Risk budget allocation.#

Suppose an amount \(x_i>0\) is invested in \(n\) assets, labeled \(i=1,..., n\), with asset return covariance matrix \(\Sigma \in \mathcal{S}_{++}^n\). We define the risk of the investments as the standard deviation of the total return, \(R(x) = (x^T\Sigma x)^{1/2}\).

We define the (relative) risk contribution of asset \(i\) (in the portfolio \(x\)) as

\[ \rho_i = \frac{\partial \log R(x)}{\partial \log x_i} = \frac{\partial R(x)}{R(x)} \frac{x_i}{\partial x_i}, \quad i=1, \ldots, n. \]

Thus \(\rho_i\) gives the fractional increase in risk per fractional increase in investment \(i\). We can express the risk contributions as

\[ \rho_i = \frac{x_i (\Sigma x)_i} {x^T\Sigma x}, \quad i=1, \ldots, n, \]

from which we see that \(\sum_{i=1}^n \rho_i = 1\). For general \(x\), we can have \(\rho_i <0\), which means that a small increase in investment \(i\) decreases the risk. Desirable investment choices have \(\rho_i>0\), in which case we can interpret \(\rho_i\) as the fraction of the total risk contributed by the investment in asset \(i\). Note that the risk contributions are homogeneous of degree zero, i.e., scaling \(x\) by a positive constant does not affect \(\rho_i\).

In the risk budget allocation problem, we are given \(\Sigma\) and a set of desired risk contributions \(\rho_i^\mathrm{des}>0\) with \(\bf{1}^T \rho^\mathrm{des}=1\); the goal is to find an investment mix \(x\succ 0\), \(\bf{1}^Tx =1\), with these risk contributions. When \(\rho^\mathrm{des} = (1/n)\bf{1}\), the problem is to find an investment mix that achieves so-called risk parity.


Explain how to solve the risk budget allocation problem using convex optimization.

Hint. Minimize \((1/2)x^T\Sigma x - \sum_{i=1}^n \rho_i^\mathrm{des} \log x_i\).


Find the investment mix that achieves risk parity for the return covariance matrix \(\Sigma\) below.

import numpy as np
import cvxpy as cp

Sigma = np.array(
        """6.1  2.9 -0.8  0.1;
           2.9  4.3 -0.3  0.9;
          -0.8 -0.3  1.2 -0.7;
           0.1  0.9 -0.7  2.3"""
rho = np.ones(4) / 4

# TODO: your code here