Raising the bar

What is cointegration? The discussions here and here  are bad because they link the definition to differencing a time series. In fact, to understand cointegration, you need two notions: stationary processes  (please read before continuing) and linear dependence.

Definition. We say that vectors $latex X_1,...,X_n$ are linearly dependent if there exist numbers $latex a_1,...,a_n$, not all of which are zero, such that the linear combination $latex a_1X_1+...+a_nX_n$ is a zero vector.

Recall from this post that stationary processes play the role of zero in the set of all processes. Replace in the above definition "vectors" with "processes" and "a zero vector" with "a stationary process" and - voilà - you have the definition of cointegration:

Definition. We say that processes $latex X_1,...,X_n$ are cointegrated if there exist numbers $latex a_1,...,a_n$, not all of which are zero, such that the linear combination $latex a_1X_1+...+a_nX_n$ is a stationary process. Remembering that each process is a collection of random variables indexed with time moments $latex t$, we obtain a definition that explicitly involves time: processes $latex \{X_{1,t}\},...,\{X_{n,t}\}$ are cointegrated if there exist numbers $latex a_1,...,a_n$, not all of which are zero, such that $latex a_1X_{1,t}+...+a_nX_{n,t}=u_t$ where $latex \{u_t\}$ is a stationary process.

To fully understand the implications, you need to know all the intricacies of linear dependence. I do not want to plunge into this lengthy discussion here. Instead, I want to explain how this definition leads to a regression in case of two processes.

If $latex \{X_{1,t}\},\{X_{2,t}\}$ are cointegrated, then there exist numbers $latex a_1,a_2$, at least one of which is not zero, such that $latex a_1X_{1,t}+a_2X_{2,t}=u_t$ where $latex \{u_t\}$ is a stationary process. If $latex a_1\ne 0$, we can solve for $latex X_{1,t}$ obtaining $latex X_{1,t}=\beta X_{2,t}+v_t$ with $latex \beta=-a_2/a_1$ and $latex v_t=1/a_1u_t$. This is almost a regression, except that the mean of $latex v_t$ may not be zero. We can represent $latex v_t=(v_t-Ev_t)+Ev_t=w_t+\alpha$, where $latex \alpha=Ev_t$, $latex w_t=v_t-Ev_t$. Then the above equation becomes $latex X_{1,t}=\alpha+\beta X_{2,t}+w_t$, which is simple regression. The case $latex a_2\ne 0$ leads to a similar result.

Practical recommendation. To see if $latex \{X_{1,t}\},\{X_{2,t}\}$ are cointegrated, regress one of them on the other and test the residuals for stationarity.