# Interpretation of intercept term in ECM

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 $$x_t$$?

• Did you mean $y_t$ rather than $x_t$ in your last sentence? – Richard Hardy Jan 22 at 18:51
• 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 $\alpha_0$. – Richard Hardy Jan 22 at 18:56
• 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 $\Delta x_t \ne 0$. – Dayne Jan 22 at 19:11
• 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 $y_t$. – Richard Hardy Jan 22 at 19:36
• 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. – Richard Hardy Jan 24 at 17:09