralmoment????22squareofcentralmoment4??????==?Foranormaldistribution,theK3.kurtosis=kurtosis–3LeptokurticMesokurticPlatykurticKurtosis>3>0=3=0<3<0kurtosis(assumingsamefattailnormalthintail31-150Let?=?+??,aandbconstants:TheY??=?+???TheY??=?2?[?]????22Ifb>0,thekurtosisYthekurtosisX.購課后務(wù)必加唯一售后微：xuebajun888sIf?=?+ahasthesamemagnitudetheoppositesign.32-150TheAsampleestimatorusedthepopulation,thebestestimatorshouldbeunbiased,andlinear(BLUE).Linear:estimatorafunctionthesampledata.Unbiased:thetheestimatorshouldequalthepopulation.Best:Iftheestimatorhastheminimumamongtheotherunbiasedestimators.ThemeanestimatortheBestLinearUnbiasedEstimatorthepopulationthedata.33-150Question1Athelosses(L)assumesthefollowingdistribution:losswith20%;onelosswith30%;twolosseswith30%;losseswith10%;andfourlosseswith10%.Whatthenumberlossandthedeviationthenumberlossevents?A.E(L)=1.2andσ=1.44B.E(L)=1.6andσ=1.20C.E(L)=1.8andσ=2.33D.E(L)=2.2andσ=9.60CorrectAnswer:B34-150Question2Ananalystconcernedwiththesymmetryandadistributionaperiodtimeforasheexamining.Shedoessomecalculationsandfindsthatthemedian4.2%,themean3.7%,andthemode4.8%.Shealsofindsthatthekurtosis2.Basedonthisinformation,thecharacterizationthedistributiontimeA.C.NegativeNegativeKurtosisLeptokurticPlatykurticPlatykurticLeptokurticCorrectAnswer:C35-150Question3Whichonethefollowingstatementsaboutthecoefficientfalse?A.Italways-1+1.B.AcoefficientmeansthattworandomC.ItatworandomD.ItscalingthetworandomCorrectAnswer:B36-150Distribution37-150DistributionBernoulliDistributionP(X=1)=pP(X=0)=1–pBinomialDistributionThexsuccessesntrails?!???????1???????=??=?=??1??=??!???!ExpectationsandvariancesExpectationBernoullirandomBinomialrandomp–p)npnp(1–p)38-150DistributionDistributionWhenanumbertrialsbutasmallsuccess,BinomialcalculationsIfsubstitute?t/?for?,andnverytheBinomialDistributiontheDistribution.??????=??=?=??=???!Xthenumbersuccessunit.λthetherandomevents;i.e.,ushowmanyoccuronunittime.Meanλλ39-150DistributionNormalDistributionAsnthebinomialdistributionNormalDistribution.Thenormalcurvesymmetrical.The11????22?????=curve2??curve∞.+∞.Themean,andmodeequal.PropertiesX~N(μ,σ2),describeditstwoμandσ2.Bell-shaped,symmetricaldistribution:=0;kurtosis=3.Alinearcombination(function)two(orindependentnormallydistributionrandomnormallydistributed.Thetailsgetthinandgobutasymptotic.40-150DistributionNormaldistributioninpractice①Sumsnormallydistributedrandomalsonormally②③④andZ-tableApplicationquantilesnormaldistributionrandomnormalrandom41-150Distribution①SumsindependentnormallydistributedrandomvariablesarealsonormallydistributedIfX~?(?,?)Y~??,?2)andtheyindependent,then2,11222222aX+bY~?(??+??,??+??)1212Whenportfolio,aandbusuallyassetExample:A$50prudentfund(PF)witha$200fund(AF).ThePF~N(0.03,0.072)andtheAF~N(0.07,0.152Assumingtheindependent,whatthedistributiontheportfolioAnswer:?~?(0.062,0.1208)2?42-150Distribution②StandardizationandZ-tableApplication?~?μ,??then????=～?0,1?CheckZ-tablefindCDF???Φ?Example:If?~?(70,9),computetheX≤64.12.Solution:?=(???)/?=(64.12?70)/3=?1.96?(?≤?1.96)=0.0250?(?≤64.12)=0.025043-150Distribution③quantilesinnormaldistribution68%observationsfalltheinterval?±?90%observationsfalltheinterval?±1.65?95%observationsfalltheinterval?±1.96?99%observationsfalltheinterval?±2.58?σσμ+1.65σ+2.58σ-1.96σ+1.96σ90%95%99%44-150Distribution④randomvariablesnormalrandomvariableAbinomialrandomanormalrandom?