Robust optimization and estimation risk

Definition Let $\mu \in \mathbb{R}^n$ be the true vector of expected one-period excess returns, in decimals, and let $\widehat{\mu}$ be an estimate computed from data. The estimation error is $\widehat{\mu} - \mu$. It is random, because $\widehat{\mu}$ depends on the sample. If the excess returns $R_1, \dots, R_N$ are independent and identically distributed with mean $\mu$ and covariance $\Sigma$, and $\widehat{\mu}$ is their sample mean, then

so coordinate $i$ has standard error $\sigma_i/\sqrt{N}$, where $\sigma_i^2 = \Sigma_{ii}$. This sampling variability of the estimate is a different quantity from $\Sigma$, the variability of returns around their mean. A plug-in optimizer treats $\widehat{\mu}$ as if it were $\mu$. With one risky asset of known variance $\sigma^2 > 0$, risk aversion $\gamma > 0$ and the mean-variance utility

of the weight $w$ under the mean $m$, the true optimum is $w = \mu/(\gamma\sigma^2)$ with utility $\mu^2/(2\gamma\sigma^2)$, and the plug-in weight $\widehat{w} = \widehat{\mu}/(\gamma\sigma^2)$ has the error $(\widehat{\mu} - \mu)/(\gamma\sigma^2)$. If $\widehat{\mu}$ is unbiased with variance $s^2$, then $\mathbb{E}[\widehat{\mu}^2] = \mu^2 + s^2$, and expanding the two utilities gives

On average the optimized utility computed from the estimate exceeds the true optimum by $s^2/(2\gamma\sigma^2)$, and the utility the plug-in weight actually delivers falls short of it by the same amount.

Definition A mean uncertainty set $\mathcal{U} \subseteq \mathbb{R}^n$ is a nonempty set of mean vectors that the investor treats as plausible, fixed before the portfolio is chosen. The two standard shapes are an ellipsoid and a box,

where $A$ is an $n \times k$ matrix in units of mean return, $\varepsilon \ge 0$ is a radius and each half-width $\delta_i$ is nonnegative. With one asset both shapes reduce to an interval $[\widehat{\mu} - \delta, \widehat{\mu} + \delta]$. The matrix $A$ describes the shape of the estimation error, not the variability of returns, so it need not be a square root of $\Sigma$. One standard choice, for independent, identically distributed normal excess returns with a known positive definite $\Sigma$, takes $A = (\Sigma/N)^{1/2}$ and $\varepsilon^2$ equal to the quantile at level $\alpha$ of the chi-square distribution with $n$ degrees of freedom. Then $\mu \in \mathcal{U}$ exactly when $N (\widehat{\mu} - \mu)^{\top} \Sigma^{-1} (\widehat{\mu} - \mu) \le \varepsilon^2$, and the random set $\mathcal{U}$ contains the true mean with probability $\alpha$. This lesson takes the set as given and does not calibrate it. A decision optimized against $\mathcal{U}$ carries a guarantee for every mean in the set and none for means outside it. In this lesson $\Sigma$ is known. A set of possible covariance matrices must contain only symmetric positive semidefinite matrices, and uncertainty about a whole distribution needs further restrictions, such as fixed moments or a ball in a stated distance between distributions.

Definition Robust optimization chooses a decision to maximize the worst value of its objective over an uncertainty set. For portfolios, let $w \in \mathbb{R}^n$ hold wealth fractions in the risky assets, with a cash account of zero excess return taking the rest unless the feasible set $\mathcal{W} \subseteq \mathbb{R}^n$ imposes a budget. Let $\Sigma$ be a known symmetric positive semidefinite covariance matrix and $\gamma > 0$, and write $U(w; \mu) = \mu^{\top} w - \tfrac{\gamma}{2} w^{\top} \Sigma w$ for the mean-variance utility under the mean $\mu$. The worst-case objective is

since the variance term does not depend on $\mu$, and the robust optimization problem is

The infimum is taken separately for each $w$, so the least favorable mean depends on the position. Each function $w \mapsto \mu^{\top} w - \tfrac{\gamma}{2} w^{\top} \Sigma w$ is concave, and an infimum of concave functions is concave, so $J$ is concave. It is finite when $\mathcal{U}$ is bounded, and maximizing it is a convex optimization problem when $\mathcal{W}$ is convex. Because $J(w) \le U(w; \mu)$ for every $\mu \in \mathcal{U}$, the optimal worst-case value is at most the plug-in optimal value at any mean in the set, in particular at $\widehat{\mu}$, which lies in both standard shapes. A larger set gives a smaller objective, since $\mathcal{U}_1 \subseteq \mathcal{U}_2$ implies $J_1(w) \ge J_2(w)$ for every $w$.

