Deliberation and the wisdom of crowds

Does pre-voting group deliberation improve majority outcomes? To address this question, we develop a probabilistic model of opinion formation and deliberation. Two new jury theorems, one pre-deliberation and one post-deliberation, suggest that deliberation is beneficial. Successful deliberation mitigates three voting failures: (1) overcounting widespread evidence, (2) neglecting evidential inequality, and (3) neglecting evidential complementarity. Formal results and simulations confirm this. But we identify four systematic exceptions where deliberation reduces majority competence, always by increasing Failure 1. Our analysis recommends deliberation that is ‘participatory’, ‘neutral’, but not necessarily ‘equal’, i.e., that involves substantive sharing, privileges no evidences, but might privilege some persons.

1. Strictly 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.

2. How 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.

3. Simplicity could be weakened considerably, to Independent Sources.

4. This 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).

5. Technical detail: A probability conditional on a random variable (such as \(Pr(\textbf{x}=1|(\textbf{e} _{s})_{s\in \cup _{i}\textbf{S}_{i}})\), where the variable is the vector \((\textbf{e}_{s})_{s\in \cup _{i}\textbf{S}_{i}}\) 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 \(Pr(\textbf{x}=1|(\textbf{e}_{s})_{s\in \cup _{i}\textbf{S}_{i}})\) that depends continuously on the \(\textbf{e}_{s}\)’s, assuming a simple opinion structure with non-zero access probabilities. So we need not bother with non-uniqueness problems.

6. i.e., with \(\left| \{s\in \cup S_{i}:e_{s}>0\}\right| ,\left| \{s\in \cup S_{i}:e_{s}<0\}\right| \ge 2\), where \(((e_{s})_{s\in {\text {Si}}})\) denotes the evidence profile.

Authors and Affiliations

1. Paris School of Economics & CNRS, Paris, France

   Franz Dietrich

2. London School of Economics, London, UK

   Kai Spiekermann

Corresponding author

Correspondence to Kai Spiekermann.

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

We thank anonymous referees for excellent suggestions. We are also grateful for helpful feedback received when presenting earlier versions of this work in various economic, political or philosophical seminars, and at the conferences Freedom and Reason (LSE, 2021), Aggregation across Disciplines (University Cergy-Pontoise, 2021), Social Choice under Risk and Uncertainty (Warwick Business School, 2022), Rationality and Democracy (Mainz, 2022), Advances in Economic Design (Conservatoire National des Arts et Métiers, Paris, 2023) and the 7th International Conference on Economic Philosophy (University of Reims, 2024). Franz Dietrich acknowledges support by the French National Research Agency through three grants (ANR-17-CE26-0003, ANR-16-FRAL-0010 and ANR-17-EURE-0001).

Appendix

A Proof of Theorems 1 and 2 about rationality

We first prove Theorem 1, in fact generalised to quasi-simple opinion structures \((\textbf{x},(\textbf{e}_{s}),(\textbf{S}_{i}))\). 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 ‘\(s\in \textbf{S}_{i}\)’ (where \(s\in S\) and \(i\in N\)) are independent across sources s and jointly independent of the state and the evidences, i.e., of \((\textbf{x},(\textbf{e}_{s}))\). 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 \((\textbf{x},(\textbf{e}_{s}),(\textbf{S}_{i}))\) into a quasi-simple one \((\textbf{x},(\textbf{e}_{s}),(\textbf{S}_{i}^{+}))\)—‘quasi’ because of interpersonal source dependencies.

We begin by proving an astonishing fact about Gaussian evidences:

Lemma 1

If an opinion structure \((\textbf{x},(\textbf{e}_{s}),(\textbf{S} _{i}))\) satisfies Simple Gaussian Evidence (e.g., is quasi-simple), then each evidence is proportional to its own log-likelihood-ratio, more precisely

where \(f(\cdot |x)\) denotes the Gaussian density (‘likelihood’) function of each evidence \(\textbf{e}_{s}\) (\(s\in S\)) given state x (\(\in \{\pm 1\}\)).

Proof

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

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

\(\square \)

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

We prove this by showing that \(\mathbb {E}(u(\textbf{o}_{i},\textbf{x} )|(\textbf{e}_{s})_{s\in \textbf{S}_{i}})\ge \mathbb {E}(u(\textbf{o},\textbf{x})|(\textbf{e}_{s})_{s\in \textbf{S}_{i}})\).^{Footnote 1} So, we fix any value \((e_{s})_{s\in S_{i}}\) of \((\textbf{e}_{s})_{s\in \textbf{S}_{i}}\) and, writing \(o_{i}\) (resp. o) for the value of \(\textbf{o}_{i}\) (resp. \(\textbf{o}\)) under \((e_{s})_{s\in S_{i}}\), we must prove that

We prove this in two steps.

Claim 1: \(Pr(\textbf{x}=1|(e_{s})_{s\in S_{i}})>(<,=)\) \(\frac{1}{2}\Leftrightarrow \sum _{s\in S_{i}}e_{s}>(<,=)\) 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 \(\textbf{e}_{s}\) given x, and \(g(\cdot |x)\) for the \(|S_{i}|\)-dimensional Gaussian density function on \(\mathbb {R}^{S_{i}}\) of the vector \((\textbf{e}_{s})_{s\in S_{i}}\), we have

Here, the first equivalence follows easily from Bayes’ rule, using that \(Pr(\textbf{x}=1)=Pr(\textbf{x}=-1)\) and also that i’s source set is independent of the state and the evidences.^{Footnote 2} 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 \(\mathbb {E}(u(o_{i},\textbf{x})|(e_{s})_{s\in S_{i}})=Pr(\textbf{x} =1|(e_{s})_{s\in S_{i}})>\frac{1}{2}\), 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 \(\mathbb {E}(u(o,\textbf{x})|(e_{s})_{s\in S_{i}})=Pr(\textbf{x}=-1|(e_{s} )_{s\in S_{i}})=1-Pr(\textbf{x}=1|(e_{s})_{s\in S_{i}})<\frac{1}{2}\), where the last inequality holds as \(Pr(\textbf{x}=1|(e_{s})_{s\in S_{i}})>\frac{1}{2}\).

Subcase 1.3: \(o=0\). Then (7) holds (with ‘>’) because \(\mathbb {E}(u(o,\textbf{x})|(e_{s})_{s\in S_{i}})=\frac{1}{2}\).

