Using Sherman–Morrison to Invert a CAPM Covariance Matrix

Summary

The document derives an inverse covariance matrix for a single-factor CAPM model with a zero risk-free rate. It represents total asset covariance as the sum of a diagonal matrix of idiosyncratic variances and a rank-one market-factor contribution. The Sherman–Morrison identity then gives the inverse as the inverse diagonal matrix minus a rank-one adjustment based on asset betas and residual variances.

The explanation identifies the matrix identity that produces the stated result, but it does not show the substitution and algebra in detail. It offers no empirical test or portfolio application, so the result is a mathematical tool rather than evidence for a trading strategy. The formula also relies on the stated one-factor structure and an invertible diagonal residual-variance matrix; other factor structures require different or extended matrix identities.

Key ideas

• The CAPM covariance matrix is a diagonal residual-risk matrix plus a rank-one market-risk term.
• The Sherman–Morrison identity provides a direct inverse for a matrix plus an outer product.
• The inverse depends on asset betas, residual variances, and market variance.
• The derivation assumes the residual-variance matrix is invertible.

Tags

# Inverse Covariance Matrix Transformation from CAPM


# Inverse Covariance Matrix Transformation from CAPM












Beginning with the CAPM model we have (with a risk free rate of 0%):

$r_i=\beta_i (r_m)+\varepsilon_i$

with $\varepsilon_i$ the diversifiable risks per assets

The variance matrix:

$\Omega = \beta'\beta \sigma_m^2 + Diag(\sigma_e^2)$

With $\sigma_m$ a constant, $Diag(\sigma_e^2)$ an N $\times$ N matrix, $\beta$ an 1 $\times$ N matrix.

Inverting the matrix we get the following result:

$\Omega^{-1} = Diag(\frac{1}{\sigma_e^2})-\frac{(\frac{\beta}{\sigma_e^2})(\frac{\beta}{\sigma_e^2})'}{\frac{1}{\sigma_m^2}+(\frac{\beta}{\sigma_e^2})'\beta}$

I don't understand how by using the inverse matrix transformation we find this result.

Thank you for your help

## Answer by Kermittfrog (score 8, accepted)

https://quant.stackexchange.com/a/58696

This is the result of the Sherman-Morrison inversion for the sum of an invertible matrix and an outer product. You will find this (and many other helpful methods) in the Matrix Cookbook. Specifically, this is equation 160 on p 18:

$$ \left(\boldsymbol{A}+\boldsymbol{bc}^T\right)^{-1}=\boldsymbol{A}^{-1}-\frac{\boldsymbol{A}^{-1}\boldsymbol{bc}^T\boldsymbol{A}^{-1}}{1+\boldsymbol{c^TA}^{-1}b} $$

HTH

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