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working paper
department
of economics
A NOTE ON SPURIOUS BREAK AND REGIME SHIFT
IN COINTEGRA TING RELA TIONSHIP
Jushan Bai
May 1996
massachusetts
institute of
technology
50 memorial drive
Cambridge, mass. 02139
A NOTE ON SPURIOUS BREAK AND REGIME SHIFT
IN COINTEGRA TING RELA TIONSHIP
Jushan
96-13
Bat
May 1996
MASSACHUSETTS INSTITUTE
JUL 18
1996
LSBRAHltS
96-13
A
Note on Spurious Break and Regime
Shift in
Cointegrating Relationship
by
Jushan Bai 1
Massachusetts Institute of Technology 2
May, 1996
1
Partial financial support
is
acknowledged from the National Science Foundation under
Grant SBR-9414083.
2
Please send correspondence to Jushan Bai, Department of Economics, E52-274B,
Cambridge,
MA
02139. Phone: (617) 253-6217. Email: jbaiQmit.edu.
MIT,
96-13
Abstract
The simulation
result of
Nunes, Kuan, and Newbold suggests that
it is
possible to
estimate a spurious break for a regression model with 1(1) disturbances. In this note,
we provide a
rigorous proof for this
phenomenon.
We
also
applies to integrated regressors, so that a spurious regression
break. However,
if
is
in
terms of the integer index rather than
This rapid rate of convergence
1(0) regressors.
may
their finding
lead to a spurious
two integrated processes are cointegrated with a structural change
in the cointegrating relationship, the break point can
consistency
show that
is
be consistently estimated. The
in
terms of the sample fraction.
not attainable for stationary or, more generally, for
Furthermore, the consistency holds even when magnitudes of breaks
are small but do not converge to zero too fast.
These consistency
results are also
obtained for a broken trend model.
Key words and
phrases.
Spurious break, spurious regression, change point, cointe-
gration, broken trend.
Running head: Spurious Break
Introduction
1
Newbold (1995) pointed out that when the disturbances
Recently, Nunes, Kuan, and
of a regression
model follow an
1(1) process there
is
a tendency to estimate a break
point in the middle of the sample, even though a break point does not exist.
phenomenon
is
by the authors and was discovered by a
called a "spurious break"
simulation experiment. In this note,
we provide a
may
We
phenomenon.
rigorous proof for this
Furthermore, we show that a spurious break occurs
a spurious regression
This
for 1(1) regressors as well, so that
lead to a spurious break.
then consider the problem in which the dependent variable and the 1(1)
re-
gressors are cointegrated but the cointegrating relationship undergoes a shift. This
a more general notion of cointegration because the cointegrating vector
invariant.
We ask
is
not time-
the question whether the break point can be consistently estimated.
It is
shown that the estimated break point converges quickly to the true break
The
rate of convergence
is
faster
given the same magnitude of
A
than the corresponding result for
pirical applications.
may be
shift.
A
em-
Cointegration describes a long-run-equilibrium condition.
An
disturbed by policy regime changes, resulting in a
special case
which can be expressed
This
shift is
model
in
rium, so that a different cointegrating vector
equilibrium.
point.
1(0) regressors,
structural change in a cointegrating relationship can be a useful
equilibrium
is
is
a shift in the
may be needed
mean
new
equilib-
to characterize this
new
level of the long-run equilibrium,
as a shift in the intercept of a cointegrating-regression
model.
exhibited graphically as a change in the "gap" between two cointegrated
series.
Although not a concern
which allows
Hansen
Ericsson, and
Hendry
is
and
Phillips (1993),
(1996).
that one
we
point out that testing for cointegration
change has been studied by a number of authors;
for a structural
(1992), Quintos
relationship
of this paper,
may
One
see, e.g.,
Gregory and Hansen (1996), and Campos,
implication of a structural change in a cointegrating
not be able to reject the null of no cointegration
if
conventional tests are used, even though a long-run relationship between two series
does, in fact, exist. This situation calls for use of the test statistics proposed
by the
aforementioned authors.
