The importance of expertise in group decisions

Abstract

Prior to a collective binary choice, members of a group receive binary signals correlated with the better option. A larger group size may produce less accurate decisions, but expertise is everywhere beneficial. If a group accounts for correlation in signals, a relatively expert member puts an upper bound on the probability of a false belief. The bound holds for any group size and signal distribution. Furthermore, a population investing in expertise is better off cultivating a small mass of elites than adopting an egalitarian policy of education.

Appendix

Proof of Proposition 1

See Boland (1989). $\square$

Proof of Lemma 1

Assume (A1)–(A4). Denote the ordered elements of ${\mathbf {g}} \in {\mathbb {G}}_{\mathbf {n}}$ by $(g_1,g_2,\ldots ,g_{|{\mathbf {g}}|})$, where $|{\mathbf {g}}| > 1$ is the cardinality of ${\mathbf {g}}$. Let $\alpha _i = a_{g_i}$ and $\beta _i = b_{g_i}$ for $i \in \{1,\ldots ,|{\mathbf {g}}|\}$. Define $\delta _0 = \mathbb {1}(g_1 > 1) \sum _{j=1}^{g_1-1} d_j$, where $\mathbb {1}(\cdot )$ is the indicator function. Define $\delta _1 = \sum _{j=1}^{g_1} d_j$. Next, recursively define $\delta _{i+1}=\sum _{j=1}^{g_{i+1}-1} d_j - \delta _i$ for $i \in \{1,\ldots ,|{\mathbf {g}}|-1\}$. Then for $i, m \in \{1,\ldots ,|{\mathbf {g}}|\}$ (with $i < m$, $i+m \le |{\mathbf {g}}|$),

$$\begin{aligned} P(\alpha _{i+m} \cap (\underset{j \le i}{\cap } \alpha _j)|A) = P(a_{g_{i+m}} \cap (\underset{j \le i}{\cap } a_{g_j})|A) = p- \sum _{j=a_{g_i}}^{a_{g_{i+m}}-1} d_j=p-\sum _{j=i}^{i+m-1} \delta _j. \end{aligned}$$

Similarly,

$$\begin{aligned} P(\beta _{i+m} \cap (\underset{j \le i}{\cup } \beta _i)|A) = P(b_{g_{i+m}} \cap (\underset{j \le i}{\cup } a_{g_j})|A) = \sum _{j=a_{g_i}}^{a_{g_{i+m}}-1} d_j= \sum _{j=i}^{i+m-1} \delta _j. \end{aligned}$$

Lastly,

$$\begin{aligned} \sum _{j=1}^{|{\mathbf {g}}|}\delta _j=\sum _{j=1}^{g_{|{\mathbf {g}}|}}d_j \le \sum _{j=1}^{n-1} d_j \le \sum _{j=1}^{\infty } d_j \le 1-p \end{aligned}$$

by construction. $\square$

Proof of Lemma 2

Assume (A1)–(A4). Note (A3) implies

$$\begin{aligned} P(\mathbf {s}_{{\mathbf{g}}}|A)=P({\bar{\mathbf{s}}}_\mathbf{g}|B) \;\; \forall {\mathbf {g}} \in {\mathbb {G}}_{\mathbf {n}}, \; {{\mathbf{s}}}_{{\mathbf{g}}} \in {\mathbb {S}}_{\mathbf {g}}. \end{aligned}$$

We seek to show

$$\begin{aligned} P({{\mathbf{s}}}_{{\mathbf{g}}}|A)=P({\bar{\mathbf{s}}}_{{\mathbf{g}}}|A) \; \forall {\mathbf {g}} \in {\mathbb {G}}_{\mathbf {n}}, \;{{\mathbf{s}}}_{{\mathbf{g}}} \in {\mathbb {S}}_{\mathbf {g}} \; \text {s.t.} \; \exists \, i,j \in {\mathbf {g}} \; \text {for which} \; s_i \ne s_j.\qquad \qquad (\hbox {L2}^\prime ) \end{aligned}$$

By (A3), (L2$^\prime$) implies $P({{\mathbf{s}}}_{{\mathbf{g}}}|A)=P({{\mathbf{s}}}_{{\mathbf{g}}}|B)$.

