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

Question

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}

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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.

Answer 1

Try to check whether the variance is a convex function of $X$.

Answer 2

Can you check, for some nontrivial set of $a_i$, and $\alpha=0$, whether the vertices of $\Delta^n$ lie in $S$?

Answer 3

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