
\appendix
\section{Additional Supporting lemmas and proofs}
\subsection{Proof of Remark~\ref{remark:gneiting}}
\begin{remark}
~\label{remark:gneiting}
    Scoring rule $s$ is (strictly) proper if and only if the corresponding (strictly) convex function $G(q)= \sum_{o\in\cO} s(q,o)q(o)$ 
\end{remark}
\begin{proof}
It follows from Theorem ~\ref{theorem:gneiting} that regular scoring rule $s$ is (strictly) proper if and only if there exists a corresponding (strictly) convex function $G$ on $\Delta(\cO)$ such that 
\begin{align}
    s(q,o)= G(q) - \sum_{o\in\cO} G'(q)(o) q(o) + G'(q)(o).
\end{align}
$(\Rightarrow)$ 
Let us assume that there exists a strictly proper scoring rule $s$. Then, according to Theorem ~\ref{theorem:gneiting} there exists a convex function $G:\Delta(\cO)\rightarrow \mathbb{R}$ such that 
\begin{align*}
    s(q,o) &= G(q) -  \sum_{o\in\cO} G'(q)(o) dq(o) + G'(q)(o) \\
    \mathbb{E}_{o\sim p}[s(q,o)] &= \mathbb{E}_{o\sim p}\bigg[G(q) -  \sum_{o\in\cO} G'(q)(o) dp(o) + G'(q)(o)\bigg] \tag*{(expert's true belief \mbox{$p$})} \\
    &= G(q) -  \sum_{o\in\cO} G'(q)(o) q(o) +  \sum_{o\in\cO} G'(q)(o) p(o),
\end{align*}
where $q$ is the true belief of the forecaster. Then, we consider the maximum expected score
\begin{align*}
     \sum_{o\in\cO} s(q,o)q(o) &:= \max_{p\in \Delta(\cO)} u_q(p)\tag*{(\mbox{$s$} is strictly proper)}\\
     &= \max_{p\in \Delta(\cO)} \mathbb{E}_{o\sim q}[s(p,o)]\\
    &= \max_{p\in \Delta(\cO)} \left\{G(p) -  \sum_{o\in\cO} G'(p)(o) p(o) +  \sum_{o\in\cO} G'(p)(o) q(o) \right\} \\
    &=G(p^*) -  \sum_{o\in\cO} G'(p^*)(o) p^*(o) +  \sum_{o\in\cO} G'(p^*)(o) q(o) \tag*{(\mbox{$p^*$} is the maximizer)}\\
    &= G(q) -  \sum_{o\in\cO} G'(q)(o) q(o) +  \sum_{o\in\cO} G'(q)(o) q(o) \tag*{(\mbox{$s$} is strictly proper so \mbox{$p^*=q$})}\\
    &= G(q).
\end{align*}

$(\Leftarrow)$

We define the (strictly) convex function $G$ using the expected score of some scoring rule $s$, i.e., $G(p)= \sum_{o\in\cO} s(p,o)p(o)$ and the subgradient $G'(p)=s(p,o)$. Then,
\begin{align*}
    G(p) - \sum_{o\in\cO} G'(p)(o) p(o) + G'(p)(o) &:=   \sum_{o\in\cO} s(p,o)p(o) -  \sum_{o\in\cO} s(p,o)p(o) + s(p,o)\\
                                    &= s(p,o)\\ 
\end{align*}
This implies that $s$ is a strictly proper scoring rule as a consequence of Theorem~\ref{theorem:gneiting}.
\end{proof}
\subsection{On existence of extreme points of $\cP$}
\label{appendix:existence-of-extreme-points}
If the set of probability distributions $\cP$ is infinite, the extreme points may not always exist. Therefore, we need to identify the conditions under which $\cP$ has a valid set of extreme points. We argue that for extreme points to exist for $\cP$, $\co(\cP)$ must equal $\co(\overline{\cP})$, where $\overline{\cP}$ is the closure of set $\cP$. 
\begin{table}[h]
    \centering
    \caption{A toy example on existence of $\ext(\cP)$ for $\cP$ made using $p_1,p_2\in\Delta(\cO)$}
    \begin{tabular}{c|c|c|c}
          \hline
          $\cP$ & Closed? & $ext(\cP)$ exists? & $\co(\cP)=\co(\overline{\cP})$\\
          \hline 
          $\{p_1,p_2\}$ & closed & yes & yes \\
         $\{p| p:=wp_1+(1-w)p_2\text{ }\forall w \in[0,1]-\{\frac{1}{2}\}\}$ & open & yes & yes\\
         $\{p| p:=wp_1+(1-w)p_2\text{ }\forall w \in(0,1)\}$ & open & no & no\\
         $\{p| p:=wp_1+(1-w)p_2\text{ }\forall w \in[0,1]\}$ & closed & yes & yes\\
         \hline 
    \end{tabular}
    \label{tab:my_label}
\end{table}

To show this we precisely define the extreme points of $\cP$, independent of the convex hull of $\cP$ as follows  
\begin{definition}
    Given a set $\cP$, we define the extreme points as $\ext(\cP)$ as the collection of $p\in \cP$ for which there does not exist a set of points $C\subseteq\cP\setminus\{p\}$ and a probability measure $w:\Delta\rightarrow[0,1]$ such that $p=\int_{C} w(q)dq$.
\end{definition}

For extreme points of a set to exist in general spaces, its convex hull must be compact according to Choquet's  Theorem~\cite{bishop1959representations}. To establish compactness of $\cP$ we first show that with an appropriate notion of distance $\Delta(\cO)$ can form a bounded metric space.  
\begin{proposition}
\label{prop:boundedspace}
    The metric space $(\Delta(\cO),d_{TV})$ is bounded, where $d_{TV}$ denotes the total variational distance between two probability distributions $p,q$ in terms of their corresponding probability measures $P,Q$ is defined as 
    \begin{equation*}
        d_{TV}(p,q):= \sup_{A\subseteq\cO}|P(A)-Q(A)|=\frac{1}{2}\int|p(o)-q(o)|do
    \end{equation*}
\end{proposition}
\begin{proof}
    As defined above, the total variational distance is half the L1 distance ~\cite{levin2017markov}. This allows us to express the total variation distance directly using densities. To show $(\Delta(\cO),d_{TV})$ is bounded, let $p,q$ be arbitrary distributions in $\Delta(\cO)$. 
    Then 
    \begin{align*}
        d_{TV}(p,q)&:= \frac{1}{2}\int|p(o)-q(o)|do\\
                    &< \frac{1}{2}\int |p(o)|+|-q(o)|do && (\text{ Triangle Inequality})\\
                    &= \frac{1}{2}\int |p(o)|do + \frac{1}{2}\int |q(o)| do \\
                    &= \frac{1}{2}\int p(o) do + \frac{1}{2}\int q(o) do && (p(o)\geq 0\text{ and } q(o)\geq 0)\\
                    &= \frac{1}{2}+\frac{1}{2} = 1
    \end{align*}
    Thus $(\Delta(\cO),d_{TV})$ is a bounded metric space.  
\end{proof}

