Wald Decomposition Theorem Any zero mean covariance stationary processxt can be represented in the form of xt = 1X j =0 dj t j + j ; where d0 = 1 ; 1X j =0 d 2 j < 1 The termt is white noise and represents the prediction error defined to bet = xt P[xt j xt 1;]. The value oft is uncorrelated witht j for anyj, thought can be predicted perfectly byxt 1;xt 2;. Proof Lett = xt P[xt j xt 1;], whereP[xt j xt 1;] is the prediction ofxt based on a linear function ofxt 1;xt 2;. In order thatP[xt j xt 1;] is a linear projection,t and xt 1;xt 2;must not be correlated with one another so that we have, E(txt j ) = 0 for j 1 (1) And since t s = xt s P[xt s j xt s 1;], then we get E(tt s) = E t(xt s P[xt s j xt s 1;]) = E(txt s) E(t P[xt s j xt s 1;]) = 0 from Eq.(1) (2) Hence we have proved thatftg is serially uncorrelated process. Now consider the projection ofxt against t;t 1;;t m for sufficiently largem. Letting ˆxt (m) denote the linear projection ofxt againstt;;t m, the typical projection ofxt is given by, ˆxt (m) = mX j =0 dj t j Applying Hamilton[1994](Chap4, Eq.(4.1.13)) to this problem and noticing thatt are serially uncorre- lated, each coefficientsdj is given by, dj = E(xtt j ) E(2 t) (3) Now since t = xt P[xt j xt 1;]; E(txt j ) = 0(j 1), we have E( 2 t) = E t(xt P[xt j xt 1;]) = E(txt) E(t P[xt j xt 1;]) = E(txt) from Eq.(1) ) d0 = E(xtt) E(2 t) = E( 2 t) E(2 t) = 1 (4) 1 And letting E( 2 t) = 2, the variance of the prediction error can be calculated as, E xt mX j =0 dj t j 2 = E