(??,??(1??)both??≥10?(1??)≥10.n=20xxxp=0.1p=0.3p=0.5TherandomanormalrandomN(λ，λ)λcommonly?≥1000.45-150tdistributiontdistributionItsimilarthenormal,exhibitsslightlytails.Itsymmetrical.????????=~??1IthasAsthethet-distributionwiththenormaldistribution.Boththenormalandtdistributionthesamplingdistributionthesamplemean.Thethatthenormalusedknowthepopulation.ThetusedmustonthesampleIndon’tknowthepopulationsothetdistributiontypically46-150DistributionLognormalDistributionIflnXnormal,thenXlognormal;alognormal,itsnaturallognormal.ItusefulforassetRightChi-Square??)DistributionF-Distribution47-150LawLargeNumbers(LLN)TheLawLargeNumbersstatesthat?asequencerandomwith??≡????then:?.s.1n?=∑Xμni=1in?.?.ThesymbolconvergesalmostThe?n(formally,n→∞)thetruepopulationμ.48-150TheCentralLimitTheoremIfX,X,…,X,ai.i.d.randomsampleanypopulationregardless12nanydistributions)witha?and?2,thenthe?n,normallywith?2?andthesamplesize??2?d?=?→??,nor????d→?0,1???49-150Question11.Onamultiplechoiceexamwithfourchoicesforsixquestions,whatthethatastudentgetslessthantwoquestionssimplyguessing?A.0.46%B.23.73%C.35.60%D.53.39%CorrectAnswer：Dp(X=0)=(3/4)6p(X=1)=6(1/4)(3/4)5Thegettinglessthantwoquestionsp(X=0)p(X=1)=53.39%.=17.80%=35.59%+50-150Question22.AtwophoneThethatthey208-hourclosestA.5.59%B.16.56%C.3.66%D.6.40%CorrectAnswer：Asolvethisquestion,firstthatthenumberphone8-hour16.Usingthedistribution,solveforthethatXbe20.20?1616???=20)==5.59%20!51-150Question33.WhichthefollowingstatementaboutthenormaldistributionnotA.KurtosisB.one.C.Thedistributionbebytwomoments,andD.Thenormaldensityfunctionhasthefollowing112(???)2]??=?[?2?2??2CorrectAnswer：B52-150Hypothesis53-1501.Inference?Looselyspeaking,isthestudyoftherelationshipbetweenapopulationaSamplingandEstimation購課后務(wù)必加唯一售后微：xuebajun888spopulationsamplingsamplepopulationparameterestimationsamplestatistic54-150andConfidenceIntervalEstimateEstimationUsingasinglethethepopulation.parameterS2→?2?→?estimatorConfidenceIntervalEstimatesignificanceconfidence–??????????????????=?????????????±?????????????????×?????????????55-150IntervalThepopulationhasanormaldistributionwithaknownvariance.Confidenceinterval:??±??2?Thepopulationhasanormaldistributionwithaunknownvariance.Confidenceinterval:??±t??2?Whensamplinga:SmallSample(N<30)Sample(N≥30)NormalNormaloror56-150IntervalExample:Assumetheratiofollowsnormaldistribution,calculatethe95%confidenceintervaltheratio.Pricetoearning(P/E)ratiosof28companiesontheNewstockexchangeCompanyP/ECompanyINTCIBMP/E27.9622.908.3036.0222.9412.1022.4322.1316.4833.7526.0512.2124.4914.8714.8727.8437.10TCJPMJNJ49.7824.8814.5526.2228.2134.7112.9921.899.86MCDMRKMMMMO20.2623.36WMTDISMean=23.25=90.13Standarddeviation=9.4957-150123nullandhypothesesasignificanceIdentifythestatisticDonotadecisionadecisionrule5458-1501:null)andhypotheses)0aH:?=18.5,H:?≠18.50?A?One-tailedvs.One-tailed0:μ=0a:μ≠00:μ≥00:μ≤0a:μ<0a:μ>02:thestatisticandthedistributionstatistic=(samplestatistic–hypothesizedthesamplestatistic)??????23.25?18.5?===2.65~??1?9.49?2859-1503:ChoosethesignificanceαandobtainsignificanceCriticalThedistributionstatistic(z,t,χ2,t(n-1),χ2(n-1),-1,n-1)12One-tailedortwo-tailedFortheAlpha=5%t??1=2.052?260-150Formulateadecisionrule0|teststatistic|>0|teststatistic|<regionliesonsamesideHA2.5%2.5%5%95%95%-2.0522.052RejectH01.645RejectH0RejectH0FailtoRejectH0FailtoRejectH05:Arriveatdecisionsay“accept”0.theconclusion:μ(not)significantlyμ0.61-150SingleH:μ=μ0z-testvs.t-testNormalpopulation,n<30z-testn≥30Knownpopulation(σ2)z-testUnknownpopulationt-testt-testorz-testz-statistic???0?????