统计学习[The-Elements-of-Statistical-Learning]第四章习题

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2020-11-07

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TheElementofStatisticalLearning{Chapter4oxstar@SJTUJanuary6,2011Ex.4.1ShowhowtosolvethegeneralizedeigenvalueproblemmaxaTBasubjecttoaTWa=1bytransformingtoastandardeigenvalueproblem.AnswerWisthecommoncovariancematrix,andit'spositive-semidenite,sowecandeneb=W12a;a=W12b;aT=bTW12Hencethegeneralizedeigenvalueproblemmaxa(aTBa)=maxb(bTW12BW12b)subjecttoaTWa=bTW12WW12b=1Sotheproblemistransformedtoastandardeigenvalueproblem.Ex.4.2Supposewehavefeaturesx2Rp,atwo-classresponse,withclasssizesN1;N2,andthetargetcodedasN=N1;N=N2.1.ShowthattheLDAruleclassiestoclass2ifxT^1(^2^1)12^T2^1^212^T1^1^1+logN1NlogN2N;andclass1otherwise.2.ConsiderminimizationoftheleastsquarescriterionNXi=1(yi0Txi)2:Showthatthesolution^satises(N2)^+N1N2N^B=N(^2^1)(aftersimplication),where^B=(^2^1)(^2^1)T.3.Henceshowthat^Bisinthedirection(^2^1)andthus^/^1(^2^1)ThereforetheleastsquaresregressioncoecientisidenticaltotheLDAcoecient,uptoascalarmultiple.4.Showthatthisresultholdsforany(distinct)codingofthetwoclasses.15.Findthesolution^0,andhencethepredictedvalues^f=^0+^Tx.Considerthefollowingrule:classifytoclass2if^yi0andclass1otherwise.ShowthisisnotthesameastheLDAruleunlesstheclasseshaveequalnumbersofobservations.(Fisher,1936;Ripley,1996)Proof1.Considerthelog-ratioofeachclassdensity(equation4.9intextbook)logPr(G=2jX=x)Pr(G=1jX=x)=log2112(2+1)T1(21)+xT1(2+1)Whenit0,theLDArulewillclassifyxtoclass2,meanwhile,weneedtoestimatetheparametersoftheGaussiandistributionsusingourtrainingdataxT^1(^2^1)12(^2+^1)T^1(^2^1)+log(1)log(2)=12^T2^1^212^T1^1^1+logN1NlogN2N2.Let0=(;0)TandcomputethepartialdeviationoftheRSS(0),thenwehave@RSS(0)@0=2NXi=1(yi0Txi)=0(1)@RSS(0)@=2NXi=1xi(yi0Txi)=0(2)Wecanalsoderivethat0=1NNXi=1(yiTxi)//from(1)(3)NXi=1xi[T(xix)]=NXi=1xiyi1NNXj=1yj//from(2)(3)(4)=2Xk=1Xgi=kxiykN1y1+N2y2N(5)=2Xk=1Nk^kNyk(N1y1+N2y2)N//Xgi=kxi=Nk^k(6)=N1N2N(y2y1)(^2^1)(7)=N(^2^1)//y1=N=N1;y2=N=N2(8)2Wealsohave(N2)^=2Xk=1Xgi=k(xi^k)(xi^k)T(9)=2Xk=1Xgi=k(xixTi2xi^Tk+^k^Tk)//xTi^k=xi^Tk(10)=2Xk=1Xgi=kxixTiNk^k^Tk//Xgi=kxi=Nk^k(11)=NXi=1xixTi(N1^1^T1+N2^2^T2)(12)NXi=1xixT=2Xk=1Xgi=kxkP2k=1Pgi=kxkNT(13)=1N(N1^1+N2^2)(N1^1+N2^2)T(14)MeanwhileNXi=1xi[T(xix)]=NXi=1xi[(xix)T]=NXi=1xixTiNXi=1xixT(15)=h(N2)^+(N1^1^T1+N2^2^T2)1N(N1^1+N2^2)(N1^1+N2^2)Ti//from(12)(14)(16)=(N2)^+N1N2N(^2^T2^1^T2^2^T1+^1^T1)(17)=(N2)^+N1N2N^B=N(^2^1)//from(8)(18)3.