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\title{Is the Volume of a Credal Set a Good Measure for Epistemic Uncertainty?\\(Supplementary Material)}

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\author[1,3]{\href{mailto:<yusuf.sale@ifi.lmu.de>?Subject=Your UAI 2023 paper}{Yusuf Sale}{}}
\author[2]{Michele Caprio}
\author[1,3]{Eyke Hüllermeier}
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\affil[1]{%
Institute of Informatics\\
University of Munich (LMU)\\
Germany
}

\affil[2]{%
    PRECISE Center\\
    Department of Computer and Information Science\\
    University of Pennsylvania\\
    USA
  }

\affil[3]{%
    Munich Center for Machine Learning\\
    Germany

}

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\appendix
\section{Proofs}\label{proofs}
\begin{proof}[Proof of Proposition \ref{prop-1}]
Let $\mathcal{P},\mathcal{Q} \subset \Delta(\mathcal{Y},\sigma(\mathcal{Y}))$ be credal sets, and assume $|\mathcal{Y}|=2$. Then we have the following.
    \begin{itemize}
        \item $\text{Vol}(\mathcal{P}) \geq 0$ and $\text{Vol}(\mathcal{P}) \leq \text{Vol}(\Delta^{2-1})=\sqrt{2}$. Hence $\text{Vol}(\cdot)$ satisfies A1.
        \item The volume being a continuous functional is a well-known fact that comes from the continuity of the Lebesgue measure, so $\text{Vol}(\cdot)$ satisfies A2.
        \item $\mathcal{Q}\subset \mathcal{P} \implies \text{Vol}(\mathcal{Q}) \leq \text{Vol}(\mathcal{P})$. This comes from the fundamental property of the Lebesgue measure, so $\text{Vol}(\cdot)$ satisfies A3.
        \item Consider a sequence $(\mathcal{P}_n)$ of credal sets on $(\mathcal{Y},\sigma(\mathcal{Y}))$ such that $\lim_{n\rightarrow \infty} [\overline{P}_n(A)-\underline{P}_n(A)]=0$, for all $A\in\sigma(\mathcal{Y})$. Then, this means that there exists $N\in\mathbb{N}$ such that for all $n\geq N$, the geometric representation of $\mathcal{P}_n$ is a subset of the geometric representation of $\mathcal{P}_{n+1}$. In addition, the limiting element of $(\mathcal{P}_n)$ is a (multi)set $\mathcal{P}^\star$ whose elements are all equal to $P^\star$, so its geometric representation is a point and its volume is $0$. Hence, probability consistency is implied by continuity A3, so $\text{Vol}(\cdot)$ satisfies A4'.
        \item The volume is invariant to rotation and translation. This is a well-known fact that comes from the fundamental property of the Lebesgue measure, so $\text{Vol}(\cdot)$ satisfies A7.
    \end{itemize}
Let us now show that the volume operator satisfies sub-additivity A5. Let $\mathcal{Y}=\mathcal{Y}_1\times\mathcal{Y}_2$. In addition, suppose we are in the general case in which $|\mathcal{Y}|=|\mathcal{Y}_1|=|\mathcal{Y}_2|=2$. In particular, let $\mathcal{Y}=\{(y_{1},y_{2}),(y_{3},y_{4})\}$, so that $\mathcal{Y}_1=\{y_{1},y_{3}\}$ and $\mathcal{Y}_2=\{y_{2},y_{4}\}$. Suppose also $y_1\neq y_3$ and $y_2\neq y_4$. Now, pick any probability measure $P$ on $\mathcal{Y}$. In general, we would have that its marginal $\text{marg}_{\mathcal{Y}_1}(P)=P^\prime$ on $\mathcal{Y}_1$ is such that  $P^\prime(y_i)=\sum_j P((y_i,y_j))$. Similarly for marginal $\text{marg}_{\mathcal{Y}_2}(P)=P^{\prime\prime}$ on $\mathcal{Y}_2$. In our case, though, the computation is easier. To see this, fix $y_1$. Then, we should sum over $j$ the probability of $(y_1,y_j)$, $y_j\in\mathcal{Y}_2$. But the only pair $(y_1,y_j)$ is $(y_1,y_2)$. A similar argument holds if we fix $y_3$, or any of the elements of $\mathcal{Y}_2$. Hence, we have that
$$P^\prime(y_1)=P((y_1,y_2))=P^{\prime\prime}(y_2) \quad \text{and} \quad P^\prime(y_3)=P((y_3,y_4))=P^{\prime\prime}(y_4).$$

