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Let $Y_{t}$, for $t=1,2, \dots$ be AR(1) process. $$ Y_{t+1} = c_{1} + \phi Y_{t} + \varepsilon_{t} $$

Next, assume that for some $X_{t}$ we have $$ X_{t} - \beta Y_{t} = u_{t}, $$ where $u_{t}$ is stationary. Therefore, we can say that $X_{t}$ and $Y_{t}$ are cointegrated.

Is it possible that $X_{t}$ is also AR(1)?

Attempt:

Assume that $u_{t}$ is i.i.d. Then, in this simplest case $X_{t}$ is $$ X_{t} = \beta Y_{t} + u_{t}, $$ therefore, it is ARMA(1,1). Though, it is not a proof of the original statement.

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    $\begingroup$ To be co-integrated you have to first be integrated. It is not possible for two AR(1) processes to be cointegrated unless the coefficient is 1 in both cases. This is just the definition of cointegration. $\endgroup$
    – Chris Haug
    Jan 27, 2021 at 15:32
  • $\begingroup$ Dear @ChrisHaug , If I understand correctly, in order to speak about co-integration of the process $Y$ to another process, Y mast be $I(d)$, with $d > 1$. Is this what you mean? $\endgroup$
    – ABK
    Jan 27, 2021 at 15:40
  • $\begingroup$ $d=1$ is fine and is the basic example. Regarding your attempt, the process is not ARMA(1,1) as expressed. For ARMA(1,1) you need lag of $X_t$ and lag of $u_t$ on the right hand side but you have neither of these. You do have $Y_t$, though, and it does not belong in an ARMA(1,1) for $X_t$. @ChrisHaug, why not post a short answer based on the comment? $\endgroup$
    – Richard Hardy
    Jan 27, 2021 at 15:43
  • $\begingroup$ Dear @RichardHardy, oh, I see. In order to have ARMA, one would need $X_{t} = \beta Y_{t} + u_{t-1}$, but in the current case we have $X_{t} = \beta Y_{t} + u_{t}$. $\endgroup$
    – ABK
    Jan 27, 2021 at 15:54
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    $\begingroup$ @ABK, you may be right (especially based on the reference), but you have not demonstrated that you are. My comment was about that (before you provided the reference). $\endgroup$
    – Richard Hardy
    Jan 27, 2021 at 16:29

1 Answer 1

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Cointegration literally means "to be integrated, together" (see the usual "common trend" interpretation). It cannot logically apply to processes which are not integrated. So if you have $|\phi| < 1$ for both AR processes, they cannot be cointegrated by definition.

More precisely, the general definition of cointegration requires $X_t$ and $Y_t$ to be integrated of order $d_1$ (the same order for both), and that there exist a linear combination $Y_t - \beta X_t$ which is integrated of order $d_2$, with $d_1 > d_2 \geq 0$. Explicitly, the order of integration of the linear combination $d_2$ must be strictly smaller than the order of integration of the individual processes $d_1$, which precludes $d_1$ from being zero.

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