The Uniqueness of the Cointegrating Vector

Suppose that we have (only) two non-stationary time-series, X_1t and X_2t (t = 1, 2, 3, .....). More specifically, suppose that both of these series are integrated of order one (i.e., I(1)). Then there are two possibilities - either X₁ and X₂ are cointegrated, or they aren't.

You'll recall that if they are cointegrated, then there is a linear combination of X₁ and X₂ that is stationary. Let's write this linear combination as Z_t = (X_1t + αX_2t). (We can normalize the first "weight" to the value "one" without any loss of generality.) The vector whose elements are 1 and α is the so-called "cointegrating vector".

You may be aware that if such a vector exists, then it is unique.

Recently, I was asked for a simple proof of this uniqueness. Here goes.........

Suppose that there are two linear combinations, Z_t = (X_1t + αX_2t), and Y_t = (X_1t + βX_2t), both of which are I(0), where α ≠ β.

Now consider their difference, W_t = (Z_t - Y_t) = (α - β)X_2t.

W_t must be I(0), because both Z_t and Y_t are I(0). However, W_t is proportional to X_2t, which is I(1), which implies that  W_t  is I(1). We have an immediate contradiction!

The only resolution is if α = β, meaning that the cointegrating vector is unique.    QED

Keep in mind that this result applies to the case where there are only two I(1) time-series. What we've seen is that the number of cointegrating vectors is either zero (no cointegration) or one. More generally, if we have time-series, all of which are I(1), then there are at most (N - 1) cointegrating vectors.

© 2018, David E. Giles