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From what I understood,
ECM is for two variables and apply OLS to estimate EC term and
VECM is for multi-variables (vector form) and apply VAR to estimate EC term.

But as I read other papers, I think I may have misunderstood.
Is VECM just an expansion of ECM?
However, in equation form, how could it be justified?

Suppose that $y_t =\alpha +\beta x_t +e_t$.
ECM: $\Delta y_t = \alpha+\sum_{j=1}^{k} \phi{_j} \Delta y_{t-j} +\sum_{j=1}^{k} \pi_j \Delta x_{t-j} +\delta z_{t-1} +v_t$ where $z_{t-1}=y_t-a-b x_t$
Suppose that $y_t =\alpha +\beta y_t +e_t$ (vector form).
VECM: $\Delta y_t = \alpha+\phi y_{t-1} +\sum_{j=1}^{p-1} \pi_j \Delta y_{t-j} +v_t$ for VAR(p).
VECM's EC term $y_{t-1}$ (vector) is just an expansion of ECM's EC term $z_{t-1}$? (I don't think so.)

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ECM consists of a single equation for a single dependent variable (that is cointegrated with another variable), while VECM consists of multiple equations for multiple dependent variables (that are cointegrated). You could take one equation from a VECM, analyze it separately, and you could call that an ECM.

Regarding

ECM is for two variables and apply OLS to estimate EC term and VECM is for multi-variables (vector form) and apply VAR to estimate EC term

VAR is a model, not an estimation technique. A VECM has an equivalent restricted-VAR representation. A VECM can be estimated using equation-by-equation OLS with the error correction term(s) taken as given. The coefficients of the error correction terms are estimated beforehand using OLS or maximum likelihood.

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  • $\begingroup$ Now I understand it. Thank you so much! $\endgroup$
    – guest
    Commented Dec 23, 2021 at 6:53
  • $\begingroup$ @guest, you are welcome! $\endgroup$ Commented Dec 23, 2021 at 6:53
  • $\begingroup$ Regarding to your edited answer, why $y_{t-1}$ should be $z_{t-1}$? I refer to this page 454, eq(12.21). $\endgroup$
    – guest
    Commented Dec 23, 2021 at 6:57
  • $\begingroup$ @guest, I think I had misunderstood you. If $\phi$ is a coefficient matrix and $y_{t-1}$ is a vector, then it is OK. $\endgroup$ Commented Dec 23, 2021 at 11:25

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