《模式识别》(边肇祺)习题答案

TMY(=)5s22=%23qf984LYf105dLYf166#BKlBKlZ|4|1897T+4|2010dSZTMY(=)5s12=% T5qKlpq=%?5VU$!BCB5qP(i = 1,.,CbNHKlpq=%?5= 5x wib qEqT=T/TpP(wi|x) = p(x|(wi)p(x) .P(wi|x) = P(wi,x)p(x)= p(x|(wi)p(x) f/P(wi|x) + P(w2|x) = 1bP(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 2.4sY/f1. P(x|= P(x|. P(= P(Klpq=%?5bP(x|= P(x|HTP( P(5x 5x w2bP(= P(HTP(x| P(x|5x 5x w2b cfw P(wj|x)B= iHx wiTMY(=)5scfKlpq=%?5TP(wi|x) = ;:;cP(wj|x)5x wib=V|5qHqqMTTp(x|(= ;:;cp(x|(5x wib 5KlB=%?5VVUp(x|w1)p(x|(12 22)P(21 11)P(5x Q5w2b9Hq1|x) =2j=11jP(wj|x)= 11P(w1|x) + 12P(w2|x)R(2|x) =2j=12jP(wj|x)= 21P(w1|x) + 22P(w2|x)1|x) (12 22)P(w2|x)(21 11)P(w1)p(x| (12 22)P(w2)p(x|w2)p(x|w1)p(x|(12 22)P(21 11)P(Tp(x|w1)p(x| (12 22)P(21 11)P(5x w1bQ5x w2b 11 = 22 = 0,12 = 21NHKlKv%1pqMbKlKv%H(11 22) + (21 11)x|w1)12 22)x|w2) 0x|w2)x|w1)NHKlKv%PP1(e) = P2(e) B%?5YfVlTV7%Z%uMTMY(=)5sB%?5Klpq=%?5YfVj =;:;cP(wj|x)5x b6BTj = ;:;cp(x|(5x bIn5s%P(w1|x) = P(w2|x)p(x|(= p(x|(Mb f/KlB=%Yf%Zb Ml(x)ll(x) = p(x|w1)p(x|l(x)1k (1) Eln(x)|= E(x)| (2) El(x)|= 1 (3) El(x)|E2l(x)|= l(x)|55(1)Eln(x)|=ln(x)p(x|w1) (p(x|n+1(p(x|n E(x)|=p(x|w2) (p(x|n+1(p(x|n Eln(x)|= E(x)|(2)El(x)|=l(x)p(x|w2)p(x|w1) 1(3)El(x)|E2l(x)|= El2(x)|E2l(x)|= l(x)| xj(j = 1,2,.,n)nMExj|= xj|= 911 = 22 = 0#12 = 21 = 1f/=%pqbK01/KlB=%Klpq=%Nb T=TbP(wi|x) = P(x|(x) P(x|( fKlpq=%w (x(x5x Q5s w2b|xK*b 1 = 2 1 =1 1212 1 2 = 1 1212 11%?5TMY(=)5sh(x) = lnl(x)= x|+ x| 12( 11 ( 12( 12 (+ 12 1| 2|=1211 12 )x( 11 12 x+12(11 12 1| 2|)= 43 43(= 0b%x1(1) = 0s% f/Inf11 = 22 = 0,12 = 211pqW1b (1)pP(e) = H9pq P(e)P(P( (2)pKlKv%9pqb1f0fHq/KlB=%NKlpq=%b10f/pqKlbP(= 7TMY(=)5s2KlKv%H(1122)+(2111)x|w1)1222)x|w2)0Vx|w1)x|w2) (x1,x1(1) 0 (x1,x1(1) f=Qf( P,P) = ( P P)2#P5(sf(P) = 1,0 P 1bL!Hq/=9PbP(X|P) =Ni=1P)1=TpP|X) = P(X|P)f(P) P(X|P)f(P) P)10 P)1X = x1,.,1p(x|)0 6 x 6 Hp(x|) = 150bKv9bKv9SKvffL() = |) (1)= N 2)9TMY(=)5s1PL()KvH1 xb*NKv9 = 3) T(A1 + B1)1 = A(A + B)1B = B(A + B)1A(A1 + B1)A(A + B)1B = (I + B1A)(A + B)1B= B1(B + A)(A + B)1B= B1B= (A1 + B1)1 = A(A + B)1B(A1 + B1)1 = B(A + B)1A p(x) N(u,2)3f(x) N(0,1)9pN(x) = 1i=1(xl/ (1) E pN(x) N(u,2 + (2) pN(x) = 1p(x)(1) EpN(x) =pN(x)p(x)wVU=UW4LYf 1g(x) = 0r = |g(x)|w|g(= 0Hq/P|x2rl 2g x g(x)|w|2w1!xg(x) 0xg5 0b!xr5x r w|w| r|w|r = w| =g(x)|w|TMY(=)5Hg(x) 0) 2uHWb|b1!a b5B9 0u 0b2 buH b 0uH 0 b 0b|b +5f1$ps%b5fJ(a) =yY(b%Z 0bypsH 0%bps%yY(5fb mbZ5fJ(a) = |Yab|2 =Nn=1(Kl=Ia = (Y 1Y 0/EpabJ(a)0J(a) =2Y T(Yab)50/EV a(1)a(k + 1) = a(k)(Yab)4k =1kT1iTMY(=)5sBhl9i%V|E a(1)a(k + 1) = a(k)k(a(k)bk)kk9F7vhlElbB4k = 1k-5fEBKpsb 1T(A + 1 = A1 A11 + x2kTT4b1(A + A1 A11 + x)= (A + I A1 x)A1=(A + x x)A1= = (A + 1 = A1 A11 + x2R(k + 1)1 = R(k)1 + TVR(k + 1) = R(k) R(k)(k)1 + (k)n5fJ(a) =yY (a)(b)2Y (a) bb!Y (a)B5J(a)0J(a) =2( b)=D = 2bNK4k =|J(a)|2JT(a)DJ(a)H0/ET = b2 Y (a)B55fJ(a) =yY (a)(b)2 = (b)2J(a) = 2(b)=D = 2b0/T = kJ(k = 4(b)2|28(b)212|2|k0/T = b2 TMY(=)5s |b =.,z ., z NbY = 11 2 a = w0,wT5Y Y 1 1 11 2 1 1 ! 1N1i 1N2iTV N ( ( 0N(T2i=1jCi(O( = m =Ni=1TVsTB |=TV 1N ( + w = N( 1w = iywZ_TV (Mb (1)T(4VU_y w|20wVXbWgb (2)gXbWb15_XbWy (y w|20w)_XbWE_bXbWZg(x) = 14TMY(=)5 1, 0|_g(x) w|2w= w|2|w|2 = 0yy (y w|20w)= w|20w 5TBFV$BLs5FLVsbT?Bs7598LVsb98LVsALVsQba 98LVsALVsLVsA98LVs BFbic(c 1)/2s75FLVsbLVsBLVsbLVsBLVTMY(=)5s5dLYf sLsV/=YfwbsLfB+ydLf%FV/wb 5UV Vb (1)sP (2)sMH (3)L!MsH3z(x)B5s:1USVU(4,6)(5,5)(3,4)(3,2)(2,5)(1,3)9bU2U6#E K#%sLbsLf%FbFyNB T(614) (618)b:ci=1P2(wi|x) = P2(wm|x) +i=wi|x)16TMY(=)5 If/K#(Pr 7=_ (1,0)T, (0