1. INTRODUCTION
A vector autoregressive (VAR) process has often been used to model a
multivariate economic time series and, following the seminal work of Engle
and Granger (1987), a cointegrating relation has
been incorporated into the VAR model. A typical n-dimensional VAR
model of order m is
for t = 1,…,T, where
{εt} is independently and identically
distributed (i.i.d.) with mean zero and a positive definite covariance
matrix Σ and det(In −
A1 z − ··· −
Am zm) has all
roots outside the unit circle or equal to 1. The model (1) can be written
in the error correction (EC) format,
where α and β are n × r matrices with
rank r, [utri ] = 1 − L, and L denotes the
lag operator. We assume 0 < r < n, and then there
are r cointegrating relations. The exact condition of the
existence of cointegration is given by Johansen (1991, 1992). We also assume
that the cointegrating rank r is known or estimated by some
testing procedure, such as the likelihood ratio (LR) test proposed by
Johansen (1988, 1991)
or the Lagrange multiplier (LM) test by Lütkepohl and Saikkonen
(2000) and Saikkonen and Lütkepohl (2000). Other testing procedures of the cointegrating
rank are reviewed by Hubrich, Lütkepohl, and Saikkonen (2001) and papers therein.
In this paper, we investigate the tests of the rank of
β1, the submatrix of β, and the rank of
β⊥,1, the submatrix of β⊥,
where β =
[β1′,β2′]′ and
β⊥ =
[β⊥,1′,β⊥,2]′,
with β⊥ being an orthogonal complement to β. In
practical analysis, we sometimes encounter cases where we need to know the
rank of β1 and/or β⊥,1. For
example, the cointegrating matrix is sometimes normalized as β* =
β(a′β)−1, as proposed by Johansen
(1988, 1991) and
Paruolo (1997), where a is an
n × r matrix with full column rank and the
prototype normalization is represented by a =
[Ir,0]′. However, there is no
guarantee that a′β is of full rank. In such a
situation, we would like to know whether the first r rows of
β have full rank. The second example is the Granger noncausality test.
As shown in Toda and Phillips (1993), when there
is a cointegrating relationship, in general the Wald statistic of the
Granger noncausality test from the last n3 variables
of xt to the first n1
variables has a nonstandard limiting distribution, depending on nuisance
parameters. However, if either the last n3 rows of
β or the first n1 rows of α have full row
rank, the Wald statistic is asymptotically χ2 distributed.
Then, the testing procedure in this paper is useful to check the rank of
the submatrix of β, whereas the existing testing procedure may be
available for the test of the rank of the submatrix of α. The third
example is the test of long-run Granger noncausality proposed by Yamamoto
and Kurozumi (2001, 2003). In a usual sense, Granger causality is
concerned with the one period ahead forecast. This concept can be extended
to the predictability of h period ahead horizon, and long-run
Granger causality is defined when the forecast horizon h goes to
infinity. See, for example, Bruneau and Jondeau (1999) and Dufour and Renault (1998). Yamamoto and Kurozumi (2003) proposed the test for long-run block
noncausality, in which it is shown that the ranks of the submatrices of
β and β⊥ play an important role in constructing
the test statistic. See Yamamoto and Kurozumi (2003) for more details.
Tests of the rank of a matrix have been investigated in the
literature, and recent econometric developments can be seen in works by
Camba-Mendez, Kapetanios, Smith, and Weale (2003), Cragg and Donald (1996, 1997), and Robin and
Smith (2000), among others. Although these
papers proposed tests of the rank of a matrix, they assumed that the
estimator of the matrix is T1/2 consistent and has
a limiting normal distribution with a nonstochastic variance matrix.
However, the estimator of the cointegrating matrix is T (or
T3/2) consistent and has an asymptotic nonstandard
distribution. As a result, we cannot apply existing testing procedures to
the cointegrating matrix.
The paper is organized as follows. In Section 2, we propose tests of
the rank of β1 and β⊥,1 for
nontrending data. We will show that the two test statistics proposed have
asymptotically a χ2 distribution and a distribution of the
maximum eigenvalue of the product of normal random matrices. Section 3
considers the case of trending data. In this case, the test statistics do
not necessarily converge to a χ2 distribution and a
distribution of the maximum eigenvalue. To overcome this situation, we
propose two testing procedures. Section 4 investigates the finite sample
properties of the tests. Section 5 concludes the paper.
In regard to notation, we use vec(A) to stack the
rows of a matrix A into a column vector,
[x], to denote the largest integer ≤ x,
a = a(a′a)−1
for a full column rank matrix a, and
signify convergence in probability, convergence in distribution, and weak
convergence of the associated probability measures. We denote the rank of
A by rk(A) and the column space of A
by sp(A). We write integrals such as
simply as ∫XdY′ to achieve notational economy, and all
integrals are from 0 to 1 except where otherwise noted.
2. TEST OF THE RANK OF THE SUBMATRIX FOR
NONTRENDING DATA
2.1. The Model with d = 0
In this section we consider a test of rank for nontrending data with
d = 0. The model considered in this section is
We estimate the model (3) by the maximum likelihood (ML) method
assuming that {εt} is Gaussian, although
asymptotic properties are preserved under more general assumptions. We
denote the ML estimator with [caret ]. For example, the ML estimator of
β is denoted by
.
