Mean variance optimization and efficient frontiers

Definition Expectations on this page are under the physical measure $\mathbb{P}$ of a one-period model, and the inputs $\mu$ and $\Sigma$ are treated as known. Let $R = (R_1, \dots, R_d)^{\top}$ be the simple returns of $d$ assets over one period, with $E[R_i^2] < \infty$ for every $i$, mean vector $\mu = E[R]$ and covariance matrix $\Sigma = \operatorname{Cov}(R)$. An investor with initial wealth $V_0 > 0$ holds $w_i V_0$ currency units of asset $i$ at the start of the period and does not trade until its end, for a weight vector $w \in \mathbb{R}^d$ with $\mathbf{1}^{\top} w = 1$, the budget constraint of the next card. The final wealth is $V_1 = V_0 (1 + w^{\top} R)$, so the portfolio return is $R_p = w^{\top} R$, and linearity gives the portfolio moments

More generally $\operatorname{Cov}(w^{\top} R, v^{\top} R) = w^{\top} \Sigma v$ for two weight vectors. Written out, the variance is $\sum_i w_i^2 \Sigma_{ii} + \sum_{i \ne j} w_i w_j \Sigma_{ij}$, so every pair of assets contributes a cross term. These are one-period identities for simple returns and fixed weights. They do not hold for log returns, because the portfolio log return $\ln(1 + w^{\top} R)$ is not $\sum_i w_i \ln(1 + R_i)$ in general, and they do not describe the compounded return of a strategy whose weights change over several periods.

Definition The budget constraint is $\mathbf{1}^{\top} w = \sum_i w_i = 1$, where $\mathbf{1}$ is the vector of ones. It says that the holdings account for all of the initial wealth, and it is what makes $w^{\top} R$ the portfolio return in the previous card. A negative weight $w_i < 0$ is a short position, in which the investor sells borrowed units of asset $i$ worth $\lvert w_i \rvert V_0$ and the proceeds fund the other holdings. In this frictionless model the full proceeds are available, and there are no margin requirements, borrowing fees or position limits. With unrestricted shorts the feasible weights form the affine hyperplane $\{w : \mathbf{1}^{\top} w = 1\}$ of dimension $d - 1$. Adding a target mean $\mu^{\top} w = m$ leaves a nonempty affine set of dimension $d - 2$ whenever $\mu$ is not a multiple of $\mathbf{1}$, because the two equations are then independent. A long-only investor also requires $w \ge 0$ componentwise. The feasible weights then form the simplex, on which $w^{\top} \mu$ is a convex combination of the $\mu_i$ and so lies between $\min_i \mu_i$ and $\max_i \mu_i$. The gross exposure $\sum_i \lvert w_i \rvert$ is at least $\sum_i w_i = 1$, with equality exactly when no weight is negative.

Definition A quadratic program is an optimization problem of the form

with $Q$ a symmetric $n \times n$ matrix, $c \in \mathbb{R}^n$, matrices $G$ and $E$ with $n$ columns, and the inequality read componentwise. It is convex exactly when $Q$ is positive semidefinite. The following standard results are stated without proof. If $Q$ is positive definite and the feasible set is nonempty, a minimizer exists and is unique. For a convex quadratic program, a feasible point is a global minimizer if and only if it satisfies the KKT conditions with some multipliers, and because every constraint is linear, no further constraint qualification such as Slater’s condition is needed. If $Q$ is only positive semidefinite, minimizers can fail to be unique, and the problem can be unbounded below. The minimum-variance problem of the next card is the case $x = w$, $Q = \Sigma$, $c = 0$, the two rows $\mathbf{1}^{\top}$ and $\mu^{\top}$ of $E$ and $b = (1, m)$, and its long-only version adds $G = -I$ and $h = 0$. The risk-aversion form, which maximizes $\mu^{\top} w - \tfrac{\delta}{2} w^{\top} \Sigma w$ for a risk aversion $\delta > 0$, is the case $Q = \delta \Sigma$ and $c = -\mu$.

