The shape of a Gaussian mixture is characterized by the probability density of the distance between two samples

TL;DR

This paper demonstrates that for Gaussian mixture models with identity covariance, the shape of the distribution can be uniquely reconstructed from the distribution of distances between two independent samples, up to a rigid motion.

Contribution

It establishes that Gaussian mixtures with identity covariance are uniquely determined by the distribution of inter-sample distances, up to Euclidean transformations.

Findings

Shape of Gaussian mixtures is recoverable from distance distributions.

Two Gaussian mixtures with the same distance distribution differ only by a rigid motion.

Results extend to distances defined by symmetric bilinear forms.

Abstract

Let $\bf{x}$ be a random variable with density $\rho(x)$ taking values in ${\mathbb R}^d$. We are interested in finding a representation for the shape of $\rho(x)$, i.e. for the orbit $\{ \rho(g\cdot x) | g\in E(d) \}$ of $\rho$ under the Euclidean group. Let $x_1$ and $x_2$ be two random samples picked, independently, following $\rho(x)$, and let $\Delta$ be the squared Euclidean distance between $x_1$ and $x_2$. We show, if $\rho(x)$ is a mixture of Gaussians whose covariance matrix is the identity, and if the means of the Gaussians are in generic position, then the density $\rho(x)$ is reconstructible, up to a rigid motion in $E(d)$, from the density of $\bf{\Delta}$. In other words, any two such Gaussian mixtures $\rho(x)$ and $\bar{\rho} (x)$ with the same distribution of distances are guaranteed to be related by a rigid motion $g\in E(d)$ as $\rho(x)=\bar{\rho} (g\cdot x)$. We also show that a similar result holds when the distance is defined by a symmetric bilinear form.