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I did cointegration test on two identical time series, and the result shows that they are not cointegrated, but intuitively, I think they are.

Can anyone share some thoughts on this? Thanks!

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    $\begingroup$ What kind of test did you use to check for co-integration? $\endgroup$
    – chrisaycock
    Apr 15, 2011 at 16:56
  • $\begingroup$ I used the matlab function egcitest() for Engle-Granger test, with both 'tau test' and 'z test'. $\endgroup$
    – user752
    Apr 15, 2011 at 17:12
  • $\begingroup$ @David Do you get the expected results from this example? $\endgroup$
    – chrisaycock
    Apr 15, 2011 at 17:17
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    $\begingroup$ Am I missing something? If the timeseries are identical how would calculating any statistical comparison beyond a simple check that they are in fact identical be meaningful? $\endgroup$
    – Joshua Chance
    Apr 15, 2011 at 22:13
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    $\begingroup$ I'm not commenting on the value of the question. I'm saying that it is incoherent. Two identical time series will never diverge and will never revert so it seams to me that asking whether they're cointegrated is more suited for philosophy than math. $\endgroup$
    – Joshua Chance
    Apr 17, 2011 at 1:09

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Let us test that $x$ and $y$ are co-integrated, say that $x_t, y_t \sim I(1)$. In the Engle-Granger we test stationarity of the error term in $$y_t = \alpha + \beta x_t + u_t$$ which we estimate as $$\hat u_t = y_t - \hat \alpha - \hat \beta x_t$$ and find that $\hat \alpha =0$, $\hat \beta = 1$, and $\hat u_t = 0 \; \forall t$.

So now when we Dickey-Fuller test residuals in something like $$\Delta \hat u_t = \gamma_0 + \gamma_1 \hat u_{t-1} + \epsilon_t$$ nothing will be significant and we won't find any co-integration.

I am not precisely schooled in this theory, so I'm not sure if this means these series can't be referred to as "co-integrated" (clearly they have the same drift) or if this is just a trivial case where the test fails,

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Two integrated series $X_t$ and $Y_t$ are cointegrated if their linear combination (some, not any) $\alpha X_t+\beta Y_t$ is stationary. If you have $P(X_t=Y_t)=1$ for all $t$, then $P(\alpha X_t+\beta Y_t=(\alpha+\beta) X_t)=1$. So according to definition of cointegration $(\alpha+\beta) X_t$ should be stationary, which is identical to $X_t$ being stationary. And here we get the contradiction, since $X_t$ is integrated, hence not stationary.

This was a basic explanation why you received your result. However a lot depends on how the actual statistic is computed. For other statistics or their software implementations you might get that two identical series are cointegrated, but that will not mean that they are. Two identical time series are the degenerate case which no-one checks against, and with degenerate cases you can always get unexpected results.

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  • $\begingroup$ The first paragraph is wrong, in particular the definition of cointegration. Stationarity needs not be achieved for any linear combination, but only for some. $\endgroup$
    – Ryogi
    Apr 13, 2012 at 20:36
  • $\begingroup$ @Ryogi, well I would have used word any if I wanted to mean any. If you do not prepend word any, then it is assumed that linear combination is specific. At least that is the convention in mathematical texts. But I can see that it might seem ambiguous. I'll change the wording to eliminate that. $\endgroup$
    – mpiktas
    Oct 11, 2012 at 7:17
  • $\begingroup$ You didn't use any quantifier in the original answer. That's wrong. Using some is acceptable. A mathematical text would state that there exist $\alpha$ and $\beta$ such that $\alpha X + \beta Y$ is stationary. $\endgroup$
    – Ryogi
    Oct 16, 2012 at 4:03
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    $\begingroup$ More importantly, since the cointegrating linear combination in this case requires $\alpha = -\beta$, the conclusion that $(\alpha + \beta) X$ must be stationary does not lead to any contradiction as $\alpha + \beta$ equals zero and $X$ could be anything. $\endgroup$
    – Ryogi
    Oct 16, 2012 at 5:01
  • $\begingroup$ If $alpha=-\beta$ then $P((\alpha+\beta)X=0)=1$. I.e. we have a constant, which technically is a stationary process with variance zero. We can always exclude this case. $\endgroup$
    – mpiktas
    Oct 16, 2012 at 6:59
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Your intuition is correct. $X_t$ and $Y_t$ are cointegrated if there exists some linear combination $\alpha X_t + \beta Y_t$ that is stationary (or more generally, of lower cointegration index --- see for example, Hamilton, pag 571). If $X_t = Y_t$, the above linear combination is zero (hence stationary) whenever $\alpha = -\beta$.

On the other hand, most tests exclude this particular case. The exact reasons depend on the specific test you are using.

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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.
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  • $\begingroup$ This post is identical to this and this. Don't repost. $\endgroup$
    – chrisaycock
    Apr 13, 2012 at 2:48
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Identical? So $Y_t=X_t$ for all $t$?

Then the difference is zero which is more than just a stationary time-series. They are perfectly collinear.

Now, it depends on the cointegration test you use whether high collinearity will show up as cointegration. If you use Engle-Granger, you do a regression first, which will find $\alpha=0,\beta=1$ and $\epsilon_t=0$. I'm not sure what an ADF test on $0$ will actually do, and this could lead to a numeric instability.

I believe that a Johansen test will also lead to a numeric instability.

If this is a serious problem for some automated system, then probably the first thing to do is look for collinearity (take the timeseries covariance matrix and look at the condition number). If they are highly collinear or identical, you should catch it before running it through the cointegration engine.

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