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Identity in public goods contribution

Abstract

Agents’ decision whether to join a group, and their subsequent contribution to a public good, depend on the group’s ideals. Agents have different preference for this public good, e.g. reductions in greenhouse gas emissions. People who become “climate insiders” obtain identity utility, but suffer disutility if they deviate from the group ideal. That ideal might create a wide but shallow group, having many members but little effect on behavior, or a narrow but deep group. Greater heterogeneity of preferences causes the contribution-maximizing ideal to create narrow but deep groups. The contribution-maximizing ideal maximizes welfare if the population is large.

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Notes

  1. 1.

    Kotchen and Moore (2008) find that environmentalists are more likely to voluntarily restrain their consumption of goods and services that generate negative externalities. Kahn (2007) finds that those who vote for green policies and register for liberal/environmental political parties live a greener lifestyle, commuting by public transit more often, favoring hybrid vehicles, and consuming less gasoline than non-environmentalists.

  2. 2.

    For example, authors Peter Nearing and Janet Luhrs advocate sustainable living through vivid illustrations of simple lifestyles; in 2007, the United Nations published “Sustainable Consumption and Production, Promoting Climate-Friendly Household Consumption Patterns,” to promote sustainable lifestyles in communities and homes; Nobel Laureate Al Gore further propelled the green lifestyle movement through his movie “An Inconvenient Truth”.

  3. 3.

    Examples include Akerlof and Kranton (2005, 2008, 2010)), Benjamin et al. (2010), Benjamin et al. (2016) and Hiller and Verdier (2014). These studies extend Akerlof and Kranton’s (2000) framework to analyze behavior in workplaces, schools, churches, and families.

  4. 4.

    Similarly, Iannaccone (1992) study sacrifice and seemingly inefficient prohibitions as a screening device where individuals sort themselves into different religions. Carvalho (2016) models an identity-based religious organization that sets religious strictness to maximize participation in its activities. In an industrial organization context, Kosfeld and von Siemens (2011) consider how workers with different willingness to cooperate self select into firms having different monetary incentives and level of worker cooperation, leading to heterogeneous corporate cultures.

  5. 5.

    In a related paper by Shayo (2009), identity utility is also derived from social status while identity cost comes from social distance from other group members rather than an ideal. Shayo (2009) studies the formation of national identities and preferences for redistribution. Costa-i-Font and Cowell (2014) review the related literature on social identity and redistribution.

  6. 6.

    In our model, V can also be viewed as the difference in identity utility between insiders and outsiders, if the outsiders also enjoy certain identity utility.

  7. 7.

    The literature on identity and economics addresses the determination of group stereotypes or prescriptions (“ideals” in our model) in varied ways. Benjamin et al. (2010; 2016) treat group prescriptions as exogenous and given. In Shayo (2009) and Bernard et al. (2016), the group stereotype is determined by the average type or behavior of the group members. In Akerlof (2017), individuals choose their own values. Akerlof and Kranton (2002) discusses the choice of a group ideal in a school setting. In this approach, an ideal setter (the school) interacts with a continuum of agents (students) who decide whether to identify with the ideal. We follow this approach in the context of public goods contribution.

  8. 8.

    For example, Aidt (1998) and Conconi (2003) assume that environmentalists organize a green lobby group that only cares about the environment. List and Sturm’s (2006) model also assumes that environmentalists’ payoff depends only on which environmental policy is undertaken.

  9. 9.

    This is an agent behind the veil of ignorance under stage 0 and the social leader has not learnt the agent’s type yet. With a fixed group size N, maximizing a representative agent’s welfare is equivalent to maximizing the total welfare.

  10. 10.

    By Eq. (4) and Lemma 2, the largest increase in contribution implemented by an ideal \(a^{*}\) is \(\frac{\theta }{1+\theta }\left[ a^{*}-\left( a^{*}-B\right) \right] =\) \(\frac{\theta B}{1+\theta }\equiv \left( \frac{2V\theta }{1+\theta }\right) ^{1/2}\).

  11. 11.

    See Bagnoli and Bergstrom (2005) for more examples and applications.

  12. 12.

    By Lemma 2, any ideal \(a^{*}>\overline{\beta }+B\) does not attract any insider.

  13. 13.

    In Appendix A.3.2, we consider non-monotonic f, and characterize the contribution-maximizing ideal when \(\gamma \) converges to 0.

  14. 14.

    Suppose \(f^{\prime }\le 0\) for all \(\beta _{i}\), then \(f\left( \underline{\beta }\right) \ge \frac{1}{\bar{\beta }-\underline{\beta }}\), implying \(\frac{1}{f\left( \underline{\beta }\right) }\le \bar{\beta }-\underline{\beta }\le B\) under small preference heterogeneity. By Proposition 3, \(\bar{a}=\underline{\beta }+B\).

  15. 15.

    In this subsection, we focus on the standard welfare measure \(M\left( a^{*}\right) \).

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Correspondence to Tat-How Teh.

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We thank the editor, an associate editor, two anonymous reviewers, Roland Bénabou, Parkash Chander, In-Koo Cho and conference and seminar participants in the 2016 EAERE Conference (Zurich), the 2017 AERE Conference (Pittsburgh), the 2018 PET Conference (Hue), Arne Ryde Workshop on Identity, Image and Economic Behavior (Lund), NTU Workshop on Environmental Economics, 4th Hanyang-Kobe-NTU Symposium in Economics, Hong Kong Baptist University, Hong Kong University of Science and Technology, Kobe University, National University of Singapore, Peking University HSBC Business School, Renmin University of China, SUFE International Workshop on “Frontiers of Environmental and Resource Economics” (Shanghai), Tsinghua University, University of Groningen, Xi’an Jiao Tong University, and Xiamen University for helpful comments.

Appendices

Appendix A: Proofs

A.1: Lemmas

Proof

(Lemma 1). The insider’s optimization problem is equivalent to

$$\begin{aligned} \max _{a_{i}}U_{i}\left( a_{i}|a^{*}\right) =\left\{ \begin{array} [c]{c} \beta _{i}a_{i}-\frac{1}{2}a_{i}^{2}-\frac{\theta }{2}\left( a_{i}-a^{*}\right) ^{2}+V\text { if }a_{i}\le a^{*}\\ \beta _{i}a_{i}-\frac{1}{2}a_{i}^{2}-\frac{\gamma }{2}\left( a_{i}-a^{*}\right) ^{2}+V\text { if }a_{i}>a^{*} \end{array} \right. . \end{aligned}$$
(14)

Because both alternative forms of \(U_{i}\left( a_{i}|a^{*}\right) \) in the right side of (14) are concave, by first order conditions, we have

$$\begin{aligned} \underset{a_{i}}{\arg \max }\beta _{i}a_{i}-\frac{1}{2}a_{i}^{2}-\frac{\theta }{2}\left( a_{i}-a^{*}\right) ^{2}+V=\beta _{i}+\frac{\theta \left( a^{*}-\beta _{i}\right) }{1+\theta }, \end{aligned}$$
(15)

and

$$\begin{aligned} \underset{a_{i}}{\arg \max }\beta _{i}a_{i}-\frac{1}{2}a_{i}^{2}-\frac{\gamma }{2}\left( a_{i}-a^{*}\right) ^{2}+V=\beta _{i}+\frac{\gamma \left( a^{*}-\beta _{i}\right) }{1+\gamma }. \end{aligned}$$
(16)

If \(\theta =\gamma \), then trivially the solution is given by either of these two expressions, which are the same. Now suppose \(\theta \ne \gamma \). Observe that

$$\begin{aligned} \beta _{i}+\frac{\eta \left( a^{*}-\beta _{i}\right) }{1+\eta }\le a^{*}\iff \beta _{i}\le a^{*} \end{aligned}$$
(17)

for both \(\eta \in \left\{ \theta ,\gamma \right\} \). Suppose \(\beta _{i}\le a^{*}\). Then the right side of (15) is no higher than \(a^{*}\) and thus optimal given \(a_{i}\le a^{*}\). Note that when \(a_{i}=a^{*}\), the two functional forms in the right side of (14) are the same, so the utility function is continuous at \(a_{i}=a^{*}\). We then claim that \(\frac{dU_{i}\left( a_{i}|a^{*}\right) }{da_{i}}\le 0\) for all \(a_{i}\ge a^{*}\): Concavity of \(U_{i}\left( a_{i}|a_{i}\ge a^{*}\right) \) implies that \(\frac{dU_{i}\left( a_{i}|a^{*}\right) }{da_{i} }\le 0\) for all \(a_{i}\ge \beta _{i}+\frac{\gamma \left( \beta _{i}-a^{*}\right) }{1+\gamma }\) by (16), and meanwhile \(a_{i}\ge a^{*}\) and \(\beta _{i}\le a^{*}\) imply \(a_{i}\ge a^{*}\ge \beta _{i} +\frac{\gamma \left( \beta _{i}-a^{*}\right) }{1+\gamma }\) by ( 17) . Therefore, \(\frac{dU_{i}\left( a_{i}|a^{*}\right) }{da_{i}}\le 0\) for all \(a_{i}\ge a^{*}\). Consequently the right side of (15) is optimal if \(\beta _{i}\le a^{*}\).

Suppose \(\beta _{i}>a^{*}\), then the right side of (16) is no lower than \(a^{*}\) and thus optimal for \(a_{i}\in [a^{*} ,\infty )\,\). A logic similar to the above shows that \(\frac{dU_{i}\left( a_{i}|a^{*}\right) }{da_{i}}\ge 0\) for all \(a_{i}\le a^{*}\) and consequently the right side of (16) is optimal if \(\beta _{i} >a^{*}\). \(\square \)

Proof

(Lemma 2). Comparing (5) and (6) while using the definitions of B and D, we have the following.

$$\begin{aligned}&\frac{1}{2}\frac{\beta _{i}\left( \beta _{i}+2\theta a^{*}\right) -\theta a^{*2}}{1+\theta }+V\ge \frac{\beta _{i}^{2}}{2}\nonumber \\&\quad \Leftrightarrow 2V\left( 1+\theta \right) \ge \theta \beta _{i}^{2}-2\theta a^{*}\beta _{i}+\theta a^{*2}\nonumber \\&\quad \Leftrightarrow \frac{2V\left( 1+\theta \right) }{\theta }\ge \left( \beta _{i}-a^{*}\right) ^{2}\nonumber \\&\quad \Leftrightarrow a^{*}-B\le \beta _{i}\le a^{*}+B. \end{aligned}$$
(18)

Similarly,

$$\begin{aligned}&\frac{1}{2}\frac{\beta _{i}\left( \beta _{i}+2\gamma a^{*}\right) -\gamma a^{*2}}{1+\gamma }+V\ge \frac{1}{2}\beta _{i}^{2}\\&\quad \Leftrightarrow a^{*}-D\le \beta _{i}\le a^{*}+D. \end{aligned}$$

If \(\beta _{i}\le a^{*}\), the agent will be an under-contributor as an insider, and if Inequality (18) is satisfied, the utility of being an (under-contributing) insider will be not lower than that of being an outsider. Combining \(\beta _{i}\le a^{*}\) and Inequality (18), an agent will identify with the ideal and become an under-contributing insider if and only if \(a^{*}-B\le \beta _{i}\le a^{*}\). Similar logic with \(\beta _{i}\ge a^{*}\) shows that an agent will identify with the ideal and become an over-contributing insider if and only if \(a^{*}<\beta _{i}\le a^{*}+D\). \(\square \)