The proof uses the following two results. First, suppressing the conditional probability notation,

$$\begin{aligned} P({{\mathbf{s}}}_{{\mathbf{g}}})=0 \;\; \text {if} \;\; \exists \, i<j<k \;\text {for which} \; s_i = s_k \ne s_j. \end{aligned}$$

(L2.A)

Second, defining ${\mathbf {g}}_j^{+}=\{i \in {\mathbf {g}}|i\ge j\}$ and ${\mathbf {g}}_j^{-}=\{i \in {\mathbf {g}}|i < j\}$,

$$\begin{aligned} \begin{aligned} P({{\mathbf{s}}}_{{\mathbf{g}}})=\delta _{j-1} \;\, \text {if} \;\, \exists j \in {\mathbf {g}} \;\; \text {s.t.} \;&(s_i=a_i \, \forall i \in {\mathbf {g}}_j^{-}) \; \text {and} \; (s_i=b_i \, \forall i \in {\mathbf {g}}_j^{+}) \\ \text {or} \;&(s_i=b_i \, \forall i \in {\mathbf {g}}_j^{-}) \; \text {and} \; (s_i=a_i \, \forall i \in {\mathbf {g}}_j^{+}), \end{aligned} \end{aligned}$$

(L2.B)

where $\delta _{j-1}$ is constructed per Lemma 1. Since (L2$^\prime$) implies at least two of the signals are different, (L2.A) and (L2.B) cover all relevant cases. Suppose $\exists \, i<j<k \;\text {for which} \; s_i = s_k \ne s_j$. Then $P({{\mathbf{s}}}_{{\mathbf{g}}})=0$ by (L2.A). But $s_i = s_k \ne s_j \Rightarrow {\bar{s}}_i = {\bar{s}}_k \ne {\bar{s}}_j$. Then $P({\bar{\mathbf{s}}}_{{\mathbf{g}}})=0$ by (L2.A) again, so $P({{\mathbf{s}}}_{{\mathbf{g}}})=P({\bar{\mathbf{s}}}_{{\mathbf{g}}})$. For the remaining sequences in the class of (L2.B), the result is immediate. $\square$

Proof of L2.A

Without loss of generality, suppose $s_i=a_i$, $s_j=b_j$, and $s_k=a_k$. Assumption (A4) applies to the redefined system with $i=1$, $j=2$, and $k=3$, per Lemma 1. Note that

$$\begin{aligned} P(\alpha _1 \cap \alpha _3) = P(\alpha _1 \cap \alpha _2 \cap \alpha _3) + P (\alpha _1 \cap \beta _2 \cap \alpha _3). \end{aligned}$$

By (A4), $P(\alpha _1 \cap \alpha _2 \cap \alpha _3) = p - \delta _1 - \delta _2$. By another application of Lemma 1, then, $P(\alpha _1 \cap \alpha _3)=p - \delta _1 - \delta _2$. Thus, $P(\alpha _1 \cap \beta _2 \cap \alpha _3)=0$.

Proof of L2.B

The proof is by induction. Apply Lemma 1 to subgroup $\mathbf {g+1}$ nesting ${\mathbf {g}}$ (with $|\mathbf {g+1}|=|{\mathbf {g}}|+1$). For $|{\mathbf {g}}|=2$, by the application of (A4) to the redefined system, $P(\alpha _1 \cap \beta _2)=\delta _1$. Then

$$\begin{aligned} P(\beta _1 \cap \alpha _2)=1-P(\overline{\beta _{1} \cap \alpha _{2}}) =1-P(\alpha _1 \cup \beta _2)&=1-[P(\alpha _1)+P(\beta _2)-P(\alpha _1 \cap \beta _2)]\\&=1-[p+1-p - \delta _1]\\&=\delta _1, \end{aligned}$$

where the second equality follows from De Morgan’s Law, so the induction hypothesis holds for $|{\mathbf {g}}|=2$. For $|{\mathbf {g}}|+1$, define ${\mathbf {g}}_j^{+}=\{i|j \le i \le |{\mathbf {g}}|\}$ and ${\mathbf {g}}_j^{-}=\{i|1 \le i < j\}$ for $j \in {\mathbf {g}}$. Assume $s_i=\alpha _i \; \forall i \in {\mathbf {g}}_j^{-}$ and $s_i=\beta _i \, \forall i \in {\mathbf {g}}_j^{+}$ (the proof for the opposite case is similar). That is, consider a sequence of the form $(\alpha _1,\alpha _2,\ldots ,\alpha _{j-1},\beta _j,\beta _{j+1},\ldots , \beta _{|{\mathbf {g}}|})$ for some $j \in \{2,3,\ldots ,|{\mathbf {g}}|\}$. Then

