Sharpe Maximization under Quadratic Constraints

Question

When doing Sharpe optimization

$$ \max_x \frac{\mu^T x}{\sqrt{x^T Q x}} $$

there is a common trick (section 5.2) used to put the problem in convex form. You add a variable $\kappa$ such that $x = y/\kappa$ choose $\kappa$ s.t. $\mu^T y=1$. Changing the problem to the simple convex problem

$$ \min_{y,\kappa} y^T Q y \; \text{where} \; \mu^T y = 1, \kappa > 0 $$

which is easy to solve.

Unfortunately, my problem also has a second-order constraint that becomes non-convex in $(y,\kappa)$ $$ x^T P x \leq \sigma^2 \implies y^T P y \leq \kappa^2 \sigma^2 $$

Is there a trick to keep this problem convex and allow the use of second-order cone programming algorithms?

Answer 1

There is no generic solution. However, the KKT conditions are of the forms \begin{align*} \begin{cases} Qy + \lambda_1 \mu +\lambda_2 Py = 0,\\ \mu^T y = 1,\\ y^TPy \leq k^2 \sigma^2,\\ \lambda_2 \big( y^TPy - k^2 \sigma^2\big) = 0. \end{cases} \end{align*} Here, the condition $$\lambda_2 \big( y^TPy - k^2 \sigma^2\big) = 0 $$ means that two cases need to considered, that is, the one on the boundary $y^TPy = k^2 \sigma^2$ and the one inside the domain $y^TPy < k^2 \sigma^2.$

On the boundary, it is the standard Lagrange problem with conditions \begin{align*} \begin{cases} Qy + \lambda_1 \mu +\lambda_2 Py = 0,\\ \mu^T y = 1,\\ y^TPy = k^2 \sigma^2. \end{cases} \end{align*}

Inside the domain, the constraints are \begin{align*} \begin{cases} Qy + \lambda_1 \mu = 0,\\ \mu^T y = 1,\\ y^TPy < k^2 \sigma^2. \end{cases} \end{align*}

The final solution $(y^T, k)$ is the one so that the global overall minimum is reached.