Case 2: \(o_{i}=-1\), i.e., \(\sum _{s\in S_{i}}e_{s}<0\). Then \(\mathbb {E}(u(o_{i},\textbf{x})|(e_{s})_{s\in S_{i}})=Pr(\textbf{x} =-1|(e_{s})_{s\in S_{i}})=1-Pr(\textbf{x}=1|(e_{s})_{s\in S_{i}})>\frac{1}{2} \), where the inequality holds because \(Pr(\textbf{x}=1|(e_{s})_{s\in S_{i} })<\frac{1}{2}\) 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 \(\mathbb {E}(u(o_{i},\textbf{x})|(e_{s})_{s\in S_{i}})=\frac{1}{2}\). Further, \(Pr(\textbf{x}=1|(e_{s})_{s\in S_{i}})=Pr(\textbf{x}=-1|(e_{s})_{s\in S_{i} })=\frac{1}{2}\), by Claim 1 as \(\sum _{s\in S_{i}}e_{s}=0\).

Subcase 3.1: \(o=0\). Then (7) holds (with ‘\(=\)’) because \(o_{i}=o\).

Subcase 3.2: \(o=1\). Then (7) holds (with ‘\(=\)’) because \(\mathbb {E}(u(o,\textbf{x})|(e_{s})_{s\in S_{i}})=Pr(\textbf{x}=1|(e_{s} )_{s\in S_{i}})=\frac{1}{2}\).

Subcase 3.3: \(o=-1\). Then (7) holds (with ‘\(=\)’) because \(\mathbb {E}(u(o,\textbf{x})|(e_{s})_{s\in S_{i}})=Pr(\textbf{x}=-1|(e_{s} )_{s\in S_{i}})=\frac{1}{2}\). \(\square \)

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

Lemma 2

If an opinion structure \((\textbf{x},(\textbf{e}_{s}),(\textbf{S} _{i}))\) satisfies Generalised Gaussian Evidence, then for each \(\varnothing \ne S^{\prime }\subseteq S\) the vector \((\textbf{e}_{s})_{s\in S^{\prime }}\) has a log-likelihood-ratio given by

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

Proof

Let \((\textbf{x},(\textbf{e}_{s}),(\textbf{S}_{i}))\) satisfy Generalised Gaussian Evidence. Fix \(\varnothing \ne S^{\prime }\subseteq S\). Write \(\mu ^{\prime }\) for \(\mu |_{S^{\prime }}\) and \(\Sigma ^{\prime }\) for \(\Sigma |_{S^{\prime }}\). Given any x \(\in \{\pm 1\}\), \((\textbf{e}_{s})_{s\in S^{\prime }}\) has an \(N(x\mu ^{\prime },\Sigma ^{\prime })\) distribution, with density function given by:

So, for all \(e\in \mathbb {R}^{S^{\prime }}\),

\(\square \)

Proof of Theorem 2 generalised to generalised quasi-simple opinion structures

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

First assume \(S_{i}=\varnothing \). Then (8) reduces to \(\mathbb {E} (u(o_{i},\textbf{x}))\ge \mathbb {E}(u(o,\textbf{x}))\), which holds with ‘\(=\)’ because, firstly, \(\mathbb {E}(u(o_{i},\textbf{x}))=\mathbb {E}(u(0,\textbf{x} ))=\mathbb {E}(0)=0\), and secondly, \(\mathbb {E}(u(o,\textbf{x}))=\frac{1}{2}(u(o,1)+u(o,0))\), 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 \(S_{i}\ne \varnothing \). The proof proceeds in two steps.

Claim 1: \(Pr(\textbf{x}=1|(e_{s})_{s\in S_{i}})>(<,=)\) \(\frac{1}{2}\Leftrightarrow g((e_{s})_{s\in S_{i}})>(<,=)\) 0.

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 \((\textbf{e}_{s})_{s\in S_{i}}\), we have

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. \(\square \)

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 \((\textbf{x},(\textbf{e}_{s}),(\textbf{S}_{i}))\) (we could have used a general opinion structure), the share-absorb process with parameters \((p_{s,i\rightarrow },p_{s,i\leftarrow })_{s\in S,i\in N}\) 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 \(\textbf{S}_{i}^{+}=\textbf{S}_{i}\cup \{s\in S:\) ‘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 \(p_{s,i\rightarrow }\) if \(s\in S_{i}\) and 0 otherwise\(_{i}\); 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 [\(s\not \in S_{i}\) 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 \((\textbf{S}_{i}^{+})\) 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 \((\textbf{x},(\textbf{e}_{s}),(\textbf{S}_{i}^{+}))\), as we already know how \((\textbf{x},(\textbf{e}_{s}))\) is distributed and that \((\textbf{S}_{i}^{+})\) is independent of \((\textbf{x},(\textbf{e}_{s}))\). The second answer implies an alternative (and equivalent) definition of the share-absorb process as a deliberation process in the general sense of Sect. 7, i.e., a mapping D from initial evidence profiles to lotteries over new source profiles.

Fix a simple^{Footnote 3} opinion structure \((\textbf{x},(\textbf{e} _{s}),(\textbf{S}_{i}))\) and a share-absorb process with sharing and absorbing probabilities \((p_{s,i\rightarrow },p_{s,i\leftarrow })_{s\in S,i\in N}\), generating a new source profile \((\textbf{S}_{i}^{+})\). The probability of any new source profile \((S_{i}^{+})\) (any value of \((\textbf{S}_{i}^{+})\)) is

where, for each source \(s\in S\), \(\pi _{s}\) is the probability that the new set of owners of s is \(I_{s}=\{i:s\in S_{i}^{+}\}\), and equals

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

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

Proof of (9)

For each \(s\in S\), define two random subgroups, the old set of owners \(\textbf{J}_{s}=\{i:s\in \textbf{S}_{i}\}\) and the new one \(\textbf{I}_{s}=\{i:s\in \textbf{S}_{i}^{+}\}\). Fix any \((S_{i} ^{+})\), and define each \(I_{s}\) (\(s\in S\)) as above. Note that \((\textbf{S} _{i}^{+})\) takes the value \((S_{i}^{+})\) if and only if \((\textbf{I}_{s})\) (\(=(\textbf{I}_{s})_{s\in S}\)) takes the value \((I_{s})\) (\(=(I_{s})_{s\in S} )\). Hence, \(Pr((S_{i}^{+}))=Pr((I_{s}))\). The sets \(\textbf{I}_{s}\) are independent across sources s. So, \(Pr((I_{s}))=\prod _{s\in S}Pr(I_{s}),\) and thus

