They are obviously all greater than 1 in absolute value. Therefore the process is stationary We next obtain the MA(oo)coefficients from this VAR(1) process using the relation as described in equation(10) 0.500 0.250 0=I3,业1=更1业0=0.10.10.3,亚2=更1业=0.010.070.12 00.20.3 0.020.080.15 0.1250 0 ya=更1y 0.0370.0310.057,亚 0.0180.0380.069 We finally calculate the autocovariance of this process. From(15)we have vec ec(Io)=[I-重1⑧重]ec(!2) 0.750 0 0 0.050.95-0.150000 0 0-0.10.850 0 0 0 0 0000 0.050 0.95 0 0 0.150 -0.01-0.01-0.03-0.010.99-0.03-0.03-0.03-0.0 0.02-0.030 0.020.970-0.06-0.09 0 0 0 0.010 0.02-0.02-0.06-0.030.97-0.09 0 0 0 0.04-0.060 0.0600.91 2.25 3.000 0 0.019 1.0 1.172 0.5 0.674 0.019 0.5 0.674They are obviously all greater than 1 in absolute value. Therefore the process is stationary. We next obtain the MA(∞) coefficients from this V AR(1) process using the relation as described in equation (10). Ψ0 = I3, Ψ1 = Φ1Ψ0 =   0.5 0 0 0.1 0.1 0.3 0 0.2 0.3   , Ψ2 = Φ1Ψ1 =   0.25 0 0 0.01 0.07 0.12 0.02 0.08 0.15   , Ψ3 = Φ1Ψ2 =   0.125 0 0 0.037 0.031 0.057 0.018 0.038 0.069   , Ψ4 = .... We finally calculate the autocovariance of this process. From (15) we have vec(Σ) = vec(Γ0) = [I9 − Φ1 ⊗ Φ1] −1 vec(Ω) =               0.75 0 0 0 0 0 0 0 0 −0.05 0.95 −0.15 0 0 0 0 0 0 0 −0.1 0.85 0 0 0 0 0 0 −0.05 0 0 0.95 0 0 −0.15 0 0 −0.01 −0.01 −0.03 −0.01 0.99 −0.03 −0.03 −0.03 −0.09 0 −0.02 −0.03 0 −0.02 0.97 0 −0.06 −0.09 0 0 0 −0.01 0 0 0.85 0 0 0 0 0 −0.02 −0.02 −0.06 −0.03 0.97 −0.09 0 0 0 0 −0.04 −0.06 0 −0.060 0.91               −1               2.25 0 0 0 1.0 0.5 0 0.5 0.74               =               3.000 0.161 0.019 0.161 1.172 0.674 0.019 0.674 0.954               . 12