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Time-varying cointegration, identification, and cointegration spaces

  • Luis Filipe Martins EMAIL logo and Vasco J. Gabriel


We derive the conditions under which time-varying cointegration leads to cointegration spaces that may be time-invariant or, in contrast, time-varying. The model of interest is a vector error correction model with arbitrary time-varying cointegration parameters. We clarify the role of identification and normalization restrictions and show that structural breaks in error-correction models may actually correspond to stable long-run economic relationships, as opposed to a single-equation setup, in which an identification restriction is imposed. Moreover, we show that, in a time-varying cointegrating relationship with a given number of variables and cointegration rank, there is a minimum number of orthogonal Fourier functions that most likely guarantees time-varying cointegrating spaces.

Corresponding author: Luis Filipe Martins, ISCTE – Business School, Av-das Forças Armadas, 1649-026 Lisboa, Portugal

4 Appendix: Proofs of Lemma 1 and Theorems

Proof of Lemma 1. For a pair t, s, the null space

is of dimension dim(Nts(ξ))=(m+1)rrank(ξ)<(m+1)r because ξ≠0. See Strang (1988), for example. Hence, rank(ξ)<(m+1)r and the result follows noting that 0<rank(ξ) ≤ min {k, (m+1)r}.

So now the question is whether any of these solutions

of Lemma 1 correspond to λt, λs≠0 that results in
To prove Theorems 1 and 2 we need some new definitions and two auxiliary lemmas. Take
as a generic solution to the system
and let the number of free variables in xts≠0 be l=(m+1)rrank(ξ). In the case of k<(m+1)r, we have l=(m+1)rk, …,(m+1)r–1; whereas in the case of k≥(m+1)r and rank(ξ)<(m+1)r, we have l=1, …,(m+1)r–1. The system is now Pts λts=xts where xts≠ 0 is
is 2r×1 and
is a matrix (m+1)r × 2r, with r× 2r elements
By defining
the solutions λt, λs are found from the equations


in ℜr for i=1, …,m, where

is given.

LEMMA A.1. For all ts, t, s=1, …,T, P1,T(t)≠ P1,T(s). When I=2, …,m<T, Pi,T(t)=Pi,T(s) for at least one pair t, s with ts. Moreover, for all t≠ s,t,s=1,…,T and(i=1, …,m – 1, Pi,T(t)≠ Pi+1,T(s).

PROOF: The time periods such that Pi,T(t)=Pi,T(s) are

with p=1,2, …, that is,
with p=1,2, …. These are integer numbers as long as
If i=1 then
which rules out all cases since |st| and s+t cannot be greater or equal than 2T. The number of pairs t, s that satisfy Pi,T(t)=Pi,T(s) increases with i. If i=2, then Pi,T(t)=Pi,T(s) for all s+t=1+T. If i≥3, then
with p=1,2, …, such that
The last result follows from 0.5/i ∉ℵ’ for any i integer. Note that, as a corollary, if λt, λs≠ 0 then
but, if
then λt, λs are not necessarily both different from zero.

Lemma A.2. For any ts and fixed r,m > 0,rank(Pts)=2r

PROOF: Without loss of generality, take r=m=1. Then,

has rank equal to 2 because P1,T(t)≠ P1,T(s) for any ts (see Lemma A.1.).

Proof of Theorem 1. In the system Pts λts=xts, where xts≠ 0, the number of unknowns does not exceed the number of equations, (m + 1)r≥2r and rank(Pts)=2r for all ts, m≥1, by Lemma A.2. By the same reasoning, rank(Pts, xts)=rank(Pts)=2r for all ts, if m=1. Thus, when m=1, in the cases of k<(m+1)r=2r and k≥(m+1)r=2r with rank(ξ)<(m+1)r=2r there is one solution

to the system. Consequently,
if both λt, λs≠ 0 with λt,≠ λs or
if either λt or λs equals zero. By (21), λt=0 for some t,s,t≠s, if
where ηts ∈{P1,T(s), P1,T(t)} and
(where λt=λs≠ 0). On the contrary, λt, λs≠0 with λtλs, for any t,s,ts, if
Note that, in the previous case,
can have up to r free variables. When
the number of free variables in xts is at most r, whereas if
for all r × r matrices ϒts the number of free variables in xts is at most 2r. Given that
for ηts ∉ {P1,t(s), P1,t(t)} implies βt(1)≠ 0 with rank(ξ) ≤ r the result then follows from the necessary conditions in Lemma 1 for the system

Proof of Theorem 2. With m≥2,rank(Pts, xts) is either 2r or 2r+1 Whenever rank(Pts, xts)=2r=rank(Pts) there is one solution

to the system and consequently
if both λt, λs≠ 0 with λtλs or
if either λt or λs equals zero. Whenever rank(Pts, xts)=2r+1>rank(Pts) there is no solution to the system and therefore
So that rank(Pts, xts)=2r, we need

for some

with at least one different from zero for each i. With (21), we conclude that
depend on the free variables
for m≥2. Hence, rank(Pts, xts)=2r, if

for some non zero r × r matrices

and by setting
are given by (14). Similar to the previous Theorem, λt=0 for some t, s where ts, if
where ηts ∈ {P1,T(s), P1,T(t)} and
Hence, given that the solutions (24) imply βt(m)≠ 0 and that
depend on the free variable
for TI spaces, we have rank(ξ) ≤ r and the result follows from the necessary conditions in Lemma 1 for the system

  1. 1

    The identification restrictions can be made more general,

    given the usual notation, and taking any k and any r:

  2. 2

    One such case is the Purchasing Power Parity TV cointegration analysis in Bierens and Martins (2010). It most probably has TV cointegrating spaces since r=1, k=3 and m was always >5 according to the Hannan-Quinn criterion.

  3. 3

    This will form a unique solution by applying the transversality condition

  4. 4

    US data (from 1900 to 2006) is available from Robert Shiller’s webpage (, where stock prices are January values for the Standard and Poor Composite Index, dividends are year-averages and both series are deflated by January values of the producer price index. Following several other authors, we do not include the latest available sampling period, as the deviations from the implied relationship are unusually large and persistent, albeit temporary.


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Published Online: 2013-04-11

©2013 by Walter de Gruyter Berlin Boston

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