Rational exaggeration and counter-exaggeration in information aggregation games

Abstract

We study an information aggregation game in which each of a finite collection of “senders” receives a private signal and submits a report to the center, who then makes a decision based on the average of these reports. The integration of three features distinguishes our framework from the related literature: players’ reports are aggregated by a mechanistic averaging rule, their strategy sets are intervals rather than binary choices, and they are ex ante heterogeneous. In this setting, players engage in a “tug-of-war,” as they exaggerate and counter-exaggerate in order to manipulate the center’s decision. While incentives to exaggerate have been studied extensively, the phenomenon of counter-exaggeration is less well understood. Our main results are as follows. First, the cycle of counter-exaggeration can be broken only by the imposition of exogenous bounds on the space of admissible sender reports. Second, in the unique pure-strategy equilibrium, all but at most one player is constrained with positive probability by one of the report bounds. Our third and fourth results hold for a class of “anchored” games. We show that if the report space is strictly contained in the signal space, then welfare is increasing in the size of the report space, but if the containment relation is reversed, welfare is independent of the size of the space. Finally, the equilibrium performance of our heterogeneous players can be unambiguously ranked: a player’s equilibrium payoff is inversely related to the probability that her exaggeration will be thwarted by the report bounds.

Appendix: Proofs

Proof of Proposition 1

To prove the proposition, we apply Theorems 1 and 2 of Athey (2001). The first of these theorems is used to establish existence for finite-action aggregation games. The second implies existence for general aggregation games. To apply Athey’s first theorem, we define a finite-action aggregation game to be one in which players are restricted to choose actions from a finite subset of $A$. In all other respects, finite-action aggregation games are identical to (infinite action) aggregation games. We now check that $u$ satisfies Athey’s Assumption A1. Clearly, our types have joint density w.r.t. Lebesgue measure, which is bounded and atomless. Moreover, the integrability condition in Athey’s A1 is trivially satisfied, since $u$ is bounded. Moreover, inequality (6) implies that the SCC holds. Therefore, every finite-action aggregation game has an MPE in which player ${r}$’s equilibrium strategy ${s_{r}}$ is non-decreasing. By Athey’s Theorem 2, the restricted game has an MPE, call it ${\mathbf {s} }^*$. To show that ${\mathbf {s} }^*$ will also be an equilibrium for the original, unrestricted game, it suffices to show that for all ${r}$, all $\theta$ and all $a> \bar{a}$, $\frac{\partial U_{r}( a, \theta ; \mathbf {s}_{-{r}}^*)}{\partial a} < 0$. To establish this, note that $\mathbf {s}_{-{r}}^*\ge \mathbf {\underline{s}}_{-{r}}\ge 0$, so that since $t$ is strictly increasing, $a> \bar{a}$ implies

$$\begin{aligned} U_{{r}}^\prime ( a, \theta ; \mathbf {s}_{-{r}}^*) \,{ < } \, U_{{r}}^\prime ( \bar{a}, \theta ; \mathbf {s}_{-{r}}^*) \, { \le } \, U_{{r}}^\prime ( \bar{a}, \theta ; \mathbf {\underline{s}}_{-{r}}) \, { \le } \, U_{{r}}^\prime ( \bar{a}, \theta ; 0 ) \, { \le } \, 0 \end{aligned}$$

Finally, to establish that $s_{r}$ is strictly increasing and continuously differentiable on , note that $U_{{r}}^\prime ( {s_{r}}(\cdot ) , \cdot ; \mathbf {s}_{-{r}}) = 0$ on . From (6), Assumption A2 and the implicit function theorem, we have, for all , $\frac{\mathrm {d}{s_{r}}( \theta )}{\mathrm {d}\theta } = - { \frac{\partial ^2 U_{r}( {s_{r}}(\theta ) , \theta ; \mathbf {s}_{-{r}})}{\partial a \partial \theta } }\Big / { \frac{\partial ^2 U_{r}( {s_{r}}(\theta ) , \theta ; \mathbf {s}_{-{r}})}{\partial a^2}} > 0$. $\square$