$$\begin{aligned} P(\alpha _1,\alpha _2,\ldots ,\alpha _{j-1},\beta _j,\beta _{j+1},\ldots , \beta _{|{\mathbf {g}}|})&=P(\alpha _1,\alpha _2,\ldots ,\alpha _{j-1},\beta _j, \beta _{j+1},\ldots ,\beta _{|{\mathbf {g}}|},\beta _{|{\mathbf {g}}|+1})\\&\quad +P(\alpha _1,\alpha _2,\ldots ,\alpha _{j-1},\beta _j,\beta _{j+1},\ldots , \beta _{|{\mathbf {g}}|},\alpha _{|{\mathbf {g}}|+1})\\&=P(\alpha _1,\alpha _2,\ldots ,\alpha _{j-1},\beta _j,\beta _{j+1},\ldots , \beta _{|{\mathbf {g}}|},\beta _{|{\mathbf {g}}|+1})\\&\quad +0 \end{aligned}$$

by (L2.A), so

$$\begin{aligned} P(\alpha _1,\alpha _2,\ldots ,\alpha _{j-1},\beta _j,\beta _{j+1},\ldots , \beta _{|{\mathbf {g}}|},\beta _{|{\mathbf {g}}|+1})=P(\alpha _1,\alpha _2,\ldots , \alpha _{j-1},\beta _j,\beta _{j+1},\ldots ,\beta _{|{\mathbf {g}}|})=\delta _{j-1}. \end{aligned}$$

$\square$

Proof of Proposition 2

Let $\forall \mathbf {g'} \subseteq {\mathbf {g}} \in {\mathbb {G}}_{\mathbf {n}}$, and assume (S1) and (A1)–(A4). Without loss of generality, let $S=A$, and suppress the conditional notation for the state of the world. Per Lemma 1,

$$\begin{aligned} P(\mu ^+|{\mathbf {g}})=P\left( \underset{i \in {\mathbf {g}}}{\cap } \alpha _i\right) \le P\left( \underset{i \in \mathbf {g'}}{\cap } \alpha _i\right) =P(\mu ^+|\mathbf {g'}), \end{aligned}$$

where the equalities follow from Lemma 2 and the inequality from $\mathbf {g'} \subseteq {\mathbf {g}}$. Similarly,

$$\begin{aligned} P(\mu ^-|{\mathbf {g}})=P\left( \underset{i \in {\mathbf {g}}}{\cap } \beta _i\right) \le P\left( \underset{i \in \mathbf {g'}}{\cap } \beta _i\right) =P(\mu ^-|\mathbf {g'}), \end{aligned}$$

which establishes (1). For (2), note that (A4) and Lemma 2 imply

$$\begin{aligned} P(\mu ^+|{\mathbf {n}})=P\left( \underset{i \in {\mathbf {n}}}{\cap } \alpha _i\right) =P\left( \underset{i \in {\mathbf {g}}}{\cap } a_i\right) =p-\sum\nolimits_{j=1}^{n-1} d_j. \end{aligned}$$

Likewise,

$$\begin{aligned} P(\mu ^-|{\mathbf {n}})=P\left( \underset{i \in {\mathbf {n}}}{\cap } b_i\right) =1-P\left( \underset{i \in {\mathbf {n}}}{\cup } a_i\right) =(1-p)-\sum _{j=1}^{n-1} d_j. \end{aligned}$$

To establish the last equality, note $P(a_1 \cup a_2) = P(a_1) + P(a_2) - P(a_1 \cap a_2) = 2p - (p - d_1)$ by (A4), so induction with Lemma 1 confirms $P\left(\bigcup\nolimits_{{i \in {\mathbf{n}}}} {a_{i} } \right) = p + \sum\limits_{{j = 1}}^{{n - 1}} {d_{j} }$. $\square$

Proof of Proposition 3

For ${\mathbf {g}} \in {\mathbb {G}}_{\mathbf {n}}$ s.t. $|{\mathbf {g}}| \in \{2x-1|x\in {\mathbb {N}}^+\}$, assume (S2), (A1)–(A4), and, without loss of generality, $S={A}$. Apply Lemma 1, and suppress the conditional notation for the binary state. Let $\mathbb {1}(\cdot )$ denote the indicator function. Define

$$\begin{aligned} {\mathbb {M}}_\alpha = \{ {\mathbf {s}}_{\mathbf {g}} \in {\mathbb {S}}_{\mathbf {g}} | \sum _{i \in {\mathbf {g}}} \mathbb {1}(s_i =\alpha _i) > \sum _{i \in {\mathbf {g}}} \mathbb {1}(s_i =\beta _i)\}, \end{aligned}$$

with ${\mathbb {M}}_\beta$ its complementary set. Because $|{\mathbf {g}}|$ is odd, the two sets cover all signal profiles.

