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The Wold Decomposition , Study notes of Time and Motion Studies

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The Wold Decomposition
Herman J. Bierens
February 1, 2012
Abstract
In Chapter 7 in Bierens (2004) the Wold decomposition was motivated
by claiming that every zero-mean covariance stationary process Xtcan
be written as Xt=P
j=1 βjXtj+Ut,where E[Ut.Xtj]=0for all
j1,and P
j=1 βjXtjis the projection of Xton its past. However,
in general this claim is incorrect. In this note I will give a more general
(and hopefully correct) proof of the Wold decomposition.
1 Projections on spaces spanned by a sequence
The fundamental projection theorem states that:
Theorem 1. Given a sub-Hilbert space Sof a Hilbert space Hand an
element yH,there exists a unique element bySsuch that ||yby|| =
infzS||yz||.Moreover the residual u=ybyis orthogonal to any zS:
hu, zi=0.
Proof : See for example Bierens (2004, Th. 7.A.3, p. 202).
This result is the basis for the famous Wold (1938) decomposition for
covariance stationary time series, which in its turn is the basis for time series
analysis.
Thanks to Peter Boswijk (University of Amsterdam) for pointing out an error in a
previous version of this note. Moreover, the queries of the students in my graduate time
series courses have led to substantial improvements of the proof of the Wold decomposition.
1
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pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a

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The Wold Decomposition

Herman J. Bierens

February 1, 2012

Abstract

In Chapter 7 in Bierens (2004) the Wold decomposition was motivated

by claiming that every zero-mean covariance stationary process Xt can

be written as Xt =

P∞

j=1 βj^ Xt−j^ +^ Ut,^ where^ E[Ut.Xt−j^ ] = 0^ for all j ≥ 1 , and

P∞

j=

βj Xt−j is the projection of Xt on its past. However,

in general this claim is incorrect. In this note I will give a more general

(and hopefully correct) proof of the Wold decomposition.

1 Projections on spaces spanned by a sequence

The fundamental projection theorem states that:

Theorem 1. Given a sub-Hilbert space S of a Hilbert space H and an

element y ∈ H, there exists a unique element by ∈ S such that ||y − by|| =

infz∈S ||y − z||. Moreover the residual u = y − by is orthogonal to any z ∈ S:

hu, zi = 0.

Proof : See for example Bierens (2004, Th. 7.A.3, p. 202).

This result is the basis for the famous Wold (1938) decomposition for

covariance stationary time series, which in its turn is the basis for time series

analysis.

∗ Thanks to Peter Boswijk (University of Amsterdam) for pointing out an error in a

previous version of this note. Moreover, the queries of the students in my graduate time

series courses have led to substantial improvements of the proof of the Wold decomposition.

The proof of the Wold decomposition in Anderson (1994) is more trans-

parent than the original proof by Wold (1938). However, rather than follow-

ing Anderson’s proof, I will in this note derive first a general Wold decompo-

sition for a regular sequence

1 in a general Hilbert space, and then specialize

this result to the Wold decomposition for covariance stationary time series.

First, we need to define sub-Hilbert spaces spanned by a sequence in a

Hilbert space, as follows.

Let {xk}

∞ k=1 be a sequence of elements of a Hilbert space^ H,^ and let

Mm = span({xj }

m j=1)

be the space spanned by x 1 , ..., xm, i.e., Mm consists of all linear combinations

of x 1 , ..., xm. Then

Lemma 1. Mm is a Hilbert space.

Proof : Without loss of generality we may assume that the m × m matrix

Σm with elements hxi, xj i , i, j = 1, ..., m, is non-singular, as otherwise we can

remove one or more xj ’s from the list {xj }

m j=1 and still span the same space.

For example, suppose that rank(Σm) = m − 1 , and let c = (c 1 , ..., cm)

0 be

the eigenvector corresponding to the zero eigenvalue. Then

Pm

j=

cj xj

2

c

0 Σmc = 0, hence

Pm

j=

cj xj = 0 (the latter being the zero element of Mm).

Since at least one component of c is non-zero, for example ci, we can write

xi =

Pm

j=2(cj^ /c^1 )xj^ if^ i^ = 1,

Pm− 1

j=1 (cj^ /cm)xj^ if^ i^ =^ m,

Pi− 1

j=1(cj^ /ci)xj^ −^

Pm

j=i+1(cj^ /ci)xj^ if^1 < i < m,

so that

Mm =

span({xj }

m j=2)^ if^ i^ = 1,

span({xj }

m− 1 j=1 )^ if^ i^ =^ m,

span(x 1 , ..., xi− 1 , xi+1, ..., xm) if 1 < i < m.

