by the likelihood-ratio, or equivalently (after changing to a logarithmic scale) by the
logarithm of its likelihood-ratio. We adopt this approach. Thus, we define an *information*
function induced by the opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i), g)$ (or simply by $(\mathbf{x}, (\mathbf{e}_s))$) as a
function *info* that maps each evidence bundle $(e_s)_{s \in S'}$ in $\cup_{S' \subseteq S} \Pi_{s \in S'} \mathcal{E}_s$ to a real number
given by

$$
info((e_s)_{s \in S'}) = \log \frac{f_{S'}((e_s)_{s \in S'} | 1)}{f_{S'}((e_s)_{s \in S'} | -1)} \quad (6)
$$

if $f_{S'}((e_s)_{s \in S'}|1)$, $f_{S'}((e_s)_{s \in S'}|-1) \neq 0$, where, for each source set $S' \subseteq S$, $f_{S'}(\cdot|\mathbf{x})$ is a joint probability density or mass function of the bundle $(\mathbf{e}_s)_{s \in S'}$ conditional on the state $\mathbf{x}$ (with respect to some measure over $\Pi_{s \in S'} \mathcal{E}_s$). For instance, $f_{S'}(\cdot|\mathbf{x})$ could be a multivariate Gaussian density function over $\mathbb{R}^{S'}$, if evidences are Gaussian. The likelihood-ratio captures how much more likely state 1 makes the bundle than state -1. If this ratio is above (resp. below, equal to) 1, then $info((e_s)_{s \in S'})$ is positive (resp. negative, zero), indicating support for 1 (resp. for -1, for neither). When $S'$ is a singleton set $\{s\}$, then we simply write '$info(e_s)$' and talk of the '$information in e_s$'.

We call the source set $S'$ or the evidence bundle $(e_s)_{s \in S'}$ complementary if the com-
bined information $info((e_s)_{s \in S'})$ differs from the total isolated information $\sum_{s \in S'} info(e_s)$.
Such complementarity is ruled out if evidences are state-conditionally independent.<fnref n="28" /> This
is why simple opinion structures rule out Failure 3.

The complementarity of a set $S'$ can be improper, being merely inherited from that
of some subset $S'' \subsetneq S'$. If for instance three evidences are complementary merely be-
cause the first two are complementary while the third is independent, then the triple
is improperly complementary. A formal definition of Failure 3 must rule out improper
complementarity of $S'$, since there is obviously no real neglect of complementarity if an
improperly complementary set $S'$ is not fully accessed whilst its properly complementary
subsets are fully accessed. Formally, a complementary set $S'$ is properly complement-
ary if each subset $S''$ of $S'$ is complementary with the rest of $S'$, i.e., $info((e_s)_{s \in S'}) \neq$
$info((e_s)_{s \in S''}) + info((e_s)_{s \in S' \setminus S''})$.

We are ready to state the above notions and result formally:<fnref n="29" />

**Definition 1** Given an opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i), g)$ (inducing an information function info) and a source set $S' \subseteq S$,

* Dispersion of $S'$ is the event that $S' \subseteq \cup_i \mathbf{S}_i$ and $S' \not\subseteq \mathbf{S}_i$ for some $i$, i.e., $S'$ is available to the group but not to all members,

* Complementarity of $S'$ is the event that $info((e_s)_{s \in S'}) \neq \sum_{s \in S'} info(e_s)$, i.e., information in combined evidence differs from total isolated information,

* Proper complementarity of $S'$ is the event that $S'$ is complementary and moreover each subset $S'' \subseteq S'$ is complementary with $S' \setminus S''$, i.e., $info((e_s)_{s \in S'}) \neq info((e_s)_{s \in S''}) + info((e_s)_{s \in S' \setminus S''})$.

* Failure 3 over $S'$ is the event that $S'$ is properly complementary and dispersed.

<footnote n="28"><sup>28</sup>Under state-conditional independence, the joint likelihood function factorises: $f_{S'}((e_s)_{s \in S'}|\mathbf{x}) = \prod_{s \in S'} f_{\{s\}}(e_s|\mathbf{x})$ for each state $\mathbf{x} \in \{\pm 1\}$. Taking the ratio across states and then the logarithm on both sides yields $info((e_s)_{s \in S'}) = \sum_{s \in S'} info(e_s)$.</footnote>

<footnote n="29"><sup>29</sup>Strictly speaking, density function need neither exist nor be unique. But in practical applications, there usually exists a unique natural version of $f(\cdot|\mathbf{x})$, and hence of the information function info. Hence, the information function info corresponding to the opinion structure, and thus the events defined based on it (events of complementarity and of Failure 3), will exist and be essentially unique in practical applications.</footnote>
Instead of defining a measure of Failure 3 (as done for Failures 1 and 2), let us define
what an instance of Failure 3 is. The definition will first be stated loosely:

*Informal Definition:* Failure 3 occurs over a set of sources $S'$ if (the evidence from) $S'$ is complementary and dispersed.<fnref n="27" />

What means ‘complementarity’ and ‘dispersed’? We leave ‘complementary’ unspecified for a moment. We call $S'$ *dispersed* if all sources in $S'$ are available to the group but not to all members, i.e., $S' \subseteq \cup_i S_i$ but $S' \not\subseteq S_i$ for at least one person $i$. So, someone or even everyone does not fully access $S'$, hence fails use the complementarity in $(e_s)_{s \in S'}$ (if any). Worse, fewer persons use this complementarity than use any given evidence in $(e_s)_{s \in S'}$, because anyone who fully accesses $S'$ must access any given $s \in S'$. So, the complementarity is *underused* compared to isolated evidences. Exactly this constitutes the failure. The parallel to Failure 1 is clear: While in Failure 1 some evidence is less accessed and thus underappreciated, in Failure 3 some complementarity is less accessed and thus underappreciated.

Now assume deliberation creates a new source profile ($S_i^+$), following any monotonic deliberation process, e.g., a share-absorb process or one of its three extensions in Section 7.3. ‘Monotonic’ means that nobody loses sources: $S_i \subseteq S_i^+$ for all persons $i$. How is Failure 3 affected? The good news is: Failure 3 occurs over fewer source sets $S'$. Why? While deliberation does not affect the complementarity of $S'$, it may eliminate the dispersion of $S'$ by transforming people’s partial access to $S'$ into full access. The opposite effect is impossible, since monotonic deliberation never reduces anyone’s access to sources.

There is a caveat to the claim that monotonic deliberation reduces Failure 3: if de-
liberation can discover new sources – so that $\cup_i S_i^+$ grows beyond $\cup_i S_i$, as is possible for
certain deliberation processes such as share-discover-absorb processes – then deliberation
can create a Failure 3 over particular sets $S'$, namely sets $S'$ that contain newly discovered
source from $\cup_i S_i^+ \setminus \cup_i S_i$. But this fact can hardly count against deliberation, since the
group’s initial avoidance of Failure 3 over such sets $S'$ came merely from not yet having
discovered part of $S'$.

The complementarity of an evidence bundle $(e_s)_{s \in S'}$ can be given a precise statistical meaning. How? The central tool is that of *information*. For any evidence bundle $(e_s)_{s \in S'}$, consider a real number *info*((e_s)_{s∈S'}) representing the information in (e_s)_{s∈S'} about the state, i.e., the evidential support for state 1 against state -1. This number could be *g*((e_s)_{s∈S'}), the aggregate evidence according to the opinion structure, or it could be ‘statistical information’ according to one of the powerful approaches developed in statistics – it will be both if the opinion structure models ‘statistically rational’ opinion formation. According to the most canonical statistical approach, the information in (e_s)_{s∈S'} is defined by the likelihood-ratio, or equivalently (after changing to a logarithmic scale) by the logarithm of its likelihood-ratio. Formally, adopting this approach, let an *information function* for the opinion structure (x, (e_s), (S_i), g) (or simply for (x, (e_s))) be a function

<footnote>usually comes with unawareness of complementarities between owned and unowned evidence. In creating
awareness of new evidence, deliberation also brings the complementarity between old and new evidence
to awareness.</footnote>

<footnote n="27">²⁷In saying ‘evidence’ rather than more generally ‘influence’, we anticipate that a (proper) complementarity can only exist between evidences, not noises, as will become clear.</footnote>
**Proposition 5** Given any opinion structure $(\mathbf{x}, (e_s), (\mathbf{S}_i), g)$ (inducing an information function info), every monotonic deliberation process reduces Failure 3 in the sense that, for all $S' \subseteq S$, if Failure 3 occurs over $S'$ post-deliberation then Failure 3 already occurred over $S'$ pre-deliberation or $S'$ was unavailable pre-deliberation, formally $F3_{S'}^+ \subseteq F3_{S'} \cup \{S' \not\subseteq \cup_i \mathbf{S}_i\}$, where $F3_{S'}$ and $F3_{S'}^+$ are the events of Failure 3 over $S'$ pre- resp. post-deliberation and $\{S' \subseteq \cup_i \mathbf{S}_i\}$ is the event that $S' \not\subseteq \cup_i \mathbf{S}_i$.

*Proof sketch.* Given the assumptions, we must show that $F3_{S'}^+ \subseteq F3_{S'} \cup \{S' \not\subseteq \cup_i \mathbf{S}_i\}$, or equivalently that $F3_{S'}^+ \cap \{S' \subseteq \cup_i \mathbf{S}_i\} \subseteq F3_{S'}$. Let $C_{S'}$ and $D_{S'}$ denote the events that $S'$ is complementary resp. dispersed pre-deliberation, and $C_{S'}^+$ and $D_{S'}^+$ the corresponding events post-deliberation. By definition, $F3_{S'} = C_{S'} \cap D_{S'}$ and $F3_{S'}^+ = C_{S'}^+ \cap D_{S'}^+$. Now $C_{S'} = C_{S'}^+$, as events of complementarity are unaffected by individual source sets, hence by deliberation. So it suffices to show that $C_{S'} \cap D_{S'}^+ \cap \{S' \subseteq \cup_i \mathbf{S}_i\} \subseteq C_{S'} \cap D_{S'}$, or simply that $D_{S'}^+ \cap \{S' \subseteq \cup_i \mathbf{S}_i\} \subseteq D_{S'}$. In words: if $S'$ is dispersed post-deliberation and initially already available, then $S'$ is already dispersed pre-deliberation. This holds as $\mathbf{S}_i \subseteq \mathbf{S}_i^+$ for all $i$ by monotonicity. ■

The result simplifies if attention is restricted to non-discovering deliberation processes, in which no new sources are discovered, i.e., for which $\cup_i \mathbf{S}_i^+ \subseteq \cup_i \mathbf{S}_i$:

**Corollary 1** Given any opinion structure $(\mathbf{x}, (e_s), (\mathbf{S}_i), g)$ (with an induced information function info), every monotonic and non-discovering deliberation process reduces Failure 3 in the sense that, for all $S' \subseteq S$, if Failure 3 occurs over $S'$ post-deliberation then Failure 3 already occurred over $S'$ pre-deliberation, formally $F3_{S'}^+ \subseteq F3_{S'}$.

*Proof.* Note $F3_{S'}^+ \subseteq \{S' \subseteq \cup_i \mathbf{S}_i^+\} = \{S' \subseteq \cup_i \mathbf{S}_i\}$; the second equality holds because $\cup_i \mathbf{S}_i^+ = \cup_i \mathbf{S}_i$ by ‘non-discovering’ and ‘monotonic’. So, in Proposition 5, ‘$F3_{S'}^+ \subseteq F3_{S'} \cup \{S' \not\subseteq \cup_i \mathbf{S}_i\}$’ is equivalent to ‘$F3_{S'}^+ \subseteq F3_{S'}$’. ■

The above analysis suggests measuring the degree of complementarity of an evidence bundle $(e_s)_{s \in S'}$ by the gap between information in combined evidence and total isolated information:

$$comp((e_s)_{s \in S'}) = info((e_s)_{s \in S'}) - \sum_{s \in S'} info(e_s).$$

This measure captures the ‘relational’ rather than ‘isolated’ information in a bundle, i.e., the information contained in the relationship between evidences rather than in each evidences intrinsically. The evidence is complementary simpliciter whenever the measure takes a non-zero value.

# 9 Concluding Remarks

Knowledge that groups hold in a dispersed fashion is often used poorly because voting is bad at aggregating the knowledge underlying votes. The fundamental tension between respecting voter's equality and achieving factually correct decisions is therefore hard to resolve. Giving up the principle of one-person-one-vote is unpalatable, but holding on to it is short of rare symmetries – epistemically suboptimal.

We show that deliberation can mitigate the tension and enable electoral democracies and other groups to make better use of evidence. The effect of deliberation on collective
info that maps each bundle $(e_s)_{s \in S'}$ in $\cup_{S' \subseteq S} \Pi_{s \in S'} \mathcal{E}_s$ to a real number given by

$$ \mathrm{info}((e_s)_{s \in S'}) = \log \frac{f_{S'}((e_s)_{s \in S'}|1)}{f_{S'}((e_s)_{s \in S'}|-1)} \quad (7) $$

if $f_{S'}((e_s)_{s \in S'}|1), f_{S'}((e_s)_{s \in S'}|-1) \neq 0$, where, for each source set $S' \subseteq S$, $f_{S'}(\cdot|\mathbf{x})$ is a joint probability density or mass function of the bundle $(\mathbf{e}_s)_{s \in S'}$ conditional on the state $\mathbf{x}$ (with respect to some measure over $\Pi_{s \in S'} \mathcal{E}_s$). For instance, $f_{S'}(\cdot|\mathbf{x})$ could be a multivariate Gaussian density function over $\mathbb{R}^{S'}$, if evidences are Gaussian. The likelihood-ratio tells how much more likely state 1 makes the bundle than state $-1$. If this ratio is above (resp. below, equal to) 1, then info((e_s)_{s \in S'}) is positive (resp. negative, zero), indicating support for 1 (resp. for $-1$, for neither). When $S'$ is a singleton set $\{s\}$, then we simply write 'info(e_s)' and talk of the 'information in e_s'.

We call the source set $S'$ or the evidence bundle $(e_s)_{s \in S'}$ complementarity if the combined information info((e_s)_{s \in S'}) differs from the total isolated information $\sum_{s \in S'} info(e_s)$. Such complementarity is ruled out if evidences are state-conditionally independent.<fnref n="28" /> This is why simple opinion structures rule out Failure 3.

The complementarity of a set $S'$ can be improper, being merely inherited from that of some subset $S'' \subsetneq S'$. If for instance three evidences are complementary merely because the first two are complementary while the third is independent, then the triple is improperly complementary. A formal definition of Failure 3 must rule out improper complementarity of $S'$, since there is obviously no real neglect of complementarity if an improperly complementary set $S'$ is not fully accessed whilst its properly complementary subsets are fully accessed. Formally, a complementary set $S'$ is properly complementary if each subset $S''$ of $S'$ is complementary with the rest of $S'$, i.e., $\mathrm{info}((\mathbf{e}_s)_{s \in S'}) \neq \mathrm{info}((\mathbf{e}_s)_{s \in S''}) + \mathrm{info}((\mathbf{e}_s)_{s \in S' \setminus S''})$.

We are ready to state the above notions and result formally:<fnref n="29" />

**Definition 1** Given an opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i), g)$ (with an information function info) and a source set $S' \subseteq S$,

* Dispersion of $S'$ is the event that $S' \subseteq \cup_i \mathbf{S}_i$ and $S' \not\subseteq \mathbf{S}_i$ for some $i$, i.e., $S'$ is available to the group but not to all members,

* Complementarity of $S'$ is the event that $\mathrm{info}((\mathbf{e}_s)_{s \in S'}) \neq \sum_{s \in S'} \mathrm{info}(\mathbf{e}_s)$, i.e., information in combined evidence differs from total isolated information,

* Proper complementarity of $S'$ is the event that $S'$ is complementary and moreover each subset $S'' \subseteq S'$ is complementary with $S' \setminus S''$, i.e., $\mathrm{info}((\mathbf{e}_s)_{s \in S'}) \neq \mathrm{info}((\mathbf{e}_s)_{s \in S''}) + \mathrm{info}((\mathbf{e}_s)_{s \in S' \setminus S''})$.

* Failure 3 over $S'$ is the event that $S'$ is properly complementary and dispersed.

**Proposition 5** Given any opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i), g)$ (with an information function info), every monotonic deliberation process reduces Failure 3 in the sense that, for all $S' \subseteq S$, if Failure 3 occurs over $S'$ post-deliberation then Failure 3 already occurred over

<footnote n="28">²⁸Under state-conditional independence, the joint likelihood function factorises: $f_{S'}((e_s)_{s \in S'}|\mathbf{x}) = \prod_{s \in S'} f_{\{s\}}(e_s|\mathbf{x})$ for each state $\mathbf{x} \in \{\pm 1\}$. Taking the ratio across states and then the logarithm on both sides yields $\mathrm{info}((e_s)_{s \in S'}) = \sum_{s \in S'} \mathrm{info}(e_s)$.</footnote>

<footnote n="29">²⁹Strictly speaking, density function need neither exist nor be unique. But in practical applications, there usually exists a single natural version of $f(\cdot|\mathbf{x})$, and hence of the information function info. Hence, the events to be defined based on info, i.e., events of complementarity and Failure 3, will exist and be essentially unique in practical applications.</footnote>
decisions can be studied at two levels: the general level of overall correctness probability of outcomes, or the level of specific failures that threaten collective correctness. At the general level, we have presented the (to our best knowledge) first jury theorems about the effect of deliberation on majority decisions. They suggest that deliberation can increase group competence, though not overcoming the objective limits of available evidence. This message is different from that of orthodox jury theorems, which might suggest that deliberation is unnecessary (because large group performance is very well anyway) or even harmful (because voter independence is undermined).

We have studied three information processing failures: overcounting widespread evidence, neglecting evidential inequality, and neglecting evidential complementarity. A mixture of theoretic results and simulations supports an overall positive picture: deliberation tends to mitigate these failures. But there are systematic exceptions. A typology of harmful deliberation has been presented. This analysis led to a recommendation for deliberation processes that are *participatory and neutral*: everyone contributes substantially and no evidences are privileged over others.

Interestingly, while our epistemic approach to democracy suggests participatory and neutral deliberation, a procedural-fairness approach to democracy might instead suggest participatory and equal deliberation, because participation adds legitimacy to outcomes and equality is a central fairness requirement (e.g., Manin 1987, Christiano 1996, Habermas 1996, Knight and Johnson 1997, Cohen 1997, Young 2002). A hybrid epistemic-procedural approach might therefore recommend participatory, neutral and equal deliberation.

Taken together, our results provide a robust argument in favor of pre-ballot deliberation on epistemic grounds. Deliberation is not only valuable because democratic citizens owe one another reasons, or because the practice of deliberation is intrinsically valuable, or because deliberation helps structuring voter preferences such as to escape voting paradoxes. It is also valuable because it helps groups make better use of their evidence when voting.

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Christiano, Thomas. 1996. "Deliberative Equality and Democratic Order". *Nomos* 38: 251–87.
S' pre-deliberation or S' was unavailable pre-deliberation, formally $F3_{S'}^+ \subseteq F3_{S'} \cup \{S' \not\subseteq \cup_i S_i\}$, where $F3_{S'}$ and $F3_{S'}^+$ are the events of Failure 3 over S' pre- resp. post-deliberation and $\{S' \subseteq \cup_i S_i\}$ is the event that $S' \not\subseteq \cup_i S_i$.

*Proof sketch.* Given the assumptions, we must show that $F3_{S'}^+ \subseteq F3_{S'} \cup \{S' \not\subseteq \cup_i S_i\}$, or equivalently that $F3_{S'}^+ \cap \{S' \subseteq \cup_i S_i\} \subseteq F3_{S'}$. Let $C_{S'}$ and $D_{S'}$ denote the events that S' is complementary resp. dispersed pre-deliberation, and $C_{S'}^+$ and $D_{S'}^+$ the corresponding events post-deliberation. By definition, $F3_{S'} = C_{S'} \cap D_{S'}$ and $F3_{S'}^+ = C_{S'}^+ \cap D_{S'}^+$. Now $C_{S'} = C_{S'}^+$, as complementarity events are unaffected by individual source sets, hence by deliberation. So it suffices to show that $C_{S'} \cap D_{S'}^+ \cap \{S' \subseteq \cup_i S_i\} \subseteq C_{S'} \cap D_{S'}$, or simply that $D_{S'}^+ \cap \{S' \subseteq \cup_i S_i\} \subseteq D_{S'}$. In words: if S' is dispersed post-deliberation and initially already available, then S' is already dispersed pre-deliberation. This holds as $S_i \subseteq S_i^+$ for all i by monotonicity. ■

The result simplifies if attention is restricted to non-discovering deliberation processes, in which no new sources are discovered, i.e., for which $\cup_i S_i^+ \subseteq \cup_i S_i$:

**Corollary 1** Given any opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i), g)$ (with an information function info), every monotonic and non-discovering deliberation process reduces Failure 3 in the sense that, for all $S' \subseteq S$, if Failure 3 occurs over $S'$ post-deliberation then Failure 3 already occurred over $S'$ pre-deliberation, formally $F3_{S'}^+ \subseteq F3_{S'}$.