Assumption (A4) implies $P(\cap _{i \in {\mathbf {g}}} \alpha _i)=p-\sum _{j=1}^{|{\mathbf {g}}|-1}\delta _j$ and $P(\cap _{i \in {\mathbf {g}}} \beta _i)=(1-p)-\sum _{j=1}^{|{\mathbf {g}}|-1}\delta _j$, where the proof of Proposition 2 confirms the latter equality. Define ${\mathbf {g}}_j^{+}=\{i \in {\mathbf {g}}|i\ge j\}$ and ${\mathbf {g}}_j^{-}=\{i \in {\mathbf {g}}|i < j\}$. By Lemma2, $P({{\mathbf{s}}}_{{\mathbf{g}}})=0$ if $\exists i<j<k$ for which $s_i = s_k \ne s_j$. Also,

$$\begin{aligned} \begin{aligned} P({{\mathbf{s}}}_{{\mathbf{g}}})=\delta _{j-1} \;\, \text {if} \;\, \exists j \in {\mathbf {g}} \;\; \text {s.t.} \;&(s_i=\alpha _i \, \forall i \in {\mathbf {g}}_j^{-}) \cap (s_i=\beta _i \, \forall i \in {\mathbf {g}}_j^{+}) \\ \text {or} \;&(s_i=\beta _i \, \forall i \in {\mathbf {g}}_j^{-}) \cap (s_i=\alpha _i \, \forall i \in {\mathbf {g}}_j^{+}). \end{aligned} \end{aligned}$$

But then

$$\begin{aligned} P(\mu ^+|{\mathbf {g}})&=P({{\mathbf{s}}}_{{\mathbf{g}}} \in {\mathbb {M}}_\alpha )= p-\sum _{j=1}^{|{\mathbf {g}}|-1} \delta _j + \sum _{j=1}^{|{\mathbf {g}}|-1}\delta _j = p, \; \text {and} \\ P(\mu ^-|{\mathbf {g}})&=P({{\mathbf{s}}}_{{\mathbf{g}}} \in {\mathbb {M}}_\beta )= (1-p)-\sum _{j=1}^{|{\mathbf {g}}|-1} \delta _j+\sum _{j=1}^{|{\mathbf {g}}|-1}\delta _j=(1-p).\; \end{aligned}$$

$\square$

Proof of Proposition 4

The proof is by induction. Assume (S1), (A1)–(A3), and $S=A$ (the proof for $S=B$ is similar). The induction hypothesis clearly holds for $n=1$. For illustration, consider the general case of $n=2$ presented in Table 1 of Sect. 4.

By definition, $x+y=p_2$, and $x>1-p_1-y$ since $p_1,p_2>1/2$. If $p_2>p_1$ (equivalently, $p_1-x<y$), then $P(\mu ^-|{\mathbf {n}})=p_1-x+1-p_1-y=1-p_2=1-\text {max}\{p_1,p_2\}$. If $p_2<p_1$ (equivalently, $p_1-x>y$), then $P(\mu ^-|{\mathbf {n}})=y+1-p_1-y=1-p_1=1-\text {max}\{p_1,p_2\}$. Lastly, if $p_2=p_1$, then $P(\mu ^-|{\mathbf {n}})=1-p_1-y<1-p_1=1-\text {max}\{p_1,p_2\}$. Therefore, $P(\mu ^-|{\mathbf {n}})\le 1-\text {max}\{p_1,p_2\}$, and the induction hypothesis holds for $n=2$.