We now discuss the conditions on $\cP$ such that $\ext(\cP)\subseteq\cP$. 
\begin{proposition}
\label{prop:whatconditionsonPforgoodbehaviour}
    There exists a probability measure $w\in\Delta(\ext(\cP))$ for all $p\in\cP$ such that 
    \begin{equation*}
        p = \int_{p\in\ext(\cP)} w(p)dp
    \end{equation*}
    iff $\co(\cP)=\co(\overline{\cP})$, where $\co(\overline{\cP})$ denotes the convex hull of the closure of $\cP$ when $\cO$ is finite. And for cases where $\cO$ is an infinite continuous set, $\co(\cP)$ must additionally be totally bounded.
\end{proposition}
\begin{proof}
    The above result is a direct implication of the Heine-Borel Theorem (Theorem 2.41, \citep{rudin1976}) and Choquet's theorem~\citep{bishop1959representations}. First we discuss the proof for the case where $\cO$ is finite. Since $\cP\subseteq\Delta(\cO)$, using ~\ref{prop:boundedspace} we can say that $\cP$ is bounded. This means that $\co(\cP)$ is also bounded. Now, we know that the convex hull of a closed set $\overline{\cP}$ is also closed. Therefore, $\co(\overline{\cP})$ is closed and since $\co(\cP)=\co(\overline{\cP})$, $\co(\cP)$ is also closed. This makes $\co(\cP)$ compact as it is both bounded and closed by Heine-Borel Theorem. Now we can directly apply Choquet's theorem to obtain a probability measure $w$ for every $p\in\cP$ such that $p = \int_{p\in\ext(\cP)} w(p)dp$. In case when $\cO$ is an infinite continuous set, we are dealing with $\cP\subseteq\Delta(\cO)$, where $\Delta(\cO)$ may not have Heine-Borel Property, thus $\co(\cP)$ being totally bounded in addition to closed ensures that $\co(\cP)$ is compact and therefore Choquet's theorem is applicable. 
\end{proof}
The proposition ~\ref{prop:whatconditionsonPforgoodbehaviour} tries to identify what conditions should $\cP$ satisfy so that we can interpret $\co(\cP)$, i.e., the convex hull of $\cP$ as a credal set with valid extreme points $\ext(\cP)$. The general condition is that $\co(\cP)=\co(\overline{\cP})$ as a condition on $\cP$. Equivalently, the condition on $\co(\cP)$ is that it is closed. Notice that for finite $\cP$, it is trivially satisfied. This allows us to exclude $\cP=(0,\frac{1}{2})$ type of open sets from our discussion since they are open sets and its convex hull will violate closedness i.e. $\co(\cP)=\cP=(0,\frac{1}{2})$. Depending on the convention, if credal sets for $\cP$ are defined as the closure of their convex hulls, i.e., $\overline{\co}(\cP)$, then credal sets are compact (Heine-Borel Theorem) and Proposition ~\ref{prop:whatconditionsonPforgoodbehaviour} is applicable. Thus for our , we will restrict our discussion to $\cP$ such that $\co(\cP)=\co(\overline{\cP})$.
\subsection{Lower and Upper probabilities are always extreme points}
\begin{lemma}
\label{lemma:pinfandpsupinextP}
Let $\cP\subseteq\Delta(\cO)$ be the forecaster's belief and $\ext(\cP)$ the extreme points of the convex hull generated by $\cP\subseteq\Delta(\cO)$. Given any scoring rule $s:2^{\Delta(\cO)}\times \cO\rightarrow\mathbb{R}$ and forecasts $\cQ\subseteq\Delta(\cO)$, let
\begin{equation*}
    p_{\text{L}}^{(s,\cQ)} := \arg\inf_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)], \quad\quad p_{\text{U}}^{(s,\cQ)} := \arg\sup_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)] .
\end{equation*}
Then, both $p_{\text{L}}^{(s,\cQ)}$ and $p_{\text{U}}^{(s,\cQ)}$ belong to $\ext(\cP)$ for all pairs of $s$ and $\cQ$. In addition, $\cP\simeq \ext(\cP)$.
\end{lemma}
\begin{proof}
Firstly, for all $p \in \cP$, either $p \in \ext(\cP)$ or $p \not\in \ext(\cP)$. This follows trivially from the definition of extreme points of a convex hull in section \ref{subsection:Imprecise-probability-and-credal-sets}. 

The proof proceeds as follows. In \textbf{(i)} and \textbf{(ii)}, we show that $p_{\text{L}}^{(s,\cQ)},p_{\text{U}}^{(s,\cQ)}\in \ext(\cP)$ for all pairs of $s$ and $\cQ$, respectively, with a contradiction. 
Then, given \textbf{(i)} and \textbf{(ii)}, $\cP\simeq \ext(\cP)$ follows from  Definition \ref{def:equivalenceofimpreciseforecasts} for the equivalence of imprecise beliefs.

\textbf{(i) Lower probability:} For all $s$ and $\cQ$, $p_{\text{L}}^{(s,\cQ)}\in \ext(\cP)$.

We prove this by contradiction. Let us first assume there exists a pair of $s,\cQ$ such that $p_{\text{L}}^{(s,\cQ)}\in \cP\setminus \ext(\cP)$. Since we have fixed $s$ and $\cQ$, we drop the superscript from $p_{\text{L}}^{(s,\cQ)}$ for readability and treat $p_{\text{L}}$ as a distribution in $\cP$. 
Next, given $p_{\text{L}}\in \cP\setminus \ext(\cP)$, it implies that there exists a second order distribution $w\in \Delta(\ext(\cP))$ such that $w(p)>0$ for all $p\in \ext(\cP)$.  
\begin{align*}
    p_{\text{L}} &= \int_{p\in \ext(\cP)}w(p)dp \\
(\implies)\quad \mathbb{E}_{p_{\text{L}}}[s(\cQ,o)]&=  \mathbb{E}_{\int_{p\in \ext(\cP)}w(p)dp}[s(\cQ,o)] \\
&= \int_{p\in \ext(\cP)}w(p)\mathbb{E}_{p}[s(\cQ,o)]dp\\
&> \inf_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)]. && (w(p)>0 \text{ for all } p\in \ext(\cP))
\end{align*}
This results in a contradiction because $\ext(\cP)\subseteq \cP$. Therefore, $p_{\text{L}}\in ext(\cP)$.Since our choice of $\cQ$ and $s$ was arbitrary, the contradiction holds for all $\cQ\subseteq\Delta(\cO)$ and $s$. Therefore,  $p_{\text{L}}^{(s,\cQ)}\in \ext(\cP)$ for all and $s$.


\textbf{(ii) Upper probability:} For all $s$ and $\cQ$, $p_{\text{U}}^{(s,\cQ)}\in \ext(\cP)$.

Similarly, we show that $p_{\text{U}}\in \ext(\cP)$. Suppose that $p_{\text{U}}\in \cP\setminus \ext(\cP)$. This implies that there exists a second order distribution $w\in \Delta(\ext(\cP))$ such that $w(p)>0$ for all $p\in \ext(\cP)$ and 
\begin{align*}
    p_{\text{U}} &= \int_{p\in \ext(\cP)}w(p)dp\\
(\implies)\quad \mathbb{E}_{p_{\text{U}}}[s(\cQ,o)] &=  \mathbb{E}_{\int_{p\in \ext(\cP)}w(p)dp}[s(\cQ,o)] \\
&= \int_{p\in \ext(\cP)}w(p)\mathbb{E}_{p}[s(\cQ,o)]dp\\
&< \sup_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)]. && (w(p)>0 \text{ for all } p\in \ext(\cP))
\end{align*}
This also results in a contradiction since $\ext(\cP)\subseteq \cP$. Hence, both $p_{\text{L}}$ and $p_{\text{U}}$ belong to $\ext(\cP)$. 

