Comparison of the voluntary contribution and Pareto-efficient mechanisms under voluntary participation

Abstract

We compare the voluntary contribution mechanism with any mechanism attaining Pareto-efficient allocations when each agent can choose whether he/she participates in the mechanism for the provision of a non-excludable public good. We find that, in our participation game, the voluntary contribution mechanism, because of its higher participation probability in the unique symmetric mixed strategy Nash equilibrium, may perform better than any Pareto-efficient mechanism in terms of the equilibrium expected provision level of the public good and the equilibrium expected payoff of each agent. Our results suggest that the voluntary contribution mechanism, which cannot realize Pareto-efficient allocations under compulsory participation, might be superior to any Pareto-efficient mechanism if we allow agents to voluntarily choose participation in the mechanism.

Appendices

A. Appendix: Proof of Theorem 1

Let

$$\begin{aligned} h^{{\text{PE}}}(p,\alpha ,n)&\equiv \left[ U_{\textsf{P}}^{{\text{PE}}}(p,\alpha ,n)- U_{\textsf{NP}}^{{\text{PE}}}(p,\alpha ,n)\right] \frac{1}{(1-\alpha )^{1-\alpha }} \\&=\sum _{k=0}^{n-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) p^{k}(1-p)^{n-1-k}g^{{\text{PE}}}(\alpha , k), \end{aligned}$$

where

$$\begin{aligned} g^{{\text{PE}}}(\alpha ,k)\equiv \alpha ^{\alpha }(k+1)^{1-\alpha }-k^{1-\alpha }. \end{aligned}$$

Here, $h^{{\text{PE}}}(p,\alpha ,n)$ and $g^{{\text{PE}}}(\alpha ,k)$ measure the expected incentive to participate under any PEM when there are n agents and the incentive to participate under any PEM given the participation of k agents, respectively.¹³

1.1 A.1 Outline of the proof

In this subsection, we provide the outline of the proof. We first establish the following preliminary result (Lemma 1): $g^{{\text{PE}}}$ is strictly decreasing in the number k of participants. This indicates that the incentive to participate declines as the number of participants increases.

To prove statement (i), by using Lemma 1, we first show that $h^{{\text{PE}}}$ is strictly decreasing in p. That is, the expected incentive to participate decreases as the probability of other agents choosing P increases. Moreover, note that $h^{{\text{PE}}}(0,\alpha ,n) > 0$, that is, an agent has a positive incentive to participate when the probability of other agents choosing P is zero. By using these facts and the continuity of $h^{{\text{PE}}}$ with respect to p, we can show the existence and uniqueness of a symmetric mixed strategy Nash equilibrium.

To prove statement (ii), suppose $p^{{\text{PE}}}(\alpha , n+1) <1$, which implies that there is $\hat{p} \in \ ]0,1[$ with $h^{{\text{PE}}}(\hat{p}, \alpha ,n+1) = 0$. From Lemma 1 and with some calculations, we first show $h^{{\text{PE}}}(\hat{p},\alpha ,n) > 0$ (i.e., $U_{\textsf{P}}^{{\text{PE}}}(\hat{p},\alpha ,n) > U_{\textsf{NP}}^{{\text{PE}}}(\hat{p},\alpha , n)$). By the strict decreasingness and continuity of $h^{{\text{PE}}}$ with respect to p, this fact yields that the equilibrium participation probability in the case of n agents is larger than that in the case of $n+1$ agents; that is, $p^{{\text{PE}}}(\alpha , n+1) < p^{{\text{PE}}}(\alpha , n)$.

1.2 A.2 Preliminary result

Lemma 1

For each $\alpha \in \ ]0, 1[$, $\frac{\partial g^{{\tiny \text{PE}}}(\alpha , k)}{\partial k}< 0$.

