Convex Hull of Outer Products of (Normalised) Nonnegative Vectors

Question

If I define $\mathcal{A} = \{ xx^T : x \in \mathbb{R}^d, \| x \|_2 \leqslant 1 \}$, then (assuming I recall correctly) it is known that the convex hull of $\mathcal{A}$ is given by

\begin{align} \text{conv} (\mathcal{A}) = \{ M \in \text{Mat}_{d \times d} (\mathbf{R}): M = M^T, \| M \|_* = 1\}, \end{align}

where $\| \cdot \|_*$ is the nuclear norm.

Suppose I now define $\mathcal{A}_+ = \{ xx^T : x \in \mathbb{R}_+^d, \| x \|_2 \leqslant 1\}$, i.e. I restrict to taking outer products of nonnegative vectors $x$ with themselves.

My question is: Does there exist a similar characterisation of $\text{conv} (\mathcal{A}_+)$ ?

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If context is useful: I'm interested in understanding and characterising when it is possible to write a symmetric probability distributions on two variables as a mixture of i.i.d. distributions, i.e. if $p(x,y) = p(y, x)$, when does there exist

• some parameter space $\Theta$,
• some family of distributions $\{ p(\cdot | \theta) \}_{\theta \in \Theta}$, and
• some (prior) distribution $\pi$ such that

\begin{align} p(x, y) = \int_{\Theta} p(x|\theta) \, p(y|\theta) \, \pi(\theta) \, d\theta? \end{align}

I recognise that this has some links to de Finetti's theorem, but I'm not yet sure whether that link can be turned into an answer.

Answer 1

Your characterization of $\text{conv} (\mathcal{A})$ needs one additional restriction---that $M$ is positive semidefinite (the equivalence of these two sets follows fairly quickly from the spectral decomposition).

For $\text{conv} (\mathcal{A}_+)$, the convex hull is the exact same, but with the positive semidefiniteness requirement replaced by a complete positivity requirement. Since checking complete positivity is NP-hard, don't expect an easy way of determining membership in this convex hull.