实用多元统计分析答案08

164Chapter 88.1 Eigenvalues of * are À1 = 6. À2 = 1. The principal campenents areYl = ..894X1 + .447X2Y2 = .~47X1 - .894XZ, .Vareyi) = À1 '= 6. Therefore. proporti()n of tatal population varianceexplained by ~l is 6/(15+1) = .86.8.2£ = (1 . .6325J.6325 1(a) Y, = ,.707L, + .7!J7L¿Var(Yi) = À, = 1.6325'2 = .7n7Z, ~ .707Z2Proportien of total populationvarianceexpl ained by Yl is1.6325/(1+1. = ..816(b) No. The two (standardized) variables contribute ~qually to theprincipal components in 8.2(a). The two variables contributeunequally to the principal components in 8.1 because 'Of theirunequal varian~es.(c) Py L = .903;1 1 .PYl Zz = .903;.Py Z = .4292 ,8.3 Ei~envalues of tare 2.' 4. 4. E;genvect~rs assaciate with the ei~en-values 4. 4 are not unique. One choi~~ is =i = iO 1 O)çnd:~ =(0 0 1). With these assignments .the 'principal components arey 1 = Xl' Y 2 = X2 and Y 3 = X3 .8.4 figenvalues of * are selutions.of 1;-À11 = (a2-Àp-2t~2_ÀH.a2,p)2 = 0Thus ~0'2-À)H.a2_À)2_2cr4p2J = 0 S'O À = 0'2 'Or À ='"2~lt~hl,). ForÀ1 = (12,~i = (l/ff,.o,-l/I2J. 'For À2 =O'2(l+i'Ii; ~ fj/Z.)1NEiZ). ferÀ~'=a2'tl-pI2). ~~ = 0/". -1/12; 1/121.65Pri nci'fa 1ComponentVari ancePropert ion of TotalVariance Explained1 1Y 1 = a Xl - 12 X31 1 1y 2 = "2 Xl + 12 X2 + 2" X31 1 1y 3 = '2 Xl - /ž X2 + 2" X3021/30'2 (1-.p12)1 (1+1'12)1 (l-pl2). a2(l+pm8.5 (a) Eigenva.l ues of 2 satisfyIE-ul = (l-À)3 + 2.p3 - 3(1:'À)p1 = 0or (l +29-À)(1-p-À)2 = O. Hence À1 = 1 + 2.p; À2 = À3 = 1 - ?and results are consistent with (8-16) for p = 3.(b) By direct multiplication1 1.ø ( c 1) = (1 + (P-1)9 H c 1 ).i y.p - y~ -thus varidying the fir~t eig~nvalu~-eigenv~ct~~ pair. further~ :i= (l-p)~i' ; = 2...