Now let zn =

Pm

j=1 βj,nxj^ be a Cauchy sequence in^ Mm,^ and denote

βn = (β 1 ,n, ..., βm,n)

0

. Then for each j, βj,n is a Cauchy sequence in R because

0 = lim min(n 1 ,n 2 )→∞

||zn 1 − zn 2 ||

2 = lim min(n 1 ,n 2 )→∞

m X

j=

(βj,n 1 − βj,n 2 ) xj

2

(^1) See Definition 4 below.

assumption implies that for each n we can select a zn ∈ Mn such that

lim n→∞

||bz − zn|| = 0. (1)

Let ||z − bz|| = δ and ||z − bzn|| = δn, and note that δn ≥ δ. Since

δ

2 n =^ ||z^ −^ zbn||

2 ≤ ||z − zn||

2 = ||z − zb + bz − zn||

2

= ||z − zb||

2

  • ||bz − zn||

2

  • 2 hz − z,b bz − zni

= δ

2

  • ||bz − zn||

2

it follows from (1) that

lim n→∞

δn = δ. (2)

Recall that z = bz + u, where hu, xi = 0 for all x ∈ M∞. Hence

||bz − bzn||

2 = ||z − zbn − u||

2 = ||z − bzn||

2

  • ||u||

2 − 2 hz − zbn, ui

= ||z − zbn||

2

  • ||u||

2 − 2 hz, ui = δ

2 n −^ δ

2 (3)

where the last equality follows from

hz, ui − hu, ui = hbz, ui = 0 (4)

and hu, ui = ||u||

2 = δ

2

. The theorem now follows from (2) and (3). Q.E.D.

Remark 1. Although each projection bzn is a linear combination of x 1 , ..., xn,

in general the result of Theorem 2 does not imply that there exists a sequence

{θj }

∞ j=1 such that^ bz^ =^

P∞

j=

θj xj.

As a counter example, consider the Hilbert space R 0 of zero-mean ran-

dom variables with finite second moments, endowed with the inner product

hX, Y i = E[X.Y ] and associated norm and metric. Let

Xt = Vt − Vt− 1 ,

where the Vt’s are independent N(0, 1) distributed. This is clearly a zero-

mean covariance stationary process, with covariance function γ(0) = 2, γ(1) =

− 1 , γ(m) = 0 for m ≥ 2. Hence Xt ∈ R 0 for all t.

For given t, let

M

t− 1 −∞ =^ span^ ({Xt−m}

∞ m=1)^ ,^ M

t− 1 t−n =^ span^ (Xt−^1 , ...., Xt−n)^.

The projection Xbt,n of Xt on M

t− 1 t−n takes the form

X^ b t,n =

X^ n

j=

θn,j Xt−j

where the coefficients θn,j are the solutions of the normal equations

γ(m) =

X^ n

k=

γ(|k − m|)θn,k, m = 1, ..., n.

hence for n ≥ 3 ,

− 1 = 2 .θn, 1 − θn, 2

0 = −θn, 1 + 2θn, 2 − θn, 3

0 = −θn, 2 + 2θn, 3 − θn, 4

0 = −θn,n− 2 + 2θn,n− 1 − θn,n

0 = −θn,n− 1 + 2θn,n

The solutions of these normal equations are

θn,j =

j

n + 1

− 1 , j = 1, ...., n,

hence

X^ b t,n =

n X

j=

μ j

n + 1

Xt−j (5)

Next, let Xbt be the projection of Xt on M

t− 1 −∞,^ and suppose that there

exists a sequence {θj }

∞ j=1 such that^ Xb t =^

P∞

j=1 θj^ Xt−j^.^ Note that the latter

is merely a short-hand notation for

lim n→∞

Xb t −

X^ n

j=

θj Xt−j

2

= lim n→∞

E

Ã

Xb t −

X^ n

j=

θj Xt−j

If so, it follows from Theorem 2 and (5) that

0 = lim n→∞

E

Ã

Xn

j=

θj Xt−j −

X^ n

j=

μ j

n + 1

Xt−j

2 Projections on the span of an orthonormal

sequence

On the other hand,

Theorem 3. If a sequence {xj }

∞ j=1 in a Hilbert space^ H^ is orthonormal, i.e.,

hxi, xj i =

1 if i = j,

0 if i 6 = j,

then any projection bz of z ∈ H on span({xj }

P^ j=1)^ takes the form^ bz^ = ∞ j=1 θj^ xj^ (in the sense that^ limn→∞^ ||bz^ −

Pn

j=1 θj^ xj^ ||^ = 0),^ where^ θj^ =^ hz, xj^ i

and

P∞

j=

θ

2 j <^ ∞.

