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Not-for-Publication Appendix to
Evaluating Direct Multi–Step Forecasts *
Todd E. Clark Federal Reserve Bank of Kansas City
Michael W. McCracken Board of Governors of the Federal Reserve System
October 2005
Abstract
This not-for-publication appendix contains proofs of Theorems 3.1 - 3.4 as discussed in the text. It also contains lemmas used to prove the theorems.
*Clark (corresponding author): Economic Research Dept.; Federal Reserve Bank of Kansas City; 925 Grand; Kansas City, MO 64198; [email protected]. McCracken: Board of Governors of the Federal Reserve System; 20th and Constitution N.W.; Mail Stop #61; Washington, D.C. 20551; [email protected]. Earlier versions of this paper were titled “Evaluating Long–Horizon Forecasts.” McCracken thanks LSU and the University of Missouri–Columbia for financial support during work on substantial portions of this paper. The views expressed herein are solely those of the authors and do not necessarily reflect the views of the Federal Reserve Bank of Kansas City, Board of Governors, Federal Reserve System, or any of its staff.
1
0. Table of Contents
Section Page Number
0. Table of Contents 1
1. Introduction 2
2. Notation 2
3. Statistics 3
4. Assumptions 4
5. Lemmas A1 - A13
A1 5
A2 6
A3 7
A4 10
A5, A6 11
A7, A8 14
A9 15
A10 20
A11 21
A12 24
A13 25
6. Theorems 3.1 – 3.4
Theorem 3.1 29
Theorem 3.2 30
Theorem 3.3 31
Theorem 3.4 32
7. References 33
8. Referenced Lemmas and Theorems 34
9. Summary Table of Limiting Distributions 36
2
1. Introduction
This not-for-publication appendix contains proofs of Theorems 3.1 - 3.4 as discussed in the
text. It also contains lemmas used to prove the theorems. Herein we focus on an environment in
which the models are linear and either 0 < < or = 0. The proofs provide results for each of
the recursive, rolling and fixed schemes. Note that, for simplicity, the P + 1 terms that
appear in the text formulas are replaced by P in the theoretical results below, without any
consequence. Throughout, the null is maintained.
2. Notation
The following notation will be used. Forecasts of the scalar yt+ , t = R,…,T- , 1 < are
generated using a (k1 + k2 = k 1) vector of covariates x2,t = ' ' '1,t 22,t(x ,x ) and two linear parametric
models, ' *i,t ix , i = 1,2, each of which is estimated. Under the null, model 2 is unrestricted and
nests the restricted model 1. We denote the -step ahead forecast errors as 1,t+u = 't+ 1,t 1,t
ˆy -x
and 2,t+u = 't+ 2,t 2,t
ˆy -x for models 1 and 2, respectively.
Let i,t+ ih ( ) = 't+ i,t i i,t(y -x )x , hi,t+ = *
i,t+ ih ( ) , qi,t = 'i,t i,tx x and Bi = -1
i,t(Eq ) . Let J denote
the selection matrix 'k k k k1 1 1 2
(I ,0 ) , supt denote R t Tsup , and for matrices A and C defined in
Lemma A4, 2,t+h = -1 ' 1/22 2,t+A CB h and 2H (t) = -1 ' 1/2
2 2A CB H (t) . For any (m n) matrix G with
elements gi,j and column vectors gj let vec(G) denote the (mn 1) vector ' ' ' '1 2 n[g ,g ,...,g ] and let |G|
denote i, j i, jmax | g | . For the sequence Ut+ defined in Assumption 2, U(t) is defined analogous to
H(t) in Assumption 1. For tc = 21,t+ 1,t+ 2,t+ˆ ˆ ˆu -u u , td = 2 2
1,t+ 2,t+ˆ ˆu -u , c = -1 T-t=R+j t+ˆ(P- +1) c , d
= -1 T-t=R+j t+
ˆ(P- +1) d and 0 j j < , ccˆ (j) = -1 T-
t=R+j t+ t+ -jˆ ˆ(P- +1) (c -c)(c -c) , ddˆ (j) =
3
-1 T-t=R+j t+ t+ -j
ˆ ˆ(P- +1) (d -d)(d -d) , ccˆ (j) = '
ccˆ (-j) and dd
ˆ (j) = 'dd
ˆ (-j) . Throughout, t tz
denotes T-t=R tz and Szz denotes limVar( 1/ 2
t tP z ) for any sequence of variables zt.
In the text we use the index s to denote the time argument of Brownian motion W(s). In this
appendix we instead use . We do so in order to preserve s as a generic index on summations
throughout the appendix. Having made that change, let W( ; ) denote a vector Brownian
motion with covariance kernel , W( ; I) = W( ) and let denote weak convergence.
3. Statistics
For ease of reference, the encompassing and forecast accuracy statistics are presented below.
MSE-T = 1/2
jddj=-j
d(P- +1)ˆK(j/M) (j)
= 2 2
t 1,t+ 2,t+
2 2 2 2j T-t=R+j 1,t+ 2,t+ 1,t+ j 2,t+ jj=- j
ˆ ˆ(u -u )
ˆ ˆ ˆ ˆK(j/M)[ (u -u -d)(u -u -d)]
ENC-T = 1/2
jccj=-j
c(P- +1)ˆK(j/M) (j)
= 2
t 1,t+ 1,t+ 2,t+
2 2j T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ j 1,t+ j 2,t+ jj=-j
ˆ ˆ ˆ(u -u u )
ˆ ˆ ˆ ˆ ˆ ˆK(j/M)[ (u -u u -c)(u -u u -c)]
MSE-F = 2
d(P- +1)×MSE
=2 2
t 1,t+ 2,t+-1 2
t 2,t+
ˆ ˆ(u -u )ˆ(P- +1) u
(or 1/2
2
dR( ) ×(P- +1)×P- +1 MSE)
ENC-NEW = 2
c(P- +1)×MSE
= 2
t 1,t+ 1,t+ 2,t+-1 2
t 2,t+
ˆ ˆ ˆ(u -u u )ˆ(P- +1) u
(or 1/2
2
cR( ) ×(P- +1)×P- +1 MSE).
4
4. Assumptions
Assumption 1: The parameter estimates i,tˆ , i = 1,2, t = R,…,T- , satisfy *
i,t iˆ - = i iB (t)H (t) where
i iB (t)H (t) equals -1 -1 -1t- t-s=1 s=1i,s i,s+(t q ) (t h ) , -1 -1 -1t- t-
s=t-R+1 s=t-R+1i,s i,s+(R q ) (R h ) ,
-1 -1 -1R- R-s=1 s=1i,s i,s+(R q ) (R h ) for the recursive, rolling and fixed schemes respectively.
Assumption 2: (a) Ut+ = ' ' ' 't+ 2,t 2,t 2,t+[u ,x -Ex ,h ] is covariance stationary, (b) EUt+ = 0, (c) Eq2,t <
and is positive definite, (d) For some r > 8, Ut+ is uniformly Lr bounded, (e) For some r > d >
2, Ut+ is strong mixing with coefficients of size rd/(r d), (f) With t+U denoting the vector of
nonredundant elements of Ut+ , -1 'T- T-s=1 s=1T s+ s+lim T E( U )( U ) = < is positive definite.
Assumption 3: (a) Let K(x) be a continuous kernel such that for all real scalars x, |K(x)| 1, K(x)
= K(-x) and K(0) = 1, (b) For some bandwidth M and constant i (0, 0.5), M = O(Pi), (c) For all
j > 1, '2,t+ 2,t+ -jEh h = 0, (d) The number of covariances j , used to estimate the long-run
covariances Scc and Sdd defined in Section 3.1, satisfies 1 j .
Assumption 4: P,Rlim P/R = (0, ), with = -1(1+ ) .
Assumption 4 : P,Rlim P/R = 0, with = 1.
5
5. Lemmas
Lemma A1: Let Assumptions 1, 2 and either 4 or 4 hold. For each i = 1,2, (a) 1/2tsup T |U(t)| =
Op(1), (b) 1/2t i isup T |vec[B (t)]-vec[B ]| = Op(1), (c) 1/2 *
t i,t iˆsup T | - | = Op(1).
Proof of Lemma A1: (a) We will show this for the recursive scheme. The fixed scheme follows
immediately from the recursive and the rolling follows from a proof similar to that for the
recursive. Note that by definition T1/2U(t) = -1/2 t-s=1 s+(T/t)T U . Recall that |.| denotes the max
norm and hence for t+U defined in Assumption 2, |Ut+ | = | t+U |. Therefore suptT1/2|U(t)|
(T/R) -1/2 t-s=1t s+sup |T U |. Both Assumption 4 and 4 imply that T/R is bounded. Given
Assumption 2 and Corollary 29.19 of Davidson (1994) we know that -1/2 t-s=1 s+T U 1/2W( ).
Following arguments similar to those in Lemma 2.1 of Corradi, Swanson and Olivetti (2001) we
obtain -1/2 t-s=1t s+sup |T U | d
1/ 21sup | W( ) | = Op(1) and the proof is complete.
(b) We will show this for the recursive scheme. The fixed scheme follows immediately and the
rolling follows from a proof similar to that for the recursive.
First note that -1 -1i if(vec[B (t)])-f(vec[B ]) = i ivec[B (t)]-vec[B ] for a continuously
differentiable function f(.) with f ( )/ f( ) f ( ). Second note that there exists an open
neighborhood -1iN(vec[B ]) of -1
ivec[B ] = i,tvec[Eq ] and a finite positive constant D such that
-1N(vec[B ])isup |f ( )| < D. Taking a first order Taylor expansion of f(.) around -1
ivec[B ] we obtain
1/2t i isup T |vec[B (t)]-vec[B ]| = 1/2 ' -1 -1
t t i isup T |f ( ) (vec[B (t)]-vec[B ])|
= ' -1/2 t-s=1t t i,s i,ssup |f ( ) ((T/t)T vec[q -Eq ])|
2 -1/2 t-s=1t t t i,s i,sk (T/R)(sup |f ( )|)(sup |T vec[q -Eq ]|)
6
for some t with elements that lie on the line between the corresponding elements of -1ivec[B (t)]
and -1ivec[B ] . Both Assumption 4 and 4 imply that T/R is bounded. That
-1/2 t-s=1t i,s i,ssup |T vec[q -Eq ]| = Op(1) follows from Lemma A1 (a). The result will follow if
t tsup |f ( )| = Op(1). By Lemma A1 (a) we know that -1ivec[B (t)] a.s.
-1ivec[B ]and hence for all
> 0 there exists R sufficiently large that -1t iProb( N(vec[B ])) > 1 . This in turn implies
that for large enough R, t tsup |f ( )| -1N(vec[B ])isup |f ( )| < D with probability greater than 1
and the proof is complete.
(c) If we add and subtract Bi and apply the triangle inequality we obtain
1/2t i isup T |B (t)H (t)| 1/2
t i i t ik(sup |B (t)-B |)(sup T |H (t)|) + 1/2i t ik|B |(sup T |H (t)|) .