Next consider the general case for $n>2$. Index the $n^2$ possible signal vectors by $j=1,2,\ldots ,n^2$. Without loss of generality, order the signal vectors according to whether they create an “incorrect” or “correct” inference, with the “uninformative” cases at the end:

$$\begin{aligned} P({\mathbf {s}}_j|A) = {\left\{ \begin{array}{ll} \mu ^-_j \quad \text {for} \quad j=1,2,\ldots ,k \\ \mu _j \quad \text {for} \quad j=k+1,\ldots ,n^2/2 \\ \mu ^+_j \quad \text {for} \quad j=n^2/2+1,\ldots ,n^2/2+k+1 \\ \mu _j \quad \text {for}\quad j=n^2/2+k+2,\ldots ,n^2 \end{array}\right. } \end{aligned}$$

for some $k \in \{1,2,\ldots ,n^2/2\}$ (Proposition 3 shows $k\ge 1$), where it is understood that if $k=n^2/2$, no event is “uninformative.” Furthermore, let ${\mathbf {s}}_j=\mathbf {{\bar{s}}}_{j+n^2/2}$ for $j=1,2,\ldots ,n^2/2$. Define $\mu ^-_{aj}$, $\mu _{aj}$, and $\mu ^+_{aj}$ for the respective probabilities of intersections with $a_{n+1}$. Likewise, define $\mu ^-_{bj}$ for the conditional intersection of $b_{n+1}$ and so on. Let $\mathbb {1}(\cdot )$ denote the indicator function. Then, for $\mathbf {n+1}$ nesting ${\mathbf {n}}$ (with $|\mathbf {n+1}|=n+1$),

$$\begin{aligned} P(\mu ^-|\mathbf {n+1})&=\sum _{j=1}^{k} \mathbb {1}(\mu ^-_{aj}<\mu ^+_{bj+n^2/2})\mu ^-_{aj}\\&\quad +\mathbb {1} (\mu ^-_{bj}<\mu ^+_{aj+n^2/2})\mu ^-_{bj} \\&\quad + \mathbb {1}(\mu ^+_{aj+n^2/2}<\mu ^-_{bj})\mu ^+_{aj+n^2/2}+\mathbb {1} (\mu ^+_{bj+n^2/2}<\mu ^-_{aj})\mu ^+_{bj+n^2/2}\\&\quad +\sum _{j=k+1}^{n^2/2} \mathbb {1}(\mu _{aj}<\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1}(\mu _{bj}<\mu _{aj+n^2/2})\mu _{bj}\\&\quad + \mathbb {1}(\mu _{aj+n^2/2}<\mu _{bj})\mu _{aj+n^2/2}+\mathbb {1}(\mu _{bj +n^2/2}<\mu _{aj})\mu _{bj+n^2/2}.\\ \end{aligned}$$

Because

$$\begin{aligned}&\mathbb {1}(\mu ^-_{aj}<\mu ^+_{bj+n^2/2})\mu ^-_{aj} +\mathbb {1}(\mu ^-_{bj}<\mu ^+_{aj+n^2/2})\mu ^-_{bj} + \mathbb {1}(\mu ^+_{aj+n^2/2}<\mu ^-_{bj})\mu ^+_{aj+n^2/2}\\&\qquad +\mathbb {1}(\mu ^+_{bj+n^2/2}<\mu ^-_{aj})\mu ^+_{bj+n^2/2} \\&\quad \le \mu ^-_j, \end{aligned}$$

and

$$\begin{aligned}&\mathbb {1}(\mu _{aj}<\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1}(\mu _{bj}<\mu _{aj+n^2/2}) \mu _{bj} +\mathbb {1}(\mu _{aj+n^2/2}<\mu _{bj})\mu _{aj+n^2/2}\\&\qquad +\mathbb {1} (\mu _{bj+n^2/2}<\mu _{aj})\mu _{bj+n^2/2} \\&\quad \le \mu _j, \end{aligned}$$

the induction hypothesis implies (via Lemma 1):

$$\begin{aligned} P(\mu ^-|\mathbf {n+1})\le \sum _{j=1}^{k} \mu ^-_j +\sum _{j=k+1}^{n^2/2} \mu _j\le 1-\max \limits _{i \in \mathbf {n+1}} p_i, \end{aligned}$$

where the last inequality follows from Lemma 5. $\square$

Lemma 5

Assume (S1) and (A1)–(A3). Adopting the terminology and ordering of Proposition 4,

$$\begin{aligned} \sum _{j=1}^{k} \mu ^-_j +\sum _{j=k+1}^{n^2/2} \mu _j \le 1-\max \limits _{i \in \mathbf {n+1}} p_i. \end{aligned}$$