Proof of Proposition 2

Suppose that ${\mathbf {s} }$ is a non-monotone pure-strategy profile, that is, for some ${r}$, and some $\theta _{r}$, as well as $\updelta > 0$ with and $\theta _{r}+ \updelta < {\tilde{\theta }_{{r}}}({\mathbf {s} })$, we have $da = {s_{r}}(\theta + \updelta ) - {s_{r}}( \theta ) \le 0$. Since, clearly, $U_{r}( \cdot , \cdot ; \mathbf {s}_{-{r}})$ is twice continuously differentiable, it follows from the Taylor–Lagrange theorem that there exists $\upzeta \in [ 0 , 1 ]$ such that

$$\begin{aligned}&U_{{r}}^\prime \big ( {s_{r}}(\theta + \updelta ) , \theta + \updelta ; \mathbf {s}_{-{r}}\big ) -U_{{r}}^\prime \big ( {s_{r}}(\theta ) , \theta ; \mathbf {s}_{-{r}}\big ) \\&\quad \!=\! \bigtriangledown U_{{r}}^\prime \big ( {s_{r}}(\theta ) \!+\! \upzeta da , \theta + \upzeta \updelta ; \mathbf {s}_{-{r}}\big ) \cdot ( da , \updelta ) \!\ge \! \frac{\partial ^2 U_{r}\big ( {s_{r}}(\theta ) \!+\! \upzeta da , \theta + \upzeta \updelta ; \mathbf {s}_{-{r}}\big ) }{\partial a \partial \theta } \updelta \!\ge \! 0 \end{aligned}$$

The weak inequality holds because $da \le 0$, and from Assumption A2 and (4), $U_{r}( \cdot , \cdot ; \mathbf {s}_{-{r}})$ is strictly concave in ${r}$’s action. The strict inequality follows from (6), given that $\updelta > 0$. But optimality requires that $U_{{r}}^\prime \big ( {s_{r}}(\cdot ) , \cdot ; \mathbf {s}_{-{r}}\big ) \equiv 0$ on . Hence, ${\mathbf {s} }$ cannot be an equilibrium profile. $\square$

Proof of Proposition 5

We will prove uniqueness only for non-degenerate equilibrium profiles. Uniqueness for other profiles is ensured by restriction (9), but we omit the details. Let $\pmb {\lambda }^*$ be an NMPE for the aggregation game, and let $\pmb {\lambda }$ be any other profile of strategies such that for some ${{ j}}$, $\lambda _{{ j}}\ne \lambda _{{ j}}^*$. We will show that if $\pmb {\lambda }$ satisfies the necessary condition (16), then it fails the other necessary condition (7′).

We first establish a property of EDFA that will be useful. Differentiating (14) w.r.t. $\lambda _{r}$ and inferring from (10) that , we obtain

(29)

Suppose w.l.o.g. that $\lambda _{{ j}}> \lambda _{{ j}}^*$. From (29), $\mathbb {E}\xi _{{ j}}(\lambda _{{ j}}) < \mathbb {E}\xi _{{ j}}(\lambda _{{ j}}^*)$. For all ${r}\ne {{ j}}$, (16) implies that $\mathbb {E}\xi _{r}(\lambda _{r}) < \mathbb {E}\xi _{r}(\lambda _{r}^*)$, and (29), in turn, implies that $\lambda _{r}> \lambda _{r}^*$. To establish that $\pmb {\lambda }$ cannot satisfy (7′), it suffices to show that

$$\begin{aligned} \left( \sum _{i}\lambda _{i}+ \sum _{{i}\ne {{ j}}} \mathbb {E}\xi _{i}(\lambda _{i}) \right) >&\left( \sum _{i}\lambda _{i}^*+ \sum _{{i}\ne {{ j}}} \mathbb {E}\xi _{i}(\lambda _{i}^*) \right) = n k_{{ j}}\end{aligned}$$

or, equivalently,

$$\begin{aligned} \lambda _{{ j}}- \lambda _{{ j}}^*+ \sum _{{i}\ne {{ j}}} \left( \lambda _{i}- \lambda _{i}^*\right) >&\sum _{{i}\ne {{ j}}} \left( \mathbb {E}\xi _{i}(\lambda _{i}^*) - \mathbb {E}\xi _{i}(\lambda _{i}) \right) \end{aligned}$$

This last inequality is indeed satisfied, since by assumption, $\lambda _{{ j}}> \lambda _{{ j}}^*$, while (29) implies that for all ${i}\ne {{ j}}$, $\lambda _{i}- \lambda _{i}^*> \mathbb {E}\xi _{i}(\lambda _{i}^*) - \mathbb {E}\xi _{i}(\lambda _{i})$.