\textbf{Equivalence of $\cP$ and $\ext(\cP)$:} Next, we show that $\cP$ and $\ext(\cP)$ are equivalent by applying Definition \ref{def:equivalenceofimpreciseforecasts}. For any reported set of beliefs $Q\subseteq \Delta(\cO)$ and scoring rule $s$, 
\begin{align*}
    \text{im}(\bm{V}^{\cP}(\cQ))&=\left[\inf_{p\in \cP}\mathbb{E}_{p}[s(\cQ,o)], \;\sup_{p\in \cP}\mathbb{E}_{p}[s(\cQ,o)]\right]\\
    &= \left[\mathbb{E}_{p_{\text{L}}}[s(\cQ,o)], \; \mathbb{E}_{p_{\text{U}}}[s(\cQ,o)]\right]\\
    &= \left[\inf_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)],\sup_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)]\right]  && (p_{\text{L}},p_{\text{U}} \in \ext(\cP)\text{ and } \ext(\cP)\subseteq \cP)\\
    &= \text{im}(\bm{V}^{\ext(\cP)}(\cQ)).
\end{align*}
This completes the proof.
\end{proof}
\subsection{Equivalence of extreme points for elicitation}
\begin{lemma}
\label{lemma:equivalentPimpliessameexteremepoints}
If two beliefs $\cP,\cP'\subseteq\Delta(\cO)$ are equivalent, i.e., $\cP\simeq\cP'$, then $\ext(\cP)=\ext(\cP')$.    
\end{lemma}
\begin{proof}
    By Definition~\ref{def:equivalenceofimpreciseforecasts}, two imprecise beliefs $\cP,\cP'\subseteq\Delta(\cO)$ are equivalent if for all scoring rule $s$ and forecast $\cQ\subseteq\Delta(\cO)$, $\text{im}(\bm{V}^{\cP}(\cQ))=\text{im}(\bm{V}^{\cP'}(Q))$. This means that, 
    \begin{align*}
        \inf_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)]=\inf_{p'\in\cP'}\mathbb{E}_{p'}[s(\cQ,o)]\quad \text{ and }\quad \sup_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)]=\sup_{p'\in\cP'}\mathbb{E}_{p'}[s(\cQ,o)].
    \end{align*}
    %Assuming all possible $s,\cQ$ and $\rho$, cover the space of all gambles $f:\cO\rightarrow\mathbb{R}$.
    ($\Rightarrow$) For the first part of the proof, we show that $\ext(\cP)\subseteq \ext(\cP')$. Let $q\in \ext(\cP)$, we know that for all $s,\cQ$,
    \begin{align*}
         \inf_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)] & \leq \mathbb{E}_{q}[s(\cQ,o)]\leq\sup_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)] \\
        \inf_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)] & \leq \mathbb{E}_{q}[s(\cQ,o)]\leq \sup_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)] && (\cP\simeq \ext(\cP) \text{ from Lemma}~\ref{lemma:pinfandpsupinextP})\\
        \inf_{p'\in\cP'}\mathbb{E}_{p'}[s(\cQ,o)] & \leq \mathbb{E}_{q}[s(\cQ,o)]\leq \sup_{p'\in\cP'}\mathbb{E}_{p'}[s(\cQ,o)] && (\cP\simeq\cP'\text{ by definition})\\
        \inf_{p'\in \ext(\cP')}\mathbb{E}_{p'}[s(\cQ,o)]&\leq \mathbb{E}_{q}[s(\cQ,o)]\leq \sup_{p'\in \ext(\cP')}\mathbb{E}_{p'}[s(\cQ,o)]. && (\cP'\simeq \ext(\cP') \text{ from Lemma}~\ref{lemma:pinfandpsupinextP})
    \end{align*}
    The last inequalities imply that $q\in \ext(\cP')$. ($\Leftarrow$) Next, we show that $\ext(\cP') \subseteq \ext(\cP)$. Let $q'\in \ext(\cP')$. Then, we know that for all $s,\cQ$,
    \begin{align*}
        \inf_{p'\in \ext(\cP')}\mathbb{E}_{p}[s(\cQ,o)]&\leq \mathbb{E}_{q'}[s(\cQ,o)]\leq \sup_{p'\in \ext(\cP')}\mathbb{E}_{p'}[s(\cQ,o)] \\
        \inf_{p'\in\cP'}\mathbb{E}_{p'}[s(\cQ,o)]&\leq \mathbb{E}_{q'}[s(\cQ,o)]\leq \sup_{p'\in\cP'}\mathbb{E}_{p}[s(\cQ,o)] && (\cP'\simeq \ext(\cP') \text{ from Lemma}~\ref{lemma:pinfandpsupinextP})\\
        \inf_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)]&\leq \mathbb{E}_{q'}[s(\cQ,o)]\leq \sup_{p\in\cP}\mathbb{E}_{p}[s(\cQ,o)] && (\cP\simeq\cP'\text{ by definition})\\
        \inf_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)]&\leq \mathbb{E}_{q'}[s(\cQ,o)]\leq \sup_{p\in \ext(\cP)}\mathbb{E}_{p}[s(\cQ,o)]. && (\cP\simeq \ext(\cP) \text{ from Lemma}~\ref{lemma:pinfandpsupinextP})
    \end{align*}
    The last inequalities imply that $q' \in \ext(\cP)$. Since both $\ext(\cP)\subseteq \ext(\cP')$ and $\ext(\cP')\subseteq \ext(\cP)$, we can conclude that $\ext(\cP) = \ext(\cP')$.
\end{proof}

\subsection{Preference relation in the subset of a credal set}
The lemma argues that the dominance induced by the preference relation associated with a credal set can only be refined by considering its subsets. Formally, 
\begin{lemma} 
    \label{lemma:additional-subset-q-preference}
    For any pair of imprecise forecasts $\cP, \cQ\subseteq\Delta(\cO)$ such that $co(\cQ)\subset co(\cP)$
    \begin{align*}
      a\succeq_{\cP} a' \implies a \succeq_{\cQ} a' \quad \forall\text{ }a,a'\in\cA
    \end{align*}
    where $\succeq_\cP,\succeq_\cQ$ are the partial preference relations over the space of actions induced by the corresponding expected utility profiles $\{\mathbb{E}_p[u(\cdot,o)]\}_{p\in\cP}$ and $\{\mathbb{E}_q[u(\cdot,o)]\}_{q\in\cQ}$. 
\end{lemma}
\begin{proof}
    Let us assume an arbitrary $\cQ$ and $\cP$ such that $\cQ\subset \cP$. Now let us consider a pair of inputs $x,x'\in\cX$ such that $x\succeq_{\cP} x'$. This implies that  
    \begin{align*}
\mathbb{E}_p[u(x,o)]&\geq\mathbb{E}_p[u(x',o)]\quad \forall p\in\cP\\
      \implies \quad \mathbb{E}_p[u(x,o)]&\geq \mathbb{E}_p[u(x',o)] \quad \forall p\in co(\cP)\\
      \implies \quad \mathbb{E}_q[u(x,o)]&\geq \mathbb{E}_q[u(x',o)]\quad \forall q\in co(\cQ) &&(co(\cQ)\subset co(\cP))\\
      \implies \quad \mathbb{E}_q[u(x,o)]&\geq \mathbb{E}_q[u(x',o)] \quad \forall q\in \cQ\\
      \implies \quad x &\succeq_{\cQ} x'
    \end{align*}
\end{proof}
\section{Proof of Results in Section \ref{section:setup}}
\subsection{Proof of Proposition ~\ref{prop:equivalenceofpreciseforecasts}}
\begin{proof}
    ($\Leftarrow$) 
    Let us assume that there are two identical distributions $p,q\in\Delta(\cO)$, i.e., $p=q$ that implies $\mathbb{E}_{p}[s(\cQ,o)]=\mathbb{E}_{q}[s(\cQ,o)]$ for all $\cQ\subseteq\Delta(\cO)$ and IP scoring rule $s$. Therefore, $\{p\}\simeq\{q\}$. 