Spurious Break
2
Consider the model:
Vt
(
=
x'tPi
+
x't /32
+e
{
{
et
t
Let $\{k) be the least squares estimator of
t
=
1,2,...,*
t
=
k
+
l,...,T
based on the
f3\
first
T—
$2(k) be the least squares estimator of #2 based on the last
k observations, and
k observations,
i.e.,
k
Pi( k )
= {J2 x
t
x 't)
T
&(*)
=
.
E
(
(J2 x tyt),
_,
**<)'
sum
E
(
t=k+l
Define the
T
.
*&)•
t=k+l
of squared residuals for the full
k
sample
as
T
2
E
sT (k) = Y,(yt-x'Mk)) +
(yt-x'Mk))
t=k+l
t=\
and define the break point estimator
k
=
as
,
argmin 1 < < T S T (A:).
fc
Finally, let
AT
where
<
A
<
A
<
1.
= min{A
:
The behavior
A
=
of
argminue[AiI] 5T ([Tu])
Aj
is
considered for two cases: 1(0) and 1(1)
error processes.
For an 1(0) error process,
let
Q(X) and R(\) be defined as
Kuan, and Newbold (1995, hereafter NKN),
the limit of
the limit of
Dj
DT
(Y/t=i x t x't )D^
1
In
for
some
Q>
More
and [A3] of Nunes,
Q(X)
is
an appropriate scaling matrix Dj, and R(\)
is
respectively.
specifically,
Xrp \ 1
52t=i x t^t-
strictly increasing.
for
in [Al]
The
and
The matrix Q(X)
process R(X)
-R(A)
is
is
is
Gaussian.
a Brownian motion.
assumed
When
x
to
t
be positive- definite and
is
stationary,
Q(X)
= \Q
we assume
In this section,
Theorem
their
there
is
no break,
i.e.
=
j3\
NKN
/?2-
show that
3.1b)
A T -i argmax Ae[A
TiM(A)
\e[A,A]
where M(X)
For an 1(1) error process e t
-
[i?(l)
R(X)]'[Q(1)
NKN
1
),
for
some a >
-
we assume that T~ 2
,
>
being a standard Wiener process, and c
[A3'] in
(1)
a stochastic process given by
is
M(A) = R{\)'Q{X)- X R{\) +
W(u)
(see
Q^m) -
J2 t =i £
2
/2(A)].
(2)
c/ A W^ 2 (u)g?u with
=>•
a constant. Further assume (see
0,
[XA]
T-*/
where G(A)
is
Gaussian process.
2
/^
1/2
£> ^ G(A)
(3)
<£t
NKN
prove that
A T -i argmax Ae[AjX] M*(A)
(4)
where
M*(A)
=
1
G(A)'g(A)" G(A)
Examining
(2)
and
(5)
we
+
[G(l)
find that,
said
is
a
random
Andrews
M(A) —>
oo as A
(1993). In their
-
^(A)]-
in the
—
>
or
Remark
However,
1,
thus At
1 (p.
to characterize the limiting behavior of
742),
—
{0, 1}, if A
for A near
Their original assumption
current form allows
process.
j/<
is
stated in terms of
yt
—
or
not diverge to infinity as A decreases to zero or increases to
The
(5)
1(0) or 1(1), the
for /(0) error process e t
NKN pointed out
M*(X)
is
G(A)].
Not much further can be
.
they find that M*(A) behaves differently from M(X). More
1
^!) -
absence of a break, the estimated
,
[A, A].
1
whether the error process
variable with support in [A A]
on the compact interval
prove that
G(X)]'[Q(l)
Namely,
results are essentially the same.
break point At
-
rather than
et,
,
and A
NKN
—
>
1;
further
also see
that they were unable
1.
Through simulation,
specifically,
M*(X) does
1.
which applies to
yt
being
1(1).
to depend on deterministic regressors as well as on an additive 1(1) error
we
In the following,
shall
prove that M*(X)
uniformly bounded in probability over
of
T' a
MT ([TA]), where a
k
t
t=l
T
k
_j
(E
<=1
x t)
Note that M*(X)
[0, 1].
