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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.

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    $\begingroup$ Yes indeed if $x_t$ is $I(1)$ then $x_t -x_{t-d} $ is stationary $I(0)$. $\endgroup$
    – Yves
    Jan 22 at 10:49
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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.
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  • $\begingroup$ 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 $\endgroup$
    – Dayne
    Jan 22 at 15:14
  • $\begingroup$ @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. $\endgroup$ Jan 22 at 15:47
  • $\begingroup$ 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). $\endgroup$
    – Dayne
    Jan 22 at 16:14
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    $\begingroup$ @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! $\endgroup$ Jan 22 at 16:39
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    $\begingroup$ @Dayne, I +1'ed your other question yesterday already but was too lazy/tired to attempt answering it. $\endgroup$ Jan 22 at 16:44

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