Suppose $x\in\mathbb{R}^n$ is a random variable with mean $\mu$ and covariance $ \Sigma$. Consider a stochastic convex optimization problem, i.e. an optimization problem with chance constraints, meaning there is a small, but finite probability, $\Delta\leq 0.5$, of violating the constraints.

In all of the cases I've encountered so far, you assume that the constraint space, $\mathcal{X}$, is a polytope, meaning it can be written as

$$ \mathcal{X} \triangleq \bigcap_{j=1}^{M} \ \{x:\alpha_j^\intercal x \leq \beta_j\} $$

Qualitatively, this represents a finite intersection of linear inequality constraints, which is a convex region. In 2D, this is simply a polygon, with $M$ vertices. For example, if $M = 3$, then the intersection of the three lines would form a triangle. If $M = 4$, this would be a square, and so on. The reason people assume the constraint space is a convex polytope is because, using Boole's inequality (which gives an upper bound on the union of sets), the chance constraints can be written as

$$ \begin{align} \text{Pr}(x\notin&\mathcal{X}) \leq \Delta\\ &\Updownarrow\\ \text{Pr}(\alpha_j^\intercal x \leq \beta_j) &\geq 1 - \delta_j, \ \forall j = 1,...,M\\ \sum_{j=1}^{M} \delta_j &\leq \Delta, \end{align} $$

where the joint probability of violating the constraints is split up into the individual probability of violating each $j$th constraint. This is extremely useful, because the second expression is nothing more than the probability of a random variable with mean $\alpha_j^\intercal \mu$ and covariance $\alpha_j^\intercal \Sigma \alpha_j$. Thus, this probability can be written in terms of the standard normal CDF ($\Phi$) as

$$ \Phi\Bigg[\frac{\beta_j - \alpha_j^\intercal \mu}{\sqrt{\alpha_j^\intercal \Sigma \alpha_j}} \Bigg] \geq 1 - \delta_j \Rightarrow \alpha_j^\intercal \mu + \|\Sigma^{1/2} \alpha_j\|^2 \Phi^{-1}(1-\delta_j) \leq \beta_j, $$ since $\Sigma > 0$ is always positive definite, as it represents a standard deviation. The above inequality constraint is a second order cone constraint, and the resulting optimization problem is a SOCP.

However, what if the constraint space is now not an polytope (or polygon), but rather a cone, specifically a convex cone. In that case, $\mathcal{X}$ would be defined as

$$ \mathcal{X} = \{x : \|Ax+b\|_2 \leq c^\intercal x + d\}. $$ Is it possible, in any way, to calculate $\text{Pr}(x\notin\mathcal{X})$, or something like that as in the case of a polytope? You would have to make some kind of approximation or relaxation, such as Markov's inequality or Chebyshev inequality, to get rid of the probability and turn it into an expectation. However, I can't seem to figure out a solution. For my purposes, the cone is centered at the origin, so $b = d = 0$ if that makes it simpler to work with. This type of constraint is more natural in a physical setting, especially in controls, where you want to steer distributions from some initial $x\sim\mathcal{N}(\mu_0,\Sigma_0)$, to the origin for example.

I haven't found any other literature on this subject, so if anyone has any insights, it would be appreciated!

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    $\begingroup$ Are you assuming that the random variable $x$ is normally distributed with mean $\mu$ and covariance $\Sigma$? It is not clear from the post, but the derivation for the polyhedral case only works under this assumption as far as I'm aware $\endgroup$ Apr 2, 2020 at 22:19
  • $\begingroup$ Yea, I'm assuming exactly that. $\endgroup$ Apr 2, 2020 at 22:38
  • $\begingroup$ Another thing that is confusing me is what the decision variables and random variables are in your formulation... You say that $x$ is a random variable, but also state that $\mathcal{X}$ is the feasible region (implying that $x$ is the decision variable). Are $\alpha$ and $\beta$ actually the decision variables for the polyhedral case? Is $x$ the random variable and $A$, $b$, $c$, and $d$ the decision variables for the SOCP setting? $\endgroup$ Apr 2, 2020 at 23:32

1 Answer 1


This problem of deriving a tractable approximation to $\text{Pr}(\left\lVert Ax + b \right\rVert \leq c^{\text{T}} x + d) \geq 1 - \Delta$ is known to be computationally intractable in general, see this paper for reference. Section 2.2 of the above paper also provides a tractable approximation to this chance constraint for your setting.

Edit: Section 6.1 of this paper shows that SOCP-based chance constraints with normal random variables may be nonconvex.

Edit 2: Another approach for constructing a convex approximation of the chance-constraint is to use scenario approximation, which can yield some feasibility guarantees. In this approach, you basically sample the random variable many times and replace the chance constraint by the set of sampled random constraints (which are enforced deterministically). If you tune the number of samples, you can get solutions that satisfy the original chance constraint with high probability.

  • $\begingroup$ I don't think I necessarily agree with that. I see your confusion from the previous comment, let me try to explain a little bit better. I am considering a control problem where the position of the body, $\mathbf{x} = [x,y,z]^\intercal$ is part of the state vector. The state is a random variable, and the whole problem is that it starts with some initial Gaussian distribution $\mathbf{x}_0 \sim \mathcal{N}(\mu_0,\Sigma_0)$ and must end at some final distribution $\mathbf{x}_f \sim \mathcal{N}(\mu_f,\Sigma_f)$. The constraints on the state are that they must remain with a convex cone. $\endgroup$ Apr 3, 2020 at 1:32
  • $\begingroup$ Additionally, the system is stochastic, meaning there is noise in it. So, I have to formulate the constraints as chance constraints. In the case of a polytope region, as I mentioned in the problem statement, it is common to turn the chance constraints into second-order cone constraints. But for the case of a convex cone, I am almost sure there has to be a way to approximate or simplify the chance constraints. So in the chance constraints, there are no decision variables in them directly, but the state follows some dynamics $\dot{x} = Ax + Bu + Dw$, so the decision variables are the controls. $\endgroup$ Apr 3, 2020 at 1:35
  • $\begingroup$ @JoshPilipovsky In that case, the chance constraints will determine the feasible controls in your problem. Therefore, strictly speaking, you should also consider the equations that define how the controls affect the states within the chance constraint (I know that some authors in the control literature don't write it that way), i.e., something like $\text{Pr}_w(g(x,u) \leq 0, \: h(x,u) = 0) \geq 1 - \Delta$, where $g$ and $h$ are suitably defined. Anyway, I'm fairly confident my answer still stands - I will try to dig up a result on computational hardness $\endgroup$ Apr 3, 2020 at 2:11
  • $\begingroup$ If you plug-in the dependence of $x$ on $u$, then you still get a SOCP-type chance constraint on $u$. One reason I'm fairly certain that this is a hard problem is that there is a lot of literature on tractable approximations to chance constraints defined by convex inequalities. There are entire sections on the book on robust optimization that present tractable approximations to such chance constraints $\endgroup$ Apr 3, 2020 at 2:14
  • $\begingroup$ Thanks for the book, I'll look into it. So you don't think there is a way to approximate this analytically? I've just worked out a result using Markov's inequality, it leads to a quadratic constraint on the control (decision variable), so the whole problem becomes a QCQP, but it's not working when I try to solve on MATLAB. $\endgroup$ Apr 3, 2020 at 2:35

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