*Proof.* Note $F3_{S'}^+ \subseteq \{S' \subseteq \cup_i S_i^+\} = \{S' \subseteq \cup_i S_i\}$; the second equality holds because $\cup_i S_i^+ = \cup_i S_i$ by ‘non-discovering’ and ‘monotonic’. So, in Proposition 5, ‘$F3_{S'}^+ \subseteq F3_{S'} \cup \{S' \not\subseteq \cup_i S_i\}$’ is equivalent to ‘$F3_{S'}^+ \subseteq F3_{S'}$’. ■

The above analysis suggests measuring the degree of complementarity of an evidence bundle $(e_s)_{s \in S'}$ by the gap between information in combined evidence and total isolated information:

$$comp((e_s)_{s \in S'}) = info((e_s)_{s \in S'}) - \sum_{s \in S'} info(e_s).$$

This measure captures the ‘relational’ rather than ‘isolated’ information in a bundle, i.e., the information contained in the relationship between evidences rather than in each evidences intrinsically. The evidence is complementary simpliciter whenever the measure takes a non-zero value.

# 9 Concluding Remarks

Knowledge that groups hold in a dispersed fashion is often used poorly because voting is bad at aggregating the knowledge underlying votes. The fundamental tension between respecting voter equality and achieving factually correct decisions is therefore hard to resolve. Giving up the principle of one-person-one-vote is unpalatable, but holding on to it is – short of rare symmetries – epistemically suboptimal.

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Hoek, Daniel, and Richard Bradley. 2022. "Million dollar questions: why deliberation
general level, we have presented the (to our best knowledge) first jury theorems about the
effect of deliberation on majority decisions. They suggest that deliberation can increase
group competence, though not overcoming the objective limits of available evidence. This
message is different from that of orthodox jury theorems, which might suggest that delib-
eration is unnecessary (because large group perform very well anyway) or even harmful
(because voter independence is undermined).

We have studied three information processing failures: overcounting widespread evidence, neglecting evidential inequality, and neglecting evidential complementarity. A mixture of theoretic results and simulations supports an overall positive picture: deliberation tends to mitigate these failures. But there are systematic exceptions. A typology of harmful deliberation has been presented. This analysis led to a recommendation for deliberation processes that are *participatory and neutral*: everyone contributes substantially and no evidences are privileged over others.

Interestingly, while our epistemic approach to democracy suggests participatory and neutral deliberation, a procedural-fairness approach to democracy might instead suggest participatory *and equal* deliberation, because participation adds legitimacy to outcomes and equality is a central fairness requirement (e.g., Manin 1987, Christiano 1996, Habermas 1996, Knight and Johnson 1997, Cohen 1997, Young 2002). A hybrid epistemic-procedural approach might therefore recommend participatory, neutral and equal deliberation.

Taken together, our results provide a robust argument in favor of pre-ballot deliberation on epistemic grounds. Deliberation is not only valuable because democratic citizens owe one another reasons, or because the practice of deliberation is intrinsically valuable, or because deliberation helps structuring voter preferences such as to escape voting paradoxes. It is also valuable because it helps groups make better use of their evidence when voting.

References

Aristotle, 1988. *The Politics*. Cambridge Texts in the History of Political Thought.
Cambridge: Cambridge University Press.

Austen-Smith, David, and Jeffrey S. Banks. 1996. "Information Aggregation, Rationality, and the Condorcet Jury Theorem," *American Political Science Review* 90 (1): 34–45.

Austen-Smith, David, and Timothy J. Feddersen. 2006. “Deliberation, Preference Uncertainty, and Voting Rules.” *American Political Science Review* 100 (2): 209–17.
https://doi.org/10.1017/S0003055406062113.

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https://doi.org/10.3982/ECTA16376.

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# Appendix

## A Proof of Theorems 1 and 2 about rationality

We first prove Theorem 1, in fact generalised to quasi-simple opinion structures $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$. Such opinion structures are defined exactly like simple ones except that Independent Sources is weakened to Quasi-Independent Sources, the condition that the source-access events $s \in S$ and $i \in N$) are independent across sources $s$ and jointly independent of the state and the evidences, i.e., of $(\mathbf{x}, (\mathbf{e}_s))$. This condition weakens Independent Sources by no longer requiring independence across persons of the source-access events. The generalisation ensures that the theorem also captures post-deliberation opinions. Indeed, a share-absorb process transforms a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ into a quasi-simple one $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i^+))$ – ‘quasi’ because of interpersonal source dependencies.

We begin by proving an astonishing fact about Gaussian evidences:

**Lemma 1** If an opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfies Simple Gaussian Evidence (e.g., is quasi-simple), then each evidence is proportional to its own log-likelihood-ratio, more precisely

$$ \mathbf{e}_s = \frac{\sigma^2}{2} \log \frac{f(\mathbf{e}_s|1)}{f(\mathbf{e}_s|-1)} \quad \text{for each } s \in S, $$

where $f(\cdot|x)$ denotes the Gaussian density ('likelihood') function of each evidence $\mathbf{e}_s$ ($s \in S$) given state $x$ ($\in \{\pm 1\}$).

*Proof.* Let $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfy Simple Gaussian Evidence. Let $s \in S$. Conditional on a state $x$ ($\in \{\pm 1\}$), $\mathbf{e}_s$ is normally distributed with mean $x$ and variance $\sigma^2$, hence has a Gaussian density function given by

$$ f(e|x) = \frac{1}{\sigma(2\pi)^{1/2}} e^{-\frac{1}{2} \left(\frac{e-x}{\sigma}\right)^2} \quad \text{for all } e \in \mathbb{R}. $$

We have $\mathbf{e}_s = \frac{\sigma^2}{2} \log \frac{f(\mathbf{e}_s|1)}{f(\mathbf{e}_s|-1)}$ because, for all values $e \in \mathbb{R}$ of $\mathbf{e}_s$,

$$ \begin{aligned} \log \frac{f(e|1)}{f(e|-1)} &= \log \frac{\exp\left(-\frac{1}{2}\left(\frac{e-1}{\sigma}\right)^2\right)}{\exp\left(-\frac{1}{2}\left(\frac{e+1}{\sigma}\right)^2\right)} = \log\left[\exp\left(\frac{1}{2}\left(\frac{e+1}{\sigma}\right)^2 - \frac{1}{2}\left(\frac{e-1}{\sigma}\right)^2\right)\right] \\ &= \frac{1}{2\sigma^2} \left[(e^2 + 2e + 1) - (e^2 - 2e + 1)\right] = \frac{2}{\sigma^2}e. \blacksquare \end{aligned} $$

*Proof of Theorem 1 generalised to quasi-simple opinion structures.* Fix a quasi-simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$, a person $i$, and a possible opinion of $i$, i.e., a random variable $\mathbf{o}$ generating values in $\{1, 0, -1\}$ based on $i$'s evidence bundle $(\mathbf{e}_s)_{s \in \mathbf{S}_i}$. We must show that $\mathbb{E}(u(\mathbf{o}_i, \mathbf{x})) \ge \mathbb{E}(u(\mathbf{o}, \mathbf{x}))$, where the utility of any opinion-state pair $(o, x)$ in $\{1, 0, -1\} \times \{1, -1\}$ is the correctness level, given by

$$ u(o, x) = \begin{cases} 1 & \text{if } o = x \text{ (correct opinion)} \\ 0 & \text{if } o = -x \text{ (false opinion)} \\ \frac{1}{2} & \text{if } o = 0 \text{ (neutral opinion).} \end{cases} $$
Hong, Lu, and Scott E Page. 2004. "Groups of Diverse Problem Solvers Can Outperform Groups of High-Ability Problem Solvers." *Proceedings of the National Academy of Sciences of the United States of America* 101 (46): 16385-89.

Hong, Lu, and Scott E Page. 2012. "Some Microfoundations of Collective Wisdom." In *Collective Wisdom: Principles and Mechanisms*, edited by Hélène Landemore and Jon Elster, 56-71. Cambridge: Cambridge University Press.

Knight, Jack, and James Johnson. 1997. "What Sort of Political Equality Does Deliberative Democracy Require?" In *Deliberative Democracy: Essays on Reason and Politics*, edited by James Bohman and William Rehg, 279-319. Cambridge MA: MIT Press.

Ladha, Krishna K. 1992. "The Condorcet Jury Theorem, Free Speech, and Correlated Votes." *American Journal of Political Science* 36 (3): 617-34.

Landa, Dimitri, and Adam Meirowitz. 2009. "Game Theory, Information, and Deliberative Democracy." *American Journal of Political Science* 53 (2): 427-44.
https://doi.org/10.1111/j.1540-5907.2009.00379.x.

List, Christian, and Robert E. Goodin. 2001. "Epistemic Democracy: Generalizing the Condorcet Jury Theorem." *Journal of Political Philosophy* 9 (3): 277-306.

Manin, Bernard. 1987. "On Legitimacy and Political Deliberation". *Political Theory* 15 (3): 338-68. https://doi.org/10.1177/0090591787015003005.

McLean, Iain, and Fiona Hewitt, eds. 1994. *Condorcet: Foundations of Social Choice and Political Theory*. Aldershot, Hants, England; Brookfield, VT: Elgar.

Min, John B., and James K. Wong. 2018. "Epistemic Approaches to Deliberative Democracy." *Philosophy Compass* 13 (6): e12497.
https://doi.org/10.1111/phc3.12497.

Müller, Julian F. 2018. "Epistemic Democracy: Beyond Knowledge Exploitation." *Philosophical Studies* 175 (5): 1267-88. https://doi.org/10.1007/s11098-017-0910-9.

Page, Scott E. 2007. *The Difference: How the Power of Diversity Creates Better Groups, Firms, Schools, and Societies*. Princeton (NJ): Princeton University Press.

Perote-Peña, Juan, and Ashley Piggins. 2015. "A Model of Deliberative and Aggregative Democracy." *Economics & Philosophy* 31 (1): 93-121.
https://doi.org/10.1017/S0266267114000418.

Pivato, Marcus. 2016. "Statistical Utilitarianism." In *The Political Economy of Social Choices*, edited by Maria Gallego and Norman Schofield, 187-204. Studies in Political Economy. Cham: Springer International Publishing.
https://doi.org/10.1007/978-3-319-40118-8_8.

Pivato, Marcus. 2017. "Epistemic Democracy with Correlated Voters." *Journal of Mathematical Economics* 72: 51-69. https://doi.org/10.1016/j.jmateco.2017.06.001.

Rafiee Rad, Soroush, and Olivier Roy. 2021. "Deliberation, Single-Peakedness, and Coherent Aggregation." *American Political Science Review* 115 (2): 629-48.
https://doi.org/10.1017/S0003055420001045.

Rousseau, Jean-Jacques. 1972. *Du Contrat Social*. Oxford: Oxford University Press.

Young, Iris Marion. 2002. *Inclusion and Democracy*. Oxford: Oxford University Press.
We prove this by showing that $\mathbb{E}(u(o_i, x)|(e_s)_{s \in S_i}) \ge \mathbb{E}(u(o, x)|(e_s)_{s \in S_i})$.<fnref n="30" /> So, we fix any value $(e_s)_{s \in S_i}$ of $(e_s)_{s \in S_i}$ and, writing $o_i$ (resp. $o$) for the value of $o_i$ (resp. $o$) under $(e_s)_{s \in S_i}$, we must prove that

$$ \mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) \ge \mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}). \quad (7) $$

We prove this in two steps.

**Claim 1:** $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > (<, =) \frac{1}{2} \Leftrightarrow \sum_{s \in S_i} e_s > (<, =) 0$.

We only prove the equivalence for ‘>’, as those for ‘<’ and ‘=’ are analogous. For any state $x \in \{\pm 1\}$, writing $f(\cdot|x)$ for the Gaussian density function on $\mathbb{R}$ of any $e_s$ given $x$, and $g(\cdot|x)$ for the $|S_i|$-dimensional Gaussian density function on $\mathbb{R}^{S_i}$ of the vector $(e_s)_{s \in S_i}$, we have

$$ \begin{align*} \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2} &\Leftrightarrow \frac{g((e_s)_{s \in S_i}|1)}{g((e_s)_{s \in S_i}|-1)} > 1 \\ &\Leftrightarrow \prod_{s \in S_i} \frac{f(e_s|1)}{f(e_s|-1)} > 1 \\ &\Leftrightarrow \sum_{s \in S_i} \log \frac{f(e_s|1)}{f(e_s|-1)} > 0 \\ &\Leftrightarrow \sum_{s \in S_i} e_s > 0. \end{align*} $$

Here, the first equivalence follows easily from Bayes' rule, using that $\mathrm{Pr}(\mathbf{x} = 1) = \mathrm{Pr}(\mathbf{x} = -1)$ and also that $i$'s source set is independent of the state and the evidences.<fnref n="31" /> The second equivalence holds by state-conditional independence of the evidences. The third equivalence holds by applying the logarithm on both sides of the previous inequality. The fourth equivalence holds by Lemma 1. Q.e.d.

**Claim 2:** The inequality (7) holds (completing the proof).

We proceed case by case.

**Case 1:** $o_i = 1$, i.e., $\sum_{s \in S_i} e_s > 0$. Then $\mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2}$, where the inequality holds by Claim 1 as $\sum_{s \in S_i} e_s > 0$.

**Subcase 1.1:** $o = 1$. Then (7) holds (with ‘=’) because $o_i = o$.

**Subcase 1.2:** $o = -1$. Then (7) holds (with ‘>’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = 1 - \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) < \frac{1}{2}$, where the last inequality holds as $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2}$.

**Subcase 1.3:** $o = 0$. Then (7) holds (with ‘>’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \frac{1}{2}$.

**Case 2:** $o_i = -1$, i.e., $\sum_{s \in S_i} e_s < 0$. Then $\mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = 1 - \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2}$, where the inequality holds because $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) < \frac{1}{2}$ by Claim 1 as $\sum_{s \in S_i} e_s < 0$. An argument similar to that in Case 1 then implies (7).

**Case 3:** $o_i = 0$, i.e., $\sum_{s \in S_i} e_s = 0$. Then $\mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) = \frac{1}{2}$. Further, $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = \frac{1}{2}$, by Claim 1 as $\sum_{s \in S_i} e_s = 0$.

**Subcase 3.1:** $o = 0$. Then (7) holds (with ‘=’) because $o_i = o$.

<footnote n="30">³⁰Strictly speaking, we must show that this inequality holds for some versions of the conditional expectations on both sides. This qualification is necessary because conditional expectations are random variables that are not unique, but still ‘essentially unique’ in that any two versions of a conditional expectation coincide outside a zero-probability event.</footnote>

<footnote n="31">³¹How is this independence condition used here? Informally, it ensures that the identity of the source set $S_i$ is of no extra information, i.e., that only the evidences from those sources carry information. This explains why the numerator and denominator of the likelihood-ratio each features the joint (state-conditional) likelihood of the evidences only, not of the evidences and the source set $S_i$. To be slightly more explicit, conditionalising on the evidence bundle $(e_s)_{s \in S_i}$ is equivalent to conditionalising first on the source set $S_i$ and then on the evidences from these sources; which however reduces to conditionalising only on the evidences, by the independence condition.</footnote>
# Appendix

## A Proof of Theorems 1 and 2 about rationality

We first prove Theorem 1, in fact generalised to quasi-simple opinion structures $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$. Such opinion structures are defined exactly like simple ones except that Independent Sources is weakened to Quasi-Independent Sources, the condition that the source-access events $s \in S$ and $i \in N$) are independent across sources $s$ and jointly independent of the state and the evidences, i.e., of $(\mathbf{x}, (\mathbf{e}_s))$. This condition weakens Independent Sources by no longer requiring independence across persons of the source-access events. The generalisation ensures that the theorem also captures post-deliberation opinions. Indeed, a share-absorb process transforms a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ into a quasi-simple one $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i^+))$ – ‘quasi’ because of interpersonal source dependencies.

We begin by proving an astonishing fact about Gaussian evidences:

**Lemma 1** If an opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfies Simple Gaussian Evidence (e.g., is quasi-simple), then each evidence is proportional to its own log-likelihood-ratio, more precisely

$$ \mathbf{e}_s = \frac{\sigma^2}{2} \log \frac{f(\mathbf{e}_s|1)}{f(\mathbf{e}_s|-1)} \quad \text{for each } s \in S, $$

where $f(\cdot|x)$ denotes the Gaussian density ('likelihood') function of each evidence $\mathbf{e}_s$ ($s \in S$) given state $x$ ($\in \{\pm 1\}$).

*Proof.* Let $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfy Simple Gaussian Evidence. Let $s \in S$. Conditional on a state $x$ ($\in \{\pm 1\}$), $\mathbf{e}_s$ is normally distributed with mean $x$ and variance $\sigma^2$, hence has a Gaussian density function given by

$$ f(e|x) = \frac{1}{\sigma(2\pi)^{1/2}} e^{-\frac{1}{2} \left(\frac{e-x}{\sigma}\right)^2} \quad \text{for all } e \in \mathbb{R}. $$

We have $\mathbf{e}_s = \frac{\sigma^2}{2} \log \frac{f(\mathbf{e}_s|1)}{f(\mathbf{e}_s|-1)}$ because, for all values $e \in \mathbb{R}$ of $\mathbf{e}_s$,

$$ \log \frac{f(e|1)}{f(e|-1)} = \log \frac{\exp\left(-\frac{1}{2}\left(\frac{e-1}{\sigma}\right)^2\right)}{\exp\left(-\frac{1}{2}\left(\frac{e+1}{\sigma}\right)^2\right)} = \log \left[ \exp\left(\frac{1}{2}\left(\frac{e+1}{\sigma}\right)^2\right) - \exp\left(\frac{1}{2}\left(\frac{e-1}{\sigma}\right)^2\right) \right] \\ = \frac{1}{2\sigma^2} \left[ (e^2 + 2e + 1) - (e^2 - 2e + 1) \right] = \frac{2}{\sigma^2}e. \blacksquare $$

*Proof of Theorem 1 generalised to quasi-simple opinion structures.* Fix a quasi-simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$, a person $i$, and a possible opinion of $i$, i.e., a random variable $\mathbf{o}$ generating values in $\{1, 0, -1\}$ based on $i$'s evidence bundle $(\mathbf{e}_s)_{s \in \mathbf{S}_i}$. We must show that $\mathbb{E}(u(\mathbf{o}_i, \mathbf{x})) \ge \mathbb{E}(u(\mathbf{o}, \mathbf{x}))$, where the utility of any opinion-state pair $(o, x)$ in $\{1, 0, -1\} \times \{1, -1\}$ is the correctness level, given by

$$ u(o, x) = \begin{cases} 1 & \text{if } o = x \text{ (correct opinion)} \\ 0 & \text{if } o = -x \text{ (false opinion)} \\ \frac{1}{2} & \text{if } o = 0 \text{ (neutral opinion).} \end{cases} $$
Subcase 3.2: $o = 1$. Then (7) holds (with ‘=’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \text{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) = \frac{1}{2}$.

Subcase 3.3: $o = -1$. Then (7) holds (with ‘=’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \text{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = \frac{1}{2}$. ■

We now prove Theorem 2, again starting with a lemma.

**Lemma 2** If an opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfies Generalised Gaussian Evidence, then for each $\emptyset \neq S' \subseteq S$ the vector $(\mathbf{e}_s)_{s \in S'}$ has a log-likelihood-ratio given by

$$ \log \frac{f((e_s)_{s \in S'}|1)}{f((e_s)_{s \in S'}|-1)} = 2g((e_s)_{s \in S_i}) \quad \text{for all } (e_s)_{s \in S'} \in \mathbb{R}^{S'}, $$

where $f(\cdot|x)$ denotes the density function of the vector $(\mathbf{e}_s)_{s \in S'}$ given state $x$ ($\in \{\pm 1\}$), i.e., the multivariate Gaussian density function with mean $x\mu|_{S'}$ and covariance matrix $\Sigma|_{S'}$.

*Proof.* Let $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfy Generalised Gaussian Evidence. Fix $\emptyset \neq S' \subseteq S$. Write $\mu'$ for $\mu|_{S'}$ and $\Sigma'$ for $\Sigma|_{S'}$. Given any $x \in \{\pm 1\}$, $(\mathbf{e}_s)_{s \in S'}$ has an $N(x\mu', \Sigma')$ distribution, with density function given by:

$$ f(e|x) = \frac{1}{(\det \Sigma')^{1/2}(2\pi)^{|S'|/2}} \exp \left( -\frac{1}{2}(e - x\mu')^T \Sigma'^{-1} (e - x\mu') \right) \quad \text{for all } e \in \mathbb{R}^{S'}. $$

So, for all $e \in \mathbb{R}^{S'}$,

$$ \begin{aligned} \log \frac{f(e|1)}{f(e|-1)} &= \log \frac{\frac{1}{(\det \Sigma')^{1/2}(2\pi)^{|S'|/2}} \exp \left(-\frac{1}{2}(e-\mu')^T \Sigma'^{-1} (e-\mu')\right)}{\frac{1}{(\det \Sigma')^{1/2}(2\pi)^{|S'|/2}} \exp \left(-\frac{1}{2}(e+\mu')^T \Sigma'^{-1} (e+\mu')\right)} \\ &= \log \exp \left( \frac{1}{2} \left[ (e+\mu')^T \Sigma'^{-1} (e+\mu') - (e-\mu')^T \Sigma'^{-1} (e-\mu') \right] \right) \\ &= \frac{1}{2} \left[ e^T \Sigma'^{-1} e + 2\mu'^T \Sigma'^{-1} e + \mu'^T \Sigma'^{-1} \mu' - (e^T \Sigma'^{-1} e - 2\mu'^T \Sigma'^{-1} e + \mu'^T \Sigma'^{-1} \mu') \right] \\ &= 2\mu'^T \Sigma'^{-1} e = 2g(e). \blacksquare \end{aligned} $$

*Proof of Theorem 2 generalised to generalised quasi-simple opinion structures.* Let $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ be generalised quasi-simple, i.e., generalised simple except that ‘Independent Sources’ is weakened to ‘Quasi-Independent Sources’. Let $\mathbf{o}$ be a possible opinion of a given person $i$, i.e., a random variable into $\{1, 0, -1\}$ based on $(\mathbf{e}_s)_{s \in \mathbf{S}_i}$. We show $\mathbb{E}(u(\mathbf{o}_i, \mathbf{x})) \ge \mathbb{E}(u(\mathbf{o}, \mathbf{x}))$, with $u$ defined as before. As in the proof of Theorem 1, it suffices to fix a value $(e_s)_{s \in \mathbf{S}_i}$ of $(\mathbf{e}_s)_{s \in \mathbf{S}_i}$, and, writing $o_i$ and $o$ for the value of $\mathbf{o}_i$ resp. $\mathbf{o}$ under $(e_s)_{s \in \mathbf{S}_i}$, to prove that

$$ \mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in \mathbf{S}_i}) \ge \mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in \mathbf{S}_i}). \quad (8) $$

First assume $S_i = \emptyset$. Then (8) reduces to $\mathbb{E}(u(o_i, \mathbf{x})) \ge \mathbb{E}(u(o, \mathbf{x}))$, which holds with ‘=’ because, firstly, $\mathbb{E}(u(o_i, \mathbf{x})) = \mathbb{E}(u(0, \mathbf{x})) = \mathbb{E}(0) = 0$, and secondly, $\mathbb{E}(u(o, \mathbf{x})) = \frac{1}{2}(u(o, 1) + u(o, 0))$, which equals 0 regardless of the value of $o$, since $o = 1 \Rightarrow [u(o, 1) = 1 \& u(o, -1) = -1]$, $o = -1 \Rightarrow [u(o, 1) = -1 \& u(o, -1) = 1]$, and $o = 0 \Rightarrow u(o, 1) = u(o, -1) = 0$.