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

First assume \(I_{s}=\varnothing \). Note that \(\textbf{I}_{s}\) takes the value \(\varnothing \) if and only if \(\textbf{J}_{s}\) takes the value \(\varnothing \). The probability of the latter is \(\prod _{i}\overline{p_{s\rightarrow i}}\). So, \(Pr(I_{s})=\prod _{i}\overline{p_{s\rightarrow i}}\). This is what had to be proved, since the claimed expression for \(Pr(I_{s})\) (\(=\pi _{s}\)) indeed reduces to \(\prod _{i}\overline{p_{s\rightarrow i}}\) if \(I_{s}=\varnothing \).

From now on assume \(I_{s}\ne \varnothing \). Then it is certain that \(\textbf{J}_{s}\) takes a value \(J_{s}\) that satisfies \(\varnothing \ne J_{s}\subseteq I_{s}\), and each such value \(J_{s}\) has non-zero probability (as each \(p_{s\rightarrow i}\) is strictly between 0 and 1), so can be conditionalised on. By implication,

In this expression, the term \(Pr(J_{s})\) can be written as

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 \(\overline{!_{s}}\) has non-zero probability (as each \(p_{s,i\rightarrow }\) is strictly between 0 and 1), so can be conditionalised on. Hence, \(Pr(I_{s} |J_{s})\) is writable as \(Pr(!_{s}|J_{s})Pr(I_{s}|!_{s},J_{s})+Pr(\overline{!_{s}}|J_{s})Pr(I_{s}|\overline{!_{s}},J_{s})\), where \(Pr(I_{s} |\overline{!_{s}},J_{s})\) is 0 if \(J_{s}\ne I_{s}\) and 1 if \(J_{s}=I_{s} \). So,

Note that if \(J_{s}=I_{s}\) then

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

In this,

So, after rearranging and relabelling the index ‘\(J_{s}\)’ into ‘I’,

\(\square \)

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 \(\textbf{I}_{s}\) and \(\textbf{J}_{s}\) defined in the proof of (9). Since the events \((\textbf{S}_{i}^{+})=(S_{i}^{+})\) and \((\textbf{I}_{s})=(I_{s})\) are equivalent, and the events \((\textbf{S} _{i})=(S_{i})\) and \((\textbf{J}_{s^{\prime }})=(J_{s^{\prime }})\) are also equivalent,

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 \(p_{s,i\rightarrow }\) is strictly between 0 and 1. (The generalisation to extreme parameters follows by continuity.)

If \(J_{s}=I_{s}=\varnothing \), then \(Pr(I_{s}|J_{s})=1\), because if no one initially owns s, then certainly no one shares or absorbs s.

If \(J_{s}=\varnothing \) and \(I_{s}\ne \varnothing \), 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 \(\varnothing \ne J_{s}\subseteq I_{s}.\) As in the proof of (9), denote by \(!_{s}\) the event that at least someone shares s. Given \((S_{i})\), each of \(!_{s}\) and \(\overline{!_{s}}\) has non-zero probability (because the parameters \(p_{s,i\rightarrow }\) are neither 0 nor 1, and in case of \(!_{s}\) also because \(J_{s}\ne \varnothing \)). So we can conditionalise on \(!_{s}\) and on \(\overline{!_{s}}\), and write

Hence, as \(Pr(I_{s}|\overline{!_{s}},J_{s})\) is 0 if \(J_{s}\ne I_{s}\) and 1 if \(J_{s}=I_{s}\),

In this,

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

Of these six cases, the first can be subsumed under the last, as the formula in the last reduces to 1 if \(J_{s}=I_{s}=\varnothing \); and the second can be subsumed under the fourth, as the formula in the fourth reduces to 0 if \(\varnothing =J_{s}.\) This yields formula (11). \(\square \)

C The Jury Theorems: Proofs

Proof of equation (2)

Under the given assumptions, \(\textbf{o}_{ideal}=\textbf{x}\) holds if and only if total evidence \(\sum _{s\in S}\textbf{e}_{s}\) has the same sign as \(\textbf{x}\). The probability of this event equals the conditional probability that \(\sum _{s\in S}\textbf{e}_{s}>0\) given \(\textbf{x}=1\), by Simple Gaussian Evidence. Given \(\textbf{x}=1\), \(\sum _{s\in S}\textbf{e}_{s}\) is the sum of \(\left| S\right| \) independent Gaussian variables of mean 1 and variance \(\sigma ^{2}\), hence is itself a Gaussian variable, with mean \(\left| S\right| \) and variance \(\left| S\right| \sigma ^{2}\). The probability that such a variable is positive equals the probability that a standard-Gaussian variable is below \(\frac{\sqrt{\left| S\right| } }{\sigma }\), by a simple linear transformation. \(\square \)

Proof of the Pre-Deliberation Jury Theorem

Assume a simple opinion structure\((\textbf{x},(\textbf{e}_{s} ),(\textbf{S}_{i}))\) 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 \(\textbf{o}_{maj}\) for \(\textbf{o}_{maj,n}\). We first show that \(p_{maj}\le p_{ideal}\), i.e., that \(Pr(\textbf{o}_{maj}=\textbf{x})\le Pr(\textbf{o} _{ideal}=\textbf{x})\). We begin by proving a general claim:

Claim: For every discrete random variable \(\textbf{z}\) that is independent of the state-evidence combination \((\textbf{x},(\textbf{e}_{s}))\) (e.g., for \(\textbf{z}=(\textbf{S}_{i})\)),

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

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

Now choose \(\textbf{z}=(\textbf{S}_{i})\). Then

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

The ‘Claim’ and (13) jointly imply that, still for \(\textbf{z} =(\textbf{S}_{i})\),

