see that if the ground-truth ${\boldsymbol{\varTheta}}_\natural$ satisfies , solving  even without  will produce a solution ${\boldsymbol{\varTheta}}_\star$ that satisfies , thanks to uniqueness. In practice when we are given a less accurate ${\boldsymbol{\varOmega}}$, such “redundant” information will help us improve the estimation error, but that goes beyond identifiability consederations.

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The proof of Theorem \[thm:unique\] is referred to [@huang2016nips]. Here we provide some insights on this geometry-inspired sufficiently scattered condition, and discuss why it is a reasonable (and thus practical) assumption. The notation $\operatorname{cone}({\boldsymbol{M}}^{{\!\top\!}})^*=\{{\boldsymbol{x}}:{\boldsymbol{M}}{\boldsymbol{x}}\geq0\}$ comes from the convention in convex analysis that it is the *dual cone* of the conical hull of the row vectors of ${\boldsymbol{M}}$, i.e., $\operatorname{cone}({\boldsymbol{M}}^{{\!\top\!}})=\{{\boldsymbol{M}}^{{\!\top\!}}\bm{\alpha}:\bm{\alpha}\geq0\}$. Similarly, we can derive that the dual cone of ${\mathcal{C}}$ is ${\mathcal{C}}^*=\{{\boldsymbol{x}}:\|{\boldsymbol{x}}\|\leq{\boldsymbol{\mathit{1}}}^{{\!\top\!}}{\boldsymbol{x}}/\sqrt{K-1}\}$. A useful property of the dual cone is that for two convex cones $\mathcal{A}$ and $\mathcal{B}$, $\mathcal{A}\subseteq\mathcal{B}$ iff $\mathcal{B}^*\subseteq\mathcal{A}^*$. Therefore, the first requirement of sufficiently scattered in Definition \[def:suf\_scat\] equivalently means $${\mathcal{C}}^*\subseteq\operatorname{cone}({\boldsymbol{M}}^{{\!\top\!}}).$$ We give a geometric illustration of the sufficiently scattered condition in Figure \[fig:geo-scattered\] for $K=3$, and we focus on the 2-dimensional plane ${\boldsymbol{\mathit{1}}}^{{\!\top\!}}{\boldsymbol{x}}=1$. The intersection between this plane and the nonnegative orthant is the probability simplex, which is the triangle in Figure \[fig:geo-scattered\]. The outer circle 