Proof

By differentiating $g^{{\text{PE}}}$ with respect to k, we obtain

$$\begin{aligned} \frac{\partial g^{{\text{PE}}}(\alpha , k)}{\partial k}&= \alpha ^{\alpha } (1-\alpha )(k+1)^{-\alpha } - (1-\alpha )k^{-\alpha } \\&=(1-\alpha ) \left[ \left( \frac{\alpha }{k+1} \right) ^{\alpha } - \left( \frac{1}{k} \right) ^{\alpha } \right] . \end{aligned}$$

Since $\alpha \in \ ]0,1[$, we obtain $\left( \frac{\alpha }{k+1} \right) ^{\alpha } < \left( \frac{1}{k} \right) ^{\alpha }$, which implies $\frac{\partial g^{{\tiny \text{PE}}}(\alpha , k)}{\partial k}< 0$. $\square$

1.3 A.3 Proof of statement (i) 

We start showing that $h^{{\text{PE}}}$ is strictly decreasing in p. Differentiation of $h^{{\text{PE}}}$ with respect to p yields

$$\begin{aligned} \frac{\partial h^{{\text{PE}}}(p,\alpha ,n)}{\partial p} = \sum _{k = 0}^{n-2} (n-1) \left( {\begin{array}{c}n-2\\ k\end{array}}\right) p^k(1-p)^{n-2-k} \left[ g^{{\text{PE}}}(\alpha , k+1) - g^{{\text{PE}}}(\alpha , k) \right] . \end{aligned}$$

(1)

The derivation of (1) requires some calculations and is thus relegated to Online Appendix E. It then follows from Lemma 1 that $g^{{\text{PE}}}(\alpha , k+1) - g^{{\text{PE}}}(\alpha , k) < 0$. This implies $\frac{\partial h^{{\tiny \text{PE}}}(p,\alpha ,n)}{\partial p} < 0$. Now we have either

• (a) for each $p \in \ ]0, 1[$, $h^{{\text{PE}}}(p,\alpha ,n) \ne 0$ or

• (b) there is $\bar{p} \in \ ]0,1[$ such that $h^{{\text{PE}}}(\bar{p}, \alpha ,n) = 0$.

Suppose (a) holds. Note that $h^{{\text{PE}}}(0,\alpha ,n) =\binom{n-1}{0}g^{{\text{PE}}}(\alpha , 0) = \alpha ^{\alpha } > 0$. By using this fact and the continuity of $h^{{\text{PE}}}$ with respect to p, we obtain that for each $p \in [0, 1[$, $h^{{\text{PE}}}(p, \alpha ,n) > 0$ and $h^{{\text{PE}}}(1, \alpha ,n) \ge 0$. This implies $p^{{\text{PE}}}(\alpha , n) = 1$.

Suppose (b) holds. Then, $\bar{p}$ is a symmetric mixed strategy Nash equilibrium. Since $h^{{\text{PE}}}(0,\alpha ,n) > 0$ and $h^{{\text{PE}}}$ is continuous and strictly decreasing in p, we see that such $\bar{p}$ uniquely exists.

Hence, in each case, we have the desired result. $\square$

1.4 A.4 Proof of statement (ii)

Suppose that $p^{{\text{PE}}}(\alpha , n+1) <1$. This implies that there is $\hat{p} \in \ ]0,1[$ such that $h^{{\text{PE}}}(\hat{p}, \alpha ,n+1) = 0$.

The proof is in two steps.Step 1: ${{\varvec{h}}}^{{\textbf{PE}}}(\varvec{\hat{p}}, {\varvec{\alpha }},{{\varvec{n}}}){\textbf{ > 0}}$. Let $k^{*} > 0$ be such that $g^{{\text{PE}}}(\alpha , k^{*}) < 0$ and $g^{{\text{PE}}}(\alpha , k^{*}-1) \ge 0$. Then, Lemma 1, together with $h^{{\text{PE}}}(\hat{p}, \alpha , n+1) = 0$, implies that such $k^{*}$ exists and the following conditions hold:

GE1.:

For each $k \in \{0, \dots , k^{*}-2\}$, $g^{{\text{PE}}}(\alpha , k) > 0$;

GE2.:

$g^{{\text{PE}}}(\alpha , k^{*}-1) \ge 0$;

GE3.:

For each $k \in \{k^{*}, \dots , n\}$, $g^{{\text{PE}}}(\alpha , k) < 0$.