Lemma A1 (b) implies that t i isup |B (t)-B | = op(1). Since Lemma A1 (a) implies that
1/2t isup T |H (t)| = Op(1) the proof is complete.
Lemma A2: Let Assumptions 1, 2 and 4 hold. For i = 1,2, 't i,t+ i ih B (t)H (t) = '
t i,t+ i ih B H (t) +
op(1).
Proof of Lemma A2: Add and subtract Bi to obtain
(A1) 't i,t+ i ih B (t)H (t) = '
t i,t+ i ih B H (t) + 't i,t+ i i ih (B (t)-B )H (t) .
We must then show that the last right-hand side term in (A1) is op(1). We do so for the recursive
scheme; the arguments are similar for the other schemes. Note that
't i,t+ i i ih (B (t)-B )H (t) = -1/2 1/2 ' -1/2 -1/2 t-
t s=1i i i,t+ i,s+T (T/t)vec[T (B (t)-B )] [T h (T h )] .
7
Given Assumption 2, Lemmas A1 (a)-(b) and Corollary 29.19 of Davidson (1994), that
1/2 ' -1/2 -1/2 t-t s=1i i i,t+ i,s+(T/t)vec[T (B (t)-B )] [T h (T h )] is Op(1) follows from Theorem 4.1 of
Hansen (1992). Since T-1/2 = o(1) the proof is complete.
Lemma A3: Let Assumptions 1, 2 and 4 hold. For i,j = 1,2 and 0 m j
(a) ' 'T-t=R+m i i i,t+ j,t+ -m j jH (t)B (t)h h B (t-m)H (t-m) = ' '
t i i i,t+ j,t+ -m j jH (t)B E(h h )B H (t) + op(1).
(b) ' 't i i i,t j,t j jH (t)B (t)x x B (t)H (t) = ' '
t i i i,t j,t j jH (t)B E(x x )B H (t) + op(1).
Proof of Lemma A3: We will show the two results for the recursive scheme. Proofs for the
rolling and fixed schemes are similar. Since m and the arguments of the summations are finite,
we immediately dispense with the summation Tt R m (.) and replace it with t (.) . (a) The proof
is conducted in two stages. The first stage consists of showing that
' 't i i i,t+ j,t+ -m j jH (t)B (t)h h B (t-m)H (t-m) = ' '
t i i i,t+ j,t+ -m j jH (t)B h h B H (t) + op(1).
In this proof only let a1 = Bi, a2 = Bi(t) Bi, b1 = Bj, b2 = Bj(t m) Bj, c1 = Hj(t) and c2 =
Hj(t m) Hj(t). Using this notation, if we add and subtract Bi, Bj and Hj(t) we obtain the identity
(A2) ' 't i i i,t+ j,t+ -m j jH (t)B (t)h h B (t-m)H (t-m) = ' '
t i v i,t+ j,t+ -m w x1 v,w,x 2
[ H (t)a h h b c ].
Note that the outer summation in the right-hand side of (A2) contains 8 terms corresponding to
different combinations of (v, w, x). When v = w = x = 1 the argument takes the value
' 't i i i,t+ j,t+ -m j jH (t)B h h B H (t) . To obtain the result we must show that the remaining seven pieces
in (A2) are each op(1). The proof of each is very similar. Here we only show that the term
' 't i 2 i,t j,t m 2 2H (t)a h h b c = ' '
t i i i i,t+ j,t+ -m j j j jH (t)(B (t)-B )h h (B (t-m)-B )(H (t-m)-H (t)) is op(1).
8
Taking absolute values we immediately have
' 't i i i i,t+ j,t+ -m j j j j| H (t)(B (t)-B )h h (B (t-m)-B )(H (t-m)-H (t))|
4 -1 ' 1/4 1/4 1/4t i,t+ j,t+ m t i t j t jk (P/T)(P |h h |)(sup |T H (t)|)(sup |T H (t-m)|+sup |T H (t)|)
1/4 1/4t i i t j j(sup T |B (t)-B |)(sup T |B (t-m)-B |) .
Given Assumption 2, that -1 't i,t+ j,t+ -mP |h h | = Op(1) follows from Markov’s inequality. That
suptT1/4|Hi(t)|, suptT1/4|Hj(t)|, suptT1/4|Hj(t m)|, 1/4t i isup T |B (t)-B | and 1/4
t j jsup T |B (t-m)-B | are op(1)
follows from Lemma A1 (a)-(b) and the fact that m is finite.
The second stage of the proof consists of showing that
' 't i i i,t+ j,t+ -m j jH (t)B h h B H (t) = ' '
t i i i,t+ j,t+ -m j jH (t)B E(h h )B H (t) + op(1).
To do so add and subtract 'i,t+ j,t+ -mE(h h ) to obtain
(A3) ' 't i i i,t+ j,t+ -m j jH (t)B h h B H (t)
= ' 't i i i,t+ j,t+ -m j jH (t)B E(h h )B H (t) + ' ' '
t i i i,t+ j,t+ -m i,t+ j,t+ -m j jH (t)B (h h -E(h h ))B H (t) .
It then suffices to show that the second right-hand side term in (A3) is op(1). Rearranging terms
we obtain
' ' 't i i i,t+ j,t+ -m i,t+ j,t+ -m j jH (t)B (h h -E(h h ))B H (t)
= -1/2 2 -1/2 ' -1/2 ' -1/2 ' 't- t-t s=1 s=1j,s+ j i,s+ i i,t+ j,t+ -m i,t+ j,t+ -mT (T/t) [T h B T h B )]vec[T (h h -E(h h ))] .
That 2 -1/2 ' -1/2 ' -1/2 ' 't- t-t s=1 s=1j,s+ j i,s+ i i,t+ j,t+ -m i,t+ j,t+ -m(T/t) [T h B T h B )]vec[T (h h -E(h h ))] is Op(1)
follows from Assumption 2, Corollary 29.19 of Davidson (1994) and Theorem 4.1 of Hansen
(1992). Since T-1/2 is o(1) the proof is complete.
9
(b) The proof is conducted in two stages. The first stage consists of showing that
' 't i i i,t j,t j jH (t)B (t)x x B (t)H (t) = ' '
t i i i,t j,t j jH (t)B x x B H (t) + op(1).
In this proof only let a1 = Bi, a2 = Bi(t) Bi, b1 = Bj and b2 = Bj(t) Bj. Using this notation, if we
add and subtract Bi and Bj we obtain the identity
(A4) ' 't i i i,t j,t j jH (t)B (t)x x B (t)H (t) = ' '
t i v i,t j,t w j1 v,w 2
[ H (t)a x x b H (t)] .
Note that the outer summation in the right-hand side of (A4) contains four terms corresponding
to different combinations of (v, w). When v = w = 1 the argument takes the value
' 't i i i,t j,t j jH (t)B x x B H (t) . To obtain the result we must show that the remaining three pieces in
(A4) are each op(1). The proof of each is very similar. Here we only show that the term
' 't i 2 i,t j,t 2 jH (t)a x x b H (t) = ' '
t i i i i,t j,t j j jH (t)(B (t)-B )x x (B (t)-B )H (t) is op(1). Taking absolute
values we immediately have
' 't i i i i,t j,t j j j| H (t)(B (t)-B )x x (B (t)-B )H (t) |
4 -1 ' 1/4 1/4t i,t j,t t i t jk (P/T)(P |x x |)(sup |T H (t)|)(sup |T H (t)|)
1/4 1/4t i i t j j(sup T |B (t)-B |)(sup T |B (t)-B |) .
Given Assumption 2, that -1 't i,t j,tP |x x | = Op(1) follows from Markov’s inequality. That
suptT1/4|Hi(t)|, suptT1/4|Hj(t)|, 1/4t i isup T |B (t)-B | and 1/4
t j jsup T |B (t)-B | are op(1) follows from
Lemma A1 (a)-(b).
The second stage of the proof consists of showing that
' 't i i i,t j,t j jH (t)B x x B H (t) = ' '
t i i i,t j,t j jH (t)B E(x x )B H (t) + op(1).
10
To do so add and subtract 'i,t j,tE(x x ) to obtain
(A5) ' 't i i i,t j,t j jH (t)B x x B H (t)
= ' 't i i i,t j,t j jH (t)B E(x x )B H (t) + ' ' '
t i i i,t j,t i,t j,t j jH (t)B (x x -E(x x ))B H (t) .
It then suffices to show that the second right-hand side term in (A5) is op(1). Rearranging terms
we obtain
' ' 't i i i,t j,t i,t j,t j jH (t)B (x x -E(x x ))B H (t)
= -1/2 2 -1/2 ' -1/2 ' -1/2 ' 't- t-t s=1 s=1j,s+ j i,s+ i i,t j,t i,t j,tT {(T/t) [T h B T h B )]vec[T (x x -E(x x ))]} .
That 2 -1/2 ' -1/2 ' -1/2 ' 't- t-t s=1 s=1j,s+ j i,s+ i i,t j,t i,t j,t{(T/t) [T h B T h B )]vec[T (x x -E(x x ))]} is Op(1) follows
from Assumption 2, Corollary 29.19 of Davidson (1994) and Theorem 4.1 of Hansen (1992).
Since T-1/2 is o(1) the proof is complete.
Lemma A4: Let Assumptions 1-2 hold. (a) Let '1 2-J B J+B = D and -1/2 -1/2
2 2B DB = Q, then Q is
idempotent. (b) Let A be a (k k2) matrix with k k2 2I on the upper diagonal (k2 k2) block and
zeroes elsewhere. There exists a symmetric orthonormal matrix C such that Q = 'CAA C .
Proof of Lemma A4: (a) For i,j = 1,2, let (i, j)2,tq denote the (ki kj) ij-block of the matrix q2,t. Since
(1,2) (2,1) (1,2)1 2,t 2 2,t 1 1 2,t 2
2 (2,1)2 2,t 1 2
B (I+[Eq ]N [Eq ]B ) -B [Eq ]NB =
-N [Eq ]B N,
(2,2) (2,1) (1,2) -12 2,t 2,t 1 2,tN = [Eq -Eq B Eq ] and
(1,2) (2,1) (1,2)1 2,t 2 2,t 1 1 2,t 2
(2,1)2 2,t 1 2
B [Eq ]N [Eq ]B -B [Eq ]ND =
-N [Eq ]B N
11
the result follows from straightforward algebra. (b) Since Q is idempotent the result follows from
Schur’s Decomposition Theorem (Magnus and Neudecker, 1988, p.16).
Lemma A5: Let Assumptions 1, 2 and 4 hold. For [ ,1] , (a) -1/2 1/2t-s=1 2,s+ hhT h S W( ) ,
(b) -1/2 -1 1/2t-s=1 2,s+ hh(T/t)T h S W( ) , (c) -1/2 t-
s=t-R+1 2,s+(T/R)T h -1 1/2hhS {W( )-W( - )} .