From now on let $S_i \neq \emptyset$. The proof proceeds in two steps.

*Claim 1:* $\text{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > (<, =) \frac{1}{2} \Leftrightarrow g((e_s)_{s \in S_i}) > (<, =) 0$.
We prove this by showing that $\mathbb{E}(u(o_i, x)|(e_s)_{s \in S_i}) \ge \mathbb{E}(u(o, x)|(e_s)_{s \in S_i})$.<fnref n="30" /> So, we fix any value $(e_s)_{s \in S_i}$ of $(e_s)_{s \in S_i}$ and, writing $o_i$ (resp. $o$) for the value of $o_i$ (resp. $o$) under $(e_s)_{s \in S_i}$, we must prove that

$$ \mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) \ge \mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}). \quad (8) $$

We prove this in two steps.

*Claim 1:* $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > (<, =) \frac{1}{2} \Leftrightarrow \sum_{s \in S_i} e_s > (<, =) 0$.

We only prove the equivalence for ‘>’, as those for ‘<’ and ‘=’ are analogous. For any state $x \in \{\pm 1\}$, writing $f(\cdot|x)$ for the Gaussian density function on $\mathbb{R}$ of any $e_s$ given $x$, and $g(\cdot|x)$ for the $|S_i|$-dimensional Gaussian density function on $\mathbb{R}^{S_i}$ of the vector $(e_s)_{s \in S_i}$, we have

$$ \begin{align*} \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2} &\Leftrightarrow \frac{g((e_s)_{s \in S_i}|1)}{g((e_s)_{s \in S_i}|-1)} > 1 \\ &\Leftrightarrow \prod_{s \in S_i} \frac{f(e_s|1)}{f(e_s|-1)} > 1 \\ &\Leftrightarrow \sum_{s \in S_i} \log \frac{f(e_s|1)}{f(e_s|-1)} > 0 \\ &\Leftrightarrow \sum_{s \in S_i} e_s > 0. \end{align*} $$

Here, the first equivalence follows easily from Bayes’ rule, using that $\mathrm{Pr}(\mathbf{x} = 1) = \mathrm{Pr}(\mathbf{x} = -1)$ and also that $i$’s source set is independent of the state and the evidences.<fnref n="31" /> The second equivalence holds by state-conditional independence of the evidences. The third equivalence holds by applying the logarithm on both sides of the previous inequality. The fourth equivalence holds by Lemma 1. Q.e.d.

*Claim 2:* The inequality (8) holds (completing the proof).

We proceed case by case.

*Case 1:* $o_i = 1$, i.e., $\sum_{s \in S_i} e_s > 0$. Then $\mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2}$, where the inequality holds by Claim 1 as $\sum_{s \in S_i} e_s > 0$.

*Subcase 1.1:* $o = 1$. Then (8) holds (with ‘=’) because $o_i = o$.

*Subcase 1.2:* $o = -1$. Then (8) holds (with ‘>’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = 1 - \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) < \frac{1}{2}$, where the last inequality holds as $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2}$.

*Subcase 1.3:* $o = 0$. Then (8) holds (with ‘>’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \frac{1}{2}$.

*Case 2:* $o_i = -1$, i.e., $\sum_{s \in S_i} e_s < 0$. Then $\mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = 1 - \mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2}$, where the inequality holds because $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) < \frac{1}{2}$ by Claim 1 as $\sum_{s \in S_i} e_s < 0$. An argument similar to that in Case 1 then implies (8).

*Case 3:* $o_i = 0$, i.e., $\sum_{s \in S_i} e_s = 0$. Then $\mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in S_i}) = \frac{1}{2}$. Further, $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) = \mathrm{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = \frac{1}{2}$, by Claim 1 as $\sum_{s \in S_i} e_s = 0$.

*Subcase 3.1:* $o = 0$. Then (8) holds (with ‘=’) because $o_i = o$.

<footnote n="30">³⁰Strictly speaking, we must show that this inequality holds for some versions of the conditional expectations on both sides. This qualification is necessary because conditional expectations are random variables that are not unique, but still ‘essentially unique’ in that any two versions of a conditional expectation coincide outside a zero-probability event.</footnote>

<footnote n="31">³¹How is this independence condition used here? Informally, it ensures that the identity of the source set $S_i$ is of no extra information, i.e., that only the evidences from those sources carry information. This explains why the numerator and denominator of the likelihood-ratio each features the joint (state-conditional) likelihood of the evidences only, not of the evidences and the source set $S_i$. To be slightly more explicit, conditionalising on the evidence bundle $(e_s)_{s \in S_i}$ is equivalent to conditionalising first on the source set $S_i$ and then on the evidences from these sources; which however reduces to conditionalising only on the evidences, by the independence condition.</footnote>
We only show the equivalence for ‘>’, as that for ‘<’ or ‘=’ is analogous. For any state $x \in \{\pm 1\}$, writing $f(\cdot|x)$ for the Gaussian density function on $\mathbb{R}^{S_i}$ of the vector $(\mathbf{e}_s)_{s \in S_i}$, we have

$$ 
\begin{align*}
\Pr(\mathbf{x} = 1|(e_s)_{s \in S_i}) > \frac{1}{2} &\Leftrightarrow \frac{f((e_s)_{s \in S_i}|1)}{f((e_s)_{s \in S_i}|-1)} > 1 \\
&\Leftrightarrow \log \frac{f((e_s)_{s \in S_i}|1)}{f((e_s)_{s \in S_i}|-1)} > 0 \\
&\Leftrightarrow g((e_s)_{s \in S_i}) > 0.
\end{align*}
$$

Here, the first equivalence holds in analogy to the corresponding equivalence in the proof of Theorem 1. The second equivalence holds trivially. The third equivalence holds Lemma 2. Q.e.d.

Claim 2: The inequality (8) holds (completing the proof).

This claim follows from Claim 1 through an argument analogous to that underlying Claim 2 in the proof of Theorem 1. ■

## B Formal Analytics of Share-Absorb Processes

The definition of share-absorb processes has been stated informally. The formalisation is obvious. In short, given a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ (we could have used a general opinion structure), the share-absorb process with parameters $(p_{s,i \to}, p_{s,i \leftarrow})_{s \in S, i \in N}$ assumes that there exist events ‘i shares s’ and ‘i absorbs s’ for any person $i \in N$ and source $s \in S$; that the new source set of any person $i$ is $\mathbf{S}_i^+ = \mathbf{S}_i \cup \{s \in S : \text{ 'i absorbs s'}\}$, the set of initially accessed or later absorbed sources; that, for any person $i$ and source $s$, the probability of ‘i shares s’ given any initial source profile $(S_j)$ is $p_{s,i \to}$ if $s \in S_i$ and 0 otherwise; that, for any person $i$ and source $s$, the probability of ‘i absorbs s’ given any initial source profile $(S_j)$ and any sharing profile is $p_{s,i \leftarrow}$ if $s \notin S_i$ and someone shares $s$ in the sharing profile] and 0 otherwise (where a ‘sharing profile’ is a combination of truth values of the sharing events across persons and sources); and, finally, that the access, sharing, and absorbing events are jointly independent of the state and the evidences.

How is the new source profile $(\mathbf{S}_i^+)$ distributed? And how is it distributed conditional on the initial evidence profile? We now answer both questions. The first answer completes the description of the new opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i^+))$, as we already know how $(\mathbf{x}, (\mathbf{e}_s))$ is distributed and that $(\mathbf{S}_i^+)$ is independent of $(\mathbf{x}, (\mathbf{e}_s))$. The second answer implies an alternative (and equivalent) definition of the share-absorb process as a deliberation process in the general sense of Section 7, i.e., a mapping $D$ from initial evidence profiles to lotteries over new source profiles.

Fix a simple<fnref n="32" /> opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ and a share-absorb process with sharing and absorbing probabilities $(p_{s,i \to}, p_{s,i \leftarrow})_{s \in S, i \in N}$, generating a new source profile $(\mathbf{S}_i^+)$. The probability of any new source profile $(S_i^+)$ (any value of $(\mathbf{S}_i^+)$) is

$$ \Pr((S_i^+)) = \prod_{s \in S} \pi_s \qquad (9) $$

where, for each source $s \in S$, $\pi_s$ is the probability that the new set of owners of $s$ is

<footnote n="32">³²Simplicity could be weakened considerably, to Independent Sources.</footnote>
Subcase 3.2: $o = 1$. Then (8) holds (with ‘=’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \text{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) = \frac{1}{2}$.

Subcase 3.3: $o = -1$. Then (8) holds (with ‘=’) because $\mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in S_i}) = \text{Pr}(\mathbf{x} = -1|(e_s)_{s \in S_i}) = \frac{1}{2}$. ■

We now prove Theorem 2, again starting with a lemma.

**Lemma 2** If an opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfies Generalised Gaussian Evidence, then for each $\emptyset \neq S' \subseteq S$ the vector $(\mathbf{e}_s)_{s \in S'}$ has a log-likelihood-ratio given by

$$ \log \frac{f((e_s)_{s \in S'}|1)}{f((e_s)_{s \in S'}|-1)} = 2g((e_s)_{s \in S_i}) \quad \text{for all } (e_s)_{s \in S'} \in \mathbb{R}^{S'}, $$

where $f(\cdot|x)$ denotes the density function of the vector $(\mathbf{e}_s)_{s \in S'}$ given state $x$ ($\in \{\pm 1\}$), i.e., the multivariate Gaussian density function with mean $x\mu|_{S'}$ and covariance matrix $\Sigma|_{S'}$.

*Proof.* Let $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ satisfy Generalised Gaussian Evidence. Fix $\emptyset \neq S' \subseteq S$. Write $\mu'$ for $\mu|_{S'}$ and $\Sigma'$ for $\Sigma|_{S'}$. Given any $x \in \{\pm 1\}$, $(\mathbf{e}_s)_{s \in S'}$ has an $N(x\mu', \Sigma')$ distribution, with density function given by:

$$ f(e|x) = \frac{1}{(\det \Sigma')^{1/2}(2\pi)^{|S'|/2}} \exp \left( -\frac{1}{2}(e - x\mu')^T \Sigma'^{-1} (e - x\mu') \right) \quad \text{for all } e \in \mathbb{R}^{S'}. $$

So, for all $e \in \mathbb{R}^{S'}$,

$$ \begin{aligned} \log \frac{f(e|1)}{f(e|-1)} &= \log \frac{\frac{1}{(\det \Sigma')^{1/2}(2\pi)^{|S'|/2}} \exp \left(-\frac{1}{2}(e-\mu')^T \Sigma'^{-1} (e-\mu')\right)}{\frac{1}{(\det \Sigma')^{1/2}(2\pi)^{|S'|/2}} \exp \left(-\frac{1}{2}(e+\mu')^T \Sigma'^{-1} (e+\mu')\right)} \\ &= \log \exp \left( \frac{1}{2} \left[ (e+\mu')^T \Sigma'^{-1} (e+\mu') - (e-\mu')^T \Sigma'^{-1} (e-\mu') \right] \right) \\ &= \frac{1}{2} \left[ e^T \Sigma'^{-1} e + 2\mu'^T \Sigma'^{-1} e + \mu'^T \Sigma'^{-1} \mu' - (e^T \Sigma'^{-1} e - 2\mu'^T \Sigma'^{-1} e + \mu'^T \Sigma'^{-1} \mu') \right] \\ &= 2\mu'^T \Sigma'^{-1} e = 2g(e). \blacksquare \end{aligned} $$

*Proof of Theorem 2 generalised to generalised quasi-simple opinion structures.* Let $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ be generalised quasi-simple, i.e., generalised simple except that ‘Independent Sources’ is weakened to ‘Quasi-Independent Sources’. Let $\mathbf{o}$ be a possible opinion of a given person $i$, i.e., a random variable into $\{1, 0, -1\}$ based on $(\mathbf{e}_s)_{s \in \mathbf{S}_i}$. We show $\mathbb{E}(u(\mathbf{o}_i, \mathbf{x})) \ge \mathbb{E}(u(\mathbf{o}, \mathbf{x}))$, with $u$ defined as before. As in the proof of Theorem 1, it suffices to fix a value $(e_s)_{s \in \mathbf{S}_i}$ of $(\mathbf{e}_s)_{s \in \mathbf{S}_i}$, and, writing $o_i$ and $o$ for the value of $\mathbf{o}_i$ resp. $\mathbf{o}$ under $(e_s)_{s \in \mathbf{S}_i}$, to prove that

$$ \mathbb{E}(u(o_i, \mathbf{x})|(e_s)_{s \in \mathbf{S}_i}) \ge \mathbb{E}(u(o, \mathbf{x})|(e_s)_{s \in \mathbf{S}_i}). \quad (9) $$

First assume $S_i = \emptyset$. Then (9) reduces to $\mathbb{E}(u(o_i, \mathbf{x})) \ge \mathbb{E}(u(o, \mathbf{x}))$, which holds with ‘=’ because, firstly, $\mathbb{E}(u(o_i, \mathbf{x})) = \mathbb{E}(u(0, \mathbf{x})) = \mathbb{E}(0) = 0$, and secondly, $\mathbb{E}(u(o, \mathbf{x})) = \frac{1}{2}(u(o, 1) + u(o, 0))$, which equals 0 regardless of the value of $o$, since $o = 1 \Rightarrow [u(o, 1) = 1 \& u(o, -1) = -1]$, $o = -1 \Rightarrow [u(o, 1) = -1 \& u(o, -1) = 1]$, and $o = 0 \Rightarrow u(o, 1) = u(o, -1) = 0$.

From now on let $S_i \neq \emptyset$. The proof proceeds in two steps.

*Claim 1:* $\text{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) > (<, =) \frac{1}{2} \Leftrightarrow g((e_s)_{s \in S_i}) > (<, =) 0$.
$I_s = \{i : s \in S_i^+\}$, and equals

$$
\pi_s = \left( \prod_{i \in I_s} p_{s \to i} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s \to i}} \right) \left( \prod_{i \in I_s} \overline{p_{s, i \to}} \right) \\
+ \left( \prod_{i \in \overline{I_s}} \overline{p_{s, i \leftarrow}} \right) \sum_{\emptyset \neq I \subseteq I_s} \left( \prod_{i \in I} p_{s \to i} \right) \left( \prod_{i \in \overline{I}} \overline{p_{s \to i}} \right) \left( \overline{\prod_{i \in I} \overline{p_{s, i \to}}} \right) \left( \prod_{i \in I_s \setminus I} p_{s, i \leftarrow} \right).
$$

Further, the conditional probability of any new source profile ($S_i^+$) given any initial evidence profile (($e_s)_{s \in S_i}$), or given just ($S_i$), is

$$
\mathrm{Pr}((S_i^+)|((e_s)_{s \in S_i}) = \mathrm{Pr}((S_i^+)|(S_i)) = \prod_{s \in S} \gamma_s \quad (10)
$$

where, for each source $s \in S$, $\gamma_s$ is the probability that the new set of owners of source $s$
is $I_s = \{i : s \in S_i^+\}$ given that the initial one is $J_s = \{i : s \in S_i\}$, which equals

$$
\gamma_s = \begin{cases}
\left( \overline{\prod_{i \in J_s} p_{s, i \to}} \right) \left( \prod_{i \in I_s \setminus J_s} p_{s, i \leftarrow} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s, i \to}} \right) & \text{if } J_s \subsetneq I_s \\
\left( \overline{\prod_{i \in J_s} p_{s, i \to}} \right) \left( \prod_{i \in \overline{J_s}} \overline{p_{s, i \to}} \right) + \prod_{i \in J_s} \overline{p_{s, i \to}} & \text{if } J_s = I_s \\
0 & \text{otherwise.}
\end{cases} \tag{11}
$$

Proof of (9). For each $s \in S$, define two random subgroups, the old set of owners $\mathbf{J}_s = \{i : s \in \mathbf{S}_i\}$ and the new one $\mathbf{I}_s = \{i : s \in \mathbf{S}_i^+\}$. Fix any $(S_i^+)$, and define each $I_s$ ($s \in S$) as above. Note that $(\mathbf{S}_i^+)$ takes the value $(S_i^+)$ if and only if $(\mathbf{I}_s) (= (\mathbf{I}_s)_{s \in S})$ takes the value $(I_s) (= (I_s)_{s \in S})$. Hence, $\mathrm{Pr}((S_i^+)) = \mathrm{Pr}((I_s))$. The sets $\mathbf{I}_s$ are independent across sources $s$. So, $\mathrm{Pr}((I_s)) = \prod_{s \in S} \mathrm{Pr}(I_s)$, and thus

$$
\mathrm{Pr}((S_i^+)) = \prod_{s \in S} \mathrm{Pr}(I_s).
$$

Now fix a source $s$. We calculate $\Pr(I_s) (= \pi_s)$. We do this under the assumption that all parameters $p_{s \to i}$ and $p_{s, i \to}$ are strictly between 0 and 1. This is sufficient since the formula generalises to extreme parameter values by a continuity argument.

First assume $I_s = \emptyset$. Note that $\mathbf{I}_s$ takes the value $\emptyset$ if and only if $\mathbf{J}_s$ takes the value $\emptyset$. The probability of the latter is $\prod_i \overline{p_{s \to i}}$. So, $\Pr(I_s) = \prod_i \overline{p_{s \to i}}$. This is what had to be proved, since the claimed expression for $\Pr(I_s)$ ($= \pi_s$) indeed reduces to $\prod_i \overline{p_{s \to i}}$ if $I_s = \emptyset$.

From now on assume $I_s \neq \emptyset$. Then it is certain that $\mathbf{J}_s$ takes a value $J_s$ that satisfies
$\emptyset \neq J_s \subseteq I_s$, and each such value $J_s$ has non-zero probability (as each $p_{s \to i}$ is strictly
between 0 and 1), so can be conditionalised on. By implication,

$$
\mathrm{Pr}(I_s) = \sum_{\emptyset \subsetneq J_s \subseteq I_s} \mathrm{Pr}(J_s)\mathrm{Pr}(I_s|J_s). \quad (12)
$$

In this expression, the term $\mathrm{Pr}(J_s)$ can be written as

$$
\mathrm{Pr}(J_s) = \left( \prod_{i \in J_s} p_{s \to i} \right) \left( \prod_{i \in \overline{J_s}} \overline{p_{s \to i}} \right).
$$
We only show the equivalence for ‘>’, as that for ‘<’ or ‘=’ is analogous. For any state $x \in \{\pm 1\}$, writing $f(\cdot|x)$ for the Gaussian density function on $\mathbb{R}^{S_i}$ of the vector $(\mathbf{e}_s)_{s \in S_i}$, we have

$$ 
\begin{align*}
\Pr(\mathbf{x} = 1 | (e_s)_{s \in S_i}) > \frac{1}{2} &\Leftrightarrow \frac{f((e_s)_{s \in S_i} | 1)}{f((e_s)_{s \in S_i} | -1)} > 1 \\
&\Leftrightarrow \log \frac{f((e_s)_{s \in S_i} | 1)}{f((e_s)_{s \in S_i} | -1)} > 0 \\
&\Leftrightarrow g((e_s)_{s \in S_i}) > 0.
\end{align*}
$$

Here, the first equivalence holds in analogy to the corresponding equivalence in the proof of Theorem 1. The second equivalence holds trivially. The third equivalence holds Lemma 2. Q.e.d.

*Claim 2:* The inequality (9) holds (completing the proof).