Proof : Let bzn be the projection of z on span({xj }

n j=1).^ Then

||z − bzn||

2 = min c 1 ,...,cn

z −

n X

j=

cj xj

2

= min c 1 ,...,cn

||z||

2 − 2

X^ n

j=

cj hz, xj i +

X^ n

i=

X^ n

j=

cicj hxi, xj i

= min c 1 ,...,cn

||z||

2 − 2

n X

j=

cj hz, xj i +

n X

j=

c

2 j

hence,

bzn =

n X

j=

θj xj , where θj = hz, xj i. (10)

Moreover, denoting un = z − zbn, it follows from (9) and (10) that

||un||

2

z −

X^ n

j=

θj xj

2

= ||z||

2 − 2

X^ n

j=

θj hz, xj i +

X^ n

j=

X^ n

i=

θj θi hxj , xii

= ||z||

2 −

n X

j=

θ

2 j ≥^0 (11)

so that

Pn

j=1 θ

2 j ≤^ ||z||

2 for all n and thus

P∞

j=1 θ

2 j <^ ∞.^ Finally, it follows

from Theorem 2 that

lim n→∞

bz −

X^ n

j=

θj xj

= lim n→∞

||bz − zbn|| = 0.

Q.E.D.

3 The general Wold decomposition

3.1 Preliminary definitions and results

Let S 1 , S 2 be a pair of subspaces of a Hilbert space H. We say that:

Definition 2. S 1 and S 2 are orthogonal, denoted by S 1 ⊥S 2 , if for each

x 1 ∈ S 1 and each x 2 ∈ S 2 , hx 1 , x 2 i = 0.

Lemma 4. Let S 1 and S 2 be sub-Hilbert spaces satisfying S 1 ⊥S 2. Then

span(S 1 , S 2 ) = {y = x 1 + x 2 : x 1 ∈ S 1 , x 2 ∈ S 2 }

is a Hilbert space.

Proof : Let yn be a Cauchy sequence in span(S 1 , S 2 ). Then yn = x 1 ,n+x 2 ,n,

where x 1 ,n ∈ S 1 and x 2 ,n ∈ S 2. Since x 1 ,n − x 1 ,m ∈ S 1 and x 2 ,n − x 2 ,m ∈ S 2 it

follows from the orthogonality condition S 1 ⊥S 2 that

||yn − ym||

2 = ||(x 1 ,n − x 1 ,m) + (x 2 ,n − x 2 ,m)||

2

= ||x 1 ,n − x 1 ,m||

2

  • ||x 2 ,n − x 2 ,m||

2

+2 hx 1 ,n − x 1 ,m, x 2 ,n − x 2 ,mi

= ||x 1 ,n − x 1 ,m||

2

  • ||x 2 ,n − x 2 ,m||

2 ,

hence limmin(n,m)→∞ ||yn −ym|| = 0 implies that limmin(n,m)→∞ ||x 1 ,n −x 1 ,m|| =

0 and limmin(n,m)→∞ ||x 2 ,n − x 2 ,m|| = 0. Because S 1 and S 2 are Hilbert spaces

there exist an x 1 ∈ S 1 and an x 2 ∈ S 2 such that limn→∞ ||x 1 ,n − x 1 || = 0 and

limn→∞ ||x 2 ,n − x 2 || = 0, hence limn→∞ ||yn − y|| = 0, where y = x 1 + x 2 ∈

span(S 1 , S 2 ). Q.E.D.

in the sense that limn→∞ kx − w −

Pn

k=1 αkekk^ = 0,^ where^ {ek}

∞ k=1 is an

orthonormal sequence in M∞, αk = hx, eki ,

P∞

k=

α

2 k <^ ∞,^ and

w ∈ S∞ ∩ U

⊥ ∞,^ (13)

with S∞ = ∩

∞ n=1span({xk}

∞ k=n)^ and^ U

⊥ ∞ the orthogonal complement of^ U∞^ =

span({ek}

∞ k=1).^ Note that^ (13)^ implies that^ w^ is orthogonal to all the^ ek’s:

hek, wi = 0 for k = 1, 2 , 3 , ....

