a case where ${\boldsymbol{M}}$ is not sufficiently scattered, and it also happens to be a case where ${\boldsymbol{M}}$ is not identifiable.

As we can see, the elliptical cone ${\mathcal{C}}^*$ is tangent to all the facets of the nonnegative orthant. As a result, for ${\boldsymbol{M}}$ to be sufficiently scattered, it is necessary that there are enough rows of ${\boldsymbol{M}}$ lie on the boundary of the nonnegative orthant, i.e., ${\boldsymbol{M}}$ is relatively sparse. Specifically, if ${\boldsymbol{M}}$ is sufficiently scattered, then each column of ${\boldsymbol{M}}$ contains at least $K-1$ zeros [@huang2014tsp]. This is a very important insight, as exactly checking whether a matrix is sufficiently scattered may be computationally hard. In the present paper we further show the following result.

\[ppst:volume\] The ratio between the volume of the hyperball obtained by intersecting ${\boldsymbol{\mathit{1}}}^{{\!\top\!}}{\boldsymbol{x}}=1$ and ${\mathcal{C}}^*$ and the probability simplex is $$\label{eq:ratio}
\frac{1}{\sqrt{\pi K}}\left(\frac{4\pi}{K(K-1)}\right)^{\frac{K-1}{2}}
\Gamma\left(\frac{K}{2}\right).$$

The proof is given in the supplementary material. As $K$ grows larger, the volume ratio goes to zero at a super-exponential decay rate. This implies that the volume of the inner sphere quickly becomes negligible compared to the volume of the probability simplex, as $K$ becomes moderately large. The take home point is that, for a practical choice of 