4. Equivalence with the standard I(1) and I(2) conditions
We want to prove the equivalence with the standard conditions in
Johansen (1996) and derive the explicit expression of H(1). We intro-
duce the following notation: let A⊥ be the orthogonal complement of
an m × n matrix A of rank n < m ,let A = A ( AA )- 1 and write the
Taylor expansion of Π(z) at z = 1 as
Π(г) = Π(1) + Π(1)(г - 1) + Π21)(г - 1)2 + (г - l)3Π3(г).
The order of integration is established (Johansen, 1996) by some re-
duced and full rank conditions on specific matrices: Xt is I(1) if and
only if IΠ(1) I = 0 and '/ Π(1)(l∣ = 0; similarly, the I(2) condition
states that the order of integration is two if and only if ∣Π(1)∣ = 0,
∣(0lΠ(1)(l∣ = 0 and ∣α2θβ21 = 0 where θ = -ɑ + ∏(1)βa0∏(1),
α2 = (l^l, β2 = (LnL, (0lΠ(1)(l = ξη' and ξ, η are p — r × s matrices
of full rank s < p - r. Using (3.1) and Theorem 3.3 we rewrite the
identity Π(г)Πa(г) = Πa(г)Π(г) = ∣Π(г)∣Ip as
(4.1) Π( г ) H ( г ) = ( г — 1) dg ( г ) Ip
and H(г)Π(г) = (г — 1)dg(г)Ip. At г = 1 they read αβ0H(1) =
H (1)(β 0 = 0p and mean that
H(1) =βlζd(0l,
for some p — r × p — r matrix ζd of rank 0 < rd ≤ p — r.
Proposition 4.1 (I (1) case). A necessary and sufficient condition for
∣(0lΠ(1)(l∣ = 0 is that
a=m—1
in Πa(г) = (г — 1)aH(г) and H(1) = 0p. The explicit expression for
H(1) is
H (1) = g (1) (l ( (0l Π(1) (l ) -1 αL.
Proof. Assume a = m — 1 so that d = 1; differentiate (4.1) at г = 1
to get ∏(1)H(1) — αβ0H(1) = g(1)Ip and thus
αL ∏(1) β±ζ 1 = g (1) Ip-r.
Then g(1) = 0 implies ∣ζ 11 = 0, ∣αL∏(1)(l∣ = 0 and ζ 1 = g(1)((0lΠ(1)(l)-1,
and thus the I(1) condition is satisfied.
Assume now the I(1) condition holds and suppose d = m — a > 1;
differentiating (4.1) at г = 1 we get αL∏(1)( ,zd = 0p-r ; since ζd = 0p-r
this contradicts ∣(0l∏(1)(l∣ = 0 and implies m — a = 1. ■