^B=(^2^1)(^2^1)T(^2^1)Tisascalar,therefore^Bisinthedirection(^2^1),and^=1N2^1N(^2^1)N1N2N^B//from(18)(19)=1N2NN1N2N(^2^1)T^1(^2^1)(20)/^1(^2^1)(21)4.ByreplacingNwithN1N2N(y2y1)(from(7)and(8))andfrom(20),westillhave^=N1N2N(N2)(y2y1)(^2^1)T^1(^2^1)/^1(^2^1)5.Theboundaryconditionisyi=0,sofrom(3)wehave^0=1NNXi=1^Txi=^T^=^T^3Whentheclasseshaveequalnumbersofobservations,i.e.N1=N2=N=2^=12(^1+^2)(22)Thenwehavepredictedvalues^f=(x^)T^/(x^)T^1(^2^1)(23)/xT^1(^2^1)12(^1+^2)T^1(^2^1)//from(22)(24)WhiletheLDAruleindicatethatthelog-ratioofeachclassdensity=zeroistheboundarycondition(N1=N2so1=2)logPr(G=2jX=x)Pr(G=1jX=x)=12(2+1)T1(21)+xT1(2+1)(25)Compare(24)with(25),wecanndthattheyarethesamerule.ButwhenN16=N2,theserulesareobviouslydierent.Ex.4.3SupposewetransformtheoriginalpredictorsXto^Yvialinearregression.Indetail,let^Y=X(XTX)1XTY=X^B,whereYistheindicatorresponsematrix.Similarlyforanyinputx2Rp,wegetatransformedvector^y=^BTx2RK.ShowthatLDAusing^YisidenticaltoLDAintheoriginalspace.ProofTransformedvector^y=^BTx,sowehave^^yk=Pgi=k^yiNk=Pgi=k^BTxiNi=^BT^xk(26)^^y`=Pgi=`^yiN`=Pgi=`^BTxiNi=^BT^x`(27)^^y=PNk=1Pgi=k(^yi^^yk)(^yi^^yk)TNK(28)=PNk=1Pgi=k^BT(xi^xk)(xi^xk)T^BNK(29)=^BT^x^B(30)Substitute(26)(27)(30)forgiandinequation(4.9)logPr(G=kj^Y=^y)Pr(G=`j^Y=^y)=logk`12(^xk+^x`)T^B(^BT^x^B)1^BT(^xk^x`)+xT^B(^BT^x^B)1^BT(^xk^x`)Infact,if^B(^BT^x^B)1^BT(^xk^x`)=^1x(^xk^x`),LDAusing^YisidenticaltoLDAintheoriginalspace.Sowewillproveitbellow.Yisaindicatorresponsematrix,thereforeNk^xk=Xgi=kxi=XTyk(31)^x=1NKNXi=1xixTiKXk=1Nk^xk(^xk)T//from(12)(32)=1NKXTXXTKXk=1(ykyTk)X(33)=1NK(XTXXTYYTX)(34)4WeneedaprojectionmatrixH=^x^B(^BT^x^B)1^BTwhichsatisesHH=H.WehireHXTY=^x^B(^BT^x^B)1^BTXTYforassistance,where(bythedenitionof^B)^BT=((XTX)1XTY)T=YTX(XTX)1^x^B=1NK(XTXXTYYTX)(XTX)1XTY=1NKXTY(IYTX(XTX)1XTY)(^BT^x^B)1=(YTX(XTX)11NKXTY(IYTX(XTX)1XTY))1=(NK)(YTX(XTX)1XTY(IYTX(XTX)1XTY))1LetP=^BTXTY=YTX(XTX)1XTY,wehave^x^B=1NKXTY(IP)(^BT^x^B)1=(NK)(P(IP))1P(IP)isinvertible,soPandIPareinvertible,hence(^BT^x^B)1=(NK)(IP)1P1HXTY=1NKXTY(IP)(NK)(IP)1P1YTX(XTX)1XTY=XTYHerewecanproveHXTY=XTY)HXTyk=XTyk)HNk^xk=Nk^xk//from(31))^x^B(^BT^x^B)1^BT^xk=^xk//denitionofH)^B(^BT^x^B)1^BT^xk=^1x^xk)^B(^BT^x^B)1^BT(^xk^x`)=^1x(^xk^x`)5