Proof of Lemma 5

Assume (S1) and (A1)–(A3). The proof is by induction. The induction hypothesis clearly holds for $n=1$. Suppose $n>1$. We seek to show

$$\begin{aligned} \Gamma \equiv \sum _{j=1}^{k} \mu ^-_j +\sum _{j=k+1}^{n^2/2} \mu _j =&\sum _{j=1}^{k} \mathbb {1}(\mu ^-_{aj}<\mu ^+_{bj+n^2/2})\mu ^-_{aj}\\&+\mathbb {1} (\mu ^-_{bj}<\mu ^+_{aj+n^2/2})\mu ^-_{bj}\\&+ \mathbb {1}(\mu ^+_{aj+n^2/2}<\mu ^-_{bj})\mu ^+_{aj+n^2/2}+\mathbb {1} (\mu ^+_{bj+n^2/2}<\mu ^-_{aj})\mu ^+_{bj+n^2/2}\\&+\mathbb {1}(\mu ^-_{aj}=\mu ^+_{bj+n^2/2})\mu ^-_{aj}+\mathbb {1} (\mu ^-_{bj}=\mu ^+_{aj+n^2/2})\mu ^-_{bj}\\&+\sum _{j=k+1}^{n^2/2} \mathbb {1}(\mu _{aj}<\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1} (\mu _{bj}<\mu _{aj+n^2/2})\mu _{bj}\\&+\mathbb {1}(\mu _{aj+n^2/2}<\mu _{bj})\mu _{aj+n^2/2}+\mathbb {1} (\mu _{bj+n^2/2}<\mu _{aj})\mu _{bj+n^2/2}\\&+\mathbb {1}(\mu _{aj}=\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1} (\mu _{bj}=\mu _{aj+n^2/2})\mu _{bj}\\ \le&1-\max \limits _{i \in \mathbf {n+1}} p_i. \end{aligned}$$

Because

$$\begin{aligned}&\mathbb {1}(\mu ^-_{aj}<\mu ^+_{bj+n^2/2})\mu ^-_{aj}+\mathbb {1} (\mu ^-_{bj}<\mu ^+_{aj+n^2/2})\mu ^-_{bj}\\&\quad +\mathbb {1}(\mu ^+_{aj+n^2/2}<\mu ^-_{bj})\mu ^+_{aj+n^2/2}+\mathbb {1} (\mu ^+_{bj+n^2/2}<\mu ^-_{aj})\mu ^+_{bj+n^2/2}\\&\quad +\mathbb {1}(\mu ^-_{aj}=\mu ^+_{bj+n^2/2})\mu ^-_{aj}+\mathbb {1} (\mu ^-_{bj}=\mu ^+_{aj+n^2/2})\mu ^-_{bj} \le \mu ^-_{bj}+\mu ^+_{bj+n^2/2},\\&\mathbb {1}(\mu _{aj}<\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1} (\mu _{bj}<\mu _{aj+n^2/2})\mu _{bj} + \mathbb {1} (\mu _{aj+n^2/2}<\mu _{bj})\mu _{aj+n^2/2}\\&\quad +\mathbb {1} (\mu _{bj+n^2/2}<\mu _{aj})\mu _{bj+n^2/2}\\&\quad +\mathbb {1}(\mu _{aj}=\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1} (\mu _{bj}=\mu _{aj+n^2/2})\mu _{bj} \le \mu _{bj}+\mu _{bj+n^2/2}, \end{aligned}$$

and

$$\begin{aligned} 1-p_{n+1}=\sum _{j=1}^{k} (\mu ^-_{bj} +\mu ^+_{bj+n^2/2}) +\sum _{j=k+1}^{n^2/2} (\mu _{bj} +\mu _{bj+n^2/2}) \end{aligned}$$

by construction, then $\Gamma \le 1-p_{n+1}$.

Next, because

$$\begin{aligned}&\mathbb {1}(\mu ^-_{aj}<\mu ^+_{bj+n^2/2})\mu ^-_{aj}+\mathbb {1} (\mu ^-_{bj}<\mu ^+_{aj+n^2/2})\mu ^-_{bj} + \mathbb {1} (\mu ^+_{aj+n^2/2}<\mu ^-_{bj})\mu ^+_{aj+n^2/2}\\&\quad +\mathbb {1} (\mu ^+_{bj+n^2/2}<\mu ^-_{aj})c\mu ^+_{bj+n^2/2}\\&\quad +\mathbb {1}(\mu ^-_{aj}=\mu ^+_{bj+n^2/2})\mu ^-_{aj}+\mathbb {1} (\mu ^-_{bj}=\mu ^+_{aj+n^2/2})\mu ^-_{bj} \le \mu ^-_j, \end{aligned}$$

