FRM一级强化段定量分析（打印版）.docx

该【FRM一级强化段定量分析（打印版） 】是由【经管专家】上传分享，文档一共【50】页，该文档可以免费在线阅读，需要了解更多关于【FRM一级强化段定量分析（打印版） 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。QuantitativeAnalysis,83а??????????Framework?FundamentalsofProbability?RandomVariablesandbasicstatistics?????DistributionHypothesisTestingLinearRegressionTimeSeriesMeasuringReturns,VolatilityandCorrelation?SimulationandBootstrapping2-150FundamentalsofProbability3-??AxiomsofProbabilityzAnyeventAintheeventspacehasPA?0???(PrA?0).,PA??A=PA?+PA?.????Extensionsplementmustbe1PA??A?=PA+PA?=1posedinto:PA??B=PA+PB?PA??B4-??UnconditionalProbability(MarginalProbability)zTheprobabilityofaneventwithoutanyrestrictions(orlackinganypriorinformation),commonlyknownasP(A).,conditionalonEventA,isgivenbyP(A??B)PBA=;PA>0P(A)?JointprobabilityzP(A??B)isthejointprobability,-??Independentevents(orunconditionalindependentevents):zIftheevent(B)isnotinfluencedbywhethertheotherevent(A)occurs,thenwesaythoseeventsareindependent,??B=P(A)×P(B)ConditionalindependencezLikeprobability,independencecanberedefinedtoholdconditionalonanotherevent(C),twoeventsAandBareconditionallyindependentif:PA??BC=PAC×PBCzNotethattwotypesofindependence—unconditionalandconditional—,-?TotalProbabilityFormulazIfaneventAmustresultinoneofthemutuallyexclusiveevents??,??,????,……,??,then?????=???????|???+???????|???+...+???????|???…A11??????=??(???)A2???(2)????=?…A3???7-’rule??Specifically,’.??????????|????????|??==×??????posteriorprobabilitypriorprobability8-’rule?Example:%%%ofallfundmanagersareexcellentmanagersandtheremaining80%-’rule?Questions:??,giventhatthenewmanageroutperformedthemarketthelastthreeyears?TIPS:P(O):(E):(A):-’rule?Answer1:P(E)=20%?Answer2:Theprobabilityofanexcellentmanageroutperformingthemarketis0%[P(O|E)=70%].Theprobabilityofanaveragemanageroutperformingthemarketis0%[P(O|A)=50%].75Next,weneedtouseBayes’heprobabilitythatthenewmanagerisexcellentgiventhatthemanageroutperformedthemarketthreeyearsinarow.?????=?=???=?=????×???+????×???=?×+?×???×?(??)×=?????====%?(?)-’rule?Answer3:,%%astheweightfortheprobabilitythatthemanagerisaverage:??=????×???+????×???=×+×=-?ThefollowingisaprobabilitymatrixforX={1,2,3}andY={1,2,3}.X1231236%12%2%15%30%5%9%18%3%YEachofthefollowingistrueexcept:(X,Y)isnon-=3conditionalonX=1is10%.=2is50%.?CorrectAnswer:B13-150RandomVariableandbasicstatistics14-??DefinitionofarandomvariablezArandomvariableisafunction?(.,aneventinthesamplespace?),,(PMF)Cumulativedistributionfunction(CDF):????=PX????(PDF)isusedinsteadofPMF?(?=??)=ur.?15-??,(probabilitytable):Whendealingwiththejointprobabilitiesoftwovariables,:[1,X]-?Marginaldistributions:,??????17-?ConditionalDistributions:=2f?,?2,yf?,?2,?|?yx=2==f?29TheconditionaldistributionisthenY1250%50%18-?Independenceponentsofabivariaterandomvariableareindependentiff?,?x,??=f???×f??????=1×f???=1=?f?,?1,1=??????19-?Moment:Asstatedpreviously,:???=????????,(??2)?=????,(??1)Non-centralmoment:????20-??Theexpectationofafunctiong(X,Y)isaprobabilityweightedaverageoftheesg(X,Y).Theexpectationisdefinedas:????,?=????,???,?(?,?)Example,considerthefollowingjointPMF:X123415%60%10%15%YNow,considerthefunction:??,?=??.Theexpectationof??,?istherefore:????,????,??(?,?)????,?=1?×+1?×+2?×+2?×==????{?,?}???{?,?}21-?,thedistributionofYconditiononX=2f?,?2,??f?,?2,??f?|???x=2==f?%50%TheconditionalexpectationofYisthenE?X=2=1×50%+2×50%=-??ExpectedValuezAmeasureofcentraltendency–thefirstmomentE(X)=?P(x)x=P(x)x+P(x)x+...+P(x)x????????PropertiesofExpectedValuezIfbisaconstant,E(b)=,E(aX)=aE(X).zIfaandbareconstants,thenEaX+b=aEX+E(b).zE(X?)?[E(X)]?.zEX+Y=E(X)+E(Y).zIngeneral,EXY?E(X)E(Y);IfXandYareindependentrandomvariables,thenEXY=E(X)E(Y).23-?Themedianmeasuresthe50%,,????????=??(???)Τ?Whenthesamplesizeiseven,????????=1Τ2(???Τ?+???Τ???)??monlyreportedquantilesarethe25%and75%,thedataaresortedfromsmallesttolargestzSecond,calculatetheunitinterval=100%/(??1)zFindthe25%quantileand75%(?????and?????)arefrequentlyusedtogethertoestimatetheinter-quartilerange:?????=???????????24-???VariancezAmeasureofdispersion–thesecondmomentσ???(X???)????=?=EX???,weusethefollowingformula:??X=EX??[E(X)]?????,??,isknownasthestandarddeviation,-,then:??(aX)=a???(X).zIfbisaconstant,then:??(X+b)=??(X).zIfaandbareconstant,then:??(aX+b)=a???(X).zIfXandYareindependentrandomvariablesandaandbareconstants,then??aX+bY=a???X+b???(Y).zIfXandYaretwoindependentrandomvariables,then:??X+Y=??X+??(Y)zTherelationshipbetweenexpectationandvariance:??X=EX??EX??X?Y=??X+??(Y)?26-?????,?=????????????=????????????1?????????=???????????1???:xuebajun888s27-?PropertiesofCovariancezIfXandYareindependentrandomvariables,:9???±?=??+??±2CovX,Y??zIfa,bandcareconstant,thenCov(a+bX,cY)=Cov(a,cY)+Cov(bX,cY)=b×c×Cov(X,Y)zThecovarianceofXanditselfisthevarianceofXCovX,X=EX?EXX?EX=??X9928-??CorrelationcoefficientCovX,Y???????????[(????)(????)]??????===?????????????PropertiesofCorrelationcoefficientzCorrelationhasnounits,rangesfrom-1to+,theircovarianceiszero,therefore,,however,,Y=:????=???+???±2??????±???????29-??SkewnesszAmeasureofasymmetryofaPDF–thethirdmoment.???????????thirdcentralmomentcubeofstandarddeviation?==?SymmetricalandnonsymmetricaldistributionszPositivelyskewed(rightskewed)andnegativelyskewed(leftskewed)Negative-skewedSymmetricPositive-skewedMeanMedianModeMean=Median=ModeModeMedianMeanzPositiveskewed:Mode<median<mean,havingarightfattailzNegativeskewed:Mode>media>mean,havingaleftfattail30-150