This claim follows from Claim 1 through an argument analogous to that underlying Claim 2 in the proof of Theorem 1. ■

## B Formal Analytics of Share-Absorb Processes

The definition of share-absorb processes has been stated informally. The formalisation is obvious. In short, given a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ (we could have used a general opinion structure), the share-absorb process with parameters $(p_{s,i \to}, p_{s,i \leftarrow})_{s \in S, i \in N}$ assumes that there exist events ‘i shares s’ and ‘i absorbs s’ for any person $i \in N$ and source $s \in S$; that the new source set of any person $i$ is $\mathbf{S}_i^+ = \mathbf{S}_i \cup \{s \in S : \text{ 'i absorbs s'}\}$, the set of initially accessed or later absorbed sources; that, for any person $i$ and source $s$, the probability of ‘i shares s’ given any initial source profile $(S_j)$ is $p_{s,i \to}$ if $s \in S_i$ and 0 otherwise; that, for any person $i$ and source $s$, the probability of ‘i absorbs s’ given any initial source profile $(S_j)$ and any sharing profile is $p_{s,i \leftarrow}$ if $s \notin S_i$ and someone shares $s$ in the sharing profile] and 0 otherwise (where a ‘sharing profile’ is a combination of truth values of the sharing events across persons and sources); and, finally, that the access, sharing, and absorbing events are jointly independent of the state and the evidences.

How is the new source profile $(\mathbf{S}_i^+)$ distributed? And how is it distributed conditional on the initial evidence profile? We now answer both questions. The first answer completes the description of the new opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i^+))$, as we already know how $(\mathbf{x}, (\mathbf{e}_s))$ is distributed and that $(\mathbf{S}_i^+)$ is independent of $(\mathbf{x}, (\mathbf{e}_s))$. The second answer implies an alternative (and equivalent) definition of the share-absorb process as a deliberation process in the general sense of Section 7, i.e., a mapping $D$ from initial evidence profiles to lotteries over new source profiles.

Fix a simple<fnref n="32" /> opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ and a share-absorb process with sharing and absorbing probabilities $(p_{s,i \to}, p_{s,i \leftarrow})_{s \in S, i \in N}$, generating a new source profile $(\mathbf{S}_i^+)$. The probability of any new source profile $(S_i^+)$ (any value of $(\mathbf{S}_i^+)$) is

$$ \Pr((S_i^+)) = \prod_{s \in S} \pi_s \qquad (10) $$

where, for each source $s \in S$, $\pi_s$ is the probability that the new set of owners of $s$ is

<footnote n="32">³²Simplicity could be weakened considerably, to Independent Sources.</footnote>
We now calculate $Pr(I_s|J_s)$. Denote by !_s the event that at least someone shares s. Given the (non-empty) event $J_s$, each of !_s and !_s has non-zero probability (as each $p_{s,i}$ is strictly between 0 and 1), so can be conditionalised on. Hence, $Pr(I_s|J_s)$ is writable as $Pr(!_s|J_s)Pr(I_s|!,J_s) + Pr(\overline{!_s}|J_s)Pr(I_s|\overline{!_s},J_s)$, where $Pr(I_s|\overline{!_s},J_s)$ is 0 if $J_s \neq I_s$ and 1 if $J_s = I_s$. So,

$$Pr(I_s|J_s) = \begin{cases} Pr(!_s|J_s)Pr(I_s|!,J_s) & \text{if } J_s \neq I_s \\ Pr(!_s|J_s)Pr(I_s|!,J_s) + Pr(\overline{!_s}|J_s) & \text{if } J_s = I_s. \end{cases}$$

Note that if $J_s = I_s$ then

$$Pr(\overline{I_s}|J_s) = \prod_{i \in I_s} \overline{p_{s,i}}.$$

Upon inserting the derived expressions into (12) and rearranging,

$$Pr(I_s) = \left( \prod_{i \in I_s} p_{s \to i} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s \to i}} \right) \left( \prod_{i \in I_s} \overline{p_{s,i \to}} \right) \\ + \sum_{\emptyset \neq J_s \subseteq I_s} \left( \prod_{i \in J_s} p_{s \to i} \right) \left( \prod_{i \in \overline{J_s}} \overline{p_{s \to i}} \right) Pr(!_s|J_s)Pr(!_s|!,J_s).$$

In this,

$$Pr(!_s|J_s)Pr(I_s|!,J_s) = \left(\overline{\prod_{i \in J_s} \overline{p_{s,i \to}}}\right) \left(\prod_{i \in I_s \setminus J_s} p_{s,i \leftarrow}\right) \left(\overline{\prod_{i \in \overline{I_s}} p_{s,i \leftarrow}}\right). \blacksquare$$

So, after rearranging and relabelling the index 'J_s' into 'I',

$$Pr(I_s) = \left( \prod_{i \in I_s} p_{s \to i} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s \to i}} \right) \left( \prod_{i \in I_s} \overline{p_{s,i \to}} \right) \\ + \left( \prod_{i \in \overline{I_s}} \overline{p_{s,i \to}} \right) \sum_{\emptyset \neq I \subseteq I_s} \left( \prod_{i \in I} p_{s \to i} \right) \left( \prod_{i \in \overline{I}} \overline{p_{s \to i}} \right) \left( \overline{\prod_{i \in I} \overline{p_{s,i \to}}} \right) \left( \prod_{i \in I_s \setminus I} p_{s,i \leftarrow} \right). \blacksquare$$

*Proof of (10).* Fix any initial evidence profile $((e_s)_{s \in S_i})$ and new source profile $(S_i^+)$. Notation is as above. By definition of share-absorb processes, $Pr((S_i^+)|((e_s)_{s \in S_i})) = Pr((S_i^+)|(S_i))$. $I_s$ and $J_s$ are instances of the random variables $\mathbf{I}_s$ and $\mathbf{J}_s$ defined in the proof of (9). Since the events $(\mathbf{S}_i^+) = (S_i^+)$ and $(\mathbf{I}_s) = (I_s)$ are equivalent, and the events $(\mathbf{S}_i) = (S_i)$ and $(\mathbf{J}_s') = (J_s')$ are also equivalent,

$$Pr((S_i^+)|(S_i)) = Pr((I_s)|(J_s')) = \prod_{s \in S} Pr(I_s|(J_s')) = \prod_{s \in S} \underbrace{Pr(I_s|J_s)}_{\gamma_s},$$

where the second and third equalities hold by construction of share-absorb processes.

Now fix a source $s \in S$. It remains to prove that $Pr(I_s|J_s) (= \gamma_s)$ is given by (11). We do this under the assumption that each $p_{s,i}$ is strictly between 0 and 1. (The generalisation to extreme parameters follows by continuity.)

If $J_s = I_s = \emptyset$, then $Pr(I_s|J_s) = 1$, because if no one initially owns $s$, then certainly no one shares or absorbs $s$.
$I_s = \{i : s \in S_i^+\}$, and equals

$$
\begin{aligned}
\pi_s = & \left( \prod_{i \in I_s} p_{s \to i} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s \to i}} \right) \left( \prod_{i \in I_s} \overline{p_{s, i \to}} \right) \\
& + \left( \prod_{i \in \overline{I_s}} \overline{p_{s, i \leftarrow}} \right) \sum_{\emptyset \neq I \subseteq I_s} \left( \prod_{i \in I} p_{s \to i} \right) \left( \prod_{i \in \overline{I}} \overline{p_{s \to i}} \right) \left( \overline{\prod_{i \in I} \overline{p_{s, i \to}}} \right) \left( \prod_{i \in I_s \setminus I} p_{s, i \leftarrow} \right).
\end{aligned}
$$

Further, the conditional probability of any new source profile ($S_i^+$) given any initial evidence profile (($e_s)_{s \in S_i}$), or given just ($S_i$), is

$$
\mathrm{Pr}((S_i^+)|((e_s)_{s \in S_i})) = \mathrm{Pr}((S_i^+)|(S_i)) = \prod_{s \in S} \gamma_s \quad (11)
$$

where, for each source $s \in S$, $\gamma_s$ is the probability that the new set of owners of source $s$
is $I_s = \{i : s \in S_i^+\}$ given that the initial one is $J_s = \{i : s \in S_i\}$, which equals

$$
\gamma_s = \begin{cases}
\left( \overline{\prod_{i \in J_s} p_{s,i \to}} \right) \left( \prod_{i \in I_s \setminus J_s} p_{s,i \leftarrow} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s,i \leftarrow}} \right) & \text{if } J_s \subsetneq I_s \\
\left( \overline{\prod_{i \in J_s} p_{s,i \to}} \right) \left( \prod_{i \in \overline{J_s}} \overline{p_{s,i \to}} \right) + \prod_{i \in J_s} \overline{p_{s,i \to}} & \text{if } J_s = I_s \\
0 & \text{otherwise.}
\end{cases} \tag{12}
$$

Proof of (10). For each $s \in S$, define two random subgroups, the old set of owners $\mathbf{J}_s = \{i : s \in \mathbf{S}_i\}$ and the new one $\mathbf{I}_s = \{i : s \in \mathbf{S}_i^+\}$. Fix any $(S_i^+)$, and define each $I_s$ ($s \in S$) as above. Note that $(\mathbf{S}_i^+)$ takes the value $(S_i^+)$ if and only if $(\mathbf{I}_s) (= (\mathbf{I}_s)_{s \in S})$ takes the value $(I_s) (= (I_s)_{s \in S})$. Hence, $\mathrm{Pr}((S_i^+)) = \mathrm{Pr}((I_s))$. The sets $\mathbf{I}_s$ are independent across sources $s$. So, $\mathrm{Pr}((I_s)) = \prod_{s \in S} \mathrm{Pr}(I_s)$, and thus

$$
\mathrm{Pr}((S_i^+)) = \prod_{s \in S} \mathrm{Pr}(I_s).
$$

Now fix a source $s$. We calculate $\mathrm{Pr}(I_s) (= \pi_s)$. We do this under the assumption that all parameters $p_{s \to i}$ and $p_{s, i \to}$ are strictly between 0 and 1. This is sufficient since the formula generalises to extreme parameter values by a continuity argument.

First assume $I_s = \emptyset$. Note that $\mathbf{I}_s$ takes the value $\emptyset$ if and only if $\mathbf{J}_s$ takes the value $\emptyset$. The probability of the latter is $\prod_i \overline{p_{s \to i}}$. So, $\mathrm{Pr}(I_s) = \prod_i \overline{p_{s \to i}}$. This is what had to be proved, since the claimed expression for $\mathrm{Pr}(I_s) (= \pi_s)$ indeed reduces to $\prod_i \overline{p_{s \to i}}$ if $I_s = \emptyset$.

From now on assume $I_s \neq \emptyset$. Then it is certain that $\mathbf{J}_s$ takes a value $J_s$ that satisfies
$\emptyset \neq J_s \subseteq I_s$, and each such value $J_s$ has non-zero probability (as each $p_{s \to i}$ is strictly
between 0 and 1), so can be conditionalised on. By implication,

$$
\mathrm{Pr}(I_s) = \sum_{\emptyset \subsetneq J_s \subseteq I_s} \mathrm{Pr}(J_s) \mathrm{Pr}(I_s|J_s). \quad (13)
$$

In this expression, the term $\mathrm{Pr}(J_s)$ can be written as

$$
\mathrm{Pr}(J_s) = \left( \prod_{i \in J_s} p_{s \to i} \right) \left( \prod_{i \in \overline{J_s}} \overline{p_{s \to i}} \right).
$$
If $J_s = \emptyset$ and $I_s \neq \emptyset$, then $Pr(I_s|J_s) = 0$, because a source that no one owns is never shared, hence never acquired.

If $J_s \not\subseteq I_s$, i.e., if $J_s$ is not a subset of $I_s$, then $Pr(I_s|J_s) = 0$, because during deliberation no one loses any initially held sources.

Now assume the remaining case that $\emptyset \neq J_s \subseteq I_s$. As in the proof of (9), denote by $!_s$ the event that at least someone shares $s$. Given $(S_i)$, each of $!_s$ and $!_s$ has non-zero probability (because the parameters $p_{s,i\to}$ are neither 0 nor 1, and in case of $!_s$ also because $J_s \neq \emptyset$). So we can conditionalise on $!_s$ and on $!_s$, and write

$$Pr(I_s|J_s) = Pr(\!_s|J_s)Pr(I_s|!_s, J_s) + Pr(\overline{!_s}|J_s)Pr(I_s|\overline{!_s}, J_s).$$

Hence, as $Pr(I_s|\overline{!_s}, J_s)$ is 0 if $J_s \neq I_s$ and 1 if $J_s = I_s$,

$$Pr(I_s|J_s) = \begin{cases} Pr(\!_s|J_s)Pr(I_s|!_s, J_s) & \text{if } \emptyset \neq J_s \subsetneq I_s \\ Pr(\!_s|J_s)Pr(I_s|!_s, J_s) + Pr(\overline{!_s}|J_s) & \text{if } \emptyset \neq J_s = I_s \end{cases}$$

In this,

$$Pr(\overline{!_s}|J_s) = \prod_{i \in J_s} \overline{p_{s,i \to}}$$

$$Pr(\!_s|J_s) = \overline{\prod_{i \in J_s} \overline{p_{s,i \to}}}$$

$$Pr(I_s|!_s, J_s) = \left( \prod_{i \in I_s \setminus J_s} p_{s,i \leftarrow} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s,i \leftarrow}} \right).$$

Here, $Pr(I_s|!_s, J_s)$ reduces to $\prod_{i \in \overline{J_s}} \overline{p_{s,i \leftarrow}}$ if $J_s = I_s$. In sum, we have shown that

$$Pr(I_s|J_s) = \begin{cases} 1 & \text{if } J_s = I_s = \emptyset \\ 0 & \text{if } \emptyset = J_s \subsetneq I_s \\ 0 & \text{if } J_s \not\subseteq I_s \\ \left(\overline{\prod_{i \in J_s} \overline{p_{s,i \to}}}\right) \left(\prod_{i \in I_s \setminus J_s} p_{s,i \to}\right) \left(\overline{\prod_{i \in \overline{I_s}} \overline{p_{s,i \to}}}\right) & \text{if } \emptyset \neq J_s \subsetneq I_s \\ \left(\overline{\prod_{i \in J_s} \overline{p_{s,i \to}}}\right) \left(\prod_{i \in \overline{J_s}} \overline{p_{s,i \to}}\right) + \prod_{i \in J_s} \overline{p_{s,i \to}} & \text{if } \emptyset \neq J_s = I_s \end{cases}$$

Of these six cases, the first can be subsumed under the last, as the formula in the last reduces to 1 if $J_s = I_s = \emptyset$; and the second can be subsumed under the fourth, as the formula in the fourth reduces to 0 if $\emptyset = J_s$. This yields formula (11). ■

# C The Jury Theorems: Proofs

*Proof of equation (2).* Under the given assumptions, $o_{ideal} = x$ holds if and only if total evidence $\sum_{s \in S} e_s$ has the same sign as $x$. The probability of this event equals the conditional probability that $\sum_{s \in S} e_s > 0$ given $x = 1$, by Simple Gaussian Evidence. Given $x = 1$, $\sum_{s \in S} e_s$ is the sum of $|S|$ independent Gaussian variables of mean 1 and variance $\sigma^2$, hence is itself a Gaussian variable, with mean $|S|$ and variance $|S|\sigma^2$. The probability that such a variable is positive equals the probability that a standard-Gaussian variable is below $\frac{\sqrt{|S|}}{\sigma}$, by a simple linear transformation. ■
We now calculate $Pr(I_s|J_s)$. Denote by !_s the event that at least someone shares s. Given the (non-empty) event $J_s$, each of !_s and !_s has non-zero probability (as each $p_{s,i}$ is strictly between 0 and 1), so can be conditionalised on. Hence, $Pr(I_s|J_s)$ is writable as $Pr(!_s|J_s)Pr(I_s|!,J_s) + Pr(\overline{!_s}|J_s)Pr(I_s|\overline{!_s},J_s)$, where $Pr(I_s|\overline{!_s},J_s)$ is 0 if $J_s \neq I_s$ and 1 if $J_s = I_s$. So,

$$Pr(I_s|J_s) = \begin{cases} Pr(!_s|J_s)Pr(I_s|!,J_s) & \text{if } J_s \neq I_s \\ Pr(!_s|J_s)Pr(I_s|!,J_s) + Pr(\overline{!_s}|J_s) & \text{if } J_s = I_s. \end{cases}$$

Note that if $J_s = I_s$ then

$$Pr(\overline{I_s}|J_s) = \prod_{i \in I_s} \overline{p_{s,i}}.$$

Upon inserting the derived expressions into (13) and rearranging,

$$Pr(I_s) = \left( \prod_{i \in I_s} p_{s \to i} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s \to i}} \right) \left( \prod_{i \in I_s} \overline{p_{s,i \to}} \right) \\ + \sum_{\emptyset \neq J_s \subseteq I_s} \left( \prod_{i \in J_s} p_{s \to i} \right) \left( \prod_{i \in \overline{J_s}} \overline{p_{s \to i}} \right) Pr(!_s|J_s)Pr(!_s|!,J_s).$$

In this,

$$Pr(!_s|J_s)Pr(I_s|!,J_s) = \left( \overline{\prod_{i \in J_s} \overline{p_{s,i \to}}} \right) \left( \prod_{i \in I_s \setminus J_s} p_{s,i \leftarrow} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s,i \leftarrow}} \right). \blacksquare$$

So, after rearranging and relabelling the index 'J_s' into 'I',

$$Pr(I_s) = \left( \prod_{i \in I_s} p_{s \to i} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s \to i}} \right) \left( \prod_{i \in I_s} \overline{p_{s,i \to}} \right) \\ + \left( \prod_{i \in \overline{I_s}} \overline{p_{s,i \to}} \right) \sum_{\emptyset \neq I \subseteq I_s} \left( \prod_{i \in I} p_{s \to i} \right) \left( \prod_{i \in \overline{I}} \overline{p_{s \to i}} \right) \left( \overline{\prod_{i \in I} \overline{p_{s,i \to}}} \right) \left( \prod_{i \in I_s \setminus I} p_{s,i \leftarrow} \right). \blacksquare$$

**Proof of (11).** Fix any initial evidence profile $((e_s)_{s \in S_i})$ and new source profile $(S_i^+)$. Notation is as above. By definition of share-absorb processes, $Pr((S_i^+)|((e_s)_{s \in S_i})) = Pr((S_i^+)|(S_i))$. $I_s$ and $J_s$ are instances of the random variables $I_s$ and $J_s$ defined in the proof of (10). Since the events $(S_i^+) = (S_i^+)$ and $(I_s) = (I_s)$ are equivalent, and the events $(S_i) = (S_i)$ and $(J_{s'}) = (J_{s'})$ are also equivalent,

$$Pr((S_i^+)|(S_i)) = Pr((I_s)|(J_{s'}) = \prod_{s \in S} Pr(I_s|(J_{s'}) = \prod_{s \in S} \underbrace{Pr(I_s|J_s)}_{\gamma_s},$$

where the second and third equalities hold by construction of share-absorb processes.

Now fix a source $s \in S$. It remains to prove that $Pr(I_s|J_s) (= \gamma_s)$ is given by (12). We do this under the assumption that each $p_{s,i}$ is strictly between 0 and 1. (The generalisation to extreme parameters follows by continuity.)

If $J_s = I_s = \emptyset$, then $Pr(I_s|J_s) = 1$, because if no one initially owns s, then certainly no one shares or absorbs s.
*Proof of the Pre-Deliberation Jury Theorem.* Assume a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ for an infinite population $N = \{1, 2, ...\}$. Notation is as usual.

(a) To prove the non-asymptotic claim, we fix a group size $n$ and write $\mathbf{o}_{maj}$ for $\mathbf{o}_{maj,n}$. We first show that $p_{maj} \le p_{ideal}$, i.e., that $\mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}) \le \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$. We begin by proving a general claim:

*Claim:* For every discrete random variable $\mathbf{z}$ that is independent of the state-evidence combination $(\mathbf{x}, (\mathbf{e}_s))$ (e.g., for $\mathbf{z} = (\mathbf{S}_i)$),

$$ \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) > \frac{1}{2} $$

except in a zero-probability event (i.e., except if the combination $((\mathbf{e}_s), \mathbf{z})$ falls into a set into which it falls with zero probability).

To show the claim, note first that such a variable $\mathbf{z}$ is independent of the event $\mathbf{o}_{ideal} = \mathbf{x}$ conditional on $(\mathbf{e}_s)$, because $\mathbf{o}_{ideal}$ is a function of $(\mathbf{e}_s)$. So, $\mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z})$ can be replaced by $\mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s))$, which, by construction of the ideal opinion $\mathbf{o}_{ideal}$, indeed exceeds $\frac{1}{2}$, except in the zero-probability event that $\sum_s \mathbf{e}_s = 0$ (i.e., except if $\mathbf{o}_{ideal}$ is zero, hence certainly distinct from $\mathbf{x}$). Q.e.d.

Now choose $\mathbf{z} = (\mathbf{S}_i)$. Then

$$ \mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) = \begin{cases} \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) & \text{if } \mathbf{o}_{maj} = \mathbf{o}_{ideal} \\ 1 - \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) & \text{if } \mathbf{o}_{maj} = -\mathbf{o}_{ideal} \\ 0 & \text{if } \mathbf{o}_{maj} = 0. \end{cases} \quad (13) $$

Here, ‘if $\mathbf{o}_{maj} = \mathbf{o}_{ideal}$’ of course means ‘if $((\mathbf{e}_s), \mathbf{z})$ takes a value such that $\mathbf{o}_{maj} = \mathbf{o}_{ideal}$’, which is indeed a well-defined condition because the value of $((\mathbf{e}_s), \mathbf{z})$ determines the values of $\mathbf{o}_{maj}$ and $\mathbf{o}_{ideal}$, hence determines whether $\mathbf{o}_{maj} = \mathbf{o}_{ideal}$. The meanings of ‘if $\mathbf{o}_{maj} = -\mathbf{o}_{ideal}$’ and ‘if $\mathbf{o}_{maj} = 0$’ are analogous.

