Does I(1) imply a process is cointegrated with its lag?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.