A.2: Remarks

Proof

(Remark 1). This follows directly from Lemma 1, Eqs. (1) and (4): these imply that when \(\beta _{i}\le a^{*}\), \(\beta _{i}\le a_{i}^{I}\le a^{*}\) and when \(\beta _{i}>a^{*}\), \(\beta _{i}>a_{i}^{I}>a^{*}\). \(\square \)

Proof

(Remark 2). Part (i): Using Equations (1 ) and (3), we have the following. If \(a^{*}\ge \beta _{j}>\beta _{i}\), meaning that both agents with \(\beta _{i}\) and \(\beta _{j}\) will be under-contributing insiders by the proof of Remark 1, then

$$\begin{aligned} a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) =\beta _{j}-\beta _{i}>\frac{\beta _{j}-\beta _{i}}{1+\theta }=a^{I}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) . \end{aligned}$$

If \(a^{*}<\beta _{i}<\beta _{j}\), meaning that both agents with \(\beta _{i}\) and \(\beta _{j}\) will be over-contributing insiders by the proof of Remark 1, then

$$\begin{aligned} a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) =\beta _{j}-\beta _{i}>\frac{\beta _{j}-\beta _{i}}{1+\gamma }=a^{I}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) . \end{aligned}$$

If \(\beta _{j}>a^{*}\ge \beta _{i}\), meaning that the agent with \(\beta _{i}\) will be an under-contributor while the other an over-contributor as insiders, then

$$\begin{aligned}&a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) \\&\quad =\beta _{j}-\beta _{i}\\&\quad >\left[ \beta _{j}+\frac{\gamma \left( a^{*}-\beta _{j}\right) }{1+\gamma }\right] -\left[ \beta _{i}+\frac{\theta \left( a^{*}-\beta _{i}\right) }{1+\theta }\right] \\&\quad =a^{I}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) , \end{aligned}$$

because \(\frac{\gamma \left( a^{*}-\beta _{j}\right) }{1+\gamma } -\frac{\theta \left( a^{*}-\beta _{i}\right) }{1+\theta }<0.\)

Part (ii): If \(\beta _{i}<a^{*}<\beta _{j}\), then \(\left| a^{B}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) \right| =a^{B}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) <a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) =\) \(\left| a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) \right| \) because \(a^{I}\left( \beta _{i}\right) >a^{B}\left( \beta _{i}\right) \). If \(\beta _{j}<a^{*}<\beta _{i}\), then \(\left| a^{B}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) \right| =-a^{B}\left( \beta _{j}\right) +a^{I}\left( \beta _{i}\right) <-a^{B}\left( \beta _{j}\right) +a^{B}\left( \beta _{i}\right) =\) \(\left| a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) \right| \) because \(a^{I}\left( \beta _{i}\right) <a^{B}\left( \beta _{i}\right) \). If \(\beta _{j}<\beta _{i}<a^{*}\), \(\left| a^{B}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) \right| =-a^{B}\left( \beta _{j}\right) +a^{I}\left( \beta _{i}\right) >-a^{B}\left( \beta _{j}\right) +a^{B}\left( \beta _{i}\right) =\left| a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) \right| \) because \(a^{I}\left( \beta _{i}\right) >a^{B}\left( \beta _{i}\right) \). If \(\beta _{j}>\beta _{i}>a^{*}\), \(\left| a^{B}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) \right| =a^{B}\left( \beta _{j}\right) -a^{I}\left( \beta _{i}\right) >a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) =\left| a^{B}\left( \beta _{j}\right) -a^{B}\left( \beta _{i}\right) \right| \) because \(a^{I}\left( \beta _{i}\right) <a^{B}\left( \beta _{i}\right) \). \(\square \)

Proof

(Remark 3). All results follow from direct inspections. The only exception is the result on \(g\left( \bar{a}\right) \) being (weakly) increasing in \(\theta \). Clearly, if \(\overline{\beta }-\underline{\beta }>B\) then \(g(\bar{a})=\frac{V}{\overline{\beta }-\underline{\beta }}\) is independent of \(\theta \). If \(\overline{\beta }-\underline{\beta }\le B\), then we can use \(B=\left( \frac{2V\left( 1+\theta \right) }{\theta }\right) ^{1/2}\) to calculate the derivative of \(g(\bar{a})=\frac{\theta }{1+\theta }\left( B-\frac{\overline{\beta }-\underline{\beta }}{2}\right) \) with respect to \(\theta \) as

$$\begin{aligned} \frac{dg(\bar{a})}{d\theta }=\left( \frac{1}{2}\left( 2V\right) ^{1/2}\left( \frac{\theta }{1+\theta }\right) ^{-1/2}-\frac{\overline{\beta }-\underline{\beta }}{2}\right) \frac{1}{(1+\theta )^{2}}=\frac{B-(\overline{\beta }-\underline{\beta })}{2(1+\theta )^{2}}\ge 0. \end{aligned}$$

\(\square \)

A.3: Contribution maximization (Propositions 1, 3, 4)

This subsection proves Propositions 1, 3, and 4.

Before proving the propositions, we first establish some claims. We rewrite \(g\left( a^{*}\right) \) in (7) as

$$\begin{aligned}g\left( a^{*}\right) &=\frac{\theta }{1+\theta }\int _{\max \{a^{*}-B,\underline{\beta }\}}^{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}\\&\quad +\frac{\gamma }{1+\gamma }\int _{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }^{\min \{a^{*}+D,\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}. \end{aligned}$$

Note that \(g\left( a^{*}\right) \), being composed of continuous functions (integrals and min/max functions), is continuous. We first narrow down the possible range for \(\bar{a}\) in the following series of claims.

Claim 1

\(\bar{a}\in \left( \underline{\beta },\overline{\beta }+B\right) \).

Proof

For all \(a^{*}\le \underline{\beta }\), \(g\left( a^{*}\right) \le 0\) because all insiders (if any) are over-contributors. For all \(a^{*} \ge \overline{\beta }+B\), \(g\left( a^{*}\right) =0\) because no agents join as insiders by Lemma 2. All these ideals are dominated by \(a^{*}=\overline{\beta }\), where

$$\begin{aligned} g\left( \overline{\beta }\right) =\frac{\theta }{1+\theta }\int _{\max \{\overline{\beta }-B,\underline{\beta }\}}^{\overline{\beta }}\left( \overline{\beta }-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}>0\text {.} \end{aligned}$$

Hence, any \(a^{*}\notin \left( \underline{\beta },\overline{\beta }+B\right) \) cannot be contribution-maximizing. \(\square \)

Claim 2

Suppose \(\overline{\beta }-\underline{\beta }\le B\), then \(\bar{a}\ge \underline{\beta }+B\).

Proof

By Claim 1, any \(a^{*}<\underline{\beta }\) can never be optimal. For all \(\underline{\beta }\le a^{*}<\underline{\beta } +B\),

$$\begin{aligned}&g\left( a^{*}\right) |_{\underline{\beta }\le a^{*} <\underline{\beta }+B}\nonumber \\&\quad =\frac{\theta }{1+\theta }\int _{\underline{\beta }}^{\min \{a^{*} ,\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i} \nonumber \\&\quad \quad +\frac{\gamma }{1+\gamma }\int _{\min \{a^{*} ,\overline{\beta }\}}^{\min \{a^{*}+D,\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i} \end{aligned}$$
(19)
$$\begin{aligned}&\quad \le \frac{\theta }{1+\theta }\int _{\underline{\beta }}^{\min \{a^{*},\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}. \nonumber \\&\quad <\frac{\theta }{1+\theta }\int _{\underline{\beta }}^{\overline{\beta }}\left( \underline{\beta }+B-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}\nonumber \\&\quad =g\left( \underline{\beta }+B\right) . \end{aligned}$$
(20)

The first inequality follows from the fact that the second integral in (19) is nonpositive. The second inequality uses the fact that (20) is increasing in \(a^{*}\). The last equality follows from the definition of g, using \(\underline{\beta }+B\ge \overline{\beta }\). Since \(g\left( a^{*}\right) |_{\underline{\beta }\le a^{*}<\underline{\beta }+B}<g\left( \underline{\beta }+B\right) \) and by Claim 1, any \(a^{*}<\underline{\beta }+B\) cannot be optimal, and so \(\bar{a}\ge \underline{\beta }+B\). \(\square \)

Claim 3

Suppose \(\overline{\beta }-\underline{\beta }>B\). If \(f^{\prime }\ge 0\), then \(\bar{a}\ge \overline{\beta }\). If \(f^{\prime }\le 0\) or \(\gamma \) converges to 0 then \(\bar{a}\ge \underline{\beta }+B\).

Proof

By Claim 1, any \(a^{*}<\underline{\beta }\) can never be optimal. Suppose \(f^{\prime }\ge 0\) and consider \(a^{*}\in \left[ \underline{\beta },\overline{\beta }\right) \),

$$\begin{aligned} g\left( a^{*}\right) |_{a^{*}\in \left[ \underline{\beta } ,\overline{\beta }\right) }&=\frac{\theta }{1+\theta }\int _{\max \{a^{*}-B,\underline{\beta }\}}^{a^{*}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}\nonumber \\&\quad +\frac{\gamma }{1+\gamma }\int _{a^{*}} ^{\min \{a^{*}+D,\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}. \end{aligned}$$
(21)
$$\begin{aligned}&<\frac{\theta }{1+\theta }\int _{\max \{a^{*}-B,\underline{\beta } \}}^{a^{*}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}\nonumber \\&\le \frac{\theta }{1+\theta }\int _{\overline{\beta }-B}^{\overline{\beta } }\left( \overline{\beta }-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}\nonumber \\&=g(\overline{\beta }). \end{aligned}$$
(22)

The first inequality follows from the fact that the second integral in (21) is negative (since \(\beta _{i}\ge a^{*}\) in the term and \(\min \{a^{*}+D,\overline{\beta }\}>a^{*}\)). For the second inequality, we note that if \(a^{*}\le \underline{\beta }+B\), then (22) has derivative \(\frac{\theta }{1+\theta }F(a^{*})\ge 0\); while if \(a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }\right) \) then (22) has derivative

$$\begin{aligned} \frac{\theta }{1+\theta }\left( F\left( a^{*}\right) -F(a^{*}-B)-Bf\left( a^{*}-B\right) \right) \ge 0 \end{aligned}$$

given that \(f^{\prime }\ge 0\) (hence F is convex). The final equality follows from the definition of g, using \(\overline{\beta }>\underline{\beta }+B\). Since \(g\left( a^{*}\right) |_{a^{*}\in \left[ \underline{\beta },\overline{\beta }\right) }<g(\overline{\beta })\) and by Claim 1, any \(a^{*}<\overline{\beta }\) cannot be optimal, and so \(\bar{a}\ge \overline{\beta }\).

