(5.1)--线性模型习题课（习题三）.pdf

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1、5?.?SK4.65?.y=X+e,e (0,2I),rk(Xnp)=p,yV ar(i)2/x0ixi,1 i p,pxiLX?1i?.?x0ixj=0i 6=j.)i=C0i,C0i=0,.,0,1,0,.,0=1i?1?0,rk(X)=pX0X_,?V ar(i)=V ar(C0i)=C0i(X0X)1X0Cov(y)X(X0X)1Ci=2C0i(X0X)1X0X(X0X)1Ci=2C0i(X0X)1Ci-A=X0X=x01.x0p?x1.xp?=x01x1x01x2.x01xpx02x1x02x2.x02xp.x0px1x0px2.x0pxp,?A 0,|?(C0iCi)2=(A12C

2、i)0A12Ci2(A12Ci)0A12Ci(A12Ci)0A12Ci=C0iACi C0iA1Ci 1 x0ixi C0iA1Ci1x0ixi C0iA1Ci,V ar(i)=2C0iA1Ci 2/x0ixi.75?C0iA1Ci=1x0ixi?k(A12Ci)0=ki(A12Ci)0,A12Ci=kiA12Ci,?C0iA12=kiC0iA12 C0iA=kiC0i?x0ix1x0ix2.x0ixi.x0ixp?=?00.ki.0?x0ixj=0,j 6=i;x0ixi=ki.C0i(A kiIp)=01px0ixj=0,i 6=j.5x0ixj=0,i 6=jA=x01x1.x0pxp

3、C0iA1Ci=1x0ixi V ar(i)=2C0iA1Ci=2/x0ixi.4.1.1?i=c0i,i=1,.,k?,i,i=1,.,k,K=Pki=1ii?=Pki=1ii=Pki=1ic0i?BLU?Oyi=c0i,=Pki=1ii=Pki=1ic0i,i,i=1,.,kqi=c0i,i=1,.,k?ci M(X0),(Pki=1ic0i)0=Pki=1ici M(X0),?E()=E(Pki=1ic0i)=Pki=1ic0iE()=Pki=1ic0i=,?O,55w,?.ey?.?L0y?,?OukE(L0y)=L0X=Pki=1ic0iL0X=Pki=1ic0i.uV ar(L0y

4、)V ar(kXi=1ic0i)=2L0L L0XCov()X0L=2L0L 2L0X(X0X)X0L=2L0(In PX)L 0w,?y.?BLU?O.5.1 Ou?81,?9?u,:=?y1i=1+1x1i+e1i,e1i N(0,2),i=1,.,m,y2j=2+2x2j+e2j,e2j N(0,2),j=1,.,n,ke1i,e2jpOu?en?b?(1)H1:1=2,81(2)H2:1=2,8?(3)H3:,x0,1+1x0=2+2x0,8?u:(x0,y0),y0=1+1x0.y?.?/Xey1=y11.y1m=1x11.1x1m 11!+e1,e1=e11.e1m Nm(0,2I

5、)y2=y21.y2n=1x21.1x2n 22!+e2,e2=e21.e2n Nn(0,2I)y1y2!=X100X2!12!+e1e2!,X1=1x11.1x1m,X2=1x21.1x2n,i=ii!,i=1,2.rk X100X2!=4,=12!?=1 2!=X01X100X02X2!1 X0100X02!y1y2!=(X01X1)1X01y1(X02X2)1X02y2!u?.?SSe=y01y1+y02y2 10X01y 20X02y(1)u?H1:1=2,=u?H1=(0,1,0,1)1122=0,H1=(0,1,0,1),u?u?OF1=(H1)0(H1(X0X)1H01)1(H1

6、)/1SSe/(n+m 4),X=X100X2!.KkF1 F1,n+m43?wY(0 F1,n+m4(),K?b?1=2;eF1 F1,n+m4(),K?b?1=2.(2)u?H2:1=2,=u?H2=(1,0,1,0)1122=0,H2=(1,0,1,0),u?u?OF2=(H2)0(H2(X0X)1H02)1(H2)/1SSe/(n+m 4),X=X100X2!,KkF2 F1,n+m43?wY(0 F1,n+m4(),K?b?1=2;eF2 F1,n+m4(),K?b?1=2.(3)u?H3,x0,1+1x0=2+2x0,=u?H3=0,H3=?1x01x0?.u?u?OF3=(H3)

7、0(H3(X0X)1H03)1(H3)/1SSe/(n+m 4),X=X100X2!,KkF3 F1,n+m43?wY(0 F1,n+m4(),K?b?;eF3 F1,n+m4(),K?b?.6.3u?58?.y=01+X+e,e N(0,2I),Xn (p 1)?.?b?H1:1=.=p1=cH2:1=.=p1H3:1+.+p1=c?FOpc?yy=01+X+e,e N(0,2I),-X=(1,X),=(0,0)0?y=X+e,u=(X0X)1X0y,RSS()=0X0y=y0PXy(1)u?b?1=.=p1=c,?uu?H=c1,H=(0,Ip1),b?“?.?.yi=0+cp1j=1xi

