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What is your intuition / interpretation of a distribution of eigenvalues of a correlation matrix? I tend to hear that usually 3 largest eigenvalues are the most important, while those close to zero are noise. Also, I've seen a few research papers investigating how naturally occuring eigenvalue distributions differ from those calculated from random correlation matrices (again, distinguising noise from signal).

Please feel free to elaborate on your insights.

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  • $\begingroup$ Do you have in mind any particular application, that is do you seek for general advices about how many EVs we need to consider apart from any application (i.e. on a pure mathematical side) or should it apply to a specific context (e.g. factor analysis, PCA, etc.)? $\endgroup$ – chl Sep 20 '10 at 10:32
  • $\begingroup$ I'm interested more in the mathematical side, ie eigenvalues as a property of the data underlying a correlation matrix. If it makes sense to discuss this in terms of specific context, feel free to do so too. $\endgroup$ – Eduardas Sep 20 '10 at 11:00
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I tend to hear that usually 3 largest eigenvalues are the most important, while those close to zero are noise

You can test for that. See the paper linked in this post for more detail. Again if your dealing with financial times series you might wanna correct for leptokurticity first (i.e. consider the series of garch-adjusted returns, not the raw returns).

I've seen a few research papers investigating how naturally occuring eigenvalue distributions differ from those calculated from random correlation matrices (again, distinguising noise from signal).

Edward:> Usually, one would do it the other way arround: look at the multivariate distribution of eigenvalues (of correlation matrices) coming from the application you want. Once you have identified a credible candidate for the distribution of eigenvalues, it should be fairly easy to generate from them.

The best procedure on how to identify the multivariate distribution of your eigenvalues depends on how many assets you want to consider simultaneously (i.e. what are the dimensions of your correlation matrix). There is a neat trick if $p\leq 10$ ($p$ being the number of assets).

Edit (comments by Shabbychef)

four step procedure:

  1. Suppose you have $j=1,...,J$ sub samples of multivariate data. You need an estimator of the variance-covariance matrix $\tilde{C}_j$ for each sub-sample $j$ ( you could use the classical estimator or a robust alternative such as the fast MCD, which is well implemented in matlab, SAS, S,R,...). As usual, if your dealing with financial times series you would want to consider the series of garch-adjusted returns, not raw returns.
  2. For each sub sample $j$, compute $\tilde{\Lambda}_j=$ $\log(\tilde{\lambda}_1^j)$ ,..., $\log(\tilde{\lambda}_p^j)$ , the eigen values of $\tilde{C}_j$.
  3. Compute $CV(\tilde{\Lambda})$, the convex hull of the $J \times p$ matrix whose j-th entry is $\tilde{\Lambda}_j$ (again, this is well implemented in Matlab, R,...).
  4. Draw points at random from inside $CV(\tilde{\Lambda})$ (this done by giving weight $w_i$ to each of the edges of $CV(\tilde{\Lambda})$ where $w_i=\frac{\gamma_i}{\sum_{i=1}^{p}\gamma_i}$, where $\gamma_i$ is a draw from an unit exponential distribution (more details here).

A limitation is that fast computation of the convex hull of a series of points becomes extremely slow when the number of dimensions is larger than 10. $J\geq2$

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    $\begingroup$ I'm curious: what is the trick? $\endgroup$ – shabbychef Sep 20 '10 at 21:20
  • $\begingroup$ Do you mean the eigenvectors of $\tilde{C}$ in 3? not values? $\endgroup$ – shabbychef Sep 21 '10 at 0:16
  • $\begingroup$ no. $\lambda_1$ is a scalar. $\endgroup$ – user603 Sep 21 '10 at 0:27
  • $\begingroup$ This is a very odd procedure; has it been published somewhere? $\endgroup$ – shabbychef Sep 21 '10 at 2:42
  • $\begingroup$ @Shabbychev:> no, but i had the opportunity to work on a related problem (just not one involving time series) a while ago (same problem as this one stats.stackexchange.com/questions/2572/… ) $\endgroup$ – user603 Sep 21 '10 at 6:45
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Eigenvalues give magnitudes of principle components of data spread.


(source: yaroslavvb.com)
First dataset was generated from Gaussian with covariance matrix $\left(\matrix{3&0\\\\0&1}\right)$ second dataset is the first dataset rotated by $\pi/4$

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One way I have studied this problem in the past is to construct the 'eigenportfolios' of the correlation matrix. That is, take the eigenvector associated with the $k$th largest eigenvalue of the correlation matrix and scale it to a gross leverage of 1 (i.e. make the absolute sum of the vector equal to one). Then see if you can find any real physical or financial connection between the stocks which have large representation in the portfolio.

Usually the first eigenportfolio is almost equal weighted in every name, which is to say the 'market' portfolio consisting of all assets with equal dollar weights. The second eigenportfolio may have some semantical meaning, depending on which time period you look over: e.g. mostly energy stocks, or bank stocks, etc. In my experience, you would be hard pressed to make any story out of the fifth eigenportfolio or beyond, and this depends in some part universe selection and the time period considered. This is just fine because usually the fifth eigenvalue or so is not too far beyond the limits imposed by the Marchenko-Pastur distribution.

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Each value of your $N$ variables defines a point in an $N$ dimensional space. This cloud of points is often ellipsoid-like (if it is not, then you should not consider the variables as linearly related and the correlation does not mean much). The axis of the ellipsoid correspond to the eigenvectors of the correlation matrix, and their "strength" to their eigenvalues. The proof can be found in any time series analysis textbook that covers Principal Component Analysis. The loose intuition of why PCA or other eigenvalue-based methods matter is that you have some process that has some "main" causes, and the rest is "noise". If we make the ansatz that the noise is loosely equal in every dimension (because we might not know anything about it we assume it is not particularly directed).

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