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I am now reading a paper by Sommers, H. J., et al. "Spectrum of large random asymmetric matrices." Physical Review Letters 60.19 (1988): 1895-1898., it claims a mathematical statement (equation (2) in the paper) as following:

Given an ensemble of $N\times N$ real asymmetric random matrix $J_{ij}$ defined by a Gaussian distribution with zero mean and correlations

$N[J_{ij}^{2}]_J = 1$ and $N[J_{ij}J_{ji}]_J = \tau$ for $i \neq j$ and $-1\leq \tau \leq 1$

these correlations can be derived from a Gaussian measure

$P(J) \propto \exp\big[ -\frac{N}{2(1 - \tau^2)}\mathrm{Tr}(JJ^T - \tau JJ) \big]$, where $J_{ij}^{T} = J_{ji}$.

But I am not sure how this derivation to be calculated, and in addition it says this measure implies for the diagonal elements $N[J_{ii}^2]_J = 1 + \tau$. Could someone help me to complete the detailed derivation ?

In short for questions:

(1) Why Gaussian measure $P(J)$ has this specific formula shown above (e.g. trace etc.)? How it can be related or even calculated from the definition of multivariate normal distribution or matrix normal distribution $$p(\mathbf{X}\mid\mathbf{M}, \mathbf{U}, \mathbf{V}) = \frac{\exp\left( -\frac{1}{2} \, \mathrm{tr}\left[ \mathbf{V}^{-1} (\mathbf{X} - \mathbf{M})^{T} \mathbf{U}^{-1} (\mathbf{X} - \mathbf{M}) \right] \right)}{(2\pi)^{np/2} |\mathbf{V}|^{n/2} |\mathbf{U}|^{p/2}}$$ as it is shown in the Wikipedia page.

(2) How to derive the correlations value (i.e. 1 and $\tau$) from the defined Gaussian measure ?

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    $\begingroup$ just calculate the marginal distribution $P(J_{12},J_{21})$ and then the answer follows $\endgroup$
    – Carlo Beenakker
    Commented Dec 27, 2015 at 16:11
  • $\begingroup$ @CarloBeenakker This formula of Gaussian measure is quite strange to me, is it somehow related to so-called Matrix Normal Distribution, as from Wikipedia page for the definition ? Since it is with trace, quite different from the definition of multivariate normal distribution. $\endgroup$
    – Xingdong Zuo
    Commented Dec 27, 2015 at 18:09
  • $\begingroup$ it's a multivariate normal distribution for the entire set of matrix elements of $J$ $\endgroup$
    – Carlo Beenakker
    Commented Dec 27, 2015 at 18:15
  • $\begingroup$ @CarloBeenakker Does it mean that when we insert the $M = 0$ (zero mean) and $V, U$ based on $[J_{ij}^{2}]_J = \frac{1}{N}$ and $[J_{ij}J_{ji}]_J = \frac{1}{\tau}$, then we can get the formula $P(J) \propto \exp\big[ -\frac{N}{2(1 - \tau^2)}\mathrm{Tr}(JJ^T - \tau JJ) \big]$ ? $\endgroup$
    – Xingdong Zuo
    Commented Dec 28, 2015 at 22:54
  • $\begingroup$ I would just vectorize the elements of $J$ into a vector $V$ of length $N^2$, and then you have a usual vector-normal distribution $P(V)\propto\exp(-V^{\rm T}\Sigma V)$ for a suitable $\Sigma$ ; I'm not sure you can write $P(J)$ in the form of a matrix-normal distribution. $\endgroup$
    – Carlo Beenakker
    Commented Dec 28, 2015 at 23:10

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Just expanding a bit on the comment by Beenakker. If you write the Gaussian measure in terms of the matrix elements,

$$ \prod_{ij}\exp[−\frac{N}{2(1−τ^2)}(J_{ij}^2-\tau J_{ij}J_{ji})]$$

You can see that an element $J_{ij}$ with $i\neq j$ is only correlated with $J_{ji}$. It is easy to compute the average values mentioned in the paper (consider for concreteness $i=1$, $j=2$ as mentioned in the comment).

You can also notice that the diagonal elements are distributed according to

$$ \prod_{i}\exp[−\frac{N}{2(1−τ^2)}J_{ii}^2(1-\tau)]=\prod_{i}\exp[−\frac{N}{2(1+τ)}J_{ii}^2]$$,

where you can pretty much see the answer to your second question.

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  • $\begingroup$ Thanks for the answer. I am a bit confused why the Gaussian measure defined by this specific formula, i.e. $P(J) \propto \exp\big[ -\frac{N}{2(1 - \tau^2)}\mathrm{Tr}(JJ^T - \tau JJ) \big]$, where $J_{ij}^{T} = J_{ji}$. Is it possible to explain it from standard multivariate normal distribution ? $\endgroup$
    – Xingdong Zuo
    Commented Dec 27, 2015 at 17:42
  • $\begingroup$ Writing it with the trace and the transpose is just a shortcut for the distribution which appears in my first equation (indeed, a multivariate normal distribution for the matrix elements). $\endgroup$
    – Marcel
    Commented Dec 27, 2015 at 17:54
  • $\begingroup$ Yeah, I see your first equation is to expend the one in the question above. Well what confuses me is that the distribution formula is strange to me, since it is quite different from the multivariate normal distribution, i.e. $P(J) = \det(2\pi\Sigma)^{-\frac{1}{2}}(x - \mu)^T \Sigma^{-1} (x - \mu)$. But the formula we have in the question(and in paper) is the one with trace, transpose. Even if it is similar to the formula of matrix normal distribution $\endgroup$
    – Xingdong Zuo
    Commented Dec 27, 2015 at 18:02
  • $\begingroup$ i.e. $p(\mathbf{X}\mid\mathbf{M}, \mathbf{U}, \mathbf{V}) = \frac{\exp\left( -\frac{1}{2} \, \mathrm{tr}\left[ \mathbf{V}^{-1} (\mathbf{X} - \mathbf{M})^{T} \mathbf{U}^{-1} (\mathbf{X} - \mathbf{M}) \right] \right)}{(2\pi)^{np/2} |\mathbf{V}|^{n/2} |\mathbf{U}|^{p/2}}$ from Wikipedia page of matrix normal distribution. It is still quite different. $\endgroup$
    – Xingdong Zuo
    Commented Dec 27, 2015 at 18:05
  • $\begingroup$ After searching Wikipedia, it seems the original formula of $P(J)$ looks similar to the definition of Wishart distribution. Not standard multivariate normal distribution, Is it true ? $\endgroup$
    – Xingdong Zuo
    Commented Dec 30, 2015 at 15:03

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