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If i have a matrix of multivariate asset returns for $N$ stocks, and i compute from it the covariance matrix and then the correlation matrix, can I always know which of the two will have the higher condition number (higher to infinity means more ill-conditioned, as opposed to near 1 for well-conditioned)? or is the condition number of two different (types of) matrices completely incomparable?

If one is always more well-conditioned than the other, is there a mathematical proof for this? other criteria besides the condition number are welcome

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Yes, you can compare matrix condition numbers if evaluating them for the same problem, for example taking the matrix's inverse. For L2:

enter image description here

For the additional mathematical characterization of conditioning and its impact, check out the first half of these lecture notes from a class I took: https://github.com/mandli/intro-numerical-methods/blob/master/12_LA_conditioning_stability.ipynb

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  • $\begingroup$ attempts one part of the question, but have no idea what $\kappa$ and $\lambda$ are. for the first half of the question, i am consistently seeing the correlation matrix of randomly generated numbers at least to be more well-conditioned than the covariance matrix. wonder if this fact remains for asset returns $\endgroup$
    – develarist
    Aug 17, 2020 at 14:08
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    $\begingroup$ Kappa is the condition number of matrix A and lambdas are the eigenvalues of A. Sorry for the lack of clarity, the link should provide more insight. $\endgroup$
    – Quantoisseur
    Aug 17, 2020 at 14:17
  • $\begingroup$ i still don't follow what this has to do with the question since the formula only evaluates one matrix. what's the difference with using it versus just calculating the condition number the regular way for the two different matrices? $\endgroup$
    – develarist
    Aug 17, 2020 at 17:03
  • $\begingroup$ I was answering the part of the question that asks if the condition numbers of two matrices are comparable. I don't know if one matrix will always have a higher condition number. $\endgroup$
    – Quantoisseur
    Aug 17, 2020 at 17:37
  • $\begingroup$ if you were answering that part, where does the second matrix enter the answer? Only one matrix is shown $\endgroup$
    – develarist
    Aug 17, 2020 at 18:04
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After having tried this with randomly generated vectors, I am consistently seeing the correlation matrix of randomly generated numbers, regardless of which distribution they are sampled from, are always more well-conditioned than the covariance matrix. Which is strange because the covariance matrix exists before the correlation matrix: the correlation matrix must be computed from the covariance matrix, and the other way around cannot be done.

In other words, the covariance matrix, being more ill-conditioned, in fact is transformed into a more well-conditioned, stable, matrix when it is converted to the correlation matrix.

which makes me wonder if all the financial models that rely on the covariance matrix would be better of using the correlation matrix as an input instead, given all the animosity towards the instability and ill-conditioning of the covariance. I know that the covariance possess variance, or risk, so slanting models to strictly interpret correlations instead would result in missing out on the more relevant measure, which is risk, not correlation, so it seems that we are putting interpretability first compared to other highly-related options, which comes at the price of numerical instability and estimation error

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