Definition For a fixed $w$, the worst-case mean return $\inf_{\mu \in \mathcal{U}} \mu^{\top} w$ equals $-h_{\mathcal{U}}(-w)$, where $h_{\mathcal{U}}(y) = \sup_{\mu \in \mathcal{U}} y^{\top} \mu$ is the support function of the set. For the ellipsoid it is

By the Cauchy-Schwarz inequality, $(Au)^{\top} w = u^{\top}(A^{\top} w) \ge -\lVert u \rVert_2 \lVert A^{\top} w \rVert_2 \ge -\varepsilon \lVert A^{\top} w \rVert_2$. If $A^{\top} w \ne 0$, the choice $u^{*} = -\varepsilon A^{\top} w / \lVert A^{\top} w \rVert_2$ attains the bound, and the worst mean is

If $A^{\top} w = 0$, every $u$ gives $0$ and the penalty is $0$. For the box each coordinate is minimized separately. Over $\mu_i \in [\widehat{\mu}_i - \delta_i, \widehat{\mu}_i + \delta_i]$ the term $\mu_i w_i$ is smallest at the lower endpoint when $w_i > 0$ and at the upper endpoint when $w_i < 0$, because multiplying by a negative weight reverses the order, so

The robust counterparts of the mean-variance problem are therefore deterministic problems,

for the ellipsoid, and the same problem with $\sum_i \delta_i \lvert w_i \rvert$ in place of $\varepsilon \lVert A^{\top} w \rVert_2$ for the box. Both penalties are convex and positively homogeneous of degree one in $w$, so subtracting them keeps the objective concave.

Definition Suppose there is one risky asset with estimated mean excess return $\widehat{\mu}$, known variance $\sigma^2 > 0$ and risk aversion $\gamma > 0$, the mean uncertainty set is $[\widehat{\mu} - \delta, \widehat{\mu} + \delta]$ with $\delta \ge 0$, cash has zero excess return, and the weight $w \in \mathbb{R}$ is unrestricted, so short positions and borrowing are allowed. Then the worst-case objective

is strictly concave, and its unique maximizer and maximum are

The kink at $0$ needs one-sided derivatives. For $w > 0$ the derivative $\widehat{\mu} - \delta - \gamma\sigma^2 w$ vanishes at $(\widehat{\mu} - \delta)/(\gamma\sigma^2)$, which is positive exactly when $\widehat{\mu} > \delta$. For $w < 0$ the derivative $\widehat{\mu} + \delta - \gamma\sigma^2 w$ vanishes at $(\widehat{\mu} + \delta)/(\gamma\sigma^2)$, which is negative exactly when $\widehat{\mu} < -\delta$. A concave function on $\mathbb{R}$ is maximized at $0$ exactly when its right derivative there, $\widehat{\mu} - \delta$, is at most $0$ and its left derivative, $\widehat{\mu} + \delta$, is at least $0$, that is, when $\lvert \widehat{\mu} \rvert \le \delta$. If $w$ must lie in an interval $[\ell, h]$, concavity makes the constrained maximizer the point of $[\ell, h]$ closest to $w^{*}$. The hypothesis $\sigma^2 > 0$ matters. With $\sigma^2 = 0$ and $\lvert \widehat{\mu} \rvert > \delta$ the objective is piecewise linear and grows without bound in the direction of the estimate, so it has no maximizer.

Definition Regularization adds to an objective a penalty on the size of the decision. The worst-case penalties are regularizers produced by the uncertainty set, $\varepsilon \lVert A^{\top} w \rVert_2$ from an ellipsoid and the weighted absolute-value penalty $\sum_i \delta_i \lvert w_i \rvert$ from a box. Their effect depends on the feasible set. Let $P$ be the penalty, $w_0$ a maximizer of the estimated utility $U(w; \widehat{\mu})$ over $\mathcal{W}$ and $w_1$ a maximizer of $U(w; \widehat{\mu}) - P(w)$ over the same set. Adding the optimality inequalities $U(w_0; \widehat{\mu}) \ge U(w_1; \widehat{\mu})$ and $U(w_1; \widehat{\mu}) - P(w_1) \ge U(w_0; \widehat{\mu}) - P(w_0)$ gives $P(w_1) \le P(w_0)$, so the penalty is never larger at the penalized optimum than at the plug-in optimum. That guarantee is for the penalty as a whole, not for each holding. Because the penalties are not differentiable at zero, they can also set positions exactly to zero. The box penalty has a kink wherever a single coordinate is zero, so it can remove individual holdings. With an invertible $A$ the ellipsoidal penalty has its only kink at $w = 0$, so it can remove the whole risky portfolio at once. Under the long-only full-investment constraint $w \ge 0$, $\mathbf{1}^{\top} w = 1$, the box penalty equals $\sum_i \delta_i w_i$, which is linear, so it acts exactly as replacing each $\widehat{\mu}_i$ by $\widehat{\mu}_i - \delta_i$. With a common half-width $\delta$ it is the constant $\delta$ and leaves the optimal weights unchanged. Under the budget $\mathbf{1}^{\top} w = 1$ with short positions allowed, writing $w_i^{-} = \max(-w_i, 0)$,

so the penalty charges gross short exposure. The link to statistical regularization is mathematical. A penalty chosen for statistical reasons describes a particular uncertainty about means only when its weights match a set the investor is prepared to defend.