I have two questions:

  1. Why test $u_t$ (see below) for stationarity instead of any other linear combination?
  2. And to confirm, if $u_t$ is not stationary does it mean that $x_t$ and $y_t$ are not cointegrated?

If $x_t$ and $y_t$ are cointegrated, then there exists a linear combination of them that is stationary. So in Engle-Granger they decide the linear combination using a linear regression:

$$ y_t - \beta x_t = u_t $$

Why won't I just test the following instead? $$ y_t - x_t = z_t $$

Is $u_t$ somehow superior to $z_t$? In what way?


2 Answers 2


Purpose of the first step in Engle-Granger cointegration test

...is to find a stationary combination of the integrated variables at hand. If there are just two variables, then it will be the stationary combination as it must be unique.

Why test $u_t$ (see below) for stationarity instead of any other linear combination?

Because when the stationary combination is unique, any other linear combination will be nonstationary regardless of whether the integrated variables at hand are cointegrated or not.

And to confirm, if $u_t$ is not stationary does it mean that $x_t$ and $y_t$ are not cointegrated?

Yes. If $u_t$ obtained from the first step of the Engle-Granger procedure is nonstationary, then $x_t$ and $y_t$ are not cointegrated.

Why won't I just test the following instead? $y_t-x_t=z_t$

Because that is very restrictive and need not hold in presence of cointegration. If two series $y_t$ and $x_t$ are cointegrated, there exists a $\beta$ such that $y_t-\beta x_t=u_t$ where $u_t$ is stationary. But there is no requirement that $\beta=1$.

Is $u_t$ somehow superior to $z_t$? In what way?

See the argumentation above. But when could $z_t$ be more useful than $u_t$? Say we have the following subject-matter knowledge of $\beta$: if cointegration is present (which we are not sure about and want to test for), $\beta=1$. Then we can just take $\beta=1$ which implies using $z_t$ in place of $u_t$ and test $z_t$ for a unit root. If, on the other hand, we ignored the knowledge and used $u_t$ rather than $z_t$, we could by chance end up finding a fit that is too good such that while $y_t$ and $x_t$ are not cointegrated in reality, $u_t$ still appears stationary (due to a small sample).


These are two different hypotheses:

  1. $y$ and $x$ are cointegrated for any vector (hence estimate vector $(1, -\beta)$

  2. $y$ and $x$ are cointegrated for the vector $(1,-1)$

Usually, you would start with the first, and if you find cointegration, you could then test if $\beta=1$

  • $\begingroup$ I see ... so why is the Engle-Granger method choice of coefficients so popular? I guess there is no test to see if there exists any a, b such that aX+bY is stationary? $\endgroup$ May 13, 2016 at 0:38
  • $\begingroup$ And I understand beta is the choice to minimize u (defined above), but it seems there should be some breakthrough in this two step procedure since it has been given a name. $\endgroup$ May 13, 2016 at 0:41
  • 1
    $\begingroup$ ad 1: If the cointegration vector is $(1,-\beta)$, then OLS consistently estimates it. Among just two variables, there can be at most one cointegrating relationship, so there is no point to search for another one than the one that would be identified by OLS if it exists. $\endgroup$ May 13, 2016 at 6:26
  • 1
    $\begingroup$ ad 2: the Engle/Granger (Econometrica 1987) paper suggesting this procedure was the first paper developing rigorous tests for cointegration at all (they partly got a Nobel for that!), so, yes, it was a breakthrough. $\endgroup$ May 13, 2016 at 6:27
  • $\begingroup$ First, I want to thank you and let you know I really appreciate your response. ad 1: This seems to contradict what Matifou says that $y$ and $x$ can be cointegrated for any vector. Or maybe I am misunderstanding Matifou. ad2: Why is this test more rigorous? Is the distribution of the test statistic resulting from Engle-Granger somehow more well-understood since we are testing $u_t$ the residual of a regression? $\endgroup$ May 13, 2016 at 13:53

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