==t-statistic???0???62-150theH:μ=μY0XwhetherthetwoIfthenullhypothesistrue,then:EZ=EX?EY=μ?μ=0.iiiXYThestatisticconstructed:?Z?X??YT==22?+??2?2??n??XYZnWhenXandYbothandmutuallyindependent,thestatisticiifortestingthatthemeansequal:?X??YT=22?n?nXY+XY63-150theExample:AssumetwodistributionsthattherainfallCityXandCityY,.Their,,samplesizeand22?=10CM;?=14CM;?=4;?=6,n=n=12,?=0.3XYXYXYXYQuestion:WhethertherainfallCityXsignificantlythatCityY?H:μ=μYThethestatistic0X?X??Y10?14T===?5.2122?+??2?4+6?2×0.3×4×6XY??n12,thenullwouldbea99.99%confidence.64-150H:?=???0?The??1)?2?2=??=??12?0?=samplesize?2=samplevariance2?=hypothesizedvalueforthepopulationvariance065-150H:?=???0??TheF-test2?1?==??1=??111222?2Alwaysputthethe?>?221266-150SummarytypeAssumptionsH0distributionNormallypopulation,population???0????=0?=?=?(0,1)Normallypopulation,unknownpopulation???0????=0???1)???1)Mean?ZT==??n2theequalityofmeansZ1=2?X??Y22?+??2?XY??n??1?2Normallypopulation?2=02?2=?2??1)2?0independentnormallypopulations2222?=??=????(??1,??1)12121267-150P①PApproachPthesmallestsignificanceforwhichthenullhypothesisourP/E???????23.25?18.5?===2.65~??(27)→?≈0.0159.49?282TheRuleP0thelessthanthesignificancethehypothesistest.Donot0thethanthesignificance68-150P2.5%2.5%theteststatisticCriticalforPositivetheteststatistic<alpha,nullhypothesis.69-150IIIErrorsIthenullhypothesisactuallytrue.IIthenullhypothesisactuallyfalse.Significance(α)ThemakingaIerror:Significance=IaThethenullhypothesisfalse:a=1–II70-150Question1AnalystJohnconcernedthatthedayssalesoutstandinghasits27days.asample36heasampleDSO29dayswithsampledeviation7.one-sidedhypothesis,statedwith95%confidence,thatDSOthan27.Doesshethenull?A.No,notone-sidednullthet-statisticlessthan1.65.B.No,notone-sidednullthet-statisticlessthan1.96.C.one-sidednullthet-statisticthan1.65.D.one-sidednullthet-statisticthan1.96.Correctanswer：C71-150Question2IftheP/E30stocksacertainindustrialis18andthesampledeviation3.5,theclosestA.0.12B.0.34C.0.64D.1.56Correctanswer：C72-150Question3Assumerandomnormallywith10anda25.WithoutusingathethatXfallswithin1.75and21.65?A.68%B.94%C.95%D.99%CorrectAnswer:B73-150Question4Assumethepopulationfundhasunknowndistributionwith8%and10%.asample410funds,whatthethesample0.6%?A.5%B.8%C.10%D.12%:A74-150Question5WhichthefollowingstatementshypothesistestingA.IIthetheaactuallyfalse.B.Hypothesistestingusedabouttheapopulationonthestatisticsforasamplethatanotherpopulation.C.AllelsethethechancemakingatypeIthecostthemakingatypeIID.Ifthethanthesignificancethenthestatisticstheintervals.Correctanswer：C75-15076-150InterpretationregressionfunctionAnslopecoefficient1wouldthatthechangeb1unitsforevery1unitchangetheindependentinterceptterm0canbeinterpretedtomeanthatthe?0.erroristheportionthatcan’tbeexplainedbyindependentThisshockisassumedtohavemean0sothatE[Y]=E[b+bX+01ε]=bE[X]01Slopecoefficient?=b+??+?01???coefficientErrorterm77-150OLSanumberassumptions.Mostthemajorassumptionspertainthe(i.e.,AlinearXandIndependentnotrandom,independentnotindependentnotmultipleThethe(i.e.,??