Now consider \(a^{*}\in \left[ \underline{\beta },\underline{\beta }+B\right) \),

$$\begin{aligned}g\left( a^{*}\right) |_{a^{*}<\underline{\beta }+B}&=\frac{\theta }{1+\theta }\int _{\underline{\beta }}^{a^{*}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i} \nonumber \\&\quad +\frac{\gamma }{1+\gamma } \int _{a^{*}}^{\min \{a^{*}+D,\overline{\beta }\}}\left( a^{*} -\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}. \end{aligned}$$
(23)

The first term in the right side of (23) is increasing in \(a^{*}\). As for the second term, if \(a^{*}+D>\overline{\beta }\) the derivative is \(\frac{\gamma }{1+\gamma }\left( 1-F(a^{*})\right) \ge 0\); if \(a^{*}+D\le \beta \), the derivative is

$$\begin{aligned} \frac{\gamma }{1+\gamma }\left( F(a^{*}+D)-F(a^{*})-Df\left( a^{*}+D\right) \right) \ge 0, \end{aligned}$$

provided that \(f^{\prime }\le 0\) (i.e. F is concave). Therefore, (23) is increasing in \(a^{*}\in \left[ \underline{\beta },\underline{\beta }+B\right) \) as long as either the second term in the right side of (23) is increasing (a sufficient condition of which is \(f^{\prime }\le 0\)) or converges to 0 (a sufficient condition of which is \(\gamma \) converging 0). Thus, if \(f^{\prime }\le 0\) or \(\gamma \) converges 0, then \(g\left( a^{*}\right) |_{a^{*}\in \left[ \underline{\beta },\underline{\beta }+B\right) }<g\left( \underline{\beta }+B\right) \), so any \(a^{*}\in \left[ \underline{\beta },\underline{\beta }+B\right) \) cannot be optimal, which, together with Claim 1, implies \(\bar{a} \ge \underline{\beta }+B\). \(\square \)

A.3.1: Completing the proofs

Proposition 1 is a special case of Propositions 3 and 4. Therefore, we first prove Propositions 3 and 4 and then get back to Proposition 1.

Proof

(Proposition 3) By Claims 1 and 2, when \(\overline{\beta }-\underline{\beta }\le B\), we can focus on \(a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] \), in which

$$\begin{aligned} g\left( a^{*}\right) =\frac{\theta }{1+\theta }\int _{a^{*}-B} ^{\overline{\beta }}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}, \end{aligned}$$
(24)

the derivative of which is

$$\begin{aligned} \frac{dg\left( a^{*}\right) }{da^{*}}=\frac{\theta }{1+\theta }\left( \frac{1-F(a^{*}-B)}{f\left( a^{*}-B\right) }-B\right) f\left( a^{*}-B\right) . \end{aligned}$$

The increasing hazard rate assumption (Assumption 1) implies \(\frac{dg\left( a^{*}\right) }{da^{*}}\) is single-crossing from above, i.e. if \(\frac{dg\left( a^{*}\right) }{da^{*}}|_{a^{*}=a_{1}}\le 0\) then \(\frac{dg\left( a^{*}\right) }{da^{*}}|_{a^{*}=a_{2}}<0\) for all \(a_{2}>a_{1}\). Therefore, \(g\left( a^{*}\right) \) is quasiconcave, meaning that the solution to the first-order condition indeed maximizes g whenever it exists. If \(\frac{1-F(\underline{\beta })}{f(\underline{\beta } )}=\frac{1}{f(\underline{\beta })}\le B\), then the single-crossing property implies \(\frac{dg\left( a^{*}\right) }{da^{*}}<0\) for all \(a^{*}>\underline{\beta }+B\), so \(\bar{a}=\underline{\beta }+B\). If \(\frac{1}{f(\underline{\beta })}>B\), by the intermediate value theorem there exists a solution \(\bar{a}\in (\underline{\beta }+B,\overline{\beta }+B]\) that solves the first-order condition \(\frac{1-F(\bar{a}-B)}{f\left( \bar{a}-B\right) } -B=0\), or equivalently, condition (12). The uniqueness of the solution to the first-order condition follows from the single-crossing property. \(\square \)

Proof

(Proposition 4) (i) Suppose \(f^{\prime }\ge 0\). By Claims 1 and 3, when \(\overline{\beta }-\underline{\beta }>B\) and \(f^{\prime }\ge 0\), we can focus on \(a^{*}\in \left[ \overline{\beta },\overline{\beta }+B\right] \), in which the expression of \(g\left( a^{*}\right) \ \)takes the same form as (24). Using the same argument in the proof of Proposition 3, we have the following: If \(\frac{1-F(\overline{\beta }-B)}{f(\overline{\beta }-B)}\le B\), then \(\frac{dg\left( a^{*}\right) }{da^{*}}<0\) for all \(a^{*} >\overline{\beta }\), so \(\bar{a}=\overline{\beta }\). If \(\frac{1-F(\overline{\beta }-B)}{f(\overline{\beta }-B)}>B\), we have \(\bar{a}\in (\overline{\beta },\overline{\beta }+B]\), which is the solution of the first-order condition \(\frac{1-F(\bar{a}-B)}{f\left( \bar{a}-B\right) }-B=0\).

(ii) Suppose \(f^{\prime }<0\). By Claims 1 and 3, when \(\overline{\beta }-\underline{\beta }>B\) and \(f^{\prime }<0\), we can focus on \(a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] \), in which

$$\begin{aligned}g\left( a^{*}\right) &=\frac{\theta }{1+\theta }\int _{a^{*}-B} ^{\min \{a^{*},\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i} \nonumber \\&\quad +\frac{\gamma }{1+\gamma }\int _{\min \{a^{*},\overline{\beta }\}}^{\min \{a^{*}+D,\overline{\beta } \}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}. \end{aligned}$$
(25)

Let \(\chi _{1}(a^{*})\) denote the derivative of the first term in the right side of (25). We have

$$\begin{aligned} \chi _{1}(a^{*})&= \left\{ \begin{array}{l} \frac{\theta }{1+\theta }\left( F(a^{*})-F(a^{*}-B)-Bf\left( a^{*}-B\right) \right) \text { if }\underline{\beta }+B\le a^{*}\le \overline{\beta }\\ \frac{\theta }{1+\theta }\left( 1-F(a^{*}-B)-Bf\left( a^{*}-B\right) \right) \text { if }\overline{\beta }<a^{*}\le \overline{\beta }+B \end{array}\right. , \nonumber \\&<0, \end{aligned}$$
(26)

where the inequality is due to \(f^{\prime }<0\) (i.e. F is strictly concave). Let \(\frac{1+\gamma }{\gamma }\chi _{2}(a^{*})\) denote the derivative of the second term in the right side of (25). We have

$$\begin{aligned} \chi _{2}(a^{*})&=\left\{ \begin{array} [c]{l} 0\text { if }a^{*}+D>\overline{\beta }\text { and }a^{*}>\bar{\beta }\\ 1-F(a^{*})\text { if }a^{*}+D>\overline{\beta }\text { and }a^{*} \le \bar{\beta }\\ F(a^{*}+D)-F(a^{*})-Df\left( a^{*}+D\right) \text { if }a^{*}+D\le \overline{\beta } \end{array} \right. \\&\in \left[ 0,1\right] , \end{aligned}$$

where \(F(a^{*}+D)-F(a^{*})-Df\left( a^{*}+D\right) \ge 0\) due to \(f^{\prime }<0\) (i.e. F is strictly concave). Further define threshold \(\bar{\gamma }\) such that

$$\begin{aligned} \frac{\bar{\gamma }}{1+\bar{\gamma }}\equiv -\max _{a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] }\{\chi _{1}(a^{*})\}, \end{aligned}$$
(27)

in which the maximizer exists because \(\chi _{1}\) is continuous in the compact interval. Note that \(\bar{\gamma }>0\) since \(\chi _{1}(a^{*})<0\) for all \(a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] \).

To prove \(\bar{a}=\underline{\beta }+B\), by Claim 1, it is sufficient to show that \(\frac{dg\left( a^{*}\right) }{da^{*}}<0\) for all \(\bar{a}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] \) whenever \(\gamma <\bar{\gamma }\):

$$\begin{aligned} \frac{dg\left( a^{*}\right) }{da^{*}}&=\chi _{1}(a^{*} )+\frac{1+\gamma }{\gamma }\chi _{2}(a^{*})\\&\le \max _{a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] }\left\{ \chi _{1}(a^{*})\right\} +\frac{\gamma }{1+\gamma }\\&<\max _{a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] }\left\{ \chi _{1}(a^{*})\right\} +\frac{\bar{\gamma }}{1+\bar{\gamma }}\\&=0. \end{aligned}$$

The first inequality uses \(\chi _{2}\left( a^{*}\right) \le 1\), while the second inequality uses \(0\le \gamma <\bar{\gamma }\) and the fact that \(\frac{\gamma }{1+\gamma }\) is increasing in \(\gamma \) for any nonnegative \(\gamma \); the final equality uses the definition of \(\bar{\gamma }\). \(\square \)

Proof

(Proposition 1) When F is uniform, for any \(\beta \in \left[ \underline{\beta },\overline{\beta }\right] \), \(\frac{1-F(\beta )}{f(\beta )}=\overline{\beta }-\beta \). Therefore, if \(\overline{\beta }-\underline{\beta }\le B\), then \(\frac{1}{f(\underline{\beta })}=\overline{\beta }-\underline{\beta }\le B\); hence Proposition 3 implies \(\bar{a}=\underline{\beta }+B\). If \(\overline{\beta }-\underline{\beta }>B\), then \(\frac{1-F(\overline{\beta }-B)}{f(\overline{\beta }-B)}=B\); hence Proposition 4(i) implies \(\bar{a}=\overline{\beta }\). \(\square \)

In what follows, we explain the intuition of the proof of Proposition 1. Claim 1 implies that we should focus on \(a^{*}\in \left( \underline{\beta },\overline{\beta }+B\right) \), where there is a positive mass of under-contributing insiders.

Suppose \(\overline{\beta }-\underline{\beta }\le B\). Claim 2 further restricts the possible range of the contribution-maximizing ideal to \(a^{*}\in [\underline{\beta }+B,\overline{\beta }+B)\). This is because there are no overcontributing insiders for all \(a^{*}\le \underline{\beta }+B\) (given \(\overline{\beta }-\underline{\beta }\le B\)), and \(a^{*}=\underline{\beta }+B\) leads to higher contribution of undercontributing insiders than any \(a^{*}<\underline{\beta }+B\). Next, for all \(a^{*}\in [\underline{\beta }+B,\overline{\beta } +B)\), we have

$$\begin{aligned} g\left( a^{*}\right) =\frac{\theta }{1+\theta }\int _{a^{*}-B} ^{\overline{\beta }}\left( \frac{a^{*}-\beta _{i}}{\overline{\beta }-\underline{\beta }}\right) d\beta _{i}. \end{aligned}$$

The derivative is \(\frac{dg\left( a^{*}\right) }{da^{*} }=\frac{\theta }{1+\theta }\left( \frac{\overline{\beta }-a^{*}}{\overline{\beta }-\underline{\beta }}\right) \le 0\), where the inequality follows from \(a^{*}\ge \underline{\beta }+B\ge \overline{\beta } \) (given \(\overline{\beta }-\underline{\beta }\le B\)). The strict inequality holds as long as \(a^{*}>\underline{\beta }+B\). Therefore, starting from \(a^{*}=\underline{\beta }+B\), a marginal increase in the ideal decreases contribution.