Then, $h^{{\text{PE}}}(\hat{p}, \alpha , n+1) = 0$ is equivalent to

$$\begin{aligned} \underbrace{\sum _{k = 0}^{k^{*}-1}\left( {\begin{array}{c}n\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-k}g^{{\text{PE}}}(\alpha , k)}_{{> 0 \text{ by } \text{ GE1 } \text{ and } \text{ GE2 }}} = \underbrace{-\sum _{k = k^{*}}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-k}g^{{\text{PE}}}(\alpha , k)}_{{ > 0 \text{ by } \text{ GE3 }}}. \end{aligned}$$

This equality can be rearranged to give

$$\begin{aligned}&n(1-\hat{p})\sum _{k = 0}^{k^{*}-1}\frac{1}{n-k}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) \\&\quad = -n(1-\hat{p}) \sum _{k = k^{*}}^{n-1}\frac{1}{n-k}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) - \hat{p}^{n}g^{{\text{PE}}}(\alpha , n). \end{aligned}$$

Multiplying both sides of the above by $\frac{1}{n(1-\hat{p})}$,

$$\begin{aligned} \begin{aligned}&\sum _{k = 0}^{k^{*}-1}\frac{1}{n-k}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) \\&\quad = \underbrace{- \sum _{k = k^{*}}^{n-1}\frac{1}{n-k}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k)}_{{> 0 \text{ by } \text{ GE3 }}} \underbrace{- \frac{\hat{p}^{n}g^{{\text{PE}}}(\alpha , n)}{n(1-\hat{p})}}_{{ > 0 \text{ by } \text{ GE3 }}}. \end{aligned} \end{aligned}$$

(2)

There are two cases.

• Case 1: ${{\varvec{k}}}^{*} {\varvec{\ne }} {{\varvec{n}}}$. Note that if ($n>$) $j> k^{*} > \ell$, then $\frac{1}{n-j}> \frac{1}{n-k^{*}} > \frac{1}{n-\ell }$. Using this fact and (2), we obtain

  $$\begin{aligned}&\sum _{k = 0}^{k^{*}-1}\frac{1}{n-k^{*}}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) \\&\quad > - \sum _{k = k^{*}}^{n-1}\frac{1}{n-k^*}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) - \frac{1}{n-k^{*}} \cdot \frac{\hat{p}^{n}g^{{\text{PE}}}(\alpha , n)}{n(1-\hat{p})}. \end{aligned}$$

  Multiplying both sides of the above by $n-k^*$,

  $$\begin{aligned}&\sum _{k = 0}^{k^{*}-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k)\\&\quad > - \sum _{k = k^{*}}^{n-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) - \frac{\hat{p}^{n}g^{{\text{PE}}}(\alpha , n)}{n(1-\hat{p})}, \end{aligned}$$

  which implies that

  $$\begin{aligned} \sum _{k = 0}^{n-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) > - \frac{\hat{p}^{n}g^{{\text{PE}}}(\alpha , n)}{n(1-\hat{p})}. \end{aligned}$$

  (3)

  Since the left-hand side of (3) is equal to $h^{{\text{PE}}}(\hat{p}, \alpha , n)$, (3) becomes

  $$\begin{aligned} h^{{\text{PE}}}(\hat{p}, \alpha , n)> - \frac{\hat{p}^{n}g^{{\text{PE}}}(\alpha , n)}{n(1-\hat{p})} > 0, \end{aligned}$$

  where the second inequality follows from GE3. This gives the desired conclusion.

