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

Question

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.

Answer 1

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.