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

Finally, assume Imperfect Access. Then with non-zero probability the variable \(\textbf{z}=(\textbf{S}_{i})\) 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 \(Pr(\textbf{o} _{maj}=\textbf{x})<Pr(\textbf{o}_{ideal}=\textbf{x})\), 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 \(p_{s\rightarrow i} \ge 2^{-1/\left| S\right| }+\epsilon \) for all s and i. Consider a person i. The probability of having full source set S satisfies \(Pr(\textbf{S}_{i}=S)\ge \frac{1}{2}+\epsilon ^{\left| S\right| }\), because

Since the full-access events ‘\(\textbf{S}_{i}=S\)’ (\(i=1,2,...\)) are mutually independent (by Independent Sources) and each of probability at least \(\frac{1}{2}+\epsilon ^{\left| S\right| }\), the probability that the proportion of members with full access exceeds \(\frac{1}{2}\) (the event \(\frac{\#\{i\in \{1,...,n\}:\textbf{S}_{i}=S\}}{n}>\frac{1}{2}\)) tends to one as \(n\rightarrow \infty \), by the law of large numbers. In other words, the probability of a majority with full access (the event \(\#\{i\in \{1,...,n\}:\textbf{S}_{i}=S\}>\frac{n}{2}\)) tends to 1 as \(n\rightarrow \infty \). Meanwhile, full access implies an ideal opinion (i.e., \(\textbf{S} _{i}=S\) implies \(\textbf{o}_{i}=\textbf{o}_{ideal}\)). So a majority with full access implies a majority with the ideal opinion (i.e., \(\#\{i\in \{1,...,n\}:\textbf{S}_{i}=S\}>\frac{n}{2}\) implies \(\textbf{o}_{maj,n} =\textbf{o}_{ideal}\)). Hence, also the probability of an ideal majority opinion converges to one: \(Pr(\textbf{o}_{maj,n}=\textbf{o}_{ideal} )\rightarrow 1\). This implies that \(Pr(\textbf{o}_{maj,n}=\textbf{x} )\rightarrow Pr(\textbf{o}_{ideal}=\textbf{x})\), i.e., that \(p_{maj,n} \rightarrow p_{ideal}\). \(\square \)

Proof of the Post-Deliberation Jury Theorem

Assume a simple opinion structure \((\textbf{x},(\textbf{e}_{s}),(\textbf{S}_{i}))\) 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 \(\textbf{o}_{maj} ^{+}\) for \(\textbf{o}_{maj}\), and apply the ‘Claim’ with \(\textbf{z} =(\textbf{S}_{i}^{+})\) rather than \(\textbf{z}=(\textbf{S}_{i})\), which is possible since also \((\textbf{S}_{i}^{+})\) is independent of \((\textbf{x},(\textbf{e}_{s}))\).

(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

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 \(p_{s\rightarrow j} p_{s,j\rightarrow }\), 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 \(p_{s,i,n} p_{s,i\leftarrow }.\)

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., \(p_{s,i\leftarrow }\). Q.e.d.

Claim b3: For any \(s\in S\), \(n\in \{1,2,...\}\), and \(i\in \{1,...,n\}\), \(Pr(s\in \textbf{S}_{i,n}^{+})=\overline{\overline{p_{s\rightarrow i}} \times \overline{p_{s,i,n}p_{s,i\leftarrow }}}\).

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

Claim b4: For any \(s\in S\) and \(i\in \{1,2,...\}\), \(Pr(s\in \textbf{S}_{i,n}^{+})\rightarrow \overline{\overline{p_{s\rightarrow i}} \times \overline{p_{s,i\leftarrow }}}\) as \(n\rightarrow \infty \).

Fix s and i. By Claim b3, we just show \(p_{s,i,n}\rightarrow 1\), i.e., \(\prod _{j\in \{1,...,n\}\backslash \{i\}}\overline{p_{s\rightarrow j}p_{s,j\rightarrow }}\rightarrow 0\). By Non-Vanishing Participation, \(p_{s\rightarrow j}p_{s,j\rightarrow }\not \rightarrow 0\) as \(j\rightarrow \infty \), and hence \(\overline{p_{s\rightarrow j}p_{s,j\rightarrow } }\not \rightarrow 1\) as \(j\rightarrow \infty \). In consequence, \(\prod _{j\in \{1,...,n\}\backslash \{i\}}\overline{p_{s\rightarrow j}p_{s,j\rightarrow }}\rightarrow 0\) as \(n\rightarrow \infty \). Q.e.d.

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

Fix a person i. We have \(Pr(\textbf{S}_{i,n}^{+}=S)=\prod _{s\in S} Pr(s\in \textbf{S}_{i,n}^{+})\), because the access events ‘\(s\in \textbf{S} _{i,n}^{+}\)’ are independent across sources s as a consequence of the fact that the pre-deliberation access events ‘\(s\in \textbf{S}_{i}\)’ are independent across sources (by Source Independence) and the share-absorb process operates independently across sources. So, by Claim b4, \(Pr(\textbf{S}_{i,n} ^{+}=S)\rightarrow \prod _{s\in S}\overline{\overline{p_{s\rightarrow i}} \times \overline{p_{s,i\leftarrow }}}\) as \(n\rightarrow \infty \). Now choose \(\epsilon >0\) as in Acquisition Competence. Then, for all s, \(\overline{p_{s\rightarrow i}}\times \overline{p_{s,i\leftarrow }}\le 1-2^{-1/\left| S\right| }-\epsilon \), i.e., \(\overline{\overline{p_{s\rightarrow i}} \times \overline{p_{s,i\leftarrow }}}\ge 2^{-1/\left| S\right| }+\epsilon \). So,

Hence \(\lim _{n\rightarrow \infty }Pr(\textbf{S}_{i,n}^{+}=S)\ge \frac{1}{2}+\epsilon ^{\left| S\right| }.\) 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

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

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

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

where the first equality holds by independence between ‘\(s\in \textbf{S} _{i,n}^{+}\)’ and ‘\(s\in \textbf{S}_{i,n}^{+}\)’ 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: \(\overline{p_{s\rightarrow k}}\)) and not absorbed ex-post (probability: \(\overline{p_{s,i\leftarrow }}\)). Q.e.d.

Claim b7:: \(\min _{s\in S,k\le n}p_{s,k,n}\rightarrow 1\) as \(n\rightarrow \infty \).