For the monotonicity of MPE, we prove $\lambda _{i}^*> \lambda _{{ j}}^*$ by contradiction. Suppose $\lambda _i^* \le \lambda _j^*$. The fact that and ${\tilde{\theta }_{{r}}}(\lambda _r)=\bar{a}- \lambda _r$ imply that and ${\tilde{\theta }}_i(\lambda _i^*) \ge {\tilde{\theta }}_j(\lambda _j^*)$. Then, from (14), we know that $\mathbb {E}\xi _i(\lambda _i^*) \ge \mathbb {E}\xi _j(\lambda _j^*)$. But (16) and $k_i > k_j$ imply that $\mathbb {E}\xi _i(\lambda _i^*) < \mathbb {E}\xi _j(\lambda _j^*)$. $\square$

Proof of Proposition 6

We first establish a property of $\mathbb {E}\mathbf {\upxi }$. Aggregating the identity in (13) across players and rearranging, we obtain

$$\begin{aligned} \mathbb {E}_{\mathbf {\upvartheta }}\big ( \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }- \mu {\left( {\mathbf {\upvartheta }}\right) }\big ) = {\mu {\left( \pmb {\lambda }^*\right) }}+ \mu {\left( \mathbb {E}\mathbf {\upxi }\right) }. \end{aligned}$$

(30)

Let $\xi _{r}^*= \xi _{r}( \lambda _{r}^*)$. Expanding the left-hand side of (20), we obtain

$$\begin{aligned} \mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {\upvartheta }}\right) }+ k_{r}- \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }\right) ^2= & {} \mathbb {E}_{\mathbf {\upvartheta }}(\mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }- \mu {\left( {\mathbf {\upvartheta }}\right) }- k_{r})^2 \nonumber \\= & {} \mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }- \mu {\left( {\mathbf {\upvartheta }}\right) }\right) ^2 - 2k_{r}\mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }\right. \nonumber \\&\left. -\mu {\left( {\mathbf {\upvartheta }}\right) }\right) + k_{r}^2 \nonumber \\= & {} \mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }- \mu {\left( {\mathbf {\upvartheta }}\right) }\right) ^2 - 2k_{r}\big ( \mu {\left( \mathbb {E}\mathbf {\upxi }^*\right) } \nonumber \\&+ {\mu {\left( \pmb {\lambda }^*\right) }}\big ) + k_{r}^2 \end{aligned}$$

(31)

The last equality follows from (30). Expanding the first term on the right-hand side of (31),

$$\begin{aligned} \mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }- \mu {\left( {\mathbf {\upvartheta }}\right) }\right) ^2= & {} \mathbb {E}_{\mathbf {\upvartheta }}\Big ( \mu \big ( \underbrace{ {\mathbf {s}^*}( {\mathbf {\upvartheta }})- ({\mathbf {\upvartheta }}+\pmb {\lambda }^*)} _{\mathbf {\xi ^*}}\big ) +{\mu {\left( \pmb {\lambda }^*\right) }}\Big )^2 \nonumber \\= & {} \mathbb {E}_{\mathbf {\upvartheta }}\big (\mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})- ({\mathbf {\upvartheta }}+ \pmb {\lambda }^*) \right) } \big )^2 \nonumber \\&+ 2 {\mu {\left( \pmb {\lambda }^*\right) }}\mu {\left( \mathbb {E}\mathbf {\upxi }^*\right) } + {\mu {\left( \pmb {\lambda }^*\right) }}^2 \end{aligned}$$

(32)

The first equality merely adds and subtracts ${\mu {\left( \pmb {\lambda }^*\right) }}$ and rearranges terms; the second averages both sides of the identity in (13). Now, expand the first term in (32) to obtain

$$\begin{aligned} \mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})- ({\mathbf {\upvartheta }}+ \pmb {\lambda }^*) \right) } \right) ^2= & {} \left( \sum _{i}\mathbb {E}_{{\upvartheta }_{i}}\left( s_{i}^*\left( {\upvartheta }_{i}\right) - \left( {\upvartheta }_{i}+ \lambda _{i}^*\right) \right) ^2 \right. \nonumber \\&\left. + {\sum \sum }_{i \ne j} \mathbb {E}\xi _{i}^*\mathbb {E}\xi _{{ j}}^*\right) /n^2 \nonumber \\= & {} \left( \sum _{i}V\xi _{i}^*+ \sum _{i}\left( \mathbb {E}\xi _{i}^*\right) ^2 + {\sum \sum }_{i \ne j} \mathbb {E}\xi _{i}^*\mathbb {E}\xi _{{ j}}^*\right) /n^2\nonumber \\= & {} \left( \sum _{i}V\xi _{i}^*+ \left[ \sum _{i}\mathbb {E}\xi _{i}^*\right] ^2\right) /n^2 \nonumber \\= & {} \mu {\left( \mathbf {V\upxi }^*\right) }/n + \left( \mu {\left( \mathbb {E}\mathbf {\upxi }^*\right) }\right) ^2 \end{aligned}$$