Proof of Lemma A5: (a) Given Assumption 2 and the fact that -1 'T Ts=1 s=12,s+ 2,s+ hhT E( h )( h ) S
< the result follows from Corollary 29.19 of Davidson (1994). (b) Given (a) the result follows
from the Continuous Mapping Theorem. (c) That T/R -1 is immediate. Write
-1/2 t-s=t-R+1 2,s+T h as -1/2 -1/2t- t-R
s=1 s=12,s+ 2,s+T h -T h . That -1/2 1/2t-s=1 2,s+ hhT h S W( ) follows from
(a). For the second piece, if we define = then Corollary 29.19 of Davidson (1994)
implies -1/2 t-Rs=1 2,s+T h 1/2
hhS W( ') .
Lemma A6: Let Assumptions 1, 2 and 4 hold. 't 2 2,t+H (t)h d 1 where 1 equals
1 -1 'hhW ( )S dW( ) , -1 '
hh{W(1)-W( )}S W( ) and 1-1 'hh{W( )-W( - )}S dW( ) for the
recursive, fixed and rolling schemes respectively.
Proof of Lemma A6: The result is simple for the fixed. We will show the result for the
recursive scheme. For the rolling scheme if we note that 't 2 2,t+H (t)h can be rewritten as
-1/2 ' -1/2t-t s=1 2,s+ 2,t+(T/R) (T h ) (T h ) -1/2 ' -1/2t-R
t s=1 2,s+ 2,t+(T/R) (T h ) (T h ) then the result will
follow from an argument similar to that for the recursive but repeated for each of the two
components. Throughout assume that > 1. If = 1 the argument is made simpler since many
of the terms do not exist.
12
The results are modifications of those in Hansen (1992). As such we will use his notation
throughout. Let the operator EiX denote E(X| i) where t-1/2 i
s=1 2,s 2,i t,T 1)(T h ,h :i is the
smallest sigma-field containing the past history of { -1/2 ts=1 2,s 2,tT h ,h } for all T. Define t+ =
i=0 i 2,t+ +i i-1 2,t+ +i(E h -E h ) and zt+ = i=1 i 2,t+ +iE h . Then 2,t+h = t+ + zt+ -1 zt+ .
In the above notation
-1/2 ' -1/2t-t s=1 2,s+ 2,t+(T/t)(T h ) (T h )
= -1/2 ' -1/2t+ -1t s= +1 2,s 2,t+(T/t)(T h ) (T h ) '-1
t j=1 2,t+j 2,t+(1/t)( h ) h
= -1/2 ' -1/2t+ -1t s= +1 2,s t+(T/t)(T h ) (T ) + 't+ -1
t s= +1 2,s t+ -1 t+(1/t)( h ) (z -z )
'-1t j=1 2,t+j 2,t+(1/t)( h ) h
= -1/2 ' -1/2t-t s=1 2,s+ t+(T/t)(T h ) (T ) + -1 'R+ -2
s=1+ 2,s R+ -1R ( h )z -1 'T+ -1s=1+ 2,s T+T ( h )z
2 -1 'T-1 t+ -1t=R s=1+ 2,s t+(t +t) ( h ) z + 'T-1
t=R-1 2,t+ t+(1/t)h z '-1t j=1 2,t+j 2,t+(1/t)( h ) h .
That -1/2 ' -1/2t-t s=1 2,s+ t+(T/t)(T h ) (T ) d
1 -1 'hhW ( )S dW( ) follows from Theorem 4.1 of
Hansen (1992). The result will follow if the sum of the remaining five terms is op(1).
Consider the second and third right-hand side terms. If we take the absolute value of each we
obtain both -1 'R+ -2s=1+ 2,s R+ -1|R ( h )z | -1/2 -1/2R+ -2
s=1+2 2,s R+ -1(T/R)k |T h ||T z | and -1 'T+ -1s=1+ 2,s T+|R ( h )z |
-1/2 -1/2T+ -1s=1+2 2,s T+(T/R)k |T h ||T z | . Assumption 4 implies that (T/R) is bounded while Lemma A1
(a) implies that both -1/2 R+ -2s=1+ 2,s|T h | and -1/2 T+ -1
s=1+ 2,s|T h | are Op(1). That the second and third
right-hand side terms are op(1) follows from (A.3) of Hansen (1992) wherein he shows that both
-1/2R+ -1|T z | and -1/2
T+|T z | are op(1).
13
Consider the fourth right-hand side term. If we take its absolute value we obtain
2 -1 'T-1 t+ -1t=R s=1+ 2,s t+| (t +t) ( h ) z | 2 -1/2 -1/2t+ -1
s=1+2 t 2,s t T t+[(T-1-R)/(R +R)]k (sup |T h |)(sup |T z |) . Lemma
A1 (a) implies that -1/2 t+ -1s=1+t 2,s(sup |T h |) is Op(1). That -1/2
t T t+(sup |T z |) = op(1) follows from
(A.3) of Hansen (1992). The result follows since by Assumption 4, 2(T-1-R)/(R +R) is o(1).
Consider the fifth right-hand side term. We show that it converges in probability to
'-1j=1 2,t+j 2,t+-ln( ) Eh h . To do so, add and subtract 'T-1
t=R-1 2,t+ t+(1/t)E(h z ) and obtain
' 'T-1t=R-1 2,t+ t+ 2,t+ t+(1/t)(h z -Eh z ) + 'T-1
t=R-1 2,t+ t+(1/t)(Eh z ) . Since '2,t+ 2,t+ -jEh h = 0 all j it is clear
that 'T-1t=R-1 2,t+ t+(1/t)(Eh z ) = 'T-1
t=R-1 i=12,t+ i 2,t+ +i(1/t)(Eh [ E h ]) = -1 'T-1 -1t=R-1 j=1 2,t+j 2,t+(T (T/t))( Eh h ) .
Since for large enough T, -1 Tt=RT (T/t) ~ 1 1d = -ln( ), the result will follow if
' 'T-1t=R-1 2,t+ t+ 2,t+ t+(1/t)(h z -Eh z ) = -1 ' 'T-1
t=R-1 2,t+ t+ 2,t+ t+T (T/t)(h z -Eh z ) = op(1). If we define UTt
(T/t) and et' '2,t+ t+ 2,t+ t+h z -Eh z then the result follows from Theorem 3.2 of Hansen (1992).
Because of the minus sign, the proof will be complete if the final right-hand side term
converges in probability to '-1j=1 2,t+j 2,t+-ln( ) Eh h . To show this add and subtract '-1
j=1 2,t+j 2,t+Eh h
to obtain '-1t j=1 2,t+j 2,t+(1/t)( h ) h = -1/2 -1/2 ' '-1
j=1 t 2,t+j 2,t+ 2,t+j 2,t+T T (T/t)(h h -Eh h ) +
-1 '-1t j=1 2,t+j 2,t+(T (T/t))( Eh h ) . Given Assumption 2, Corollary 29.11 of Davidson (1994) implies
that -1/2 ' '-1j=1 t 2,t+j 2,t+ 2,t+j 2,t+T (T/t)(h h -Eh h ) = Op(1). Since T-1/2 = o(1) the result is obtained
because -1 '-1t j=1 2,t+j 2,t+(T (T/t))( Eh h ) = '-1
j=1 2,t+j 2,t+-ln( ) Eh h + o(1) was established in the
preceding paragraph.
14
Lemma A7: Let Assumptions 1, 2 and 4 hold. 't 2 2 d 2H (t)H (t) where 2 equals
1 -2 'hhW ( )S W( )d , 1-2 '
hh{W( )-W( - )}S {W( )-W( - )}d and -1 'hhW ( )S W( ) for
the recursive, rolling and fixed schemes respectively.
Proof of Lemma A7: The result is immediate for the fixed. Given Lemma A5 the results for the
recursive and rolling follow from the Continuous Mapping Theorem.
Lemma A8: Let Assumptions 1, 2 and 4 hold. For i = 1,2, 1/2t i,t i,R
ˆ ˆsup R | - |= op(1).
Proof of Lemma A8: We will show the result for the recursive scheme. The fixed is immediate
and the rolling follows from a proof similar to that for the recursive. First note that
i,t i,Rˆ ˆ- = *
i,t iˆ( - ) *
i,R iˆ( - ) = -1t- t-
s=1 s=1i,s i,s+( q ) ( h ) Bi(R)Hi(R)
= -1 -1 -1 -1t- t-s=R- +1 s=R- +1i i,s i i,s+(B (R)+R q ) (H (R)+R h ) Bi(R)Hi(R)
= -1 t-s=R- +1i i,s+B (R)(R h ) -1 -1 -1t- t-
s=R- +1 s=R- +1i i,s i i,s i iB (R)(R q )[I+B (R)(R q )] B (R)H (R)
-1 -1 -1 -1t- t- t-s=R- +1 s=R- +1 s=R- +1i i,s i i,s i i,s+B (R)(R q )[I+B (R)(R q )] B (R)(R h )
where the final equality follows from the formula for the inverse of a sum of matrices as
presented in Greene (2000, p. 32). Rearranging terms and using the triangle inequality we obtain
(A6) 1/2t i,t i,R
ˆ ˆsup R | - | = 1/2 -1/2 t-s=R- +1t i i,s+(P/R) sup |B (R)(P h )|
+ -1 t-s=R- +1t i i,s(P/R)sup |B (R)(P q ) -1 -1 -1/2t- R-
s=R- +1 s=1i i,s i i,s+[I+(P/R)B (R)(P q )] B (R)(R h )|
+ 3/2 -1 t-s=R- +1t i i,s(P/R) sup |B (R)(P q )
-1 -1 -1/2t- t-s=R- +1 s=R- +1i i,s i i,s+[I+(P/R)B (R)(P q )] B (R)(P h )| .
We must then show that the three right-hand side terms in (A6) are each op(1). First note that
both Bi(R) and -1/2 R-s=1 i,s+R h are invariant to the index t. Given Lemmas A1 (a) and (b), each of
15
these terms can be treated as Op(1) constants with respect to the operator supt.
Consider the first right-hand side term in (A6). Given Assumption 2 and Corollary 29.19 of
Davidson (1994) we know that -1/2 t-s=R- +1 i,s+P h 1/ 2
hhS W( ). Then following arguments similar
to those in Lemma A1 (a) we obtain -1/2 t-s=R- +1t i i,s+sup |B (R)(P h )| d
1/ 20 1 i hhsup |B S W( )| =
Op(1). Since (P/R)1/2 = o(1) the desired result follows.
Consider the second right-hand side term in (A6). In particular rewrite -1 t-s=R- +1 i,sP q as
-1/2 -1/2 t-s=R- +1 i,s i,sP [P (q -Eq )] + i,sEq (t-R-1)/P . Given Assumption 2 and Corollary 29.19 of
Davidson (1994) we know that -1/2 t-s=R- +1 i,s i,sP vec(q -Eq ) 1/2
qqS W( ). The Continuous
Mapping Theorem then implies that -1/2 -1/2 t-s=R- +1 i,s i,sP [P (q -Eq )] + i,sEq (t-R-1)/P i,sEq =
-1iB . Given Assumption 2 and Corollary 29.19 of Davidson (1994) we know that -1/2 R-
s=1 i,s+R h
d1/2hhS V , V a standard normal vector. Then following arguments similar to those in Lemma A1
(a) we obtain -1 t-s=R- +1t i i,ssup |B (R)(P q ) -1 -1 -1/2t- R-
s=R- +1 s=1i i,s i i,s+[I+(P/R)B (R)(P q )] B (R)(R h )| d
1/20 1 i hhsup | B S V| = 1/2
i hh|B S V| = Op(1). Since P/R = o(1) the desired result follows.