[0, 1]
and
is
the limiting process
is
defined in (3) and
is
k
,
M£(k)= (J2 £
a well defined process on
is
{J2 x £ t)
x<x
+
t
E
{
T
,
e t x t)
E
(
t=k+l
t=l
T
-1
x*x
(
t=k+l
E
x < e <)
t=k+l
(6)
We
assume that a >
shall
2 because this
When
regressor (e.g., a constant, or a trend).
is
possible that
Theorem
a
For
1
=
This case
1.
the
a
when x contains a nonzero mean
true
is
xt
t
is
a 7(0) process with zero mean,
it
not considered in this paper.
is
defined in (3), assume
a >
2.
We
have
= Op (l).
sup M*(A)
(7)
Ae(o,i)
Proof of Theorem
P, we have
z'
Pz <
For an arbitrary vector z and an arbitrary projection matrix
1.
z'z.
Apply
this inequality to
Mj{k)
to obtain
T
MT(k)<J2 £
a >
Since
T~ 2
2,
J2t=i £ t
M*(A),
To
is
we have T~ a M$(k) < T~ 2
h as a hmit,
T~ a Mj{k)
=
is
rule out the possibility that
and A
=
1.
As the
and
limit of
for A
is
=
obtained from
0.
(5)
(8)
[1,T].
Moreover, because
uniformly bounded in probability. Thus
—
Xt
1.
>
{0,1},
we need
0,
to further
M*(A)
is
not defined yet at
M*(0) should be defined
as
=
0,Q(A)
is
=
(9)
0,
and C(A)'g(A)- 1 G(A)
the limit of the
first
term
T~ a Now
.
HEWE^r'E^) < r^e? < r- ^
2
i=l
examine the
= GilYQil^Gil)
Note that the term G(A)'Q(A)- 1 G(A)
i=l
its limit,
(0, 1).
Strictly speaking,
by taking G{X)
the right hand side divided by
E
for all k
M*(X) when A —>
M*(0)
which
£f=1 e]
uniformly bounded in probability for A £
behavior of M*(A) for A near
A
forallJfc€[l,T].
t
t=l
t=\
4=1
22
£t
=
of (6)
on
which converges to zero
any given
in probability for
1
follows that G(A)'(5(A)- G(A) ->
in probability as A -> 0.
we can
is
the limit of M*(A) as A —*
0.
A
—
next show that the
at
»
1,
or
=
M*(l)
2
(i)
With probability
M*(0)
If
=
[TX] with A
Thus the
of
>
0.
It
definition of (9)
define, as the limit of
maximum
—
M*(X)
is
M*(A) when
not attained
1.
Theorem
fiij
We
M*(0).
Similarly,
k
k, or for
=
1,
M*(l) < M*(A),
<
/or every
A
<
1.
(10)
G{X) has a nonsingular covariance function, then with probability
M*(0)
To prove Theorem
Lemma
1
=
M*(l) < M*(A),
we need the
2,
For arbitrary
<
/or every
following
A
<
1
1.
(11)
lemma.
positive-definite matrices
A
and
B
with
-
y)
A>B
(p
x
p),
and
arbitrary vectors x and y (p x 1), we have
x
x'A~ x
Proof of
Lemma
1
:
to prove
equal to
-z'Hz
with the
first
x
y
-
(x
- y)\A - B)~
l
(x
<
0.
(12)
Define the matrix
(
^
^ ~
It suffices
- y'B~
H to
for z'
p rows
=
(/,
{A
(A-B)-i-A-*
-(A-B)-*
-(A-B)'
{A-By' + B1
be positive-semidefinite because the
(x',y').
Let
D=
(A
\
J'
left
- B)' + B' >
1
1
(A — B)~ l D~*) and second p rows
[l6)
1
0.
(0, /).
hand
Let
side of (12)
C
is
be a matrix
Using the identity
- B)- - A- = (A- B)- D~ {A - B) -l
1
1
1
1
to obtain
C'HC =
Thus
C'HC
lemma.
is
positive-semidefinite, so
diag(0,
is
D) >
0.
H because C has full rank.
This proves the
Proof of Theorem
The
2.