Definition Let $d \ge 2$, let $\Sigma$ be positive definite, and let $\mu$ not be a multiple of $\mathbf{1}$, so that the assets do not all have the same mean. For a target mean $m \in \mathbb{R}$, with shorts unrestricted, the minimum-variance problem is

The factor $\tfrac12$ simplifies the derivatives and does not change the minimizer. Write

Since $\Sigma^{-1}$ is positive definite, $\langle x, y \rangle = x^{\top} \Sigma^{-1} y$ is an inner product, so $A > 0$, and the Cauchy-Schwarz inequality gives $B^2 \le AC$, with equality exactly when $\mu$ is a multiple of $\mathbf{1}$. Hence $\Delta > 0$. With the Lagrangian

stationarity gives $\Sigma w = \lambda \mathbf{1} + \gamma \mu$, so $w = \lambda \Sigma^{-1} \mathbf{1} + \gamma \Sigma^{-1} \mu$, and the two constraints become the linear system $A\lambda + B\gamma = 1$ and $B\lambda + C\gamma = m$, whose determinant is $\Delta$. Its unique solution gives the frontier weights

This feasible stationary point is the unique global minimizer. Write $w$ for $w(m)$. Any other feasible portfolio is $w + h$ with $\mathbf{1}^{\top} h = 0$ and $\mu^{\top} h = 0$, so $h^{\top} \Sigma w = \lambda \mathbf{1}^{\top} h + \gamma \mu^{\top} h = 0$, and

which is larger unless $h = 0$. So every target $m$ is attainable and has exactly one minimum-variance portfolio. If instead all means equal a common value, the target constraint is redundant at that value and impossible at every other, and these formulas do not apply.

Definition Under the hypotheses of the previous card, multiplying $\Sigma w = \lambda \mathbf{1} + \gamma \mu$ on the left by $w^{\top}$ and using both constraints gives $w^{\top} \Sigma w = \lambda + \gamma m$. So the smallest variance at the target $m$ is

Its derivative is $2(Am - B)/\Delta = 2\gamma$, so the multiplier of the mean constraint is half the rate at which the minimum variance rises with the target. The minimum over $m$ is at $m = B/A$, with variance $1/A$ and weights $\Sigma^{-1} \mathbf{1}/A$. This is the global minimum-variance portfolio, which also solves the problem with the budget constraint alone. A feasible portfolio is mean-variance efficient if no feasible portfolio has a mean at least as high and a variance at least as low, with at least one of the two strictly better. With the budget constraint and unrestricted shorts, the efficient portfolios are exactly the $w(m)$ with $m \ge B/A$. In the plane of volatility and mean, the curve $\sigma = \sigma(m)$ is one branch of a hyperbola with its leftmost point at the global minimum-variance portfolio, and the efficient frontier is the part of that curve with $m \ge B/A$. Every portfolio on the lower part is dominated by the global minimum-variance portfolio, which has a higher mean and a lower variance. Because $w(m)$ is affine in $m$, every frontier portfolio is an affine combination of any two distinct frontier portfolios, which is two-fund separation.

Definition Let $x$ be a feasible point of a problem with inequality constraints $g_i(x) \le 0$. The constraint $i$ is active, or binding, at $x$ if $g_i(x) = 0$, and inactive if $g_i(x) < 0$. Equality constraints are active at every feasible point. The long-only minimum-variance problem adds $-w_i \le 0$ for every $i$ to the minimum-variance problem. With the sign convention of the Lagrangian above and a multiplier $\nu_i$ for each bound, its KKT conditions are $\mathbf{1}^{\top} w = 1$, $\mu^{\top} w = m$ and

For the long-only problem with the budget alone, drop $\gamma$ and the mean constraint. Both are convex quadratic programs with linear constraints, so a feasible $w$ is optimal if and only if multipliers satisfying these conditions exist. Complementary slackness forces $\nu_i = 0$ for every asset held with $w_i > 0$, so held assets satisfy $(\Sigma w)_i = \lambda + \gamma \mu_i$, while an asset at its bound may have $(\Sigma w)_i > \lambda + \gamma \mu_i$, with the difference supplied by $\nu_i \ge 0$. If the feasible set is empty, no point satisfies the conditions and the problem has no solution. Which bounds are active at the optimum is not known in advance, and finding them is the work of active-set and interior-point algorithms, which are not developed here.