Consider the third right-hand side term in (A6). Arguments as for the first two pieces imply
-1 t-s=R- +1t i i,ssup |B (R)(P q ) -1 -1 -1/2t- t-
s=R- +1 s=R- +1i i,s i i,s+[I+(P/R)B (R)(P q )] B (R)(P h )| d
1/20 1 i hhsup | B S W( )| = Op(1). Since (P/R)3/2 = o(1) the proof is complete.
Lemma A9: Let Assumptions 1, 2 and 4 hold, let i,j,m,n = 1,2 and let 0 r,s j . (a) Each of the
following are op((P/R)1/2):
(i) 't i,t+ i,t i,R
ˆ ˆh ( - ) and (ii) * ' ' *t i,R i i,t j,t j,R j
ˆ ˆ( - ) x x ( - ) .
(b) Each of the following are op(P/R) (and hence also op((P/R)1/2)):
16
(iii) * ' 't i,R i i,t j,t j,t j,R
ˆ ˆ ˆ( - ) x x ( - ) ,
(iv) ' 't i,t i,R i,t j,t j,t j,R
ˆ ˆ ˆ ˆ( - ) x x ( - ) ,
(v) * ' 't i,R i i,t+ -r j,t+ -s j,t-s j,R
ˆ ˆ ˆ( - ) h h ( - ) ,
(vi) ' 't i,t-r i,R i,t+ -r j,t+ -s j,t-s j,R
ˆ ˆ ˆ ˆ( - ) h h ( - ) ,
(vii) * ' ' ' * *t i,R i i,t+ -r m,t-s j,t-s j,R j m,R m
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) ,
(viii) * ' ' ' *t i,R i i,t+ -r m,t-s j,t-s j,t-s j,R m,R m
ˆ ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) ,
(ix) * ' ' 't i,R i i,t+ -r m,t-s j,t-s j,t-s j,R m,t-s m,R
ˆ ˆ ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) ,
(x) ' ' ' * *t i,t-r i,R i,t+ -r m,t-s j,t-s j,R j m,R m
ˆ ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) ,
(xi) ' ' ' *t i,t-r i,R i,t+ -r m,t-s j,t-s j,t-s j,R m,R m
ˆ ˆ ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) ,
(xii) ' ' 't i,t-r i,R i,t+ -r m,t-s j,t-s j,t-s j,R m,t-s m,R
ˆ ˆ ˆ ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) ,
(xiii) * ' * ' ' ' ' * *t i,R i j,R j i,t-r j,t-r n,t-s m,t-s m,R m n,R n
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - )) ,
(xiv) * ' * ' ' ' ' *t i,R i j,R j i,t-r j,t-r n,t-s m,t-s m,t-s m,R n,R n
ˆ ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - )) ,
(xv) * ' * ' ' ' 't i,R i j,R j i,t-r j,t-r n,t-s m,t-s m,t-s m,R n,t-s n,R
ˆ ˆ ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - )) ,
(xvi) ' * ' ' ' ' *t i,t-r i,R j,R j i,t-r j,t-r n,t-s m,t-s m,t-s m,R n,R n
ˆ ˆ ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ) ,
(xvii) ' * ' ' ' 't i,t-r i,R j,R j i,t-r j,t-r n,t-s m,t-s m,t-s m,R n,t-s n,R
ˆ ˆ ˆ ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - )) ,
(xviii) ' ' ' ' 't i,t-r i,R j,t-r j,R i,t-r j,t-r n,t-s m,t-s m,t-s m,R n,t-s n,R
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - )) .
Proof of Lemma A9: (a) It suffices to show that (R/P)1/2 times each of the terms is op(1).
Consider (i), 1/2 't i,t+ i,t i,R
ˆ ˆ(R/P) h ( - ) = 1/2 -1/2 ' 1/2 1/2t i,t+ i,t i,R
ˆ ˆ(P/R) (P h )[(R/P) R ( - )] . The second
paragraph of the proof of Lemma A8 implies that 1/2 1/2i,t i,R
ˆ ˆ(R/P) R ( - ) 1/ 2i hhB S W( ) . Given
17
Assumption 2, Corollary 4.1 of Hansen (1992) then implies that
-1/2 ' 1/2 1/2t i,t+ i,t i,R
ˆ ˆ(P h )[(R/P) R ( - )] = Op(1). Since (P/R)1/2 = o(1) the result follows.
Now consider the remaining term. Taking absolute values, we obtain the inequality
(ii) 1/2 * ' ' *t i,R i i,t j,t j,R j
ˆ ˆ|(R/P) ( - ) x x ( - )|
2 1/2 1/2 * 1/2 * -1 'ti,R i j,R j i,t j,t
ˆ ˆk (P/R) (R | - |)(R | - |)(P |x x |) ,
Given Assumption 2, Markov’s inequality implies that -1 't i,t j,tP |x x | = Op(1). That 1/2 *
i,R iˆR | - | =
Op(1) follows from Lemma A1 (c). Since 1/2t i,t i,R
ˆ ˆsup R | - | = op(1) by Lemma A8 and P/R = o(1)
the proof is complete.
(b) It suffices to show that (R/P) times each of the terms is op(1). Taking absolute values, we
obtain the inequalities
(iii) * ' 't i,R i i,t j,t j,t j,R
ˆ ˆ ˆ|(R/P) ( - ) x x ( - )| 2 1/2 * 1/2 -1 'ti,R i t j,t j,R i,t j,t
ˆ ˆ ˆk (R | - |)(sup R | - |)(P |x x |) ,
(iv) ' 't i,t i,R i,t j,t j,t j,R
ˆ ˆ ˆ ˆ|(R/P) ( - ) x x ( - )|
2 1/2 1/2 -1 'tt i,t i,R t j,t j,R i,t j,t
ˆ ˆ ˆ ˆk (sup R | - |)(sup R | - |)(P |x x |) .
(v) * ' 't i,R i i,t+ -r j,t+ -s j,t-s j,R
ˆ ˆ ˆ|(R/P) ( - ) h h ( - )|
2 1/2 * 1/2 -1 'ti,R i t j,t-s j,R i,t+ -r j,t+ -s
ˆ ˆ ˆk (R | - |)(sup R | - |)(P |h h |) ,
(vi) ' 't i,t-r i,R i,t+ -r j,t+ -s j,t-s j,R
ˆ ˆ ˆ ˆ|(R/P) ( - ) h h ( - )|
2 1/2 1/2 -1 'tt i,t-r i,R t j,t-s j,R i,t+ -r j,t+ -s
ˆ ˆ ˆ ˆk (sup R | - |)(sup R | - |)(P |h h |) ,
(vii) * ' ' ' * *t i,R i i,t+ -r m,t-s j,t-s j,R j m,R m
ˆ ˆ ˆ|(R/P) ( - ) h vec(x x ) (( - ) ( - ))|
-1/2 3 1/2 * 1/2 *i,R i j,R j
ˆ ˆR k (R | - |)(R | - |) 1/2 * -1 ' 'tm,R m i,t+ -r m,t-s j,t-s
ˆ(R | - |)(P |h vec(x x ) |) ,
18
(viii) * ' ' ' *t i,R i i,t+ -r m,t-s j,t-s j,t-s j,R m,R m
ˆ ˆ ˆ ˆ|(R/P) ( - ) h vec(x x ) (( - ) ( - ))|
-1/2 3 1/2 * 1/2i,R i t j,t-s j,R
ˆ ˆ ˆR k (R | - |)(sup R | - |) 1/2 * -1 ' 'tm,R m i,t+ -r m,t-s j,t-s
ˆ(R | - |)(P |h vec(x x ) |) ,
(ix) * ' ' 't i,R i i,t+ -r m,t-s j,t-s j,t-s j,R m,t-s m,R
ˆ ˆ ˆ ˆ ˆ| (R / P) ( - ) h vec(x x ) (( - ) ( - )) |
-1/2 3 1/2 * 1/2i,R i t j,t-s j,R
ˆ ˆ ˆR k (R | - |)(sup R | - |)
1/2 -1 ' 'tt m,t-s m,R i,t+ -r m,t-s j,t-s
ˆ ˆ(sup R | - |)(P |h vec(x x ) |) ,
(x) ' ' ' * *t i,t-r i,R i,t+ -r m,t-s j,t-s j,R j m,R m
ˆ ˆ ˆ ˆ|(R/P) ( - ) h vec(x x ) (( - ) ( - ))|
-1/2 3 1/2 1/2 *t i,t-r i,R j,R j
ˆ ˆ ˆR k (sup R | - |)(R | - |) 1/2 * -1 ' 'tm,R m i,t+ -r m,t-s j,t-s
ˆ(R | - |)(P |h vec(x x ) |) ,
(xi) ' ' ' *t i,t-r i,R i,t+ -r m,t-s j,t-s j,t-s j,R m,R m
ˆ ˆ ˆ ˆ ˆ|(R/P) ( - ) h vec(x x ) (( - ) ( - ))|
-1/2 3 1/2 1/2t i,t-r i,R t j,t-s j,R
ˆ ˆ ˆ ˆR k (sup R | - |)(sup R | - |)
1/2 * -1 ' 'tm,R m i,t+ -r m,t-s j,t-s
ˆ(R | - |)(P |h vec(x x ) |) ,
(xii) ' ' 't i,t-r i,R i,t+ -r m,t-s j,t-s j,t-s j,R m,t-s m,R
ˆ ˆ ˆ ˆ ˆ ˆ|(R/P) ( - ) h vec(x x ) (( - ) ( - ))|
-1/2 3 1/2 1/2t i,t-r i,R t j,t-s j,R
ˆ ˆ ˆ ˆR k (sup R | - |)(sup R | - |)
1/2 -1 ' 'tt m,t-s m,R i,t+ -r m,t-s j,t-s
ˆ ˆ(sup R | - |)(P |h vec(x x ) |) ,
(xiii) * ' * ' 't i,R i j,R j i,t-r j,t-r
ˆ ˆ|(R/P) (( - ) ( - ) )vec(x x ) ' ' * *n,t-s m,t-s m,R m n,R n
ˆ ˆvec(x x ) (( - ) ( - ))|
-1 4 1/2 * 1/2 * 1/2 *i,R i j,R j m,R m
ˆ ˆ ˆR k (R | - |)(R | - |)(R | - |)
1/2 * -1 ' ' 'tn,R n i,t-r j,t-r n,t-s m,t-s
ˆ(R | - |)(P |vec(x x )vec(x x ) |) ,
(xiv) * ' * ' ' ' ' *t i,R i j,R j i,t-r j,t-r n,t-s m,t-s m,t-s m,R n,R n
ˆ ˆ ˆ ˆ ˆ|(R/P) (( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))|
-1 4 1/2 * 1/2 * 1/2i,R i j,R j t m,t-s m,R
ˆ ˆ ˆ ˆR k (R | - |)(R | - |)(sup R | - |)
19
1/2 * -1 ' ' 'tn,R n i,t-r j,t-r n,t-s m,t-s
ˆ(R | - |)(P |vec(x x )vec(x x ) |) ,
(xv) * ' * ' ' ' 't i,R i j,R j i,t-r j,t-r n,t-s m,t-s m,t-s m,R n,t-s n,R