M*(0) < M*(A)
inequality
equivalent to
is
1
1
1
G(l)'g(l)- G(l)-G(A)'g(A)- G(A)-[G(l)-G(A)]'[Q(l)-Q(A)]- [G(l)-G(A)]<0.
Clearly, part
=
x
let
£
=
Theorem
of
=
G(l), and y
H
be defined
G(A).
in (13).
(G(l)', G(A)')'. Let
>
with Ai
the
(i)
>
A2
maximum
•
•
•
2 follows
rj
so does
the
is
implying
—
2
first
with probability
The above
NKN.
Let
(ii).
Then M*(0) < M*(A)
is
letting
A =
A=
equivalent to
A 2p where X^s are the eigenvalues of H. Since
<
component
is
B=
Q{1),
B =
Q(l) and
-£'#£ <
T be an orthogonal matrix such that T'HT
eigenvalue of i/
A1
/;
by
1
,
Thus T£
£.
Lemma
Next, consider
-?Ht =
where
from
=
H
Q(X) and
0,
diag(A l5
>
Q(X),
and
where
...,
H
A 2p )
^
0,
positive. It follows that
is
-(rO'diag(A 1 ,...,A 2p )r^
When
of T£.
<
-r)
2
^
G(X) has a nonsingular covariance matrix,
a vector of normal variables with a nonsingular covariance matrix,
with probability
1
because P(rj 2
=
0)
=
0.
That
is,
—£'H£ <
1.
analysis applies to 1(1) regressors as well, which
is
not considered by
Let
yt
where both x and e f are
t
1(1), so that
=
x't fi_+ e t
we have a spurious
regression.
Assume
[TX]
T- 2 J2x
^Q(X)
t x't
(14)
t=i
where Q(X)
i.e.
Q(u)
is
a stochastic matrix with Q(X)
— Q(v) >
with probability
1
>
for v
<
(a.s.)
u.
and Q(X)
is
strictly increasing,
Also assume
[TX]
T- 2 J2x
t
et
^G(X)
(15)
t=i
where G(-)
is
a vector of
random
processes,
and
for
each
A,
G(X) possesses a density
function and a nonsingular covariance matrix.
Theorem
3 The results of Theorem
1
and Theorem 2 apply
to 1(1) regressors satis-
fying (14) and (15), so that spurious regression leads to a spurious break.
Proof of Theorem
proof of (7) under the
new
only need to note that
if
definite with probability
the proof
is
imply
with
(3)
DT = T
A—B
A, B, and
1,
2
I and
To prove
.
we
are stochastic matrices that are positive-
then inequality (12) holds with probability
We conjecture that
responsible for
is
Theorem
1.
The
rest of
.
2.
grated
may
its
a spurious break will not occur because an 1(1) error
occurrence.
Shift in Cointegrating Relationships
In the previous section
we show
that two integrated processes that are not cointe-
What happens when
give rise to a spurious break.
the two processes are
cointegrated but the cointegrating relationship undergoes a shift?
be consistently estimated in the presence of
The
Can
the shift point
1(1) regressors?
issue of a structural change in cointegrating relationships
of considerable
is
A
interest. Cointegration describes a system's long-run equilibrium condition.
may have
A
structural change
model allows us to describe such a system. Consider,
{ai+jix + e
t
C*2
where x t
=
Xt-i
+
et
with x
=
Yl'jLo Q-jVt-j
sequences with
+
8 (8
is
we assume
that
a shift in the
fc
mean
>
=
t
and Var(e t )
=
finite 4
t=
t
+ 72Z< + £t
processes such that e t
addition,
and
et
>
=
=
0.
[Tr
]
for
(16)
+ l,...,T
fc
We
assume
Y^jLo^j(,t-ji
moments, and
0)
1,2, ...,k
some r G
X2jj| a j|
(0, 1).
<
e t are e t are 1(0) linear
where
rj t
=
- a 2 ,7i -
72)'.