Suppose \(\overline{\beta }-\underline{\beta }>B\). Claim 3 restricts the possible range of the contribution-maximizing ideal to \(a^{*}\in [\overline{\beta },\overline{\beta }+B)\). This is because there are no overcontributing insiders for all \(a^{*}<\overline{\beta }\), and \(a^{*}=\overline{\beta }\) leads to higher contribution of undercontributing insiders than any \(a^{*}<\overline{\beta }\). Next, for all \(a^{*}\in [\overline{\beta },\overline{\beta }+B)\), the expressions of \(g\left( a^{*}\right) \) and \(\frac{dg\left( a^{*}\right) }{da^{*}} \) are exactly the same as the previous case. This implies that, again, starting from \(a^{*}=\overline{\beta }\), a marginal increase in the ideal decreases contribution.

A.3.2: Large preference heterogeneity with non-monotonic f

As an extension of Proposition 4, the following proposition characterizes the contribution-maximizing ideal under large preference heterogeneity for a class of non-monotonic density functions in a special case of \(\gamma \) converging to 0. We assume that there exists a unique \(\hat{\beta }\in \left( \underline{\beta },\bar{\beta }\right) \) such that \(f^{\prime }\left( \beta _{i}\right) >0\) for \(\beta _{i}\in \left( \underline{\beta },\hat{\beta }\right) \), while \(f^{\prime }\left( \beta _{i}\right) <0\) for \(\beta _{i}\in \left( \hat{\beta },\overline{\beta }\right) \). That is, the distribution has a single peak at \(\hat{\beta }\). Some commonly used distribution functions, such as normal distributions, extreme value distributions are special cases of these single-peaked distributions.

Given \(\gamma \) converges to 0, agents with \(\beta _{i}\ge a^{*}\) identify with the ideal but contribute their baseline, and incur no disutility. Thus, the ideal does not discourage high-demand agents from contributing, and we can focus on the trade-off between attracting more under-contributing members and increasing their contributions.

Proposition 6

Suppose \(\overline{\beta }-\underline{\beta }>B\), the density function f has a unique single peak \(\hat{\beta }\in \left( \underline{\beta },\overline{\beta }\right) \), and \(\gamma \) converges to 0. If \(F(\underline{\beta }+B)\le Bf(\underline{\beta })\), then \(\bar{a}=\underline{\beta }+B\); otherwise, \(\bar{a}>\underline{\beta }+B\) and it is the unique solution to \(F\left( \bar{a}\right) -F\left( \bar{a}-B\right) =Bf\left( \bar{a}-B\right) \).

Proof

Given \(\gamma \) converges to 0, by Claims 1 and 3, we can focus on \(a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] \), in which

$$\begin{aligned} g\left( a^{*}\right) =\frac{\theta }{1+\theta }\int _{a^{*}-B} ^{\min \{a^{*},\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) f\left( \beta _{i}\right) d\beta _{i}. \end{aligned}$$

Therefore, \(\frac{dg\left( a^{*}\right) }{da^{*}}\) is the same as \(\chi _{1}(a^{*})\) defined in (26). Since \(F\left( a^{*}\right) =1\) for all \(a^{*}\ge \overline{\beta }\), \(\frac{dg\left( a^{*}\right) }{da^{*}}\) can be simplified as

$$\begin{aligned} \frac{dg\left( a^{*}\right) }{da^{*}}=\frac{\theta }{1+\theta }\left( F(a^{*})-F(a^{*}-B)-Bf\left( a^{*}-B\right) \right) , \end{aligned}$$
(28)

for all \(a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] \).

We first narrow down the possible range for \(\bar{a}\). For all \(a^{*} <\hat{\beta }\), we have \(f^{\prime }\left( \beta _{i}\right) >0\) for all \(\beta _{i}\in [a^{*}-B,a^{*}]\), which implies \(\frac{dg\left( a^{*}\right) }{da^{*}}>0\) by (28). For all \(a^{*}>\hat{\beta }+B\), we have \(f^{\prime }\left( \beta _{i}\right) <0\) for all \(\beta _{i}\in [a^{*}-B,a^{*}]\), which implies \(\frac{dg\left( a^{*}\right) }{da^{*}}<0\) by (28). These two observations imply

$$\begin{aligned} \overline{a}\in \left[ \max \left\{ \underline{\beta }+B,\hat{\beta }\right\} ,\hat{\beta }+B\right] \equiv H. \end{aligned}$$

Next, we want to show (28) is single-crossing for all \(a^{*}\in H\). Formally, we want to prove that, for all \(a_{L}\in H\) such that \(\frac{dg\left( a_{L}\right) }{da^{*}}\le 0\), we have \(\frac{dg\left( a^{*}\right) }{da^{*}}<0\) for all \(a^{*}\in H\) that satisfy \(a^{*}>a_{L}\). To prove this, we first note \(\frac{dg\left( a_{L}\right) }{da^{*}}\le 0\) implies \(f\left( a_{L}\right) \le f(a_{L}-B)\). To see this, suppose by contradiction \(f\left( a_{L}\right) >f(a_{L}-B)\); this implies \(F\left( a_{L}\right) >F(a_{L}-B)+Bf\left( a_{L}-B\right) \) given single-peakedness, which further implies \(\frac{dg\left( a_{L}\right) }{da^{*}}>0\), a contradiction. Next, we note that \(f\left( a_{L}\right) \le f(a_{L}-B)\) implies \(f\left( a^{*}\right) <f(a^{*}-B)\) given \(a_{L}\), \(a^{*}\in H\), and \(a^{*}>a_{L} \). Then, for all \(a^{*}\in H\) such that \(a^{*}>a_{L}\), we have \(\frac{dg\left( a^{*}\right) }{da^{*}}<0\) because the derivative of (28) satisfies

$$\begin{aligned} \frac{d^{2}g\left( a^{*}\right) }{da^{*2}}|_{a^{*}\in H\text { and }a^{*}>a_{L}}&=\frac{\theta }{1+\theta }\left( f(a^{*})-f(a^{*}-B)-Bf^{\prime }\left( a^{*}-B\right) \right) \\&\le \frac{\theta }{1+\theta }\left( f(a^{*})-f(a^{*}-B)\right) \\&<0, \end{aligned}$$

where the first inequality is due to \(f^{\prime }\left( a^{*}-B\right) \ge 0\) (since \(a^{*}\in H\) implies \(a^{*}-B\le \hat{\beta }\) ), while the second inequality uses the fact that \(f\left( a^{*}\right) <f(a^{*}-B)\).

We are now ready to prove the proposition. Suppose \(F(\underline{\beta }+B)\le Bf(\underline{\beta })\) so that \(\frac{dg\left( \underline{\beta }+B\right) }{da^{*}}\le 0\) by (28). The single-crossing property of (28) then implies \(\frac{dg\left( a^{*}\right) }{da^{*}}<0\) for all \(a^{*}>\underline{\beta }+B\), so the contribution maximizing ideal \(\bar{a}=\underline{\beta }+B\). Suppose instead \(F(\underline{\beta }+B)>Bf(\underline{\beta })\), then by (28) the intermediate value theorem guarantees the existence of a solution to the first-order condition \(F(\bar{a})-F(\bar{a}-B)=Bf\left( \bar{a}-B\right) \) in compact interval H. Moreover, the single-crossing property of (28) guarantees that the first-order condition is sufficient for maximizing \(g\left( a^{*}\right) \) and has a unique solution. \(\square \)

A.4: Welfare maximization (Propositions 2 and 5)

This subsection proves Propositions 2 and 5.

Denote \(n=(N-1)E(\beta _{i})\), so condition (11) implies

$$\begin{aligned} n>\left( \frac{2V\theta }{1+\theta }\right) ^{1/2}=\frac{\theta }{1+\theta }B \end{aligned}$$

by the definition of B. Since \(\beta _{i}\)’s are i.i.d, (9) can be written as \(M\left( a^{*}\right) =E\left[ \beta _{i}a_{i}-\frac{1}{2}a_{i}^{2}\right] +nE\left( a_{i}\right) \). Given an insider’s contribution in Equation (3) and an outsider’s contribution \(\beta _{i}\), after some simple algebra we obtain

$$\begin{aligned} E\left( a_{i}\right)&=\int _{\underline{\beta }}^{\bar{\beta }}\beta _{i}dF\left( \beta _{i}\right) \\&\quad +\frac{\theta }{1+\theta }\int _{\max \{a^{*}-B,\underline{\beta }\}} ^{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }\left( a^{*}-\beta _{i}\right) dF\left( \beta _{i}\right) +\frac{\gamma }{1+\gamma }\int _{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }^{\min \{a^{*}+D,\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) dF\left( \beta _{i}\right) , \end{aligned}$$

and

$$\begin{aligned} E\left[ \beta _{i}a_{i}-\frac{1}{2}a_{i}^{2}\right]&=\frac{1}{2} \int _{\underline{\beta }}^{\bar{\beta }}\beta _{i}^{2}dF\left( \beta _{i}\right) \\&\quad-\frac{1}{2}\int _{\max \{a^{*}-B,\underline{\beta }\}}^{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }\left( \frac{\theta \left( a^{*}-\beta _{i}\right) }{1+\theta }\right) ^{2}dF\left( \beta _{i}\right) \\&\quad-\frac{1}{2}\int _{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }^{\min \{a^{*}+D,\overline{\beta }\}}\left( \frac{\gamma \left( a^{*}-\beta _{i}\right) }{1+\gamma }\right) ^{2}dF\left( \beta _{i}\right) . \end{aligned}$$

Substitute both expressions into \(M\left( a^{*}\right) \) to obtain:

$$\begin{aligned} M\left( a^{*}\right) =M_{0}+M_{1}\left( a^{*}\right) +M_{2}\left( a^{*}\right) , \end{aligned}$$
(29)

where the first term

$$\begin{aligned} M_{0}\equiv \int _{\underline{\beta }}^{\bar{\beta }}\left( \frac{1}{2}\beta _{i}^{2}+n\beta _{i}\right) dF\left( \beta _{i}\right) \end{aligned}$$

is independent of \(a^{*}\), while

$$\begin{aligned} M_{1}\left( a^{*}\right)&\equiv \frac{\theta }{1+\theta }\int _{\max \{a^{*}-B,\underline{\beta }\}}^{\max \left\{ \underline{\beta } ,\min \{a^{*},\overline{\beta }\}\right\} }\left[ n\left( a^{*} -\beta _{i}\right) -\frac{1}{2}\frac{\theta \left( a^{*}-\beta _{i}\right) ^{2}}{1+\theta }\right] dF\left( \beta _{i}\right) . \end{aligned}$$
(30)
$$\begin{aligned} M_{2}\left( a^{*}\right)&\equiv \frac{\gamma }{1+\gamma }\int _{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }^{\min \{a^{*}+D,\overline{\beta }\}}\left[ n\left( a^{*}-\beta _{i}\right) -\frac{1}{2}\frac{\gamma \left( a^{*}-\beta _{i}\right) ^{2} }{1+\gamma }\right] dF\left( \beta _{i}\right) \end{aligned}$$
(31)

respectively represent the welfare change of under- and over-contributing insiders.

A.4.1: Preliminaries

We first prove a series of preliminary claims concerning \(M_{1}\) and \(M_{2}\).