Using the result that
for 0 ≤ r ≤ 1 by the functional central limit theorem,
where W(·) is an n-dimensional Brownian motion
with a variance matrix Σ, Johansen (1988,
1995) showed that
where
with
,
and G0(·) and V(·) are
independent. He also showed that
are consistent estimators of α, Σ, and
Γi, respectively.
Let us partition β as β′ =
[β1′,β2′] where
β1 and β2 are n1 ×
r and (n − n1) ×
r matrices, respectively (0 < n1 <
n). Similarly, we partition β⊥′ =
[β⊥,1′,β⊥,2′]
conformably. Note that β1′ β⊥,1
does not necessarily equal zero, whereas
β′β⊥ = β1′
β⊥,1 + β2′
β⊥,2 must be zero. Our interest lies in finding the
rank of β1, and thus we consider the following testing
problem:
Note that the rank of β1 is at most p ≡
min(n1,r).
To test the rank of β1, we follow the same strategy as
Robin and Smith (2000), who test the rank of a
matrix and investigate its quadratic form. In our situation, we construct
a quadratic form of β1. The advantage of considering a
quadratic form is that the eigenvalues are nonnegative real values, even
if those of β1 are complex values. Then, the null
hypothesis H0 becomes equivalent to the existence of
f positive real and n1 − f
zero eigenvalues.
Let Ψ and Φ be r × r and
n1 × n1 possibly stochastic
matrices that are symmetric and positive definite almost surely (a.s.).
Because they are full rank matrices (a.s.), the rank of β1
is equal to the rank of
Φ−1β1Ψβ1′
(a.s.). Therefore, the test of the rank of β1 is equivalent
to that of
Φ−1β1Ψβ1′,
and then we consider the rank of the latter matrix. Note that, although
this strategy is basically the same as that followed by Robin and Smith
(2000), we cannot directly use their result
because they assume that the estimated matrix is asymptotically normally
distributed with a convergence rate T1/2, whereas
is shown to be T-consistent and the limiting distribution is
mixed Gaussian.
For the test of the rank of β1, we define Ψ =
α′Σ−1α and
These Ψ and Φ are chosen so that the limiting distribution
of the test statistic does not depend on nuisance parameters. Other
choices of Φ may be possible because, as shown in the Appendix, the
test statistic asymptotically does not depend on
β1(β′β)−1β1′,
which appears when (6) is expanded. For example, we can use a constant
multiple of (β′β)−1 in the second term of
(6). However, as indicated in the Appendix, Φ has to be invariant to
the normalization of β. We use the definition (6) just because it
seems simplest among other choices.
Let λ1 ≥ λ2 ≥
··· ≥ λn1 be
the ordered eigenvalues of
Φ−1β1Ψβ1′,
which are the solution of the determinant equation
Then, under H0, λ1 ≥
··· ≥ λf > 0 and
λf+1 = ··· =
λn1 = 0 (a.s.).
We construct a sample analogue of (7) using the ML estimator and
investigate the limiting distributions of the eigenvalues. The sample
analogue of (7) is given by
where
is the first n1 rows of
,
and
where
,
with R1t being the regression residual of
xt−1 on
[utri ]xt−1,…,[utri ]xt−m+1,
and we denote the ordered eigenvalues of (8) as
.
Note that when n1 > r, the smallest
n1 − r eigenvalues are obviously equal
to 0, that is,
.
We can easily see from the expressions (6) and (9) that
are positive definite (a.s.), whereas the expression (10) is simpler and
may be used to construct
in practice.
To test the rank of β1, we consider the following test
statistic:
which rejects the null hypothesis when
takes large values. The second equality is established because p
= min(n1,r) and
when n1 > r.
We can also consider the null hypothesis of
rk(β1) = f against the alternative of
rk(β1) = f + 1. In this case, the test
statistic is defined by
To denote the limiting distribution of
,
we define λmax,j,k* as the
maximum eigenvalue of
where X*′ is a j × k matrix
with vec(X*′) ∼
N(0,Ij×k). The
critical points of this distribution are given in Table
1 for the case where j ≥ k. They are
calculated by simulations with 1,000,000 replications. Because the nonzero
eigenvalues of X*X*′ are the same as those of
X*′X*, we can refer to the percentage points of
λmax,k,j when j <
k.
Critical values of the
λmin,j,k* and
λmax,j,k*
THEOREM 1. Let
be given by (10). If f < p, under H0,
.
Remark 1. Because the determinant equation (8) converges to (7) in
distribution, the estimated ordered eigenvalues of (8) also converge in
distribution to those of (7). Then, under the alternative,
(a.s.), so that
goes to infinity. Therefore, the tests
are consistent.
