If four variables are cointegrated, does this mean that each two of them are cointegrated? Can we regress only two of the four variables assuming they are cointegrated, or do we need to check for cointegration of the two variables?
For me if four variables are cointegrated this means that they move close to each other, to say it in crude way. This means that if four variables are moving together then each pair should be moving together, too. Is this a valid reasoning?
Four 'variables' $x_i$ are co-integrated if one can find a linear combination of the four variables that is integrated of order zero (noise let's say) so
$x_1, x_2, x_3, x_4$ are co-integrated if there exist $\alpha_i$ such that $\sum_i \alpha_i x_i \text{ is I(0) }$.
From this you can not derive that there exists a combination of $x_1$ and $x_2$ is $I(0)$.
What you could say is that, for each $x_i$ (with $\alpha_i \ne 0$) one can write $x_i = \sum_{k,k \ne i} \frac{\alpha_k}{\alpha_i} x_k + \text{ 'noise'}$. So each $x_i$ 'moves together' with a linear combination of the other three.
Only in very exceptional cases one can say that $x_i$ and $x_j$ 'move together' namely only when there exists a linear combination with $\alpha_k=0$ for $k \not \in \{i,j\}$. This will in general not hold.
Suppose you have a system of $k$ integrated variables with $k>2$.
Suppose that taken together these variables are cointegrated.
Proposition: there may exist a pair of variables that is not cointegrated.
One example is enough to prove the proposition.
Let $x \sim I(1)$, $y \sim I(1)$, and let them be not cointegrated.
Let $\varepsilon$ be an i.i.d. sequence; $\varepsilon \sim I(0)$.
Let $z:=x+y+\varepsilon$. Obviously, $z \sim I(1)$.
Take a linear combination of the three integrated variables: $z-(x+y)$. It equals $\varepsilon$ and is $I(0)$. Therefore, $x$, $y$ and $z$ are cointegrated.
Meanwhile, $x$ and $y$ are not cointegrated by the assumption above.
Thus you have an example where the system of three integrated variables is cointegrated while a pair of these variables is not cointegrated. (Actually, the other two possible pairs, $(x,z)$ and $(y,z)$ will also be not cointegrated, but this observation is not necessary for the proof.)