The ‘Claim’ and (13) jointly imply that, still for $\mathbf{z} = (\mathbf{S}_i)$,

$$ \mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) \le \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) \quad (14) $$

with probability one. By taking expectations on both sides (thereby averaging out $(\mathbf{e}_s)$ and $\mathbf{z}$), we obtain $\mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}) \le \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$, i.e., $p_{maj} \le p_{ideal}$.

Finally, assume Imperfect Access. Then with non-zero probability the variable $\mathbf{z} = (\mathbf{S}_i)$ takes a value such that some source is not accessed by anyone, hence not accessed by a majority. This easily implies that with non-zero probability the second or third case in (13) applies. So, in (14) the ‘$\le$’ is a ‘$<$’ with non-zero probability. Hence, taking the expectation on both sides of (14) now yields $\mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}) < \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$, i.e., $p_{maj} < p_{ideal}$.

(b) We now show the convergence claim, assuming Access Competence. By this assumption, there is an $\epsilon > 0$ such that $p_{s \to i} \ge 2^{-1/|S|} + \epsilon$ for all $s$ and $i$. Consider a person $i$. The probability of having full source set $S$ satisfies $\mathrm{Pr}(\mathbf{S}_i = S) \ge \frac{1}{2} + \epsilon^{|S|}$, because

$$ \mathrm{Pr}(\mathbf{S}_i = S) = \prod_{s \in S} p_{s \to i} \ge \prod_{s \in S} (2^{-1/|S|} + \epsilon) = (2^{-1/|S|} + \epsilon)^{|S|} \ge (2^{-1/|S|})^{|S|} + \epsilon^{|S|} = \frac{1}{2} + \epsilon^{|S|}. $$

Since the full-access events $\mathbf{S}_i = S$ ($i = 1, 2, ...$) are mutually independent (by Independent Sources) and each of probability at least $\frac{1}{2} + \epsilon^{|S|}$, the probability that the proportion
If $J_s = \emptyset$ and $I_s \neq \emptyset$, then $Pr(I_s|J_s) = 0$, because a source that no one owns is never shared, hence never acquired.

If $J_s \not\subseteq I_s$, i.e., if $J_s$ is not a subset of $I_s$, then $Pr(I_s|J_s) = 0$, because during deliberation no one loses any initially held sources.

Now assume the remaining case that $\emptyset \neq J_s \subseteq I_s$. As in the proof of (10), denote by $!_s$ the event that at least someone shares $s$. Given $(S_i)$, each of $!_s$ and $!_s$ has non-zero probability (because the parameters $p_{s,i\to}$ are neither 0 nor 1, and in case of $!_s$ also because $J_s \neq \emptyset$). So we can conditionalise on $!_s$ and on $!_s$, and write

$$Pr(I_s|J_s) = Pr(\!_s|J_s)Pr(I_s|!_s, J_s) + Pr(\overline{!_s}|J_s)Pr(I_s|\overline{!_s}, J_s).$$

Hence, as $Pr(I_s|\overline{!_s}, J_s)$ is 0 if $J_s \neq I_s$ and 1 if $J_s = I_s$,

$$Pr(I_s|J_s) = \begin{cases} Pr(\!_s|J_s)Pr(I_s|!_s, J_s) & \text{if } \emptyset \neq J_s \subsetneq I_s \\ Pr(\overline{!_s}|J_s)Pr(I_s|\overline{!_s}, J_s) + Pr(\overline{!_s}|J_s) & \text{if } \emptyset \neq J_s = I_s \end{cases}$$

In this,

$$Pr(\overline{!_s}|J_s) = \prod_{i \in J_s} \overline{p_{s,i \to}}$$

$$Pr(\!_s|J_s) = \overline{\prod_{i \in J_s} \overline{p_{s,i \to}}}$$

$$Pr(I_s|!_s, J_s) = \left( \prod_{i \in I_s \setminus J_s} p_{s,i \leftarrow} \right) \left( \prod_{i \in \overline{I_s}} \overline{p_{s,i \leftarrow}} \right).$$

Here, $Pr(I_s|!_s, J_s)$ reduces to $\prod_{i \in \overline{J_s}} \overline{p_{s,i \leftarrow}}$ if $J_s = I_s$. In sum, we have shown that

$$Pr(I_s|J_s) = \begin{cases} 1 & \text{if } J_s = I_s = \emptyset \\ 0 & \text{if } \emptyset = J_s \subsetneq I_s \\ 0 & \text{if } J_s \not\subseteq I_s \\ \left(\overline{\prod_{i \in J_s} \overline{p_{s,i \to}}}\right) \left(\prod_{i \in I_s \setminus J_s} p_{s,i \to}\right) \left(\prod_{i \in \overline{I_s}} \overline{p_{s,i \to}}\right) & \text{if } \emptyset \neq J_s \subsetneq I_s \\ \left(\overline{\prod_{i \in J_s} \overline{p_{s,i \to}}}\right) \left(\prod_{i \in \overline{J_s}} \overline{p_{s,i \to}}\right) + \prod_{i \in J_s} \overline{p_{s,i \to}} & \text{if } \emptyset \neq J_s = I_s \end{cases}$$

Of these six cases, the first can be subsumed under the last, as the formula in the last reduces to 1 if $J_s = I_s = \emptyset$; and the second can be subsumed under the fourth, as the formula in the fourth reduces to 0 if $\emptyset = J_s$. This yields formula (12). ■

# C The Jury Theorems: Proofs

*Proof of equation (2).* Under the given assumptions, $o_{ideal} = x$ holds if and only if total evidence $\sum_{s \in S} e_s$ has the same sign as $x$. The probability of this event equals the conditional probability that $\sum_{s \in S} e_s > 0$ given $x = 1$, by Simple Gaussian Evidence. Given $x = 1$, $\sum_{s \in S} e_s$ is the sum of $|S|$ independent Gaussian variables of mean 1 and variance $\sigma^2$, hence is itself a Gaussian variable, with mean $|S|$ and variance $|S|\sigma^2$. The probability that such a variable is positive equals the probability that a standard-Gaussian variable is below $\frac{\sqrt{|S|}}{\sigma}$, by a simple linear transformation. ■
of members with full access exceeds $\frac{1}{2}$ (the event $\#\{i \in \{1, ..., n\} : \mathbf{S}_i = S\} > \frac{n}{2}$) tends to one as $n \to \infty$, by the law of large numbers. In other words, the probability of a majority with full access (the event $\#\{i \in \{1, ..., n\} : \mathbf{S}_i = S\} > \frac{n}{2}$) tends to 1 as $n \to \infty$. Meanwhile, full access implies an ideal opinion (i.e., $\mathbf{S}_i = S$ implies $\mathbf{o}_i = \mathbf{o}_{ideal}$). So a majority with full access implies a majority with the ideal opinion (i.e., $\#\{i \in \{1, ..., n\} : \mathbf{S}_i = S\} > \frac{n}{2}$ implies $\mathbf{o}_{maj,n} = \mathbf{o}_{ideal}$). Hence, also the probability of an ideal majority opinion converges to one: $\mathrm{Pr}(\mathbf{o}_{maj,n} = \mathbf{o}_{ideal}) \to 1$. This implies that $\mathrm{Pr}(\mathbf{o}_{maj,n} = \mathbf{x}) \to \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$, i.e., that $p_{maj,n} \to p_{ideal}$. ■

**Proof of the Post-Deliberation Jury Theorem.** Assume a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ and a share-absorb process, both for an infinite population $N = \{1, 2, ...\}$. The usual notation applies.

(a) The non-asymptotic claim holds by a version of the proof of part (a) of the Pre-Deliberation Jury Theorem. One should substitute $\mathbf{o}_{maj}^+$ for $\mathbf{o}_{maj}$, and apply the 'Claim' with $\mathbf{z} = (\mathbf{S}_i^+)$ rather than $\mathbf{z} = (\mathbf{S}_i)$, which is possible since also $(\mathbf{S}_i^+)$ is independent of $(\mathbf{x}, (\mathbf{e}_s))$.

(b) We now turn to the asymptotic claim. We shall face the difficulty of interpersonal correlations between post-deliberation source sets. The weak law of large numbers in Pivato's (2017) version for correlated variables will ultimately come to help, but first several claims must be established. We assume Acquisition Competence (needed only from Claim b5) and Non-Vanishing Participation (needed only from Claim b4).

*Claim b1:* For any source $s \in S$, group size $n \in \{1, 2, ...\}$, and group member $i \in \{1, ..., n\}$, the probability that another member shares $s$ is

$$p_{s,i,n} = \frac{\prod_{j \in \{1, ..., n\} \setminus \{i\}} \overline{p_{s \to j} p_{s,j \to}}}{\overline{p_{s \to i} \times p_{s,i,n} p_{s,i \to}}} \quad (15)$$

The probability is given by (15) because it equals the probability that it is *not* the case that each other member $j$ does *not* share $s$, where $j$ shares $s$ with probability $p_{s \to j} p_{s,j \to}$, the product of the probabilities of accessing $s$ and of sharing an accessed $s$. Q.e.d.

*Claim b2:* For any $s \in S$, $n \in \{1, 2, ...\}$, and $i \in \{1, ..., n\}$, the probability that some member other than $i$ shares $s$ and then $i$ absorbs $s$, given that $i$ has not accessed $s$ initially, is $p_{s,i,n} p_{s,i \to}$.

The claim holds because the relevant probability is the product of the probability that someone else shares $s$, i.e., $p_{s,i,n}$ by Claim b1, and the probability that $i$ absorbs a shared source $s$, i.e., $p_{s,i \to}$. Q.e.d.

*Claim b3:* For any $s \in S$, $n \in \{1, 2, ...\}$, and $i \in \{1, ..., n\}$, $\mathrm{Pr}(s \in \mathbf{S}_{i,n}^+) = \overline{\mathrm{Pr}(s \to i) \times \mathrm{Pr}(s_{i,n} p_{s,i \to})}$.

This holds because $i$ does *not* hold $s$ post-deliberation if and only if $i$ does not initially access $s$ (probability: $\overline{p_{s \to i}}$) and $i$ does not absorb $s$ (probability: $\overline{p_{s,i,n} p_{s,i \to}}$). Q.e.d.

*Claim b4:* For any $s \in S$ and $i \in \{1, 2, ...\}$, $P(s \in \mathbf{S}_{i,n}^+) \to \overline{p_{s \to i} \times p_{s,i \to}}$ as $n \to \infty$.

Fix $s$ and $i$. By Claim b3, we just show $p_{s,i,n} \to 1$, i.e., $\prod_{j \in \{1, ..., n\} \setminus \{i\}} \overline{p_{s \to j} p_{s,j \to}} \to 0$. By Non-Vanishing Participation, $p_{s \to j} p_{s,j \to} \not\to 0$ as $j \to \infty$, and hence $\overline{p_{s \to j} p_{s,j \to}} \not\to 1$ as $j \to \infty$. In consequence, $\prod_{j \in \{1, ..., n\} \setminus \{i\}} \overline{p_{s \to j} p_{s,j \to}} \to 0$ as $n \to \infty$. Q.e.d.

*Claim b5:* For any $i \in \{1, 2, ...\}$, the full-access probability $\mathrm{Pr}(\mathbf{S}_{i,n}^+ = S)$ converges to a value of at least $\frac{1}{2} + \epsilon^{|S|}$ as $n \to \infty$, where $\epsilon > 0$ is the threshold in Acquisition Competence (which is independent of $i$).

Fix a person $i$. We have $\mathrm{Pr}(\mathbf{S}_{i,n}^+ = S) = \prod_{s \in S} \mathrm{Pr}(s \in \mathbf{S}_{i,n}^+)$, because the access events $s \in \mathbf{S}_{i,n}^+$ are independent across sources $s$ as a consequence of the fact that
*Proof of the Pre-Deliberation Jury Theorem.* Assume a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ for an infinite population $N = \{1, 2, ...\}$. Notation is as usual.

(a) To prove the non-asymptotic claim, we fix a group size $n$ and write $\mathbf{o}_{maj}$ for $\mathbf{o}_{maj,n}$. We first show that $p_{maj} \le p_{ideal}$, i.e., that $\mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}) \le \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$. We begin by proving a general claim:

*Claim:* For every discrete random variable $\mathbf{z}$ that is independent of the state-evidence combination $(\mathbf{x}, (\mathbf{e}_s))$ (e.g., for $\mathbf{z} = (\mathbf{S}_i)$),

$$ \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) > \frac{1}{2} $$

except in a zero-probability event (i.e., except if the combination $((\mathbf{e}_s), \mathbf{z})$ falls into a set into which it falls with zero probability).

To show the claim, note first that such a variable $\mathbf{z}$ is independent of the event $\mathbf{o}_{ideal} = \mathbf{x}$ conditional on $(\mathbf{e}_s)$, because $\mathbf{o}_{ideal}$ is a function of $(\mathbf{e}_s)$. So, $\mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z})$ can be replaced by $\mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s))$, which, by construction of the ideal opinion $\mathbf{o}_{ideal}$, indeed exceeds $\frac{1}{2}$, except in the zero-probability event that $\sum_s \mathbf{e}_s = 0$ (i.e., except if $\mathbf{o}_{ideal}$ is zero, hence certainly distinct from $\mathbf{x}$). Q.e.d.

Now choose $\mathbf{z} = (\mathbf{S}_i)$. Then

$$ \mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) = \begin{cases} \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) & \text{if } \mathbf{o}_{maj} = \mathbf{o}_{ideal} \\ 1 - \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) & \text{if } \mathbf{o}_{maj} = -\mathbf{o}_{ideal} \\ 0 & \text{if } \mathbf{o}_{maj} = 0. \end{cases} \quad (14) $$

Here, ‘if $\mathbf{o}_{maj} = \mathbf{o}_{ideal}$’ of course means ‘if $((\mathbf{e}_s), \mathbf{z})$ takes a value such that $\mathbf{o}_{maj} = \mathbf{o}_{ideal}$’, which is indeed a well-defined condition because the value of $((\mathbf{e}_s), \mathbf{z})$ determines the values of $\mathbf{o}_{maj}$ and $\mathbf{o}_{ideal}$, hence determines whether $\mathbf{o}_{maj} = \mathbf{o}_{ideal}$. The meanings of ‘if $\mathbf{o}_{maj} = -\mathbf{o}_{ideal}$’ and ‘if $\mathbf{o}_{maj} = 0$’ are analogous.

The ‘Claim’ and (14) jointly imply that, still for $\mathbf{z} = (\mathbf{S}_i)$,

$$ \mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) \le \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x}|(\mathbf{e}_s), \mathbf{z}) \quad (15) $$

with probability one. By taking expectations on both sides (thereby averaging out $(\mathbf{e}_s)$ and $\mathbf{z}$), we obtain $\mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}) \le \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$, i.e., $p_{maj} \le p_{ideal}$.

Finally, assume Imperfect Access. Then with non-zero probability the variable $\mathbf{z} = (\mathbf{S}_i)$ takes a value such that some source is not accessed by anyone, hence not accessed by a majority. This easily implies that with non-zero probability the second or third case in (14) applies. So, in (15) the ‘$\le$’ is a ‘$<$’ with non-zero probability. Hence, taking the expectation on both sides of (15) now yields $\mathrm{Pr}(\mathbf{o}_{maj} = \mathbf{x}) < \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$, i.e., $p_{maj} < p_{ideal}$.

(b) We now show the convergence claim, assuming Access Competence. By this assumption, there is an $\epsilon > 0$ such that $p_{s \to i} \ge 2^{-1/|S|} + \epsilon$ for all $s$ and $i$. Consider a person $i$. The probability of having full source set $S$ satisfies $\mathrm{Pr}(\mathbf{S}_i = S) \ge \frac{1}{2} + \epsilon^{|S|}$, because

$$ \mathrm{Pr}(\mathbf{S}_i = S) = \prod_{s \in S} p_{s \to i} \ge \prod_{s \in S} (2^{-1/|S|} + \epsilon) = (2^{-1/|S|} + \epsilon)^{|S|} \ge (2^{-1/|S|})^{|S|} + \epsilon^{|S|} = \frac{1}{2} + \epsilon^{|S|}. $$

Since the full-access events $\mathbf{S}_i = S$ ($i = 1, 2, ...$) are mutually independent (by Independent Sources) and each of probability at least $\frac{1}{2} + \epsilon^{|S|}$, the probability that the proportion
the pre-deliberation access events 's ∈ S<sub>i</sub>' are independent across sources (by Source Independence) and the share-absorb process operates independently across sources. So, by Claim b4, Pr(S<sub>i,n</sub><sup>+</sup> = S) → ∏<sub>s∈S</sub> over<sub>s→i</sub> × over<sub>s,i←</sub> as n → ∞. Now choose ϵ > 0 as in Acquisition Competence. Then, for all s, over<sub>s→i</sub> × over<sub>s,i←</sub> ≤ 1 − 2<sup>−1/|S|</sup> − ϵ, i.e., over<sub>s→i</sub> × over<sub>s,i←</sub> ≥ 2<sup>−1/|S|</sup> + ϵ. So,

$$
\prod_{s \in S} \overline{p_{s \to i} \times p_{s, i \leftarrow}} \geq (2^{-1/|S|} + \epsilon)^{|S|} \geq (2^{-1/|S|})^{|S|} + \epsilon^{|S|} = \frac{1}{2} + \epsilon^{|S|}.
$$

Hence $\lim_{n \to \infty} \mathrm{Pr}(\mathbf{S}_{i,n}^{+} = S) \ge \frac{1}{2} + \epsilon^{|S|}$. Q.e.d.

Claim b6: For all $n \in \{1, 2, ...\}$ and distinct $i, j \in \{1, ..., n\}$, the covariance between $i$'s and $j$'s full access satisfies

$$
\mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \leq \prod_{s \in S} \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} - \prod_{s \in S} \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k,n}} \overline{p_{s,k \leftarrow}}
$$

Fix $n \in \{1, 2, ...\}$ and distinct $i, j \in \{1, ..., n\}$. Then

$$
\begin{align*}
\mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) &= \mathrm{Pr}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) - \prod_{k=i,j} \mathrm{Pr}(\mathbf{S}_{k,n}^{+} = S) \\
&= \prod_{s \in S} \mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+}) - \prod_{s \in S} \prod_{k=i,j} \mathrm{Pr}(s \in \mathbf{S}_{k,n}^{+}),
\end{align*}
$$

where the second equality holds by independence across sources of the access events. Since
$\mathrm{Pr}(s \in \mathbf{S}_{k,n}^{+}) = \overline{p_{s \to k}} \times \overline{p_{s,k,n}} \overline{p_{s,k \leftarrow}}$ by Claim b3, it suffices to show that

$$
\mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+}) \leq \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} \quad \text{for all } s \in S.
$$

This holds since, letting $E$ be the event that $s$ is shared by someone in $\{1, ..., n\}$,

$$
\begin{align*}
\mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+}) &\le \mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+} | E) \\
&= \prod_{k=i,j} \mathrm{Pr}(s \in \mathbf{S}_{k,n}^{+} | E) = \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}},
\end{align*}
$$

where the first equality holds by independence between 's ∈ S<sub>i,n</sub><sup>+</sup> and 's ∈ S<sub>i,n</sub><sup>+</sup>, given E,
and the second equality holds because s is held ex-post if and only if it is not the case
that s is not accessed ex-ante (probability: p<sub>s→k</sub>) and not absorbed ex-post (probability:
p<sub>s,i←</sub>). Q.e.d.

Claim b7: : $\min_{s \in S, k \le n} p_{s,k,n} \to 1$ as $n \to \infty$.

For all $s$ and $n$, pick $i_{s,n} \in \{1, \dots, n\}$ with $p_{s \to i_{s,n}} p_{s, i_{s,n} \to} = \max_{k \le n} p_{s \to k} p_{s, k \to}$. By construction,

$$
\min_{k \le n} p_{s,k,n} = \overline{\prod_{j \in \{1, \dots, n\} \setminus \{i_{s,k}\}} \overline{p_{s \to j} p_{s,j \to}}}.
$$

By Non-Vanishing Participation, $p_{s \to j} p_{s, j \to} \not\to 0$. So $\prod_{j \in \{1, \dots, n\} \setminus \{i_{s,k}\}} \overline{p_{s \to j} p_{s, j \to}} \to 0$.
Hence, $\min_{k \le n} p_{s, k, n} \to 1$. So, as $|S|$ is finite, $\min_{s \in S, k \le n} p_{s, k, n} \to 1$. Q.e.d.

Claim b8: $\delta_n \equiv \max_{s \in S, k \le n} (\overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} - \overline{p_{s \to k}} \times \overline{p_{s,k,n}} \overline{p_{s,k \leftarrow}}) \to 0$ as $n \to \infty$.
of members with full access exceeds $\frac{1}{2}$ (the event $\#\{i \in \{1, ..., n\} : \mathbf{S}_i = S\} > \frac{n}{2}$) tends to one as $n \to \infty$, by the law of large numbers. In other words, the probability of a majority with full access (the event $\#\{i \in \{1, ..., n\} : \mathbf{S}_i = S\} > \frac{n}{2}$) tends to 1 as $n \to \infty$. Meanwhile, full access implies an ideal opinion (i.e., $\mathbf{S}_i = S$ implies $\mathbf{o}_i = \mathbf{o}_{ideal}$). So a majority with full access implies a majority with the ideal opinion (i.e., $\#\{i \in \{1, ..., n\} : \mathbf{S}_i = S\} > \frac{n}{2}$ implies $\mathbf{o}_{maj,n} = \mathbf{o}_{ideal}$). Hence, also the probability of an ideal majority opinion converges to one: $\mathrm{Pr}(\mathbf{o}_{maj,n} = \mathbf{o}_{ideal}) \to 1$. This implies that $\mathrm{Pr}(\mathbf{o}_{maj,n} = \mathbf{x}) \to \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$, i.e., that $p_{maj,n} \to p_{ideal}$. ■

**Proof of the Post-Deliberation Jury Theorem.** Assume a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s), (\mathbf{S}_i))$ and a share-absorb process, both for an infinite population $N = \{1, 2, ...\}$. The usual notation applies.