=).Thevariancetheconstant(i.e.,thehomoskedastic).Theobservations(i.e.,theThenormallyTherearefiveassumptionsforthelinearregressionmodel:The??,haveameanconditionaltheXiConstantShocksindependentthan0(σ>).2XThesampleobservationsi.i.d.randomdrawsthepopulationoutliersexplanatoryvariablesarenotperfectlylinearlydependent.78-150OrdinaryLeastOrdinaryleastsquares(OLS)TheOLSestimationtheestimatingthepopulationbusingthebtheii(i.e.,theTheOLSsamplecoefficientsthosethat:minimize??2=?[???+?×?]2????011TheOLSsamplecoefficientsthosethat:???(?,?)?2?)????=1=xy???=????0179-150Theconfidenceintervalforthecoefficient,1andb0??±(CriticalValue×?)??11??±(?????????????×?)??00Thethecoefficient.1??=????1??2(??2+??2)???=??20????80-150CoefficientHypothesisAt-testalsobeusedthehypothesisthatthetrueslopeb1,equalsomehypothesizedk(usuallyk=0),theis:nullhypothesisandhypothesis?0:?=0?:?≠0statistic?11????11T=???1Thestatisticalsotransformeda???????=2(1?Φ?)0Torlessthansize.81-150Therelationshipeachcomponent222??????=????+????????YSum????=?+??1?????→??0??Sum???→????→????YExplainedSumb0X82-150Thereportingmatrixcomponent:AnalysisdfkSSMSSESS/kExplainedESSRSSTSSn-k-1n–1RSS/(n-k-1)-83-150JointHypothesisTheisnotapplicabletestingcomplexhypothesesthatthanone,becausetheestimatorsAnusedwhetherleastoneslopecoefficientsignificantlyH:b=b=b=…=b=0;H:leastoneb≠0=1k)0123k1j???????????1?=TheassessestheeffectivenessthemodelawholeexplainingtheDecisionH0,F(test-statistic)>Fc(critical84-150DatacollegestudentsHeightHeight85-150SummaryoutputsRRRANOVA2FaeFtPHe86-150MeasuringR2CoefficientDetermination)Aintuitivemeasurethefit”theItathetheItslimits0≤R2≤1.?????2==1?thatasimpletwo-variableregression,the2thecoefficient(r)XandY.ii222?=?→?=±?87-150MeasuringThelimitationswithR2AddingathealwaystheR2TheR2amodifiedversiontheR2thatdoesnotnecessarilywitha??????1?djusted?2=1???????1??1???????1??1=1?=1???2)????1AdjustedR2≤R2;R2lessthancannotwithR2nosuchthinga"good"for288-150MeasuringCalculatingR2andR2Theanalystrunsamonthlyontheindependentwithin60months.Thetotalsum570,thesum180.R2andR2.AssumingthattheanalystnowfourindependentfortheR270.0%.Identifythethatanalystsmostuse.Answer:570?180?2==68.42%57060?121??2=65.5%?=1??60?5?160?121??′2=64.6%′?=1??60?9?189-150theConditionalHeteroskedasticityThethe(i.e.,conditionalon)theindependentConditionaldoessignificantforstatisticalYHighresidualvarianceLowresidualvariance0XEffectHeterosonAnalysisTheusuallyThecoefficient(the)affected.Ifthesmall,butthecoefficientthemselvesnotaffected,thet-statisticswillandthenullhypothesisnostatisticalsignificanceHow90-150Multi-collinearity(Multiple-regressiononly)Multicollinearitythesituationthattwoorindependenthighlycorrelatedwithmethodsmulti-collinearity1.T-teststhatnonetheindividualcoefficientssignificantlythanthesignificanceandtheR2high.23.Theabsolutethesampleanytwothan0.7(i.e.,|r|>0.7)..CommonsenseMethodsmulti-collinearityoneorthe91-150BiasForarise,thingsmustbetrue:leastonetheincludedmustbewiththemustbethetheImplicationsBiasawhetherthesamplesizeorsmall.Whetherthisorsmallonthe?,?theandtheThelargercorrelation,largerisbias.CorrectionsBiasAddconstructamultiple92-150approachesappropriatecomplexity①②③M-fold93-150①General-to-specificmodelselection12.First.thewithwiththesmallestabsolutet-statistic(statisticallyinsignificant345.usingtheexplanatoryandunqualified.thementioneduntilthemodelcontainsnocoefficientsthatstatisticallyinsignificant..Commonchoicefor?1%and0.1%least2.57or3.29,94-150②M-foldcross-validationItathatperformsfittingobservationsnotusedthem-fold1234.Thefirstaset.Ifadatasethasnexplanatorythen2?possiblespecifications..Splittingthedatamsizedblocks,usingm-1blocks(trainingset)andwithdatathe.