# Sharpe Maximization under Quadratic Constraints

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?

• The KKT theorem can still be applied, as it does not have to have linear inequality constraint. Jun 26, 2015 at 12:41
• Can you expand? I read some about the KKT theorem but I'm not sure how this helps me solve the problem. Jun 26, 2015 at 18:29

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 $y^TPy < k^2 \sigma^2.$
The final solution $(y^T, k)$ is the one so that the global overall minimum is reached.
• Thanks. Interesting. This looks close but I'm still confused on how the quadratic form was converted to a linear form here $Qy$ by adding $\lambda_1$? Do you have a reference? Also, this final problem is still not convex, correct? What sort of numerical solver could be used in this case? Jun 29, 2015 at 19:25
• @rhaskett: The form $Qy$ is the vector derivative $\frac{\partial y^T Qy}{\partial y}$. The final problem is actually easier: you only need to find a solution from the linear system and then check whether it satisfies the inequality -- it does not have to be a convex problem. Jun 29, 2015 at 19:45
• @Gordan This is really interesting. You mention linear inequality constraints as an issue earlier. Does that mean this doesn't work if say an additional $0 \leq x_i \leq 1$ ($0 \leq y_i \leq \kappa$) constraint is imposed? Jun 29, 2015 at 21:33