3
$\begingroup$

Let $\vec{\mu}_1, \vec{\mu}_2,\ldots, \vec{\mu}_k \in \Delta^{d-1}$ be $k\ (k\geq 2)$ distinct vectors on the standard simplex, where $$\Delta^{d-1} = \{\vec{\mu}\in R^{d}:\| \vec{\mu}\|_1 = 1,\mu_j \geq 0\}.$$ (Here $\mu_j$ means the $j$-th entry of $\vec{\mu} \in R^d$.) We define $\Delta^{k-1}$ similarly.

Our goal is to find a $\vec{\mu}^* \in \Delta^{d-1}$ that satisfies $\vec{\mu}^* \notin \{\vec{\mu}_1,\ldots, \vec{\mu}_k\}$, such that for some $\vec{w} \in \Delta^{k-1}$,
$$ \sum_{i} w_i \vec{\mu}_i = \vec{\mu}^*, $$ and $$ \sum_{i} w_i \vec{\mu}_i^{\otimes 2} = (\vec{\mu}^*)^{\otimes 2}. $$ (Here $\vec{\mu}_i^{\otimes 2}$ means $\vec{\mu}_i\vec{\mu}_i^\top$.) Equivalently, one can think of constructing $\vec{\mu}_1, \vec{\mu}_2,\ldots, \vec{\mu}_k$ for a given $\vec{\mu}^*$.

Here is the question: Is there a principled way to do this construction? Or does there exist a solution at all?

In the simplest example where $k=2$ and $\vec{\mu}^* = (1/d,\ldots,1/d)^\top$, it seems that the solution doesn't exist. So what about general settings?

A variant of this problem is that we want to match $$ \sum_{i} w_i \vec{\mu}_i = \sum_{i} w_i^* \vec{\mu}_i^*, $$ and $$ \sum_{i} w_i \vec{\mu}_i^{\otimes 2} = \sum_{i} w_i^* (\vec{\mu}_i^*)^{\otimes 2}, $$ where $\vec{w}^*\in\Delta^{k-1}$, $\vec{w}^*\neq \vec{w}$, $\vec{\mu}_1^*,\ldots, \vec{\mu}_k^* \in \Delta^{d-1}$, and $\vec{\mu}_1^*,\ldots, \vec{\mu}_k^* \notin \{\vec{\mu}_1,\ldots, \vec{\mu}_k\}$.

$\endgroup$
2
  • $\begingroup$ What does the notation $\vec{v}^{\otimes 2}$ mean? $\endgroup$
    – usul
    Mar 1, 2016 at 1:03
  • $\begingroup$ @usul It means $\vec{v}\vec{v}^\top$. I have edited the notation to make it more clear. $\endgroup$
    – Minkov
    Mar 1, 2016 at 1:20

1 Answer 1

3
$\begingroup$

It is a standard result that the matrices of the form $\mu^{\otimes 2}$ for nonzero $\mu$ are the extreme rays of the positive semidefinite cone. That is to say, your condition on the second moments implies that $\mu^{\otimes 2}$ is a scalar multiple of $\mu_i^{\otimes 2}$ for some $i$. The normalization then gives $\mu = \mu_i$. So the there never exists a solution to the main question.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for the answer. Do you know whether the variant of the problem admits a solution? $\endgroup$
    – Minkov
    Mar 1, 2016 at 7:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.