Suppose there is a structural VAR model, such as: $A y_{t} = B y_{t-1} + \varepsilon_{t}$, where $\varepsilon_{t} \sim N(0, I_{n})$; then the matrix representing the contemporaneous impulse-response function to the structural shocks $\varepsilon_{t}$ is $A^{-1}$.

I am currently working on a model where the determinant of A is positive by construction, since (i) the $a_{ij}$ elements that are not restricted are always positive due to the economic structure I'm studying; (ii) the diagonal elements are all equal to one; and (iii) the last row of A is a vector of zeros but with the last element equal to one, i.e., $\left(0, \; \dots, \; 0, \; 1 \right)$.

Because the determinant of $A$ is always positive, the determinant of its inverse is also. I was wondering, however, if this should/could have an underlying economic significance.

On one hand, I cannot help but have the intuition that having $det(A^{-1}) > 0$ throughout my parameter space means that whenever structural shocks occur in my model this causes the variables in the system to act "procyclically". In other words, they were bound to go up together or down together, depending on the vector of structural shocks. The structural response of one variable to another's shock and vice-versa would make them all lever on one another.

This intuition I have also tells me that the shocks would dissipate over time only if $det(A^{-1}) < 1$, but this is just this - intuition.

On the other hand, I might just be telling non-sense by unwarrantly applying analogies to higher dimensions without actually considering the math behind it in full.

I would be very glad if anyone could share their opinion, proofs, or references. More specifically, my question is how to interpret (or if to interpret at all!) the determinant of the inverse of the structural parameters matrix $A$. Thank you.

  • $\begingroup$ Having positive entries in a matrix is insufficient to ensure a positive determinant; consider the 2 x 2 matrix with entries (1,2) on the first row and (2,1) on the second row. Are there other constraints in your system that force $A$ to be positive-definite, which would necessarily have a positive determinant? $\endgroup$ – EdM Jun 29 '15 at 18:06
  • $\begingroup$ @EdM, you're right, I forgot to mention two things: (i) the last row is a vector of zeros with the last element as 1, that is, $\left( 0, \; \dots, \; 0, \; 1 \right)$ and (ii) the diagonals are 1. Therefore, the determinant is positive. $\endgroup$ – Douglas K Jul 1 '15 at 12:56

Because the $\varepsilon_t$ have a distribution that is symmetric upon negating any single component, the model is unchanged if (say) the first component of $\varepsilon_t$ is everywhere negated. The relationship

$$Ay_t = By_{t-1}+\varepsilon_t$$

can be preserved by negating the first columns of $A$ and $B$. That negates the determinant of $A$.

Consequently, the sign of the determinant of $A$ is meaningless.


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