ˆ ˆ ˆ ˆ ˆ ˆ|(R/P) (( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))|
-1 4 1/2 * 1/2 * 1/2i,R i j,R j t m,t-s m,R
ˆ ˆ ˆ ˆR k (R | - |)(R | - |)(sup R | - |)
1/2 -1 ' ' 'tt n,t-s n,R i,t-r j,t-r n,t-s m,t-s
ˆ ˆ(sup R | - |)(P |vec(x x )vec(x x ) |) ,
(xvi) ' * ' 't i,t-r i,R j,R j i,t-r j,t-r
ˆ ˆ ˆ(R/P) (( - ) ( - ) )vec(x x ) ' ' *n,t-s m,t-s m,t-s m,R n,R n
ˆ ˆ ˆvec(x x ) (( - ) ( - )
-1 4 1/2 1/2 * 1/2t i,t-r i,R j,R j t m,t-s m,R
ˆ ˆ ˆ ˆ ˆR k (sup R | - |)(R | - |)(sup R | - |)
1/2 * -1 ' ' 'tn,R n i,t-r j,t-r n,t-s m,t-s
ˆ(R | - |)(P |vec(x x )vec(x x ) |) ,
(xvii) ' * ' 't i,t-r i,R j,R j i,t-r j,t-r
ˆ ˆ ˆ(R/P) (( - ) ( - ) )vec(x x )
' 'n,t-s m,t-s m,t-s m,R n,t-s n,R
ˆ ˆ ˆ ˆvec(x x ) (( - ) ( - ))
-1 4 1/2 1/2 * 1/2t i,t-r i,R j,R j t m,t-s m,R
ˆ ˆ ˆ ˆ ˆR k (sup R | - |)(R | - |)(sup R | - |)
1/2 -1 ' ' 'tt n,t-s n,R i,t-r j,t-r n,t-s m,t-s
ˆ ˆ(sup R | - |)(P |vec(x x )vec(x x ) |) ,
(xviii) ' ' 't i,t-r i,R j,t-r j,R i,t-r j,t-r
ˆ ˆ ˆ ˆ(R/P) (( - ) ( - ) )vec(x x )
' 'n,t-s m,t-s m,t-s m,R n,t-s n,R
ˆ ˆ ˆ ˆvec(x x ) (( - ) ( - ))
-1 4 1/2 1/2 1/2t i,t-r i,R t j,t-r j,R t m,t-s m,R
ˆ ˆ ˆ ˆ ˆ ˆR k (sup R | - |)(sup R | - |)(sup R | - |)
1/2 -1 ' ' 'tt n,t-s n,R i,t-r j,t-r n,t-s m,t-s
ˆ ˆ(sup R | - |)(P |vec(x x )vec(x x ) |) .
Given Assumption 2, Markov’s inequality implies that -1 ' ' 't i,t-r j,t-r n,t-s m,t-sP |vec(x x )vec(x x ) | and
-1 ' 't i,t+ -r m,t-s j,t-sP |h vec(x x ) | -1 '
t i,t+ -r j,t+ -sP |h h | are Op(1). That 1/2 *i,R i
ˆR | - | = Op(1) follows from
Lemma A1 (c). Since 1/2t i,t-r i,R
ˆ ˆsup R | - | = op(1) by Lemma A8 and both R-1/2 and R-1 are o(1) the
20
proof is complete.
Lemma A10: (a) Let Assumptions 1, 2 and 4 hold. 2t 1,t+ 1,t+ 2,t+ˆ ˆ ˆ(u -u u ) = 2 '
t 2 2,t+H (t)h +
op(1). (b) Let Assumptions 1, 2 and 4 hold. 2t 1,t+ 1,t+ 2,t+ˆ ˆ ˆ(u -u u ) =
1/2 2 1/2 ' -1/2t2 2,t+(P/R) [R H (R)][P h ] + op((P/R)1/2).
Proof of Lemma A10: (a) If we note that 1,t+u = ut+ - ' *1,t 1,t 1
ˆx ( - ) and 2,t+u = ut+ - ' *2,t 2,t 2
ˆx ( - )
we obtain
(A7) 2t 1,t+ 1,t+ 2,t+ˆ ˆ ˆ(u -u u ) = ' '
t 1,t+ 1 1 2,t+ 2 2{-h B (t)H (t)+h B (t)H (t)}
't 1 1 1,t 1 1{-H (t)B (t)q B (t)H (t) ' '
1 1 1,t 2,t 2 2H (t)B (t)x x B (t)H (t)} .
Consider the first bracketed right-hand side term in (A7). If we note that h1,t+ = Jh2,t+ and apply
the definition of 2,t+h , by Lemmas A2 and A4 we obtain
' 't 1,t+ 1 1 2,t+ 2 2{-h B (t)H (t)+h B (t)H (t)} = 2 '
t 2 2,t+H (t)h + op(1).
We now need only show that the second bracketed right-hand side term in (A7) is op(1). Since
'1,t 2,tEx x = 2,tJEq = 1
2JB the result follows by Lemma A3 (b).
(b) If we note that 1,t+u = ut+ - ' *1,t 1,R 1
ˆx ( - ) - '1,t 1,t 1,R
ˆ ˆx ( - ) and 2,t+u = ut+ - ' *2,t 2,R 2
ˆx ( - ) -
'2,t 2,t 2,R
ˆ ˆx ( - ) , we obtain
2t 1,t+ 1,t+ 2,t+ˆ ˆ ˆ(u -u u ) = t 0,t 1,t{r r } + 10
t j 2 j,t{ r }
' 't 1,t+ 1 1 2,t+ 2 2{-h B (R)H (R)+h B (R)H (R)}
+ 't 1,t+ 1,t 1,R
ˆ ˆ{-h ( - ) + '2,t+ 2,t 2,R
ˆ ˆh ( - ) + * ' *1,R 1 1,t 1,R 1
ˆ ˆ( - ) q ( - )
21
* ' ' *1,R 1 1,t 2,t 2,R 2
ˆ ˆ( - ) x x ( - ) + * '1,R 1 1,t 1,t 1,R
ˆ ˆ ˆ2( - ) q ( - ) * ' '1,R 1 1,t 2,t 2,t 2,R
ˆ ˆ ˆ( - ) x x ( - )
* ' '2,R 2 2,t 1,t 1,t 1,R
ˆ ˆ ˆ( - ) x x ( - ) + '1,t 1,R 1,t 1,t 1,R
ˆ ˆ ˆ ˆ( - ) q ( - ) ' '1,t 1,R 1,t 2,t 2,t 2,R
ˆ ˆ ˆ ˆ( - ) x x ( - )}
= 1/2 -1/2 ' ' 1/2t 2,t+ 1 2 2(P/R) [P h ][-JB (R)J +B (R)][R H (R)] + 10
t j=2 j,T{ r } .
Lemma A1 (b) implies that for i = 1,2, Bi(R) a.s. Bi. Given Assumption 2, Corollary 29.19 of
Davidson (1994) implies that both 1/22R H (R) and -1/2
t 2,t+P h are Op(1). That t 0,t 1,t{r +r } =
1/2 2 1/2 ' -1/2t2 2,t+(P/R) [R H (R)][P h ] + op((P/R)1/2) follows from Lemma A4 and the definition of
2,t+h . The result follows since by Lemma A9 (a) - (b), 10t j=2 j,t{ r } = op((P/R)1/2).
Lemma A11: (a) Let Assumptions 1, 2 and 4 hold. For 0 j j
2 2T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ ˆ ˆ(u -u u -c)(u -u u -c) = 4 ' '
t 2 2,t+ 2,t+ -j 2H (t)[Eh h ]H (t) + op(1). (b) Let
Assumptions 1, 2 and 4 hold. For 0 j j 2 2T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ ˆ ˆ(u -u u -c)(u -u u -c) =
4 1/2 ' ' 1/22 2,t+ 2,t+ -j 2(P/R) [R H (R)][Eh h ][R H (R)] + op(P/R).
Proof of Lemma A11: (a) Given Lemma A10 (a), Lemma A6 implies that c is Op(P-1) and
hence 2(P-j- )c = op(1). Since j is finite this also implies that both 2T-t=R+j 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆc (u -u u ) and
2T-t=R+j 1,t+ 1,t+ 2,t+ˆ ˆ ˆc (u -u u ) are op(1). It then suffices to show that
2 2T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ ˆ ˆ(u -u u )(u -u u ) = 4 ' '
t 2 2,t+ 2,t+ -j 2H (t)[Eh h ]H (t) + op(1).
If we note that 1,t+u = ut+ - ' *1,t 1,t 1
ˆx ( - ) and 2,t+u = ut+ - ' *2,t 2,t 2
ˆx ( - ) , we have
(A8) 2 2T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ ˆ ˆ(u -u u )(u -u u )
= [ ' 'T-t=R+j 1 1 1,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) ' 'T-
t=R+j 1 1 1,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j)
' 'T-t=R+j 2 2 2,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) + ' 'T-
t=R+j 2 2 2,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j) ]
22
+ [ * ' ' ' * *Tt R j 1,t 1 1,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 1,t 1 1,t+ 1,t-j 2,t-j 1,t-j 1 2,t-j 2
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 2,t 2 2,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
* ' ' ' * *T-t=R+j 2,t 2 2,t+ 1,t-j 2,t-j 1,t-j 1 2,t-j 2
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
* ' ' ' * *T-t=R+j 1,t-j 1 1,t+ -j 1,t 1,t 1,t 1 1,t 1
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 1,t-j 1 1,t+ -j 1,t 2,t 1,t 1 2,t 2
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 2,t-j 2 2,t+ -j 1,t 1,t 1,t 1 1,t 1
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
* ' ' ' * *Tt R j 2,t-j 2 2,t+ -j 1,t 2,t 1,t 1 2,t 2
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - ))
+ * ' * ' ' ' ' * *Tt R j 1,t 1 1,t 1 1,t 1,t 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))
* ' * ' ' ' ' * *Tt R j 1,t 1 1,t 1 1,t 1,t 1,t-j 2,t-j 1,t-j 1 2,t-j 2
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))
* ' * ' ' ' ' * *Tt R j 1,t 1 2,t 2 1,t 2,t 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))
+ * ' * ' ' ' ' * *Tt R j 1,t 1 2,t 2 1,t 2,t 1,t-j 2,t-j 1,t-j 1 2,t-j 2
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - )) ].