Claim 4

For all \(a^{*}\), denote the integrands of \(M_{1}\) and \(M_{2}\) as \(m_{1}\left( \beta _{i};a^{*}\right) \) and\(m_{2}\left( \beta _{i};a^{*}\right) \) respectively. Then,

  • for all \(\beta _{i}\) in the integration interval of \(M_{1}\), \(m_{1}\left( \beta _{i};a^{*}\right) >0\) if \(\beta _{i}<a^{*}\); \(m_{1}\left( a^{*};a^{*}\right) =0\); moreover, \(m_{1}\) is increasing in \(a^{*}\);

  • for all \(\beta _{i}\) in the integration interval of \(M_{2}\), \(m_{2}\left( \beta _{i};a^{*}\right) <0\) if \(\beta _{i}>a^{*}\); \(m_{2}\left( a^{*};a^{*}\right) =0\); moreover, \(m_{2}\) is increasing in \(a^{*}\).

Proof

Consider \(m_{1}\). If \(a^{*}\le \underline{\beta }\), then the integration interval of \(M_{1}\) is empty. Suppose instead \(a^{*}>\underline{\beta }\), then the integration interval is \([\max \{a^{*}-B,\underline{\beta } \},\min \left\{ a^{*},\overline{\beta }\right\} ]\), and

$$\begin{aligned} \frac{dm_{1}\left( \beta _{i};a^{*}\right) }{d\beta _{i}}|_{\beta _{i} \in [\max \{a^{*}-B,\underline{\beta }\},\min \left\{ a^{*} ,\overline{\beta }\right\} ]}&=-n+\frac{\theta }{1+\theta }\left( a^{*}-\beta _{i}\right) \\&\le -n+\frac{\theta }{1+\theta }B\\&<0 (\text {by condition }(11)), \end{aligned}$$

where the first inequality comes from \(\beta _{i}\ge a^{*}-B\). For all \(\beta _{i}<a^{*}\) in the integration interval, \(\frac{dm_{1}}{d\beta _{i} }<0\) implies \(m_{1}\left( \beta _{i};a^{*}\right) >m_{1}(a^{*};a^{*})=0\), where the equality follows from the definition of \(m_{1}\). Finally,

$$\begin{aligned} \frac{dm_{1}}{da^{*}}=-\frac{dm_{1}}{d\beta _{i}}=n-\frac{\theta }{1+\theta }\left( a^{*}-\beta _{i}\right) >0. \end{aligned}$$

Consider \(m_{2}\). If \(a^{*}\ge \overline{\beta }\), then the integration interval of \(M_{2}\) is empty. Suppose instead \(a^{*}<\overline{\beta }\), then the integration interval is \([\max \left\{ \underline{\beta },a^{*}\right\} ,\min \{a^{*}+D,\overline{\beta }\}]\), and

$$\begin{aligned} \frac{dm_{2}\left( \beta _{i};a^{*}\right) }{d\beta _{i}}|_{\beta _{i} \in [\max \left\{ \underline{\beta },a^{*}\right\} ,\min \{a^{*}+D,\overline{\beta }\}]}=-n+\frac{\gamma }{1+\gamma }\left( a^{*}-\beta _{i}\right) <0\text {,} \end{aligned}$$

where the inequality follows from \(\beta _{i}\ge a^{*}\).

For all \(\beta _{i}>a^{*}\) in the integration interval, \(\frac{dm_{2} }{d\beta _{i}}<0\) implies \(m_{2}\left( \beta _{i};a^{*}\right) <m_{2}(a^{*};a^{*})=0\), where the equality follows from the definition of \(m_{2}\). Finally, we have \(\frac{dm_{2}}{da^{*}}=-\frac{dm_{2}}{d\beta _{i}}>0\). \(\square \)

Claim 5

For all \(a^{*}\), \(M_{1}\left( a^{*}\right) \ge 0\) and \(M_{2}\left( a^{*}\right) \le 0\), and each of these inequalities is strict when the corresponding integration interval is non-empty.

Proof

The claim follows immediately from \(m_{1}\left( \beta _{i};a^{*}\right) \ge 0\) and \(m_{2}\left( \beta _{i};a^{*}\right) \le 0\) by Claim 4. \(\square \)

Claim 6

For all \(a^{*}\in (\underline{\beta },\underline{\beta }+B]\), \(\frac{dM_{1}\left( a^{*}\right) }{da^{*}}>0\).

Proof

For the specified range of \(a^{*}\),

$$\begin{aligned}&M_{1}\left( a^{*}\right) =\frac{\theta }{1+\theta }\int _{\underline{\beta } }^{\min \{a^{*},\overline{\beta }\}}m_{1}\left( \beta _{i};a^{*}\right) dF\left( \beta _{i}\right) .\\&\frac{dM_{1}\left( a^{*}\right) }{da^{*}}=\left\{ \begin{array}[c]{l} \frac{\theta }{1+\theta }\int _{\underline{\beta }}^{a^{*}}\frac{dm_{1}\left( \beta _{i};a^{*}\right) }{da^{*}}dF\left( \beta _{i}\right) +m_{1}\left( a^{*};a^{*}\right) f\left( a^{*}\right) \\ \frac{\theta }{1+\theta }\int _{\underline{\beta }}^{\overline{\beta }}\frac{dm_{1}\left( \beta _{i};a^{*}\right) }{da^{*}}dF\left( \beta _{i}\right) \end{array} \begin{array}[c]{c} \text {if }a^{*}\le \overline{\beta }\\ \text {if }a^{*}>\overline{\beta } \end{array} \right\} >0, \end{aligned}$$

where we have utilized \(m_{1}\left( a^{*};a^{*}\right) =0\) and \(\frac{dm_{1}\left( \beta _{i};a^{*}\right) }{da^{*}}>0\) from Claim 4. \(\square \)

Claim 7

Suppose f is monotonic, i.e. \(f^{\prime }\le 0\) or \(f^{\prime }\ge 0\) for all \(\beta _{i}\). For all \(a^{*}\in (\underline{\beta }+B,\overline{\beta }+B)\):

  • (i) if \(a^{*}\le \overline{\beta }\), then \(\frac{dM_{1}\left( a^{*}\right) }{da^{*}}\) has the same sign as \(f^{\prime }\);

  • (ii) if \(a^{*}\ge \overline{\beta }\), then

    $$\begin{aligned} \frac{dM_{1}\left( a^{*}\right) }{da^{*}}=\frac{1}{2}\Phi _{1} +\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) (1-F(a^{*}-B)-Bf\left( a^{*}-B\right) )\text {,} \end{aligned}$$

    where \(\Phi _{1}\) is defined in (33) and has exactly the same sign as \(f^{\prime }\).

Proof

For the specified range of \(a^{*}\),

$$\begin{aligned}&M_{1}\left( a^{*}\right) =\frac{\theta }{1+\theta }\int _{a^{*}-B} ^{\min \{a^{*},\overline{\beta }\}}m_{1}\left( \beta _{i};a^{*}\right) dF\left( \beta _{i}\right) . \\&\frac{dM_{1}\left( a^{*}\right) }{da^{*}}=\left\{ \begin{array} [c]{l} \frac{\theta }{1+\theta }\left[ \int _{a^{*}-B}^{a^{*}}\frac{dm_{1}\left( \beta _{i};a^{*}\right) }{da^{*}}dF\left( \beta _{i}\right) -m_{1}\left( a^{*}-B;a^{*}\right) f\left( a^{*}-B\right) +m_{1}\left( a^{*};a^{*}\right) f\left( a^{*}\right) \right] \\ \frac{\theta }{1+\theta }\left[ \int _{a^{*}-B}^{\overline{\beta }} \frac{dm_{1}\left( \beta _{i};a^{*}\right) }{da^{*}}dF\left( \beta _{i}\right) -m_{1}\left( a^{*}-B;a^{*}\right) f\left( a^{*}-B\right) \right] \end{array} \begin{array}[c]{c} \text {if }a^{*}\le \overline{\beta }\\ \text {if }a^{*}>\overline{\beta } \end{array} \right\} . \end{aligned}$$

Applying \(m_{1}\left( a^{*};a^{*}\right) =0\) and substituting the definition of \(m_{1}\) and \(\frac{dm_{1}}{da^{*}}\) in Claim 4 and its proof to the above, we get

$$\begin{aligned} \frac{dM_{1}\left( a^{*}\right) }{da^{*}}= & {} \frac{\theta }{1+\theta }\left( \int _{a^{*}-B}^{\min \{a^{*},\overline{\beta }\}}\left( n-\frac{\theta \left( a^{*}-\beta _{i}\right) }{1+\theta }\right) dF\left( \beta _{i}\right) \right. \nonumber \\&\left. -\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) Bf\left( a^{*}-B\right) \right) . \end{aligned}$$
(32)

Define

$$\begin{aligned} \Phi _{1}&=\left( \frac{\theta }{1+\theta }\right) ^{2}\int _{a^{*} -B}^{\min \{a^{*},\overline{\beta }\}}\left[ B-2\left( a^{*}-\beta _{i}\right) \right] dF\left( \beta _{i}\right) , \nonumber \\ \Phi _{2}&=F\left( \min \{a^{*},\overline{\beta }\}\right) -F(a^{*}-B)-Bf\left( a^{*}-B\right) . \end{aligned}$$
(33)

By adding and subtracting \(\frac{1}{2}\frac{\theta B}{1+\theta }\) to the integrand in (32) and some rearrangements, we get:

$$\begin{aligned} \frac{dM_{1}\left( a^{*}\right) }{da^{*}}&=\frac{\theta }{1+\theta }\int _{a^{*}-B}^{\min \{a^{*},\overline{\beta }\}}\left[ n-\frac{\theta \left( a^{*}-\beta _{i}\right) }{1+\theta }+\frac{1}{2}\frac{\theta B}{1+\theta }-\frac{1}{2}\frac{\theta B}{1+\theta }\right] \\&\quad -\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) Bf\left( a^{*}-B\right) \\&=\frac{\theta }{1+\theta }\int _{a^{*}-B}^{\min \{a^{*},\overline{\beta }\}}\left[ -\frac{\theta \left( a^{*}-\beta _{i}\right) }{1+\theta } +\frac{1}{2}\frac{\theta B}{1+\theta }\right] dF\left( \beta _{i}\right) \\&\quad +\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) \int _{a^{*}-B}^{\min \{a^{*},\overline{\beta }\}}dF\left( \beta _{i}\right) \\&\quad -\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) Bf\left( a^{*}-B\right) \\&=\frac{\theta }{1+\theta }\int _{a^{*}-B}^{\min \{a^{*},\overline{\beta }\}}\frac{\theta }{1+\theta }\left[ \frac{B-2\left( a^{*}-\beta _{i}\right) }{2}\right] dF\left( \beta _{i}\right) \\&\quad +\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) \left( F\left( \min \{a^{*},\overline{\beta }\}\right) -F(a^{*}-B)-Bf\left( a^{*}-B\right) \right) \\&=\frac{1}{2}\Phi _{1}+\frac{\theta }{1+\theta }\left( n-\frac{1}{2} \frac{\theta B}{1+\theta }\right) \Phi _{2}\text {.} \end{aligned}$$

Case (i): \(a^{*}\le \overline{\beta }\). By condition (11), \(n>\frac{1}{2}\frac{\theta B}{1+\theta }\). Therefore, the sign of \(\frac{dM_{1}\left( a^{*}\right) }{da^{*}}\) is pinned down whenever \(\Phi _{1}\) and \(\Phi _{2}\) have the same sign. Using integration by parts,

$$\begin{aligned} \Phi _{1}=\left( \frac{\theta }{1+\theta }\right) ^{2}\left( BF\left( a^{*}\right) +BF\left( a^{*}-B\right) -2\int _{a^{*}-B}^{a^{*} }F\left( \beta _{i}\right) d\beta _{i}\right) . \end{aligned}$$
(34)

By inspection, we find that \(\Phi _{1}=0\) if F is linear (i.e. \(f^{\prime }=0\) for all \(\beta _{i}\)); \(\Phi _{1}\le (<)0\) if F is (strictly) concave (i.e. \(f^{\prime }\le (<)0\) for all \(\beta _{i}\)); and \(\Phi _{1}\ge (>)0\) if F is (strictly) convex (i.e. \(f^{\prime }\ge (>)0\) for all \(\beta _{i}\)). Therefore, \(\Phi _{1}\) has the same sign as \(f^{\prime }\) when f is monotonic. Moreover,

$$\begin{aligned} \Phi _{2}=F\left( a^{*}\right) -F(a^{*}-B)-Bf\left( a^{*}-B\right) , \end{aligned}$$

which also has the same sign as \(f^{\prime }\) based on direct inspection. Putting the results together, we know \(\frac{dM_{1}\left( a^{*}\right) }{da^{*}}\) has the same sign as \(f^{\prime }\).