Remark 2. Although the test statistics are constructed using the
estimator of β⊥,1, we do not have to assume that it
is of full rank. We can see that the rank of β⊥,1 is
at least n1 − f under
H0, noting that the column space of
β⊥,1 must contain n1 −
f bases that are orthogonal to sp(β1)
because [β1,β⊥,1] has full
row rank n1. Because β1′
β⊥,1 is not necessarily equal to zero, it is
possible for sp(β⊥,1) to contain some of
the bases that span sp(β1), so that the rank of
β⊥,1 may be greater than n1
− f. It is shown in the Appendix that the limiting
distributions of the test statistics depend not on the rank of
β⊥,1 but on the number of the bases orthogonal to
sp(β1), n1 − f,
unless f = n1. When f =
n1, all the eigenvalues are asymptotically greater
than zero (a.s.), and then the test statistics will diverge. This case is
excluded from the theorem (f is assumed to be less than
p = min(n1,r)). In other words, our
tests cannot be applied for the null hypothesis of full rank. If we need
to check whether β1 is of full rank or not, we may test for
the null of f = n1 − 1, and if we
rejected the null hypothesis, we would conclude that it is a full row rank
matrix.
Remark 3. Because the hypothesis about the rank of β1
can be regarded as a restriction on the cointegrating matrix β, we may
consider using the LR test as proposed by, for example, Johansen (1991, 1995) and Johansen and
Juselius (1990, 1992).
In fact, when f = 0 the null hypothesis is equivalent to
β1 = 0, and this hypothesis can be expressed as a linear
restriction on β such as β = Hφ, where H =
[0,In−n1]′
and φ is an (n − n1) ×
r unknown parameter. Then, the LR test is applicable to the test
of f = 0. However, for 0 < f < p, the
null hypothesis is expressed as β1 = β11
β12′ where β11 and β12
are n1 × f matrices with full column
rank f. Then, we have to estimate the model with this
restriction. Although the LR test might be applicable to the nonlinear
hypothesis, it seems tedious to estimate the model with this nonlinear
restriction, whereas our test uses only the ML estimator without the
restriction. It is beyond our scope to investigate the applicability of
the LR test to our case, and we do not discuss this in detail.
We may represent the null hypothesis as proposed by Boswijk (1996) and apply the LR test. According to his paper,
the null hypothesis of rk(β1) = f is
expressed as β = (Hoφ,ψ) where
Ho =
[0,In2]′ and
(φ,ψ) ∈
Rn2×(r−f)
× Rn×f. As pointed out
by Boswijk (1996, p. 156), the LR test for this
hypothesis has an asymptotic χ2 distribution only when
“no linear combination of ψ lies in the column space of”
Ho. Because there is no guarantee of this
condition, we do not consider his method in this paper.
Next, we consider a test of the rank of the submatrix of
β⊥. The testing problem is
For the same reason as in the test of β1, we
investigate the rank of
,
where
are (n − r) × (n −
r) and n1 × n1 full
rank matrices (a.s.). Similar to (7), we consider the following
determinant equation:
where
and
and the sample analogue of (12) is given by
where
and
Let
be ordered eigenvalues of (12) and (13), respectively, and we construct
the following test statistics, with q =
min(n1,n − r):
THEOREM 2. Let
be given by (14). If g < q, under
H0⊥,
.
Note that the consistency of the tests is shown in a similar way as in
Remark 1. We also note that we cannot test the null of
rk(β⊥,1) = q by a similar reason
to that given in Remark 2.
Given the preceding two theorems, we can test the rank of
β1 and β⊥,1. In addition, we may
consider the procedure to decide the rank of the submatrix, as the
cointegrating rank is selected sequentially using the test of the
cointegrating rank. For example, to decide the rank of β1,
we first test the null of f = 0. If the null hypothesis is
accepted, the rank of β1 is decided to be zero. Otherwise,
we then test the hypothesis of f = 1. We sequentially continue to
test the rank of β1 until the null hypothesis is accepted.
When the null of f = p − 1 is rejected, we
consider that β1 has full rank. Similarly, the rank of
β⊥,1 can be decided by the same procedure.
2.2. The Model with d ≠ 0
In the previous section, we considered the model with d = 0
for nontrending data. However, in practice, we sometimes consider the
model (2) with d ≠ 0 but with the level of data having no
linear trend. In this case, the constant term can be expressed as
d = αρ0 where ρ0 is an
r × 1 coefficient vector, so that the model (2) becomes
where β+ =
[β′,ρ0]′ and
xt−1+ =
[xt−1′,1]′. The
ML estimator of β+ can be obtained by the reduced rank
regression of [utri ]xt on
xt−1+ corrected for
[utri ]xt−1,…,[utri ]xt−m+1,
and the estimator of the cointegrating matrix is the first n rows
of
.
To test the rank of the submatrix of β for the model (15), we use
defined by
where
are (n − r + 1) × (n −
r) and (n + 1) × (n − r +
1) matrices defined by
and
,
with R1t+ being the regression
residual of xt−1+ on
[utri ]xt−1,…,[utri ]xt−m+1.
THEOREM 3. Consider the model (15) and let
be given by (16). If f < p, under H0,
.
THEOREM 4. Consider the model (15) and let
be given by (14). If g < q, under
H0⊥,
.
In practical analysis, we will obtain
by the reduced rank regression, and we have to calculate
from
.
