I am trying to obtain the eigenvalue distribution of a finite dimensional Wishart matrix. Let $A_{n\times n}\sim\mathbb{W}(\Sigma_{n\times n},m)$ where $\mathbb{W}(\Sigma_{n\times n},m)$ denotes the Wishart distribution with covariance $\Sigma_{n\times n}$ and degrees of freedom $m$. I know that the result in the asymptotic case as $n,m\to\infty$ and $n/m\to c$ is given by the Marchenko–Pastur (M-P) density. My research led me to the following book:

T. W. Anderson, "An Introduction to Multivariate Statistical Analysis", Ed. 2, John Wiley & Sons, Inc., 1984

It is my understanding that this is a classical textbook on multivariate statistics and a lot of its results are very well cited. One that was of particular interest to me is in page 534, Theorem 13.3.2 (notation changed to match the usage above):

If $A\ (n\times n)$ has the distribution $\mathbb{W}(I,m)$, then the characteristic roots $(l_1\geq l_2\geq\ldots\geq l_n\geq 0)$ have the [following] density over the range when the density is not 0.

$$\frac{\pi^{n^2/2}\prod_{i=1}^n l_i^{(m-n-1)/2}\exp\left(-\frac{1}{2}\sum_{i=1}^n l_i\right)\prod_{i\lt j}(l_i-l_j)}{2^{nm/2}\Gamma_n(m/2)\Gamma_n(n/2)}$$

where $\Gamma_n(x)$ is the multivariate Gamma function defined as

$$\Gamma_n(x)=\pi^{n(n-1)/4}\prod_{i=1}^n \Gamma\left(x+(1-i)/2\right)$$

I have the following questions:

  1. The expression for the density is a bit strange to me. Am I right in understanding that if there is an underlying continuous density $\rho(x)$, the expression above gives the density at a particular point $l_j$ as $\rho(l_j)$? If so, how do I flesh out $\rho(x)$ from the complicated expression?
  2. It seems to me that the expression requires knowledge of the eigenvalues $l_1,\ldots, l_n$ to calculate the density, and as such is less useful than a function $\rho(x)$ (like the M-P density), which is free of the individual eigenvalues. For example, if I needed to know the density for $n=10000$, I'll have to first evaluate the eigenvalues for a $10000\times 10000$ matrix before using the expression. In that case, bin counting or histogram would be simpler to use than this expression.
  3. Coming to evaluating the above density numerically, the denominator seems to be humungous ($10^{5000}$ish) when compared to the numerator ($10^{300}$ish) for nominal values of say, $n=30$ and $m=100$. As such, the density evaluates to almost 0 everywhere. I haven't posted the code as I believe it is my interpretation of the expression above (questions 1 & 2) that are incorrect, and not my implementation. However, I can provide code in Mathematica to evaluate the above expression, if anyone wants it.
  • 1
    $\begingroup$ The formula you quoted is the joint density for the eigenvalues $l_i$, which are random variables. I don't understand what you mean by your first question. For the second, "less useful" depends on what you're trying to do, which you are unclear about. For the third question, note that you're quoting a formula which holds for a different normalization than the M-P result: for large $n$ the eigenvalues spread out a lot. $\endgroup$ Jan 5 '12 at 16:31

Hi, I think you should have a look at this:

Zanella, A., M. Chiani and M.Z. Win, "On the marginal distribution of the eigenvalues of wishart matrices" IEEE Transactions on Communications 57 (2009):1050-1060

Cheers, FP

  • $\begingroup$ For the record, Zanella et al considers the complex Wishart distribution, while Anderson's book focusses on the real case (which is harder). I have yet to see a tractable formula for the marginal distribution of the eigenvalues of real Wishart matrices (anyone? :-)). $\endgroup$
    – llude
    Sep 4 '19 at 12:19

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