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?

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

1 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.

  • $\begingroup$ 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? $\endgroup$
    – rhaskett
    Commented Jun 29, 2015 at 19:25
  • $\begingroup$ @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. $\endgroup$
    – Gordon
    Commented Jun 29, 2015 at 19:45
  • $\begingroup$ @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? $\endgroup$
    – rhaskett
    Commented Jun 29, 2015 at 21:33
  • $\begingroup$ @rhaskett, See Section 2.2 of this lecture notes homes.soic.indiana.edu/classes/spring2012/csci/b553-hauserk/… may be helpful. $\endgroup$
    – Gordon
    Commented Jun 30, 2015 at 12:37

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