Let $\mathcal{P}^\prime$ and $\mathcal{P}^{\prime\prime}$ denote the marginal convex sets of probability distributions on $\mathcal{Y}_1$ and $\mathcal{Y}_2$, respectively, and let $\mathcal{P}$ denote the convex set of joint probability distributions on $\mathcal{Y}=\mathcal{Y}_1\times\mathcal{Y}_2$ \citep{couso}. Then, given our argument above, we have that $\text{Vol}(\mathcal{P}) < \text{Vol}(\mathcal{P}^\prime) + \text{Vol}(\mathcal{P}^{\prime\prime}) = 2 \text{Vol}(\mathcal{P})$. So in the general $|\mathcal{Y}|=|\mathcal{Y}_1|=|\mathcal{Y}_2|=2$ case where $y_1\neq y_3$ and $y_2\neq y_4$, the volume is subadditive.
\end{proof}


\begin{proof}[Proof of Proposition \ref{prop-2}]
Immediate from the assumption on the instance of SI.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main_theorem}]
Pick any compact set $\mathcal{P}\subset\mathcal{M}(\Omega,\mathcal{F})$ and any set $\mathcal{Q}$ satisfying (a)-(c). Let $B^d_r \subset \mathbb{R}^d$ denote a generic ball in $\mathbb{R}^d$ of radius $r>0$. Notice that $N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)=N^{\text{pack}}_{r-\epsilon}(\mathcal{P})-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}) \geq 0$ because $\mathcal{P}\supset\mathcal{Q}$. Then, the proof goes as follows
    \begin{align}
        \frac{\text{Vol}(\mathcal{P})-\text{Vol}(\mathcal{Q}^\prime)}{\text{Vol}(\mathcal{P})} &= \frac{\frac{1}{c(r,d,\mathcal{P})}\text{Vol}(\tilde{\mathcal{P}}_r)-\frac{1}{c(r-\epsilon,d,\mathcal{Q}^\prime)}\text{Vol}(\tilde{\mathcal{Q}}^\prime_{r-\epsilon})}{\frac{1}{c(r,d,\mathcal{P})}\text{Vol}(\tilde{\mathcal{P}}_r)} \label{eq4}\\
        &\geq \frac{\text{Vol}(\tilde{\mathcal{P}}_r)-\text{Vol}(\tilde{\mathcal{Q}}^\prime_{r-\epsilon})}{\text{Vol}(\tilde{\mathcal{P}}_r)} \label{eq5}\\
        &=\frac{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_r)-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)\text{Vol}(B^d_{r-\epsilon})}{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_r)} \label{eq6}\\
        &=\frac{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_1)r^d-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)\text{Vol}(B^d_1)(r-\epsilon)^d}{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_1)r^d} \label{eq7}\\
        &=\frac{N^{\text{pack}}_r(\mathcal{P})r^d-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)(r-\epsilon)^d}{N^{\text{pack}}_r(\mathcal{P})r^d} \nonumber\\
        &=1-\frac{N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)}{N^{\text{pack}}_r(\mathcal{P})}\left(1-\frac{\epsilon}{r} \right)^d \nonumber\\
        &\geq 1- \left( 1-\frac{\epsilon}{r} \right)^d, \label{eq8}
    \end{align}
    where \eqref{eq4} comes from equation \eqref{volume_eq}, \eqref{eq5} comes from the fact that $r-\epsilon \leq r \implies c(r-\epsilon,d,\mathcal{Q}^\prime) \geq c(r,d,\mathcal{P})$ by \eqref{decr_c}, \eqref{eq6} comes from $\tilde{\mathcal{P}}_r$ being the union of pairwise disjoint balls of radius $r$, \eqref{eq7} comes from properties of the volume of a ball of radius $r$ in $\mathbb{R}^d$, and \eqref{eq8} comes from property (c) of $\mathcal{Q}$.
\end{proof}

