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Let $R$ be a semi-definite $N\times N$ circulant Toeplitz matrix and let $N\to \infty$. Let $D$ be an $N\times N$ diagonal matrix where the elements on the main diagonal are independent chi-square distributed random variables with two degrees of freedom. What is the expectation $Z=\mathbb{E}(R+D)^{-1}$?

The Circulant matrix is formed from a Fourer transform $R(\omega)$. So as $N$ grows, the eigenvalues converge to $R(\omega)$.

From simple experiments in Matlab, the following seems to hold:

  1. The solution is of the form $(R+\lambda I)^{-1}$ for some $\lambda$. If true, the problem reduces into finding $\lambda$.

  2. Due to 1) it holds that $R(\omega)+\frac{1}{Z(\omega)}=constant$, where $Z(\omega)$ is the induced Fourier transform of $Z$. Thus, since $R(\omega)$ is known, the problem reduces into finding a single value of $Z(\omega)$ or some other quantity such as $\int Z(\omega)d\omega$ etc.

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  • $\begingroup$ expectation of a matrix? you mean the average of every single matrix element? unlikely that the large-$N$ limit will be of much help for that.... $\endgroup$ Nov 2, 2013 at 20:55
  • $\begingroup$ yes, for each element. From Matlab, it appears as if the solution is $(R+\lambda I)^{-1}$ for some positive value $\lambda$. If true in general, the problem reduces into finding $\lambda$. And I agree, the large limit is not very helpful, but laid my complete problem down anyway. $\endgroup$
    – john stark
    Nov 3, 2013 at 11:09
  • $\begingroup$ I have some difficulty in interpreting the question. Is $R$ a function of $N$? How does it vary with $N$? By "cyclic" you mean "circulant" as in the title? $\endgroup$ Nov 3, 2013 at 23:12
  • $\begingroup$ Cyclic = circulant indeed. Also, the matrix R is formed from some Fourier transform, so the eigenvalues converge to this transform as $N$ grows. See the OP for an update of the text. $\endgroup$
    – john stark
    Nov 4, 2013 at 17:05

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