(a) The non-asymptotic claim holds by a version of the proof of part (a) of the Pre-Deliberation Jury Theorem. One should substitute $\mathbf{o}_{maj}^+$ for $\mathbf{o}_{maj}$, and apply the 'Claim' with $\mathbf{z} = (\mathbf{S}_i^+)$ rather than $\mathbf{z} = (\mathbf{S}_i)$, which is possible since also $(\mathbf{S}_i^+)$ is independent of $(\mathbf{x}, (\mathbf{e}_s))$.

(b) We now turn to the asymptotic claim. We shall face the difficulty of interpersonal correlations between post-deliberation source sets. The weak law of large numbers in Pivato's (2017) version for correlated variables will ultimately come to help, but first several claims must be established. We assume Acquisition Competence (needed only from Claim b5) and Non-Vanishing Participation (needed only from Claim b4).

*Claim b1:* For any source $s \in S$, group size $n \in \{1, 2, ...\}$, and group member $i \in \{1, ..., n\}$, the probability that another member shares s is

$$p_{s,i,n} = \frac{\prod_{j \in \{1, ..., n\} \setminus \{i\}} \overline{p_{s \to j} p_{s,j \to}}}{\prod_{j \in \{1, ..., n\} \setminus \{i\}} \overline{p_{s \to i} \times p_{s,i \to}}} \quad (16)$$

The probability is given by (16) because it equals the probability that it is *not* the case that each other member $j$ does *not* share $s$, where $j$ shares $s$ with probability $p_{s \to j} p_{s,j \to}$, the product of the probabilities of accessing $s$ and of sharing an accessed $s$. Q.e.d.

*Claim b2:* For any $s \in S$, $n \in \{1, 2, ...\}$, and $i \in \{1, ..., n\}$, the probability that some member other than $i$ shares $s$ and then $i$ absorbs $s$, given that $i$ has not accessed $s$ initially, is $p_{s,i,n} p_{s,i \to}$.

The claim holds because the relevant probability is the product of the probability that someone else shares $s$, i.e., $p_{s,i,n}$ by Claim b1, and the probability that $i$ absorbs a shared source $s$, i.e., $p_{s,i \to}$. Q.e.d.

*Claim b3:* For any $s \in S$, $n \in \{1, 2, ...\}$, and $i \in \{1, ..., n\}$, $\mathrm{Pr}(s \in \mathbf{S}_{i,n}^+) = \overline{\mathrm{Pr}(s \to i) \times \mathrm{Pr}(s_{i,n} p_{s,i \to})}$.

This holds because $i$ does *not* hold $s$ post-deliberation if and only if $i$ does not initially access $s$ (probability: $\overline{p_{s \to i}}$) and $i$ does not absorb $s$ (probability: $\overline{p_{s,i,n} p_{s,i \to}}$). Q.e.d.

*Claim b4:* For any $s \in S$ and $i \in \{1, 2, ...\}$, $P(s \in \mathbf{S}_{i,n}^+) \to \overline{p_{s \to i} \times p_{s,i \to}}$ as $n \to \infty$.

Fix $s$ and $i$. By Claim b3, we just show $p_{s,i,n} \to 1$, i.e., $\prod_{j \in \{1, ..., n\} \setminus \{i\}} \overline{p_{s \to j} p_{s,j \to}} \to 0$. By Non-Vanishing Participation, $p_{s \to j} p_{s,j \to} \not\to 0$ as $j \to \infty$, and hence $\overline{p_{s \to j} p_{s,j \to}} \not\to 1$ as $j \to \infty$. In consequence, $\prod_{j \in \{1, ..., n\} \setminus \{i\}} \overline{p_{s \to j} p_{s,j \to}} \to 0$ as $n \to \infty$. Q.e.d.

*Claim b5:* For any $i \in \{1, 2, ...\}$, the full-access probability $\mathrm{Pr}(\mathbf{S}_{i,n}^+ = S)$ converges to a value of at least $\frac{1}{2} + \epsilon^{|S|}$ as $n \to \infty$, where $\epsilon > 0$ is the threshold in Acquisition Competence (which is independent of $i$).

Fix a person $i$. We have $\mathrm{Pr}(\mathbf{S}_{i,n}^+ = S) = \prod_{s \in S} \mathrm{Pr}(s \in \mathbf{S}_{i,n}^+)$, because the access events $s \in \mathbf{S}_{i,n}^+$ are independent across sources $s$ as a consequence of the fact that
For all $s \in S$, $n \in \{1, 2, ...\}$ and $k \le n$, we have

$$
\begin{align*}
\overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} - \overline{p_{s \to k}} \times \overline{p_{s,k,n}} &= \overline{p_{s \to k}} \times \overline{p_{s,k,n}} - \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} \\
&= \overline{p_{s \to k}} (\overline{p_{s,k,n}} - \overline{p_{s,k \leftarrow}}) \\
&= \overline{p_{s \to k}} (p_{s,k \leftarrow} - p_{s,k,n}) \\
&= \overline{p_{s \to k}} p_{s,k \leftarrow} (1 - p_{s,k,n}) \\
&\ge 1 - p_{s,k,n}.
\end{align*}
$$

This lower bound implies the desired convergence via Claim b7. Q.e.d.

*Claim b9:* The average covariance of full access between group members converges to zero, i.e.,

$$
\frac{1}{n^2} \sum_{i,j=1}^{n} \mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \to 0 \text{ as } n \to \infty.
$$

It suffices to show that $Cov(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \le 1$ whenever $i = j$ ($\le n$) and that $\max_{i,j \le n, i \ne i} Cov(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \to 0$, by a simple argument (which uses that each $Cov(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S)$ is positive). The former is obvious. We now show the latter.
By Claim b6 and the positivity of all the covariances, it is enough to prove that, for any distinct $i, j$,

$$
\prod_{s \in S} \prod_{k=i,j} a_{s,k} - \prod_{s \in S} \prod_{k=i,j} a_{s,k,n} \to 0 \text{ as } n \to \infty,
$$

where $a_{s,k} = \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}}$ and $a_{s,k,n} = \overline{p_{s \to k}} \times \overline{p_{s,k,n}}$. Fix distinct $i, j$. Note that $a_{k,s} = (a_{s,k} - a_{s,k,n}) + a_{s,k,n} \le \delta_n + a_{s,k,n}$, by Claim b8. So it suffices to show that

$$
\prod_{s \in S} \prod_{k=i,j} (\delta_n + a_{s,k,n}) - \prod_{s \in S} \prod_{k=i,j} a_{s,k,n} \to 0 \text{ as } n \to \infty. \quad (16)
$$

By developing the product $\prod_{s \in S} \prod_{k=i,j} (\delta_n + a_{s,k,n})$, check that it equals a polynomial in $\delta_n$ (of order $2|S|$) whose constant term is `+ $\prod_{s \in S} \prod_{k=i,j} a_{s,k,n}$`. This constant term cancels out against `-$ $\prod_{s \in S} \prod_{k=i,j} a_{s,k,n}$`, so that the expression in (16) is a polynomial in $\delta_n$ with zero constant term. As $n \to \infty$, $\delta_n$ converges to 0 (by Claim b8), and so any polynomial in $\delta_n$ with zero constant term also converges to 0. This proves (16). Q.e.d.

*Claim b10:* $p_{maj,n}^{+} \to p_{ideal}$. (This completes the proof.)

Since every person i's full-access event $\mathbf{S}_{i,n}^{+} = S$ has probability converging to $\frac{1}{2} + \epsilon^{|S|}$
as $n \to \infty$ by Claim b5, while the average covariance of these events tends to zero by Claim
b9, the probability that the proportion of members with full access exceeds $\frac{1}{2}$ (the event
$\frac{\#\{i \in \{1, ..., n\}: \mathbf{S}_{i,n}^{+} = S\}}{n} > \frac{1}{2}$) tends to one, by the weak law of large numbers in Pivato's (2017)
version for correlated variables.<fnref n="33" /> Equivalently, the probability of a majority with full
access (the event $\#\{i \in \{1, ..., n\}: \mathbf{S}_{i,n}^{+} = S\} > \frac{n}{2}$) tends to 1. Since (a majority with) full
access implies (a majority with) an ideal opinion, also the probability of an ideal majority
opinion converges to 1: $\mathrm{Pr}(\mathbf{o}_{maj,n}^{+} = \mathbf{o}_{ideal}) \to 1$. So, $\mathrm{Pr}(\mathbf{o}_{maj,n}^{+} = \mathbf{x}) \to \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$,
i.e., $p_{maj,n}^{+} \to p_{ideal}$. ■

<footnote n="33">³³This version of the law follows from the proof of Pivato’s Theorem 5.2 (i.e., from Claim 2 in that proof, combined with Chebyshev’s Inequality). A closely related result is Proposition A2 in Pivato (2016).</footnote>
the pre-deliberation access events 's ∈ S<sub>i</sub>' are independent across sources (by Source Independence) and the share-absorb process operates independently across sources. So, by Claim b4, Pr(S<sub>i,n</sub><sup>+</sup> = S) → ∏<sub>s∈S</sub> over<sub>s→i</sub> × over<sub>s,i←</sub> as n → ∞. Now choose ϵ > 0 as in Acquisition Competence. Then, for all s, over<sub>s→i</sub> × over<sub>s,i←</sub> ≤ 1 − 2<sup>−1/|S|</sup> − ϵ, i.e., over<sub>s→i</sub> × over<sub>s,i←</sub> ≥ 2<sup>−1/|S|</sup> + ϵ. So,

$$
\prod_{s \in S} \overline{p_{s \to i} \times p_{s, i \leftarrow}} \geq (2^{-1/|S|} + \epsilon)^{|S|} \geq (2^{-1/|S|})^{|S|} + \epsilon^{|S|} = \frac{1}{2} + \epsilon^{|S|}.
$$

Hence $\lim_{n \to \infty} \mathrm{Pr}(\mathbf{S}_{i,n}^{+} = S) \ge \frac{1}{2} + \epsilon^{|S|}$. Q.e.d.

Claim b6: For all $n \in \{1, 2, ...\}$ and distinct $i, j \in \{1, ..., n\}$, the covariance between $i$'s and $j$'s full access satisfies

$$
\mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \leq \prod_{s \in S} \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} - \prod_{s \in S} \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k,n}} \overline{p_{s,k \leftarrow}}
$$

Fix $n \in \{1, 2, ...\}$ and distinct $i, j \in \{1, ..., n\}$. Then

$$
\begin{align*}
\mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) &= \mathrm{Pr}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) - \prod_{k=i,j} \mathrm{Pr}(\mathbf{S}_{k,n}^{+} = S) \\
&= \prod_{s \in S} \mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+}) - \prod_{s \in S} \prod_{k=i,j} \mathrm{Pr}(s \in \mathbf{S}_{k,n}^{+}),
\end{align*}
$$

where the second equality holds by independence across sources of the access events. Since
$\mathrm{Pr}(s \in \mathbf{S}_{k,n}^{+}) = \overline{p_{s \to k}} \times \overline{p_{s,k,n}} \overline{p_{s,k \leftarrow}}$ by Claim b3, it suffices to show that

$$
\mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+}) \leq \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} \quad \text{for all } s \in S.
$$

This holds since, letting $E$ be the event that $s$ is shared by someone in $\{1, ..., n\}$,

$$
\begin{align*}
\mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+}) &\le \mathrm{Pr}(s \in \mathbf{S}_{i,n}^{+}, s \in \mathbf{S}_{j,n}^{+} | E) \\
&= \prod_{k=i,j} \mathrm{Pr}(s \in \mathbf{S}_{k,n}^{+} | E) = \prod_{k=i,j} \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}},
\end{align*}
$$

where the first equality holds by independence between 's ∈ S<sub>i,n</sub><sup>+</sup> and 's ∈ S<sub>i,n</sub><sup>+</sup>, given E,
and the second equality holds because s is held ex-post if and only if it is not the case
that s is not accessed ex-ante (probability: p<sub>s→k</sub>) and not absorbed ex-post (probability:
p<sub>s,i←</sub>). Q.e.d.

Claim b7: : $\min_{s \in S, k \le n} p_{s,k,n} \to 1$ as $n \to \infty$.

For all s and n, pick $i_{s,n} \in \{1, ..., n\}$ with $p_{s \to i_{s,n}} p_{s, i_{s,n} \to} = \max_{k \le n} p_{s \to k} p_{s, k \to}$. By construction,

$$
\min_{k \le n} p_{s,k,n} = \overline{\prod_{j \in \{1, \dots, n\} \setminus \{i_{s,k}\}} \overline{p_{s \to j} p_{s,j \to}}}.
$$

By Non-Vanishing Participation, $p_{s \to j} p_{s, j \to} \not\to 0$. So $\prod_{j \in \{1, \dots, n\} \setminus \{i_{s,k}\}} \overline{p_{s \to j} p_{s, j \to}} \to 0$.
Hence, $\min_{k \le n} p_{s, k, n} \to 1$. So, as $|S|$ is finite, $\min_{s \in S, k \le n} p_{s, k, n} \to 1$. Q.e.d.

Claim b8: $\delta_n \equiv \max_{s \in S, k \le n} (\overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} - \overline{p_{s \to k}} \times \overline{p_{s,k,n}} \overline{p_{s,k \leftarrow}}) \to 0$ as $n \to \infty$.
# D Proofs and Clarifications for Section 5.2

In Section 5.2, some of the definitions and propositions were stated informally. We now provide the exact statements, followed by the proofs of all results.

Consider a simple opinion structure. Majority rule is epistemically monotonic if, for all evidence profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S'_i})$,

$$P(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i}) < Pr(\mathbf{x} = 1|(e'_s)_{s \in \cup_i S'_i}) \Rightarrow o_{maj} \le o'_{maj}, \quad (17)$$

where $o_{maj}$ and $o'_{maj}$ denote the majority outcomes in $\{-1, 0, 1\}$, respectively.<fnref n="34" /> We say that epistemic monotonicity is violated ‘at a (given) evidence profile’ if this profile is one of two profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S'_i})$ for which (17) is violated. The two subtypes of epistemic monotonicity are defined as follows:

* Monotonicity in evidence holds if (17) holds for those evidence profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S'_i})$ such that $S_i = S'_i$ for all persons $i \in N$.

* Monotonicity in sources holds if (17) holds for those evidence profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S'_i})$ such that $e_s = e'_s$ for all sources $s$ contained in some $S_i$ and some $S'_j$ ($i, j \in N$).

A violation of monotonicity in evidence resp. sources ‘at a (given) evidence profile’ is defined like for general epistemic monotonicity.

We now state Propositions 3 and 4 formally. We begin with Proposition 3 about violations of monotonicity in evidence. At an evidence profile $((e_s)_{s \in S_i})$, let us call an available source $s \in \cup_i S_i$ a minority source if fewer persons access $s$ than do not not access $s$ among the persons accessing at least one source, i.e., $\#\{i : s \in S_i\} < \#\{i : s \notin S_i \neq \emptyset\}$.

**Proposition 3** Given a simple opinion structure with non-zero access probabilities and at least three sources, there exist thresholds $\Delta, \Delta' > 0$ such that

(a) monotonicity in evidence is violated at each evidence profile at which Failure 1 occurs to a degree $SI \ge \Delta$ (and $\Delta$ is low enough that such profiles exist),

(b) monotonicity in evidence is violated at each evidence profile with at least one minority source, in such a way that at the other evidence profile of the violation Failure 2 occurs to a degree $II \ge \Delta'$.

To paraphrase this result, monotonicity in evidence is frequently violated in the presence of Failures 1 or 2: it is violated at all evidence profiles with large enough Failure 1, and at almost all evidence profiles when paired with some other evidence profile with large enough Failure 2. In fact, a stronger result holds with precise thresholds given by $\Delta = \frac{6|S|-1}{5|S|-5}$ and $\Delta' = \frac{4}{n+2}$, as our proof will show.

We now formally restate our result about violations of monotonicity in sources:

**Proposition 4** Given a simple opinion structure with non-zero access probabilities and at least three persons, there exist thresholds $\Delta, \Delta' > 0$ such that

<footnote n="34">³⁴ Technical detail: A probability conditional on a random variable (such as $Pr(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i})$, where the variable is the vector $(e_s)_{s \in \cup_i S_i}$ with random components and random dimensionality) is a function of that variable and is defined only ‘almost uniquely’, where two versions coincide at all values of that variable except for a set of values of probability zero. The condition (17) should be read as applying to the uniquely existing ‘natural’ version of $Pr(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i})$ that depends continuously on the $e_s$'s, assuming a simple opinion structure with non-zero access probabilities. So we need not bother with non-uniqueness problems.</footnote>
For all $s \in S$, $n \in \{1, 2, ...\}$ and $k \le n$, we have

$$ 
\begin{align*}
\overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} - \overline{p_{s \to k}} \times \overline{p_{s,k,n}} &= \overline{p_{s \to k}} \times \overline{p_{s,k,n}} - \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}} \\
&= \overline{p_{s \to k}} (\overline{p_{s,k,n}} - \overline{p_{s,k \leftarrow}}) \\
&= \overline{p_{s \to k}} (p_{s,k \leftarrow} - p_{s,k,n}) \\
&= \overline{p_{s \to k}} p_{s,k \leftarrow} (1 - p_{s,k,n}) \\
&\ge 1 - p_{s,k,n}.
\end{align*}
$$

This lower bound implies the desired convergence via Claim b7. Q.e.d.

*Claim b9:* The average covariance of full access between group members converges to zero, i.e.,

$$ \frac{1}{n^2} \sum_{i,j=1}^{n} \mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \to 0 \text{ as } n \to \infty. $$

It suffices to show that $\mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \le 1$ whenever $i = j$ ($\le n$) and that $\max_{i,j \le n, i \ne i} \mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S) \to 0$, by a simple argument (which uses that each $\mathrm{Cov}(\mathbf{S}_{i,n}^{+} = S, \mathbf{S}_{j,n}^{+} = S)$ is positive). The former is obvious. We now show the latter. By Claim b6 and the positivity of all the covariances, it is enough to prove that, for any distinct $i, j$,

$$ \prod_{s \in S} \prod_{k=i,j} a_{s,k} - \prod_{s \in S} \prod_{k=i,j} a_{s,k,n} \to 0 \text{ as } n \to \infty, $$

where $a_{s,k} = \overline{p_{s \to k}} \times \overline{p_{s,k \leftarrow}}$ and $a_{s,k,n} = \overline{p_{s \to k}} \times \overline{p_{s,k,n}}$. Fix distinct $i, j$. Note that $a_{k,s} = (a_{s,k} - a_{s,k,n}) + a_{s,k,n} \le \delta_n + a_{s,k,n}$, by Claim b8. So it suffices to show that

$$ \prod_{s \in S} \prod_{k=i,j} (\delta_n + a_{s,k,n}) - \prod_{s \in S} \prod_{k=i,j} a_{s,k,n} \to 0 \text{ as } n \to \infty. \quad (17) $$

By developing the product $\prod_{s \in S} \prod_{k=i,j} (\delta_n + a_{s,k,n})$, check that it equals a polynomial in $\delta_n$ (of order $2|S|$) whose constant term is `+ $\prod_{s \in S} \prod_{k=i,j} a_{s,k,n}$`. This constant term cancels out against `-$ $\prod_{s \in S} \prod_{k=i,j} a_{s,k,n}$`, so that the expression in (17) is a polynomial in $\delta_n$ with zero constant term. As $n \to \infty$, $\delta_n$ converges to 0 (by Claim b8), and so any polynomial in $\delta_n$ with zero constant term also converges to 0. This proves (17). Q.e.d.

*Claim b10:* $p_{maj,n}^{+} \to p_{ideal}$. (This completes the proof.)

Since every person $i$'s full-access event $\mathbf{S}_{i,n}^{+} = S$ has probability converging to $\frac{1}{2} + \epsilon|S|$ as $n \to \infty$ by Claim b5, while the average covariance of these events tends to zero by Claim b9, the probability that the proportion of members with full access exceeds $\frac{1}{2}$ (the event $\frac{\#\{i \in \{1, ..., n\}: \mathbf{S}_{i,n}^{+} = S\}}{n} > \frac{1}{2}$) tends to one, by the weak law of large numbers in Pivato's (2017) version for correlated variables.33 Equivalently, the probability of a majority with full access (the event $\#\{i \in \{1, ..., n\}: \mathbf{S}_{i,n}^{+} = S\} > \frac{n}{2}$) tends to 1. Since (a majority with) full access implies (a majority with) an ideal opinion, also the probability of an ideal majority opinion converges to 1: $\mathrm{Pr}(\mathbf{o}_{maj,n}^{+} = \mathbf{o}_{ideal}) \to 1$. So, $\mathrm{Pr}(\mathbf{o}_{maj,n}^{+} = \mathbf{x}) \to \mathrm{Pr}(\mathbf{o}_{ideal} = \mathbf{x})$, i.e., $p_{maj,n}^{+} \to p_{ideal}$. ■

<footnote>33This version of the law follows from the proof of Pivato's Theorem 5.2 (i.e., from Claim 2 in that proof, combined with Chebyshev's Inequality). A closely related result is Proposition A2 in Pivato (2016).</footnote>
(a) monotonicity in sources is violated at each evidence profile with at least two evidences supporting each option<sup>35</sup>, in such a way that at the other evidence profile of the violation Failure 1 occurs to a degree **SI** ≥ Δ,

(b) monotonicity in sources is violated at each evidence profile with at least two evidences supporting each option, in such a way that at the other evidence profile of the violation Failure 2 occurs to a degree **II** ≥ Δ'.

