《模式识别》(边肇祺)习题答案

()1222384105166167188189K-L2010221()§1§2•2.1CP(wi)i=1,...,Ci∗=maxiP(wi)x∈wi•2.2P(wi|x)=p(x|wi)P(wi)p(x).P(wi|x)=P(wi,x)p(x)=p(x|wi)P(wi)p(x)•2.3P(wi|x)+P(w2|x)=1P(w1|x)+P(w2|x)=P(w1,x)p(x)+P(w2,x)p(x)=P(w1,x)+P(w2,x)p(x)=p(x)p(x)=1•2.41.P(x|w1)=P(x|w2)2.P(w1)=P(w2)P(x|w1)=P(x|w2)P(w1)P(w2)x∈w1x∈w2P(w1)=P(w2)P(x|w1)P(x|w2)x∈w1x∈w2•2.51.c2.P(wi|x)P(wj|x)j̸=ix∈wi2()cP(wi|x)=maxj=1;:::;cP(wj|x)x∈wip(x|wi)P(wi)=maxj=1;:::;cp(x|wj)P(wj)x∈wi•2.6p(x|w1)p(x|w2)(λ12−λ22)P(w2)(λ21−λ11)P(w1),x∈w1w2R(α1|x)=2∑j=1λ1jP(wj|x)=λ11P(w1|x)+λ12P(w2|x)R(α2|x)=2∑j=1λ2jP(wj|x)=λ21P(w1|x)+λ22P(w2|x)R(α1|x)R(α2|x)x∈w1λ11P(w1|x)+λ12P(w2|x)λ21P(w1|x)+λ22P(w2|x)(λ21−λ11)P(w1|x)(λ12−λ22)P(w2|x)(λ21−λ11)P(w1)p(x|w1)(λ12−λ22)P(w2)p(x|w2)p(x|w1)p(x|w2)(λ12−λ22)P(w2)(λ21−λ11)P(w1)p(x|w1)p(x|w2)(λ12−λ22)P(w2)(λ21−λ11)P(w1)x∈w1x∈w2•2.7λ11=λ22=0,λ12=λ21(λ11−λ22)+(λ21−λ11)∫R2p(x|w1)dx−(λ12−λ22)∫R1p(x|w2)dx=0∫R1p(x|w2)dx=∫R2p(x|w1)dxP1(e)=P2(e)•2.83()j∗=maxj=1;:::;cP(wj|x)x∈wjj∗=maxj=1;:::;cp(x|wj)P(wj)x∈wjP(w1|x)=P(w2|x)p(x|w1)P(w1)=p(x|w2)P(w2)•2.9•2.10l(x)l(x)=p(x|w1)p(x|w2)l(x){(1)E{ln(x)|w1}=E{ln+1(x)|w2}{(2)E{l(x)|w2}=1{(3)E{l(x)|w1}−E2{l(x)|w2}=var{l(x)|w2}(1)E{ln(x)|w1}=∫ln(x)p(x|w1)dx=∫(p(x|w1))n+1(p(x|w2))ndxE{ln+1(x)|w2}=∫ln+1p(x|w2)dx=∫(p(x|w1))n+1(p(x|w2))ndxE{ln(x)|w1}=E{ln+1(x)|w2}(2)E{l(x)|w2}=∫l(x)p(x|w2)dx=∫p(x|w1)dx=1(3)E{l(x)|w1}−E2{l(x)|w2}=E{l2(x)|w2}−E2{l(x)|w2}=var{l(x)|w2}•2.11xj(j=1,2,...,n)nE[xj|wi]=ijηvar[xj|wi]=i2j2σ2λ11=λ22=0λ12=λ21=10−1•2.12P(wi|x)=P(x|wi)P(x)∑cj=1P(x|wi)P(wi)•2.13•2.14R(ai|x)R(ai|x)=c∑j=1λijP(wj|x)=c∑j=1λijp(x|wj)P(wj)////omitthesamepartp(x)R(ak|x)=minj=1;2;:::;NR(aj|x)ak•2.15xT−1x=CC4()•2.16