Case (ii): \(a^{*}\ge \overline{\beta }\). We note

$$\begin{aligned} \Phi _{1}&=\left( \frac{\theta }{1+\theta }\right) ^{2}\int _{a^{*} -B}^{\overline{\beta }}\left[ B-2\left( a^{*}-\beta _{i}\right) \right] dF\left( \beta _{i}\right) \\&=\left( \frac{\theta }{1+\theta }\right) ^{2}\int _{a^{*}-B}^{a^{*} }\left[ B-2\left( a^{*}-\beta _{i}\right) \right] dF\left( \beta _{i}\right) \end{aligned}$$

because distribution F has zero density for \(\beta _{i}>\overline{\beta }\). Using integration by parts, \(\Phi _{1}\) becomes (34) exactly, and so \(\Phi _{1}\) has the same sign as \(f^{\prime }\), as shown in Case (i) above. Meanwhile, \(\Phi _{2}=1-F(a^{*}-B)-Bf\left( a^{*}-B\right) \). Part (ii) of the claim is thus proved. \(\square \)

A.4.2: Narrowing down the range of \(\hat{a}\)

We now prove a series of claims to narrow down the possible range of welfare-maximizing ideal, \(\hat{a}\).

Claim 8

\(\hat{a}\in \left( \underline{\beta },\overline{\beta }+B\right) \).

Proof

For all \(a^{*}\le \underline{\beta }\), all insiders (if any) are over-contributors so \(M\left( a^{*}\right) =M_{0}+M_{2}(a^{*})\le M_{0}\), where the inequality is due to Claim 5. For all \(a^{*}\ge \overline{\beta }+B\), \(M\left( a^{*}\right) =M_{0} \) because no agents join as insiders. All these ideals are dominated by \(a^{*}=\overline{\beta }\), where \(M\left( a^{*}\right) =M_{0} +M_{1}(a^{*})>M_{0}\), where the inequality is due to Claim 5. \(\square \)

Claim 9

Suppose \(\overline{\beta }-\underline{\beta }\le B\), then \(\hat{a}\ge \underline{\beta }+B\).

Proof

By Claim 8, any \(a^{*}\le \underline{\beta }\) can never be optimal. For all \(a^{*}\in \left( \underline{\beta },\underline{\beta }+B\right) \)

$$\begin{aligned} M\left( a^{*}\right) |_{a^{*}<\underline{\beta }+B}&=M_{0} +M_{1}\left( a^{*}\right) +M_{2}\left( a^{*}\right) \\&\le M_{0}+M_{1}\left( a^{*}\right) \\&<M_{0}+M_{1}\left( \underline{\beta }+B\right) \\&=M\left( \underline{\beta }+B\right) \end{aligned}$$

where the first inequality uses \(M_{2}\left( a^{*}\right) \le 0\) by Claim 5, the second inequality uses \(M_{1}\left( a^{*}\right) \) being increasing by Claim 6, the final equality is due to \(M_{2}\left( \underline{\beta }+B\right) =0\) utilizing the supposition that \(\overline{\beta }\le \underline{\beta }+B\). \(\square \)

Claim 10

Suppose \(\overline{\beta }-\underline{\beta }>B\). If \(f^{\prime }\ge 0\) for all \(\beta _{i}\), then \(\hat{a}\ge \overline{\beta }\).

Proof

By Claim 8, any \(a^{*}\le \underline{\beta }\) can never be optimal. Suppose \(f^{\prime }\ge 0\) for all \(\beta _{i}\) and consider \(a^{*}\in \left( \underline{\beta },\overline{\beta }\right) \),

$$\begin{aligned} M\left( a^{*}\right) |_{a^{*}<\overline{\beta }}&=M_{0} +M_{1}\left( a^{*}\right) +M_{2}\left( a^{*}\right) \\&<M_{0}+M_{1}\left( a^{*}\right) \\&\le M_{0}+M_{1}\left( \overline{\beta }\right) \\&=M\left( \overline{\beta }\right) , \end{aligned}$$

where the first inequality uses \(M_{2}\left( a^{*}\right) <0\) by Claim 5 (note \(a^{*}<\overline{\beta }\) means \(M_{2}\) has a non-empty integration interval). The second inequality is due to \(M_{1}\left( a^{*}\right) \) being (weakly) increasing for all \(a^{*}\in (\underline{\beta },\overline{\beta }]\), which is because of Claim 6, 7(i) and the assumption that \(f^{\prime }\ge 0\) for all \(\beta _{i}\). The final equality follows from \(M_{2}\left( \overline{\beta }\right) =0\) by the definition of \(M_{2}\). \(\square \)

A.4.3: Completing the proofs

We assume that condition (11) is satisfied throughout this subsection, that is, \(n>\frac{1}{2}\frac{\theta B}{1+\theta }\).

Claim 11

Suppose \(\overline{\beta }-\underline{\beta }\le B\). If \(f^{\prime }\le 0\) for all \(\beta _{i}\), then \(\hat{a}=\underline{\beta }+B=\bar{a}\). If \(f^{\prime }\ge 0\) for all \(\beta _{i}\), then \(\hat{a}\ge \bar{a}\).

Proof

By Claims 8 and 9, we can focus on \(a^{*}\in \left[ \underline{\beta }+B,\overline{\beta }+B\right] \), for which \(M\left( a^{*}\right) =M_{0}+M_{1}\left( a^{*}\right) \) and so \(\frac{dM\left( a^{*}\right) }{da^{*}}=\frac{dM_{1}\left( a^{*}\right) }{da^{*}}\). For all \(a^{*}\ge \underline{\beta }+B\ge \overline{\beta }\), Claim 7(ii) implies

$$\begin{aligned}&\frac{dM\left( a^{*}\right) }{da^{*}}=\frac{dM_{1}\left( a^{*}\right) }{da^{*}}\\&\quad =\frac{1}{2}\Phi _{1}+\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) (1-F(a^{*}-B)-Bf\left( a^{*}-B\right) )\text {,} \end{aligned}$$

where \(\Phi _{1}\) has the same sign as \(f^{\prime }\). Therefore, if \(f^{\prime }\le 0\) for all \(\beta _{i}\), then \(\Phi _{1}\le 0\). Meanwhile \(f^{\prime }\le 0\) implies F is concave, so for all \(a^{*}\ge \underline{\beta }+B\),

$$\begin{aligned}&1-F(a^{*}-B)-Bf\left( a^{*}-B\right) \\&\le 1-F(a^{*}-B)-\left( \overline{\beta }-\underline{\beta }\right) f\left( a^{*}-B\right) \\&\le 1-F(a^{*}-B)-\left[ \overline{\beta }-\left( a^{*}-B\right) \right] f\left( a^{*}-B\right) \\&\le 0, \end{aligned}$$

where the first inequality follows from \(\overline{\beta }-\underline{\beta }\le B\), the second follows from \(a^{*}-B\ge \underline{\beta }\), and the third from \(f^{\prime }\le 0\) for all \(\beta _{i}\). For all \(a^{*}>\underline{\beta }+B\), the second inequality is strict and so \(1-F(a^{*}-B)-Bf\left( a^{*}-B\right) <0\). These results imply that \(\frac{dM\left( a^{*}\right) }{da^{*}}(\le )<0\) for all \(a^{*} (\ge )>\underline{\beta }+B\). Therefore, \(\hat{a}=\underline{\beta }+B\), which coincides with \(\bar{a}=\underline{\beta }+B\) in Proposition 3 (by Footnote 14).

If \(f^{\prime }\ge 0\) for all \(\beta _{i}\), then \(\Phi _{1}\ge 0\) by Claim 7(ii). Since \(f^{\prime }\ge 0\) for all \(\beta _{i}\), \(1\ge \left( \overline{\beta }-\underline{\beta }\right) f(\underline{\beta })\ge Bf(\underline{\beta })\), implying \(\frac{1}{f(\underline{\beta })}\ge B\).

If \(\frac{1}{f(\underline{\beta })}>B\), Proposition 3 and its proof show that the contribution maximizing \(\bar{a}\) is determined by

$$\begin{aligned} 1-F(\overline{a}-B)-Bf\left( \overline{a}-B\right) =0, \end{aligned}$$
(35)

while for all \(a^{*}\in \left[ \underline{\beta }+B,\bar{a}\right) \),

$$\begin{aligned} 1-F(a^{*}-B)-Bf\left( a^{*}-B\right) >0. \end{aligned}$$

These imply that \(\frac{dM\left( a^{*}\right) }{da^{*}}>0\) for all \(a^{*}\in \left[ \underline{\beta }+B,\bar{a}\right) \) and \(\frac{dM\left( a^{*}\right) }{da^{*}}\ge 0\) for \(a^{*}=\bar{a}\), and so \(\hat{a}\ge \bar{a}\).

If \(\frac{1}{f(\underline{\beta })}=B\), Proposition 3 shows that \(\bar{a}\) \(=\underline{\beta }+B\), and so \(\hat{a}\ge \bar{a}\) by Claim 9. \(\square \)

Claim 12

Suppose \(\overline{\beta }-\underline{\beta } >B\). If \(f^{\prime }=0\) for all \(\beta _{i}\) then \(\hat{a} =\overline{\beta }=\bar{a}\). If \(f^{\prime }\ge 0\) for all \(\beta _{i}\) then \(\hat{a}\ge \bar{a}\).