So, in short, monotonicity in sources is violated frequently in the presence of Failure 1 or 2: it is violated at almost all evidence profiles when paired with some other evidence profile with large enough Failure 1 or 2. The result holds in a stronger version with precise thresholds given by $\Delta = 4\frac{n-2}{n|S|}$ and $\Delta' = (3 - 2\sqrt{2})\frac{n-1}{n}$, as our proof will show.

*Proof of Remark 1.* Let (17) be violated, i.e., $\Pr(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i}) < \Pr(\mathbf{x} = 1|(e'_s)_{s \in \cup_i S'_i})$ but $o_{maj} > o'_{maj}$. Let $o_{maj}$ be efficient; we show that $o'_{maj}$ is inefficient. As $o_{maj} > o'_{maj}$, we have $o_{maj} \ge 0$ and $o'_{maj} \le 0$. As $o_{maj}$ is efficient and $o_{maj} \ge 0$, we have $\Pr(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i}) \ge 0$. So $\Pr(\mathbf{x} = 1|(e'_s)_{s \in \cup_i S'_i}) > 0$, i.e., outcome 1 is efficient at $(e'_s)_{s \in \cup_i S'_i}$. Hence, $o'_{maj} (\le 0)$ is inefficient. ■

*Notation for the proofs of Propositions 2-4:* The values of $\mathbf{o}_{maj}$, **SI**, **II**, $\mathbf{N}_s$ ($s \in S$) and $\mathbf{E}_i$ ($i \in N$) at an evidence profile $((e_s)_{s \in S_i})$ are denoted $\mathbf{o}_{maj}$, **SI**, **II**, $\mathbf{N}_s$ resp. $E_i$; and their value for an evidence profile denoted using prime(s) (such as $((e'_s)_{s \in S'_i})$ or $((e'_s)_{s \in S_i})$ or $((e_s)_{s \in S'_i})$) are denoted $o'_{maj}$, **SI'**, **II'**, $\mathbf{N}'_s$ resp. $E'_i$.

*Proof of Proposition 2.* Fix a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s)_{s \in S}, (\mathbf{S}_i)_{i \in N})$ such that $|S|, |N| \ge 2$, and $p_{s \to i} \ne 0$ for all $s \in S$ and $i \in N$.

(a) Pick distinct sources $s_1, s_2 \in S$. Let $((e_s)_{s \in S_i})$ be an evidence profile with source sets $S_1 = \{s_1, s_2\}$ and $S_2 = \cdots = S_n = \{s_2\}$ and with evidences $e_{s_1} = -2$ and $e_{s_2} = 1$. At $((e_s)_{s \in S_i})$, all persons $i$ have the same evidence strength of $|E_i| = 1$, so that $\mathbf{II} = 0$, whereas $\mathbf{SI} \ne 0$ as $\#N_{s_1} = 1$ while $\#N_{s_2} = n \ne 1$. Finally, the majority opinion $\mathbf{o}_{maj}$ is inefficient, because the efficient opinion is $-1$ (as $\sum_{s \in \cup_i S_i} e_s = e_{s_1} + e_{s_2} = -1 < 0$ so that $\Pr(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i}) < \frac{1}{2}$) but the majority opinion satisfies $\mathbf{o}_{maj} \ge 0$ as only person 1 votes for $-1$ while the other $n-1$ ($\ge 1$) persons vote for 1.

(b) Let $m = |S|$, say $S = \{s_1, \dots, s_m\}$. Consider two cases.

*Case 1:* $m \ge n$. Let $((e_s)_{s \in S_i})$ be the evidence profile with $S_i = \{s_i\}$ for all $i = 1, \dots, n-1$, $S_n = \{s_n, \dots, s_m\}$, $e_{s_1} = \dots = e_{s_{m-1}} = 1$ and $e_{s_m} = -m$. At $((e_s)_{s \in S_i})$, each source is accessed by exactly one person, so that $\mathbf{SI} = 0$, whereas $\mathbf{II} \ne 0$ because persons $i = 1, \dots, n-1$ have evidence strength $|E_i| = 1$ while person $n$ has $|E_n| = m - (m-n) \times 1 = n \ne 1$. The majority opinion $\mathbf{o}_{maj}$ is inefficient, because the efficient opinion is $-1$ (as $\sum_{s \in \cup_i S_i} e_s = (m-1) \times 1 - m < 0$ so that $\Pr(\mathbf{x} = 1|((e_s)_{s \in S_i})) < \frac{1}{2}$) but the majority opinion satisfies $\mathbf{o}_{maj} \ge 0$ as $n-1$ ($\ge 1$) persons vote for 1 while only person $n$ votes for $-1$.

*Case 2:* $m < n$. Let $((e_s)_{s \in S_i})$ be evidence profile with $S_i = \{s_i\}$ for all $i = 1, \dots, m$, $S_{m+1} = \dots = S_n = \emptyset$, $e_{s_1} = \dots = e_{s_{m-1}} = 1$, and $e_{s_m} = -m$. At $((e_s)_{s \in S_i})$, each source is accessed by exactly one person, so that $\mathbf{SI} = 0$, whereas $\mathbf{II} \ne 0$ because persons $i = 1, \dots, m-1$ have evidence strength $|E_i| = 1$ while person $m$ has $|E_m| = m > 1$. The majority opinion $\mathbf{o}_{maj}$ is again inefficient, since it satisfies $\mathbf{o}_{maj} \ge 0$ while the efficient

<footnote n="35"><sup>35</sup>i.e., with $|\{s \in \cup S_i : e_s > 0\}|, |\{s \in \cup S_i : e_s < 0\}| \ge 2$, where $((e_s)_{s \in S_i})$ denotes the evidence profile</footnote>
D Proof of Propositions 2–4

Notation for the proofs of Propositions 2–4: The values of $\mathbf{o}_{maj}$, $\mathbf{SI}$, $\mathbf{II}$, $\mathbf{N}_s$ ($s \in S$) and $\mathbf{E}_i$ ($i \in N$) at an evidence profile $((e_s)_{s \in S_i})$ are denoted $\mathbf{o}_{maj}$, $\mathbf{SI}$, $\mathbf{II}$, $\mathbf{N}_s$ resp. $E_i$; and their value for an evidence profile denoted using prime(s) (such as $((e'_s)_{s \in S'_i})$ or $((e'_s)_{s \in S_i})$ or $((e_s)_{s \in S'_i})$) are denoted $\mathbf{o}'_{maj}$, $\mathbf{SI}'$, $\mathbf{II}'$, $\mathbf{N}'_s$ resp. $E'_i$.

*Proof of Proposition 2.* Fix a simple opinion structure $(\mathbf{x}, (\mathbf{e}_s)_{s \in S}, (\mathbf{S}_i)_{i \in N})$ such that $|S|, |N| \ge 2$, and $p_{s \to i} \ne 0$ for all $s \in S$ and $i \in N$.

(a) Pick distinct sources $s_1, s_2 \in S$. Let $((e_s)_{s \in S_i})$ be an evidence profile with source sets $S_1 = \{s_1, s_2\}$ and $S_2 = \cdots = S_n = \{s_2\}$ and with evidences $e_{s_1} = -2$ and $e_{s_2} = 1$. At $((e_s)_{s \in S_i})$, all persons $i$ have the same evidence strength of $|E_i| = 1$, so that $\mathbf{II} = 0$, whereas $\mathbf{SI} \ne 0$ as $\#N_{s_1} = 1$ while $\#N_{s_2} = n \ne 1$. Finally, the majority opinion $\mathbf{o}_{maj}$ is inefficient, because the efficient opinion is $-1$ (as $\sum_{s \in \cup_i S_i} e_s = e_{s_1} + e_{s_2} = -1 < 0$ so that $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i}) < \frac{1}{2}$) but the majority opinion satisfies $\mathbf{o}_{maj} \ge 0$ as only person 1 votes for $-1$ while the other $n-1$ ($\ge 1$) persons vote for 1.

(b) Let $m = |S|$, say $S = \{s_1, \dots, s_m\}$. Consider two cases.

*Case 1:* $m \ge n$. Let $((e_s)_{s \in S_i})$ be the evidence profile with $S_i = \{s_i\}$ for all $i = 1, \dots, n-1$, $S_n = \{s_n, \dots, s_m\}$, $e_{s_1} = \dots = e_{s_{m-1}} = 1$ and $e_{s_m} = -m$. At $((e_s)_{s \in S_i})$, each source is accessed by exactly one person, so that $\mathbf{SI} = 0$, whereas $\mathbf{II} \ne 0$ because persons $i = 1, \dots, n-1$ have evidence strength $|E_i| = 1$ while person $n$ has $|E_n| = m - (m-n) \times 1 = n \ne 1$. The majority opinion $\mathbf{o}_{maj}$ is inefficient, because the efficient opinion is $-1$ (as $\sum_{s \in \cup_i S_i} e_s = (m-1) \times 1 - m < 0$ so that $\mathrm{Pr}(\mathbf{x} = 1|(e_s)_{s \in S_i}) < \frac{1}{2}$) but the majority opinion satisfies $\mathbf{o}_{maj} \ge 0$ as $n-1$ ($\ge 1$) persons vote for 1 while only person $n$ votes for $-1$.

*Case 2:* $m < n$. Let $((e_s)_{s \in S_i})$ be evidence profile with $S_i = \{s_i\}$ for all $i = 1, \dots, m$, $S_{m+1} = \dots = S_n = \emptyset$, $e_{s_1} = \dots = e_{s_{m-1}} = 1$, and $e_{s_m} = -m$. At $((e_s)_{s \in S_i})$, each source is accessed by exactly one person, so that $\mathbf{SI} = 0$, whereas $\mathbf{II} \ne 0$ because persons $i = 1, \dots, m-1$ have evidence strength $|E_i| = 1$ while person $m$ has $|E_m| = m > 1$. The majority opinion $\mathbf{o}_{maj}$ is again inefficient, since it satisfies $\mathbf{o}_{maj} \ge 0$ while the efficient opinion is $-1$, as can be checked (the argument uses that the voters $i = m+1, \dots, n$ vote 0, not affecting the majority outcome). ■

*Proof of Proposition 3 in its strengthened form with* $\Delta = \frac{6|S|-1}{5|S|-5}$ and $\Delta' = \frac{4}{n+2}$. Let the assumptions hold.

(a) Consider an evidence profile $((e_s)_{s \in S_i})$ at which $\mathbf{SI} > \frac{6|S|-1}{5|S|-5}$. Write the spread of a source $s \in S$ as $x_s = \#N_s$, and let $\tilde{S} = \{s: x_s \ne 0\}$ and $j = \#\tilde{S}$. We first establish the following fact:

*Claim:* There are sources $s_+, s_- \in \tilde{S}$ such that $x_{s_+} > 2x_{s_-}$.

Assume for a contradiction that $x_s \le 2x_t$ for all $s, t \in \tilde{S}$. We write

$$ \mathbf{SI} = \frac{1}{|S|(|S| - 1)} \sum_{(s,t) \in S^2: s \ne t} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)} = \frac{1}{|S|(|S| - 1)} (\Sigma_0 + \Sigma_1 + 2\Sigma_2) $$

where we define

$$ \Sigma_0 = \sum_{(s,t) \in (S \setminus \tilde{S})^2: s \ne t} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}, \quad \Sigma_1 = \sum_{(s,t) \in \tilde{S}^2: s \ne t} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}, \quad \Sigma_2 = \sum_{(s,t) \in \tilde{S} \times (S \setminus \tilde{S})} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}. $$
opinion is -1, as can be checked (the argument uses that the voters $i = m + 1, \dots, n$ vote 0, not affecting the majority outcome). ■

*Proof of Proposition 3 in its strengthened form with $\Delta = \frac{6|S|-1}{5|S|-5}$ and $\Delta' = \frac{4}{n+2}$. Let the assumptions hold.*

(a) Consider an evidence profile $((e_s)_{s \in S_i})$ at which $SI > \frac{6|S|-1}{5|S|-5}$. Write the spread of a source $s \in S$ as $x_s = \#N_s$, and let $\tilde{S} = \{s : x_s \neq 0\}$ and $j = \#\tilde{S}$. We first establish the following fact:

*Claim:* There are sources $s_+, s_- \in \tilde{S}$ such that $x_{s_+} > 2x_{s_-}$.

Assume for a contradiction that $x_s \le 2x_t$ for all $s, t \in \tilde{S}$. We write

$$SI = \frac{1}{|S|(|S| - 1)} \sum_{(s,t) \in S^2: s \neq t} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)} = \frac{1}{|S|(|S| - 1)} (\Sigma_0 + \Sigma_1 + 2\Sigma_2)$$

where we define

$$\Sigma_0 = \sum_{(s,t) \in (S \setminus \tilde{S})^2: s \neq t} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}, \quad \Sigma_1 = \sum_{(s,t) \in \tilde{S}^2: s \neq t} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}, \quad \Sigma_2 = \sum_{(s,t) \in \tilde{S} \times (S \setminus \tilde{S})} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}.$$

In $\Sigma_0$, each term $\frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}$ is 0; so $\Sigma_0 = 0$. In $\Sigma_1$, each term satisfies $\frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)} \le \frac{2}{3}$, as can be shown using that $x_s \le 2x_t$ and $x_t \le 2x_s$; so $\Sigma_1 \le \frac{2}{3}\#\{(s, t) \in \tilde{S}^2: s \neq t\} = \frac{2}{3}j(j-1)$. Finally, in $\Sigma_2$ each term satisfies $\frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)} = 2$, since $x_s > 0$ and $x_t = 0$; so $\Sigma_2 = 2|\tilde{S} \times (S \setminus \tilde{S})| = 2|\tilde{S}||S \setminus \tilde{S}| = 2j(|S| - j)$. It follows that

$$SI \le \frac{1}{|S|(|S| - 1)} \left[ 0 + \frac{2}{3}j(j-1) + 4j(|S| - j) \right] = \frac{1}{|S|(|S| - 1)} \left[ -\frac{10}{3}j^2 + (4|S| - \frac{2}{3})j \right].$$

By basic algebra, the second-order polynomial $-\frac{10}{3}j^2+(4|S|-\frac{2}{3})j$ in $j$ (regarded as a real number) takes a maximum at $j^* = \frac{6|S|-1}{10}$, where its value is $\frac{6}{5}(|S|-1/6)^2$. Since $j^* \notin \mathbb{N}$ but the actual $j$ belongs to $\mathbb{N}$, the maximum is not reached. Thus, $-\frac{10}{3}j^2+(4|S|-\frac{2}{3})j < \frac{6}{5}(|S|-1/6)^2$, and so

$$SI < \frac{6}{5} \times \frac{(|S| - 1/6)^2}{|S|(|S| - 1)} \le \frac{6}{5} \times \frac{|S| - 1/6}{|S| - 1} = \frac{6|S| - 1}{5|S| - 5}.$$

This contradicts the assumption that $SI \ge \frac{6|S|-1}{5|S|-5}$. Q.e.d.

Now fix $s_+, s_- \in \tilde{S}$ as in the 'Claim'. We construct another evidence profile $((e'_s)_{s \in S_i})$ with same source sets $S_i$ such that monotonicity in evidence is violated for $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$.

*Case 1:* $o_{maj} \ge 0$. Choose $e'_{s_+} = -1$, $e'_{s_-} > \max\{\sum_{s \in \cup_i S_i} e_s, 0\} + 1$, and $e'_s = 0$ for all $s \in (\cup_i S_i) \setminus \{s_+, s_-\}$. Then $o'_{maj} = -1$, because at $((e'_s)_{s \in S_i})$ option 1 receives only $\#N_{s-} = x_{s-}$ votes while option -1 receives $\#(N_{s+} \setminus N_{s-}) \ge x_{s+} - x_{s-} > 2x_{s-} - x_{s-} = x_{s-}$ votes. So $o_{maj} > o'_{maj}$. This implies a violation of (17) for the pair of profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$, because $Pr(\mathbf{x} = 1|(e_s)_{s \in S_i}) < Pr(\mathbf{x} = 1|(e'_s)_{s \in S_i})$ as $\sum_{s \in \cup_i S_i} e'_s > -1 + \max\{\sum_{s \in \cup_i S_i} e_s, 0\} + 1 \ge \sum_{s \in \cup_i S_i} e_s$.

*Case 2:* $o_{maj} \le 0$. Choose $e'_{s_+} = 1$, $e'_{s_-} < \min\{\sum_{s \in \cup_i S_i} e_s, 0\} - 1$, and $e'_s = 0$ for all $s \in (\cup_i S_i) \setminus \{s_+, s_-\}$. Arguments analogous to those under Case 1 show that $o'_{maj} = 1$ ($> o_{maj}$) and that (17) is violated for the pair of profiles $((e'_s)_{s \in S_i})$ and $((e_s)_{s \in S_i})$ (in this order).
In $\Sigma_0$, each term $\frac{|x_s-x_t|}{2(x_s+x_t)}$ is 0; so $\Sigma_0 = 0$. In $\Sigma_1$, each term satisfies $\frac{|x_s-x_t|}{2(x_s+x_t)} \le \frac{2}{3}$, as can be shown using that $x_s \le 2x_t$ and $x_t \le 2x_s$; so $\Sigma_1 \le \frac{2}{3}\#\{(s,t) \in \tilde{S}^2 : s \ne t\} = \frac{2}{3}j(j-1)$. Finally, in $\Sigma_2$ each term satisfies $\frac{|x_s-x_t|}{2(x_s+x_t)} = 2$, since $x_s > 0$ and $x_t = 0$; so $\Sigma_2 = 2|\tilde{S} \times (S\setminus\tilde{S})| = 2|\tilde{S}||S\setminus\tilde{S}| = 2j(|S|-j)$. It follows that

$$SI \le \frac{1}{|S|(|S|-1)} \left[ 0 + \frac{2}{3}j(j-1) + 4j(|S|-j) \right] = \frac{1}{|S|(|S|-1)} \left[ -\frac{10}{3}j^2 + (4|S| - \frac{2}{3})j \right].$$

By basic algebra, the second-order polynomial $-\frac{10}{3}j^2+(4|S|-\frac{2}{3})j$ in $j$ (regarded as a real number) takes a maximum at $j^* = \frac{6|S|-1}{10}$, where its value is $\frac{6}{5}(|S|-1/6)^2$. Since $j^* \notin \mathbb{N}$ but the actual $j$ belongs to $\mathbb{N}$, the maximum is not reached. Thus, $-\frac{10}{3}j^2+(4|S|-\frac{2}{3})j < \frac{6}{5}(|S|-1/6)^2$, and so

$$SI < \frac{6}{5} \times \frac{(|S|-1/6)^2}{|S|(|S|-1)} \le \frac{6}{5} \times \frac{|S|-1/6}{|S|-1} = \frac{6|S|-1}{5|S|-5}.$$

This contradicts the assumption that $SI \ge \frac{6|S|-1}{5|S|-5}$. Q.e.d.

Now fix $s_+, s_- \in \tilde{S}$ as in the 'Claim'. We construct another evidence profile $((e'_s)_{s \in S_i})$ with same source sets $S_i$ such that monotonicity in evidence is violated for $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$.

**Case 1:** $o_{maj} \ge 0$. Choose $e'_{s_+} = -1$, $e'_{s_-} > \max\{\sum_{s \in \cup_i S_i} e_s, 0\} + 1$, and $e'_s = 0$ for all $s \in (\cup_i S_i) \setminus \{s_+, s_-\}$. Then $o'_{maj} = -1$, because at $((e'_s)_{s \in S_i})$ option 1 receives only $\#N_{s-} = x_{s-}$ votes while option $-1$ receives $\#(N_{s_+} \setminus N_{s-}) \ge x_{s_+} - x_{s-} > 2x_{s-} - x_{s-} = x_{s-}$ votes. So $o_{maj} > o'_{maj}$. This implies a violation of (3) for the pair of profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$, because $\Pr(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i}) < \Pr(\mathbf{x} = 1|(e'_s)_{s \in \cup_i S_i})$ as $\sum_{s \in \cup_i S_i} e'_s > -1 + \max\{\sum_{s \in \cup_i S_i} e_s, 0\} + 1 \ge \sum_{s \in \cup_i S_i} e_s$.

**Case 2:** $o_{maj} \le 0$. Choose $e'_{s_+} = 1$, $e'_{s_-} < \min\{\sum_{s \in \cup_i S_i} e_s, 0\} - 1$, and $e'_s = 0$ for all $s \in (\cup_i S_i) \setminus \{s_+, s_-\}$. Arguments analogous to those under Case 1 show that $o'_{maj} = 1$ ($> o_{maj}$) and that (3) is violated for the pair of profiles $((e'_s)_{s \in S_i})$ and $((e_s)_{s \in S_i})$ (in this order).