Mahalanobisr{(1)r(a,b)=r(b,a){(2)a=br(a,b)=0{(3)r(a,c)≤r(a,b)+r(b,c)(1)r(a,b)=(a−b)T−1(a−b)=(b−a)T−1(b−a)=r(b,a)(2)r(a,b)=(a−b)T−1(a−b)≥0a=br(a,b)=0(3)−1−1=PPT•2.17−1−1=h11h12···h1dh12h22···h2d............h1dh2d···hddMahalanobisγ2=d∑i=1d∑j=1hij(xi−ui)(xj−uj)γ2=(x−u)Th11h12···h1dh12h22···h2d............h1dh2d···hdd(x−u)=d∑i=1d∑j=1hij(xi−ui)(xj−uj)•2.18d=2,d=3MahalanobisγV=Vd||12γd•2.19xmmxp(x|m)=(2π)12σ−1exp{−12(x−m)2/σ2}mm0σ2mp(m|x)=(σ3+σm)12(2π)12σσmexp[−12σ2+σ2mσ2σ2m(m−σ2mx+m0σ2σ2+σ2m)2]5()p(m|x)=p(x|m)p(m)p(x)=p(x|m)p(m)∫p(x|m)p(m)dm=(2π)12σ−1exp{−12(x−m)2/σ2}(2π)12σ−1mexp{−12(m−m0)2/σ2m}∫(2π)12σ−1exp{−12(x−m)2/σ2}(2π)12σ−1mexp{−12(m−m0)2/σ2m}dm=(σ3+σm)12(2π)12σσmexp[−12σ2+σ2mσ2σ2m(m−σ2mx+m0σ2σ2+σ2m)2]•2.20i=σ2I{(1)P(wi)̸=P(wj){(2)(1)P(wi)=P(wj)x0=12(ui+uj)(2)•2.21i=uiuj•2.24•2.23u1=(−1,0)T,u2=(1,0)T,1=2=I,P(w1)=P(w2)h(x)=−ln[l(x)]=−lnp(x|w1)+lnp(x|w2)=12(x1−u1)T−11(x1−u1)−12(x2−u2)T−12(x2−u2)+12ln|1||2|=12[(x−u1)T(x−u1)−(x−u2)T(x−u2)]ln[P(w1)P(w2)]=0(x−u1)T(x−u1)(x−u2)T(x−u2)x∈w1,s∈w2xui•2.242.231̸=21=[112121]2=[1−12−121]6()h(x)=−ln[l(x)]=−lnp(x|w1)+lnp(x|w2)=12(x1−u1)T−11(x1−u1)−12(x2−u2)T−12(x2−u2)+12ln|1||2|=12xT(−11−−12)x−(−11ui−−12uj)Tx+12(uT1−11u1−uT2−12u2+ln|1||2|)=−43x1x2+43x1ln[P(w1)P(w2)]=0x1(x2−1)=01xy11:•2.252.24λ11=λ22=0,λ12=λ21{(1)P(e)=0.05Neyman-PearsonP(e)P(e1)P(e2){(2)10-10P(e1)=0.057()2(λ11−λ22)+(λ21−λ11)∫R2p(x|w1)dx−(λ12−λ22)∫R1p(x|w2)dm=0∫R2p(x|w1)dx=∫R1p(x|w2)dmR1={(x1,x2)|x1(x2−1)0}R2={(x1,x2)|x1(x2−1)0}§3•3.1N(u,1)−∞u+∞X={x1,x2,...,xN}^uup(u)∼N(0,1)L(u)=lnp(X|u)=N∑i=1lnp(xi|u)=−12N∑i=1(xi−u)2+Cu∂L(u)∂u=N∑i=1xi−Nu=0u^u=1NN∑i=1xiMAP(maximumaposterior)p(u|X)=p(X|u)p(u)∫p(X|u)p(u)du=αN∏i=1p(xi|u)p