(33)

The first equality is obtained by expanding $\mu {\left( {\mathbf {\upvartheta }}+\pmb {\lambda }^*-{\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }$; the second is from the relationship $\mathbb {E}(X^2) = \hbox {Var}(X)+(EX)^2$ for a random variable $X$. Now, substituting (33) back into (32),

$$\begin{aligned}&\mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }- \mu {\left( {\mathbf {\upvartheta }}\right) }\right) ^2 = \mu {\left( \mathbf {V\upxi }^*\right) }/n + \big ( \mu {\left( \mathbb {E}\mathbf {\upxi }^*\right) } + {\mu {\left( \pmb {\lambda }^*\right) }}\big )^2 \end{aligned}$$

(34)

Finally, substitute (34) back into (31) to obtain

$$\begin{aligned} \mathbb {E}_{\mathbf {\upvartheta }}\left( \mu {\left( {\mathbf {\upvartheta }}\right) }+ k_{r}- \mu {\left( {\mathbf {s}^*}( {\mathbf {\upvartheta }})\right) }\right) ^2= & {} \mu {\left( \mathbf {V\upxi }^*\right) }/n + \big ( \mu {\left( \mathbb {E}\mathbf {\upxi }^*\right) } + {\mu {\left( \pmb {\lambda }^*\right) }}- k_{r}\big )^2 \\= & {} \mu {\left( \mathbf {V\upxi }^*\right) }/n + (\mathbb {E}\xi _{r}^*/n)^2 \end{aligned}$$

The last equality is obtained by adding $\mathbb {E}\xi _{r}^*/n$ to both sides of (7′) and substituting for $k_{r}$. $\square$

Proof of Remark 1

To verify that A3(iii) guarantees non-degeneracy, it suffices to check that $\mathbb {E}\xi _{r}(\lambda _{r}^*) = n ( k_{{ j}}- k_{r})$ is consistent with $\lambda _{r}^*\in \text {int}(\varLambda )$. Assuming w.l.o.g. that $\lambda _{r}^*> 0$, (14) implies

$$\begin{aligned} \mathbb {E}\xi _{r}(\lambda _{r}^*)= & {} \int _{\bar{a}- \lambda _{r}^*}^{\overline{\theta }}\left( {\upvartheta }_{r}+ \lambda _{r}^*- \bar{a}\right) \hbox {d}h({\upvartheta }_{r}) = 0.5h\Big ( \lambda _{r}^*+ {\overline{\theta }}- \bar{a}\Big )^2 \nonumber \\ \text {so that} \quad \lambda _{r}^*+ {\overline{\theta }}- \bar{a}= & {} \sqrt{ \frac{2}{h} \mathbb {E}\xi _{r}} = \sqrt{ \frac{2}{h} n(k_{{ j}}-k_{r}) } \quad \text { if }\,\lambda _{r}^*\in \text {int}(\varLambda ) \end{aligned}$$

(35)

The last equality follows from Prop. 8. Also, from A3(iii),

$$\begin{aligned} 2n (k_{{ j}}-k_{r})/h{\le } \frac{4n}{h} || {\mathbf {k}}||_\infty < \frac{4n}{h} ({\overline{\theta }}-{\underline{\theta }})/4n = ({\overline{\theta }}-{\underline{\theta }})^2 \end{aligned}$$

so that $\lambda _{r}= \sqrt{ \frac{2}{h} n(k_{{ j}}-k_{r})} - ({\overline{\theta }}- \bar{a}) < \left( ({\overline{\theta }}- {\underline{\theta }}) - ({\overline{\theta }}- \bar{a})\right) = \bar{a}- {\underline{\theta }}$, verifying that $\lambda \in \text {int}(\varLambda )$. $\square$