If
can be easily obtained as explained in Johansen (1995, p. 95). When d =
αρ0, one of the methods to calculate
is as follows. First we calculate the orthogonal projection matrix of
.
Then, by the singular value decomposition, M is expressed as
Ml Mλ
Mr′ where
Ml and Mr are
n × (n − r) orthogonal matrices and
Mλ is an (n − r) ×
(n − r) diagonal matrix with positive diagonal
elements. Because sp(M) =
sp(Ml) and they are orthogonal to
,
we can use Ml as
.
3. THE TEST OF THE RANK OF THE SUBMATRIX FOR
TRENDING DATA
When data are trending, xt can be
expressed as the sum of the stochastic trend, the deterministic trend, and
the I(0) component such that
where C =
β⊥(α⊥′Γβ⊥)−1α⊥′
as defined in Section 2.1, τ = Cd,
C1(L) = (C(L) −
C(1))/(1 − L) with C(L)
being a lag polynomial when [utri ]xt is
represented as the vector moving-average process like
[utri ]xt = C(L)(d
+ εt), and x0* is a
stochastic component such that β′x0* = 0.
See Johansen (1991, 1995) for more details. In this case,
β⊥ is decomposed to τ, the coefficient of a
linear trend in (17), and γ, an n × (n −
r − 1) matrix that is orthogonal to τ. We partition
γ and τ into
[γ1′,γ2′]′ and
[τ1′,τ2′]′ in
the same way as β. As shown in Chapter 13.2 of Johansen (1995),
can be expressed as
where
where G(r) =
[G1′(r),G2′(r)]′
with G1(r) =
G0(r) − ∫G0
ds, G0(r) =
γ′CW(r) and
G2(r) = r − ½. We
denote Ω = ∫GG′ ds and partition it into
2 × 2 blocks conformably with
[U1′,U2′]′.
We express the (i,j) block element of
(∫GG′ ds)−1 as
Ωij for i,j = 1 and 2. In this
section, we need the estimator of Ω11, which is given
by
and S11 is defined in the same way as in the
previous section, with R1t being the
regression residual of xt−1 on a
constant and
[utri ]xt−1,…,[utri ]xt−m+1.
Convergence of
is proved in Lemma 2(iii) in the Appendix, whereas the consistency of
other ML estimators, such as
,
is shown by Johansen (1991, 1995).
In the following discussion, we will show that the limiting
distribution of
depends on whether the rank of
[β1,γ1] is n1
− 1 or n1, or equivalently, whether
τ2 = 0 or not. We will propose two testing procedures to
cope with this problem.
Let us consider the testing problem (5). Under the null hypothesis, we
can find the f linearly independent column vectors in
β1, and we define β1* as an
n1 × f matrix whose columns consist of
those f vectors. We also define an n1 ×
(n1 − f) matrix δ* as an orthogonal
complement to β1*, so that δ*′β1*
= 0. We show that the direction of δ* is important in deciding the
convergence rate of
and it also affects the limiting property of the test statistic.
Let us consider the case where r < n − 1.
Because
is the first n1 rows of
,
it is expressed from (18) as
Suppose that an n1 × 1 vector
τ1* exists that is orthogonal to γ1
(τ1*′γ1 = 0) and belongs to the column
space of δ*. Here, note that, because the n ×
n matrix [β,γ,τ] is of full rank, the first
n1 rows of this matrix,
[β1,γ1,τ1], must be
of full row rank, which implies that
a′[β1,γ1,τ1]
≠ 0 for any nonzero vector a. Then, because τ1*
is orthogonal to both β1 and γ1 by the
assumption, we have
τ1*′[β1,γ1,τ1]
= [0,0,τ1*′τ1] ≠ 0, so
that τ1*′τ1 ≠ 0. This implies
whereas for an n1 × (n −
r − 1) matrix δ0* whose columns span the
orthogonal complement to τ1* in sp(δ*),
On the other hand, if there exists no vector in sp(δ*)
that is orthogonal to γ1, we have
Therefore, the convergence rate of
depends on whether a vector τ1* orthogonal to
γ1 exists in sp(δ*).
The existence of τ1* indicates that the column space of
[β1,γ1] does not include
τ1* because τ1*′ β1 = 0
and τ1*′γ1 = 0. We also note that the
rank of [β1,γ1] must be
n1 − 1 or n1 because
[β1,γ1,τ1] has full
rank n1. Then, from another point of view, we can say
that the rank of [β1,γ1] is
n1 − 1 if a vector τ1* exists,
whereas the nonexistence of τ1* is equivalent to
rk([β1,γ1]) =
n1. Thus, we have to consider the asymptotic property
separately according to the two cases where the rank of
[β1,γ1] is n1
and n1 − 1 when r < n
− 1.
For further investigation, let us consider the case where the rank of
[β1,γ1] equals
n1 − 1. In this case, this matrix is expressed
as [Θ11,0] by some nonsingular transformation
from the right-hand side, where Θ11 is an
n1 × (n1 − 1) matrix
with rank n1 − 1. Then, using the same
nonsingular transformation, [β,γ] becomes
Let τ1* be the orthogonal complement to the
column space of Θ11. Then, because
τ1*′Θ1 = 0 and using the
expression (22), we can see that the n × 1 vector
[τ1*′,0]′ is orthogonal to
[β,γ]. Therefore, in this case, the trend parameter
τ, which is orthogonal to β and γ, is a constant multiple of
[τ1*′,0]′. In other words,
when rk[β1,γ1] =
n1 − 1, τ2 must be equal to zero.