    ($\Rightarrow$) Next, let us assume that $\{p\}\simeq \{q\}$, which means that
    \begin{align*}
         \mathbb{E}_{p}[s(\cQ,o)]=\mathbb{E}_{q}[s(\cQ,o)], \quad \forall s, \forall \cQ \subseteq\Delta(\cO).
    \end{align*}
    Since the above holds for all $s$ and $\cQ$, we choose $s$ to be strictly proper for precise forecasts and $\cQ:=\{p\}$. Hence,
    \begin{align*}
        \mathbb{E}_{p}[s(\{p\},o)]&=\mathbb{E}_{q}[s(\{p\},o)]\\
       \implies \quad p&=q. &&(s\text{ is strictly proper for precise forecasts})
    \end{align*}
    This completes the proof.
\end{proof}

%%%
\subsection{Proof of Proposition ~\ref{prop:credalsets}}
\begin{proposition}
   For imprecise beliefs $\cP,\cP'\subseteq \Delta(\cO)$ with non-empty extreme points, $\cP\simeq \cP'$ if and only if $co(\cP)=co(\cP')$.
\end{proposition}
\begin{proof}

($\Leftarrow$) First, we assume that $\cP$ and $\cP'$ induce the same credal set, i.e., 
 \begin{align*}
     \co(\cP) &= \co(\cP') \\
      \ext(\cP) &= \ext(\cP') && \text{(credal sets are convex hulls)} \\
       \ext(\cP)\simeq\cP & \quad\text{ and }\quad \ext(\cP')\simeq\cP'&&\text{(Lemma \ref{lemma:pinfandpsupinextP})}\\
     \implies \quad \cP&\simeq \cP'
 \end{align*}
Hence, $\cP$ and $\cP'$ are equivalent. 

($\Rightarrow$) Next, we assume that $\cP$ and $\cP'$ are equivalent. Then, it follows from Lemma~\ref{lemma:equivalentPimpliessameexteremepoints} that $\ext(\cP)=\ext(\cP')$. 

Let us assume that there exists a $P\in\co(\cP)$. Since credal sets are convex sets, $P$ can be expressed as a convex combination of the extreme points. Therefore, there exists some $w\in\Delta(\ext(\cP))$ such that  
\begin{align*}
    P&= \int_{p\in\ext{\cP}}w(p)dp \underset{(\vardiamondsuit)}{=} \int_{p\in\ext{\cP'}}w(p)dp &&(\vardiamondsuit:\text{Lemma~\ref{lemma:equivalentPimpliessameexteremepoints}})
\end{align*}
Thus $P\in\co(\cP')$ and therefore, $\co(\cP')\subseteq\co(\cP)$. 

Similarly, let us assume that there exists a $P'\in\co(\cP')$. Now $P'$ can also be expressed as a convex combination of the extreme points. Therefore, there exists some $w'\in\Delta(\ext(\cP'))$ such that  
\begin{align*}
    P'&= \int_{p'\in\ext{\cP'}}w'(p')dp' \underset{(\vardiamondsuit)}{=} \int_{p'\in\ext{\cP}}w(p')dp' &&(\vardiamondsuit:\text{Lemma~\ref{lemma:equivalentPimpliessameexteremepoints}})
\end{align*}
Thus $P\in\co(\cP)$ and therefore, $\co(\cP)\subseteq\co(\cP')$. Since $\co(\cP')\subseteq\co(\cP)$ and $\co(\cP)\subseteq\co(\cP')$, therefore $\co(\cP)=\co(\cP')$
\end{proof}
\subsection{Proof of Theorem~\ref{theorem:nosaneproperscoringrule}}

\textbf{Part I:} We first show that for any IP scoring rule $s$, it must give a constant score to all forecasts.
\begin{proof}    
Let us assume there exists a proper scoring rule $s:2^{\Delta(\cO)}\times \cO\rightarrow\mathbb{R}\cup\{-\infty\}$. Then, according to the definition of proper IP scoring rules for an imprecise forecaster with a vacuous belief $\Delta(\cO)$, we must have, 
\begin{equation*}
    \Delta(\cO) \succeq_{\Delta(\cO)} \cQ, \quad \forall \cQ\not\simeq\Delta(\cO).
\end{equation*}
This follows from the fact that for a proper score $s$, $\bm{V}^{\Delta(\cO)}(\Delta(\cO))$ dominates $\bm{V}^{\Delta(\cO)}(\cQ)$. Consequently, it follows from Definition \ref{def:propernessofimprecisescoringrule} that
\begin{equation}\label{eq:vacuous-dominance}
    \mathbb{E}_p[s(\Delta(\cO),o)] \geq \mathbb{E}_p[s(\cQ,o)], \quad \forall \cQ\not\simeq\Delta(\cO),\; \forall p\in\Delta(\cO).
\end{equation}
Let $\tilde{\cQ} := \{\cQ \,|\, \cQ\not\simeq\Delta(\cO)\}$ be the set of all forecasts not equivalent to the forecaster's belief ($\Delta(\cO)$), then we can rewrite \eqref{eq:vacuous-dominance} as 
    \begin{align}
        \label{eq:theorem-1-eq-1}
        \mathbb{E}_p[s(\Delta(\cO),o)]&\geq  \mathbb{E}_p[s(\cQ,o)], \quad \forall \cQ\in\tilde{\cQ},\; \forall p\in\Delta(\cO).
    \end{align}
Also, $\{q\}_{q\in\Delta(\cO)}\subseteq\tilde{\cQ}$ since $q\not\simeq\Delta(\cO)$. Combining this with Equation~\eqref{eq:theorem-1-eq-1} yields
\begin{align}
    \mathbb{E}_p[s(\Delta(\cO),o)] &\geq  \mathbb{E}_p[s(\{q\},o)], \quad \forall q\in\Delta(\cO),\; \forall p\in\Delta(\cO) \nonumber \\
    \label{eq:theorem-1-eq-2}
    \implies \quad \mathbb{E}_p[s(\Delta(\cO),o)]&\geq  \mathbb{E}_p[s(\{p\},o)], \quad \forall p\in\Delta(\cO), && 
\end{align}
where the second inequalities follow by selecting the inequalities such that $q=p$.
Similarly, let us analyse the incentives for all precise forecasters with belief $p\in\Delta(\cO)$ given a proper IP scoring rule $s$. 
Then, for all precise forecasters we must have, 
\begin{align*}
    \{p\}&\succeq_{\{p\}} \cQ, \quad \forall p\in \Delta(\cO),\; \forall \cQ\in\tilde{\cQ} && (\tilde{\cQ}:=2^{\Delta(\cO)}\setminus \{p\})\\
    \implies \quad \{p\}&\succeq_{\{p\}} \Delta(\cO), \quad \forall p \in \Delta(\cO) && (\Delta(\cO)\in \tilde{\cQ})\\
    \implies \quad \mathbb{E}_p[s(\{p\},o)]&\geq  \mathbb{E}_p[s(\Delta(\cO),o)], \quad \forall p\in\Delta(\cO).
\end{align*}    
However, it follows from Equation \eqref{eq:theorem-1-eq-2} that $\mathbb{E}_p[s(\Delta(\cO),o)]\geq \mathbb{E}_p[s(\{p\},o)]$ and $\mathbb{E}_p[s(\{p\},o)]\geq\mathbb{E}_p[s(\Delta(\cO),o)]$ for all $p\in\Delta(\cO)$. This implies that $\mathbb{E}_p[s(\Delta(\cO),o)]=\mathbb{E}_p[s(\{p\},o)]$ for all $p\in\Delta(\cO)$.