(b) Let $((e_s)_{s \in S_i})$ be an evidence profile with a minority source $t \in S$. We construct an evidence profile $((e'_s)_{s \in S_i})$ with same source sets $S_i$ such that monotonicity in evidence is violated for $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$ and $II' > \frac{4}{n+2}$.

**Case 1:** $o_{maj} \ge 0$. Choose $e'_s = -1$ for all $s \in (\cup_i S_i) \setminus \{t\}$ and $e'_t = (n+2)|S| + \max\{\sum_{s \in \cup_i S_i} e_s, 0\}$. Then $o'_{maj} = -1$ ($< o_{maj}$), because at $((e'_s)_{s \in S_i})$ the option 1 receives $\#\{i : t \in S_i\}$ votes and the option $-1$ receives $\#\{i : t \notin S_i \ne \emptyset\}$ votes, where $\#\{i : t \in S_i\} < \#\{i : t \notin S_i \ne \emptyset\}$ as $t$ is a minority source. It follows that (3) is violated for the pair of profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$, because $\Pr(\mathbf{x} = 1|(e_s)_{s \in \cup_i S_i}) < \Pr(\mathbf{x} = 1|(e'_s)_{s \in \cup_i S_i})$ as

$$\sum_{s \in \cup_i S_i} e'_s = (-1) \underbrace{|\cup_i S_i \setminus \{t\}|}_{\le |S|} + (n+2) |S| + \max \left\{ \sum_{s \in \cup_i S_i} e_s, 0 \right\} > \sum_{s \in \cup_i S_i} e_s.$$

Write a person $i$'s evidence strength at $((e'_s)_{s \in S_i})$ as $y_i = |E'_i|$. For all $j \in N \setminus N_t$ we have $y_j = |(-1)|S_j| = |S_j| < |S|$, while for all $j \in N_t$, writing $z$ for $\max\{\sum_{s \in \cup_i S_i} e_s, 0\}$, we have

$$y_j = (-1)|S_j \setminus \{t\}| + (n+2)|S| + z > z + (n+1)|S|.$$
(b) Let $((e_s)_{s \in S_i})$ be an evidence profile with a minority source $t \in S$. We construct an evidence profile $((e'_s)_{s \in S_i})$ with same source sets $S_i$ such that monotonicity in evidence is violated for $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$ and $II' > \frac{4}{n+2}$.

Case 1: $o_{maj} \ge 0$. Choose $e'_s = -1$ for all $s \in (\cup_i S_i) \setminus \{t\}$ and $e'_t = (n+2)|S| + \max\{\sum_{s \in \cup S_i} e_s, 0\}$. Then $o'_{maj} = -1$ ($< o_{maj}$), because at $((e'_s)_{s \in S_i})$ the option 1 receives $\#\{i : t \in S_i\}$ votes and the option $-1$ receives $\#\{i : t \notin S_i \ne \emptyset\}$ votes, where $\#\{i : t \in S_i\} < \#\{i : t \notin S_i \ne \emptyset\}$ as $t$ is a minority source. It follows that (17) is violated for the pair of profiles $((e_s)_{s \in S_i})$ and $((e'_s)_{s \in S_i})$, because $Pr(\mathbf{x} = 1|(e_s)_{s \in \cup S_i}) < Pr(\mathbf{x} = 1|(e'_s)_{s \in \cup S_i})$ as

$$ \sum_{s \in \cup_i S_i} e'_s = (-1) \underbrace{|\cup_i S_i) \setminus \{t\}|}_{\le |S|} + (n+2) |S| + \max \left\{ \sum_{s \in \cup S_i} e_s, 0 \right\} > \sum_{s \in \cup S_i} e_s. $$

Write a person i's evidence strength at $((e'_s)_{s \in S_i})$ as $y_i = |E'_i|$. For all $j \in N \setminus N_t$ we have $y_j = |(-1)|S_j| = |S_j| < |S|$, while for all $j \in N_t$, writing $z$ for $\max\{\sum_{s \in \cup S_i} e_s, 0\}$, we have

$$ y_j = (-1) |S_j \setminus \{t\}| + (n+2) |S| + z > z + (n+1) |S|. $$

Now for all $(i, j) \in N_t \times (N \setminus N_t)$, since $y_i > z + (n+1)|S| > |S| > y_j \ge 0$, we have

$$ \frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} = 2 \frac{y_i - y_j}{y_i + y_j} > 2 \frac{z + (n+1)|S| - |S|}{z + (n+1)|S| + |S|} > 2 \frac{n|S|}{(n+2)|S|} = \frac{2n}{n+2}. $$

In $II' = \frac{1}{n(n-1)} \sum_{(i,j) \in N^2: i \ne j} \frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$, the term $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$ is 0 if $(i, j) \in N^2$ or $(i, j) \in (N \setminus N_t)^2$, and it exceeds $\frac{2n}{n+2}$ if $(i, j) \in N_t \times (N \setminus N_t)$ or $(i, j) \in (N \setminus N_t) \times N_t$. So,

$$ II' > \frac{1}{n(n-1)} 2 \sum_{(i,j) \in N_t \times (N \setminus N_t)} \frac{2n}{n+2} = \frac{4|N_t| |N \setminus N_t|}{(n-1)(n+2)} \ge \frac{4 \times 1 \times (n-1)}{(n-1)(n+2)} = \frac{4}{n+2}. $$

Case 2: $o_{maj} \le 0$. Here choose $e'_s = 1$ for all $s \in (\cup_i S_i) \setminus \{t\}$ and $e'_t = -(n+2)|S| + \min\{\sum_{s \in \cup S_i} e_s, 0\}$. For reasons like those under Case 1, $o'_{maj} = 1$ ($> o_{maj}$), (17) is violated for the pair of profiles $((e'_s)_{s \in S_i})$ and $((e_s)_{s \in S_i})$ (in this order), and $II' > \frac{2n}{n+2}$. ■

*Proof of Proposition 4 in its strengthened form with $\Delta = 4^{\frac{n-2}{n|S|}}$ and $\Delta' = (3 - 2\sqrt{2})^{\frac{n-1}{n}}.*
Let the assumptions hold. For the proof of both parts, we fix an evidence profile $(e_s)_{s \in S_i}$ such that $|\{s \in \cup S_i : e_s > 0\}|, |\{s \in \cup S_i : e_s < 0\}| \ge 2$, and construct another profile $((e_s)_{s \in S'_i})$ with $e'_s = e_s$ for all $s \in (\cup_i S_i) \cap (\cup_i S'_i)$ such that monotonicity in sources is violated for this pair of profiles and $SI' \ge \frac{4(n-2)}{|S|n}$ resp. $II' > (3 - 2\sqrt{2})^{\frac{n-1}{n}}$. The proof assumes w.l.o.g. that $o_{maj} \ge 0$ (an analogous proof works if $o_{maj} \le 0$). To construct $((e_s)_{s \in S'_i})$, we choose a subgroup $M \subseteq N$ such that $0 < |M| < \frac{n}{2}$ (it exists as $n \ge 3$) and a source set $T \ne \emptyset$ that is a strict subset of $\{s \in \cup_i S_i : e_s < 0\}$ (it exists as $|\{s \in \cup S_i : e_s < 0\}| \ge 2$), and we define

$$ S'_i = \begin{cases} \{s \in \cup_i S_i : e_s > 0\} & \text{if } i \in M \\ T & \text{if } i \in N \setminus M. \end{cases} $$

Then $o'_{maj} = -1$, as all $i \in N \setminus M$ vote for $-1$ and $|N \setminus M| > \frac{n}{2}$. Monotonicity in sources is violated, because $o_{maj} > o'_{maj}$ although option 1 has gained evidence support
Now for all $(i, j) \in N_t \times (N \setminus N_t)$, since $y_i > z + (n+1) |S| > |S| > y_j \ge 0$, we have

$$ \frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} = 2 \frac{y_i - y_j}{y_i + y_j} > 2 \frac{z + (n+1) |S| - |S|}{z + (n+1) |S| + |S|} > 2 \frac{n |S|}{(n+2) |S|} = \frac{2n}{n+2} $$

In $II' = \frac{1}{n(n-1)} \sum_{(i,j) \in N^2: i \ne j} \frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$, the term $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$ is 0 if $(i, j) \in N^2$ or $(i, j) \in (N \setminus N_t)^2$, and it exceeds $\frac{2n}{n+2}$ if $(i, j) \in N_t \times (N \setminus N_t)$ or $(i, j) \in (N \setminus N_t) \times N_t$. So,

$$ II' > \frac{1}{n(n-1)} 2 \sum_{(i,j) \in N_t \times (N \setminus N_t)} \frac{2n}{n+2} = \frac{4|N_t| |N \setminus N_t|}{(n-1)(n+2)} \ge \frac{4 \times 1 \times (n-1)}{(n-1)(n+2)} = \frac{4}{n+2} $$

Case 2: $o_{maj} \le 0$. Here choose $e'_s = 1$ for all $s \in (\cup_i S_i) \setminus \{t\}$ and $e'_t = -(n+2) |S| + \min \{\sum_{s \in \cup S_i} e_s, 0\}$. For reasons like those under Case 1, $o'_{maj} = 1$ ($> o_{maj}$), (3) is violated for the pair of profiles $((e'_s)_{s \in S_i})$ and $((e_s)_{s \in S_i})$ (in this order), and $II' > \frac{2n}{n+2}$. ■

*Proof of Proposition 4 in its strengthened form with $\Delta = 4^{\frac{n-2}{n|S|}}$ and $\Delta' = (3 - 2\sqrt{2})^{\frac{n-1}{n}}.*
Let the assumptions hold. For the proof of both parts, we fix an evidence profile $(e_s)_{s \in S_i}$ such that $|\{s \in \cup S_i : e_s > 0\}|, |\{s \in \cup S_i : e_s < 0\}| \ge 2$, and construct another profile $((e_s)_{s \in S'_i})$ with $e'_s = e_s$ for all $s \in (\cup_i S_i) \cap (\cup_i S'_i)$ such that monotonicity in sources is violated for this pair of profiles and $SI' \ge \frac{4(n-2)}{|S|n}$ resp. $II' > (3 - 2\sqrt{2})^{\frac{n-1}{n}}$. The proof assumes w.l.o.g. that $o_{maj} \ge 0$ (an analogous proof works if $o_{maj} \le 0$). To construct $((e_s)_{s \in S'_i})$, we choose a subgroup $M \subseteq N$ such that $0 < |M| < \frac{n}{2}$ (it exists as $n \ge 3$) and a source set $T \ne \emptyset$ that is a strict subset of $\{s \in \cup_i S_i : e_s < 0\}$ (it exists as $|\{s \in \cup_i S_i : e_s < 0\}| \ge 2$), and we define

$$ S'_{i} = \begin{cases} \{s \in \cup_{i} S_{i} : e_{s} > 0\} & \text{if } i \in M \\ T & \text{if } i \in N \setminus M. \end{cases} $$

Then $o'_{maj} = -1$, as all $i \in N \setminus M$ vote for $-1$ and $|N \setminus M| > \frac{n}{2}$. Monotonicity in sources is violated, because $o_{maj} > o'_{maj}$ although option 1 has gained evidence support as $\sum_{s \in \cup_i S'_i} e_s = \sum_{s \in \cup_i S_i : e_s > 0} e_s + \sum_{s \in T} e_s > \sum_{s \in \cup_i S_i} e_s$. From here on, the proof of parts (a) and (b) diverge.

*Remaining proof for (a).* For (a), choose $M = \{m\}$ and $T = \{s \in \cup_i S_i : e_s < 0\} \setminus \{t\}$ for some $m \in N$ and some $t \in \cup_i S_i$ with $e_t < 0$. Write the spread of a source $s \in S$ as $x_s = \#N_s$. Since $x_s = n-1$ if $s \in T$ and $x_s \le 1$ if $s \in S \setminus T$ (note that $x_s = 0$ if $s \notin \cup_i S'_i$), the spread imbalance between any $s \in T$ and any $t \in S \setminus T$ satisfies $\frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)} = 2 \frac{x_s - x_t}{x_s + x_t} \ge 2^{\frac{(n-1)-1}{(n-1)+1}} = 2^{\frac{n-2}{n}}$.

Now recall that $SI' = \frac{1}{|S|(|S|-1)} \sum_{(s,t) \in S^2: s \ne t} \frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}$. Here, each $\frac{|x_s - x_t|}{\frac{1}{2}(x_s + x_t)}$ is 0 if $(s, t) \in T^2$ or $(s, t) \in (S \setminus T)^2$ and is at least $2^{\frac{n-2}{n}}$ if $(s, t) \in T \times (S \setminus T)$ or $(s, t) \in (S \setminus T) \times T$. So, $SI' \ge \frac{1}{|S|(|S|-1)} 2|T| |S \setminus T| \times 2^{\frac{n-2}{n}}$. Hence, as $|T| |S \setminus T| = |T| (|S| - |T|) \ge |S| - 1$, we have $SI' \ge 4^{\frac{n-2}{n|S|}}$. Q.e.d.

*Remaining proof for (b).* For (b), let $M$ be a maximal minority $M \subseteq N$, i.e., $|M| = \frac{n-2}{2}$ if $n$ is even and $|M| = \frac{n-1}{2}$ if $n$ is odd. Pick distinct sources $t, t' \in \cup_i S_i$ with $e_{t'} \le e_t < 0$ (they exist by assumption) and let

$$ T = \begin{cases} \{t\} & \text{if } |e_t| \le \frac{1}{\sqrt{2}} \sum_{s \in \cup_i S_i : e_s > 0} e_s \\ \{t, t'\} & \text{if } |e_t| > \frac{1}{\sqrt{2}} \sum_{s \in \cup_i S_i : e_s > 0} e_s. \end{cases} $$
as $\sum_{s \in \cup_i S'_i} e_s = \sum_{s \in \cup_i S_i: e_s > 0} e_s + \sum_{s \in T} e_s > \sum_{s \in \cup_i S_i} e_s$. From here on, the proof of parts (a) and (b) diverge.

*Remaining proof for (a).* For (a), choose $M = \{m\}$ and $T = \{s \in \cup_i S_i : e_s < 0\} \setminus \{t\}$ for some $m \in N$ and some $t \in \cup_i S_i$ with $e_t < 0$. Write the spread of a source $s \in S$ as $x_s = \#N_s$. Since $x_s = n-1$ if $s \in T$ and $x_s \le 1$ if $s \in S \setminus T$ (note that $x_s = 0$ if $s \notin \cup_i S'_i$), the spread imbalance between any $s \in T$ and any $t \in S \setminus T$ satisfies $\frac{|x_s - x_t|}{2(x_s + x_t)} = 2 \frac{x_s - x_t}{x_s + x_t} \ge 2 \frac{(n-1)-1}{(n-1)+1} = 2 \frac{n-2}{n}.$

Now recall that $SI' = \frac{1}{|S|(|S|-1)} \sum_{(s,t) \in S^2: s \ne t} \frac{|x_s - x_t|}{2(x_s + x_t)}$. Here, each $\frac{|x_s - x_t|}{2(x_s + x_t)}$ is 0 if $(s,t) \in T^2$ or $(s,t) \in (S \setminus T)^2$ and is at least $2 \frac{n-2}{n}$ if $(s,t) \in T \times (S \setminus T)$ or $(s,t) \in (S \setminus T) \times T$. So, $SI' \ge \frac{1}{|S|(|S|-1)} 2|T| |S \setminus T| \times 2 \frac{n-2}{n}$. Hence, as $|T| |S \setminus T| = |T| (|S| - |T|) \ge |S| - 1$, we have $SI' \ge 4 \frac{n-2}{n|S|}$. Q.e.d.

*Remaining proof for (b).* For (b), let $M$ be a maximal minority $M \subseteq N$, i.e., $|M| = \frac{n-2}{2}$ if $n$ is even and $|M| = \frac{n-1}{2}$ if $n$ is odd. Pick distinct sources $t, t' \in \cup_i S_i$ with $e_{t'} \le e_t < 0$ (they exist by assumption) and let

$$T = \begin{cases} \{t\} & \text{if } |e_t| \le \frac{1}{\sqrt{2}} \sum_{s \in \cup_i S_i: e_s > 0} e_s \\ \{t, t'\} & \text{if } |e_t| > \frac{1}{\sqrt{2}} \sum_{s \in \cup_i S_i: e_s > 0} e_s. \end{cases}$$

Any person $i$'s evidence strength is written $y_i = |E'_i| = \left| \sum_{s \in S'_i} e_s \right|$; it equals $\sum_{s \in \cup_i S_i: e_s > 0} e_s$ if $i \in M$ and equals $|e_t|$ or $|e_t + e_{t'}|$ if $i \in N \setminus M$. Now, the interpersonal imbalance between an $i \in M$ and a $j \in N \setminus M$ satisfies

$$\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} \ge 2(3 - 2\sqrt{2}). \qquad (18)$$

Why? If $|e_t| \le \frac{1}{\sqrt{2}}y_i$, then $T = \{t\}$ and so $y_j = |e_t| \le \frac{1}{\sqrt{2}}y_i$; thus $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} = 2 \frac{y_i - y_j}{y_i + y_j}$, where

$$\frac{y_i - y_j}{y_i + y_j} \ge \frac{y_i - \frac{1}{\sqrt{2}}y_i}{y_i + \frac{1}{\sqrt{2}}y_i} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = \frac{(\sqrt{2} - 1)^2}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = 3 - 2\sqrt{2}. \qquad (19)$$

If instead $|e_t| > \frac{1}{\sqrt{2}}y_i$, then $T = \{t, t'\}$, and so $y_j = |e_t + e_{t'}| = |e_t| + |e_{t'}| \ge 2|e_t| > \frac{2}{\sqrt{2}}y_i = \sqrt{2}y_i$, an thus $y_i < y_j/\sqrt{2}$; so, $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} = 2 \frac{y_j - y_i}{y_i + y_j}$, where $\frac{y_j - y_i}{y_i + y_j} > 3 - 2\sqrt{2}$ by a calculation analogous to (19).

Recall that $II' = \frac{1}{n(n-1)} \sum_{(i,j) \in N^2: i \ne j} \frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$. Here, each $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$ is 0 if $(i, j) \in M^2$ or $(i, j) \in (N \setminus M)^2$, and it is at least $2(3 - 2\sqrt{2})$ if $(i, j) \in M \times (N \setminus M)$ or $(i, j) \in (N \setminus M) \times M$. So, $II' \ge \frac{1}{n(n-1)} 2|M| |N \setminus M| \times 2(3 - 2\sqrt{2})$. The definition of $M$ and the fact that $n \ge 3$ imply that $|M| (n - |M|) > \frac{(n-1)^2}{4}$. Therefore,

$$II' > \frac{1}{n(n-1)} \left[ 2 \frac{(n-1)^2}{4} \times 2(3 - 2\sqrt{2}) \right] = (3 - 2\sqrt{2}) \frac{n-1}{n}. \blacksquare$$
Any person i's evidence strength is written $y_i = |E'_i| = |\sum_{s \in S'_i} e_s|$; it equals $\sum_{s \in \cup_i S_i: e_s > 0} e_s$ if $i \in M$ and equals $|e_t|$ or $|e_t + e_{t'}|$ if $i \in N \setminus M$. Now, the interpersonal imbalance between an $i \in M$ and a $j \in N \setminus M$ satisfies

$$ \frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} \geq 2(3 - 2\sqrt{2}). \quad (18) $$

Why? If $|e_t| \leq \frac{1}{\sqrt{2}}y_i$, then $T = \{t\}$ and so $y_j = |e_t| \leq \frac{1}{\sqrt{2}}y_i$; thus $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} = 2\frac{y_i - y_j}{y_i + y_j}$, where

$$ \frac{y_i - y_j}{y_i + y_j} \geq \frac{y_i - \frac{1}{\sqrt{2}}y_i}{y_i + \frac{1}{\sqrt{2}}y_i} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = \frac{(\sqrt{2} - 1)^2}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = 3 - 2\sqrt{2}. \quad (19) $$

In instead $|e_t| > \frac{1}{\sqrt{2}}y_i$, then $T = \{t, t'\}$, and so $y_j = |e_t + e_{t'}| = |e_t| + |e_{t'}| \geq 2|e_t| > \frac{2}{\sqrt{2}}y_i = \sqrt{2}y_i$, an thus $y_i < y_j/\sqrt{2}$; so, $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)} = 2\frac{y_j - y_i}{y_i + y_j}$, where $\frac{y_j - y_i}{y_i + y_j} > 3 - 2\sqrt{2}$ by a calculation analogous to (19).

Recall that $II' = \frac{1}{n(n-1)} \sum_{(i,j) \in N^2: i \neq j} \frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$. Here, each $\frac{|y_i - y_j|}{\frac{1}{2}(y_i + y_j)}$ is 0 if $(i, j) \in M^2$ or $(i, j) \in (N \setminus M)^2$, and it is at least $2(3 - 2\sqrt{2})$ if $(i, j) \in M \times (N \setminus M)$ or $(i, j) \in (N \setminus M) \times M$. So, $II' \geq \frac{1}{n(n-1)} 2|M| |N \setminus M| \times 2(3 - 2\sqrt{2})$. The definition of $M$ and the fact that $n \geq 3$ imply that $|M|(n - |M|) > \frac{(n-1)^2}{4}$. Therefore,

$$ II' > \frac{1}{n(n-1)} \left[ 2 \frac{(n-1)^2}{4} \times 2(3 - 2\sqrt{2}) \right] = (3 - 2\sqrt{2}) \frac{n-1}{n}. \quad \blacksquare $$