# Prove or disprove that the set of probabilities $\{p \in \Delta^n \mid \mbox{Var}(X) \leq \alpha\}$ is a convex set.

Prove or disprove that the set $$S:= \left\{p \in \Delta^n \mid \mbox{Var}(X) \leq \alpha \right\}$$ is convex. $$p \in \Delta^n$$ denotes that $$p$$ is a probability distribution (i.e., belongs to the probability simplex). $$X$$ is a random variable such that $$p(X = a_i) = p_i$$, where $$a_1 \leq a_2 \leq \cdots \leq a_n\in \mathbb{R}$$.

The question was taken from the exercise 2.15 of Boyd & Vandenberghe's Convex Optimization

My current attempts are to pick two arbitrary points from $$S$$ and show that the convex combination stays in $$S$$. I realized that I could use the formula $$\mbox{Var}(X) = E[(X - E(X))^2]$$, but I don't know how to proceed because the $$(\cdot)^2$$ in the variance prevents me from distributing the convex combination inside $$(\cdot)^2$$. Also, $$E(X)$$ in $$X - E(X)$$ depends on the probability distribution to provide.

Could anyone provide some pointers? I have been stuck for two days.

Try to check whether the variance is a convex function of $$X$$.
• Hi, I am a bit confused about why "variance" could/couldn't be a convex function of $X$. In particular, proving a function is convex often requires us to take two points on the convex function and show that the convex combination of the two points satisfies the Jenson's Inequality, but $X$ is a random variable, how could we take two points?
• You should check whether $\mathbf{var}(X)$ is convex in $p$. A convex function has sublevel sets which are convex sets. In this case your set is in terms of $p$ so you need to check whether variance is a convex function of $p$ Sep 23 '18 at 22:18
Can you check, for some nontrivial set of $$a_i$$, and $$\alpha=0$$, whether the vertices of $$\Delta^n$$ lie in $$S$$?
To get an intuitive feel, maybe think about some distributions that you know the variance of (i.e. if $$X$$ is Bernoulli then $$\mathbf{var}(X)= p(1-p)$$) then check if it's convex. If you find one counter example then you're done. If the statement is true this won't prove it but the intuition might help