• Case 2: ${{\varvec{k}}}^{*}{\textbf{= }} {{\varvec{n}}}$. Note that if $n-1 > k$, then $1 > \frac{1}{n-k}$. Thus,

  $$\begin{aligned} \sum _{k = 0}^{n-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k) > \sum _{k = 0}^{n-1}\frac{1}{n-k}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \hat{p}^{k}(1-\hat{p})^{n-1-k}g^{{\text{PE}}}(\alpha , k). \end{aligned}$$

  (4)

  Note that the left-hand side of (4) is equal to $h^{{\text{PE}}}(\hat{p}, \alpha , n)$. Also, by (2), the right-hand side of (4) is equal to $- \frac{\hat{p}^{n}g^{{\tiny \text{PE}}}(\alpha , n)}{n(1-\hat{p})}$. Hence, (4) becomes

  $$\begin{aligned} h^{{\text{PE}}}(\hat{p}, \alpha , n)>- \frac{\hat{p}^{n}g^{{\text{PE}}}(\alpha , n)}{n(1-\hat{p})} >0, \end{aligned}$$

  where the second inequality follows from GE3. This gives the desired conclusion.

Step 2: Conclusion. Now we have either

• (a) for each $p \in \ ]0, 1[$, $h^{{\text{PE}}}(p, \alpha , n) \ne 0$ or

• (b) there is $\bar{p} \in \ ]0, 1[$ such that $h^{{\text{PE}}}(\bar{p}, \alpha , n) = 0$.

Suppose (a) holds. Since $h^{{\text{PE}}}$ is continuous and strictly increasing in p, $h^{{\text{PE}}}(\hat{p}, \alpha , n) > 0$ (Step 1) implies that for each $p \in [0,1[$, $h^{{\text{PE}}}(p, \alpha , n) > 0$ and $h^{{\text{PE}}}(1, \alpha , n) \ge 0$. Hence, $p^{{\text{PE}}}(\alpha , n)=1 > p^{{\text{PE}}}(\alpha , n+1)$.

Suppose (b) holds. Then, by the continuity and strict decreasingness of $h^{{\text{PE}}}$ with respect to p, $h^{{\text{PE}}}(\hat{p}, \alpha , n) > 0$ (Step 1) implies that $\bar{p} > \hat{p}$, that is, $p^{{\text{PE}}}(\alpha , n) > p^{{\text{PE}}}(\alpha , n+1)$. $\square$

B. Appendix: Proof of Theorem 2

Let

$$\begin{aligned} h^{{\text{V}}}(p,\alpha ,n)&\equiv \left[ U_{\textsf{P}}^{{\text{V}}} (p,\alpha ,n)-U_{\textsf{NP}}^{{\text{V}}}(p,\alpha ,n)\right] \frac{1}{(1-\alpha )^{1-\alpha }} \\&=\sum _{k=0}^{n-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) p^{k}(1-p)^{n-1-k}g^{{\text{V}}}(\alpha ,k), \end{aligned}$$

where

$$\begin{aligned} g^{{\text{V}}}(\alpha ,k)\equiv \frac{\alpha ^{\alpha }(k+1)}{1+k\alpha }-\left[ \frac{k}{1+ \alpha (k-1)}\right] ^{1-\alpha }. \end{aligned}$$

Here, $h^{{\text{V}}}(p,\alpha ,n)$ and $g^{{\text{V}}}(\alpha ,k)$ measure the expected incentive to participate under the VCM when there are n agents and the incentive to participate under the VCM given the participation of k agents, respectively.

1.1 B.1 Outline of the proof

Fig. 4

Graphs of $g^{{\tiny \text{V}}}(\alpha ,k)$

In this subsection, we provide the outline of the proof. As in the proof of Theorem 1, we first establish preliminary results on the form of $g^{{\text{V}}}$. However, in contrast to function $g^{{\text{PE}}}$, function $g^{{\text{V}}}$ is not decreasing in k and has a somewhat complicated form (Fig. 4a illustrates the form of $g^{{\text{V}}}(0.5,k)$). Due to this fact, we cannot prove Theorem 2 by applying exactly the same proof techniques as in Theorem 1. Thus, we provide three useful lemmas regarding the form of $g^{{\text{V}}}$.