For all s and n, pick \(i_{s,n}\in \{1,...,n\}\) with \(p_{s\rightarrow i_{s,n}}p_{s,i_{s,n}\rightarrow }=\max _{k\le n}p_{s\rightarrow k} p_{s,k\rightarrow }\). By construction,

By Non-Vanishing Participation, \(p_{s\rightarrow j}p_{s,j\rightarrow }\not \rightarrow 0\). So \(\prod _{j\in \{1,...,n\}\backslash \{i_{s,k} \}}\overline{p_{s\rightarrow j}p_{s,j\rightarrow }}\rightarrow 0\). Hence, \(\min _{k\le n}p_{s,k,n}\rightarrow 1\). So, as \(\left| S\right| \) is finite, \(\min _{s\in S,k\le n}p_{s,k,n}\rightarrow 1\). Q.e.d.

Claim b8: \(\delta _{n}\equiv \max _{s\in S,k\le n}\left( \overline{\overline{p_{s\rightarrow k}}\times \overline{p_{s,k\leftarrow }}} -\overline{\overline{p_{s\rightarrow k}}\times \overline{p_{s,k,n} p_{s,k\leftarrow }}}\right) \rightarrow 0\) as \(n\rightarrow \infty \).

For all \(s\in S\), \(n\in \{1,2,...\}\) and \(k\le n\), we have

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.,

It suffices to show that \(Cov(\textbf{S}_{i,n}^{+}=S,\textbf{S}_{j,n} ^{+}=S)\le 1\) whenever \(i=j\) (\(\le n\)) and that \(\max _{i,j\le n,i\ne i}Cov(\textbf{S}_{i,n}^{+}=S,\textbf{S}_{j,n}^{+}=S)\rightarrow 0\), by a simple argument (which uses that each \(Cov(\textbf{S}_{i,n}^{+}=S,\textbf{S} _{j,n}^{+}=S)\) 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,

where \(a_{s,k}=\overline{\overline{p_{s\rightarrow k}}\times \overline{p_{s,k\leftarrow }}}\) and \(a_{s,k,n}=\overline{\overline{p_{s\rightarrow k} }\times \overline{p_{s,k,n}p_{s,k\leftarrow }}}.\) 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

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 \(2\left| S\right| \)) 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 \(n\rightarrow \infty \), \(\delta _{n}\) 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: \(p_{maj,n}^{+}\rightarrow p_{ideal}\). (This completes the proof.)

Since every person i’s full-access event \(\textbf{S}_{i,n}^{+}=S\) has probability converging to \(\frac{1}{2}+\epsilon ^{\left| S\right| }\) as \(n\rightarrow \infty \) 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 \(\frac{\#\{i\in \{1,...,n\}:\textbf{S}_{i,n}^{+}=S\}}{n}>\frac{1}{2}\)) tends to one, by the weak law of large numbers in Pivato’s (2017) version for correlated variables.^{Footnote 4} Equivalently, the probability of a majority with full access (the event \(\#\{i\in \{1,...,n\}:\textbf{S}_{i,n}^{+}=S\}>\frac{n}{2}\)) 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: \(Pr(\textbf{o} _{maj,n}^{+}=\textbf{o}_{ideal})\rightarrow 1\). So, \(Pr(\textbf{o}_{maj,n} ^{+}=\textbf{x})\rightarrow Pr(\textbf{o}_{ideal}=\textbf{x})\), i.e., \(p_{maj,n}^{+}\rightarrow p_{ideal}\). \(\square \)

D Proofs and Clarifications for Sect. 5.2

In Sect. 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 \(((e_{s}^{\prime })_{s\in S_{i}^{\prime }})\),

where \(o_{maj}\) and \(o_{maj}^{\prime }\) denote the majority outcomes in \(\{-1,0,1\}\), respectively.^{Footnote 5} 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 \(((e_{s}^{\prime })_{s\in S_{i}^{\prime }})\) 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 \(((e_{s}^{\prime })_{s\in S_{i}^{\prime }})\) such that \(S_{i}=S_{i}^{\prime }\) for all persons \(i\in N\).

• Monotonicity in sources holds if (17) holds for those evidence profiles \(((e_{s})_{s\in S_{i}})\) and \(((e_{s}^{\prime })_{s\in S_{i}^{\prime }})\) such that \(e_{s}=e_{s}^{\prime }\) for all sources s contained in some \(S_{i}\) and some \(S_{j}^{\prime }\) (\(i{,}j\in N\)).

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., \(\#\{i:s\in S_{i}\}<\#\{i:s\not \in S_{i}\ne \varnothing \}\).

Proposition 3Given a simple opinion structure with non-zero access probabilities and at least three sources, there exist thresholds \(\Delta ,\Delta ^{\prime }>0\) such that

1. (a)

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

2. (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 \(\textbf{II}\ge \Delta ^{\prime }\).

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 \(\Delta =\frac{6\left| S\right| -1}{5\left| S\right| -5}\) and \(\Delta ^{\prime }=\frac{4}{n+2}\), as our proof will show.

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

Proposition 4Given a simple opinion structure with non-zero access probabilities and at least three persons, there exist thresholds \(\Delta ,\Delta ^{\prime }>0\) such that^{Footnote 6}

1. (a)

   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 1 occurs to a degree \(\textbf{SI}\ge \Delta \),

2. (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 \(\textbf{II}\ge \Delta ^{\prime }\).

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 \(\Delta =4\frac{n-2}{n\left| S\right| }\) and \(\Delta ^{\prime }=(3-2\sqrt{2})\frac{n-1}{n}\), as our proof will show.

