<p>You are confusing concept of co-integration with concept of integration. If a series</p>
<blockquote>
<p><span class="math-container">$y_t−y_{t−1}=u_t$</span> is stationary</p>
</blockquote>
<p>then series is integrated of order 1 not co-integrated. The term <strong>co</strong>-integration is specifically used when we have some <span class="math-container">$y_t−\beta x_{t}=u_t$</span> and <span class="math-container">$u_t$</span> is stationary, have a look at <a href="https://www.rba.gov.au/publications/rdp/1995/9510/integration-and-cointegration.html" rel="noreferrer">this tutorial by Reserve Bank of Australia on time series</a>.</p>

CC BY-SA 4.0
2021-08-13T16:57:40.063Z
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.
Michael Gmeiner
CC BY-SA 4.0
2021-08-13T16:48:42.413Z
2021-08-13T16:48:42.413Z
csilvia
<p>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) (<a href="https://digidownload.libero.it/rocco.mosconi/JohansenJuselius1990.pdf" rel="nofollow noreferrer">https://digidownload.libero.it/rocco.mosconi/JohansenJuselius1990.pdf</a>)</p>
<p>Consider the bivariate vector <span class="math-container">$x_t = \left\{ \begin{array}{c} y_{t}\\ y_{t-1} \end{array}\right\}$</span></p>
<p>Which we can assume follows an autoregressive with lag 1 process as</p>
<p><span class="math-container">$x_t = cx_{t-1} + v_t$</span></p>
<p>EDIT: where if <span class="math-container">$c=I_2$</span> then we are in the special case of interest. Furthermore, we leave the residuals unrestricted such that <span class="math-container">$v_t=(u_t, u_{t-1})'$</span>.</p>
<p>We can rewrite this without a loss of information as</p>
<p><span class="math-container">$\Delta x_t = b x_{t-1} + v_t$</span> where <span class="math-container">$b=c-I_2$</span></p>
<p>The system is stationary in levels when the rank is full; i.e. <span class="math-container">$rank(b)=2$</span>.</p>
<p>The system is difference stationary when <span class="math-container">$rank(b)=0$</span>; That is <span class="math-container">$b=0$</span>.</p>
<p>Cointegration is the case between these two cases where there is a reduced rank restriction such that <span class="math-container">$rank(b)=1$</span>; i.e. one can decompose <span class="math-container">$b=\alpha\beta$</span>.</p>
<p>Based off of these definitions it is clear that your case where <span class="math-container">$b=0$</span> is definitely not cointegrated.</p>

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2021-08-15T18:43:03.560Z
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?
Richard Hardy
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2021-08-16T12:09:09.603Z
@RichardHardy Good to know I'm not alone in being originally misguided.
Michael Gmeiner
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2021-08-16T12:30:54.413Z
@MichaelGmeiner, I still do not know what the correct answer is, but see [here](https://stats.stackexchange.com/questions/540337/is-a-random-walk-cointegrated-with-its-own-lag?noredirect=1#comment991993_540337).
Richard Hardy
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2021-08-16T13:56:38.103Z
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?
Chris Haug
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2021-08-16T19:21:54.450Z
@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.
Andrew M
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2021-08-16T19:47:23.090Z
@ChrisHaug I think you would see by reading Johansen and Juselius (1990) that $v_t=(u_t, u_{t-1})'$ which is left unrestricted. Imposing additional restrictions on the residuals changes the model completely and is not what I had in mind with my answer. If you need this clarification then I am happy to edit my answer.
Andrew M
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2021-08-16T19:53:32.217Z
@AndrewM, I did look at it yesterday and I think I found what you were citing, and then I posted my comment. I may need another look to answer you precisely. I think Chris Haug's argument is that you can found a $b$ of rank $0$ while he found another $b$ of rank $1$. What do we do when two such $b$s exist, each yielding a different implication w.r.t. cointegration?
Richard Hardy
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2021-08-16T19:59:24.267Z
@AndrewM, *can found --> have found.
Richard Hardy
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2021-08-16T20:09:18.553Z
@RichardHardy I have tried to make my answer more clear on this. A) Chris Haug formulated a completely different model by imposing restrictions on the residuals. This restriction implies that this is no longer a joint two variable system but an identity and a single equation. This fundamentally changes the problem and is not valid for thinking about the reduced rank restriction. B) I am also not saying that every reduced rank restriction implies cointegration (also require that both series are I(1) on their own which is not satisfied in Chris' parameterization of the model).
Andrew M
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2021-08-14T15:18:44.903Z
2021-08-16T19:55:06.510Z
2021-08-16T19:55:06.510Z
Andrew M
<blockquote>
<p>Does I(1) imply a process is cointegrated with its lag?</p>
</blockquote>
<p>No. By <em>definition</em> cointegration is a property of <em>multiple</em> 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.</p>
<p>In fact let me quote from Hamilton [emphasis mine]:</p>
<blockquote>
<p>An <span class="math-container">$(n \times 1$</span>) vector time series <span class="math-container">$\textbf{y}_t$</span> is said to be <em>cointegrated</em> <strong>if each of the series taken individually</strong> is <span class="math-container">$I(1)$</span>, that is, nonstationary with a unit root, while some linear combination of the series <span class="math-container">$\textbf{a'y}_t$</span> is stationary, or <span class="math-container">$I(0)$</span>, for some nonzero <span class="math-container">$(n \times 1)$</span> vector <span class="math-container">$\textbf{a}$</span>.</p>
</blockquote>
<p>By definition variable cannot simply be cointegrated with itself. You need at least two separate series <span class="math-container">$y_{1t}$</span>, <span class="math-container">$y_{2t}$</span> to be able to talk about cointegration.</p>
<p>Mind you lags of a variable do not constitute <em>new series</em>. By definition, following Hamilton pp 43:</p>
<blockquote>
<p><span class="math-container">$\{y_t\}^\infty_{t=-\infty}= \{...y_{-1},y_0, y_2, ...,  y_{t},y_{t+1}, y_{t+2}...\},$</span></p>
<p>the infinite sequence <span class="math-container">$\{y_t\}^\infty_{t=-\infty}$</span> would still be viewed as a single realization from a time series process.</p>
</blockquote>
<p>lags of a time series do not represent a new time series process.</p>

