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    多元统计分析第03章.pdf

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    多元统计分析第03章.pdf

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2(r)A2=A(A?).(?4?X Nn(,2In),A=A0,K12X0AX 2(r,),?=0A/2 A2=A?rank(A)=r(r n)1n?!?oN?b?u?3.1 A?O?!?C?g.?(?5?g.?5?5:?X Nn(,2In),A?n?,B?m n?,-=X0AX,Z=BX,eBA=0,KBX?X0AX?p?.(?6?g.?p?:?X Nn(,2In),A,B?n?,K:AB=0 X0AX?X0BX?p?.1n?!?oN?b?u?3.1 A?O?!?C?g.?(?5?g.?5?5:?X Nn(,2In),A?n?,B?m n?,-=X0AX,Z=BX,eBA=0,KBX?X0AX?p?.(?6?g.?p?:?X Nn(,2In),A,B?n?,K:AB=0 X0AX?X0BX?p?.1n?!?oN?b?u?3.1 A?O?!?C?g.?2.?p?ggg.p?g.?ke(?:(?1?X Np(,),0,KX01X 2(p,),?=01.(?2?X Np(,),0,A?,?rank(A)=r,K(X )0A(X )2(r)AA=A(?3?X Np(,),0,A,B?p?,K(X )0A(X )?(X )0B(X )?AB=0pp1n?!?oN?b?u?3.1 A?O?!?C?g.?2.?p?ggg.p?g.?ke(?:(?1?X Np(,),0,KX01X 2(p,),?=01.(?2?X Np(,),0,A?,?rank(A)=r,K(X )0A(X )2(r)AA=A(?3?X Np(,),0,A,B?p?,K(X )0A(X )?(X )0B(X )?AB=0pp1n?!?oN?b?u?3.1 A?O?!?C?g.?2.?p?ggg.p?g.?ke(?:(?1?X Np(,),0,KX01X 2(p,),?=01.(?2?X Np(,),0,A?,?rank(A)=r,K(X )0A(X )2(r)AA=A(?3?X Np(,),0,A,B?p?,K(X )0A(X )?(X )0B(X )?AB=0pp1n?!?oN?b?u?3.1 A?O?!?C?g.?3.?%t?%F?3.1.2?X N(,1)?Y 2(n)?p?,-T=XnY,KT?kn?gd?,?%?%t?,P?T t(n,).3.1.3?X 2(m,)?Y 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1)555?2T2?F?X:?T2 T2(p,n),Kn p+1npT2 F(p,n p+1)1n?!?oN?b?u?3.1 A?O?n!A?(Hotelling)T2?1.AAA?T2?555?555?1?oNNp(,),?X()(=1,n),X?A O?l?,K?O?T2=n(n 1)(X )0A1(X )T2(p,n 1)555?2T2?F?X:?T2 T2(p,n),Kn p+1npT2 F(p,n p+1)1n?!?oN?b?u?3.1 A?O?n!A?(Hotelling)T2?555?3?oNNp(,),?X()(=1,n),X?A O?l?.PT2=n(n 1)X0A1XKn ppT2n 1 F(p,n p,)?=n01555?4T2?O?p,nk,?.555?5T2?O?zC?C.1n?!?oN?b?u?3.1 A?O?n!A?(Hotelling)T2?555?3?oNNp(,),?X()(=1,n),X?A O?l?.PT2=n(n 1)X0A1XKn ppT2n 1 F(p,n p,)?=n01555?4T2?O?p,nk,?.555?5T2?O?zC?C.1n?!?oN?b?u?3.1 A?O?n!A?(Hotelling)T2?555?3?oNNp(,),?X()(=1,n),X?A O?l?.PT2=n(n 1)X0A1XKn ppT2n 1 F(p,n p,)?=n01555?4T2?O?p,nk,?.555?5T2?O?zC?C.1n?!?oN?b?u?3.1 A?O?o!%?d(Wilks)?O?9?ooo!%?ddd(Wilks)?OOO?999?1.