Proof : Denote

Sn = span({xk}

∞ k=n).

Note that M∞ = S 1. Project each xk on Sk+1, so that xk = bxk + uk with

projection bxk ∈ Sk+1 and residual uk. Recall that by the regularity condition,

||uk|| > 0 , hence ek = uk/||uk|| is well defined. It is not hard to verify that

the residuals uk are orthogonal, so that the ek’s are orthonormal, and that

U∞ ⊂ M∞. It follows now from Theorem 3 that (12) holds with αk = hx, eki , P∞

k=1 α

2 k <^ ∞, and^ w^ ∈^ U

⊥ ∞,^ where the latter follows from the fact that^ w

the residual of the projection of x on U∞. Therefore, the actual contents of

Theorem 4 is that w ∈ S∞.

The theorem under review will be proved in six steps:

Step 1. As before, let Mn = span({xk}

n k=1).^ I will show^ first that

Mn ⊂^ span^

Un, U

⊥ n ∩^ S^2

[c.f. Lemma 4], where Un = span(e 1 , ..., en) = span(u 1 , ..., un) and U

⊥ n is the

orthogonal complement of Un.

Proof. Let z ∈ Mn be arbitrary. Recall that z takes the form z = Pn

k=1 ckxk.^ Substituting^ xk^ =^ bxk^ +^ uk^ =^ bxk^ +^ ||uk||ek^ we can write^ z^ as

z =

n X

k=

ck (bxk + uk) =

n X

k=

ckuk +

n X

k=

ck xbk

n X

k=

ck||uk||ek +

n X

k=

ck bxk

n X

k=

ck||uk||ek + z 2

where

z 2 =

n X

k=

ck bxk

Note that

z 2 =

n X

k=

ck bxk ∈ S 2 (15)

because xbk ∈ Sk+1 ⊂ S 2 for k = 1, 2 , ..., n.

Next, project z 2 on Un. This projection takes the form

bpn =

n X

k=

dkek, where dk = hz 2 , eki ,

with residual

wn+1 ∈ U

⊥ n.^ (16)

However, e 1 is orthogonal to any element of S 2 , and z 2 ∈ S 2. Therefore,

d 1 = hz 2 , e 1 i = 0 and thus

bpn =

X^ n

k=

dkek ∈ span ({ek}

n k=2)^ ⊂^ S^2 ,

where the latter follows from ek ∈ Sk ⊂ S 2 for k = 2, 3 , ..., n. Because

wn+1 = z 2 − pbn where both terms are elements of S 2 , it follows that

wn+1 ∈ S 2. (17)

Combining (16) and (17) now yields

wn+1 ∈ U

⊥ n ∩^ S^2.

Thus, denoting α 1 = c 1 ||u 1 ||, αk = ck||uk|| + dk for k = 2, 3 , ..., n, we can

write

z =

n X

k=

αkek + wn+1, where wn+1 ∈ U

⊥ n ∩^ S^2 ,

hence z ∈ span(Un, U

⊥ n ∩^ S^2 ).^ This proves (14).

Step 2. Next, it will be shown that

span

Un, U

⊥ n ∩^ S^2

= span

Un, U

⊥ n ∩^ Sn+

which implies (20).

However, Sn+1,m ⊂ S 2 ,m and therefore

U

⊥ n ∩^ Sn+1,m^ ⊂^ U

⊥ n ∩^ S^2 ,m.^ (21)

Combining (20) and (21) now yields

U

⊥ n ∩^ S^2 ,m^ =^ U

⊥ n ∩^ Sn+1,m^ for^ m > n,

which in its turn implies that

U

⊥ n ∩^

∞ m=n+1S^2 ,m

= U

⊥ n ∩^

∞ m=n+1Sn+1,m

Finally, note that S 2 = ∪

∞ m=n+1S^2 ,m^ and^ Sn+1^ =^ ∪

∞ m=n+1Sn+1,m,^ hence it

follows from (22) that (18) holds.

Step 3. Denote Rn = span

Un, U

⊥ n ∩^ Sn+

. Then

M∞ = ∪

∞ n=1Rn.^ (23)

Proof. It follows from (19) that Mn ⊂ Rn, hence

M∞ = ∪

∞ n=1Mn^ ⊂ ∪

∞ n=1Rn.^ (24)

However, we also have Rn ⊂ M∞, as is not hard to verify, hence

∞ n=1Rn^ ⊂^ M∞.^ (25)

Thus, the result (23) follows from (24) and (25).