Proof of Proposition 9

The existence of a unique MPE was established in Prop. 5. Non-degeneracy is implied by Assumption A3(iii) (see p. 22). Consider $\pmb {\lambda }^*$ such that $\mathbb {E}\xi _{r}(\lambda _{r}^*) = - n k_{r}$ for all ${r}$ with $\lambda _r \in \text {int}(\varLambda )$, and Parts ) and ) of the Proposition are satisfied. Our symmetry conditions ensure that such a vector exists, i.e., that if ${\bar{{r}}}$ and ${\underline{{ {r}}}}$ are matched players, if ${\lambda }^*_{\underline{{ {r}}}}= -{\lambda }^*_{\bar{{r}}}$, and $\mathbb {E}\xi _{\bar{{r}}}(\lambda _{\bar{{r}}}^*) = - n k_{\bar{{r}}}$, it follows from symmetry, (13), and (14) that $\mathbb {E}\xi _{\underline{{ {r}}}}(\lambda _{\underline{{ {r}}}}^*) = - n k_{\underline{{ {r}}}}$. With the restrictions in (8), we only need to verify that (7′) is satisfied by $\pmb {\lambda }^*$. Since $\sum _{i}k_{i}=0$ (Assumption A1), we have

$$\begin{aligned} - n k_{r}= \sum _{{i}\ne {r}} n k_{i}= - \sum _{{i}\ne {r}} \mathbb {E}\xi _{i}(\lambda ^*_{i}) \end{aligned}$$

(36)

Moreover, from Parts ) and ) of the proposition, $\sum _{i}\lambda _{i}^*= 0$. Substituting this property and (36) into the right-hand side of (7′), we obtain

$$\begin{aligned} \sum _{i}\lambda _{i}^*+ \sum _{{i}\ne {r}} \mathbb {E}\xi _{i}(\lambda _{i}^*) = n k_{r}, \end{aligned}$$

verifying that (7′) is indeed satisfied. $\square$

Proof of Proposition 10

From Prop. 9, we have

(37)

where, in the first integration, we substituted in $\underline{a}=-\bar{a}$. Totally differentiating both sides with respect to $\bar{a}$ and $\lambda _r$ and noting that and ${\tilde{\theta }_{{r}}}= \bar{a}- \lambda _r$, we obtain

Hence, . When ${r}$ is bi-constrained, and $H({\tilde{\theta }_{{r}}})$ are both nonzero, so that $\frac{\mathrm {d}\lambda _{r}}{\mathrm {d}\bar{a}} \in ( 0 , 1 )$. When ${r}$ is up-constrained (respectively, down-constrained), (respectively, $H({\tilde{\theta }_{{r}}}) = 1$), so that $\frac{\mathrm {d}\lambda _{r}}{\mathrm {d}\bar{a}}$ reduces to $1$ (respectively, $-$1). If ${r}$ is the middle player, $\lambda _{r}=0$, and, since everything is symmetric, so that $\frac{\mathrm {d}\lambda _{r}}{\mathrm {d}\bar{a}}=0$. $\square$

Proof of Proposition 11

Since Part i) of the Proposition follows immediately from the discussion below Prop. 10, we need only to prove in detail Part ii). Specifically, letting $I^*$ be the set of players who are bi-constrained in equilibrium, we will show that player ${r}$’s expected payoff increases by $- \frac{1}{n^2} \sum _{{i}\in I^*} \frac{\mathrm {d}V\xi _{i}}{\mathrm {d}\bar{a}}$, where

Suppose there is a player ${i}$ whose strategy is bi-constrained. (If ${i}$ is not the middle player, her matched player is also bi-constrained.) We will show that as $\bar{a}$ increases by $da$, the variance term $V\xi _{i}$ decreases, which, from (20′), induces the same increase in $- \frac{1}{n^2} \frac{\mathrm {d}V\xi _{i}}{\mathrm {d}\bar{a}} da$ in the expected payoff of each player. Let the distribution function of player ${i}$’s deviation from affine, $\xi _{i}$, be denoted as $F_{i}(\cdot )$. Obviously, $F_{i}(\cdot )$ is derived from the distribution function of $\theta$, $H(\cdot )$, as well as from ${i}$’s strategy and the announcement bounds. The random variable $\xi _{i}$ can be considered as a function of $\theta _{i}$:

(38)