and

$$\begin{aligned}&\mathbb {1}(\mu _{aj}<\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1} (\mu _{bj}<\mu _{aj+n^2/2})\mu _{bj} + \mathbb {1} (\mu _{aj+n^2/2}<\mu _{bj})\mu _{aj+n^2/2}+\mathbb {1} (\mu _{bj+n^2/2}<\mu _{aj})\mu _{bj+n^2/2}\\&\quad +\mathbb {1}(\mu _{aj}=\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1} (\mu _{bj}=\mu _{aj+n^2/2})\mu _{bj} \le \mu _j, \end{aligned}$$

the induction hypothesis implies

$$\begin{aligned} \Gamma \le \sum _{j=1}^{k} \mu ^-_j +\sum _{j=k+1}^{n^2/2} \mu _j\le 1 -\max \limits _{i \in {\mathbf {n}}} p_i. \end{aligned}$$

Thus, $\Gamma \le 1-\max \nolimits _{i \in \mathbf {n+1}} p_i$. $\square$

Proof of Proposition 5

Assume (S1) and (A1)–(A3). Following the same approach and terminology of Proposition 4, the induction hypothesis clearly holds for $n=1$. Let $\mathbf {n+1}$ nest ${\mathbf {n}}$ (with $|\mathbf {n+1}|=n+1$). For $n+1$, then,

$$\begin{aligned} P(\mu |\mathbf {n+1})&=\sum _{j=1}^{k} \mathbb {1}(\mu ^-_{aj}=\mu ^+_{bj+n^2/2})\mu ^-_{aj}+\mathbb {1} (\mu ^-_{bj}=\mu ^+_{aj+n^2/2})\mu ^-_{bj}\\&\quad + \mathbb {1}(\mu ^+_{aj+n^2/2}=\mu ^-_{bj})\mu ^+_{aj+n^2/2} +\mathbb {1}(\mu ^+_{bj+n^2/2}=\mu ^-_{aj})\mu ^+_{bj+n^2/2}\\&\quad +\sum _{j=k+1}^{n^2/2} \mathbb {1}(\mu _{aj}=\mu _{bj+n^2/2})\mu _{aj} +\mathbb {1}(\mu _{bj}=\mu _{aj+n^2/2})\mu _{bj}\\&\quad + \mathbb {1}(\mu _{aj+n^2/2}=\mu _{bj})\mu _{aj+n^2/2}+\mathbb {1} (\mu _{bj+n^2/2}=\mu _{aj})\mu _{bj+n^2/2}. \end{aligned}$$

Because

$$\begin{aligned}\mathbb {1}(\mu ^-_{aj} & =\mu ^+_{bj+n^2/2})\mu ^-_{aj}+\mathbb {1} (\mu ^-_{bj}=\mu ^+_{aj+n^2/2})\mu ^-_{bj} + \mathbb {1}(\mu ^+_{aj+n^2/2}= \mu ^-_{bj})\mu ^+_{aj+n^2/2}\\&\qquad +\mathbb {1}(\mu ^+_{bj+n^2/2} =\mu ^-_{aj})\mu ^+_{bj+n^2/2} \\&\quad \le 2\mu ^-_{bj}+2\mu ^+_{bj+n^2/2},\\\mathbb {1}(\mu _{aj} & =\mu _{bj+n^2/2})\mu _{aj}+\mathbb {1}(\mu _{bj} =\mu _{aj+n^2/2})\mu _{bj} + \mathbb {1}(\mu _{aj+n^2/2} =\mu _{bj})\mu _{aj+n^2/2}\\&\qquad +\mathbb {1}(\mu _{bj+n^2/2}=\mu _{aj})\mu _{bj+n^2/2} \\&\quad \le 2\mu _{bj}+2\mu _{bj+n^2/2}, \end{aligned}$$

and

$$\begin{aligned} 1-p_{n+1}=\sum _{j=1}^{k} (\mu ^-_{bj} +\mu ^+_{bj+n^2/2})+\sum _{j=k+1}^{n^2/2} (\mu _{bj} +\mu _{bj+n^2/2}) \end{aligned}$$

by construction, then $P(\mu |\mathbf {n+1}) \le 2(1-p_{n+1})$. An exact analogy to Lemma 5 then establishes $P(\mu |\mathbf {n+1}) \le 2(1-\max \nolimits _{i \in {\mathbf {n}}} p_i)$. Thus, $P(\mu |\mathbf {n+1}) \le 2(1-\max \nolimits _{i \in \mathbf {n+1}} p_i)$. $\square$

