Deriving the Minimum-Risk Portfolio with Unit Attribute Exposure

Summary

The document derives a portfolio that has unit exposure to a chosen asset attribute while minimizing portfolio variance. Given an attribute vector and an invertible covariance matrix, the optimization minimizes the quadratic risk expression subject to a linear exposure constraint. The Lagrange conditions imply that the covariance matrix times the portfolio weights is proportional to the attribute vector.

Multiplying by the inverse covariance matrix expresses weights as a scalar multiple of the inverse-covariance-weighted attributes. Enforcing unit exposure determines the scalar, producing weights equal to that vector divided by its attribute-weighted normalization. The derivation assumes the covariance matrix is invertible and symmetric, as needed for the stated steps. The result is a mathematical construction; the document does not discuss estimation error, constraints on holdings, transaction costs, or empirical performance.

Key ideas

• The objective is to minimize portfolio variance subject to unit exposure to an attribute.
• The first-order condition makes the covariance-weighted holdings proportional to the attribute vector.
• Applying the inverse covariance matrix gives holdings proportional to the covariance-adjusted attribute vector.
• The unit-exposure constraint fixes the proportionality constant through a normalization term.
• The derivation assumes an invertible, symmetric covariance matrix and does not address implementation constraints.

Tags

# Characteristic Portfolio for an Attribute


# Characteristic Portfolio for an Attribute












Given a vector of attributes(eg.E/P ratios, betas) for N assets

$a^T = {a_1,a_2,...,a_N}$ The exposure of portfolio $h_P$ to attribute a is

$a = \sum_{n}a_n h_{P,n}$

Proposition: There is a unique portfolio $h_a$ that has minimum risk and unit exposure to a. The holdings(weights) of the characteristic portfolio $h_a$ are given by

$h_a = \frac{V^{-1}a}{a^TV^{-1}a}$

For the prrof we write:

Minimise $h^TVh$ subject to constriant : $h^Ta=1$

Using Langrange multiplier we get the equations:

a. $h^Ta = 1$

b. $Vh - \lambda a = 0$

Question: How does substituting a in b yields the result of the proposition ?

## Answer by Alex C (score 2, accepted)

https://quant.stackexchange.com/a/41202

From b. we get $Vh = \lambda a$, so $h=\lambda V^{-1}a$ (assuming V is invertible).

Using this to evaluate a. we get $h^Ta = \lambda a^T V^{-1}a=1$ (assuming $V^{-1}$ is symmetric). We can solve this for lambda: $\lambda=\frac{1}{a^T V^{-1}a}$

Now we can use this lambda in the previous expression for h to find the final explicit expression for h:

$$h=\frac{V^{-1}a}{a^T V^{-1}a}$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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