Consider the first bracketed right-hand side term in (A8). If we apply the definition of 2,t+h then
by Lemmas A3 (a) and A4 we obtain
[ ' 'T-t=R+j 1 1 1,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) ' 'T-
t=R+j 1 1 1,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j)
' 'T-t=R+j 2 2 2,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) + ' 'T-
t=R+j 2 2 2,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j) ]
= 4 ' 't 2 2,t+ 2,t+ -j 2H (t)[Eh h ]H (t) + op(1).
We now need only show that the second bracketed term on the right-hand side of (A8) is op(1).
To do so we show that * ' ' ' * *Tt R j 1,t 1 1,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) is op(1). The
remaining terms follow similar arguments. Taking the absolute value we obtain
23
* ' ' ' * *Tt R j 1,t 1 1,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ| ( - ) h vec(x x ) (( - ) ( - )) |
3 1/3 * 1 ' ' 1/3 * 2tt 1,t 1 1,t 1,t j 1,t j t 1,t j 1
ˆ ˆk (P / T)(sup T | |)(P | h vec(x x ) |)(sup T | |) .
Assumption 4 implies that P/T is bounded. Given Assumption 2, that
1 ' 't 1,t 1,t j 1,t jP | h vec(x x ) | is Op(1) follows from Markov’s inequality. Since j is finite, that
1/3 *t 1,t 1
ˆ(sup T | |) and 1/3 * 2t 1,t j 1
ˆ(sup T | |) are op(1) follows from Lemma A1 (c) and the
proof is complete.
(b) Given Lemma A10 (b), Lemma A6 implies that c is Op(P-1) and hence 2(P-j)c = op(1).
Since j is finite this also implies that both 2T-t=R+j 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆc (u -u u ) and 2T-
t=R+j 1,t+ 1,t+ 2,t+ˆ ˆ ˆc (u -u u )
are op(1). It then suffices to show that 2 2T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ ˆ ˆ(u -u u )(u -u u ) =
4 1/ 2 ' ' 1/ 22 2,t 2,t j 2(P / R) [R H (R)][Eh h ][R H (R)] + op(P/R).
First note that, in the notation of Lemma A10 (b), 2 2T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ ˆ ˆ(u -u u )(u -u u ) =
T 10 10t R j i 2 i 20,t 1,t i,t 0,t j 1,t j i,t j{r r r }{r r r }. If we take the product within the argument we
obtain 121 distinct crossproduct terms. Rather than list each explicitly, we will simply note that
the important terms are z0,t = 0,t 0,t jr r , z1,t = 0,t 1,t jr r , z2,t = 1,t 0,t jr r and z3,t = 1,t 1,t jr r . The remaining
terms are each of the form in Lemma A9 (b) for some choice of indices i,j,m,n {1, 2} and 0
r,s j . Using this notation we obtain
2 2T-t=R+j 1,t+ 1,t+ 2,t+ 1,t+ -j 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ ˆ ˆ(u -u u )(u -u u ) = T
t R j 0,t 1,t 2,t 3,t{z z z z } + T 121t R j i 4 i,t{ z }
* ' ' *T-t=R+j 1,R 1 1,t+ 1,t+ -j 1,R 1
ˆ ˆ{( - ) h h ( - ) * ' ' *1,R 1 1,t+ 2,t+ -j 2,R 2
ˆ ˆ( - ) h h ( - )
* ' ' *2,R 2 2,t+ 1,t+ -j 1,R 1
ˆ ˆ( - ) h h ( - )} + * ' ' *2,R 2 2,t+ 2,t+ -j 2,R 2
ˆ ˆ( - ) h h ( - )} + T 121t R j i 4 i,t{ z }
= 1/2 ' -1 ' ' -1 'T- T-t=R+j t=R+j2 1 1,t+ 1,t+ -j 1 1 1,t+ 2,t+ -j 2(P/R)[R H (R)][JB (R)(P h h )B (R)J -JB (R)(P h h )B (R)
24
+ -1 ' ' -1 ' 1/2T- T-t=R+j t=R+j2 2,t+ 1,t+ -j 1 2 2,t+ 2,t+ -j 2 2B (R)(P h h )B (R)J +B (R)(P h h )B (R)][R H (R)]
+ T 121t R j i 4 i,t{ z } .
Lemma A1 (b) implies that both B1(R) a.s. B1 and B2(R) a.s. B2. Given Assumption 2,
Corollary 29.19 of Davidson (1994) implies that 1/ 2 '2R H (R) is Op(1). Assumption 2 also suffices
for -1 'Tt R j 1,t+ 1,t+ -jP h h a.s.
'1,t+ 1,t+ -jEh h , -1 'T
t R j 1,t+ 2,t+ -jP h h a.s.'
1,t+ 2,t+ -jEh h ,
-1 'Tt R j 2,t+ 1,t+ -jP h h a.s.
'2,t+ 1,t+ -jEh h and -1 'T
t R j 2,t+ 2,t+ -jP h h a.s.'
2,t+ 2,t+ -jEh h . That
Tt R j 0,t 1,t 2,t 3,t{z z z z } = 4 1/ 2 ' ' 1/ 2
2 2,t 2,t j 2(P / R) [R H (R)][Eh h ][R H (R)] + op(P/R) then
follows from Lemma A4 and the definition of 2,t+h . The result follows since by Lemma A9 (b),
T 121t R j i 4 i,t{ z } = op(P/R).
Lemma A12: (a) Let Assumptions 1, 2 and 4 hold. 2 2t 1,t+ 2,t+ˆ ˆ(u -u ) = 2 '
t 2 2,t+2 H (t)h
2 ' 't 2 2H (t)H (t) + op(1). (b) Let Assumptions 1, 2 and 4 hold. 2 2
t 1,t+ 2,t+ˆ ˆ(u -u ) =
1/2 2 1/2 ' -1/2t2 2,t+2(P/R) [R H (R)][P h ] + op((P/R)1/2).
Proof of Lemma A12: (a) If we note that 1,t+u = ut+ - ' *1,t 1,t 1
ˆx ( - ) and 2,t+u = ut+ - ' *2,t 2,t 2
ˆx ( - )
we obtain
(A9) 2 2t 1,t+ 2,t+ˆ ˆ(u -u ) = ' '
t 1,t+ 1 1 2,t+ 2 22 {-h B (t)H (t)+h B (t)H (t)}
't 1 1 1,t 1 1{-H (t)B (t)q B (t)H (t) '
1 1 2,t 2 2H (t)B (t)q B (t)H (t)}.
If we note that h1,t+ = Jh2,t+ , Eqi,t = -1iB and apply the definition of 2,t+h , the result follows from
Lemmas A2, A3 and A4.
25
(b) If we note that 1,t+u = ut+ - ' *1,t 1,R 1
ˆx ( - ) - '1,t 1,t 1,R
ˆ ˆx ( - ) and 2,t+u = ut+ - ' *2,t 2,R 2
ˆx ( - ) -
'2,t 2,t 2,R
ˆ ˆx ( - ) , we obtain
2 2t 1,t+ 2,t+ˆ ˆ(u -u ) = t 0,t 1,t2 {r +r } + 13
t j 112,t 3,t 4,t 6,t 9,t j,t{2r 2r r r r r }
' 't 1,t+ 1 1 2,t+ 2 22 {-h B (R)H (R)+h B (R)H (R)}
+ 't 1,t+ 1,t 1,R
ˆ ˆ{-2h ( - ) + '2,t+ 2,t 2,R
ˆ ˆ2h ( - ) + * ' *1,R 1 1,t 1,R 1
ˆ ˆ( - ) q ( - )
* '1,R 1 1,t 1,t 1,R
ˆ ˆ ˆ2( - ) q ( - ) + '1,t 1,R 1,t 1,t 1,R
ˆ ˆ ˆ ˆ( - ) q ( - ) * ' *2,R 2 2,t 2,R 2
ˆ ˆ( - ) q ( - )
* '2,R 2 2,t 2,t 2,R
ˆ ˆ ˆ2( - ) q ( - ) '2,t 2,R 2,t 2,t 2,R
ˆ ˆ ˆ ˆ( - ) q ( - )}
= 1/2 -1/2 ' ' 1/2t 2,t+ 1 2 22(P/R) [P h ][-JB (R)J +B (R)][R H (R)]
+ 13t j 112,T 3,T 4,T 6,T 9,T j,T{2r 2r r r r r } .
Lemma A1 (b) implies that for i = 1,2, Bi(R) a.s. Bi. Assumption 2 and Corollary 29.19 of
Davidson (1994) imply that both 1/22R H (R) and -1/2
t 2,t+P h are Op(1). That t 0,T 1,T2 {r +r } =
1/2 2 1/2 ' -1/2t2 2,t+2(P/R) [R H (R)][P h ] + op((P/R)1/2) follows by Lemma A4 and the definition of
2,t+h . The result follows since by Lemma A9 (a)-(b), 13t j=112,T 3,T 4,T 6,T 9,T j,T{2r +2r +r +r +r + r } =
op((P/R)1/2).
Lemma A13: (a) Let Assumptions 1, 2 and 4 hold. 2 2 2 2T-t=R+j 1,t+ 2,t+ 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ(u -u -d)(u -u -d) =
4 ' 't 2 2,t+ 2,t+ -j 24 H (t)[Eh h ]H (t) + op(1). (b) Let Assumptions 1, 2 and 4 hold.
2 2 2 2T-t=R+j 1,t+ 2,t+ 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ(u -u -d)(u -u -d) = 4 1/2 ' ' 1/2
2 2,t+ 2,t+ -j 24(P/R) [R H (R)][Eh h ][R H (R)] + op(P/R).
Proof of Lemma A13: (a) Given Lemma A12 (a), Lemmas A6 and A7 imply that d is Op(P-1)
and hence 2(P- -j)d = op(1). Since j is finite this also implies that both 2 2T-t=R+j 1,t+ -j 2,t+ -jˆ ˆd (u -u ) and
26
2 2T-t=R+j 1,t+ 2,t+ˆ ˆd (u -u ) are op(1). It then suffices to show that 2 2 2 2T-
t=R+j 1,t+ 2,t+ 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ(u -u )(u -u ) =
4 ' 't 2 2,t+ 2,t+ -j 24 H (t)[Eh h ]H (t) + op(1).