Proof of Remark 1

Let (17) be violated, i.e., \(Pr(\textbf{x}=1|(e_{s})_{s\in \cup _{i}S_{i}})<Pr(\textbf{x}=1|(e_{s}^{\prime })_{s\in \cup _{i}S_{i}^{\prime }})\) but \(o_{maj}>o_{maj}^{\prime }\). Let \(o_{maj}\) be efficient; we show that \(o_{maj}^{\prime }\) is inefficient. As \(o_{maj}>o_{maj}^{\prime }\), we have \(o_{maj}\ge 0\) and \(o_{maj}^{\prime }\le 0\). As \(o_{maj}\) is efficient and \(o_{maj}\ge 0\), we have \(Pr(\textbf{x} =1|(e_{s})_{s\in \cup _{i}S_{i}})\ge 0\). So \(Pr(\textbf{x}=1|(e_{s}^{\prime })_{s\in \cup _{i}S_{i}^{\prime }})>0\), i.e., outcome 1 is efficient at \((e_{s}^{\prime })_{s\in \cup _{i}S_{i}^{\prime }}\). Hence, \(o_{maj}^{\prime }\) (\(\le 0\)) is inefficient. \(\square \)

Notation for the proofs of Propositions 2–4

The values of \(\textbf{o}_{maj}\), \(\textbf{SI}\), \(\textbf{II}\), \(\textbf{N} _{s}\) (\(s\in S\)) and \(\textbf{E}_{i}\) (\(i\in N\)) at an evidence profile \(((e_{s})_{s\in S_{i}})\) are denoted \(o_{maj}\), SI, II, \(N_{s}\) resp. \(E_{i}\); and their value for an evidence profile denoted using prime(s) (such as \(((e_{s}^{\prime })_{s\in S_{i}^{\prime }})\) or \(((e_{s}^{\prime })_{s\in S_{i}})\) or \(((e_{s})_{s\in S_{i}^{\prime }})\)) are denoted \(o_{maj}^{\prime }\), \(SI^{\prime }\), \(II^{\prime }\), \(N_{s}^{\prime }\) resp. \(E_{i}^{\prime }\).

Proof of Proposition 2

Fix a simple opinion structure \((\textbf{x},(\textbf{e}_{s})_{s\in S},(\textbf{S}_{i})_{i\in N})\) such that \(\left| S\right| ,\left| N\right| \ge 2\), and \(p_{s\rightarrow i}\ne 0\) 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 \(\left| E_{i}\right| =1\), so that \(II=0\), whereas \(SI\ne 0\) as \(\#N_{s_{1}}=1\) while \(\#N_{s_{2}}=n\ne 1\). Finally, the majority opinion \(o_{maj}\) 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 \(Pr(\textbf{x}=1|(e_{s})_{s\in \cup _{i}S_{i}})<\frac{1}{2}\)) but the majority opinion satisfies \(o_{maj}\ge 0\) as only person 1 votes for \(-1\) while the other \(n-1\) (\(\ge 1\)) persons vote for 1.

(b) Let \(m=\left| S\right| \), say \(S=\{s_{1},...,s_{m}\}\). 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 \(i=1,...,n-1\), \(S_{n}=\{s_{n},...,s_{m}\}\), \(e_{s_{1}}=\cdots =e_{s_{m-1}}=1\) and \(e_{s_{m}}=-m\). At \(((e_{s})_{s\in S_{i} })\), each source is accessed by exactly one person, so that \(SI=0\), whereas \(II\ne 0\) because persons \(i=1,...,n-1\) have evidence strength \(\left| E_{i}\right| =1\) while person n has \(\left| E_{n}\right| =m-(m-n)\times 1=n\ne 1\). The majority opinion \(o_{maj}\) 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 \(Pr(\textbf{x}=1|((e_{s})_{s\in S_{i}} ))<\frac{1}{2}\)) but the majority opinion satisfies \(o_{maj}\ge 0\) 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 \(i=1,...,m\), \(S_{m+1}=\cdots =S_{n}=\varnothing \), \(e_{s_{1}}=\cdots =e_{s_{m-1}}=1\), and \(e_{s_{m}}=-m\). At \(((e_{s})_{s\in S_{i}})\), each source is accessed by exactly one person, so that \(SI=0\), whereas \(II\ne 0\) because persons \(i=1,...,m-1\) have evidence strength \(\left| E_{i}\right| =1\) while person m has \(\left| E_{m}\right| =m>1\). The majority opinion \(o_{maj}\) is again inefficient, since it satisfies \(o_{maj}\ge 0\) while the efficient opinion is \(-1\), as can be checked (the argument uses that the voters \(i=m+1,...,n\) vote 0, not affecting the majority outcome). \(\square \)

Proof of Proposition 3 in its strengthened form with

\(\Delta =\frac{6\left| S\right| -1}{5\left| S\right| -5} \) and \(\Delta ^{\prime }=\frac{4}{n+2}\). Let the assumptions hold.

(a) Consider an evidence profile \(((e_{s})_{s\in S_{i}})\) at which \(\textbf{SI}>\frac{6\left| S\right| -1}{5\left| S\right| -5}\). 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

where we define

In \(\Sigma _{0}\), each term \(\frac{\left| x_{s}-x_{t}\right| }{\frac{1}{2}\left( x_{s}+x_{t}\right) }\) is 0; so \(\Sigma _{0}=0\). In \(\Sigma _{1}\), each term satisfies \(\frac{\left| x_{s}-x_{t}\right| }{\frac{1}{2}\left( x_{s}+x_{t}\right) }\le \frac{2}{3}\), 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 \(\frac{\left| x_{s}-x_{t}\right| }{\frac{1}{2}\left( x_{s}+x_{t}\right) }=2\), since \(x_{s}>0\) and \(x_{t}=0\); so \(\Sigma _{2}=2|\tilde{S}\times (S\backslash \tilde{S})|=2|\tilde{S}||S\backslash \tilde{S}|=2j(\left| S\right| -j)\). It follows that

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

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

Now fix \(s_{+},s_{-}\in \tilde{S}\) as in the ‘Claim’. We construct another evidence profile \(((e_{s}^{\prime })_{s\in S_{i}})\) with same source sets \(S_{i}\) such that monotonicity in evidence is violated for \(((e_{s})_{s\in S_{i}})\) and \(((e_{s}^{\prime })_{s\in S_{i}})\).

