Thursday, May 31, 2018

The Uniqueness of the Cointegrating Vector

Suppose that we have (only) two non-stationary time-series, X1t and X2t (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 Zt = (X1t + αX2t). (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, Zt = (X1t + αX2t), and Yt = (X1t + βX2t), both of which are I(0), where αβ.

Now consider their difference, Wt = (Zt - Yt) = (α - β)X2t.

Wt must be I(0), because both Zt and Yare I(0). However, Wt is proportional to X2t, which is I(1), which implies that  Wt  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

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