Therefore, any IP scoring rule $s$ that satisfies properness sets up incorrect incentives for the forecaster. For example, the expected score for honestly reporting a precise forecast is the same as reporting the vacuous set of all distributions, i.e., 
\begin{equation}
    \label{eq:theorem-1-eq-3}
    \mathbb{E}_p[s(\Delta(\cO),o)]=\mathbb{E}_p[s(\{p\},o)], \quad \forall p\in\Delta(\cO).
\end{equation}
While the above equation is sufficient to discard any proper scoring rule, we show that the only IP scoring rule possible is a constant function. For $s$ to be proper for imprecise forecasts, the following must hold true for all $\cP\subseteq\Delta(\cO)$:
\begin{align}
    \cP&\succeq_\cP \{q\}, \quad \forall q\in \Delta(\cO)\nonumber \\
    \mathbb{E}_p[s(\cP,o)]&\geq\mathbb{E}_p[s(\{q\},o)], \quad \forall q\in \Delta(\cO),\; \forall p\in\cP \nonumber \\
    \mathbb{E}_p[s(\cP,o)]&\geq\mathbb{E}_p[s(\{q\},o)], \quad \forall q\in \cP, \; \forall p\in\cP \nonumber \\
    \label{eq:theorem-1-eq-4}
    \implies \mathbb{E}_p[s(\cP,o)]&\geq\mathbb{E}_p[s(\{p\},o)], \quad \forall p\in\cP.
\end{align}
Similarly, for any $p\in\Delta(\cO)$, the following must hold: 
\begin{align}
    \{p\}&\succeq_p \cP, \quad \forall p\in \Delta(\cO) \nonumber \\
    \label{eq:theorem-1-eq-5}
    \implies \mathbb{E}_p[s(\{p\},o)]&\geq \mathbb{E}_p[s(\cP,o)].
\end{align}
Combining Equations~\ref{eq:theorem-1-eq-3}, ~\ref{eq:theorem-1-eq-4} and ~\ref{eq:theorem-1-eq-5} yields 
\begin{align}
\label{eq:theorem-1-eq-6}
    \mathbb{E}_p[s(\Delta(\cO),o)]&=\mathbb{E}_p[s(\{p\},o)]=\mathbb{E}_p[s(\cP,o)], \quad \forall p\in\Delta(\cO)
\end{align}
Given Equation~\ref{eq:theorem-1-eq-6} is valid for all $p\in\Delta(\cO)$, we consider the a subset of $\Delta(\cO)$. To be precise, the set of all Dirac distributions associated with each outcome, i.e. $p\in\{\delta_{o}\}_{o\in\cO}$
\begin{align*}
    \mathbb{E}_p[s(\Delta(\cO),o)]&=\mathbb{E}_p[s(\{p\},o)]=\mathbb{E}_p[s(\cP,o)], \quad p\in \{\delta_o\}_{o\in \cO}  && (\{\delta_o\}_{o\in \cO}\subseteq \Delta(\cO))\\
    \implies s(\Delta(\cO),o)&=s(\{p\},o)=s(\cP,o), \quad \forall o\in \cO.
\end{align*}
Hence, $s$ needs to be a constant score for it to be a proper IP scoring rule.  
\end{proof}
\textbf{Part II:} There exists no strictly proper IP scoring rule $s$.
\begin{proof}
Assume that there exists a strictly proper IP scoring rule $s$. Consider a precise forecaster with belief $q\in\Delta(\cO)$. Then, we have 
\begin{align}
    \{q\}\succ_{q} \cQ, \quad \forall \cQ\not\simeq{q} \nonumber \\
    \implies \mathbb{E}_{q}[s(\{q\},o)]> \mathbb{E}_{q}[s(\cQ,o)] \nonumber \\
    \label{eq:theorem-1-eq-7}
    \implies \mathbb{E}_{q}[s(\{q\},o)]> \mathbb{E}_{q}[s(\Delta(\cO),o)]. && (\Delta(\cO)\text{ is one possible }\cQ)
\end{align}
Since $s$ is strictly proper, it satisfies Equation~\ref{eq:theorem-1-eq-3}. However, this results in a contradiction to Equation ~\ref{eq:theorem-1-eq-7}. Hence, no $s$ can be strictly proper.  
\end{proof}
%%%%
\section{Proof of Results in Section \ref{sec:imprecisescoringrules}}

\subsection{Why is Non-dictatorship Desirable?}
\label{appendix:dictatorships}
%First, we establish an equivalence between our setup of scoring an imprecise forecaster and the classical social choice problem. We begin by defining what it means for an aggregation function to be non-dictatorial:
%\begin{definition}
%    An aggregation rule $\rho$ is said to be non-dictatorial if for a profile of preferences $\{\succeq^{(i)}\}_{i\in[n]}$ , there exists no $i\in[n]$ (dictator) such that for all $x,y\in \cX$, $x\succeq_{i} y$ implies $x\succeq_{\rho}y$.
%\end{definition}
Let us assume that $\rho$ violates non-dictatorship, then $\rho$ is dictatorial. For clarity, we also define a dictatorship.
\begin{definition}(Dictatorship)
An aggregation rule $\rho$ is a dictatorial if there exists a $P_\rho\in\cP$ (dictator), that depends on $\rho$, such that for any pair of reports $\cQ,\cQ'\subseteq\Delta(\cO)$,
\begin{equation*}
    \cQ\succeq_{P_{\rho}} \cQ' \implies \cQ\succeq_{\rho[\cP]}\cQ'.
\end{equation*}
\end{definition}
A dictatorial $\rho$ not only allows the forecaster to remove indecision in their decision-making problem about which $\cQ$ to report, it also allows the forecaster to precisely resolve their epistemic uncertainty, i.e., by reducing the credal set $\cP$ to only the dictator $P_\rho$. 

