Utility based portfolio optimization deals with the problem of finding the optimal portfolio $x_T$ by maximizing the utility/objective function $O(x_T,x_0)$ where $x_0$ is the current portfolio.

In the particular case of mean-variance the objective function is equal to $$ O(x,x_0) = \mathrm{AlphaTerm}(x,x_0) - \lambda\cdot \mathrm{Var}(x) - \kappa\cdot \mathrm{TC}(x,x_0) $$ and we want to find the portfolio $x$ that maximizes this objective subject to a set of constraints such as:

  • $\mathrm{Risk}(x)\leq MR$
  • $\mathrm{Gross}(x)\leq MG$
  • $-MN\leq \mathrm{Net}(x)\leq MN$
  • $-M\beta\leq \mathrm{Beta}(x)\leq M\beta$
  • $\mathrm{TC}(x,x_0)\leq MTC$
  • Others

where $\mathrm{TC}(x,x_0)$ is the total transaction cost (i.e. Fees + Market Impact) of trading from $x_0$ to $x$ and $\mathrm{Var}(x)=\mathrm{Risk}(x)^2$ is the portfolio variance.

Is there any study about the advantages/disadvantages of the more general case such as $U_\theta(x,x_0)$ where we are maximizing $$ O_{\theta}(x,x_0) = \mathrm{AlphaTerm}(x,x_0) - \lambda\cdot \mathrm{Risk}(x)^{\theta} - \kappa\cdot\mathrm{TC}(x,x_0) $$ subject to same constraints?

What about the difference between $\theta=2$ vs. $\theta=1$?

  • $\begingroup$ How is $\lambda$ set? $\endgroup$ Commented Nov 6, 2017 at 22:48
  • $\begingroup$ What do you mean? It’s a parameter regulating the risk aversion. $\endgroup$
    – ght
    Commented Nov 7, 2017 at 10:13

3 Answers 3


Something to perhaps realize is that your two problems may not be as different as you think if $\lambda$ is an ad-hoc parameter.

For any solution to your 2nd problem (where $\theta > 1$), there exists a $\lambda$ for problem 1 which gives you the same solution as problem 2.


Let $f$ and $g$ be convex functions and let $\mathbf{x}$ denote a vector. To build some intuition, consider the two problems (stylized versions of your problem 1 and problem 2).

Problem A:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $\mathbf{x}$)} & f(\mathbf{x}) + \lambda_1g(\mathbf{x}) \end{array} \end{equation}

Problem B:

Let $h(x)$ be convex and non-decreasing over the range of $g$, hence composition $h(g(\mathbf{x}))$ is also convex. Consider \begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $\mathbf{x}$)} & f(\mathbf{x}) + \lambda_2 h\left( g(\mathbf{x}) \right) \end{array} \end{equation}

First order conditions

Both are convex optimization problems and the first order conditions are necessary and sufficient.

  • The first order condition for problem A is $\frac{\partial f}{\partial \mathbf{x}} + \lambda_1 \frac{\partial g}{\partial \mathbf{x}} = 0$.
  • The first order condition for problem B is $\frac{\partial f}{\partial \mathbf{x}} + \lambda_2 h'(g(\mathbf{x}))\frac{\partial g}{\partial \mathbf{x}} = 0$.

Let $\mathbf{x}^*$ be the solution to either problem A or problem B. If $\lambda_1 = \lambda_2 h'(g(\mathbf{x}^*))$ then problem A and problem B have the same solution.

Your penalty is $h(x) = x^\theta$ (which is convex and increasing on the region $x >0$ if $\theta > 1$). Observe $h'(x) = \theta x^{\theta - 1}$. Also observe that to achieve the same solution $\mathbf{x}^*$, you could have $\lambda_1 > \lambda_2$ or $\lambda_1 < \lambda_2$ depending on the specific parameterization. As you've formulated the problem, increasing $\theta$ doesn't necessarily increase the penalty for larger risk (where risk is defined as the standard deviation of returns)! For example, let's say your solution $\mathbf{x}^*$ for $\theta = 1$ gives a standard deviation of returns of .01. Simply raising $\theta$ to $1000$ would increase the convexity of the risk penalty but it would decrease the level of the penalty to numerical irrelevance.

Your $g(\mathbf{x}) = \sqrt{\mathbf{x}'\Sigma \mathbf{x}}$ where $\Sigma$ is the covariance matrix of returns.

  • $\begingroup$ @ght This is not to say your two problems are the same; just that if $\lambda$ is arbitrary, letting $\theta$ vary too in some sense doesn't increase the set of possible solutions. For any particular velocity $v$ over the speed limit, a speeding ticket based upon $v$ or $v^2$ can give the same deterrence effect. At 10 mph over, paying a $10v$ fine or a $\frac{1}{2} v^2$ fine has the same marginal cost of higher speed (since the first derivative is the same). $\endgroup$ Commented Nov 7, 2017 at 17:33

I doubt that there are papers that study exactly the set of constraints that you have listed. But there are papers that look, for instance, into specifications for risk and reward other than variance and mean. For instance, in https://ssrn.com/abstract=1365167 and https://ssrn.com/abstract=2975529 we have looked whether alternative risk functions provide any advantage. In a nutshell, these studies find that emphasising risk works well, and that using risk functions that only look at downside risk/losses work better than a symmetric risk function (i.e. variance).

  • $\begingroup$ Thanks I’ll take a look at the references. To clarify my question is more about theta than the constraints. The constraints where just an example. $\endgroup$
    – ght
    Commented Oct 31, 2017 at 22:43

Using $\theta$ variation in place of variance have the following effect:

  • if you want to be protected “against” super-diffusivity (ie dynamics which variance / risk grows faster than linearly with time), use a $\theta>1$
  • if you believe dynamics you are working against are sub-diffusive, use $\theta<1$

The later case can be reallistic for intraday dynamics (negative auto-correlations, or presence of mean-reversion), see for instance Optimal starting times, stopping times and risk measures for algorithmic trading (by Labadie and L), while the former can be good if you fear some “bubbles”.

  • $\begingroup$ If $\theta$ is less than 1, the problem is no longer guaranteed to be convex and may be more difficult to solve. The penalty term is essentially $\left( \| \mathbf{x} \|_2 \right) ^\theta$. $\endgroup$ Commented Nov 6, 2017 at 17:54
  • $\begingroup$ Is what your referring to more in the spirit of $\operatorname{E}[(r-\mu)^\theta]$ rather than $\sqrt{\operatorname{E}[(r-\mu)^2]}^\theta$ (which is what the OP is doing...)? $\endgroup$ Commented Nov 6, 2017 at 20:07
  • $\begingroup$ What I have in mind is to write $\mathbb{V}(r-\mu)^\theta$ @MatthewGunn $\endgroup$
    – lehalle
    Commented Nov 6, 2017 at 20:36

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