Proof

By Claims 8 and 10, we can focus on \(a^{*}\in \left[ \overline{\beta },\overline{\beta }+B\right] \), for which \(M\left( a^{*}\right) =M_{0}+M_{1}\left( a^{*}\right) \). For all \(a^{*}\ge \overline{\beta }\), Claim 7(ii) implies

$$\begin{aligned}&\frac{dM\left( a^{*}\right) }{da^{*}}=\frac{dM_{1}\left( a^{*}\right) }{da^{*}}=\frac{1}{2}\Phi _{1}\\&\quad +\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) (1-F(a^{*}-B)-Bf\left( a^{*}-B\right) )\text {,} \end{aligned}$$

where \(\Phi _{1}\) has the same sign as \(f^{\prime }\). If \(f^{\prime }=0\) for all \(\beta _{i}\), then \(\Phi _{1}=0\). Meanwhile \(f^{\prime }=0\) implies \(1-F(a^{*}-B)-Bf\left( a^{*}-B\right) =\frac{\overline{\beta }-\left( a^{*}-B\right) }{\overline{\beta }-\underline{\beta }}-\frac{B}{\overline{\beta }-\underline{\beta }}=\frac{\overline{\beta }-a^{*}}{\overline{\beta }-\underline{\beta }}\le 0\) for all \(a^{*}\ge \overline{\beta }\), where the inequality is strict for all \(a^{*}>\overline{\beta }\). Therefore, \(\frac{dM\left( a^{*}\right) }{da^{*}}<0\) for all \(a^{*} >\overline{\beta }\) and so \(\hat{a}=\overline{\beta }\), which coincides with \(\bar{a}=\overline{\beta }\) as stated in Proposition 1 when \(\overline{\beta }-\underline{\beta }>B\).

Suppose \(f^{\prime }\ge 0\) for all \(\beta _{i}\). If \(\frac{1-F(\overline{\beta }-B)}{f(\overline{\beta }-B)}\le B\), Proposition 4(i) shows that \(\bar{a}=\overline{\beta }\); by Claim 10, \(\hat{a}\ge \overline{\beta }=\bar{a}\). Otherwise, Proposition 4(i) and its proof show that \(\bar{a}>\overline{\beta }\), which is determined by \(1-F(\bar{a}-B)=Bf(\bar{a}-B)\), and that for all \(a^{*}\in \left[ \overline{\beta },\bar{a}\right) \), \(1-F(a^{*}-B)-Bf\left( a^{*}-B\right) >0\). Moreover, \(\Phi _{1}\ge 0\) since \(f^{\prime }\ge 0\) for all \(\beta _{i}\) by Claim 7(ii). These results show that \(\frac{dM\left( a^{*}\right) }{da^{*}}>0\,\) for all \(a^{*}\in \left[ \overline{\beta },\bar{a}\right) \) and so \(\hat{a}\ge \bar{a}\). \(\square \)

Claim 13

Suppose \(\overline{\beta }-\underline{\beta } >B\). If \(f^{\prime }<0\) for all \(\beta _{i}\) and \(\gamma \) coverges to 0 then \(\hat{a}=\underline{\beta }+B=\bar{a}\).

Proof

By Claim 8, we can focus on \(a^{*}\in \left[ \underline{\beta },\overline{\beta }+B\right] \). Given \(\lim _{\gamma \rightarrow 0}M_{2}\left( a^{*}\right) =0\), we have \(M\left( a^{*}\right) =M_{0}+M_{1}\left( a^{*}\right) \). For \(a^{*}\le \underline{\beta }+B\), \(M_{1}\left( a^{*}\right) \) is increasing in \(a^{*}\) by Claim 6. For \(a^{*}\in (\underline{\beta }+B,\overline{\beta }]\), \(\frac{dM_{1}\left( a^{*}\right) }{da^{*}}\) has the same sign as \(f^{\prime }<0\) for all \(\beta _{i}\) by Claim 7(i). For \(a^{*}\ge \overline{\beta }\),

$$\begin{aligned}&1-F(a^{*}-B)-Bf\left( a^{*}-B\right) \\&\quad \le 1-F(a^{*}-B)-\left[ \overline{\beta }-\left( a^{*}-B\right) \right] f\left( a^{*}-B\right) <0, \end{aligned}$$

where the first inequality follows from \(a^{*}\ge \overline{\beta }\), and the second from \(f^{\prime }<0\) for all \(\beta _{i}\). Therefore,

$$\begin{aligned} \frac{dM_{1}\left( a^{*}\right) }{da^{*}}=\frac{1}{2}\Phi _{1} +\frac{\theta }{1+\theta }\left( n-\frac{1}{2}\frac{\theta B}{1+\theta }\right) (1-F(a^{*}-B)-Bf\left( a^{*}-B\right) )\text { }<0, \end{aligned}$$

where \(\Phi _{1}<0\) since \(f^{\prime }<0\) for all \(\beta _{i}\) by Claim 7(ii). Therefore, \(M_{1}\left( a^{*}\right) \) is maximized at \(\hat{a}=\underline{\beta }+B\), which coincides with \(\bar{a}=\overline{\beta }\) in the second part of Proposition 4. \(\square \)

We are now ready to prove Propositions 2 and 5.

Proof

(Proposition 2) Given \(f^{\prime }=0\) for all \(\beta _{i}\), combining Claims 11 and 12 immediately yields the proposition. \(\square \)

Proof

(Proposition 5) For the first part, we divide (30) and (31) by n so that \(M_{1}+M_{2}\) is proportional to

$$\begin{aligned}&\frac{\theta }{1+\theta }\int _{\max \{a^{*}-B,\underline{\beta }\}} ^{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }\left[ \left( a^{*}-\beta _{i}\right) -\frac{1}{2n}\frac{\theta \left( a^{*}-\beta _{i}\right) ^{2}}{1+\theta }\right] dF\left( \beta _{i}\right) \\&\quad +\frac{\gamma }{1+\gamma }\int _{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }^{\min \{a^{*}+D,\overline{\beta }\}}\left[ \left( a^{*}-\beta _{i}\right) -\frac{1}{2n}\frac{\gamma \left( a^{*}-\beta _{i}\right) ^{2}}{1+\gamma }\right] dF\left( \beta _{i}\right) . \end{aligned}$$

When n coverges to \(\infty \), the expression above converges to

$$\begin{aligned}&\frac{\theta }{1+\theta }\int _{\max \{a^{*}-B,\underline{\beta }\}} ^{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }\left( a^{*}-\beta _{i}\right) dF\left( \beta _{i}\right) \\&\quad +\frac{\gamma }{1+\gamma }\int _{\max \left\{ \underline{\beta },\min \{a^{*},\overline{\beta }\}\right\} }^{\min \{a^{*}+D,\overline{\beta }\}}\left( a^{*}-\beta _{i}\right) dF\left( \beta _{i}\right) \equiv g\left( a^{*}\right) . \end{aligned}$$

It follows that the maximizer of \(M\left( a^{*}\right) \) is the same as the maximizer of \(g\left( a^{*}\right) \) when \(n\ \)coverges to \(\infty \), which is implied by N converging to \(\infty \). The first part of the proposition is thus proved. Combining Claims 11 - 13 yields the last two parts of the proposition. \(\square \)

Appendix B: Extension: alternative functional forms

In this appendix, we discuss two extensions of Sect. 5.1 using alternative functional forms. Section B.1 replaces quadratic cost functions by alternative convex functions and show that the results in Sect. 5.1 still hold. Section B.2 replaces the linear benefit function by a concave function, and uses an example to show that the main insights in Sect. 5.1 remain unchanged. We maintain the assumption of uniform distribution of types.

B.1: Alternative cost functions

Let the quadratic cost function of contributing to the public good \(\frac{1}{2}a^{2}\) be replaced by \(\frac{1}{p}a^{p}\), and let the disutility of an undercontributing insider i be \(\frac{\theta }{p}\left( a^{*} -a_{i}\right) ^{p}\) and that of an overcontributing insider be \(\frac{\gamma }{p}\left( a_{i}-a^{*}\right) ^{p}\), where \(p>1\).

The contribution of an outsider with type \(\beta _{i}\), denoted by \(a_{i} ^{B}\equiv a^{B}\left( \beta _{i}\right) \), is determined by the following first order condition:

$$\begin{aligned} \left( a_{i}^{B}\right) ^{p-1}=\beta _{i}. \end{aligned}$$

Calculation similar to that in the proof of Lemma 1 shows that the contribution of an insider with type \(\beta _{i}\), denoted by \(a_{i} ^{I}\equiv a^{I}\left( \beta _{i}\right) \), is determined by the following first order conditions.

$$\begin{aligned} \left( a_{i}^{I}\right) ^{p-1}&=\beta _{i}+\theta \left( a^{*} -a_{i}^{I}\right) ^{p-1}\text { if }\beta _{i}<\left( a^{*}\right) ^{p-1}\nonumber \\ \left( a_{i}^{I}\right) ^{p-1}&=\beta _{i}-\gamma \left( a_{i} ^{I}-a^{*}\right) ^{p-1}\text { if }\beta _{i}\ge \left( a^{*}\right) ^{p-1} \end{aligned}$$
(36)

It is straightforward to verify from (36) that

$$\begin{aligned} \left. \begin{array}[c]{c} a_{i}^{I}>a_{i}^{B}\\ a_{i}^{I}=a_{i}^{B}\\ a_{i}^{I}<a_{i}^{B} \end{array} \begin{array}[c]{c} \text {if }\beta _{i}<\left( a^{*}\right) ^{p-1}\\ \text {if }\beta _{i}=\left( a^{*}\right) ^{p-1}\\ \text {if }\beta _{i}>\left( a^{*}\right) ^{p-1} \end{array} \right. . \end{aligned}$$

When \(p=2\,\), we get Eq. (3) from Equation (36).

For any agent i, substituting the optimal contributions as an insider and as an outsider back to the utility functions and using the envelope theorem, we can show that the difference in the utility of being an insider and that of being an outsider, denoted by \(U_{i}^{I}-U_{i}^{O}\), satisfies

$$\begin{aligned} \frac{d\left[ U_{i}^{I}-U_{i}^{O}\right] }{d\beta _{i}}=\beta _{i}\left[ a_{i}^{I}-a_{i}^{B}\right] , \end{aligned}$$

which is positive if and only if \(\beta _{i}<\left( a^{*}\right) ^{p-1}\), implying that \(U_{i}^{I}-U_{i}^{O}\) is increasing in \(\beta _{i}\) for \(\beta _{i}<\left( a^{*}\right) ^{p-1}\) and decreasing in \(\beta _{i}\) for \(\beta _{i}>\left( a^{*}\right) ^{p-1}\). Therefore,

$$\begin{aligned} \max _{\beta _{i}}\left( U_{i}^{I}-U_{i}^{O}\right) =\left( U_{i}^{I} -U_{i}^{O}\right) |_{\beta _{i}=\left( a^{*}\right) ^{p-1}}=V>0. \end{aligned}$$

Suppose \(\left( a^{*}\right) ^{p-1}\in \left[ \underline{\beta } ,\overline{\beta }\right] \). Then, we can define \(I^{-}\) as the root of \(U_{i}^{I}-U_{i}^{O}=0\) over interval \(\left[ \underline{\beta },\left( a^{*}\right) ^{p-1}\right] \), which is unique (whenever it exists) because \(U_{i}^{I}-U_{i}^{O}\) is monotonic over this interval. If the root does not exist then we define \(I^{-}=\underline{\beta }\). Likewise, we can define \(I^{+}\) as the root of \(U_{i}^{I}-U_{i}^{O}=0\) over interval \(\left[ \left( a^{*}\right) ^{p-1},\bar{\beta }\right] \), and if the root does not exist then we define \(I^{+}=\bar{\beta }\). We thus have

$$\begin{aligned} U_{i}^{I}\ge U_{i}^{O}\iff \beta _{i}\in \left[ I^{-},I^{+}\right] . \end{aligned}$$

This implies that any agent with type \(\beta _{i}\) will choose to be an insider if and only if \(\beta _{i}\in \left[ I^{-},I^{+}\right] \). This result corresponds to Lemma 2.

