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Suppose two $I(1)$ series $x_t, y_t$ are cointegrated. Therefore $\mu_t$ in following equation is stationary:

\begin{align} y_t = \beta_0 + \beta_1x_t + \mu_t \tag{1} \end{align} Now consider the ECM representation: \begin{align} \Delta y_t = \alpha_0+\gamma\mu_{t-1}+\alpha_1\Delta x_t+\nu_t \tag{2} \end{align}

If we take first difference on $(1)$, we get:

\begin{align} \Delta y_t = \beta_1\Delta x_t+\Delta\mu_{t} \end{align}

Comparing this with equation (2), we get that:

\begin{align} \Delta\mu_t &= \alpha_0+\gamma\mu_{t-1}+(\alpha_1-\beta_1)\Delta x_t +e_t \\ \implies \mu_t &= \alpha_0+(1+\gamma)\mu_{t-1} + (\alpha_1-\beta_1)\Delta x_t +e_t \end{align} Now, if we assume that $x_t$ is a random walk model (without drift), $\mathbb E(\Delta x_t)=0$, then in above equation (with $-2<\gamma <0$) it would imply that $\mathbb E(\mu_t)\ne0$. Is this correct?

If yes (of course highly unlikely), does incorporating a constant term in ECM imposes restriction on DGP of $x_t$?

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  • $\begingroup$ Did you mean $y_t$ rather than $x_t$ in your last sentence? $\endgroup$ – Richard Hardy Jan 22 at 18:51
  • $\begingroup$ Did you mean $y_t$ rather than $x_t$ in your last sentence? It seems to me in your setup with $\alpha_0\neq 0$ we get $\mu_t$ to include a linear trend. This is because in each time period, $y_t$ departs from $x_t$ by one additional $\alpha_0$. $\endgroup$ – Richard Hardy Jan 22 at 18:56
  • $\begingroup$ I meant $x_t$ only because if expected value of $\Delta x_t = 0$ then expected value of $\mu_t$ cant be, which by definition it is supposed to be. So that means $\Delta x_t \ne 0$. $\endgroup$ – Dayne Jan 22 at 19:11
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    $\begingroup$ Still I would suggest $y_t$ there. You do not need $x_t$ to have drift, it is sufficient that $\mu_t$ includes a time trend. Then $x_t$ is cointegrated with a detrended $y_t$. $\endgroup$ – Richard Hardy Jan 22 at 19:36
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    $\begingroup$ I guess so. I might have overlooked some other case compatible with the models, but the one we covered is summarized quite well in your comment. $\endgroup$ – Richard Hardy Jan 24 at 17:09

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