







  • $\begingroup$ Nice.... What about an intuition for multiple time series and simultaneous cointegration? $\endgroup$ – user40 Feb 7 '11 at 16:27
  • $\begingroup$ +1: I didn't know this article - thank you for the link! $\endgroup$ – vonjd Feb 7 '11 at 16:32
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    $\begingroup$ @user40 A Drunk, Her Dog and A Boyfriend econ.canterbury.ac.nz/downloads/amstat.pdf $\endgroup$ – RockScience Jan 2 '12 at 7:28



  • $\begingroup$ Okay... Maybe little bit more technical please! Does it mean that in cointegration one is concerned about the difference between two price time-series? $\endgroup$ – user40 Feb 7 '11 at 16:23
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    $\begingroup$ I find it unfair to downvote this answer. It was asked to give a non-technical explanation! This one an investor will understand! The same association came Shane to mind... $\endgroup$ – vonjd Feb 7 '11 at 16:27
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    $\begingroup$ I agree completely and gave you a +1. Your answer was spot on for the question. $\endgroup$ – Shane Feb 7 '11 at 16:29
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    $\begingroup$ This is really fantastic. $\endgroup$ – Jase Dec 28 '12 at 16:35










Let $u_t$ be the random walk $$ u_t = u_{t-i} + \varepsilon_t $$ where $\mathrm{E}[\varepsilon_t]=0$ and $\mathrm{var}[\varepsilon_t]=\sigma^2$ , i.e. $\varepsilon_t$ is stationary.

Now let $$X_t = \alpha u_t +\nu_t$$ and $$Y_t = \beta u_t + \eta_t$$ where $\nu_t$ and $\eta_t$ are stationary processes similar to $\varepsilon_t$

Then both $X_t$ and $Y_t$ are non stationary because they are linear functions of the non-stationary (stochastic trend) variable $u_t$.

However $$\beta X_t - \alpha Y_t = \beta \nu_t - \alpha_t \eta_t $$ is a linear combination of the stationary disturbances and is therefore stationary. When this happens $X_t$ and $Y_t$ are said to be cointegrated. $X_t$ and $Y_t$ are said to contain the same stochastic trend.

The idea behind the Dickey-Fuller test is to estimate a regression which estimates the ratio of $\alpha$ and $\beta$ and test if the estimated residuals are stationary. These residuals do not follow a standard distribution.


Here is an empirical strategy to test for cointegration.



  • $\begingroup$ You have two identical posts. What gives? $\endgroup$ – chrisaycock Apr 13 '12 at 2:33
  • $\begingroup$ that is true, but I think that the strategy is useful in answering (for a part) the two questions right? $\endgroup$ – JohnAndrews Apr 13 '12 at 2:36
  • $\begingroup$ No, don't do that. The spam filters went off and alerted me (a moderator) that you had done this. I also notice this post is identical to this one $\endgroup$ – chrisaycock Apr 13 '12 at 2:49
  • $\begingroup$ ok, I will try to customize my answers more in the following... $\endgroup$ – JohnAndrews Apr 13 '12 at 2:51

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