8、j+e,i=1,.,n3?.-zi=yi cp1j=1xij,?.C/zi=0+e,?Kn0=ni=1zi,u03?.e?LS?O0=ni=1zin=y Pni=1Pp1j=1cxijn,0A?8RSSH1()=RSS(0)=n(y Pni=1Pp1j=1cxijn)2u?.?SSe=y0y RSS()=y0y y0PXym=rk(H)=p-1,u?b?u?OF=(RSS()RSSH1()/(p 1)SSe/(n p)?b?F Fp1,np.?Y,?F Fp1,np()?b?K?b?(2)u?b?1=.=p1,?uu?H=0,H=010.01001.01.000.11,b?“?.?.yi=0+

9、p1j=1xij1+e,i=1,.,nz?/y=1p1j=1x1j.1p1j=1xnj 01!+e=X11+eu13?.e?LS?O1=(X01X1)1X01y,1A?8RSSH2()=RSS(1)=y0PX1y?.?SSe=y0y RSS()=y0y y0PXym=rk(H)=p-2,u?b?u?OF=(RSS()RSSH2()/(p 2)SSe/(n p)?b?F Fp2,np.?Y,?F Fp2,np()?b?K?b?(3)u?b?1+.+p1=c,?uu?H=c,H=(0,1,.,1),urk(H)=1?b?d5.1.1?F=(H c)0(H(X0X)1H0)1(H c)/1SSe/(

10、n p),KkF F1,np3?wY(0 F1,np(),K?b?K?b?6.5 u58?.y=X+e,b?X?1?1y(1)Pni=1(yi yi)=0,(2)Pni=1 yi(yi yi)=0.y:(1)e=y X=y X(X0X)X0y=(I PX)y,ukX0 e=X0(I PX)y=0,?X?1?1KX0=10X0!,?X0 e=10 eX0 e!=00!,10 e=0,Pni=1(yi yi)=10 e=0.(2)Pni=1 yi(yi yi)=y0(y y)=(X)0(I PX)y=0.K1(160)yPXq+1 PXq,Xq+1=(Xq.b).yPXq+1=Xq+1(X0q+1

11、Xq+1)1X0q+1=?Xqb?X0qXqX0qbb0Xqb0b!1 X0qb0!dn2.2.4?X0qXqX0qbb0Xqb0b!1=(X0qXq)1+ADA0ADDA0D!,A=(X0qXq)1X0qb,D=b0(I PXq)b1,5D 0,PXq+1=Xq(X0qXq)1X0q+XqADA0X0q bDA0X0q XqADb0+bDb0=PXq+D(XqA b)(XqA b)0(XqA b)(XqA b)0 0,PXq+1 PXq.2160XJ?.y=Xqq+e,E(e)=0,Cov(e)=2Io y=Xqq(qq?LS?O)?y=X+e,Od=E(y E(y)0(y E(y)ydn6

12、.3.1E(q)=q+(X0qXq)1X0qXtt=E(y E(y)=E(y)E(y)=Xqq+Xq(X0qXq)1X0qXtt Xqq Xtt=Xq(X0qXq)1X0qXtt Xtt=(PXq I)Xtt=Cov(y E(y)=Cov(Xqq E(y)=XqCov(q)Xq=2Xq(X0qXq)1X0qd=0+tr()=(PXq I)Xtt)0(PXq I)Xtt+tr(2Xq(X0qXq)1X0q)=0tX0t(PXq I)2Xtt+2tr(X0qXq)1X0qXq)=0tX0t(I PXq)2Xtt+q2=0tX0t(I PXq)Xtt+q2=0tD1t+q2D1=X0t(I PXq)

13、Xt.390y(X0X)1X0X(X0X)1(X01X)1.y(X0X)1X0X(X0X)1(X01X)1 X0X (X0X)(X01X)1(X0X)X012In12X X01212X(X01212X)1X01212X X012In12X X012P12X12X.?y6.9?58?.y=X+e,e N(0,2I),?=Ay?5?O(1)y?MSEM()=E()()0?4?A=0X0(X0X0+2I)1.(2)y=Ay=0X0y2+0X0X.5e?O,2“O,2,B?5?O=Ay=0X0y 2+0X0X.y(1)Cov()=Cov(Ay)=2AA0,(E()(E()0=(AX)(AX)0KkMS

14、EM()=Cov()+(E()(E()0=2AA0+(AX )(AX )0mK?=?,?utr(2AA0+(AX )(AX )0)A=22A+2AX0X0 20X0-?u0?2A+AX0X0=0X0A=0X0(X0X0+2I)1.(2)(X0X0+2I)1=12I 12X0X0 121+0X012X=12I 12X0X02+0X0X=I+120X0XI 12X0X02+0X0Xu?Ay=0X0I+120X0XI 12X0X02+0X0Xy=0X0+120X00X0X 120X0X0X02+0X0Xy=0X0+12(0X0X)0X012(0X0X)0X02+0X0Xy=0X02+0X0Xy=0X0y2+0X0X