Sharpe Optimization with an Additional Quadratic Risk Constraint
Summary
The document considers maximizing a portfolio’s Sharpe ratio when the portfolio must also satisfy a quadratic risk limit. A standard change of variables turns the ratio objective into minimizing portfolio variance subject to a linear return normalization. Under that transformation, however, the additional quadratic limit depends on the scale variable and is generally nonconvex, so the usual convex formulation does not directly accommodate it.
The answer proposes applying the Karush–Kuhn–Tucker conditions and separating candidate solutions into two cases: the quadratic limit is binding, or it is slack. Each case has its own stationarity and feasibility equations; the boundary case includes a multiplier for the quadratic constraint. The suggested approach is to solve the cases and compare their feasible objective values to identify a global minimum. The document offers no generic reformulation into a convex program or guarantee that a second-order cone solver can handle the original constraint directly. It also does not provide numerical examples or discuss conditions needed to ensure the KKT candidates include a global optimum.
Key ideas
- A Sharpe ratio objective can be transformed into a quadratic minimization with a linear return normalization.
- An additional quadratic risk limit becomes nonconvex after the scaling substitution described.
- KKT conditions distinguish solutions where the extra constraint binds from solutions where it is slack.
- Candidate solutions from both cases must be compared, and the answer gives no general convex reformulation.
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# Sharpe Maximization under Quadratic Constraints
# 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?
## Answer by Gordon (score 2)
https://quant.stackexchange.com/a/18553
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.