If we note that 1,t+u = ut+ - ' *1,t 1,t 1
ˆx ( - ) and 2,t+u = ut+ - ' *2,t 2,t 2
ˆx ( - ) , we have
(A10) 2 2 2 2T-t=R+j 1,t+ 2,t+ 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ(u -u )(u -u )
= 4[ ' 'T-t=R+j 1 1 1,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) ' 'T-
t=R+j 1 1 1,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j)
' 'T-t=R+j 2 2 2,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) + ' 'T-
t=R+j 2 2 2,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j) ]
+ [ * ' ' ' * *Tt R j 1,t 1 1,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 1,t 1 1,t+ 2,t-j 2,t-j 2,t-j 1 2,t-j 2
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 2,t 2 2,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
* ' ' ' * *T-t=R+j 2,t 2 2,t+ 2,t-j 2,t-j 2,t-j 1 2,t-j 2
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
* ' ' ' * *T-t=R+j 1,t-j 1 1,t+ -j 1,t 1,t 1,t 1 1,t 1
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 1,t-j 1 1,t+ -j 2,t 2,t 2,t 2 2,t 2
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
+ * ' ' ' * *Tt R j 2,t-j 2 2,t+ -j 1,t 1,t 1,t 1 1,t 1
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
* ' ' ' * *Tt R j 2,t-j 2 2,t+ -j 2,t 2,t 2,t 2 2,t 2
ˆ ˆ ˆ2 ( - ) h vec(x x ) (( - ) ( - ))
+ * ' * ' ' ' ' * *Tt R j 1,t 1 1,t 1 1,t 1,t 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))
* ' * ' ' ' ' * *Tt R j 1,t 1 1,t 1 1,t 1,t 2,t-j 2,t-j 2,t-j 2 2,t-j 2
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))
* ' * ' ' ' ' * *Tt R j 2,t 2 2,t 2 2,t 2,t 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - ))
+ * ' * ' ' ' ' * *Tt R j 2,t 2 2,t 2 2,t 2,t 2,t-j 2,t-j 2,t-j 2 2,t-j 2
ˆ ˆ ˆ ˆ(( - ) ( - ) )vec(x x )vec(x x ) (( - ) ( - )) ].
Consider the first bracketed right-hand side term in (A10). If we apply the definition of 2,t+h
then by Lemmas A3 (a) and A4 we obtain
4[ ' 'T-t=R+j 1 1 1,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) ' 'T-
t=R+j 1 1 1,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j)
27
' 'T-t=R+j 2 2 2,t+ 1,t+ -j 1 1H (t)B (t)h h B (t-j)H (t-j) + ' 'T-
t=R+j 2 2 2,t+ 2,t+ -j 2 2H (t)B (t)h h B (t-j)H (t-j) ]
= 4 ' 't 2 2,t+ 2,t+ -j 24 H (t)[Eh h ]H (t) + op(1).
We now need only show that the second bracketed term on the right-hand side of (A10) is op(1).
To do so we show that * ' ' ' * *Tt R j 1,t 1 1,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ( - ) h vec(x x ) (( - ) ( - )) is op(1). The
remaining terms follow similar arguments. Taking the absolute value we obtain
* ' ' ' * *Tt R j 1,t 1 1,t+ 1,t-j 1,t-j 1,t-j 1 1,t-j 1
ˆ ˆ ˆ| ( - ) h vec(x x ) (( - ) ( - )) |
3 1/3 * -1 ' ' 1/3 * 2tt 1,t 1 1,t+ 1,t-j 1,t-j t 1,t-j 1
ˆ ˆk (P/T)(sup T | - |)(P |h vec(x x ) |)(sup T | - |) .
Assumption 4 implies that P/T is bounded. Given Assumption 2, that -1 ' 't 1,t+ 1,t-j 1,t-jP |h vec(x x ) | is
Op(1) follows from Markov’s inequality. Since j is finite, that 1/3 *t 1,t 1
ˆ(sup T | - |) and
1/3 * 2t 1,t-j 1
ˆ(sup T | - |) are op(1) follows from Lemma A1 (c) and the proof is complete.
(b) Given Lemma A12 (b), Lemmas A6 and A7 imply that d is Op(P-1) and hence 2(P- -j)d =
op(1). Since j is finite this also implies that both 2 2T-t=R+j 1,t+ -j 2,t+ -jˆ ˆd (u -u ) and 2 2T-
t=R+j 1,t+ 2,t+ˆ ˆd (u -u )
are op(1). It then suffices to show that 2 2 2 2T-t=R+j 1,t+ 2,t+ 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ(u -u )(u -u ) =
4 1/2 ' ' 1/22 2,t+ 2,t+ -j 24(P/R) [R H (R)][Eh h ][R H (R)] + op(1).
First note that, in the notation of Lemma A12 (b), 2 2 2 2T-t=R+j 1,t+ 2,t+ 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ(u -u )(u -u ) =
T- 13 13t=R+j j=11 j=110,t 1,t 2,t 3,t 4,t 6,t 9,t j,t 0,t-j 1,t-j 2,t-j 3,t-j 4,t-j 6,t-j 9,t-j j,t-j{2r +2r +2r +2r +r +r +r + r }{2r +2r +2r +2r +r +r +r + r } .
If we take the product within the argument we obtain 100 distinct crossproduct terms. Rather
than list each explicitly, we will simply note that the important terms are w0,t 4z0,t = 4 0,t 0,t jr r ,
w1,t 4z1,t = 4 0,t 1,t jr r , w2,t 4z2,t = 4 1,t 0,t jr r and w3,t 4z3,t = 4 1,t 1,t jr r . The remaining terms are
28
each of the form in Lemma A9 (b) for some choice of index i,j,m,n {1, 2} and 0 r,s j .
Using this notation we obtain
2 2 2 2T-t=R+j 1,t+ 2,t+ 1,t+ -j 2,t+ -jˆ ˆ ˆ ˆ(u -u )(u -u ) = T-
t=R+j 0,t 1,t 2,t 3,t{w +w +w +w } + T- 99t=R+j i=4 i,t{ w }
* ' ' *T-t=R+j 1,R 1 1,t+ 1,t+ -j 1,R 1
ˆ ˆ4 {( - ) h h ( - ) * ' ' *1,R 1 1,t+ 2,t+ -j 2,R 2
ˆ ˆ( - ) h h ( - )
* ' ' *2,R 2 2,t+ 1,t+ -j 1,R 1
ˆ ˆ( - ) h h ( - )} + * ' ' *2,R 2 2,t+ 2,t+ -j 2,R 2
ˆ ˆ( - ) h h ( - )} + T- 99t=R+j i=4 i,t{ w }
= 1/2 ' -1 ' ' -1 'T- T-t=R+j t=R+j2 1 1,t+ 1,t+ -j 1 1 1,t+ 2,t+ -j 24(P/R)[R H (R)][JB (R)(P h h )B (R)J -JB (R)(P h h )B (R)
+ -1 ' ' -1 ' 1/2T- T-t=R+j t=R+j2 2,t+ 1,t+ -j 1 2 2,t+ 2,t+ -j 2 2B (R)(P h h )B (R)J +B (R)(P h h )B (R)][R H (R)]
+ T- 99t=R+j i=4 i,t{ w }.
Since wi,t = 4zi,t i = 0,…,3, the proof of Lemma A11 (b) implies that T-t=R+j 0,t 1,t 2,t 3,t{w +w +w +w }
= 4 1/ 2 ' ' 1/ 22 2,t 2,t j 24(P / R) [R H (R)][Eh h ][R H (R)] + op(P/R). The result follows since by
Lemma A9 (b), T- 99t=R+j i=4 i,t{ w }= op(P/R).
29
6. Theorems
Theorem 3.1: (a) Let Assumptions 1-4 hold. For the recursive, fixed and rolling schemes, let 1
equal 1 -1 'hhW ( )S dW( ) , -1 '
hh{W(1)-W( )}S W( ) and 1-1 'hh{W( )-W( - )}S dW( )
respectively. Similarly, for the recursive, fixed and rolling schemes let 3 denote
1 -2 ' 2hhW ( )S W( )d , -1 ' 2
hhW ( )S W( ) and 1-2 ' 2hh{W( )-W( - )}S {W( )-W( - )}d
respectively. ENC-T d1/2
1 3/ . (b) Let Assumptions 1-3 and 4 hold and let V0 and V1 denote
(k2 1) independent standard normal vectors. ENC-T d' ' 2 1/20 1 0 0hh hhV S V /[V S V ] ~ N(0,1).
Proof of Theorem 3.1: We will prove this for the recursive scheme. The fixed follows from the
recursive while the rolling follows from a proof similar to that for the recursive (a) Given
Theorem 3.4 and the Continuous Mapping Theorem it suffices to show that jccj=-j
ˆP K(j/M) (j)
d4
3 . That
jccj=-j
ˆP K(j/M) (j) = 4 ' 'jt 2 2,t+ 2,t+ -j 2j=-j K(j/M)[ H (t)[Eh h ]H (t)] + op(1)
= 4 ' 'jt 2 2,t+ 2,t+ -j 2j=-jH (t)[ K(j/M)(Eh h )]H (t) + op(1)
= 4 ' ' 'jt 2 2 2,t+ 2,t+ -jj=-j( [H (t) H (t)])vec[ K(j/M)(Eh h )] + op(1)
follows from Lemma A11 (a) and the fact that j is finite. Given Assumption 3, we know that
'j2,t+ 2,t+ -jj=-j K(j/M)(Eh h ) hhS . Note that by Lemma A6 and the Continuous Mapping
Theorem ' 't 2 2H (t) H (t) d
1 -2 ' 1/2 ' 1/2hh hh[W ( )S W ( )S ]d . The result follows since
1 -2 ' 1/2 ' 1/2hh hh hh( [W ( )S W ( )S ]d )vec[S ] = 3.
(b) Given Lemmas A10 (b) and A11 (b) we can write ENC-T as
30
ENC-T = 1/2 2 1/2 ' -1/2 1/2
t2 2,t+ p4 1/2 ' ' 1/2 1/2j
2 t+ t+ -j 2 pj=-j
(P/R) [R H (R)][P h ]+o ((P/R) )[(P/R) [R H (R)][ K(j/M)(Eh h )][R H (R)]+o (P/R)]
= 1/2 ' -1/2
t2 2,t+ p1/2 ' ' 1/2 1/2j
2 t+ t+ -j 2 pj=-j
[R H (R)][P h ]+o (1)[[R H (R)][ K(j/M)(Eh h )][R H (R)]+o (1)]
= 1/2 ' -1/2
t2 2,t+1/2 ' ' 1/2 1/2j
2 t+ t+ -j 2j=-j
[R H (R)][P h ][[R H (R)][ K(j/M)(Eh h )][R H (R)]]
+ op(1).
Given Assumption 2, Corollary 29.19 of Davidson (1994) suffices for -1/2 ' 1/2 ' 't 2,t+ 2(P h ,R H (R))
d (V 11/2hhS , V 0
1/2hhS ) for independent (k 1) standard normal vectors V0 and V1. Given
Assumption 3, we know that 'j2,t+ 2,t+ -jj=-j K(j/M)(Eh h ) hhS . The result follows immediately
from the Continuous Mapping Theorem.
Theorem 3.2: (a) Let Assumptions 1-4 hold. For the recursive, fixed and rolling schemes let 2
denote 1 -2 'hhW ( )S W( )d , -1 '
hhW ( )S W( ) and
1-2 'hh{W( )-W( - )}S {W( )-W( - )}d respectively. MSE-T d
1/21 2 3( -(0.5) )/ . (b) Let
Assumptions 1-3 and 4 hold. MSE-T ENC-T = op(1).
