For bivariate time series cases, the Engle-Granger two step cointegration test is essentially testing the linear combination for a unit root. The format of the error term is thus: $$ y_t - \gamma x_t = \epsilon_t $$

When a lag is used in the ADF test, say of length $n$, is the mean reverting series: $$ y_t - \gamma x_{t-n} = \epsilon_t $$

Or is it something else? I am trying to understand how the lag changes the mean reverting series. Also, does this generalize to other cointegration tests and models?


Example: Let $y$ = housing prices in San Francisco and $x$ = housing prices in New York City. The approximation process is given by the linear combination $y_t-\gamma x_t = \epsilon_t$. Testing $\epsilon_t$ for a unit root via ADF results in a high probability of cointegration with a lag of $n$.

Does this mean the mean reversion process is $y_t - \gamma x_{t-n} = \epsilon_t$, meaning there is a lag in the relationship such that the housing prices in San Francisco are cointegrated with the housing prices in New York $n$ time steps ago? If not, how does the lag express itself in the relationship of the housing prices?

  • $\begingroup$ No, this does not look right. Forget $y_t$ and $x_t$ for a while and consider only $\epsilon_t$. Then look at how ADF test with lags works, and apply that directly to $\epsilon_t$. $\endgroup$ Commented Dec 9, 2015 at 8:17
  • $\begingroup$ @RichardHardy I have thought a little about your comment but I am not sure how it is helpful. Could you please be more explicit, perhaps in an answer? What is the structure of the mean reverting series when there is a lag? $\endgroup$
    – d0rmLife
    Commented Dec 9, 2015 at 18:24
  • $\begingroup$ Regarding the edited question, the mean-reversion process is described by the VECM. VECM explicitly shows how $y_t$ and $x_t$ interact. Doesn't the VECM representation answer your question? $y_t-\gamma x_{t-n}$ is not used anywhere (neither in the Engle-Granger procedure, nor in the Johansen procedure, nor in the VECM). $\endgroup$ Commented Dec 13, 2015 at 20:35
  • $\begingroup$ So the coefficients ${\delta_t, \delta_{t-1}, ..., \delta_{t-n}}$ describe the relative weights given to each previous time step up to $n$ steps ago? $\endgroup$
    – d0rmLife
    Commented Dec 13, 2015 at 20:41

1 Answer 1


Good detailed coverage of cointegration and cointegration testing is available in Chapter 12 of Zivot & Wang's "Modeling Financial Time Series with S-PLUS" (2006). I will base my answer on it, but I will skip most of the details that are not directly relevant with regards to the original question.

Engle & Granger's cointegration testing procedure consists of two stages:

  1. Obtaining the cointegration residual $\varepsilon_t=y_t-\beta x_t$
  2. Testing $\varepsilon_t$ for a unit root.

The stage relevant for your question is the second one. It is common to use the augmented Dickey-Fuller (ADF) test in this stage. The word "augmented" indicates that the test regression may include lags of the dependent variable (with $\delta$ coefficients in front):

$$ \Delta \varepsilon_t = \alpha + \beta t + \gamma \varepsilon_{t-1} + \delta_1 \Delta \varepsilon_{t-1} + \dotsc + \delta_{p-1} \Delta \varepsilon_{t-p+1} + u_t $$

($\alpha$ and $\beta t$ may or may not be included depending on the first stage of the Engle & Granger procedure; see the reference above for details.) The motivation for including the lags of the dependent variable can be found in Chapter 4 of the same textbook. Essentially, they are there to make the test regression better represent the underlying process being tested for a unit root.

In my understanding, the phrase

a lag is used in the ADF test, say of length $n$

means that there are $n$ lags of the dependent variable in the ADF test equation, i.e. $p-1=n$ or $p=n+1$. That is, you have an equation of the form

$$ \Delta \varepsilon_t = \alpha + \beta t + \gamma \varepsilon_{t-1} + \delta_1 \Delta \varepsilon_{t-1} + \dotsc + \delta_{n} \Delta \varepsilon_{n} + u_t $$

(with or without $\alpha$ and $\beta t$.)

Meanwhile, the first stage of the Engle & Granger procedure is not affected by the quoted statement and remains as indicated above.

  • $\begingroup$ (+1) This seems like a good answer, but I'm still trying to understand parts of it. Can you tell me a little more about the $\Delta\epsilon_t$ equation? In particular, what does $\Delta$ represent in this case? And also the $\delta$ coefficients? Also, I'm still not entirely clear on how the mean reverting series changes when a lag is used. $\endgroup$
    – d0rmLife
    Commented Dec 12, 2015 at 19:18
  • $\begingroup$ Sorry, I am travelling these days and have very limited time for Cross Validated. $\Delta$ represents first differencing; for some time series $x_t$, $\Delta x_t:=x_t-x_{t-1}$. Regarding terms with $\delta$ coefficients in front, the motivation for including them is in my answer. Regarding coefficients themselves, they are just coefficients, nothing special. And as I said, once you come to the second stage, you are down to ADF testing, which you can read about separately in any time series textbook. $\endgroup$ Commented Dec 12, 2015 at 20:12
  • $\begingroup$ I realize that ADF testing is the process of testing an univariate series for a unit root, but I feel like my main question is still not being answered. Does the use of a lag change the form of the mean-reverting series? I.e., with lag=0, the mean reverting series as described above is $y_t-\gamma X_t$; does this change when a lag is used? $\endgroup$
    – d0rmLife
    Commented Dec 12, 2015 at 21:59
  • $\begingroup$ In VECM models, the error correction term is a linear combination of levels of variables such as $y_t-\gamma X_t$, and this form prevails regardless of how many lags there are in (1) the VECM model that is the true data generating process (DGP) or the best approximation to the true DGP; (2) the ADF test which is the second stage of the Engle-Granger testing procedure. $\endgroup$ Commented Dec 13, 2015 at 19:37
  • $\begingroup$ The $X_t$ is actually a typo: $x_t$ is what I meant. Sorry for any confusion. I am going to edit the question to provide an example, because I am still not entirely clear on what effect the lag has on the mean-reverting process (best approximation of the DGP). $\endgroup$
    – d0rmLife
    Commented Dec 13, 2015 at 20:00

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