东北财经大学：《金融计量经济学导论》课程PPT教学课件（双语版）Chapter 5 Univariate time series modelling and forecasting

5-11 Solution (IfE(u,=0, then E(u=0Vi. So E(X)=E(u1+11+22)=E(u1)+BE(1)+2E(n12)=0 Var(X =E[ E(XIIX-E(X) ar (X) EI( I =E|(u1+日1u1+B2u12)(u1+ul1+2u12) eL uf+0fuf1+02uf-2+cross-products But Elcross-products=0, since Cov(u, u=0 for s+0. So var(xX)=y6=E[2+612-+62l1-2l =a2+b2a2+b (1+02+02)

5-11 (i) If E(ut )=0, then E(ut-i )=0  i. So E(Xt ) = E(ut +  1 ut-1+  2 ut-2 )= E(ut )+  1E(ut-1 )+  2E(ut-2 )=0 Var(Xt ) = E[Xt -E(Xt )][Xt -E(Xt )] Var(Xt ) = E[(Xt )(Xt )] = E[(ut +  1 ut-1+  2 ut-2 )(ut +  1 ut-1+  2 ut-2 )] = E[ +cross-products] But E[cross-products]=0 ,since Cov(ut ,ut-s )=0 for s0. So Var(Xt ) = 0 = E [ ] = = Solution 2 2 2 2 2 1 2 1 2 ut + ut− + ut− 2 2 2 2 2 1 2 1 2 ut + ut− + ut− 2 2 2 2 2 1 2  +  +  2 2 2 2 1 (1+ + )

5-12 Solution(contd (i) The acf ofX, Y =EXE(IXI-E( t-1 EIII E|(u2+61u1+6212)(ln1+62+213) E|(121+0,02l2) ,a2+0,a2 (O1+,02)a2 Y2 =e[X-E(IIX,2-E(X-2) EIX_2 EI(1+61n1+62l12)(u12+日113+日212) EICO 2t-2 )

5-12 (ii) The acf of Xt 1 = E[Xt -E(Xt )][Xt-1 -E(Xt-1 )] = E[Xt ][Xt-1 ] = E[(ut + 1 ut-1+  2 ut-2 )(ut-1 +  1 ut-2+  2 ut-3 )] = E[( )] = = 2 = E[Xt -E(Xt )][Xt-2 -E(Xt-2 )] = E[Xt ][Xt-2 ] = E[(ut +  1 ut-1+2 ut-2 )(ut-2 + 1 ut-3+ 2 ut-4 )] = E[( )] = Solution (cont’d) 2 1 2 2 2  1 ut−1 +  ut− 2 1 2 2 1  +   2 1 1 2 ( +  ) 2 2 t−2  u 2  2 

5-13 Solution(contd Y3 =EXIX-3I =E|1+61un1+2u 42(a,3+0,u.+01u+5 Soys =0 fors>2. now calculate the autocorrelations y (61+612)o (61+6162) yo(1+012+02)a2(1+2+2) B O (1+62+02)a2(1+02+2) rs=OVs>2 rO

5-13 3 = E[Xt ][Xt-3 ] = E[(ut + 1 ut-1+ 2 ut-2 )(ut-3 + 1 ut-4+ 2 ut-5 )] = 0 So s = 0 for s > 2. now calculate the autocorrelations: Solution (cont’d)    0 0 0 = = 1    s s = =  s  0 0 2 (1 ) ( ) (1 ) ( ) 2 2 2 1 1 1 2 2 2 2 2 1 2 1 1 2 0 1 1                + + + = + + + = = (1 ) (1 ) ( ) 2 2 2 1 2 2 2 2 2 1 2 2 0 2 2            + + = + + = =

5-14 ACF Plot (iii) For 81=-0.5 and 82=0.25, substituting these into the formulae above gives t=-0476 T,=0.190 Thus the acf plot will appear as follows: 08 06 04

5-14 (iii) For  1 = -0.5 and  2 = 0.25, substituting these into the formulae above gives  1 = - 0.476,  2 = 0.190. Thus the acf plot will appear as follows: ACF Plot -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 0 1 2 3 4 5 6 s acf

5-15 4 Autoregressive Processes An autoregressive model of order p, ar(p) can be expressed as y=+y1=1+2y-2+…+0y1n+ Or using the lag operator notation: t-1 Ly,y, y,=H+∑y+ ory,=+∑如Ly,+ln1 or O(D)y=u+u, where P(L)=1-(L+,L2+.,IP

5-15 • An autoregressive model of order p, AR(p) can be expressed as Or using the lag operator notation: Lyt = yt-1 L iyt = yt-i • or or where 4 Autoregressive Processes (L) ( L  L ... L ) p p = 1− 1 + 2 + 2 t t t p t p ut y =  +1 y −1 + 2 y −2 +...+ y − + = = + − + p i t i t i ut y y 1   = = + + p i t t i yt i L y u 1   t ut (L)y =  +