Proof of Theorem 3.2: We will prove this for the recursive scheme. The fixed follows from the
recursive while the rolling follows from a proof similar to that for the recursive (a) Given
Theorem 3.3 and the Continuous Mapping Theorem it suffices to show that jddj=-j
ˆK(j/M) (j)
d4
34 . That
jddj=-j
ˆP K(j/M) (j) = 4 ' 'jt 2 2,t+ 2,t+ -j 2j=-j4 K(j/M)[ H (t)[Eh h ]H (t)] + op(1)
= 4 ' 'jt 2 2,t+ 2,t+ -j 2j=-j4 H (t)[ K(j/M)(Eh h )]H (t) + op(1)
31
= 4 ' ' 'jt 2 2 2,t+ 2,t+ -jj=-j4 ( [H (t) H (t)])vec[ K(j/M)(Eh h )] + op(1)
follows from Lemma A13 (a) and the fact that j is finite. Given Assumption 3, we know that
'j2,t+ 2,t+ -jj=-j K(j/M)(Eh h ) hhS . Note that by Lemma A6 and the Continuous Mapping
Theorem ' 't 2 2H (t) H (t) d
1 -2 ' 1/2 ' 1/2hh hh[W ( )S W ( )S ]d . The result follows since
1 -2 ' 1/2 ' 1/2hh hh hh( [W ( )S W ( )S ]d )vec[S ] = 3.
(b) Given Lemmas A12 (b) and A13 (b) we can write MSE-T as
MSE-T = 1/2 2 1/2 ' -1/2 1/2
t2 2,t+1 p4 1/2 ' ' 1/2 1/2j
2 t+ t+ -j 2 pj=-j
2(P/R) [R H (R)][P h ]+o ((P/R) )[4(P/R) [R H (R)][ K(j/M)(Eh h )][R H (R)]+o (P/R)]
= 1/2 ' -1/2
t2 2,t+1 p1/2 ' ' 1/2 1/2j
2 t+ t+ -j 2 pj=-j
[R H (R)][P h ]+o (1)[[R H (R)][ K(j/M)(Eh h )][R H (R)]+o (1)]
= 1/2 ' -1/2
t2 2,t+11/2 ' ' 1/2 1/2j
2 t+ t+ -j 2j=-j
[R H (R)][P h ][[R H (R)][ K(j/M)(Eh h )][R H (R)]]
+ op(1).
The result follows immediately from the proof of Theorem 3.1 (b).
Theorem 3.3: (a) Let Assumptions 1, 2 and 4 hold. MSE-F d 1 22 - . (b) Let Assumptions 1,
2 and 4 hold. For the (k2 1) independent standard normal vectors V0 and V1 defined in
Theorem 3.1 (b), (R/P)1/2MSE-F d'0 1hh2V S V .
Proof of Theorem 3.3: (a) That -1 2t 2,t+ˆP u p
2 follows from Theorem 4.1 of West (1996).
Given Lemma A12 (a) the result then follows from Lemmas A6 and A7.
(b) That -1 2t 2,t+ˆP u p
2 follows from Theorem 4.1 of West (1996). Note that Lemma A12 (b)
implies that 2 2t 1,t+ 2,t+ˆ ˆ(u -u ) = 2 1/2 1/2 ' -1/2
t2 2,t+2 (P/R) [R H (R)][P h ] + op((P/R)1/2). Given
32
Assumption 2, Corollary 29.19 of Davidson (1994) suffices for -1/2 ' 1/2 ' 't 2,t+ 2(P h ,R H (R)) d
(V 11/2hhS , V 0
1/2hhS ) for the independent (k2 1) standard normal vectors V0 and V1 from Theorem
3.1. Scaling by (R/P)1/2 provides the desired result.
Theorem 3.4: (a) Let Assumptions 1, 2 and 4 hold. ENC-NEW d 1 . (b) Let Assumptions 1,
2 and 4 hold. 2(R/P)1/2ENC-NEW – (R/P)1/2MSE-F = op(1).
Proof of Theorem 3.4: (a) That -1 2t 2,t+ˆP u p
2 follows from Theorem 4.1 of West (1996).
Given Lemma A10 (a) the result then follows from Lemma A6.
(b) That -1 2t 2,t+ˆP u p
2 follows from Theorem 4.1 of West (1996). Note that Lemma A10 (b)
implies that 2t 1,t+ 1,t+ 2,t+ˆ ˆ ˆ(u -u u ) = 2 1/2 1/2 ' -1/2
t2 2,t+(P/R) [R H (R)][P h ] + op((P/R)1/2). Other than
the factor 2, this is identical to the expansion in the proof of Theorem 3.3 (b). Scaling by (R/P)1/2
provides the desired result.
33
7. References
Corradi, V., N.R. Swanson and C. Olivetti, 2001, Predictive ability with cointegrated variables, Journal of Econometrics, 104, 315-358.
Davidson, J., 1994, Stochastic Limit Theory (Oxford University Press, New York).
Hansen, B.E., 1992, Convergence to stochastic integrals for dependent heterogeneous processes, Econometric Theory, 8, 489-500.
Magnus, J. and H. Neudecker, 1988, Matrix Differential Calculus with Applications in Statistics and Econometrics, (Wiley, New York).
West, K.D., 1996, Asymptotic inference about predictive ability, Econometrica, 64, 1067-84.
34
8. Referenced Corollaries and Theorems
This section contains six of the primary technical references used throughout the appendix:
Corollaries 29.11 and 29.19 of Davidson (1994), Theorems 3.1, 3.3 and 4.1 of Hansen (1992)
and Theorem 4.1 of West (1996). For ease of reference, the notation used within this paper is
adopted as much as possible. To insure understanding of the issues, we recommend reading the
original sources.
Corollary 29.11 of Davidson (1994): Let the conditions (a), (b) and (c) of 29.4 hold, and instead
of condition 29.4(d) assume (d ) E(Xn(t)) 0 and E(Xn(t)2) (t) as n , each t [0,1].
Then Xn d B .
Corollary 29.19 of Davidson (1994): Let t+{U } be a zero-mean, uniformly Lr-bounded m-
vector sequence with each element L2-NED of size -1/2 on an -mixing process of size -r/(r-2);
and assume -1 'T- T-t=1 t=1t+ t+T E( U )( U ) . If XT(s) = -1/2 [sT]
t=1 t+T U , then XT W(s; ).
Theorem 3.1 of Hansen (1992): If Assumption 1 (from the Hansen article) holds and
1/ 2 1/ 2[sT] [sT]t 1 t 1t t(T U ,T V ) (U, V) in km mM
D [0,1] then 1/ 2 1/ 2[rT] ii 1 t 1 t i 1(T U )(T V ) d
r0 U dV with V(s) = W(s; ), and being defined in Assumption 1 (from the Hansen article).
Theorem 3.3 of Hansen (1992): Suppose Un U in kmMD [0,1] and U(.) is almost surely
continuous. For a random sequence {ei} and a sequence of nondecreasing sigma fields { ei } to
which {ei} is adapted, assume that ri i i-msup E|E(e | )| 0 as m . Then -1 [ns]
i=10 s 1 ni isup |n U e |
p 0.
35
Theorem 4.1 of Hansen (1992): If Assumption 1 (from the Hansen article) holds, then
-1/2 -1/2 '[rT] ii=1 t=1 t+ i+(T U )(T U ) d r '
0 BdB +r as T , where B W(s; ), and being
defined in Assumption 1 (from the Hansen article).
Theorem 4.1 of West (1996): (a) If = 0 or F t+1E f ( )/ = 0, 1/2 -1 *Tt=R t+1 t t+1
ˆP (P f ( )-Ef ( ))
d N(0, ) where = Sff1/2 -1 * * 2T
t=RT t+1 t+1lim E[P (P f ( )-Ef ( ))] . (b) If S = '
ff fh' 'fh hh
S S BBS BS B
'ff fh
'fh
S S BBS V
is p.d., 1/2 -1 *Tt=R t+1 t t+1
ˆP (P f ( )-Ef ( )) d N(0, ) where
= Sff + (1 -1ln(1+ ))( ' ' 'fh fhFBS +S B F ) + 2(1 -1ln(1+ ))FV F’.
36
9. Summary Table of Limiting Distributions
This section contains 4 panels, one each for the MSE-T, ENC-T, MSE-F, and ENC-NEW
statistics. Each panel contains the limiting distributions of the statistics for a given forecasting
scheme (recursive, rolling or fixed) and value of P,Rlim P/R = (0 < < or = 0).
Panel 1: MSE-T
Scheme\ 0 < <
Recursive1 1-1 ' -2 '
hh hh1 -2 ' 2 1/ 2
hh
W ( )S dW( )-(0.5) W ( )S W( )d[ W ( )S W( )d ]
Rolling1 1-1 ' -2 '
hh hh1-2 ' 2 1/ 2
hh
{W( )-W( - )}S dW( )-(0.5) {W( )-W( - )}S {W( )-W( - )}d[ {W( )-W( - )}S {W( )-W( - )}d ]
Fixed-1 ' -1 '
hh hh-1 ' 2 1/ 2
hh
{W(1)-W( )}S W( )-(0.5) W ( )S W( )[ W ( )S W( )]
Note: If = 0 MSE-T ENC-T = op(1) so the MSE-T distribution can be inferred from Panel 2.
Panel 2: ENC-T
Scheme\ = 0 0 < <
Recursive N(0, 1) 1 -1 '
hh1 -2 ' 2 1/ 2
hh
W ( )S dW( )[ W ( )S W( )d ]
Rolling N(0, 1) 1-1 '
hh1-2 ' 2 1/ 2
hh
{W( )-W( - )}S dW( )[ {W( )-W( - )}S {W( )-W( - )}d ]
Fixed N(0, 1) -1 '
hh-1 ' 2 1/ 2
hh
{W(1)-W( )}S W( )[ W ( )S W( )]
37
Panel 3: MSE-F
Scheme\ = 0 0 < <
Recursive '0 1hh2V S V 1 1-1 ' -2 '
hh hh2 W ( )S dW( )- W ( )S W( )d
Rolling '0 1hh2V S V
1-1 'hh
1-2 'hh
2 {W( )-W( - )}S dW( )
- {W( )-W( - )}S {W( )-W( - )}d
Fixed '0 1hh2V S V -1 ' -1 '
hh hh2 {W(1)-W( )}S W( )- W ( )S W( )
Note: When = 0 the definition of the MSE-F changes. It is rescaled by (R/P)1/2.
Panel 4: ENC-NEW
Scheme\ = 0 0 < <
Recursive '0 1hhV S V 1 -1 '
hhW ( )S dW( )
Rolling '0 1hhV S V 1-1 '
hh{W( )-W( - )}S dW( )
Fixed '0 1hhV S V -1 '
hh{W(1)-W( )}S W( )
Note: When = 0 the definition of the ENC-F changes. It is rescaled by (R/P)1/2.