Note that, because τ1* is orthogonal to
sp(β1) and sp(γ1), it is
essentially the same as τ1*.
On the other hand, when τ2 = 0, τ is expressed as
[τ1′,0]′ and then
τ1′[β1,γ1]
equals zero because τ′[β,γ] = 0. This implies
that the n1 × (n − 1) matrix
[β1,γ1] does not have full row
rank. Then, we have the following proposition.
PROPOSITION 1. The rank of
[β1,γ1] is n1
− 1 if and only if τ2 = 0.
When r = n − 1, there is no γ, and in this
case, rk(β1) must be n1
− 1 or n1. Then, under the null hypothesis of
rk(β1) = n1 − 1, δ*
becomes an n1 × 1 vector, and we have
In this case, the test statistics should be multiplied by T,
that is,
are the appropriate test statistics.
In the following theorem, the test statistics are constructed from the
eigenvalues of (8) using the same
as in the previous section and either
or
THEOREM 5. When r < n − 1,
(i.a) Let
be given by (24). If
rk([β1,γ1]) =
n1 and f < p, under
H0,
.
(i.b) Let
be given by (25). If
rk([β1,γ1]) =
n1 and f < p, under
H0,
converge in distribution to random variables that are bounded above
by
χ(n1−f)(r−f)2
and
λmax,n1−f,r−f*,
respectively.
(ii) Let
be given by (25). If
rk([β1,γ1]) =
n1 − 1 and f < p, under
H0,
.
When r = n − 1,
(iii) Let
be given by (25). Under the null hypothesis of f =
n1 − 1,
Remark 4. In the case of (i.b),
converges in distribution to
χ(n1−f)(r−f)2
if and only if δ*′τ1 = 0, which is equivalent to
the case where τ1 ∈ sp(β1*) =
sp(β1). See the proof in the Appendix. In general,
the test using (25) is conservative if
rk([β1,γ1]) =
n1.
From Theorem 5, if we know the rank of
[β1,γ1] when r <
n − 1, we can construct the test statistic
that converges to a χ2 distribution by appropriately
using (24) or (25). However, such information is not available in
practice. Notice that if
rk[β1,γ1] =
n1 − 1,
given by (24) may violate the condition that it is a full rank matrix,
and in that case, the test statistic converges not to the same
χ2 distribution as given by Theorem 5(ii) but to a random
variable that depends on a nuisance parameter. Then, the test using (24)
is not desirable in practice. On the other hand, if we use
given by (25), we can test the hypothesis by referring to a
χ2 distribution irrespective of the rank of
[β1,γ1], although the test may be
conservative and the degrees of freedom may change depending on the rank
of [β1,γ1]. Then, noting that the
critical value of
χ(n1−f)(r−f)2
in Theorem 5(i) is greater than that of
χ(n1−f−1)(r−f)2
in (ii), we propose to test the null of rk(β1) =
f as follows.
1. We construct the test statistic
using (25).
2.If
is greater than the critical value of
χ(n1−f)(r−f)2,
we reject the null hypothesis.
3. If
is less than the critical value of
χ(n1−f−1)(r−f)2,
we accept the null hypothesis.
The test statistic
is used in the same manner. In this procedure, we may encounter the case
where the test statistic is greater than the critical value of
χ(n1−f−1)(r−f)2
but less than that of
χ(n1−f)(r−f)2,
that is, the case where
,
where
c(n1−f−1)(r−f)
and
c(n1−f−1)(r−f)
are corresponding critical values. To cope with such a case, the following
corollary is useful.
COROLLARY 1. Let
be given by (25). Suppose that r < n − 1
and the rank of β1 is f (<
p).
(i) If
rk([β1,γ1]) =
n1,
converges in distribution to a random variable that is bounded above
by
λmin,r−f,n1−f*,
which is the smallest nonzero eigenvalue of (11) with j =
r − f and k = n1
− f.
(ii) If
converges in probability to zero.
The percentage points of
λmin,r−f,n1−f*
are tabulated in Table 1.
Using the preceding corollary, we can cope with the situation where
.
If
is less than some percentage (10, 5, or 1%) point of
λmin,r−f,n1−f*,
we reject the hypothesis of
rk([β1,γ1]) =
n1. In that case,
c(n1−f−1)(r−f)
is an appropriate critical value for
,
so that the null of rk(β1) = f is
rejected. On the other hand, if
is greater than the critical point of
λmin,r−f,n1−f*,
we accept the hypothesis of
rk([β1,γ1]) =
n1, so that the rank of β1 is decided
to be f. We call this testing procedure TEST1.
The other strategy is to use the result of Proposition 1. From
Johansen (1995),
converges in distribution to a normal random vector with mean zero and
the variance matrix given by CΣC′. Although
the Wald-type test may not be applicable to the test of τ2
= 0 because the variance matrix might be degenerate, we can test whether
each element of τ2 is zero or not by the t-test
statistic. We call the following testing procedure TEST2.