£ t are
When
71
=
72,
Let
X
t
=
(l,x t )'.
f3i
=
(0:1,71)'
When
and
<
/3 2
=
is
71
i.i.d.
co. In
but «i
of the long-run equilibrium. This intercept shift
a shift in the cointegrating relationship. Let
(ati
and
oo and J2jj\bj\
visualized as a change in the "gap" between two cointegrated series.
is
system
multiple long-run equilibria with an occasional shift from one equilibrium
to another.
and 8
(10),
t
Regime
there
The
remains an open question whether a spurious break arises when y t and x are
process
there
= 2.
a
identical to the previous proof because inequality
virtually identical to that of
is
cointegrated.
3
setting
(15)
a pure mathematical inequality and holds for arbitrary x t
(8) is
It
and
First, (14)
3.
^ a2
,
often
^
72,
(02,72)'
evident that the larger the magnitude of a
It is
The
break.
magnitude
it is
to identify the
rate of convergence for the estimated break point depends not only on the
on the magnitude of the regressors.
of the shift in the coefficients, but also
For stationary
where
the easier
shift,
X
t
the rate depends on the "effective magnitude of shift", 8'Q6/cr^,
,
Q = E{X X[).
t
In this case,
can be shown that k
it
=
+ O p (l).
k
This
the
is
best rate that can be achieved for stationary regressors; see, e.g., Bai (1994, 1995).
not consistent for ko, although in terms of the sample fraction,
Furthermore, k
itself is
k/T converges
at a rate of
k becomes consistent for
converges to
as
t
—»
infinity.
A:
More
T
to r
With
.
specifically,
—
00 as
>
break point more precisely than with
Lumsdaine and Stock (1994)
For the
4 Let k denote the
least
if
T —*
72,
then a(t)
=
k
)
=
such that P(\k
-
k
Because k
> M) <
\
P(k +
4.
—
>
Based on
{k
:
—
\k
that the event {k £
/
'
=
ko
->
+ Op (l),
>
00,
one can estimate the
this fact,
=
ko
+ O p (l);
we can
establish
1.
squares estimator ofko- Assuming that ko
)
—
o~\
=
[Tto].
we have
1.
for
any
>
e
0,
there exist an
M < 00
Thus
t.
= P{\k-ko\>M) + P(\k-ko\<M,k^ko)
ko)
<
Dm =
'E{XtX[)8
In particular, k
shifted- cointegrating relationship of (16) with 71 7^ 72,
Proof of Theorem
show that even
6'
00. Consequently,
for a proof.
P(k = k
where
^
71
1(0) regressors.
a more interesting result. Namely, P(k
Theorem
shall
This follows because the "effective magnitude of shift"
-
00. In particular, a(ko)
see Bai,
we
1(1) regressors,
k
Dm}
\
e
+ P(keD M
< M,
k
/
implies that
k
}.
(17)
)
By
definition,
{mmk e D M
Sj(k)
Srik)
<
<
J2t=i £ t-
YlJ=i £ ?}-
^
follows
Thus
T
P(k
e Dm)
v
'
<P( mm
keD
M
\
ST (k)<J2tf)j
(18)
'
We show that the right hand side of (18) converges to zero for every given M.
is
a finite
set, it is sufficient to
show that
for
8
each k G
Dm, P{ST(k) <
Since
Ylt=i £ t)
Dm
~~* 0-
We
<
prove this for k
Y1<k
=
=
=
=
Xi,k
£i,fc
^*:,A:o
Furthermore,
Then
for k
The sum
<
M
let
k
,
kQ
The
.
>
case of k
ko
(yi,...,yk)',
Ykj
(Xi,...,Xk)',
Xk,r
(ei,
£k,T
.-., £*:)',
(Xk+l,..-,Xk
is
=
=
=
=
X^ T
)',
= I - X^X^xjk)^
x
similar. Define
1Je
(y*+i,-,w)'
pGfc+i,...,Xr)'
(£fc+i,...,£x)'
(Xk+l,...,Xko ,0,...,0)'.
M
and
2
= I - X'KT {X'KT X k T )-
Xk T
1
,
.