\section{High-dimensional probability}\label{hdp}
 Since Theorem \ref{main_theorem} in Section \ref{instab} is intimately related with Carl-Pajor's Theorem \citep{pajor}, we state (a version) of the theorem here. 
 
\begin{theorem}[Carl-Pajor]
Let $B_{1,d}$ denote the $d$-dimensional unit euclidean ball, and let $\mathcal{P} \subset B_{1,d}$ be a polytope with $m \in \mathbb{N}$ vertices. Then, we have
\begin{align}
  \frac{\text{Vol}(\mathcal{P})}{\text{Vol}(B_{1,d})} \leq \left(4 \sqrt{\frac{\log m}{d}}   \right)^d. 
\end{align}
\end{theorem}
For further results connecting high-dimensional probability and data science, see \cite{vershynin2018high}.



% \appendix
% \section{Proofs}
% \subsection{Proof of Theorem 1}
% Pick any compact set $\mathcal{P}\subset\mathcal{M}(\Omega,\mathcal{F})$ and any set $\mathcal{Q}$ satisfying (a)-(c). Let $B^d_r \subset \mathbb{R}^d$ denote a generic ball in $\mathbb{R}^d$ of radius $r>0$. Notice that $N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)=N^{\text{pack}}_{r-\epsilon}(\mathcal{P})-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}) \geq 0$ because $\mathcal{P}\supset\mathcal{Q}$. Then, the proof goes as follows
%     \begin{align}
%         \frac{\text{Vol}(\mathcal{P})-\text{Vol}(\mathcal{Q}^\prime)}{\text{Vol}(\mathcal{P})} &= \frac{\frac{1}{c(r,d,\mathcal{P})}\text{Vol}(\tilde{\mathcal{P}}_r)-\frac{1}{c(r-\epsilon,d,\mathcal{Q}^\prime)}\text{Vol}(\tilde{\mathcal{Q}}^\prime_{r-\epsilon})}{\frac{1}{c(r,d,\mathcal{P})}\text{Vol}(\tilde{\mathcal{P}}_r)} \label{eq4}\\
%         &\geq \frac{\text{Vol}(\tilde{\mathcal{P}}_r)-\text{Vol}(\tilde{\mathcal{Q}}^\prime_{r-\epsilon})}{\text{Vol}(\tilde{\mathcal{P}}_r)} \label{eq5}\\
%         &=\frac{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_r)-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)\text{Vol}(B^d_{r-\epsilon})}{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_r)} \label{eq6}\\
%         &=\frac{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_1)r^d-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)\text{Vol}(B^d_1)(r-\epsilon)^d}{N^{\text{pack}}_r(\mathcal{P})\text{Vol}(B^d_1)r^d} \label{eq7}\\
%         &=\frac{N^{\text{pack}}_r(\mathcal{P})r^d-N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)(r-\epsilon)^d}{N^{\text{pack}}_r(\mathcal{P})r^d} \nonumber\\
%         &=1-\frac{N^{\text{pack}}_{r-\epsilon}(\mathcal{Q}^\prime)}{N^{\text{pack}}_r(\mathcal{P})}\left(1-\frac{\epsilon}{r} \right)^d \nonumber\\
%         &\geq 1- \left( 1-\frac{\epsilon}{r} \right)^d, \label{eq8}
%     \end{align}
%     where \eqref{eq4} comes from equation \eqref{volume_eq}, \eqref{eq5} comes from the fact that $r-\epsilon \leq r \implies c(r-\epsilon,d,\mathcal{Q}^\prime) \geq c(r,d,\mathcal{P})$ by \eqref{decr_c}, \eqref{eq6} comes from $\tilde{\mathcal{P}}_r$ being the union of pairwise disjoint balls of radius $r$, \eqref{eq7} comes from properties of the volume of a ball of radius $r$ in $\mathbb{R}^d$, and \eqref{eq8} comes from property (c) of $\mathcal{Q}$.


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