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I came across this text that seems to estimate a non-recursive SVAR in Eviews, but the restrictions are kinda different from what I expected.

In their recursive SVAR, they have the following results in Eviews which are the restrictions I expected given the model.

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The results above has the following restrictions for matrix A and B. So we know the coefficients of the contemporaneous variables.

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In the non-recursive SVAR, they have the following results

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But I expected the following retrictions for matrices A and B instead.

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As we can see, their results show the restrictions below. How can we then know the coefficients of the contemporaneous variables? I know e = (A^-1)Bu will match the model in the non-recursive example above, but I don't understand why A is an identity matrix and B is not diagonal. And why my expectation in the red matrixes above is wrong. I actually want B to be diagonal or identity matrix in the non-recursive SVAR, not matrix A. Thanks for any help.

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  • $\begingroup$ Does your confusion have something to do with the fact that the ordering of the coefficients in EViews is columnwise in the A matrix and indices with zero are skipped? for example, the columns of a lower-triangular matrix would be (1) c1, c2, c3; (2) 0, c4, c5; (3) c6. $\endgroup$
    – BrsG
    Jul 10, 2021 at 8:21
  • $\begingroup$ @BrsG My confusion is that if A matrix is diagonal, then what are the contemporaneous coefficients? When I read books on SVAR, matrix A (containing the contemporaneous restrictions) is never diagonal, yeah? Like in Ae = Bu, where u orthogonal structural shock, how can A be diagonal? I guess, then there are no contemporaneous effects in the model? Thanks! $\endgroup$
    – Emmanuel Ameyaw
    Jul 11, 2021 at 10:59
  • $\begingroup$ Indeed, when A is diagonal, this corresponds to the special case with no contemporaneous interaction between the variables. In $Ae=Bu$ case, you can impose restrictions on $A$, $B$, or $S=A^{-1}B$, each meaning different things. $\endgroup$
    – BrsG
    Jul 11, 2021 at 15:29
  • $\begingroup$ @BrsG. Thanks for the reply. Lemme ask though for a hint. When you impose restrictions directly on S, what is your goal? Only correct identification of the shocks? $\endgroup$
    – Emmanuel Ameyaw
    Jul 18, 2021 at 6:09
  • $\begingroup$ $S$ gives you the long-term restrictions. For example, by setting a zero in the appropriate position you can impose that shocks have no impact on a certain variable at an infinite horizon. $\endgroup$
    – BrsG
    Jul 19, 2021 at 8:32

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