Given that $\theta _{i}$ is distributed according to $H(\cdot )$, the distribution function $F_{i}(\cdot )$ of $\xi _{i}$ can be derived by combining $H(\cdot )$ and (38). Specifically, the support of $F_{i}$ is ; the fact that ${i}$ is bi-constrained implies that ${{\tilde{\theta }_{{i}}}- {\overline{\theta }}}< 0$ and . The values of $F_{i}$ are given by

(39)

Note, in particular, that $F_{i}(\cdot )$ jumps up at $x=0$ from $1-H({\tilde{\theta }_{{i}}})$ to . To derive the variance $V\xi _{i}$, note first that since ${i}$ is bi-constrained, $\lambda _{i}\in \text {int}(\varLambda )$. We can, therefore, invoke Prop. 9 to obtain

where the last equality is obtained after integrating by parts. Thus,

(40)

The variance of $\xi _{i}$ can now be written as

(41)

where the second equality follows from integration by parts, the third from $\mathbb {E}(\xi _{i}) = - n k_{i}$, the fourth from (40), and the fifth from (39). Now, differentiating (41) with respect to $\bar{a}$ and noting that and ${\tilde{\theta }_{{i}}}= \bar{a}- \lambda _i$, we obtain

(42)

The first inequality holds because $H({\underline{\theta }})=0$ and $H({\overline{\theta }})=1$, while and $\frac{\mathrm {d}{\tilde{\theta }_{{i}}}}{\mathrm {d}\bar{a}} \equiv \frac{\mathrm {d}(\bar{a}- \lambda _{i})}{\mathrm {d}\bar{a}} = ( 1 - \frac{\mathrm {d}\lambda }{\mathrm {d}\bar{a}})$. The second equality is obtained by substituting in the value of $d \lambda _{i}/ d \underline{a}$ using (25), changing the variables of integration from $\xi _{i}$ to $\theta _{i}= {\tilde{\theta }_{{i}}}- \xi _i$ and to in the two integrations, respectively. The term in curly brackets is negative because while ${\overline{\theta }}> {\tilde{\theta }}$. $\square$

Proof of Proposition 12

We already established that as $k_{r}$ increases, $\lambda _{r}$ also increases, raising $\mathbb {E}\xi _{r}$. We will show that $\hbox {d} V\xi _{r}/\hbox {d}|k_{r}|>0$. Then, the proposition follows from (20). By symmetry, we can, w.l.o.g., assume that $k_{r}> 0$. Using the same procedures as we used to derive (42), we differentiate the expression for $V\xi _{r}$ in (41) w.r.t. $\lambda _{r}$, to obtain

(43)

where the last equality follows from Prop. 9. Note that if ${r}$ is up-constrained, the first term in expression (43) is zero. Since $\frac{\mathrm {d}V\xi _{r}}{\mathrm {d}k_{r}} = \frac{\partial V\xi _{r}}{\partial k_{r}} + \frac{\partial V\xi _{r}}{\partial \lambda _{r}}\frac{\mathrm {d}\lambda _{r}}{\mathrm {d}k_{r}}$, we take the derivative of (41) with respect to $k_{r}$ and combine (26) with (43) to obtain

(44)

To see the effect of increasing $k_{\bar{{r}}}$ on players’ expected payoffs, we totally differentiate (20) w.r.t. $k_{\bar{{r}}}$, noting that to preserve symmetry, $\frac{\mathrm {d}k_{\underline{{ {r}}}}}{\mathrm {d}k_{\bar{{r}}}} = -1$, where ${\underline{{ {r}}}}$ is ${\bar{{r}}}$’s matched player. As $k_{\bar{{r}}}$ increases, ${\bar{{r}}}$’s and ${\underline{{ {r}}}}$’s welfare decline by $\left( \frac{2}{n^2} \frac{\mathrm {d}V\xi _{\bar{{r}}}}{\mathrm {d}k_{\bar{{r}}}} + 2k_{\bar{{r}}}\right)$; for other players, the decline is $\frac{2}{n^2} \frac{\mathrm {d}V\xi _{\bar{{r}}}}{\mathrm {d}k_{\bar{{r}}}}$. $\square$

