信号检测与估计答案

Problem4.31,0y1()23(y+1),0y1()4py1y()0,pyotherwise0()40,pyotherwisea.FindtheBaysdecisionandminimumBaysriskfortestingH1versusH0withequalaprioriprobabilities,00110110==0and==1CCCC，versusH0withequalaprioriprobabilities,bFindtheminimaxdecisionandminimaxriskforthesamecostsb.Findtheminimaxdecisionandminimaxriskforthesamecostsasinpart(a).c.FindtheNeyman-Pearsondecisionandtheforafalse-alarmprobabilityofPd011UESTC-何子述等probabilityof01fProblem4.3Solution:Thelikelihoodratioforthisproblemis1()1()=,013()pyLyypy223()4py（y+1）(a)Forthiscase,thethresholdisone.WechooseH1if21()=134Ly（y+1）or213y4Thus,thedecisionregionsare:10R:01/3andR:1/31yyTheBayes’riskis:10R:01/3andR:1/31yy1/3121312UESTC-何子述等201/3131(1)0.4519242rydydyProblem4.3(b)Forthispartwewanttofindathresholdsuchthatthetwo(b)Forthispart,wewanttofindathresholdsuchthatthetwoconditionalrisksareequal,i.e.,001203(1)=4yyydydySolvingthis,wearriveatacubicequationwhichmustsolve,namely300740yy00yyTwoofthesolutionsareimaginary,andthusunacceptable.Ifwemodelthecubicequationas30b3UESTC-何子述等30xaxbProblem4.3therealrootcanbewrittenas=xABxABwhere1/3232427bbaAand1/323B-2427bba24272427resultingin0=0.54793yTheaverageriskisr=0.45207(c)Inthiscase,wemustfindathresholdsuchthat0203y1,4yfdyor300340fyy4UESTC-何子述等Problem4.3Thesolutionis0=yABThesolutionis0yAB1/32241ffAwhere1/32241Band241ffA241ffBandThbbilitfdttiiTheprobabilityofdetectionis0yPdyy00dPdyy5UESTC-何子述等Problem4.6RepeatProblem4.3forthehypothesiswheresisafixedrealpositivenumberandnhasapdf:1:y=s+n,H0:y=s+nHwheresisafixedrealpositivenumberandnhasapdf:21()(1)pnn(1)nSolution:111211pyHys0211pyHys21()ys6UESTC-何子述等21()()1()ysLyysProblem4.6()Withliibbilitidiftk(a)Withequalaprioriprobabilitiesanduniformcosts,wemakeD1decisionifAssispositive10LyysDAssispositive10DyTheBayes’riskisthen1011111tan0D1022011111tan2[1()]2[1()]2srdydyysys7UESTC-何子述等Problem4.6(b)Accordingtothegivencost,wehave0110PP002211[1()][1()]yydydyysys[()][()]yy00y(c)Theprobabilityoffalsealarmisdeterminedbypickingathresholdsuchthat0y021[1()]fydyys0y8UESTC-何子述等0[1()]yysProblem4.60tan()2fysThedetectionprobabilityis0()2fy021[()1]dyPdyysWhichresultsin0[()1]yys111tan[tan()2]22dfPs229UESTC-何子述等Problem4.9Supposeyisarandomvariablewhich,underhypothesisH0,hasapdf:1AndunderhypothesisH1hasapdf2/201()-2ypyeyAndunderhypothesisH1,hasapdf11/5,0y5()0pyotherwise(a).FindtheBaysdecisionruleandtheminimumBaysriskforriskfor30,otherwisetestingH1versusH0withuniformcostsandaprioriprobability(b).Findtheminimaxdecisionandminimaxriskfortheuniformcosts.034()(c).FindtheNeyman-Pearsondecisionandthecorresponding10UESTC-何子述等detectionprobabilityforafalsealarmprobability01fProblem4.9Solution:(a)Thelikelihoodratiois2/22()05ypyey10(),05()5()0,y0ory5pyeyLypyUniformcostsimplythatC00=C11=0andC01=C10=1.ItfollowsthattheBayes’thresholdis101332ln(15/2)2ln(15/2)DBLyyoryTheregionforchoosingis15y189R11H,R11UESTC-何子述等15y1.89RProblem4.9Inthiscase,theBayes’riskisFurthermore,100011rPPFurthermore,25/2101.8911511.89()()0.02922222yPedyerferf1891Also,1.8901010.3785PdySotheBayes’riskisr=0.11612UESTC-何子述等10fpProblem4.9(b)WewanttofindathresholdsuchthatP10=P01.Itfollowsthat5ywewantSolvingtheseintegrals,weareleftwiththetranscendentalfunction0051001010()()yyPpydypydyPSolvingtheseintegrals,weareleftwiththetranscendentalfunction000.50.5()0.22yerfy=091andusingthisvaluetheactualriskis=08120yPP0.91andusingthisvalue,theactualriskis0.81.(c)Ifwewant,itfollowsthatwewantathreshold10fP0y0y100011PPsuchthatTheprobabilityofdetectionis0510001()d=0.2-(),or2(12).22ffyypyyerfyerf13UESTC-何子述等Theprobabilityofdetectionis051111()d10.22(12).fyPpyyerfProblem5.1Wecollectkmeasurements,eachoneindependentandidenticallydistributed.UnderhypothesisH0,eachmeasurementisarandomvariablewhosepdfis,i=1,…..k.whichhaszeromeanandvariance.UnderH1thepdfis0()ipx1()ipx22i=1,…..k,andhasmeanandvariance.(a).UsethelimittheoremandtheNeyman-Pearsoncriteriontoshowthatthe02()yasymptoticthresholdrequiredtoachieveafalse-almarmprobabilityis21()kffaskincreases.(b).Usingthethresholdfoundinpart(a),showthattheasymptoticprobabilityof212(2)fkerfc()gp()yppydetectionis11((2))dfkperfcerfc14UESTC-何子述等((2))22dfperfcerfcProblem5.1kSolution:(a)Ifweletusingthecentrallimittheorem,as，1kiiyxkusingthecentrallimittheorem,as，dk20221()exp()22kkypykkand2kk2122()1()exp()22kkykpykk22kk12212Dkkpykyk0120exp2kkkDpyypyk15UESTC-何子述等00()dfkkypyyProblem5.1Usingthechangeofvariableinsidetheintegral,weareleftwith2/2ktykareleftwith22011exp()()fytdterfcandfromthisitfollowsthatthethresholdis202/2p()()22fykfk120(2)2fyerfck(b)Theprobabilityofdetectionisdefinedas212()1()exp()kdkkkykPpydydy(b)Theproba