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Studying a Covariance-Based Maximum Sharpe Ratio Statistic

Article Quant Q&A · Author: alejandroll10

Summary

The document defines a statistic formed from a sample covariance estimate, the inverse of the true covariance matrix, and a vector of ones. It notes that the statistic arises as a maximum equilibrium Sharpe ratio under certain conditions, then asks how its distribution changes as the dimension grows or as the covariance estimate becomes less accurate. Normality is offered as a possible assumption, and simulations are mentioned as having been checked.

The post supplies no derivation, simulation results, or theoretical conclusion. It does not specify the estimator, its regularization, or a precise model for estimation error, all of which could affect the distribution. The material therefore establishes a statistical research question and some motivation, but does not give a usable formula for predicting how the statistic behaves.

Key ideas

  • The statistic combines an estimated covariance matrix with the inverse of the true covariance matrix.
  • The document connects the statistic to a maximum equilibrium Sharpe ratio under unspecified conditions.
  • It asks how dimension and covariance estimation error affect the statistic's distribution.
  • Normality and simulation checks are mentioned, but no theoretical result or numerical evidence is presented.

Tags

Full text
# Distribution of sample covariance times inverse covariance times sample covariance


# Distribution of sample covariance times inverse covariance times sample covariance












I want to understand the distribution of the random variable:

$$S_n = \frac{1}{n^2} 1'\hat \Sigma \Sigma ^{-1} \hat \Sigma 1$$.

1 is a vector of ones of size n, and the variance is of size nxn. $\hat \Sigma$ is an estimate, possibly regularized.

We can assume normality if necessary.

$S_n$ shows up as the maximum equilibrium Sharpe ratio under some conditions.

In particular, I want to understand how $S_n$ changes when either n grows or when $\hat \Sigma$ is a worse approximation of the covariance matrix.

My intuition is that n should grow in both cases, and I checked with simulations but would like theoretical results.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.