# If $y_t$ and $x_t$ are cointegrated, then are $y_t$ and $x_{t-d}$ also cointegrated?

Assume that $$x_t, y_t$$ are $$I(1)$$ series which have a common stochastic trend $$u_t = u_{t-1}+e_t$$. Particularly, consider the following DGP

\begin{align} y_t&=\alpha_y+u_t+a_t \tag{1} \\ \end{align}

\begin{align} x_t&=\alpha_x+u_t+b_t \tag{2} \\ \end{align}

Here $$a_t, b_t$$ are independent white noise processes.

Substituting $$u_t$$ from $$(2)$$ in $$(1)$$, we get:

\begin{align} y_t &= \beta_0 + x_t+\mu_t \tag{3}\\ \text{where } \beta_0 &\equiv \alpha_y-\alpha_x; \text{ and} \\ \mu_t &\equiv a_t-b_t \end{align}

Based on $$(3)$$, $$x_t$$ and $$y_t$$ are cointegrated (is this correct?). Now consider an alternate formulation:

From $$(2)$$ we have that: \begin{align} \Delta x_t &= \Delta u_t+\Delta b_t \\ \implies x_t &= x_{t-1} + \nu_t \tag{4}\\ \text{where } \nu_t &\equiv e_t + \Delta b_t \end{align}

substituting $$(4)$$ in $$(3)$$, we get:

\begin{align} y_t &= \beta_0+x_{t-1}+\eta_t \tag{5}\\ \text{where } \eta_t &\equiv a_t-b_{t-1}+e_t \end{align}

Does equation (5) means that $$x_{t-1}$$ and $$y_t$$ are cointegrated (or that cointegration tests would fail to reject the null of co-integration)?

This can be extended to more lags of course but the variance of error term in long-term equation will keep increasing with lags. Clearly there is some fundamental gap in my understanding here but I actually getting such results for some series.

• Yes indeed if $x_t$ is $I(1)$ then $x_t -x_{t-d}$ is stationary $I(0)$.
– Yves
Jan 22 at 10:49

To answer your title question: Yes, if $$y_t$$ and $$x_t$$ are cointegrated, then $$y_t$$ and $$x_{t-d}$$ are also cointegrated.

I think you got the intuition right:

• $$y_t$$ and $$x_t$$ are cointegrated and thus share a common stochastic trend.
• $$x_t-x_{t-d}$$ is a $$d$$-element sum of I(0) and thus still I(0).
• subtracting $$x_t-x_{t-d}$$ (which is I(0)) from $$x_t$$ (which is I(1)) yields $$x_{t-d}$$ which still has the same stochastic trend as $$x_t$$ because $$x_t-x_{t-d}$$ is merely I(0).
• Therefore, $$y_t$$ and $$x_{t-d}$$ are cointegrated.
• Thanks! So couple of doubts: (1) Perhaps naive, but how does showing that $x_{t-d}$ is I(0) prove that $y_t$ and $x_{t-d}$ are cointegrated? (2) Would it be true for all $d$? Intuitively, I think that as we increase $d$ the additional sum of errors ($e_{t-d+1}$ to $e_t$) that are part of $y_t$ but not $x_{t-d}$ might start creating some problems in testing for cointegration. But if so, that would mean that the variance of white noise terms of $y_t$ and $x_t$ (relative to variance of white noise of the underlying stochastic trend ($e_t$) has influence on cointegration test Jan 22 at 15:14
• @Dayne, 1. $x_{t-d}$ is not I(0), it is the same as $x_t$ which is I(1). 2. Yes, as long as $d$ is finite and fixed. (You could analyze some sort of asymptotics when $d$ is growing, that would be another matter.) Regarding the power of cointegration tests, you are right, large $d$ would reduce test power, and this would be mostly felt in small samples. I agree about the variance, too. Jan 22 at 15:47
• terribly sorry about the mistake in (1). I meant, how does the fact that $x_t-x_{t-d}$ is I(0) sufficient to claim that $y_t$ and $x_{t-d}$ are cointegrated. And thanks again about answer to (2). Jan 22 at 16:14
• @Dayne, it sometimes looks straightforward after one has been pondering a topic for a long time, not before. I had thought about the question earlier, so I found it easy. But it is a great question! Jan 22 at 16:39
• @Dayne, I +1'ed your other question yesterday already but was too lazy/tired to attempt answering it. Jan 22 at 16:44