Shrinking the sample covariance matrix
Instituto de Matemática, Estatística e Computação Científica (IMECC),
Universidade Estadual de Campinas (UNICAMP).
Which is the best estimator of the populational mean in a multivariate setting (\(N > 3\))?
Any Shrinkage estimator for the large-dimensional covariance matrix has three ingredients:
S, F and \(\delta\)
Notation
A linear shrinkage estimator is given by:
\[\hat{\Sigma} = (1 - \delta)S + \delta F\]
\(F\) | Reference |
---|---|
\(\sigma^2 \times \mathbb{I}\) | A well-conditioned estimator for large-dimensional covariance matrices (Ledoit and Wolf 2004a) |
Single factor of Sharpe (1963) (CAPM) | Improved estimation of the covariance matrix of stock returns with an application to portfolio selection (Ledoit and Wolf 2003) |
\(f_{ij} = \sqrt{\sigma_{ii}\sigma_{jj}}\rho\) | Honey, I shrunk the sample covariance matrix (Ledoit and Wolf 2004b) |
\(f_{ij} = \eta\) and \(f_{ii} = \sigma^2\) | Essays on risk and return in the stock market (Ledoit 1995) |
First Idea
Let \(\lambda_1, \cdots, \lambda_N\) be the eiganvalues of \(S\) and let
\[(1 - \delta) S + \delta F = \hat{\Sigma} = U \Lambda^{\ast} U\] be the spectral decomposition of \(\hat{\Sigma}\).
Can be proved that the elements \(\lambda_1^{\ast}, \cdots, \lambda_N^{\ast}\) of the diagonal matrix \(\Lambda^{\ast}\) are equal to \[\lambda_i^{\ast} = \delta \sigma^2 + (1 - \delta) \lambda_i\]
Carlos Trucíos (IMECC/UNICAMP) | FCM-UNMSM 2025 | Risk-Based Portfolio Allocation | ctruciosm.github.io