,
we have
of squared residuals Sr(k)
given by
is
= n'^F!,, + y iT M2 n, T
'
Sr(fc)
fc
=
£i, fc
M
l£l
,
*
2^'X T 'M2 e,, T
+ e^Maejb.T +
fc
fc
+ 8' X^'
2
Xt<T 8
T
=
E
~
et
e i,*- x i,*(^J,*- x u)~ ^i,fc £ i,fc
_
r
£ fc,T^ fc,T(-^it,r^A:,T)~
X'kT e k ,T
k
+28'
J2
X et-28X'kM Xk,k
1
(X'ktT Xk,T)~ Xk,Te k ,T
t
t=k+l
ko
+ S '(
E
X Xt)t - X XWo( Xk,T Xk,T)~
$'
t
'k
<
ko
1
X'k jCo Xk,k
8
t=k+l
T
= Yl £2 +
t
+
aT
ir
+ cT +
dT
+
eT
+
/x
t=\
We
have used the fact that
J2t=k+i
x Xv
XkT
F° r each k G Dm,
t
= EtU+i
'e k ,T
it is
xt£t
and
XkT 'Xk
,T
= X'kM Xk
easy to see that ax, 6j, dx, and /x are
all
,k
=
p (l).
Thus
«?r(A;)-E £ ?
For each
A;
<
the
ko,
for k
=
ko
—
T
.2
St(*)-E*?
t=l
1,
**£«
+
is
the
P (T)
sum
right
E
*'(
xtX?)8 +
p (l)
(19)
t=k+l
t=fc+l
term on the
first
whereas the second term
example,
E
= 2^
(=1
hand
side of (19)
and dominates
involves only one
in
is
bounded by
magnitude the
summand, and
first
Op (y/T),
term. For
=
26'Xko e ko +S'(X'kQ
=
28 A s ko
X
ko
+ Op (l)
)6
+28 2 (^e)e ko +8l +
2
e )+ 8
l(l2 e
(^
i=\
i=l
28 l 8 2
z=l
0p (62 Y:e
The
any
+
)
>
e
°v( l )
(
2 0)
j:e y
(8 2
i
term can be rewritten
last
and converges
for
l
+
>)
as
T£f(-!=Ei_i
e,)
2
,
which dominates the
to positive infinity with probability approaching
1.
first
term
This implies that
0,
P(ST (k)-^e 2 <0)<e
t
t=i
for large T.
Combining with
for all large
T
When
for ko,
from
and
(18),
only the intercept has a break
magnitude of
(20).
we have,
for every e
>
0,
P{k
^
k
)
<
2e
.
even though k/T
"effective
(17)
When
82
converges to
still
shift" stays
=
0,
(at\
/ a2
ji
,
To at
=
72), k
is
a rate of T.
no longer consistent
This
is
because the
bounded. The lack of consistency can also be seen
ST (k) - ELi
e?
=
2<
^ £ *o +
Sl
+
P (1),
which cannot be
guaranteed to be positive.
se.
The underlying reason
for k being consistent for ko
Roughly speaking,
=
if
k
+ O p (l)
ko
is
not the 1(1) regressor per
any
holds, then for
set of regressors
that the second term (19) converges to infinity and dominates the
consistent for ko- In particular, this
we
is
first
X
t
such
term, k will be
true for polynomials regressions. For simplicity,
consider the following broken-trend model:
{oi +7i< + £t
oc 2
where
et
is
+
-y 2 t
a linear process such that
<
+e
e<
t
=
Yl'jLo
t
=
1,2,
t
=
k
..., Ar
(21)
+
l,...,T
a j r]t-j with
and Y^]LoJ\ a j\ <
r\ t
a sequence of martingale
differences
and
Theorem
5 For the broken trend model (21), assume 71 =^72-
sup,- Er\\
00,
squares estimator of ko with ko
P(k
=
=
Then
[Tto]
&o)
—*
°°-
as
1,
10
T—
>
00
Let k be the least
MIT LIBRARIES
3 9080 00988 2306
Proof of Theorem
Let
5.
X
t
The
(19) can be copied here.
=
Then the proof
(1, t)'.
hand
right
side (19)
X
(with probability 1) for the newly defined
k
=
—
ko
for
t
still
of
Theorem 4 up
to equation
converges to positive infinity
each k £
Dm-
For example, for
1,
StW-^s?
