Below is a spread built with two ETFs that pass the cointegration test i.e. Adjusted Dickey Fuller, adfTest(type="nc") in R's fUnitRoots with a p-value < 0.01.

The red line is the trendline.

What test can I use to proove that: (1) both securities are cointegrated and (2) they are mean reversing and the mean is constantly 0 (i.e. stationary, not trended)?


EWA and EZA spread

  • $\begingroup$ Depends on what you mean by "prove". $\endgroup$
    – LazyCat
    Apr 12, 2012 at 20:50
  • $\begingroup$ You can include a constant and a deterministic trend in your initial regression. You then test for the presence of the trend (ie the significance of the parameter before the trend term). $\endgroup$
    – Zarbouzou
    Apr 13, 2012 at 10:27

1 Answer 1


Here is an empirical strategy to test for cointegration.

FIRST, check whether both $X_t$ and $Y_t$ contain an unit root.

  • If they are both stationary then model $Y_t$ or $X_t$ in levels (and nothing is wrong).
  • If one of the two is $I(1)$ (non-stationary for one level), then take differences to ensure stationarity.
  • If they are both non-stationary, and hence $I(1)$, then test for co-integration:

    1. if the residuals are $I(0)$, then we speak of the presence of cointegration. Estimate then an ECM model ($Y_t = \beta_0 + \beta_1 X_t + \eta_t$ obtaining $\hat{\beta_0}$ and $\hat{\beta_1}$ and using it in: $\Delta Y_t = \Delta X_t'\phi - \psi(Y_{t-1}-\hat{\beta_0} - \hat{\beta_1}X_t) + \varepsilon_t$. When $\varepsilon_t \sim N(0,1)$ then both $\psi$ and $\phi$ are asymptotically valid.
    2. if the residuals are $I(1)$ then we speak of spurious regression. In that case you should model both variables by taking the first differences.
  • 1
    $\begingroup$ This post is identical to this and this. Don't repost. $\endgroup$ Apr 13, 2012 at 2:33
  • $\begingroup$ I disagree with the latest part of you explanation that "2. (...) In that case you should model both variables by taking the first differences." $\endgroup$
    – tagoma
    Apr 14, 2012 at 10:31
  • $\begingroup$ Yes I understand why you say that, but given the two variables you have I think that is the best you can do. What do you propose? $\endgroup$ Apr 14, 2012 at 11:24

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