I was reading about cointegration (page 5) and it said

"If a linear combination of combination of a set of $I(1)$ variables is $I(0)$, then the variables are cointegrated."

Does this mean that any 2 $I(1)$ variables, no matter what they are will definitely be cointegrated? If yes, can you give the proof. If no, could you give a counterexample. I'm very curious about this topic.


2 Answers 2


The answer is no. It is said that there must be some particular linear combination of $I(1)$ variables, but there may not exist one. In that case we have so called spurious regression, you may find an example here. We say that two or more time series are cointegrated if they share a common stochastic drift, so it is possible for variables to be $I(1)$ and to have different stochastic drifts.

  • $\begingroup$ Can you please say how the book is called where you took the example explaining the cointegration? Thanks!! $\endgroup$
    – DatamineR
    Nov 22, 2013 at 21:59

Suppose you have $$x_t = x_{t-1} + \delta_t, y_t = y_{t-1} + \epsilon_t$$ with obvious assumptions ($\epsilon_t$ are i.i.d., $\delta_t$ are i.i.d. and independent of $\epsilon_t$, $x_0$ and $y_0$ are independent of each other and $\epsilon$ and $\delta$). Then no matter how hard you try, for $a^2+b^2 \neq 0$, $ax_t + by_t$ will be an $I(1)$ process: it is itself non-stationary, while its first difference is $a\delta_t + b\epsilon_t$ is white noise.

So if anything, cointegration should be viewed as a much rarer phenomenon than lack of cointegration. Examples of cointegration would require special constructions. Define $z_t = z_{t-1} + \gamma \delta_t + \nu_t$ for some $\gamma\neq0$, where $\nu_t \sim ARMA(p,q)$. Then $z_t$ and $x_t$ are cointegrated, as $z_t - \gamma x_t = z_0 - \gamma x_0 + \nu_t$ is now stationary.


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