Case 1: \(o_{maj}\ge 0\). Choose \(e_{s_{+}}^{\prime }=-1\), \(e_{s_{-} }^{\prime }>\max \left\{ \sum _{s\in \cup _{i}S_{i}}e_{s},0\right\} +1\), and \(e_{s}^{\prime }=0\) for all \(s\in (\cup _{i}S_{i})\backslash \{s_{+},s_{-}\}\). Then \(o_{maj}^{\prime }=-1\), because at \(((e_{s}^{\prime })_{s\in S_{i}})\) option 1 receives only \(\#N_{s_{-}}=x_{s_{-}}\) votes while option \(-1\) receives \(\#(N_{s_{+}}\backslash N_{s_{-}})\ge x_{s_{+}}-x_{s_{-}}>2x_{s_{-} }-x_{s_{-}}=x_{s_{-}}\) votes. So \(o_{maj}>o_{maj}^{\prime }\). This implies a violation of (17) for the pair of profiles \(((e_{s})_{s\in S_{i}})\) and \(((e_{s}^{\prime })_{s\in S_{i}})\), because \(Pr(\textbf{x}=1|(e_{s})_{s\in \cup S_{i}})<Pr(\textbf{x}=1|(e_{s}^{\prime })_{s\in \cup S_{i}})\) as \(\sum _{s\in \cup _{i}S_{i}}e_{s}^{\prime }>-1+\max \left\{ \sum _{s\in \cup _{i}S_{i}} e_{s},0\right\} +1\ge \sum _{s\in \cup _{i}S_{i}}e_{s}\).

Case 2: \(o_{maj}\le 0\). Choose \(e_{s_{+}}^{\prime }=1\), \(e_{s_{-} }^{\prime }<\min \left\{ \sum _{s\in \cup _{i}S_{i}}e_{s},0\right\} -1\), and \(e_{s}^{\prime }=0\) for all \(s\in (\cup _{i}S_{i})\backslash \{s_{+},s_{-}\}\). Arguments analogous to those under Case 1 show that \(o_{maj}^{\prime }=1\) (\(>o_{maj}\)) and that (17) is violated for the pair of profiles \(((e_{s}^{\prime })_{s\in S_{i}})\) 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 \(((e_{s}^{\prime })_{s\in S_{i}})\) with same source sets \(S_{i}\) such that monotonicity in evidence is violated for \(((e_{s})_{s\in S_{i}})\) and \(((e_{s}^{\prime })_{s\in S_{i}})\) and \(II^{\prime }>\frac{4}{n+2}\).

Case 1: \(o_{maj}\ge 0\). Choose \(e_{s}^{\prime }=-1\) for all \(s\in (\cup _{i}S_{i})\backslash \{t\}\) and \(e_{t}^{\prime }=(n+2)\left| S\right| +\max \left\{ \sum _{s\in \cup S_{i}}e_{s},0\right\} \). Then \(o_{maj}^{\prime }=-1\) (\(<o_{maj}\)), because at \(((e_{s}^{\prime })_{s\in S_{i} })\) the option 1 receives \(\#\{i:t\in S_{i}\}\) votes and the option \(-1\) receives \(\#\{i:t\not \in S_{i}\ne \varnothing \}\) votes, where \(\#\{i:t\in S_{i}\}<\#\{i:t\not \in S_{i}\ne \varnothing \}\) as t is a minority source. It follows that (17) is violated for the pair of profiles \(((e_{s} )_{s\in S_{i}})\) and \(((e_{s}^{\prime })_{s\in S_{i}})\), because \(Pr(\textbf{x} =1|(e_{s})_{s\in \cup S_{i}})<Pr(\textbf{x}=1|(e_{s}^{\prime })_{s\in \cup S_{i} })\) as

Write a person i’s evidence strength at \(((e_{s}^{\prime })_{s\in S_{i}})\) as \(y_{i}=\left| E_{i}^{\prime }\right| \). For all \(j\in N\backslash N_{t}\) we have \(y_{j}=\left| (-1)\left| S_{j}\right| \right| =\left| S_{j}\right| <\left| S\right| \), while for all \(j\in N_{t}\), writing z for \(\max \left\{ \sum _{s\in \cup S_{i}}e_{s},0\right\} \), we have

Now for all \((i,j)\in N_{t}\times (N\backslash N_{t})\), since \(y_{i}>z+(n+1)\left| S\right|>\left| S\right| >y_{j}\ge 0\), we have

In \(II^{\prime }=\frac{1}{n(n-1)}\sum _{(i,j)\in N^{2}:i\ne j}\frac{\left| y_{i}-y_{j}\right| }{\frac{1}{2}(y_{i}+y_{j})}\), the term \(\frac{\left| y_{i}-y_{j}\right| }{\frac{1}{2}(y_{i}+y_{j})}\) is 0 if \((i,j)\in N^{2}\) or \((i,j)\in (N\backslash N_{t})^{2}\), and it exceeds \(\frac{2n}{n+2}\) if \((i,j)\in N_{t}\times (N\backslash N_{t})\) or \((i,j)\in (N\backslash N_{t})\times N_{t}\). So,

Case 2: \(o_{maj}\le 0\). Here choose \(e_{s}^{\prime }=1\) for all \(s\in (\cup _{i}S_{i})\backslash \{t\}\) and \(e_{t}^{\prime }=-(n+2)\left| S\right| +\min \left\{ \sum _{s\in \cup S_{i}}e_{s},0\right\} \). For reasons like those under Case 1, \(o_{maj}^{\prime }=1\) (\(>o_{maj}\)), (17) is violated for the pair of profiles \(((e_{s}^{\prime })_{s\in S_{i}})\) and \(((e_{s})_{s\in S_{i}})\) (in this order), and \(II^{\prime }>\frac{2n}{n+2}\). \(\square \)

Proof of Proposition 4in its strengthened form with \(\Delta =4\frac{n-2}{n\left| S\right| }\) and \(\Delta ^{\prime }=(3-2\sqrt{2})\frac{n-1}{n}\). Let the assumptions hold. For the proof of both parts, we fix an evidence profile \((e_{s})_{s\in S_{i}})\) such that \(\left| \{s\in \cup S_{i}:e_{s}>0\}\right| ,\left| \{s\in \cup S_{i}:e_{s}<0\}\right| \ge 2\), and construct another profile \(((e_{s})_{s\in S_{i}^{\prime }})\) with \(e_{s}^{\prime }=e_{s}\) for all \(s\in (\cup _{i}S_{i})\cap (\cup _{i}S_{i}^{\prime })\) such that monotonicity in sources is violated for this pair of profiles and \(SI^{\prime }\ge \frac{4(n-2)}{\left| S\right| n}\) resp. \(II^{\prime }>(3-2\sqrt{2})\frac{n-1}{n}\). The proof assumes w.l.o.g. that \(o_{maj}\ge 0\) (an analogous proof works if \(o_{maj}\le 0\)). To construct \(((e_{s})_{s\in S_{i}^{\prime }})\), we choose a subgroup \(M\subseteq N\) such that \(0<\left| M\right| <\frac{n}{2}\) (it exists as \(n\ge 3\)) and a source set \(T\ne \varnothing \) that is a strict subset of \(\{s\in \cup _{i}S_{i}:e_{s}<0\}\) (it exists as \(\left| \{s\in \cup S_{i}:e_{s}<0\}\right| \ge 2\)), and we define

