The Integration Order of Vector Autoregressive Processes

Now consider the numerator in (3.2) at z = 1: when we evaluate the
adjoint matrix polynomial at the unit root we have that¹

Πa(1) = 0p when r < p - 1,

because the determinant of any p - 1 × p - 1 minor extracted from Π(1)
is zero and

Π_a(1) 6= 0_p when r = p - 1,

because there is at least one non singular p - 1 × p - 1 minor in Π(1).

It follows that z - 1 can be factored out a times from Π_a(z), for some
a > 0 when r < p - 1 and for a = 0 when r = p - 1; consequently we
have that

Πa(z) = (z - 1)^aH(z),

where H(1) 6= 0p and a ≥ 0.

The only way of having a pole at z = 1 in (3.2) is if Π_a(z) goes to
zero at a slower rate than ∣Π(z) ∣. Equivalently, we could say that it
must be the case that a < m.

This is exactly what Theorem 3.3 below makes precise.

Theorem 3.3. If Assumption 3.1 holds, the order of integration of Xt
is equal to

d = m - a,

where H(1) 6= 0_p in Π_a(z) = (z - 1)^aH(z) and m is the multiplicity of
the unit root in the characteristic equation.

Proof. Assume that a = m - d and H(1) 6= 0p; then
from which we see that Π(z)^-1 has a pole of order d at z = 1; thus Xt
is I(d) by Proposition 3.2. Assume now that X_t is I (d); by Proposition
3.2 it follows that Π(z)^-1 has a pole of order d at z = 1. Then (z -
1)^(d-1)Π(z)^-1 is not defined at z = 1 and G(z) = (z- 1)^dΠ(z)^-1 is such
that G(1) = 0_p. This defines H(z) = Πa(z)/(z — 1)(^m-d) which implies
a = m — d and H (1) = g (1) G (1) = 0_p. ■

Thus the order of integration of the process is simply equal to the
multiplicity of the unit root in the characteristic equation minus the
multiplicity of the unit root in the adjoint matrix polynomial. We know
how to calculate the roots of a polynomial; then it is clear that we can

¹We use 0p for the p × p zero matrix and Ip for the identity matrix of the same
dimension. For non square matrices we write both the row and column indexes.

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