資料紹介

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

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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;] i...

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