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In Zellner 1962 p350-351, you may see (2.4) & (2.6), and you can also verify that the Sigma(c)=Sigma(c)^-1.

Why the inverse of the covariance matrix is equal to the original covariance matrix in Seemingly Unrelated Regression Equations?

I also tried to checkout whether it is due to its property of symmetry or Kronecker product but could not find out why.

Can someone explain the reason for this?

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In the case for which $\Sigma_c$ is invertible, we have, by properties of kronecker products, that $\Sigma = (\Sigma_c \otimes \mathbb{I})^{-1} = \Sigma_c^{-1} \otimes \mathbb{I}$. This isn't evidence for or against the fact that $\Sigma = \Sigma^{-1}$.

But, you can read here or here about how $\Sigma = \Sigma^{-1}$ would suggest that $\Sigma$ is a square root of $\mathbb{I}$; an orthogonal matrix (which doesn't seem realistic in this case).

I don't think $\Sigma = \Sigma^{-1}$. You may have just gotten $H'H$ in (2.5) mixed up with $\Sigma$ when it appears to be in fact $\Sigma^{-1}$.

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  • $\begingroup$ Thank you for your answer! Sadly I found out that I mistyped 2.6 as 2.5. May I ask you to explain why V(u) = V(u)^-1 ? $\endgroup$ – Veronique Jun 19 '16 at 11:49
  • $\begingroup$ In short, I wonder why V(u) in 2.4 is equal to V(u)^-1 in 2.6. $\endgroup$ – Veronique Jun 19 '16 at 12:17
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difference between subscript & superscript

I found an answer for the problem that I was stuck into. For details, [here] (http://www.public.asu.edu/~miniahn/ecn726/cn_sur.pdf)

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  • $\begingroup$ @Mustafa Thx anyway! $\endgroup$ – Veronique Jun 19 '16 at 14:56
  • $\begingroup$ I checked your link: $\Sigma \neq \Sigma^{-1}$, as I suggested in my answer. The author just uses subscripts for $\Sigma$ and superscripts for $\Sigma^{-1}$. $\endgroup$ – Mustafa S Eisa Jun 19 '16 at 21:43

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