Proof of Proposition 13

Let $I^+$ denote the members of the right-wing faction, and let $I_-^+$ denote the moderate members of this faction. Pick ${r}\in I^+$. Let $\xi _{r}(\gamma )$ denote ${r}$’s deviation from affine in the equilibrium associated with the parameter $\gamma$. Since ${r}$ is up-bounded, we have

$$\begin{aligned} n \bar{k}_{r}^+= - \mathbb {E}\xi _{r}(0) = \int _{\tilde{\theta }_{{r}}}^{\overline{\theta }}(\theta _{r}- {\tilde{\theta }_{{r}}}) \hbox {d} H(\theta _{r}) = 0.5 \int _{\tilde{\theta }_{{r}}}^1 (\theta _{r}- {\tilde{\theta }_{{r}}}) \hbox {d}\theta = ( 1 - {\tilde{\theta }_{{r}}})^2/4 \end{aligned}$$

The first equality follows from Prop. 9 and the third from Assumption A4(i). Hence, ${\tilde{\theta }_{{r}}}= 1 - 2\sqrt{n \bar{k}_{r}^+}$. Moreover, $H({\tilde{\theta }_{{r}}}) = 0.5 \int _{-1}^{\tilde{\theta }_{{r}}}d\theta = \frac{1 +{\tilde{\theta }_{{r}}}}{2}$. Now, from (44),

$$\begin{aligned} \left. \frac{\mathrm {d}V\xi _{r}}{\mathrm {d}k_{r}}\right| _{\gamma =0}= & {} 2 n^2 \bar{k}_{r}^+\left( \frac{H({\tilde{\theta }_{{r}}})}{1- H({\tilde{\theta }_{{r}}})} \right) = 2 n^2 \bar{k}_{r}^+\left( \frac{1 + {\tilde{\theta }_{{r}}}}{1- {\tilde{\theta }_{{r}}}} \right) = 2 n^2 \bar{k}_{r}^+\left( \frac{1 - \sqrt{n \bar{k}_{r}^+}}{\sqrt{n \bar{k}_{r}^+}} \right) \\= & {} 2 n \left( \sqrt{n \bar{k}_{r}^+} - n \bar{k}_{r}^+\right) . \end{aligned}$$

Hence, as , that is, for $k' > k$, $\frac{\mathrm {d}V\xi _{r}(k')}{\mathrm {d}k} > \frac{\mathrm {d}V\xi _{r}(k)}{\mathrm {d}k}$ if , and $\frac{\mathrm {d}V\xi _{r}(k')}{\mathrm {d}k} < \frac{\mathrm {d}V\xi _{r}(k)}{\mathrm {d}k}$ if . From Prop. 7, Prop. 9, and symmetry, $\text {USW}= - 2 \sum _{{i}\in I^+} V\xi _{i}(\gamma )$, so that

$$\begin{aligned} \left. \frac{\mathrm {d}\text {USW}}{\mathrm {d}\gamma }\right| _{\gamma =0}= & {} - 2 \sum _{{i}\in I^+} \left. \frac{\mathrm {d}V\xi _{i}(\gamma )}{\mathrm {d}\gamma }\right| _{\gamma =0} \\= & {} - 2 \sum _{{i}\in I_-^+} \upalpha _{i}\left( \left. \frac{\mathrm {d}V\xi _{i+(n{-}1)/4}(\gamma )}{\mathrm {d}k_{i+(n{-}1)/4}}\right| _{\gamma =0} - \left. \frac{\mathrm {d}V\xi _{i}(\gamma )}{\mathrm {d}k_{i}}\right| _{\gamma =0} \right) . \end{aligned}$$

Since $\bar{k}_{{i}+n/4}^+> \bar{k}_{i}^+$, $\left. \frac{\mathrm {d}\text {USW}}{\mathrm {d}\gamma }\right| _{\gamma =0} > 0$ if $\min (\bar{{\mathbf {k}}}^+) > \frac{1}{4(n{-}1)}$ and $\left. \frac{\mathrm {d}\text {USW}}{\mathrm {d}\gamma }\right| _{\gamma =0} < 0$ if $\max (\bar{{\mathbf {k}}}^+) < \frac{1}{4(n{-}1)}$. $\square$

Proof of Proposition 14

Under A3, the variance of ${r}$’s deviation from affine is

$$\begin{aligned} V\xi _{r}(\lambda _{r}) = (\mathbb {E}\xi _{r})^2 \left( \sqrt{ \frac{8}{9 h|\mathbb {E}\xi _{r}| } } - 1 \right) . \end{aligned}$$

(45)

