rom the result of vec operator on p. 13 of Ch. 1, we have te(C)=(F⑧F)·tec(∑)+ec(Q)=Aec∑)+tec(Q), (14) A≡(F⑧F) Let r= kp, so that F is an(rx r) matrix and A is an(r-x r) matrix. Equation(14) has the solution vec (∑)=[12-4e(Q (15) provided that the matrix[,2-Ais nonsingular. Thus, the T, j=-p+l,p-l are obtained from(15) Consider the three-dimensional V AR(1) process 0.500 y=c+0.10.10.3y-1+et, 00.20.3 2.2500 with E(EtE=Q 01.00.5 00.50.74 For this process the reverse characteristic polynomial is 100 0.500 det[0 1 0 0.10.10.3z 0 det-0.1z1-0.1z-0.3z 0.2z1-0.3 0.5z he roots of this polynomial are easily seen to be z1=2,2=2.1525,23=-15.4858From the result of vec operator on p. 13 of Ch. 1, we have vec(Σ) = (F ⊗ F) · vec(Σ) + vec(Q) = Avec(Σ) + vec(Q), (14) where A ≡ (F ⊗ F). Let r = kp, so that F is an (r × r) matrix and A is an (r 2 × r 2 ) matrix. Equation (14) has the solution vec(Σ) = [Ir 2 − A] −1 vec(Q), (15) provided that the matrix [Ir 2−A] is nonsingular. Thus, the Γj , j = −p+1, ..., p−1 are obtained from (15). Example: Consider the three-dimensional V AR(1) process yt = c +   0.5 0 0 0.1 0.1 0.3 0 0.2 0.3   yt−1 + εt , with E(εtε 0 t ) = Ω =   2.25 0 0 0 1.0 0.5 0 0.5 0.74  . For this process the reverse characteristic polynomial is det     1 0 0 0 1 0 0 0 1   −   0.5 0 0 0.1 0.1 0.3 0 0.2 0.3   z   = det   1 − 0.5z 0 0 −0.1z 1 − 0.1z −0.3z 0 −0.2z 1 − 0.3z   = (1 − 0.5z)(1 − 0.4z − 0.03z 2 ). The roots of this polynomial are easily seen to be z1 = 2, z2 = 2.1525, z3 = −15.4858. 11