(validationset).theestimatingandcomputingforatotalmtimesusedonce.Computesumformtimesandchoosethemodelwiththesmallestsum.95-150OutliersOutlierOutliersthat,thethecoefficients.distancethesensitivitytheaasingleobservationj.It:2?ijn??∑Y?Yi=1ij=KS2?ij?Ytheobservationjandthe??estimatingn-1observations.KthenumbercoefficientstheS2thetheusingobservations.jlargerthan1thatobservationjhasalargeimpactonthe96-150Question11.Aaonanindustry(in2.012.1Industry1.90.31DegreesofFreedomSSExplained13492.64824.512117.160A.ThecoefficientbetweenXandY0.889.Theindustrycoefficientat99%interval.onindustry4%,9.7%.Theindustry21%97-150Question1Answer:Dr2=R2=92.648/117.160=79%,theindustryexplains79%thet-stat(industryindex)=1.9/0.31=6.13,sothecoefficientindustrysignificant.R=2.1%+1.9×R(industryindex)=2.1%+1.9×4%=9.7%98-150TimeSeries99-150seriesAtimeseriesasetobservationsonatimeperiods,takingtheasetmonthlyU.S.retailsalesdata.100-150seriesAtimeseriescanbedecomposeddistinctcomponents①②③ThethechangesthethetimeTheseasonalcomponent,changesthetimethetime.Thecomponent,thethedata.101-150Non-StationaryTimeSeriesCovariance-stationarytimemeans,andthatnotontime.Anytimethatnotcovariance-stationarynon-stationary.Sourcesnon-Stationaritymostpervasivetimeseries:TimetrendsSeasonalitiesUnitroots(randomwalks)102-150Non-StationaryLineartimetrendY=δ+δt+εtεt~WN0,σ2t01Thisnon-stationarybecausetheontime:??=δ+δt?01NonlinearThetimeapolynomialtimeincludinghighertime:2?Y=δ+δt+δ?+?+δ?+εt012mtLog-lineartrendConstantgrowthplausiblethanconstantmostfinancialTheseusingathatthenaturallog?a???=δ+δt+ε?01tTheconstant?:1103-150Non-StationarySeries——SeasonalitiesSeasonalitiesinducenon-stationarybehaviortimebythethethemonthorquartertheSeasonalDummyTheseasonalitysandas:?=?+??+??+?+??+??11?22???1??1???=1or0thatdummy?known??thedummyvariabletrap,occursdummyIllustratesusedummyvariables?=?+????+????+?+????+???1?2?11??=???????????????????????????????=1????????????????????,???=0???????????···???==···=???=0?????????????????????.???104-150Non-StationarySeries——RandomRandomInanon-stationarytimeoncetheandseasonalitycomponentsandstrictlythetimeshouldstaystationaryandwithAR,MAorARMA.SeasonalityLinearPolynomialLog-linearvariableARMAApplylagsARMAacontainsarandomcomponent,thencannotthe105-150Non-StationarySeries——RandomHowdoesrandomwalklooksAsimplerandomt=t?1+εtSubstitutingt?1=t?2+εt?1therandomwalk:t=t?2+εt?1+?thesubstitutionuntilobservation?Y=Y+∑εt0?=1iinitial0andeveryshockandtpermanentlyaffects1Randomwalksdispersedtime,thearandomVt=tσ2,notconstant106-150Non-StationarySeries——RandomUnitUnitrandomwalksbyaddingshort-runstationarydynamicslong-runwalk.Ausingalagpolynomial.e.g.,theAR(2)Y=1.8Y?0.8Y+εtbewrittenwithat?1t?2tpolynomial:21?1.8?+0.8?Y=?t?ThecharacteristicequationfortheAR(2)withaunitZ2?1.8Z+0.8=Z?1Z?0.8=0The1and0.8.Thentheexistenceunitroot.107-150Non-StationarySeries——RandomConsidersituationIf?arandomwalk,then:?=??1+????=???+???1??1??1Δ??=0×??1+?Theγ0thearandomwalk.forUnitDickey-Fuller(ADF)1:formtheΔ??=????1+?2:NullH:γ=0Yarandomwal