Step 4. For an x ∈ M∞, let bxn be the projection of x on Rn. Then

bxn =

n X

j=

αj ej + wn+1 (26)

where αj = hx, ej i and wn+1 is the projection of x on U

⊥ n ∩^ Sn+1.^ Moreover,

X^ ∞

j=

α

2 j <^ ∞.^ (27)

Furthermore,

lim n→∞

x −

n X

j=

αj ej − wn+

Proof. By the definition of Rn and Lemma 4, bxn =

Pn

j=1 θj^ ej^ +^ w^ for some

constants θj and a w ∈ U

⊥ n ∩^ Sn+1.^ To determine the^ θj^ ’s and^ w,^ note that

x −

X^ n

j=

θj ej − w

2

= ||x − w||

2 − 2

X^ n

j=

θj hej , xi + 2

X^ n

j=

θj hej , wi

n X

j=

θj ej

2

= ||x − w||

2 − 2

n X

j=

θj hej , xi +

n X

j=

θ

2 j

because w ∈ U

⊥ n ∩^ Sn+1^ ⊂^ U

⊥ n implies^ hej^ , wi^ = 0^ and

X^ n

j=

θj ej

2

X^ n

j=

X^ n

i=

θj θi hej , eii =

X^ n

j=

θ

2 j hej^ , ej^ i^ =

X^ n

j=

θ

2 j.

Thus

||x − bxn||

2 = inf θ 1 ,...,θn,w∈U n⊥ ∩Sn+

x −

X^ n

j=

θj ej − w

2

= inf θ 1 ,...,θn,w∈U n⊥ ∩Sn+

Ã

||x − w||

2 − 2

n X

j=

θj hej , xi +

n X

j=

θ

2 j

= inf w∈U n⊥ ∩Sn+

||x − w||

2 −

n X

j=

α

2 j

= ||x − wn+1||

2 −

n X

j=

α

2 j (29)

where αj = hx, ej i and wn+1 is the projection of x on U

⊥ n ∩^ Sn+1.

as min(m, n) → ∞. Therefore, there exists a w ∈ U

⊥ k ∩^ Sk+1^ such that

(32) holds. Since k was arbitrary we now have w ∈ ∩

∞ k=1U

⊥ k =^ U

⊥ ∞ and

w ∈ ∩

∞ k=1Sk+1^ =^ S∞,^ hence

w ∈ U

⊥ ∞ ∩^ S∞.

This completes the proof of Step 6.

The theorem now follows from (27), (31), (32) and the fact that w ∈

U

⊥ ∞ ∩^ S∞^ ⊂^ U

⊥ ∞,which implies that^ hw, eki^ = 0^ for^ k^ = 1,^2 ,^3 , ........^ Q.E.D.

4 The Wold decomposition for covariance sta-

tionary time series

In the case of the Hilbert space R 0 of zero-mean random variables with finite

second moments, with inner product hX, Y i = E[X.Y ] and associated norm

and metric, the results of Theorem 4 translate as follows:

Theorem 5. Let Xt be a regular univariate zero-mean covariance stationary

time series process. Then Xt can be written as

Xt =

∞ X

j=

αj Ut−j + Wt a.s., (33)

where the Ut is a zero-mean uncorrelated process with variance 1 ,

αj = E[XtUt−j ],

X^ ∞

j=

α

2 j <^ ∞,^ (34)

and Wt is a zero-mean covariance stationary process satisfying

Wt ∈ U

⊥ t ∩^ S−∞,^ (35)

where S−∞ = ∩nspan({Xn−k}

∞ k=1)^ and^ U

⊥ t is the orthogonal complement of

Ut = span({Ut−k}

∞ k=0)^.^ The result^ (35)^ implies that

Wt ∈ span ({Wt−m}

∞ m=1)^ ,^ (36)

which in its turn implies that Wt is perfectly predictable from its past values

Wt− 1 , Wt− 2 , Wt− 3 , ..... In other words, Wt is a deterministic process. More-

over, (35) implies that

E[WtUt−m] = 0 (37)

for all leads and lags m.

Proof : Recall that Ut = Uet/

r

E

h

Ue^2 t

i

, where Uet = Xt − Xbt with Xbt the

projection of Xt on span({Xt−j }

∞ j=1).^ The uncorrelatedness of the^ Ue t’s follows

from Theorem 4, but we still need to show that E[ Uet] = 0 and E[ Ue

2 t ] =^ σ

2

for all t.