(u)=αN∏i=11√2πσexp[−(xi−u)22σ2]·1√2πσ0exp[−(u−u0)22σ20]=α′exp[−12(N∑i=1(u−xiσ)2+(u−u0σ0)2)]=α′′exp[−12[(Nσ2+1σ20)u2−2(1σ2N∑i=1xi+u0σ20)u]]p(u|X)N(un,σ2n)1σ2n=Nσ2+1σ20unσ2n=1σ2N∑i=1xi+u0σ208()unσ2nun=Nσ20Nσ20+σ2mN+σ2Nσ20+σ2u0σ2n=σ20σ2Nσ20+σ2mN=1NN∑i=1xiun•3.3X={x1,x2,...,xN}f(x,P)=PxQ(1−x),x=0,1,0≤P≤1,Q=1−PPL(P)=N∑i=1ln(Pxi(1−P)(1−xi))=N∑i=1xilnP+Nln(1−P)+N∑i=1xilnPP∂L∂P=∑Ni=1xiP−N1−P+∑Ni=1xi1−P=0P^P=1NN∑i=1xi•3.4λ(^P,P)=(^P−P)2Pf(P)=1,0≤P≤13.3^PP(X|P)=N∏i=1Pxi(1−P)1−xiPP(P|X)=P(X|P)f(P)∫P(X|P)f(P)dP=∏Ni=1Pxi(1−P)1−xi∫10∏Ni=1Pxi(1−P)1−xidP•3.7X={x1,x2,...,xN}p(x|θ)06x6θp(x|θ)=1θ0θmaxkxkL(θ)=lnp(X|θ)(1)=−Nlnθ(2)9()L(θ)θxθθ∗=maxkxk(3)•3.8(A−1+B−1)−1=A(A+B)−1B=B(A+B)−1A(A−1+B−1)A(A+B)−1B=(I+B−1A)(A+B)−1B=B−1(B+A)(A+B)−1B=B−1B=I(A−1+B−1)−1=A(A+B)−1B(A−1+B−1)−1=B(A+B)−1A•3.15p(x)∼N(u,σ2)φ(x)∼N(0,1)Parzen^pN(x)=1NhNN∑i=1φ(x−xihN)hN{(1)E[^pN(x)]∼N(u,σ2+h2N){(2)Var[^pN(x)]=1NhN2√πp(x)(1)E[^pN(x)]=∫^pN(x)p(x)dx8.1SwSb§4•4.11xg(x)=wTx+w0=0r=|g(x)|||w||g(xq)=0||x−xq||22xp=x−g(x)||w||2w1xg(x)0xqxwTxq+w0=0xrx−xp=rw||w||wTx−wTxp=r||w||r=wTx+w0||w||=g(x)||w||10()xg(x)0r=−wTx−w0||w||=−g(x)||w||r=|g(x)|||w||2xx−xp=rw||w||=g(x)||w||2wxp=x−g(x)||w||2wxx−xp=−rw||w||=g(x)||w||2wxp=x−g(x)||w||2w•4.3g(x)=5+7x+9x2{(1){(2)1y=[y1,y2,y3]T=[1,x,x2]Ta=[5,7,9]Tg(x)=aTy2ng(x)=c0+c1x+c2x2+...+cnxny=[y1,y2,...,yn+1]T=[1,x,...,xn]Ta=[c0,c1,...,cn]Tg(x)=aTy•4.31g(x)=a1+a2x+a3x22xp(x)1y=[1,x,x2]Ta=[a1,a2,a3]Tg(x)=aTy2y=[1,x,x2]T•4.4g(x)=x1+2x2−2{1g(x)=wTx+w0g(x)=0{2g(x)=aTy{3XYaTy=0XwT+w0=0X1w=[1,2]T,x=[x1,x2]T,w0=−2g(x)=wTx+w0g(x)=022y=[y1,y2,y3]T=[1,x1,x2]Ta=[−2,1,2]Tg(x)=aTy3y1=1,y2=x1,y3=x2Yy1=1•4.5FisherwFisherwααwy=αwTx•4.6a,b∈RnA=abTabTbTaAbTa,0,0,