Abstract

The optimization of a large random portfolio under the expected shortfall risk measure with an regularizer is carried out by analytical calculation for the case of uncorrelated Gaussian returns. The regularizer reins in the large sample fluctuations and the concomitant divergent estimation error, and eliminates the phase transition where this error would otherwise blow up. In the data-dominated region, where the number of different assets in the portfolio is much less than the length of the available time series, the regularizer plays a negligible role even if its strength is large, while in the opposite limit, where the size of samples is comparable to, or even smaller than the number of assets, the optimum is almost entirely determined by the regularizer. We construct the contour map of estimation error on the versus plane and find that for a given value of the estimation error the gain in due to the regularizer can reach a factor of about four for a sufficiently strong regularizer.

Highlights

  • We present the numerical solutions for the relative estimation error

  • We recall the contour map of the relative estimation error of ES without regularization

  • The regularizer takes care of the large sample fluctuations and eliminates the phase transition that would be present in the problem without regularization

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Summary

Introduction

As explained in [24], each of the three remaining order parameters in the above set of equations, q0, ∆, and has a direct financial meaning: ∆ is related to the in-sample estimator of ES (and to the second derivative of the cost function F with respect to the Lagrange multiplyer λ associated with the budget constraint) and is the insample VaR of the portfolio optimized under the ES risk measure. In the present analytical approach we have the luxury of infinitely many samples to average over, so we can obtain the value of the coefficient of regularization by demanding a given relative error (that is a given q0) for a given confidence limit α and given aspect ratio r = N/T .

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