Worst-case risk analysis

Covariance uncertainty

In this example we do worst-case risk analysis using CVXPY. Our setting is a single period Markowitz portfolio allocation problem. We have a fixed portfolio allocation \(w \in {\bf R}^n\). The return covariance \(\Sigma\) is not known, but we believe \(\Sigma \in \mathcal S\). Here \(\mathcal S\) is a convex set of possible covariance matrices. The risk is \(w^T \Sigma w\), a linear function of \(\Sigma\).

We can compute the worst (maximum) risk, over all possible covariance matrices by solving the convex optimization problem

\[\begin{split} \begin{array}{ll} \mbox{maximize} & w^T\Sigma w \\ \mbox{subject to} & \Sigma \in \mathcal S, \quad \Sigma \succeq 0, \end{array} \end{split}\]

with variable \(\Sigma\).

If the worst-case risk is not too bad, you can worry less. If not, you’ll confront your worst nightmare

Example

In the following code we solve the portfolio allocation problem

\[\begin{split} \begin{array}{ll} \mbox{minimize} & w^T\Sigma_\mathrm{nom} w \\ \mbox{subject to} & {\bf 1}^Tw = 1, \quad \mu^Tw \geq 0.1, \quad \|w\|_1 \leq 2, \end{array} \end{split}\]

and then compute the worst-case risk under the assumption that \(\mathcal S = \left\{ \Sigma^\mathrm{nom} + \Delta \,:\, |\Delta_{ii}| =0, \; |\Delta_{ij}| \leq 0.2 \right\}\).

We might expect that \(|\Delta_{ij}| = 0.2\) for all \(i \neq j\). This does not happen however because of the constraint that \(\Sigma^\mathrm{nom} + \Delta\) is positive semidefinite.

# Generate data for worst-case risk analysis.
import numpy as np

np.random.seed(2)
n = 5
mu = np.abs(np.random.randn(n, 1)) / 15
Sigma = np.random.uniform(-0.15, 0.8, size=(n, n))
Sigma_nom = Sigma.T.dot(Sigma)
print("Sigma_nom =")
print(np.round(Sigma_nom, decimals=2))

Sigma_nom =
[[ 0.58  0.2   0.57 -0.02  0.43]
 [ 0.2   0.36  0.24  0.    0.38]
 [ 0.57  0.24  0.57 -0.01  0.47]
 [-0.02  0.   -0.01  0.05  0.08]
 [ 0.43  0.38  0.47  0.08  0.92]]

# Form and solve portfolio optimization problem.
# Here we minimize risk while requiring a 0.1 return.
import cvxpy as cp


w = cp.Variable(n)
ret = mu.T @ w
risk = cp.quad_form(w, Sigma_nom)
prob = cp.Problem(cp.Minimize(risk), [cp.sum(w) == 1, ret >= 0.1, cp.norm(w, 1) <= 2])
prob.solve()
print("w =")
print(np.round(w.value, decimals=2))

w =
[-0.01  0.13  0.18  0.88 -0.18]

# Form and solve worst-case risk analysis problem.
Sigma = cp.Variable((n, n), PSD=True)
Delta = cp.Variable((n, n), symmetric=True)
risk = cp.quad_form(w.value, Sigma)
prob = cp.Problem(
    cp.Maximize(risk),
    [Sigma == Sigma_nom + Delta, cp.diag(Delta) == 0, cp.abs(Delta) <= 0.2],
)
prob.solve()
print("standard deviation =", cp.sqrt(cp.quad_form(w.value, Sigma_nom)).value)
print("worst-case standard deviation =", cp.sqrt(risk).value)
print("worst-case Delta =")
print(np.round(Delta.value, decimals=2))

standard deviation = 0.16889492230304606
worst-case standard deviation = 0.42202002145834544
worst-case Delta =
[[ 0.    0.04 -0.2  -0.    0.2 ]
 [ 0.04 -0.    0.2   0.09 -0.2 ]
 [-0.2   0.2   0.    0.12 -0.2 ]
 [-0.    0.09  0.12  0.   -0.18]
 [ 0.2  -0.2  -0.2  -0.18  0.  ]]