We now turn to stage 0, when the social leader chooses \(a^{*}\). We rely on numerical simulations to look for \(a^{*}\) that maximizes the expected contribution. Let \(p=3\), meaning that the cost functions are cubic. Let \(\theta =1\), \(\gamma =0.5\), \(V=0.1\), and \(\underline{\beta }=0.1\). We consider two cases: (i) \(\bar{\beta }=10\) (large preference heterogeneity) and (ii) \(\bar{\beta }=1\) (small preference heterogeneity).

Figure 4 shows the case with \(\bar{\beta }=10\) (large preference heterogeneity). In both graphs of Fig. 4, the horizontal axis represents \(a^{*}\). The left graph of Fig. 4 plots expected change in contribution as \(a^{*}\) varies. In the right graph, the vertical axis indicates player type \(\beta _{i}\); the vertical interval between the two solid black curves indicates the set of insiders. Among these insiders, those with \(\beta _{i}\) above the black dashed curve are over-contributors, while those with \(\beta _{i}\) below the black dashed curve are under-contributors. In both graphs, the red dotted line corresponds to the contribution maximizing ideal. Figure 4 shows that, with large preference heterogeneity, the contribution maximizing ideal attracts only high-type agents to be insiders. Moreover, under this ideal, there are no over-contributing insiders.

Fig. 4
figure4

Cubic cost functions: large preference heterogeneity (\(\bar{\beta }=10\)). Left: expected change in contribution. Right: set of insiders

Figure 5 shows the expected change in contribution in the left graph and the set of insiders in the right graph, with \(\bar{\beta }=1\) (small preference heterogeneity). Under the contribution-maximizing ideal (denoted by the red dotted line), all agents choose to be insiders. Moreover, under this ideal, an agent with type \(\underline{\beta }\) is indifferent between becoming an insider or remaining an outsider. When the ideal increases slightly, the agent will drop out and the set of insiders shrinks.

All these results are consistent with our analysis with quadratic cost functions in Sect. 5.1.

Fig. 5
figure5

Cubic cost functions: small preference heterogeneity (\(\bar{\beta }=1\)). Left: expected change in contribution. Right: set of insiders

B.2: Concave benefits

This subsection discusses an alternative direction of extending our model in Sect. 5.1: using alternative benefit functions. The assumptions of linear benefit and convex cost of the public good adopted in our model drastically simplify our analysis, as under these assumptions, each individual has a dominant strategy in contribution. In this section, we will use an alternative benefit function for the public good, thereby introducing strategic interdependence in contribution between players. Since introducing strategic interdependence largely complicates the analysis, to keep tractability, we will consider the simplest case of only two agents and assume \(\theta =\gamma \). We will use a numerical example to show that our main results in Sect. 5.1 largely remain unchanged.

Let the benefit of the public good to individual i be a concave function \(\beta _{i}\Gamma (a_{i}+ {\sum \nolimits _{j\ne i}} a_{j})\) instead of the linear function \(\beta _{i}(a_{i}+ {\sum \nolimits _{j\ne i}} a_{j})\). Let \(\Gamma ^{\prime }\) be the derivative of \(\Gamma \). Consider a simplest case of only two agents (\(N=2\)), agent 1 and agent 2. Further assume symmetric disutility of deviating from the deal for insiders: \(\theta =\gamma \), as in Akerlof and Kranton (2002). The types of the two agents, \(\beta _{1}\) and \(\beta _{2}\), are independent and follow the uniform distribution in \(\left[ \underline{\beta },\overline{\beta }\right] \) from the stage 0’s point of view, as in Sect. 5.1. The agents observe each other’s type before their respective decisions. In what follows, we denote \(A\equiv a_{1}+a_{2}\) as the aggregate contribution.

Consider a given realized pair of \(\left( \beta _{1},\beta _{2}\right) \). Simple calculation shows that, in the contribution stage, the best response function of any outsider j is determined by \(a^{O}\left( \beta _{j}\right) =\beta _{j}\Gamma ^{\prime }\left( A\right) \), where superscript O denotes outsiders, while the best response function of any insider i is determined by

$$\begin{aligned} a_{i}^{I}=\frac{\theta }{1+\theta }\left( a^{*}-\beta _{i}\Gamma ^{\prime }\left( A\right) \right) +\beta _{i}\Gamma ^{\prime }\left( A\right) . \end{aligned}$$

Both outsiders’ and insiders’ individual contributions thus depend on the contribution of the other player. Contributions are strategic substitutes. Using the two best response functions, we can pin down the aggregate contribution in the equilibrium, which is determined by

$$\begin{aligned} A=\left( \beta _{1}+\beta _{2}\right) \Gamma ^{\prime }\left( A\right) +\frac{\theta }{1+\theta }\sum _{i\in I}\left( a^{*}-\beta _{i}\Gamma ^{\prime }\left( A\right) \right) , \end{aligned}$$
(37)

where I is the set of insiders. In other words, the equilibrium aggregate contribution is characterized by the fixed point A that solves (37). An inspection on (37) suggests that aggregate contribution is increasing in \(a^{*}\) as long as I is not affected.

In the stage of self-selection (stage 1), a player will compare the utility of becoming an insider and that of remaining an outsider, conditional on her belief about what the other player will do. This causes tricky strategic interdependence as a player may prefer to become an insider when she expects the other player to stay outside, and may prefer to stay outside when she expects the other player will become an insider. In general, for each realized pair of \(\left( \beta _{1},\beta _{2}\right) \), we have the following \(2\times 2\) payoff matrix:

Agent 1 Agent 2
In Out
In \(U_{1}\left( In,In\right) ,U_{2}\left( In,In\right) \) \(U_{1}\left( In,Out\right) ,U_{2}\left( In,Out\right) \)
Out \(U_{1}\left( Out,In\right) ,U_{2}\left( Out,In\right) \) \(U_{1}\left( Out,Out\right) ,U_{2}\left( Out,Out\right) \)

where, for example, \(U_{1}\left( In,Out\right) \) and \(U_{2}\left( In,Out\right) \) denote the utility for agent 1 and agent 2 respectively, if agent 1 chooses to become an insider and agent 2 chooses to remain an outsider.

Given a specific functional form of \(\Gamma \left( .\right) \) and parameters \(\theta \), V, \(\beta _{1}\) and \(\beta _{2}\), we will be able to solve the equilibrium numerically for any given \(a^{*}\).

We now consider the social leader’s choice at Stage 0, when she cannot observe \(\beta _{1}\) and \(\beta _{2}\) but knows that \(\beta _{1}\) and \(\beta _{2}\) are independent and each of them follows the uniform distribution in \(\left[ \underline{\beta },\overline{\beta }\right] \). Let \(\Gamma \left( A\right) =\ln \left( A\right) \), and let \(\theta =1\), \(V=0.2\), and \(\underline{\beta }=4\). We will consider two cases: \(\bar{\beta }=10\) (large preference heterogeneity) and \(\bar{\beta }=5\) (small preference heterogeneity) and look for the ideal that maximizes the expected equilibrium aggregate contribution to the public good:

$$\begin{aligned} \max _{a^{*}}\int _{\left[ \underline{\beta },\bar{\beta }\right] } \int _{\left[ \underline{\beta },\bar{\beta }\right] }\frac{A}{\left( \overline{\beta }-\underline{\beta }\right) ^{2}}d\beta _{1}d\beta _{2}. \end{aligned}$$

In the following analysis, we will focus on the equilibrium with the largest insider set, whenever there are multiple equilibria. In the case that there are equilibria where the high-type agent is the only insider and where the low-type agent is the only insider, we will focus on the former equilibrium.

B.2.1: Large preference heterogeneity (\(\overline{\beta }=10\))

Fig. 6
figure6

Logarithm benefits: large preference heterogeneity (\(\overline{\beta }=10\)). Left: expected equilibrium aggregate contribution; right: participation profile under the contribution-maximizing ideal

In the left graph of Fig. 6, the black solid curve plots the expected equilibrium aggregate contribution for each ideal level, while the (horizontal) red dashed curve plots the contribution under the baseline case without \(a^{*}\). We see that the expected contribution is maximized at \(a^{*}=2.36\). Under this ideal, the right graph of Fig. 6 shows the participation profile of agents, in which the horizontal and vertical axes represent the types of agent 1 and agent 2 respectively. In this graph (and in the right graph of Fig. 7),

  • symbol “\(\times \)” denotes realization of types where both agents are outsiders in equilibrium;

  • symbol “\(\bigcirc \)” denotes realization of types where both agents are insiders in equilibrium;

  • triangle “\(\triangleleft \)” denotes realization of types where agent 2 is the only insider in equilibrium;

  • triangle “\(\triangledown \)” denotes realization of types where agent 1 is the only insider in equilibrium; and

  • the gray (black) color on the shapes (\(\bigcirc \), \(\triangleleft \) or \(\triangledown \)) denotes under-contributing (over-contributing) insider(s).

This graph shows that, under the contribution maximizing ideal, the agents with high enough types will choose to be insiders, while agents with low types will remain outsiders. This observation is consistent with the finding of our model in the main text that the optimal ideal attracts only the high-type agents when preference heterogeneity is large. On the other hand, however, the contribution-maximizing ideal here allows the possibility of overcontributors (those in black color). In our model with linear benefits (and thus no strategic interdependence), there are no overcontributing insiders under the contribution maximizing ideal.

B.2.2: Small preference heterogeneity (\(\overline{\beta }=5\))

The left graph of Fig. 7 plots the expected equilibrium aggregate contribution as the ideal level varies. We see that the contribution is maximized at \(a^{*}=1.95\). In the right graph which shows the participation profile of the agents under the contribution-maximizing ideal, all realizations are covered by the “\(\bigcirc \)” symbol with gray color, meaning that all the agents are under-contributing insiders. Moreover, the left graph shows that the expected contribution drops sharply when \(a^{*}\) is slightly above the contribution-maximizing ideal, meaning that insiders with the lowest type drop out. Altogether these findings show that with small preference heterogenity, the contribution-maximizing ideal will make the agent with the lowest type indifferent between becoming an insider and not, and thus attract all agents as under-contributing insiders and encourage them to contribute more. These results are consistent with our model with linear benefits in the main text.

Fig. 7
figure7

Logarithm benefits: small preference heterogeneity (\(\overline{\beta }=5\)). Left: expected equilibrium aggregate contribution; right: participation profile under the contribution-maximizing ideal

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Hong, F., Karp, L. & Teh, TH. Identity in public goods contribution. Soc Choice Welf 57, 617–664 (2021). https://doi.org/10.1007/s00355-021-01322-1

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