Time-varying cointegration, identification, and cointegration spaces

4 Appendix: Proofs of Lemma 1 and Theorems

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

is of dimension dim(N_ts(ξ))=(m+1)r–rank(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 λ_t, λ_s≠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 x_ts≠0 be l=(m+1)r–rank(ξ). In the case of k<(m+1)r, we have l=(m+1)r – k, …,(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 P_ts λ_ts=x_ts where x_ts≠ 0 is is 2r×1 and is a matrix (m+1)r × 2r, with r× 2r elements By defining the solutions λ_t, λ_s are found from the equations

with

in ℜ^r for i=1, …,m, where

is given.

LEMMA A.1. For all t≠s, t, s=1, …,T, P_1,T(t)≠ P_1,T(s). When I=2, …,m<T, P_i,T(t)=P_i,T(s) for at least one pair t, s with t≠s. Moreover, for all t≠ s,t,s=1,…,T and(i=1, …,m – 1, P_i,T(t)≠ P_i+1,T(s).

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

or with p=1,2, …, that is, or with p=1,2, …. These are integer numbers as long as If i=1 then which rules out all cases since |s–t| and s+t cannot be greater or equal than 2T. The number of pairs t, s that satisfy P_i,T(t)=P_i,T(s) increases with i. If i=2, then P_i,T(t)=P_i,T(s) for all s+t=1+T. If i≥3, then or 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 λ_t, λ_s≠ 0 then but, if then λ_t, λ_s are not necessarily both different from zero.

Lemma A.2. For any t≠ s and fixed r,m > 0,rank(P_ts)=2r

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

has rank equal to 2 because P_1,T(t)≠ P_1,T(s) for any t≠ s (see Lemma A.1.).

Proof of Theorem 1. In the system P_ts λ_ts=x_ts, where x_ts≠ 0, the number of unknowns does not exceed the number of equations, (m + 1)r≥2r and rank(P_ts)=2r for all t≠ s, m≥1, by Lemma A.2. By the same reasoning, rank(P_ts, x_ts)=rank(P_ts)=2r for all t≠ s, 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 λ_t, λ_s≠ 0 with λ_t,≠ λ_s or if either λ_t or λ_s equals zero. By (21), λ_t=0 for some t,s,t≠s, if where η_ts ∈{P_1,T(s), P_1,T(t)} and Clearly, when and (where λ_t=λ_s≠ 0). On the contrary, λ_t, λ_s≠0 with λ_t≠ λ_s, for any t,s,t≠ s, if and and Note that, in the previous case, can have up to r free variables. When the number of free variables in x_ts is at most r, whereas if for all r × r matrices ϒ_ts the number of free variables in x_ts is at most 2r. Given that for η_ts ∉ {P_1,t(s), P_1,t(t)} implies β_t(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(P_ts, x_ts) is either 2r or 2r+1 Whenever rank(P_ts, x_ts)=2r=rank(P_ts) there is one solution

to the system and consequently if both λ_t, λ_s≠ 0 with λ_t≠ λ_s or if either λ_t or λ_s equals zero. Whenever rank(P_ts, x_ts)=2r+1>rank(P_ts) there is no solution to the system and therefore So that rank(P_ts, x_ts)=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 and with for m≥2. Hence, rank(P_ts, x_ts)=2r, if

for some non zero r × r matrices

and and by setting and Here, and are given by (14). Similar to the previous Theorem, λ_t=0 for some t, s where t≠ s, if where η_ts ∈ {P_1,T(s), P_1,T(t)} and Hence, given that the solutions (24) imply β_t(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