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My question is about the definition of cointegrated.

$y_t =y_{t-1}+u_t$

$u_t =\eta_t +0.5\eta_{t-1}$

where $\eta_t\sim N(0,1)$ is i.i.d. white noise.

I claim that $y_t$ and $y_{t-1}$ are cointegrated because $y_t -y_{t-1}=u_t$ is stationary and both $y_t$ and $y_{t-1}$ are non-stationary.

Is this a correct application of the definition? I know it would be more standard to just consider $y_t$ as I(1), but if I said $y_t$ and $y_{t-1}$ are cointegrated, is that a mathematically correct statement?

EDIT: Firstly, thank you to everyone that helped. The "attempt" fails as 1muflon1 describes.

EDIT2: Some people are currently not satisfied with the answers, and an ongoing discussion is here.

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    $\begingroup$ I(1) means that yt is first-difference stationary. Cointegration means that two series Y and X have a stationary relationship; not that a single series is equal to a stationary error term. $\endgroup$
    – EB3112
    Aug 13 at 15:07
  • $\begingroup$ @Eb3112 Let $X=y_{t-1}$ and $Y=y_t$. Then X and Y have a stationary relationship, correct? $\endgroup$ Aug 13 at 15:10
  • $\begingroup$ I don't think it's valid to recode y_t-1 = X. Instead, X should be another series; not a past relaisation from the same series. Recoding doesn't overcome this fact. Instead, cointegration is upheld when two distinct first-difference stationary series have a stationary relationship in levels. $\endgroup$
    – EB3112
    Aug 13 at 15:52
  • $\begingroup$ "should be another series" or "must be another series" is my question. I know what I'm doing is nonstandard. I'm asking if it is mathematically correct. $\endgroup$ Aug 13 at 15:59
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Does I(1) imply a process is cointegrated with its lag?

No. By definition cointegration is a property of multiple time series variables of which they are all integrated of the same order. See definitions of cointegrated processes in Pesaran Time Series and Panel Data Analysis pp 517, or Hamilton Time Series Analysis pp 571.

In fact let me quote from Hamilton [emphasis mine]:

An $(n \times 1$) vector time series $\textbf{y}_t$ is said to be cointegrated if each of the series taken individually is $I(1)$, that is, nonstationary with a unit root, while some linear combination of the series $\textbf{a'y}_t$ is stationary, or $I(0)$, for some nonzero $(n \times 1)$ vector $\textbf{a}$.

By definition variable cannot simply be cointegrated with itself. You need at least two separate series $y_{1t}$, $y_{2t}$ to be able to talk about cointegration.

Mind you lags of a variable do not constitute new series. By definition, following Hamilton pp 43:

$\{y_t\}^\infty_{t=-\infty}= \{...y_{-1},y_0, y_2, ..., y_{t},y_{t+1}, y_{t+2}...\},$

the infinite sequence $\{y_t\}^\infty_{t=-\infty}$ would still be viewed as a single realization from a time series process.

lags of a time series do not represent a new time series process.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – 1muflon1
    Aug 16 at 9:03
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You are confusing concept of co-integration with concept of integration. If a series

$y_t−y_{t−1}=u_t$ is stationary

then series is integrated of order 1 not co-integrated. The term co-integration is specifically used when we have some $y_t−\beta x_{t}=u_t$ and $u_t$ is stationary, have a look at this tutorial by Reserve Bank of Australia on time series.

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  • $\begingroup$ My question was why couldn't $x_t=y_{t-1}$. I've figured it out though and am about to edit the question with a clarifying response and mark 1muflon1 as correct. $\endgroup$ Aug 13 at 16:57
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Just to add to the answers already given. I think the easiest way to see that this cannot be the case is using a CVAR formulation. For a reference see Johansen and Juselius (1990) (https://digidownload.libero.it/rocco.mosconi/JohansenJuselius1990.pdf)

Consider the bivariate vector $x_t = \left\{ \begin{array}{c} y_{t}\\ y_{t-1} \end{array}\right\}$

Which we can assume follows an autoregressive with lag 1 process as

$x_t = cx_{t-1} + v_t$

EDIT: where if $c=I_2$ then we are in the special case of interest. Furthermore, we leave the residuals unrestricted such that $v_t=(u_t, u_{t-1})'$.

We can rewrite this without a loss of information as

$\Delta x_t = b x_{t-1} + v_t$ where $b=c-I_2$

The system is stationary in levels when the rank is full; i.e. $rank(b)=2$.

The system is difference stationary when $rank(b)=0$; That is $b=0$.

Cointegration is the case between these two cases where there is a reduced rank restriction such that $rank(b)=1$; i.e. one can decompose $b=\alpha\beta$.

Based off of these definitions it is clear that your case where $b=0$ is definitely not cointegrated.

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    $\begingroup$ This is a compelling answer. Contrary to the other answers, I was of the opinion that an integrated series is cointegrated with its own lags. However, if the condition you are basing your answer on can be considered part of the definition of a cointegrated system, then the answer is as you say. The remaining question is, can the condition be considered part of the definition? $\endgroup$ Aug 15 at 18:43
  • $\begingroup$ @RichardHardy Good to know I'm not alone in being originally misguided. $\endgroup$ Aug 16 at 12:09
  • $\begingroup$ @MichaelGmeiner, I still do not know what the correct answer is, but see here. $\endgroup$ Aug 16 at 12:30
  • $\begingroup$ The flaw here is that you "assume [it] follows an autoregressive process", but you don't explicitly say which one (what is $v_t$?), and you happen to choose one that doesn't work, and which by my reading isn't even a standard autoregressive process (if you mean that $v_t = [u_t, u_{t-1}]'$, those are not iid). What are you claiming must be true about $v_t$ such that the rank of $b$ determines whether there is cointegration? $\endgroup$
    – Chris Haug
    Aug 16 at 13:56
  • $\begingroup$ @RichardHardy I am not sure I understand what you mean by "can the condition be considered part of the definition?" What condition are you referring to? Also you may be interested in looking and the second page of the Johansen and Juselius (1990) paper that I posted for clarification this. $\endgroup$
    – Andrew M
    Aug 16 at 19:21

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