Then \(o_{maj}^{\prime }=-1\), as all \(i\in N\backslash M\) vote for \(-1\) and \(\left| N\backslash M\right| >\frac{n}{2}\). Monotonicity in sources is violated, because \(o_{maj}>o_{maj}^{\prime }\) although option 1 has gained evidence support as \(\sum _{s\in \cup _{i}S_{i}^{\prime }}e_{s}=\sum _{s\in \cup _{i}S_{i}:e_{s}>0}e_{s}+\sum _{s\in T}e_{s}>\sum _{s\in \cup _{i}S_{i}}e_{s}.\) From here on, the proof of parts (a) and (b) diverge.

Remaining proof for (a). For (a), choose \(M=\{m\}\) and \(T=\{s\in \cup _{i}S_{i}:e_{s}<0\}\backslash \{t\}\) 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 \(s\in S\backslash T\) (note that \(x_{s}=0\) if \(s\not \in \cup _{i}S_{i}^{\prime }\)), the spread imbalance between any \(s\in T\) and any \(t\in S\backslash T\) satisfies \(\frac{\left| x_{s}-x_{t}\right| }{\frac{1}{2}(x_{s}+x_{t} )}=2\frac{x_{s}-x_{t}}{x_{s}+x_{t}}\ge 2\frac{(n-1)-1}{(n-1)+1}=2\frac{n-2}{n}.\)

Now recall that \(SI^{\prime }=\frac{1}{\left| S\right| (\left| S\right| -1)}\sum _{(s,t)\in S^{2}:s\ne t}\frac{\left| x_{s} -x_{t}\right| }{\frac{1}{2}\left( x_{s}+x_{t}\right) }\). Here, each \(\frac{\left| x_{s}-x_{t}\right| }{\frac{1}{2}\left( x_{s} +x_{t}\right) }\) is 0 if \((s,t)\in T^{2}\) or \((s,t)\in (S\backslash T)^{2}\) and is at least \(2\frac{n-2}{n}\) if \((s,t)\in T\times (S\backslash T)\) or \((s,t)\in (S\backslash T)\times T\). So, \(SI^{\prime }\ge \frac{1}{\left| S\right| (\left| S\right| -1)}2\left| T\right| \left| S\backslash T\right| \times 2\frac{n-2}{n}\). Hence, as \(\left| T\right| \left| S\backslash T\right| =\left| T\right| (\left| S\right| -\left| T\right| )\ge \left| S\right| -1\), we have \(SI^{\prime }\ge 4\frac{n-2}{n\left| S\right| }.\) Q.e.d.

Remaining proof for (b). For (b), let M be a maximal minority \(M\subseteq N\), i.e., \(\left| M\right| =\frac{n-2}{2}\) if n is even and \(\left| M\right| =\frac{n-1}{2}\) if n is odd. Pick distinct sources \(t,t^{\prime }\in \cup _{i}S_{i}\) with \(e_{t^{\prime }}\le e_{t}<0\) (they exist by assumption) and let

Any person i’s evidence strength is written \(y_{i}=\left| E_{i}^{\prime }\right| =\left| \sum _{s\in S_{i}^{\prime }}e_{s}\right| \); it equals \(\sum _{s\in \cup _{i}S_{i}:e_{s}>0}e_{s}\) if \(i\in M\) and equals \(\left| e_{t}\right| \) or \(\left| e_{t}+e_{t^{\prime }}\right| \) if \(i\in N\backslash M\). Now, the interpersonal imbalance between an \(i\in M\) and a \(j\in N\backslash M\) satisfies

Why? If \(\left| e_{t}\right| \le \frac{1}{\sqrt{2}}y_{i}\), then \(T=\{t\}\) and so \(y_{j}=\left| e_{t}\right| \le \frac{1}{\sqrt{2} }y_{i}\); thus \(\frac{\left| y_{i}-y_{j}\right| }{\frac{1}{2} (y_{i}+y_{j})}=2\frac{y_{i}-y_{j}}{y_{i}+y_{j}}\), where

If instead \(\left| e_{t}\right| >\frac{1}{\sqrt{2}}y_{i}\), then \(T=\{t,t^{\prime }\}\), and so \(y_{j}=\left| e_{t}+e_{t^{\prime }}\right| =\left| e_{t}\right| +\left| e_{t^{\prime }}\right| \ge 2\left| e_{t}\right| >\frac{2}{\sqrt{2}}y_{i}=\sqrt{2}y_{i}\), an thus \(y_{i}<y_{j}/\sqrt{2}\); so, \(\frac{\left| y_{i}-y_{j}\right| }{\frac{1}{2}(y_{i}+y_{j})}=2\frac{y_{j}-y_{i}}{y_{i}+y_{j}}\), where \(\frac{y_{j}-y_{i}}{y_{i}+y_{j}}>3-2\sqrt{2}\) by a calculation analogous to (19).

Recall that \(II^{\prime }=\frac{1}{n(n-1)}\sum _{(i,j)\in N^{2}:i\ne j} \frac{\left| y_{i}-y_{j}\right| }{\frac{1}{2}\left( y_{i} +y_{j}\right) }\). Here, each \(\frac{\left| y_{i}-y_{j}\right| }{\frac{1}{2}\left( y_{i}+y_{j}\right) }\) is 0 if \((i,j)\in M^{2}\) or \((i,j)\in (N\backslash M)^{2}\), and it is at least \(2(3-2\sqrt{2})\) if \((i,j)\in M\times (N\backslash M)\) or \((i,j)\in (N\backslash M)\times M\). So, \(II^{\prime }\ge \frac{1}{n(n-1)}2\left| M\right| \left| N\backslash M\right| \times 2(3-2\sqrt{2})\). The definition of M and the fact that \(n\ge 3\) imply that \(\left| M\right| (n-\left| M\right| )>\frac{(n-1)^{2}}{4}\). Therefore,

\(\square \)

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and permissions