Proof of E[ Uet] = 0

Let Xbt,n be the projection of Xt on span({Xt−j }

n j=1).^ Then^ Xb t,n takes the

form

X^ b t,n =

n X

j=

βj,nXt−j ,

where the βj,n’s do not depend on t. The latter follows from the fact that

the βj,n’s are the solutions of the normal equations

n X

j=

βj,nγ(i − j) = γ(i), i = 1, 2 , ..., n,

where γ(i) = E [XtXt−i] is the covariance function of Xt. Hence E[ Xbt,n] = 0.

It follows from Theorem 2 that

lim n→∞

° Xbt,n − Xbt

2

= lim n→∞

E

Xb t,n −^ Xb t

so that by Liapounov’s inequality and E[ Xbt,n] = 0,

lim n→∞

¯E[ Xbt]

¯ = lim n→∞

¯E[ Xbt − Xbt,n]

¯ ≤ lim n→∞

E

h¯ ¯ ¯ Xbt − Xbt,n

i

s

lim n→∞

E

Xb t,n −^ Xb t

Thus E[ Xbt] = 0 and therefore E[ Uet] = E[Xt − Xbt] = 0.

Proof of (34), (35) and (37)

The result of Theorem 4 can now be translated as

lim n→∞

Xt −

X^ n

j=

αj Ut−j − Wt

where Ut is a zero-mean uncorrelated covariance stationary process with unit

variance, and αk = hXt, Ut−ki = E [XtUt−k] with

P∞

k=

α

2 k <^ ∞.

We still need to prove that the αk’s do not depend on t, as follows. Recall

from the proof of E[ Ue

2 t ] =^ σ

2 that Uet,n = Xt −

Pn

j=

βj,nXt−j , so that

E

h

Xt+k Uet,n

i

= γ (k) −

X^ n

j=

βj,nγ(k + j),

which does not depend on t. Moreover, by the Cauchy-Schwarz inequality

and (39),

lim n→∞

¯E

h

Xt+k

Ue t,n −^ Ue t

´i¯ ¯ ¯

2

≤ γ (0) lim n→∞

E

Ue t,n −^ Ue t

Thus E

h

Xt+k Uet

i

= limn→∞ E

h

Xt+k Uet,n

i

. Since the latter does not depend

on t, neither does αk = E [Xt+kUt] = E

h

Xt+k

e Ut/|| Uet||

i

.

The results (35) and (37) follow straightforwardly from Theorem 4.

Proof of (33)

The result (40) implies, by Chebyshev’s inequality, that

Xt = p lim n→∞

n X

j=

αj Ut−j + Wt. (41)

Recall that convergence in probability for n → ∞ is equivalent to a.s.

convergence along a further subsequence km of an arbitrary subsequence of

n. See for example Bierens (2004, Theorem 6.B.3, p. 168). Thus for such a

subsequence km,

Xkm

j=

αj Ut−j → Xt − Wt a.s. (42)

as m → ∞, and the same holds for any further subsequence of km.

Without loss of generality we may choose k 0 = 0. Then for each n > 0 we

can find an mn such that

kmn− 1 < n ≤ kmn. (43)

Moreover, (42) implies that

kmn X

j=

αj Ut−j → Xt − Wt a.s. as n → ∞. (44)

Due to (43),

∞ X

n=

E

Ã

kmn X

j=

αj Ut−j −

n X

j=

αj Ut−j

∞ X

n=

E

Ã

kmn X

j=n+

αj Ut−j

∞ X

n=

kmn X

j=kmn− 1 +

α

2 j ≤

∞ X

j=

α

2 j <^ ∞,

so that by Chebyshev’s inequality, for arbitrary ε > 0 ,

∞ X

n=

P

kmn X

j=

αj Ut−j −

n X

j=

αj Ut−j

ε

This result implies, by the Borel-Cantelli lemma,

2 that

kmn X

j=

αj Ut−j −

X^ n

j=

αj Ut−j → 0 a.s. as n → ∞. (45)

Combining (44) and (45) it follow now that

n X

j=

αj Ut−j → Xt − Wt a.s. as n → ∞. (46)

Since

P∞

j=0 αj^ Ut−j^ is defined as^ limn→∞

Pn

j=0 αj^ Ut−j^ ,^ (33) is equivalent to

(46).

(^2) See for example Bierens (2004, Theorem 6.B.2, p. 168).