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

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?

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 $\Sigma^{-1}$.

• 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 ? – Veronique Jun 19 '16 at 11:49
• In short, I wonder why V(u) in 2.4 is equal to V(u)^-1 in 2.6. – Veronique Jun 19 '16 at 12:17

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)

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