Let us denote the set of best reports plausible under aggregation $\rho$ by $\tilde{\cQ}^{\rho}:=\{\cQ \,|\, \cQ\succeq_{\rho} \cQ', \forall \cQ'\subseteq\Delta(\cO)\}$. Since $\succeq_{\rho}$ is complete, if the set of best reports $\tilde{\cQ}$ contains more than one report, then they must be indifferent w.r.t. $\succeq_{\rho[\cP]}$. 
Given $\rho$ is a dictatorship, there exists $P_\rho\in\cP$ such that $\succeq_{P_\rho}$ dictates the preference $\succeq_{\rho[\cP]}$. That is, the set of best reports under $P_\rho$ must be exactly the same as that under $\rho$. Therefore,  
\begin{equation*}
    \tilde{\cQ}^{P}=\tilde{\cQ}^{\rho}.
\end{equation*}
This implies that the expected scores of ${P_\rho}$ and $\cP$ with any dictatorial $\rho$ is the same, i.e.,  
\begin{equation*}
    V_{\rho}^\cP(\{P_\rho\}) = V_{\rho}^\cP(\cP).
\end{equation*}

\subsection{Proof of Proposition~\ref{prop:tailoredscoringrule}}
\begin{proof}
    We prove this result by contradiction. Let us assume that there exists a tailored scoring rule $s_\rho$ that is not proper and analyse this scoring rule for an arbitrary forecaster with an imprecise belief $\cP\subseteq\Delta(\cO)$. Since $s_\rho$ is not proper, it implies that there exists $Q\subseteq\Delta(\cO)$ where  $\cQ\not\simeq\cP$ such that 
    \begin{equation}
    \label{eq:proposition-4-point-9-eq-1}
        V_{\rho}^{\cP}(\cQ)> V_{\rho}^{\cP}(\cP).
    \end{equation}
    In other words, the forecaster strictly prefers the forecast $\cQ$ over their belief $\cP$. However, let's analyse the scenario from DM's perspective when they obtain forecast $\cP$, the optimal action according to the forecast $\cP$ is
    \begin{align}
    \label{eq:proposition-4-point-9-eq-2}
    a^*_{\cP,\rho} = \arg\max_{a\in\cA} \; \rho\left[\{\mathbb{E}_{p}[u(a,o)]\}_{p\in\cP}\right].
    \end{align}
    Since $a^*_{\cP,\rho}$ is the maximizer of DM's aggregated utility, this means that for all $a\in\cA$,
    \begin{align}
    \label{eq:proposition-4-point-9-eq-3}
        \rho\left[\{\mathbb{E}_{p}[u(a^*_{\cP,\rho},o)]\}_{p\in\cP}\right] \geq \rho\left[\{\mathbb{E}_{p}[u(a,o)]\}_{p\in\cP}\right].
    \end{align}
    However, we know that from \Cref{eq:proposition-4-point-9-eq-1}
    \begin{align*}
        V_{\rho}^{\cP}(\cQ)&> V_{\rho}^{\cP}(\cP)\\
        \rho[\{\mathbb{E}_{p}[s_{\rho}(\cQ,o)]\}_{p\in\cP}]&>\rho[\{\mathbb{E}_{p}[s_\rho(\cP,o)]\}_{p\in\cP}]\\
        \rho[\{\mathbb{E}_{p}[u(a^*_{\cQ,\rho},o)]\}_{p\in\cP}]&>\rho[\{\mathbb{E}_{p}[u(a^*_{\cP,\rho},o)]\}_{p\in\cP}].
    \end{align*}
    This results in a contradiction to \Cref{eq:proposition-4-point-9-eq-3}. Therefore, $s_\rho$ must be proper. Since this holds for any choice of $\rho$, we can conclude that $s_\rho$ must be proper for any aggregation rule $\rho$.
\end{proof}

%%%
\subsection{Proof of Lemma \ref{lemma:strictness-for-precise-distributions}}

\textbf{Part I: Strict properness of IP scoring rule for precise forecasts}

\begin{proof} ($\Rightarrow$) From \Cref{theorem:gneiting}, a regular precise scoring rule $s$ is (strictly) proper if and only if there exists a corresponding (strictly) convex function $G$ on $\Delta(\cO)$ such that 
    \begin{align*}
        s(p,o)= G(p) - \sum_{o\in\cO} G'(p)(o) p(o) + G'(p)(o).
    \end{align*}
    Moreover, it follows from \Cref{remark:gneiting} that the $G(p)=\mathbb{E}_{p}[s(p,o)]$. Hence, for a tailored scoring rule $s_\rho$ on precise distribution $p\in\Delta(\cO)$ to be strictly proper, we must have 
    \begin{align*}
        G(p)&=\mathbb{E}_{p}[s_\rho(\{p\},o)]\\
            &= k\mathbb{E}_{p}[u(a^*_p,o)]+c && (\text{Tailored scoring rule; \Cref{def:tailoredscoringrule}})\\
            &= \max_{a\in\cA}\; k\mathbb{E}_{p}[u(a,o)]+c .
    \end{align*}
    Next, for $G(p)$ to be strictly convex in $p$, we must have that for all $p,q\in\Delta(\cO)$,
    \begin{align}
    \label{eq:lemma-equation-2}
         G(q) &> G(p)+\sum_{o\in\cO} G'(p)(o)[q(o)-p(o)] 
     \end{align}
    Where $G'(p)(o)$ is the $o^{\text{th}}$ component of the gradient $G'(p)$ at $p$. Let us consider the right-hand side of \Cref{eq:lemma-equation-2}.
    \begin{align}
          G(p)+\sum_{o\in\cO} G'(p)(o)[q(o)-p(o)]  &= k\mathbb{E}_{p}[u(a^*_p,o)]+ c + \sum_{o\in\cO} ku(a^*_p,o) [q(o)-p(o)]\\
             &= k\mathbb{E}_{p}[u(a^*_p,o)]+ k\mathbb{E}_{q}[u(a^*_p,o)]-k\mathbb{E}_{p}[u(a^*_p,o)]+c\\
        \label{eq:lemma-equation-3}
             &= k\mathbb{E}_{q}[u(a^*_p,o)]+c .
    \end{align}
    Since $G(q):= k\mathbb{E}_{q}[u(a^*_q,o)]+c$, for $G$ to be strictly convex, we use \Cref{eq:lemma-equation-3} to rewrite \Cref{eq:lemma-equation-2} as follows
    \begin{align*}
          G(q) &> k\mathbb{E}_{q}[u(a^*_p,o)]+c, \quad \forall p,q\in\Delta(\cO), \\
        \implies \quad k\mathbb{E}_{q}[u(a^*_q,o)]+c &> k\mathbb{E}_{q}[u(a^*_p,o)]+c, \quad \forall p,q\in\Delta(\cO) .
    \end{align*}
    Hence, $a^*_q$ must be a unique maximizer.

    ($\Leftarrow$) 
    
    We assume that $a^*_{p} := \arg\max_{a\in\cA}\mathbb{E}_{p}[u(a,o)]$ is the unique maximizer for all $p\in\Delta(\cO)$. Then, for all $p,q\in\Delta(\cO)$ and some arbitrary $\lambda\in[0,1]$,
    \begin{align*}
        G(\lambda p+(1-\lambda)q)&= \mathbb{E}_{\lambda p+(1-\lambda)q}[s_\rho(\{\lambda p+(1-\lambda)q\},o)]\\
                                 &= \lambda \mathbb{E}_p[s_\rho(\{\lambda p+(1-\lambda)q\},o)]+(1-\lambda)\mathbb{E}_q[s_\rho(\{\lambda p+(1-\lambda)q\},o)]\\
                                 &= \lambda k\mathbb{E}_p[u(a^*_{\lambda p+(1-\lambda)q},o)]+\lambda c+(1-\lambda)k\mathbb{E}_q[u(a^*_{\lambda p+(1-\lambda)q},o)]+(1-\lambda)c \\
                                 &< \lambda k\mathbb{E}_p[u(a^*_{p},o)]+\lambda c+(1-\lambda)k\mathbb{E}_q[u(a^*_{q},o)]+(1-\lambda)c && (a^*_p \text{ and } a^*_q\text{ are unique})\\
                                 &=\lambda G(p)+(1-\lambda)G(q).
    \end{align*}
    Hence, $G$ is strictly convex.
\end{proof}
\textbf{Part II: Impossibility of strictly proper scoring rules with Pareto efficient $\rho$}
\begin{proof}
Suppose that there exists the aggregation rule $\rho$ such that the tailored scoring rule $s_\rho$ is strictly proper for both precise and imprecise forecasts. This means that for all $\cP\subseteq\Delta(\cO)$, and for all $\cQ\not\simeq\cP$,
\begin{align*}
    V_\rho^\cP(\cP)&>V_\rho^\cP(\cQ)\\
    \implies \quad \rho(\{\mathbb{E}_p[s_\rho(\cP,o)]\}_{p\in\cP})&> \rho(\{\mathbb{E}_p[s_\rho(\cQ,o)]\}_{p\in\cP})\\
    \implies \quad \rho(\{\mathbb{E}_p[u(a^*_{\rho,\cP},o)]\}_{p\in\cP})&> \rho(\{\mathbb{E}_p[u(a^*_{\rho,\cQ},o)]\}_{p\in\cP}). & (s_\rho \text{ is tailored scoring rule})
\end{align*}
The aggregation rule $\rho$ maps the set of preferences $\succeq_{\cP}:=\{\succeq_p\}_{p\in\cP}$ into a complete preference relation $\succeq_{\rho(\cP)}$ which follows the aggregated utility $\rho(\{\mathbb{E}_p[u(\cdot,o)]\}_{p\in\cP})$.