Proof of Proposition 6

First, reversing the order of the summation,

$$\begin{aligned} F(p,n)&={n \atopwithdelims ()n}p^n+{n \atopwithdelims ()n-1}p^{n-1}(1-p)+{n \atopwithdelims ()n-2}p^{n-2}(1-p)^2+\ldots \\&\quad +{n \atopwithdelims ()\frac{n+1}{2}}p^{n-(n+1)/2}(1-p)^{(n-1)/2}. \end{aligned}$$

Then

$$\begin{aligned} \frac{\partial F(p,n)}{\partial p}&=np^{n-1}+\frac{n!}{(n-1)!}\bigg [(n-1)p^{n-2}(1-p)-p^{n-1}\bigg ]+\\& \quad \times \frac{n!}{2!(n-2)!}\bigg [(n-2)p^{n-3}(1-p)^2-2p^{n-2}(1-p)\bigg ]+\ldots \\&\quad +\frac{n!}{\big (\frac{n+1}{2}\big )!\big (\frac{n-1}{2}\big )!} \bigg [((n+1)/2)p^{(n+1)/2-1}(1-p)^{(n-1)/2} \\&\quad -((n-1)/2)p^{n-1}(1-p)^{(n-1)/2-1}\bigg ] \end{aligned}$$

is a telescoping series, leaving

$$\begin{aligned} \frac{\partial F(p,n)}{\partial p}&=\frac{n!}{\big (\frac{n+1}{2}\big )! \big (\frac{n-1}{2}\big )!}\bigg [((n+1)/2)p^{(n+1)/2-1}(1-p)^{(n-1)/2}\bigg ]\\&=\frac{n!}{\big (\big (\frac{n-1}{2}\big )!\big )^2}\big [p(1-p)\big ]^{(n-1)/2}. \end{aligned}$$

Then

$$\begin{aligned} \frac{\partial ^2 F(p,n+2)}{\Delta n \partial p} = \frac{(n+2)!}{\big (\big (\frac{n+1}{2}\big )!\big )^2}\big [p(1-p)\big ]^{(n+1)/2} -\frac{n!}{\big (\big (\frac{n-1}{2}\big )!\big )^2}\big [p(1-p)\big ]^{(n-1)/2}, \end{aligned}$$

which is greater than zero if and only if $4p(1-p)>(n+1)/(n+2)$. Because $Q(p) \equiv 4p(1-p)=1$ at its maximizer of $p=1/2$, the solution to the corresponding quadratic equation yields

$$\begin{aligned} \frac{\partial ^2 F(p,n+2)}{\Delta n \partial p} {\left\{ \begin{array}{ll} \ge 0 \; \text {if} \quad p \in [1/2,1/2+\sqrt{1-(n+1)/(n+2)}] \\ \le 0 \; \text {if} \quad p \in [1/2+\sqrt{1-(n+1)/(n+2)},1] \end{array}\right. }. \end{aligned}$$

Because EU(p, n) is an affine function of F(p, n), the optimal group size is single-peaked. $\square$

Proof of Lemma 3

Kaniovski and Zaigraev (2011) show F(p, n) is an increasing function, with at most one inflection point, that lies above a chord between $p=0$ and $p=1$. Thus, F(p, n) is concave. $\square$

Proof of Lemma 4

Writing the optimal investment as a function of the group size, the first-order condition implies

$$\begin{aligned} \frac{\partial F(p^*(n),n)}{\partial p}uN-n\phi '(p^*(n))=0 \end{aligned}$$

for any n. To preserve monotonicity, suppose the group expands by two members. Then

$$\begin{aligned} \frac{\frac{\displaystyle \partial F(p^*(n+2),n+2)}{\displaystyle \partial p}}{\frac{\displaystyle \partial F(p^*(n),n)}{\displaystyle \partial p}}=\frac{(n+2)\phi '(p^*(n+2))}{n\phi '(p^*(n))}. \end{aligned}$$

(L5)

Now suppose $p^*(n+2)\ge p^*(n)$. The right-hand side of (L5) is then strictly greater than one by the convexity of $\phi (p)$. By Proposition 1, F(p, n) is increasing and approaching one as n grows. Furthermore, the difference $F(p,n+2)-F(p,n)$ is decreasing in n. From Lemma 3, F(p, n) is concave and approaches one as p grows. Thus, the left-hand side of (L5) must be strictly less than one if $p^*(n+2)\ge p^*(n)$, a contradiction. $\square$