I investigate the relationship among 3 variables (XMIN, LPIB, LIPI). I found that all variables are stationnary in their first differences. Johansen's cointegration method has been adopted to examine the long run relationship among the variables. I found that there is one cointegrating vector in the model.

Now, it is convenient to make the test based on vector error correction models (VECM), so I made the test with Eviews and I don't know how to interpret the table.

  • $\begingroup$ I don't see a table with the test results. The one included looks like estimation output for an unrestricted VECM. Also, what is the actual question? First you ask about interpreting the error correction term, then you ask about interpreting the results of a Granger causality test. $\endgroup$ – Richard Hardy Jan 9 '16 at 21:13
  • $\begingroup$ Oh, and your title does not quite match the content. Consider adapting it to make it more informative and to the point. $\endgroup$ – Richard Hardy Jan 9 '16 at 21:28
  • $\begingroup$ the table is above represent the VECM. My question is how to interpret this table $\endgroup$ – Ch.Na Jan 10 '16 at 20:01
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    $\begingroup$ You can read the table as follows: here is the first column: $\Delta PIB_t=0.0069ect_{t-1}+0.1567\Delta PIB_{t-1}+0.2781\Delta PIB_{t-2}-0.0027\Delta LXMIN_{t-1}-0.0137\Delta LXMIN_{t-2}-0.0810\Delta LIPI_{t-1}-0.0839 \Delta LIPI_{t-2}+0.0056+\varepsilon_t$ where $ect_t=1.0000\cdot LPIB_{t-1}-0.4173LXMIN_{t-1}+2.9926LIPI_{t-1}-0.0204trend-21.0302$. Is that what you need? $\endgroup$ – Richard Hardy Jan 11 '16 at 19:39
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    $\begingroup$ i appreciate your help, your answer will help me a liitle bit :) $\endgroup$ – Ch.Na Jan 12 '16 at 19:18

Still having some doubts over what your question really is, I will convert a comment that you found helpful to an answer.

You can read the middle part of the table as follows:

$$ \begin{equation} \begin{aligned} \Delta PIB_t &= 0.0069ect_{t−1} \\ &+ 0.1567 \Delta PIB_{t−1}+0.2781 \Delta PIB_{t−2} \\ &− 0.0027 \Delta LXMIN_{t−1}−0.0137 \Delta LXMIN_{t−2} \\ &− 0.0810 \Delta LIPI_{t−1}−0.0839 \Delta LIPI_{t−2} \\ &+ 0.0056+\varepsilon_t \end{aligned} \end{equation} $$


$$ \begin{equation} \begin{aligned} ect_{t-1} &= 1.0000⋅LPIB_{t−1}−0.4173LXMIN_{t−1}+2.9926LIPI_{t−1}−0.0204(t-1) \\ &- 21.0302. \end{aligned} \end{equation} $$

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