1. We test each element of τ2.
2. If some of the elements of τ2 are
significant, we use Theorem 5(i.a).
3. If none of the elements of τ2 are
significant, we use Theorem 5(ii).
Next, we investigate a test of the rank of β⊥,1.
When data are trending, β⊥,1 can be decomposed into
[γ1,τ1] where γ1
and τ1 are the first n1 rows of γ
and τ, respectively. Then, testing the rank of
β⊥,1 is equivalent to testing the rank of
[γ1,τ1], and therefore we
construct a test statistic from
.
Note that
is the first n1 rows of
and is not necessarily numerically equal to
,
although they span the same column space.
Let us consider the same determinant equation as (13) with
replaced by
and
We construct the test statistics
in the same way as in the previous section. Similar to Theorem 5, we have
to distinguish two cases where r < n − 1 and
r = n − 1. When r = n − 1,
the rank of β⊥,1 (= τ1) must be 0 or
1, and in this case, we consider the null hypothesis of g =
0.
THEOREM 6. Let
be given by (26) and (27). When r < n − 1
and g < q, under H0⊥,
converge in distribution to random variables that are bounded above
by
χ(n1−g)(n−g−r)2
and
λmax,n1−g,n−g−r*,
respectively.
When r = n − 1, under the null hypothesis of
g = 0,
4. SIMULATION RESULTS
In this section, we investigate the finite sample properties of the
tests proposed in the previous sections. We consider the following
four-dimensional EC model as a data generating process (DGP):
where {εt} ∼
i.i.d.N(0,I4).
Let
and we consider the following settings of parameters.
Here DGP1(1o), 2(2o), and 3(3o) correspond to the cases where the
cointegrating rank is 1, 2, and 3, respectively. We set the (2,1) element
of β as c1, which takes values of 0, 0.005, 0.01,
0.025, 0.05, 0.075, and 0.1, and we consider the test of the rank of the
first two rows of β. The case of c1 = 0
corresponds to the null hypothesis under which the rank of
β1 is 0, 1, and 1 for DGP1, 2, and 3, whereas it is 1, 2,
and 2 when c1 ≠ 0, which corresponds to the
alternative. For the case of nontrending data, we set d = 0 for
the zero-mean process, whereas d is defined as
αρ0 for the case of d ≠ 0, where
ρ0 is set to be 1, [1,1]′, and
[1,1,1]′ for DGP1(1o), 2(2o), and 3(3o), respectively. On
the other hand, for the case of trending data, d is set to be
d1 and d2; the former corresponds
to the case where [β1,γ1] is of
full rank (τ2 ≠ 0), whereas the rank of
[β1,γ1] is n1
− 1 (τ2 = 0) when d =
d2.
Similarly, we set the (2,1) element of β⊥ as
c2 and consider the test of the rank of the first two
rows of β⊥. In this case, c2 = 0
implies that the rank of β⊥,1 is 1, 1, and 0 for
DGPo1, o2, and o3, respectively, whereas it is 2, 2, and 1 under the
alternative of c2 ≠ 0.
We set x0 = 0 and discard the first 100
observations in all experiments. The number of replication is 5,000, and
the level of significance is set equal to 0.05. We only report the results
of the test statistics
because the performances of
are almost the same as those of
.
Table 2 shows the simulation results of the
test of rk(β1). When the cointegrating rank is 1,
the empirical size is greater than the nominal size, 0.05, for T
= 100 when data are nontrending (d = 0 or d =
αρ0), whereas it becomes closer to 0.05 for T
= 200. When data are trending, τ becomes
for d = d1, whereas it is
for d = d2. Similar to the case of
nontrending data, the testing procedure TEST2 tends to overly reject the
null of c1 = 0 for T = 100, whereas the
testing procedure TEST1 seems to be slightly conservative. Under the
alternative of c1 ≠ 0, the power increases rapidly
around c1 = 0.025 for nontrending data and for
trending data with TEST2, whereas the testing procedure TEST1 seems to be
less powerful. This is because TEST1 is a conservative test. When data are
trending, both TEST1 and TEST2 are more powerful for the model with
rk([β1,γ1]) =
n1 (d = d1) than the
model with rk([β1,γ1]) =
n1 − 1 (d =
d2).
Rejection frequencies of the tests of
rk(β1)
When the cointegrating rank is 2, the relative performance is
preserved for the cases of d = 0 and d =
αρ0. For trending data, τ becomes
for d = d1 and d2,
respectively. Note that
is numerically equal to
because the determinant equation (11) with j = k = 1
has only one eigenvalue. Then, we can see that
converge in distribution to χ12 under
H0 when
rk([β1,γ1]) =
n1 = 2, whereas they converge in probability to zero
when rk([β1,γ1]) =
n1 − 1 = 1. Then, the testing procedure TEST1
accepts the null hypothesis when it is less than the critical point of
χ12. On the other hand, the asymptotic size of
TEST1 becomes 0 when
rk([β1,γ1]) =
n1 − 1 (d = d2)
because
T converges in probability to zero when
rk([β1,γ1]) =
n1 − 1. Reflecting this fact, TEST1 is too
conservative for d = d2, and it is not
powerful when the alternative is close to the null. TEST2 also seems to
have no power when the rank of
[β1,γ1] = n1
− 1 = 1 (d = d2). This is because
τ2 is very close to zero
1 For
example, even when c1 = 0.1, the third and fourth
elements of τ are
and 3/430.
