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I hope you can help me concerning the following question. It is mainly about when to use ECM or VECM. Suppose i have two time series of daily stock prices from Company Y and X and regression a regression in the form of Y ~ X revealed that the both series are cointegrated (ADF-Test on the regression residuals with p-value at 0.02). Based on that i wanted to build an ECM in R via the ecmSymfit Function. This Function returns two models in the form dY ~ dX(t-1) + dY(t-1) +e(t-1) and dX ~ dY(t-1) + dX(t-1) + e(t-1) where e(t-1) are the residuals from the regression Y~X. As i understood now these two models simply show the reaction of dY and dX given a shock in the previous period via e(t-1)*corresponding coefficient.

But what i read many times now is that proceeding in this form would not be that feasible since one has to determine which of the both stocks is the dependent and independent variable i.e. Y is affecting X or X is affecting Y. And since it could be that both are affecting each other simultaneously a VECM would be required (eventhough i thought before that this would only be required if there are at least more than two time sieries). Could you tell me if it is correct that already here a VECM is required?

And furthermore, do the two models above not already show how the variables are reacting given a shock in the prevous period (via dY ~ dX(t-1) + dY(t-1) + e(t-1) -> shows how Y reacts given a shock in e(t-1) and dX ~ dY(t-1) + dX(t-1) + e(t-1)-> shows how X reacts given a shock in e(t-1)). Or would it be required to construct an additional ECM based on the regression residuals of X~Y?

Any help would be highly appreciated since i am becoming a bit confused on when to use what.

Thanks in advance and best

LD

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  • $\begingroup$ Looks like you got it right, i.e. I do not find mistakes in your reasoning. $\endgroup$
    – Richard Hardy
    Commented Nov 30, 2020 at 7:07

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The two models you get from the function are ECMs when viewed separately and a VECM when viewed together. You can be interested in one, another or both equations, depending on what kind of inference you want to do (examine effects of the error correction term and/or lags on $X$, on $Y$ or both). A VECM would do the job in any case, but in the first two cases the relevant ECM will be sufficient, too. If you are interested in forecasting $X$, $Y$ or both more than one step ahead, you will need the VECM unless one of the equations contains all zero coefficients.

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  • $\begingroup$ Dear Richard, Thank you very much for your feedback! One additional request. I have now a model in the form of dY ~ dX(t-1) + ... + dX(t-n) + dY(t-1) +...+ dY(t-n) + e(t-1), with n=2,...,100. So i included 100 lags of dX and dY in the ECM. As stated above the residuals from the underlying regression Y~X were identified as stationary, although highly autocorrelated. But in the final output the coeffient of the EC-term seems to be insignificant (p-value ~ 0.25). Can this be the case when the underlying residuals are stationary or is it maybe due to the high number of lags included? $\endgroup$
    – user14731396
    Commented Nov 30, 2020 at 19:52
  • $\begingroup$ @user14731396, perhaps you have included too many lags or perhaps $Y$ truly does not adjust towards $X$. How long are your time series? $\endgroup$
    – Richard Hardy
    Commented Nov 30, 2020 at 20:33
  • $\begingroup$ The time series contains 650 daily closing prices of companies Y and X. I suppose it is due to the amount of lags since when controlling for the adjustment of X the EC-term is insignificant as well- $\endgroup$
    – user14731396
    Commented Nov 30, 2020 at 20:45
  • $\begingroup$ 100 lags given 650 data points seems too much. Given that these are stock prices, I would not be surprised if the best forecasting model had zero lags. $\endgroup$
    – Richard Hardy
    Commented Nov 30, 2020 at 20:58
  • $\begingroup$ But i need to use lags to make any prediction. Otherwise i would just regress dYt ~ dXt + e(t-1) and since i dont know dXt this would not be feasible for making predictions or am i wrong? Although you are of course right that this results in the best R^2. $\endgroup$
    – user14731396
    Commented Nov 30, 2020 at 21:28

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