CAPM Residual Risk and the Market Portfolio Optimization Solution

Summary

The exchange connects mean-variance portfolio choice to the CAPM decomposition of asset returns into market exposure and idiosyncratic residuals. Under the answer’s assumptions that residuals have zero mean and are mutually independent, the covariance matrix consists of a common market factor component plus a diagonal matrix of residual variances. The first-order optimization condition then makes the portfolio direction proportional to the inverse covariance matrix applied to the vector of market betas. The Sherman–Morrison–Woodbury identity is suggested to simplify that inverse and reveal the role of residual risk.

The result depends on the independence assumption and on the CAPM representation; correlated residuals would require a fuller residual covariance matrix. The response only sketches the algebra and does not complete the matrix simplification or reconcile the question’s stated constraint with the more usual unconstrained mean-variance formulation. It provides no empirical test or portfolio performance evidence.

Key ideas

• The CAPM splits returns into a market factor exposure and an idiosyncratic residual.
• With independent residuals, residual variances form the diagonal component of the covariance matrix.
• Mean-variance optimization gives a portfolio direction based on inverse covariance applied to market betas.
• The Sherman–Morrison–Woodbury identity can simplify the inverse for a single common market factor.
• Correlated residuals or different constraints can change the resulting portfolio allocation.

Tags

# Market Portfolio Optimization


# Market Portfolio Optimization












Consider the minimization problem

$$\min\left\{\frac{1}{2}x^T\Sigma x - \lambda(\mu-r_f)^Tx\right\}$$

and assume the CAPM model, i.e.

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

Assuming $\Sigma$ is invertible, prove

$$x_i \propto \frac{\beta_i}{\textrm{Var}(\varepsilon_i)}$$

It seems like lambda must stay in the minimization problem after solving for $x$, which is probably why we're only solving for proportionality, but I still cannot find a way to go about tackling this. Solving the Lagrangian yields

$$x=\lambda\Sigma^{-1}(\mu-r_f)$$

and we know

$$(\mu-r_f)^Tx=0$$

but this doesn't seem to help me. Where does the quadratic term yielding variance in the solution come from?

## Answer by steveo'america (score 2)

https://quant.stackexchange.com/a/43669

Assuming the $\epsilon_i$ are zero mean, you should find that $$ \mu - r_f = \beta \left(E[r_m] - r_f\right). $$ Further assuming the $\epsilon_i$ are independent of each other, though possibly with different variances, let $\Gamma$ be the diagonal matrix with the variances of $\epsilon_i$ on the diagonal. Then you are to find (under the more usual MVO formulation) $$ \max_x \,\, x^{\top}\beta \left(E[r_m] - r_f\right) - \frac{1}{2\lambda} x^{\top}\left(\beta \beta^{\top}\sigma^2 + \Gamma\right)x. $$ (I am keeping your $\lambda$ associated with the mean, though usually it is risk aversion and so you would see $\lambda/2$.)

Now use the Lagrange Multiplier technique to find the solution, which should be something like $$ x \propto \left(\beta\beta^{\top}\sigma^2 + \Gamma\right)^{-1}\beta, $$ and then use the Sherman-Morrison-Woodbury formula to simplify the matrix inverse.

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