even under the alternative of
c1 ≠ 0, so that the pretest of τ
2
cannot reject the null of τ
2 = 0. When τ
2 is
judged to be zero, the rank of
[β
1,γ
1] is at most
n1 − 1 = 1. Then, because we are testing the
null hypothesis of
rk(β
1) =
f = 1, we
automatically accept the hypothesis when τ
2 = 0 is accepted
in this case.
When the cointegrating rank is 3, we can see that the first two
variables of xt are cointegrated, whereas the
last two variables are stationary. Note that we cannot generate the
process such that the rank of β1 is 1 while all the
variables are nonstationary. Because we want to investigate the property
of the test under the null hypothesis, we allowed several variables to be
stationary.
In this case, the power property seems to be improved for all the
cases compared with the cases where r = 1 and 2. For trending
data, τ becomes
for d = d1, whereas it is
for d = d2. Note that in this case the last
two rows of the impact matrix C become zero because the
corresponding variables are stationary, so that τ2, the
last two rows of Cd, become zero irrespective of the value of
d. We also note that the result of Theorem 5(iii) is applied
because r = n − 1 = 3. That is, we do not have to
use the conservative test or the pretest as in the cases where r
< n − 1. This is the reason why both the size and power
properties are improved for trending data compared with the cases where
r < 3.
Table 3 reports the results of the test of
rk(β⊥,1). From the table, the test tends to
overly reject the null hypothesis for several cases when T = 100,
whereas the size becomes reasonable when T = 200, except for the
case where r = 3 and d = d1. In that
case, the test becomes conservative as investigated in Theorem 6. As to
the power, we can see that the more complicated the deterministic term
becomes, the less powerful is the test.
Rejection frequencies of the tests of
rk(β⊥,1)
5. CONCLUSION
In this paper, we proposed tests of the rank of the submatrix of
cointegration. We can test the hypothesis straightforwardly when data are
nontrending, whereas for trending data, we have to examine whether
[β1,γ1] is of full rank or not or
we have to use the conservative test. The simulation results show that we
have to carefully use the test of rk(β1) when data
are trending and f = n1 − 1, because
the test might become too conservative to reject the null hypothesis.
APPENDIX
We use the notation H alternately for different definitions
if there is no confusion.
Proof of Theorem 1. First, note that we can replace
in (8), where
is the first n1 rows of
,
because
.
The latter relation is established because
is obtained by the nonsingular transformation of the columns of
does not depend on the normalization of
.
We also define
whose columns span the orthogonal complement to
), so that
span the same column space. This implies that
can be obtained by the nonsingular transformation of the columns of
.
Then, we can also replace
.
Under the null hypothesis, rk(β1) is
f, and then an n1 × f matrix
β1* exists with rank f such that
sp(β1) = sp(β1*). We
denote the orthogonal complement to β1* by δ*. That is,
δ* is an n1 × (n1
− f) matrix with rank (n1 −
f) such that δ*′β1* = 0.
LEMMA 1.
Proof.
(ii) As shown in Chapter 13.2 of Johansen (1995),
can be expressed as
for nontrending data, where TUT converges in
distribution to (∫G0 G0′
ds)−1∫G0
dV′. Because
is the first n1 rows of
,
we have
,
so that
(iii) holds because
from (A.1).
(iv) is proved by noting that
from Johansen (1988, 1995),
.
█
Now, let us consider the determinant equation (8). Note that (8) is
equivalent to
where H = [β1*,Tδ*] is
an n × n nonsingular matrix. Using Lemma 1, we
have
To investigate the asymptotic behavior of
,
we consider
with the same expression as (9). Note that
because
by Lemma 1. Then,
is asymptotically equivalent to
Then, the equation (A.2) is asymptotically equal to
Therefore, the eigenvalues
converge in probability to zeros and are of order
T−2.
Here, notice that, in the same way as Johansen (1988, p. 246), we can find an r ×
(r − f) matrix J with rank (r
− f) such that
with J′(β1′ β1*) =
0 and J′Ψ−1J =
Ir−f , implying that
J′(α′Σ−1α)−1J
= Ir−f because Ψ =
α′Σ−1α. Then, because
|β1*′β1Ψβ1′
β1*| ≠ 0, (A.4) becomes
The variance matrix of X0′J
conditioned on G0(·) is given by
Noting that sp(β⊥,1) must contain
δ* because [β1,β⊥,1] is
of full row rank n1 and sp(β1)
does not contain δ*, we can see that
δ*′β⊥,1 has full row rank
n1 − f irrespective of the rank of
β⊥,1, which is greater than n1
− f as explained in Remark 2. As a result, we can see that
the conditional variance matrix of X0′J
is nonsingular (a.s.). Then, by multiplying the square root of the
left-hand side of (A.7) from both sides of (A.6), the determinant equation
becomes (11) with j = n1 − f
and k = r − f, and then
converges in distribution to the solution of (11). This proves Theorem 1.