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2021-08-16T09:03:49.780Z
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/128612/discussion-on-answer-by-1muflon1-does-i1-imply-a-process-is-cointegrated-with).
1muflon1
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2021-08-13T15:18:49.233Z
2021-08-13T15:42:10.757Z
2021-08-13T15:42:10.757Z
1muflon1
<p>My question is about the definition of cointegrated.</p>
<p><span class="math-container">$y_t =y_{t-1}+u_t$</span></p>
<p><span class="math-container">$u_t =\eta_t +0.5\eta_{t-1}$</span></p>
<p>where <span class="math-container">$\eta_t\sim N(0,1)$</span> is i.i.d. white noise.</p>
<p>I claim that <span class="math-container">$y_t$</span> and <span class="math-container">$y_{t-1}$</span> are cointegrated because <span class="math-container">$y_t -y_{t-1}=u_t$</span> is stationary and both <span class="math-container">$y_t$</span> and <span class="math-container">$y_{t-1}$</span> are non-stationary.</p>
<p>Is this a correct application of the definition? I know it would be more standard to just consider <span class="math-container">$y_t$</span> as I(1), but if I said <span class="math-container">$y_t$</span> and <span class="math-container">$y_{t-1}$</span> are cointegrated, is that a mathematically correct statement?</p>
<p>EDIT: Firstly, thank you to everyone that helped.
The &quot;attempt&quot; fails as 1muflon1 describes.</p>
<p>EDIT2: Some people are currently not satisfied with the answers, and an ongoing discussion is <a href="https://stats.stackexchange.com/questions/540337/is-a-random-walk-cointegrated-with-its-own-lag?noredirect=1#comment991993_540337">here</a>.</p>

CC BY-SA 4.0
2021-08-13T15:07:03.040Z
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.
EB3112
CC BY-SA 4.0
2021-08-13T15:10:14.387Z
@Eb3112 Let $X=y_{t-1}$ and $Y=y_t$. Then X and Y have a stationary relationship, correct?
Michael Gmeiner
CC BY-SA 4.0
2021-08-13T15:52:12.230Z
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.
EB3112
CC BY-SA 4.0
2021-08-13T15:59:49.720Z
"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.
Michael Gmeiner
<time-series><co-integration>
Does I(1) imply a process is cointegrated with its lag?
Michael Gmeiner
# Does I(1) imply a process is cointegrated with its lag?

(Asked by: username_0 on 2021-08-13)

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 username_6 describes.

EDIT2: Some people are currently not satisfied with the answers, and an ongoing discussion is [here](https://stats.stackexchange.com/questions/540337/is-a-random-walk-cointegrated-with-its-own-lag?noredirect=1#comment991993_540337).

**Tags:** time-series, co-integration

> **Comments:**
> **username_1**: 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.  
> **username_0**: @Eb3112 Let $X=y_{t-1}$ and $Y=y_t$. Then X and Y have a stationary relationship, correct?  
> **username_1**: 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.  
> **username_0**: "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.  

---

## 3 answers

### Answer by username_6 on 2021-08-13. Score: 3 | Accepted answer

> 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.

> **Comments:**
> **username_6**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/128612/discussion-on-answer-by-username_6-does-i1-imply-a-process-is-cointegrated-with).  

---

### Answer by username_2 on 2021-08-13. Score: 5

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](https://www.rba.gov.au/publications/rdp/1995/9510/integration-and-cointegration.html).

> **Comments:**
> **username_0**: 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 username_6 as correct.  

---

### Answer by username_3 on 2021-08-14. Score: 5

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.

> **Comments:**
> **username_4**: 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?  
> **username_0**: @RichardHardy Good to know I'm not alone in being originally misguided.  
> **username_4**: @MichaelGmeiner, I still do not know what the correct answer is, but see [here](https://stats.stackexchange.com/questions/540337/is-a-random-walk-cointegrated-with-its-own-lag?noredirect=1#comment991993_540337).  
> **username_5**: 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?  
> **username_3**: @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.  
> **username_3**: @ChrisHaug I think you would see by reading Johansen and Juselius (1990) that $v_t=(u_t, u_{t-1})'$ which is left unrestricted. Imposing additional restrictions on the residuals changes the model completely and is not what I had in mind with my answer. If you need this clarification then I am happy to edit my answer.  
> **username_4**: @AndrewM, I did look at it yesterday and I think I found what you were citing, and then I posted my comment. I may need another look to answer you precisely. I think username_5's argument is that you can found a $b$ of rank $0$ while he found another $b$ of rank $1$. What do we do when two such $b$s exist, each yielding a different implication w.r.t. cointegration?  
> **username_4**: @AndrewM, *can found --> have found.  
> **username_3**: @RichardHardy I have tried to make my answer more clear on this. A) username_5 formulated a completely different model by imposing restrictions on the residuals. This restriction implies that this is no longer a joint two variable system but an identity and a single equation. This fundamentally changes the problem and is not valid for thinking about the reduced rank restriction. B) I am also not saying that every reduced rank restriction implies cointegration (also require that both series are I(1) on their own which is not satisfied in Chris' parameterization of the model).  

---

