1

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 $x_t$?

Dayne
  • 2,113
  • 1
  • 7
  • 24
  • Did you mean $y_t$ rather than $x_t$ in your last sentence? – Richard Hardy Jan 22 '21 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 '21 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 '21 at 19:11
  • @RichardHardy I think your second comment clears some things. By including intercept i am assuming that in short run the change is not just a correction but a fixed amount also which means assuming $y_t$ also has a linear trend. So perhaps in my set up, $x_t$ needs to be a random walk with drift and thus also in $y_t$. – Dayne Jan 22 '21 at 19:18
  • 1
    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 '21 at 19:36
  • Any further thoughts? – Richard Hardy Jan 24 '21 at 06:01
  • @RichardHardy: Sorry for late reply. Yes that makes sense. Broadly, inclusion of intercept term assumes that the underlying common trend has a time trend. So this means that we should not always include an intercept term in the ECM. If the hypothesis for true process is something like equation (1) then ECM should be estimated without the intercept term. Does this make sense? – Dayne Jan 24 '21 at 13:53
  • 1
    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 '21 at 17:09
  • @RichardHardy: so I found [this](http://staff.utia.cas.cz/barunik/files/appliedecono/Lecture7.pdf) (slide 12) where the ECM is derived from an ARDL type model. So it gives another case when long term equation doesn't have a time trend but yet ECM has an intercept. In my question I have assumed a rather simple true process. If the true process is however, like an ARDL without time trend even then it appears a constant term will come in the ECM equation. – Dayne Jan 27 '21 at 05:56
  • Thanks! Interesting. I wonder if the intercept truly does not imply a time trend, though. I can make the ARDL equation work by setting $\alpha_1=1$ and $\beta_0=\beta_1$ which implies a linear time trend in $C_t$ (not sure if this implies cointegration, though, would need to think about it more). Is there another combination of coefficients that does not imply a linear time trend? – Richard Hardy Jan 27 '21 at 06:55
  • Though perhaps the intercept does not imply a linear time trend. Rearrange the ECM $\Delta y_t=\alpha(x_{t-1}-\beta_0-\beta_1 y_{t-1})+\gamma_1\Delta x_t+\varepsilon_t$ as follows: $\alpha(x_{t-1}-\beta_0-\beta_1 y_{t-1})=\Delta y_t-\gamma_1\Delta x_t-\varepsilon_t$. Now $\mathbb{E}(\Delta y_t)=\mathbb{E}(\Delta x_t)=0$ as otherwise $y_t$ and $x_t$ would have deterministic trends. Then also $\mathbb{E}(x_{t-1}-\beta_0-\beta_1 y_{t-1})=0$. That means the levels of $x_t$ and $\beta_1 y_t$ differ by a constant $\beta_0$. When rearranged back to the standard form, you get an ECM with an intercept. – Richard Hardy Jan 27 '21 at 07:32
  • If this is correct, I think we have figured things out. Just note that if the levels of $x_t$ and $\beta_1 y_t$ differ by a constant $\beta_0$ but your intercept is not $-\alpha\beta_0$, there is a linear trend in $y_t$. – Richard Hardy Jan 27 '21 at 07:34
  • @Dayne As discussed with RichardHardy, in "Now if we assume..." the asumption seems to be that $x_t$ is a random walk and not that $\Delta x_t$ is a random walk, which would make $x_t$ be $I(2)$. – Yves Jan 27 '21 at 08:17
  • @Yves Thanks fkr pointing out this error. I did mean to write $x_t$ there. Will edit it now. – Dayne Jan 27 '21 at 12:51
  • @RichardHardy Didn't understand your last few comments very well. Will read them once again with a pen paper. – Dayne Jan 27 '21 at 12:52
  • @RichardHardy: Sorry with very late reply. In regards to your second last comment, shouldn't an ECM with intercept be written like this: $\Delta y_t = \alpha_0 + \rho \mu_{t-1} + \dots$. In this set-up the $\beta_0$ term is already incorporated in $\mu_{t-1}$ so inclusion of $\alpha_0$ is not $-\alpha \beta_0$ but an extra constant term. So perhaps, including an intercept term does mean that there is a time trend. – Dayne Mar 07 '21 at 02:27
  • I guess I get it now. in the simultaneous (VAR) modelling approach the residuals are not first estimated and then fed in the ECM equation but rather a VAR is directly estimated with lagged level variables included. In the VAR/VECM set-up therefore, one should include a constant term even if the underlying $I(1)$ variables are not suspected to have time trends. But not in the EG two-step approach. Does this sound correct? – Dayne Mar 07 '21 at 02:34

0 Answers0