█
Proof of Theorem 2. The outline of the proof is the same as
the proof of Theorem 1, and thus we omit details.
Under the null hypothesis, an n1 ×
g matrix β⊥,1* exists such that
sp(β⊥,1*) =
sp(β⊥,1) and
rk(β⊥,1*) = g, and we denote the
orthogonal complement to β⊥,1* by η*. Consider
the following determinant equation:
where H =
[β⊥,1*,Tη*]. As in the
previous proof, we replaced [caret ] by ˜. Because
is the first n1 rows of
,
we obtain, using Lemma 1(iii),
Then, similar to the previous proof, we can show that
converges in distribution to a solution of (11) with j =
n1 − g and k = n
− r − g. This proves Theorem 2. █
Proof of Theorems 3 and 4. Let
.
Exactly in the same way as the proof of Lemma 13.2 in Johansen (1995), we can show that
where G0+ =
[G0′,1]′. Then, because
is the first n rows of
,
we have
whose conditional variance is given by
L′(∫G0+G0+′
ds)−1L [otimes ]
(α′Σ−1α)−1. Because
as expressed in Johansen (1995, p. 179), we
have
We also have
,
which is proved as Lemma 1(iv), where
with
replaced by
.
Then, the theorems are proved similarly to Theorems 1 and 2. █
Proof of Theorem 5. For the case where r <
n − 1, we give the following lemma.
LEMMA 2.
Proof.
From Lemma 10.3 in Johansen (1995),
T−1[γ,T−1/2τ]′S11
[γ,T−1/2τ]
converges in distribution to Ω whereas
β′S11 β converges in probability to a
positive definite matrix, Σβ, and
[γ,T−1/2τ]′S11
β = Op(1). Then,
In addition, we can see that
because
.
Using this result, we have
From (A.10) and (A.11),
converges in distribution to Ω11. █
(i.a) Proved in the same way as Theorem 1.
(i.b) In this case, the determinant equation becomes
asymptotically equivalent to
Note that, in general, for a given symmetric and positive definite
matrix A and a vector b,
and then
for any nonzero vector c. By substituting
δ*′γ1(γ′γ)−1Ω11(γ′γ)−1γ1′δ*
and
for A and b, we obtain, for a given
G(·),
where X* is an (r − f) ×
(n1 − f) matrix with vec(X*)
∼
N(0,I(r−f)(n1−f)).
The equality is established if and only if δ*′τ1
= 0.
(ii) Let us consider the determinant equation (A.2) with H =
[β1*,Tδ0*,Tτ1*].
Using Lemma 2 and by some algebra, the determinant equation is shown to be
asymptotically equivalent to
This determinant equation implies that there are f nonzero
eigenvalues, p − f − 1 eigenvalues of order
T−2, and one eigenvalue of order smaller than
T−2. Then, we can see that
We can also show that
is of order T3 if we choose H =
[β1*,Tδ0*,T3/2τ1*].
For the case where r = n − 1, the limiting
distribution is derived similarly using (23). █
Proof of Corollary 1.
(i) Note that, in general, for a given positive definite
matrix A, a vector b, and a matrix
D,
where we used the relation (A.12). By Theorem 9 of Magnus and
Neudecker (1988, p. 208), we can see that the
p − f th eigenvalue of
D′A−1D is greater than
that of D′(A +
bb′)−1D. Then, by substituting
δ*′γ1(γ′γ)−1Ω11(γ′γ)−1γ1′δ*,
,
and X′J for A, b, and
D, the limiting distribution of
is shown to be bounded above by
λmin,r−f,n1−f*
because D′A−1D =
X*X*′ in this case. Note that
if and only if δ*′τ1 = 0.
(ii) is proved in Theorem 5(ii). █
Proof of Theorem 6. Let us define β⊥,1*
and η* as in the proof of Theorem 2.
LEMMA 3.
Proof.
(i) Because η*′γ1 = 0 and
η*′τ1 = 0, we have, using (A.14),
(ii) First, note that, because
is invariant to each normalization of
.
Then, we can express
.
From the expression (A.1), we can see that
We also have, from the definition of τ,
Because the left-hand side is zero from the orthogonality
between γ and τ, the first n − r − 1
rows of
(α⊥′Γβ⊥)−1α⊥′
μ are zero. Then, because each estimator is consistent, we
have
Combining (A.15) and (A.16), we obtain
.
█
Similar to the proof of Theorem 2, we consider the same determinant
equation as (A.8). Using Lemma 3, we have
where S1 =
[In−r−1,0],
and then, using
,
(A.8) is expressed as
for large values of T, where an (n −
r) × (n − r − g)
matrix J satisfies
.
Noting that the conditional variance of
Y′S1 J is given by
the test statistic
conditioned on G(·) converges in distribution to
where vec(Y*) ∼
N(0,I(n1−g)(n−r−g))
and J =
[J1′,J2′]′.
Because
the limiting distribution (A.18) is bounded above by