%?ddd?3.1.6?X Np(,),K?1?“|?X?222?.e?X()(=1,n),A?l?,K|1nA|1n1A|?222?.3.1.7?A1 Wp(n1,),A2 Wp(n2,),0,n1 p,?A1?A2?,K2?=|A1|A1+A2|?%?ddd?OOO?OOO?,?%?ddd?.P?(p,n1,n2).1n?!?oN?b?u?3.1 A?O?o!%?d(Wilks)?O?9?2.?OOO?T2F?OOO?XXX(?1?n2=1?,?n1=n p,K(p,n,1)d=11+1nT2(p,n)(?2?n2=2?,?n1=n pKn p+1p1 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)T2(p,n 1)F=n p(n 1)pT2 F(p,n p)K?&?1?&?T2=n(X )0S1(X )(n 1)pn pFT?&?%3X?.1n?!?oN?b?u?3.2?oN?u?9?&?n!?&?&m2.?&mmm?X Np(,),a Rp,kZ=a0X N(a0,a0a)?a?2z=a0a?,?t?O?t=n(a0X a0)a0Sa?a0?&?1?&m?a0X t/2a0San a0 a0X+t/2a0San?a=ei=(0,1,0)0K?i?&?1?&m xi t/2rsiin i xi+t/2rsiin1n?!?oN?b?u?3.2?oN?u?9?&?n!?&?&ma0?&mmmnnn3.2.2b?X(t)(t=1,n)?5gp?oNNp(,)(0,?)?,Kk?a,ma0X d,a0X+d(?d=s(n 1)pn(n p)Fa0Sa)?a0?V?1 .T2mmmi?&?1?T2m?xi crsiin i xi+crsiin?c=s(n 1)p(n p)F?sii?S?1i?.1n?!?oN?b?u?3.3 oN?u?!?oN?u?!?oooNNN?uuu?X()(=1,n)?5goNX Np(1),)?Y()(=1,m)?5goNY Np(2),)?,?p?.?b?flKH0:(1)=(2),H1:(1)6=(2).1n?!?oN?b?u?3.3 oN?u?!?oN?u?1.?oooNNN?(?)?uuu?(1=2)u?O?T2=nmn+m(X Y)0(A1+A2n+m 2)1(X Y)?A1?A2?oN?l?.3H0?,T2 T2(p,n+m 2).F=(n+m 2)p+1(n+m 2)pT2 F(p,n+m p 1)?W=F F(p,n+m p 1)1n?!?oN?b?u?3.3 oN?u?!?oN?u?2.?oooNNN?uuu?(16=2)(1)?m=n?,-Z(i)=X(i)Y(i)(i=1,n)?oN=z?oN?u?H0:(1)=(2)H0:Z=0p1n?!?oN?b?u?3.3 oN?u?!?oN?u?(2)?m 6=n?,?n m:-Z(i)=X(i)rnmY(i)+1nmnXj=1Y(j)1mmXj=1Y(j)(i=1,2,n)?y:E(Z(i)=(1)(2)Cov(Z(i),Z(j)=(1+nm2,?i=j;0?i 6=j;def=ZijZ(i)Np(1)(2),Z)(i=1,n),?p?.=z?oN?u?!1n?!?oN?b?u?3.3 oN?u?!?oN?u?!?oooNNN?uuu?kk?oNNp(t),)(t=1,k),?X(t)()(t=1,k,=1,nt)?p?,?b?flKH0:(1)=(2)=(k),H1:?3i 6=j,?(i)6=(j)Pn=kXt=1nt,X=1nkXt=1ntXj=1X(t)(j),X(t)=1ntntXj=1X(t)(j)(t=1,k)1n?!?oN?b?u?3.3 oN?u?!?oN?u?ol?T=kXi=1niXj=1(X(i)(j)X)(X(i)(j)X)0|Sl?A=kXi=1niXj=1(X(i)(j)X(i)(X(i)(j)X(i)0=kXi=1Ai|ml?B=kXi=1ni(X(i)X)(X(i)X)0l?)“T=A+B1n?!?oN?b?u?3.3 oN?u?!?oN?u?q,?n?H0?u?O?=|A|A+B|=|A|T|?1Ai Wp(ni 1,)?p?,(i=1,k),d?5?A Wp(n,)23H0e,T Wp(n 1,).3?y3H0e,B Wp(k 1,),?B?A?p?.3H0e,u?O?(p,n k,k 1)?W=

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