investmenttool-probability

InvestmentTools–ProbabilitySASFCFAQuant.Review2ProbabilityArandomvariableisaquantitywhoseoutcomeisuncertain.TwodefiningpropertiesofProbability.1.ProbabilityofanyeventEisanumberbetween0and1,p(E).2.Sumoftheprobabilitiesofanylistofmutuallyexclusiveandexhaustiveeventsequals1.•Mutuallyexclusive=oneandonlyoneeventcanoccuratanytime.•Exhaustive=oneoftheeventsmustoccur,jointlycoverallpossibleoutcomes.Empiricalprobability-probabilityofaneventoccurringisestimatedfromdata,usuallyintheformofarelativefrequency.Aprioriprobability-probabilityofaneventisdeducedbyreasoningaboutthestructureoftheproblemitself.Subjectiveprobability-probabilityofaneventisbasedonapersonalassessmentwithoutreferencetoanyparticulardata.3VisualizingSampleSpace1.ListingS={Head,Tail}2.ContingencyTable3.DecisionTreeDiagram42ndCoin1stCoinHeadTailTotalHeadHHHTHH,HTTailTHTTTH,TTTotalHH,THHT,TTSContingencyTableExperiment:Toss2Coins.NoteFaces.S={HH,HT,TH,TT}SampleSpaceOutcome(Count,Total%ShownUsually)SimpleEvent(Headon1stCoin)5TreeDiagramOutcomeS={HH,HT,TH,TT}SampleSpaceExperiment:Toss2Coins.NoteFaces.THTHTHHHTTHTTH6123456RowTotal11/361/361/361/361/361/361/621/361/361/361/361/361/361/631/361/361/361/361/361/361/641/361/361/361/361/361/361/651/361/361/361/361/361/361/661/361/361/361/361/361/361/6Col1/61/61/61/61/61/61SecondDieFirstDieProbabilitiesforTwoDice7FormingCompoundEvents1.IntersectionOutcomesinBothEventsAandB‘AND’Statement·Symbol(i.e.,AB)2.UnionOutcomesinEitherEventsAorBorBoth‘OR’Statement·Symbol(i.e.,AB)8Answer:Countthecombinationsthatyield7andaddtheindividualprobabilitiestofindtheprobabilityoftheevent,A,“Rollingaseven”P(A)=P(AB1)+P(AB2)+P(AB3)+…+P(AB6)=6/36=1/6123456611/361/361/361/361/361/3621/361/361/361/361/361/3631/361/361/361/361/361/3641/361/361/361/361/361/3651/361/361/361/361/361/3661/361/361/361/361/361/36SecondDieFirstDieJointProbabilitiesforTwoDiceWhatistheProbabilityofthrowingaSevenwiththetwoDice?9AdditionRule1.UsedtoGetCompoundProbabilitiesforUnionofEvents:P(AORB)=P(AB)=P(A)+P(B)-P(AB)2.ForMutuallyExclusiveEvents:P(AORB)=P(AB)=P(A)+P(B)10Answer:Countthecombinationsthathaveonediewith1andaddtheindividualprobabilitiesORusetheAdditionRuleforprobabilities.P(AB)=P(A)+P(B)-P(AB)=1/6+1/6-1/36=11/3612345611/361/361/361/361/361/3621/361/361/361/361/361/3631/361/361/361/361/361/3641/361/361/361/361/361/3651/361/361/361/361/361/3661/361/361/361/361/361/36SecondDieFirstDieAdditionRuleforProbabilitiesWhatistheProbabilityofthrowingatleastonedieshowing1withthetwoDice?11Answer:Countthecombinationsthathaveonediewith4orgreater&addtheindividualprobabilitiesORusetheAdditionRuleforprobabilities.P(AB)=P(A)+P(B)-P(AB)=1/2+1/2-1/4=3/412345611/361/361/361/361/361/3621/361/361/361/361/361/3631/361/361/361/361/361/3641/361/361/361/361/361/3651/361/361/361/361/361/3661/361/361/361/361/361/36SecondDieFirstDieAdditionRule-AgainWhatistheProbabilityofthrowingafourorhigheroneitherthefirstorseconddie?12ConditionalProbabilityConditionalProbability:ProbabilitythatoneEventwilloccurGiventhatAnotherEventhasalreadyoccurred.TocalculateaConditionalprobability–ReviseOriginalSampleSpacetoAccountforNewInformationEliminatesCertainOutcomesResult:P(A|B)=P(AandB)P(B)1312345611/361/361/361/361/361/3621/361/361/361/361/361/3631/361/361/361/361/361/3641/361/361/361/361/361/361/651/361/361/361/361/361/3661/361/361/361/361/361/36SecondDieFirstDieConditionalProbabilityWhatistheProbabilityofthrowingatleastan8ifthefirstdieisa4?Answer:Noteweknowweareintherowassociatedwitha4forthefirstdie.Thereareonly3waysofthepossiblesixoutcomesfortheseconddiethatyieldasum8.HenceP(A|B)==P(AB)/P(B)=(3/36)/(6/36)=.5141.EventOccurrenceDoesNotAffectProbabilityofAnotherEventTossfirstdie,whatdoesitsoutcometellyouaboutthelikelyresultsintossingtheseconddie?Answer:Nothing.Thetwodieareindependent2.CausalityNotImplied3.TestsForStatisticalIndependenceP(A|B)=P(A)P(AandB)=P(A)*P(B)StatisticalIndependence15MultiplicationRule1.UsedtoGetCompoundProbabilitiesforIntersectionofEventsCalledJointEvents2.P(AandB)=P(AB)=P(A)*P(B|A)=P(B)*P(A|B)3.ForIndependentEvents:P(AandB)=P(AB)=P(A)*P(B)16Answer:Theresultonthefirstdieisindependentoftheseconddie.WeuseThemultiplicationruletofindprob.ofAandBHenceP(AB)==P(A)*P(B)=(1/6)*(1/6)=(1/36)12345611/361/361/361/361/361/3621/361/361/361/361/361/3631/361/361/361/361/361/3641/361/361/361/361/361/3651/361/361/361/361/361/3661/361/361/361/361/361/36SecondDieFirstDieMultiplicationRulewithIndependenceWhatistheProbabilityofthrowing“snakeeyes”i.e.a1oneachdie?17DiscreteR.V.’s-SummaryMeasuresExpectedValueMeanofProbabilityDistributionWeightedAverageofAllPossibleValuesX=E(X)=XiP(Xi)=p(X1)X1+p(X2)X2+…+p(Xn)XnVarianceWeightedAverageSquaredDeviationaboutMeanX2=E[(Xi-X(Xi-XP(Xi)=p(X1)(X1-X)2+p(X2)(X2-X)2+…+p(Xn))(Xn-X)2StandarddeviationSquarerootofthevariance.Measureofriskshowsdispersionofpossibleoutcomesaroundexpectedlevelofoutcomes.18Mean/VarianceCalculationTableXiP(Xi)XiP(Xi)Xi-(Xi-)2(Xi-)2P(Xi)TotalXiP(Xi)(Xi-)2P(Xi)19AssetExpectedReturn&RiskCalculatetheExpe