=
26'Xko e ko +6'(X'ko
=
2<5ie feo
X
ko
)S
+ O p (l)
6\
+
28l 82 k
t=i
+
= O p {6 2 k +
)
This implies that, for each k
infinity.
Although Theorem 5
=
ko
+ O p (l)
(but k
DM
,
ko
and Tj2
Thus,
if
the model
is
cast in the
t
for
>
t
written as
is
ko)- If
<
0)
^
we
^r,
not unex-
it is
it is
proved that
rewrite the broken trend
model
then the new slope coefficients
The magnitude
.
^ and
0.
of shift will
of Bai (1994, 1995),
be T(~/ 2
one
—
71).
essentially
is
the consistency of k for ko
shift. In this sense,
not surprising.
The
results of
Theorem
4 and
Theorem
of shift to converge to zero. Let 8 2< j
Theorem
^2,T
~~>
0, the
model
2
still
hold even
(21),
6.
assume T8 2< j
In fact, the
>
c
>
we
allow the magnitude
>
00.
assume y/T82 ,T —* 00
•
For
Under these assumptions, even
consistent for ko- That
of shift,
is,
P(k
we
—
1.
have k
=
—
still
if
ko)
»
and Bai, Lumsdaine and Stock (1994). 2 Consider the
assumed magnitude of shift
trend model, 62,t/VT
is still
—
(16),
Under the assumed magnitude
see Bai (1994, 1995)
if
= 72 — 71.
estimated break point k
Proof of Theorem
+ O p (l);
5
6 For the cointegrated regression model
the broken trend
k
k
framework
assuming an unbounded magnitude of
is
£?
not explicitly presented in the literature,
<
for
(22)
P {1)
and converges to positive
first
P(ST (k) - ELi
(21) with the linear trend expressed in the format
become Tji
+
2
not consistent for
is
2
)
(82 T) t*
In Bai (1994, 1995), the linear trend
pected.
k
is
<E
(8 2 k
)
term above dominates the
follows that the second
+
2
(8 2 k
= Op (62 Tt ) +
It
+
26 2 k e ko
is sufficient
is
stronger than necessary. For example, for the broken-
for k
=
k to be consistent for ko-
11
ko
+ Op (l).
But the assumption
is
necessary for
cointegrating regression (16). All that
infinity
with probability
1.
As
is
we
before,
X
ST (k) -
£
right
-4=
hand
model
is
Si=i
ei
is
to prove (19) converges to positive
consider k
£
= O p (VT62 T ^=
Vi ,=i
k
—
1.
,
+ {VT6 2 T
random
converges to a normal
By
,
variable
(20),
~ ± e^
k
i
e t)
side of (23) converges to positive infinity.
similar
=
k
i
e?
t=l
Because
needed
Vi
(23)
i=i
and VT82J —*
The proof
for the
oo, the
broken trend
and thus omitted.
References
[1]
Bai, J. (1994) Estimation of Structural
Working Paper No.
[2]
94-6,
Department
Change based on Wald Type
Statistics.
of Economics, M.I.T.
Bai, J. (1995) Least Absolute Deviation Estimation of a Shift. Econometric The-
ory 11, 403-436.
[3]
Bai,
J.,
R.L. Lumsdaine, and J.H. Stock (1994) Testing for and Dating Breaks in
Time
Integrated and Cointegrated
Series.
Manuscript, Kennedy School of Gov-
ernment, Harvard University.
[4]
Gregory,
A.W. and B.E. Hansen
models with regime
[5]
Hansen, B.E. (1992) Tests
cesses.
[6]
shifts.
(1996) Residual-based tests for cointegration in
Journal of Econometrics 70 99-126.
for
parameter instability
Journal of Business and Economic Statistics
Nunes, L.C.,
CM.
Kuan,
P.
in regressions
with 1(1) pro-
10, 321-335.
Newbold (1995) Spurious Break. Econometric The-
ory, 11, 736-749.
[7]
Quintos, C.E. and P.B.C. Phillips (1993) Parameter constancy in cointegrating
regressions. Empirical
Economics
18, 675-703.
12