Since $\rho$ is Pareto efficient, for all $a,a'\in\cA$, $a\succeq_{\cP} a'$ implies $a\succeq_{\rho[\cP]} a'$. Only for actions that are incomparable to one another, i.e., $a\not\succeq_{\cP}a'$ and $a'\not\succeq_{\cP}a$, $\rho$ decides to remove indecision by completing the preference as $a\succeq_{\rho[\cP]}a'$ or $a'\succeq_{\rho[\cP]}a$. 

Without loss of generality, let us assume that $\rho$ chooses to rank $a\succeq_{\rho[\cP]}a'$ for two incomparable $a,a'\in\cA$ with respect to original credal set $\cP$.
However, based on \Cref{lemma:additional-subset-q-preference}, we can construct a $Q\subseteq\Delta(\cO)$ such that $\co(\cQ)\subset \co(\cP)$ and $a^*_{\rho,\cP}=a^*_{\rho,\cQ}$. This provides a counterexample to strictness of $s_\rho$ for all Pareto efficient $\rho$.

We now explain the counterexample in detail. We construct $\cQ$ based on its partial preference relation $\succeq_Q$. The preference relation $\succeq_Q$ must be well defined for any two pair of actions $a,a'\in\cA$. To this end we use the preference relation $\succeq_{\cP}$ to define all possible scenarios for a pair of actions $a,a'\in\cA$. Either $a,a'\in\cA$ are comparable with respect to $\succeq_{\cP}$ (\textbf{Case I}) or incomparable (\textbf{Case II}). The construction of $\succeq_{\cQ}$ is defined below
\input{uai2025/new_content/tikz}
Now we are ready to reason what happens when we aggregate the partial preference $\succeq_{\cQ}$ with $\rho$. We will reason for all the cases we defined above. 
\begin{itemize}[label={}]
    \item \textbf{Case I:} For all pairs of $ a,a' \in\cA$ that are comparable w.r.t. $\succeq_{\cP}$  (Assume w.l.o.g $a\succeq_\cP a'$).
\begin{align*}
    a\succeq_\cP a' \underset{(\clubsuit)}{\implies}   a\succeq_{\rho[\cP]} a'\quad \text{ and } \quad  a\succeq_\cP a' \underset{(\vardiamondsuit)}{\implies}   a\succeq_\cQ a' \underset{(\clubsuit)}{\implies}   a\succeq_{\rho[\cQ]} a'. && (\clubsuit:\rho \text{ is PE},\vardiamondsuit\text{: \Cref{lemma:additional-subset-q-preference}})
\end{align*}
Therefore, whenever the pair of actions $a,a' \in\cA$ are comparable w.r.t. $\succeq_{\cP}$, the aggregated preference relation is the same, i.e., $\{\succeq_{\rho[\cP]}\}\equiv\{\succeq_{\rho[\cQ]}\}$.
    \item \textbf{Case II:} Consider $a,a' \in\cA$ that are incomparable w.r.t. $\succeq_{\cP}$. (Assume w.l.o.g that $\rho$ resolves this as $a\succeq_{\rho[\cP]}a'$)
\begin{itemize}[label={}]
    \item \textit{Case II.1}: The pair of $ a,a' \in\cA$ is also comparable w.r.t. $\succeq_{\cQ}$ (Assume w.l.o.g $a\succeq_{\cQ} a'$)
 \begin{align*}
    a\succeq_\cQ a' \underset{(\clubsuit)}{\implies}   a\succeq_{\rho[\cQ]} a'\quad \text{ and } \quad  a\succeq_\cQ a'\underset{(\vardiamondsuit)}{\implies} a\succeq_{\rho[\cP]} a'  && (\clubsuit:\rho \text{ is PE}, \vardiamondsuit: \text{by construction})
\end{align*}
\item \textit{Case II.2}: The pair $ a,a' \in\cA$ is incomparable w.r.t. $\succeq_{\cQ}$.

Since $\rho$ is a function, it will resolve indecision for two inputs in the same way, given that $|\cA|$ is fixed across both these resolutions:  
\begin{align*}
    \Bigg((a\not\succeq_{\cP}a' \land a'\not\succeq_{\cP}a)\implies (a\succeq_{\rho[\cP]} a') \Bigg) \land (a\not\succeq_{\cQ}a' \land a'\not\succeq_{\cQ}a)\implies a\succeq_{\rho[\cQ]} a'.
\end{align*}
\end{itemize}
Therefore, similar to Case 1, whenever the pair of actions $a,a' \in\cA$ are incomparable w.r.t. $\succeq_{\cP}$, the aggregated preference is the same, i.e., $\{\succeq_{\rho[\cP]}\}\equiv\{\succeq_{\rho[\cQ]}\}$. Hence, $a^*_{\rho,\cP}=a^*_{\rho,\cQ}$.
\end{itemize}
This makes $s_\rho$ not strictly proper.
\end{proof}

%%%
\section{Proof for Results in Section \ref{sec:rand-tailored-rules}}

\subsection{Proof of Lemma ~\ref{lemma:unqiue-representation-of-credal-set}}

\begin{proof}
     We prove this by contradiction, let us assume that, $\cA_{\ext}$ is not a sufficient way to represent credal sets in the actions space. This implies that there exists a pair of credal sets $\cQ,\cQ'\subseteq\cO$ such that $\cQ\not\simeq\cQ'$ and $\cA^{\ext}_{\cQ'}=\cA^{\ext}_\cQ$. Since $\cQ\not\simeq\cQ'$ it implies either of the two cases 
    \begin{itemize}[]
        \item \textbf{Case 1: } There exists a $ q'\in \ext(\cQ')$ such that $q'\not\in \ext(\cQ)$. This implies that 
     \begin{align*}
         \exists \text{ }a^*_{q'}\in \cA^{\ext}_{\cQ'}\quad\text{and}\quad \exists\text{ } a^*_{q'}\in \cA^{\ext}_{\cQ} && (a^*_{q'}\text{ is unique for all } q'\in\Delta(\cO))
     \end{align*}
     This results contradicts $\cA^{\ext}_{\cQ'}=\cA^{\ext}_\cQ$.
     \item \textbf{Case 2: } There exists a $q \in \ext(\cQ)$ such that $q\not\in \ext(\cQ')$. We follow the same reasoning as Case 1, i.e., 
     \begin{align*}
         \exists\text{ } a^*_{q}\in \cA^{\ext}_{\cQ}\quad\text{and}\quad \exists\text{ } a^*_{q}\in \cA^{\ext}_{\cQ'} && (a^*_{q}\text{ is unique for all } q\in\Delta(\cO))
     \end{align*}
     resulting in a contradiction with $\cA^{\ext}_{\cQ'}=\cA^{\ext}_\cQ$.
    \end{itemize}
    Hence $\cA^{\ext}$ is a unique representation for all credal sets. 
\end{proof}

