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Here's the Sims(2002) paper. It's equal to the published version.

The author gives the following example:

enter image description here

And tries to write it in the matrix notation of (1) in the paper.

However, his solution - in (4) - puts $\epsilon (t)$ in the first equation of (2) instead of $\nu (t-1)$, and $\nu (t-1)$ instead of $\epsilon (t)$. Is this correct, and why?

Also, when I try to write the the first equation of (2) the time notation doesn't match. The other two equations in (2) do match.

Any help would be appreciated.

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1 Answer 1

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Yes the $z(t)$ column vector, as it appears in $(4)$ is written upside-down in the translation. Also the first equation in $(2)$ is translated lagged once, i.e. the translated system describes the $w(t-1)$ equation.

Finally there is a typo in the $\Gamma_1$ matrix, there is a void 6th column that should be ignored, and $\theta$ should be in the 4th column not in the 5th (because it multiplies $u(t-1)$.

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  • $\begingroup$ +1 Alecos, thanks for the help. By the way, in this paper, I don't understand how the author proposes to deal with the term $E_t W(t+2)$ (or $E_{t-1}W(t+1)$ when translated lagged once). In page 3, he suggests a solution, but I don't get the english. Why won't he just add $E_t W(t+2)$ to $y(t)$ as he did with $E_t W(t+1)$? $\endgroup$
    – An old man in the sea.
    May 23, 2018 at 14:31
  • $\begingroup$ Also, I think the C vector, should also have its 1st component multiplied by -1, since u_n and u(t) show up with different signs. $\endgroup$
    – An old man in the sea.
    May 23, 2018 at 14:33
  • $\begingroup$ You are right for the -1. $\endgroup$
    – Alecos Papadopoulos
    May 23, 2018 at 19:05

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