To see this, assuming w.l.o.g. that $\lambda _{r}> 0$, and using the fact that $\lambda _{r}\in \text {int}(\varLambda )$, we have

$$\begin{aligned} V\xi _{r}( \lambda _{r}) = - (\mathbb {E}\xi _{r})^2 + \int _{\bar{a}- \lambda _{r}}^{\overline{\theta }}\left( {\upvartheta }_{r}\!+\! \lambda _{r}- \bar{a}\right) ^2 \hbox {d}h({\upvartheta }_{r}) = (h/3) \Big ( \lambda _{r}\!+\! {\overline{\theta }}- \bar{a}\Big )^3 - (\mathbb {E}\xi _{r})^2 \end{aligned}$$

which, from (35),

$$\begin{aligned} =&\frac{h}{3} \left( \frac{2}{h} \mathbb {E}\xi _{r}\right) ^{3/2} - (\mathbb {E}\xi _{r})^2 = \sqrt{ \frac{8|\mathbb {E}\xi _{r}| ^3 }{9 h}} - (\mathbb {E}\xi _{r})^2, \end{aligned}$$

establishing (45). Note from (24) that in SAGs, $\mathbb {E}\xi _{r}= -N k_r$. Substituting this into (45) and then into $(22^{\prime \prime })$, and using the fact that $\sum _{i \in {\mathcal {N}}_{\,\,1}} \mathbb {E}\xi _i = \sum _{i \in {\mathcal {N}}_{\,\,1}} \lambda _i = 0$ in SAGs, we obtain (27). To see that the inequality holds, recall that . $\square$

Proof of Proposition 15

Together with the assumption that $\theta$ is uniformly distributed, Assumption A3(iii) puts an upper bound on $k_i$: $k_i < \frac{1}{4Nh}$ for all $i \in {\mathcal {N}}_{\,\,1}$. We first show that $\Omega (k_i) :=k_{i}^2 \left( \sqrt{ \frac{8}{9Nh|k_{i}|} } - 1 \right)$ is increasing in $k_i$; we then show that $\text {Var}(\theta )/N > \Omega (1/4Nh)$. We know that

$$\begin{aligned} \frac{\hbox {d} \Omega (k_i)}{\hbox {d} k_i} \propto \sqrt{\frac{2}{Nh|k_i|}} - 2 >\sqrt{\frac{2}{Nh/4Nh}} - 2 =\sqrt{8}-2>0 \end{aligned}$$

(46)

where the inequality follows from $k_i < \frac{1}{4Nh}$. From (46) and noting from Assumption A3(ii) that $\text {Var}(\theta )=1/(12 h^2)$, we know that

$$\begin{aligned} \text {Var}(\theta )/N - \Omega (1/4Nh) = \frac{4(N-\sqrt{2})+3}{48N^2h^2}>0, \end{aligned}$$

since $N \ge 2 > \sqrt{2}$. $\square$

Proof of Proposition 16

We first establish $\lambda ^*_h>0$, so that, from (28), $\mathbb {E}\xi _{h}(\lambda _{h}^*) = 0$, and thus, $h$ is the anchor of the game. Suppose, instead, that $\lambda ^*_h \le 0$ and $\lambda ^*_h \in \text {int}(\varLambda )$. (We can easily rule out the situation when $\lambda ^*_h=\min (\varLambda ) = \underline{a}-{\overline{\theta }}$; we omit the details.) Since $k_h > k_r$ $\forall r \ne h$, (16) implies that $\mathbb {E}\xi _h(\lambda ^*_h) < \mathbb {E}\xi _r(\lambda ^*_r)$ and thus $\lambda ^*_r < \lambda ^*_h \le 0$. Since $\mathbb {E}\xi _r(\lambda _r)=0$ when $\lambda _r=0$, (29) and $\lambda ^*_r < 0$ imply

$$\begin{aligned} \lambda ^*_{r}+ \mathbb {E}\xi _{r}(\lambda ^*_{r}) < 0 \end{aligned}$$

(47)

From (7′) and $\lambda ^*_h \in \text {int}(\varLambda )$, $\lambda ^*_h = nk_h - \sum _{r \ne h} (\lambda ^*_r + \mathbb {E}\xi _r(\lambda ^*_r)) > 0$, where the inequality is due to $k_h>0$ and (47). This contradicts our supposition that $\lambda ^*_h \le 0$. Property (28) now ensures that $\mathbb {E}\xi _{r}(\lambda _{r}) = 0$, so that single-bounded aggregation games are anchored with anchor $h$. The second part of the proposition now follows from Prop. 8. $\square$