%%%
\subsection{Proof of Theorem ~\ref{theorem:strictly-proper-imprecise-scoring-rule}}

\begin{proof}
We know that for $s_{\theta}$ to be strictly proper, the following must hold for all beliefs $\cP\subseteq\Delta(\cO)$ 
\begin{align*}
    V^{\cP}_{\theta}(\cP)=V^{\cP}_{\theta}(\cQ) \quad \text{iif} \quad \cP\simeq \cQ 
\end{align*}

$(\Rightarrow)$ Given, $\theta\in\Delta(\bm{\rho})$ has full support, $V^{\cP}_{\theta}(\cP)=V^{\cP}_{\theta}(\cQ)$ implies that, 
\begin{align*}
       \rho(\{\mathbb{E}_p[s_{\rho}(\cP,o)]\}_{p\in\cP}) &= \rho(\{\mathbb{E}_p[s_{\rho}(\cQ,o)]\}_{p\in\cP}) \quad \forall \rho\in \bm{\rho}\quad \forall \cP \subseteq\Delta(\cO)\\
      \bm{\lambda}^{\top}\{\mathbb{E}_p[u(a^*_{\bm{\lambda}^{\top}\cP},o)]\}&=\bm{\lambda}^{\top}\{\mathbb{E}_p[u(a^*_{\bm{\lambda}^{\top}\cQ},o)]\}\quad \forall \bm{\lambda} \in \Delta^{|ext(\cP)|} \quad \forall \cP\subseteq\Delta(\cO) && (\bm{\rho}\text{: fixed linear aggregation})\\
      \bm{\lambda}^{\top}\{\mathbb{E}_p[u(a^*_{\bm{\lambda}^{\top}\cP},o)]\} &=\bm{\lambda}^{\top}\{\mathbb{E}_p[u(a^*_{\bm{\lambda}^{\top}\cQ},o)]\} \quad \forall \bm{\lambda} \in\{\delta_i\}_{i\in |ext(\cP)|}\quad \forall \cP\subseteq\Delta(\cO) && (\{\delta_i\}_{i\in |ext(\cP)|} \subset \Delta^{|ext(\cP)|})\\
      \mathbb{E}_p[u(a^*_{p},o)]&=\mathbb{E}_p[u(a^*_{q},o)]\quad \forall p\in \cP \quad \forall \cP\subseteq\Delta(\cO) && (q:= \delta_i^T\cQ)\\
       a^*_{p} &= a^*_{q} \quad \forall p\in \cP \quad \forall \cP\subseteq\Delta(\cO)&& \tag*{(Using \Cref{lemma:strictness-for-precise-distributions} as \mbox{$s_\theta$} is strictly proper for precise distributions)}\\
    \implies \cA^{ext}_\cP &= \cA^{ext}_\cQ && (\text{By Definition of } \cA^{ext})\\
    \implies \cP &\simeq \cQ &&(\text{\Cref{lemma:unqiue-representation-of-credal-set}})
\end{align*}

$(\Leftarrow)$ Given that $\cP\simeq \cQ$ we show that $V_{\theta}^{\cP}(\cP)=V_{\theta}^{\cP}(\cQ)$. This is trivial since two equivalent forecasts produce the same underlying partial order on the actions $\cA$. As aggregation functions make this partial order complete, by the property of being a function, they will result in the same complete order for the same partial order. Therefore, given $\cP\simeq\cQ$ implies that 
\begin{align*}
      V^{\cP}_\rho(\cP)&=V^{\cP}_\rho(\cQ) \quad \forall \rho\in\bm{\rho}\\
       \mathbb{E}_{\theta} [V^{\cP}_\rho(\cP)] &= \mathbb{E}_{\theta} [V^{\cP}_\rho(\cQ)] \quad \forall \theta \in \Delta(\bm{\rho})\\
       V^{\cP}_{\theta}(\cP)&=V^{\cP}_{\theta}(\cQ) \quad \forall \theta \in \Delta(\bm{\rho})
\end{align*}
Therefore, the imprecise forecaster is truthful in the epistemic sense w.r.t the strictly proper IP scoring rule $s_{\theta}$.  
\end{proof}
\section{Simulations}
\label{appendix:simulation}
To test the sanity of our proposed scoring rule, we simulate a scenario where an imprecise forecaster predicts a binary outcome (e.g., chance of rain tomorrow). We assume the forecaster has an imprecise forecast $[0.4,0.6]$ and uses an imprecise scoring rule $s_\rho$ where $\rho$ is a dictatorship or some other aggregation like min-max. We compare this to our randomized imprecise scoring rule $s_\theta$. Given the binary outcome, the forecaster reports an interval $\cQ:=[q_1,q_2]$ where $q_1$ denotes the lower probability and $q_2$ the upper probability respectively.  
\begin{figure}[t]
    \centering
    \begin{subfigure}{0.32\linewidth}
        \centering
        \includegraphics[width=\linewidth]{uai2025/figs/dictatorship.pdf}
        \caption{}
    \end{subfigure}
    \begin{subfigure}{0.32\linewidth}
        \centering
        \includegraphics[width=\linewidth]{uai2025/figs/minmax.pdf}
        \caption{}
    \end{subfigure}
    \begin{subfigure}{0.32\linewidth}
        \centering
        \includegraphics[width=\linewidth]{uai2025/figs/strictly.pdf}
        \caption{}
    \end{subfigure}
    \caption{(Left-to-right) In the first figure we simulate the scoring rule where $\rho$ is a dictatorship with a fixed mixing weight of 0.5, in the middle figure we simulate the scoring rule with min-max $\rho$ (pessimistic decision maker) and in the last figure we simulate the scoring rule where the aggregation is a randomized dictatorship and the forecaster obtains a distribution $\theta=\mathcal{U}[0,1]$ over $\rho$. The lower half of the figure is not plotted since that corresponds to region $q_1>q_2$, i.e. lower probability being greater than upper probability}
    \label{fig:simulations}
\end{figure}
\Cref{fig:simulations} highlights that the randomized scoring rule $s_\theta$ is strictly proper for imprecise forecasts as it has the highest expected score for the forecaster only when the forecaster reports his true belief. While in other cases of using a deterministic imprecise scoring rule $s_\rho$, if DM provides a $\rho$ such that it is a dictatorship, such as in the case of ~\Cref{fig:simulations}(a), the scoring rule is proper; however, the forecaster can lie by reporting the dictator. This can be inferred from the contour that the point $[0.5,0.5]$, which corresponds to the precise forecast $0.5$, also has the highest expected score. With $\rho$ being a min-max rule, the scoring rule $s_\rho$  is proper but not strictly as other imprecise forecasts allow the forecaster to obtain the same expected score. For our implementation we consider $\cA=[0,1]$ and $u(a,o):=(o-a)^2$ to satisfy ~\Cref{lemma:strictness-for-precise-distributions}. 

%We release our implementation at \url{https://github.com/muandet-lab/Imprecise-Scoring-Rule}.   