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I am reading Pfaff "Analysis of Integrated and Cointegrated Time Series with R" (2008). On page 130, section 8.1.2 "Determining the Cointegration Rank" they start by determining cointegration rank. Because inferences on cointegration space spanned by its vectors are dependent on whether or not linear trends exist in the data, they argue by ocular econometrics and logical reasoning that the price series have a linear trend that is consistent with the steady-state assumption of constant nominal price growth as implied by economic theory, and therefore the vector $\mu$ can be estimated without imposing any restrictions.

What is the relation between "cointegration space spanned by the eigenvectors" and "the existence of linear trend in the data"?

The cases are:

  1. no linear trend in the data;
  2. linear trend in the data.

What is the result of these two cases on "cointegration space spanned by the eigenvectors"?

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  • $\begingroup$ Could you elaborate a little more? $\endgroup$ – Richard Hardy Sep 6 '16 at 18:36
  • $\begingroup$ Assume Yt is a vector of variables, i.e, Yt=(Y1t,Y2t,...,Y5t), t=1,...,p. Pfaff: "...Because inferences on the cointegration space spanned by its vectors are dependent on whether or not linear trends exist in the data,..." $\endgroup$ – Erdogan CEVHER Sep 6 '16 at 19:14
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    $\begingroup$ It could probably make sense to cite the passage you are having trouble with. Perhaps it will reflect the actual problem best. $\endgroup$ – Richard Hardy Sep 6 '16 at 19:28
  • $\begingroup$ PFAFF 2008 "Analysis of Integrated and Cointegrated Time Series with R". Page 130. 8.1.2 Determining the Cointegration Rank J-J start by determining cointegration rank. Because inferences on cointegration space spanned by its vectors are dependent on whether or not linear trends exist in the data, they argued by ocular econometrics and logical reasoning that the price series have a linear trend that is consistent with the steady-state assumption of constant nominal price growth as implied by economic theory, and therefore the vector μ can be estimated without imposing any restrictions. $\endgroup$ – Erdogan CEVHER Sep 6 '16 at 19:31
  • $\begingroup$ Thx a lot Hardy for your nice editing. The reason I shorthened that much at the beginning is that I did not want to direct the users in advance in any direction, and that's why, I removed the context to that degree. This made it very elaborate as you implied. $\endgroup$ – Erdogan CEVHER Sep 6 '16 at 19:42

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