Convexity of the variance of a mixture distribution

Question

$X$ is a random variable that is sampled from the mixture of uniform distributions. In other words: $$X \sim \sum_{i=1}^N w_i \cdot \mathbb{U}(x_i, x_{i+1}),$$ where $\mathbb{U}(x_i, x_{i+1})$ denotes a random variable that follows a uniform distribution in $[x_i, x_{i+1}]$. For feasibility, we need $w \geq 0, \ \sum_{i=1}^N w_i = 1$.

In an optimization problem my variables are $w_i$ for $i=1,\ldots,N$, and I would like to upper bound the variance of $X$. According to Wikipedia, the variance of $X$ is: $$\mathrm{Var}(X) = \sum_{i=1}^N w_i(\sigma_i^2+ \mu_i^2 - \mu^2) $$ where $\sigma_i^2$ and $\mu_i$ are the variance and mean of $\mathbb{U}(x_i, x_{i+1})$, respectively (which are parameters), and $\mu$ is the mean of the mixture, which is $$\mu = \sum_{i=1}^N w_i \frac{x_i + x_{i+1}}{2}.$$

Thus, if my derivation is not wrong: $$ \mathrm{Var}(X) = \sum_{i=1}^N w_i\left(\sigma_i^2+ \mu_i^2 - \left(\sum_{j=1}^N w_j \frac{x_j + x_{j+1}}{2}\right)^2\right) $$ which is very ugly and appears to be non-convex to upper bound this function (edit: I want to constrain $\mathrm{Var}(X) \leq \mathrm{constant}$).

My question is, is there any trick, or any other convex approximation of such a variance, such that I can include an upper bound on the variance constraint?

Answer 1

In order to find the best upper bound for variance, for given input values of $u_i$ and $\sigma_i^2$, you should globally maximize variance with respect to the $w_i$, subject to the constraints $w_i \ge 0, \Sigma w_i = 1$.

This can be formulated as a convex QP (Quadratic Programming problem), i.e., maximizing a concave quadratic subject to linear constraints. Hence it is easy to solve, unless $n$ is gigantic, which hardly seems likely for any reasonable mixture distribution. I leave to the OP as an exercise, whether the KKT conditions can yield a closed form solution.

The convex QP takes the form:

maximize $(\Sigma_{i=1}^n w_i (\sigma_i^2 +\mu_i^2)) - \mu^2$ with respect to $\mu, w_i$

subject to $\Sigma_{i=1}^n w_i \mu_i = \mu, w_i \ge 0 \forall i, \Sigma_{i=1}^n w_i = 1$.

If all $u_i$ are equal to each other, this would be a Linear Programming problem with compact constraints. Therefore the optimum would be at a vertex of the constraints, and in this case, that vertex would be $w_i = 1$ for the $i$ corresponding to the largest $\sigma_i^2$, and all other $w_i = 0$.

Edit: In response to edit to question: "I want to constrain Var(X) $\le$ constant)"

If the naive approach of adding the constraint Var(X) $\le $ constant to my above convex QP formulation were performed, that would add a non-convex quadratic constraint, making the problem a non-convex Quadratically-Constrained Quadratic Program (QCQP), which requires a global optimizer, such as Gurobi 9.x or BARON to solve to global optimality.

However, there is an easier, faster method: Solve the (pre-Edit) convex QP formulation. Then maximum variance, accounting for the constraint Var(X) $\le$ constant), equals

min(optimal objective value of convex QP formulation,constant).