The first lemma (Lemma 2) states that for each $\alpha \in \ ]0,1[$, $g^{{\text{V}}}(\alpha , 0)$ is positive. That is, an agent has an incentive to participate if no one participates in the VCM. The second lemma (Lemma 3) states that for each $k \ge 1$, the graph of the function $g^{{\text{V}}}(\ \cdot \ , k)$ intersects the horizontal axis only once at some value $\bar{\alpha } \in \ ]0,1[$. Moreover, the derivative of $g^{{\text{V}}}(\ \cdot \ , k)$ evaluated at $\bar{\alpha }$ is negative. That is, there is a threshold $\bar{\alpha }$ such that, given the participation of k agents, an agent has an incentive to not participate if the value of the public good relative to the private good exceeds the threshold $\bar{\alpha }$; otherwise, the agent has an incentive to participate. The third lemma (Lemma 4) states that for each $\alpha \in \ ]0,1[$ and each $k \ge 0$, $g^{{\text{V}}}(\alpha , k+1)=0$ implies $g^{{\text{V}}} (\alpha , k)>0$. That is, if participation and non-participation are indifferent for an agent when there are $k+1$ participants, then the agent has an incentive to participate when there are k participants. Figure 4b illustrates these lemmas.

To prove statement (i), we focus on the case where $h^{{\text{V}}}(\bar{p},\alpha ,n) = 0$ for some $\bar{p} \in \ ]0,1[$, that is, a non-degenerate mixed strategy Nash equilibrium exists; otherwise, by the continuity of $h^{{\text{V}}}$ with respect to p and the fact that $h^{{\text{V}}}(0,\alpha ,n) > 0$, we can easily derive the existence and uniqueness result. Then, it suffices to show that $\frac{\partial h^{{\tiny \text{V}}}(\bar{p}, \alpha , n)}{\partial p} < 0$, that is, the derivative of $h^{{\text{V}}}$ with respect to p, evaluated at that equilibrium, is negative. This is because, considering that $h^{{\text{V}}}$ is continuous in p and $h^{{\text{V}}}(0,\alpha , n) > 0$, this immediately implies that $\bar{p}$ is the unique symmetric mixed strategy Nash equilibrium.

To prove statement (ii), suppose $p^{{\text{V}}}(\alpha , n+1) <1$, which implies that there is $\hat{p} \in \ ]0,1[$ with $h^{{\text{V}}}(\hat{p},\alpha ,n+1) = 0$. As in the proof of Theorem 1(ii), we first show that $h^{{\text{V}}}(\hat{p}, \alpha ,n) > 0$ by invoking Lemmas 2–4 and after that we prove that this implies $p^{{\text{V}}}(\alpha , n+1) < p^{{\text{V}}}(\alpha , n)$.

1.2 B.2 Preliminary results

Lemma 2

For each $\alpha \in \ ]0,1[$, $g^{{\text{V}}}(\alpha , 0) > 0$.

Proof

Let $\alpha \in \ ]0,1[$. Then, $g^{{\text{V}}}(\alpha , 0) = \alpha ^{\alpha } > 0$. $\square$

Lemma 3

For each $k \ge 1$, there is a unique value $\bar{\alpha } \in \ ]0,1[$ such that $g^{{\text{V}}}(\bar{\alpha }, k) = 0$. Moreover, $\frac{\partial g^{{\tiny \text{V}}}(\bar{\alpha }, k)}{\partial \alpha } < 0$.

Proof

Let

$$\begin{aligned} g^{{\text{V}}}_{\oplus }(\alpha ,k) \equiv \frac{\alpha ^{\alpha }(k+1)}{1+k\alpha }\quad \text{ and }\quad g^{{\text{V}}}_{\ominus }(\alpha ,k) \equiv \left[ \frac{k}{1+ (k-1)\alpha }\right] ^{1-\alpha }. \end{aligned}$$

Note that $g^{{\text{V}}}(\alpha , k) = g^{{\text{V}}}_{\oplus }(\alpha ,k) - g^{{\text{V}}}_{\ominus }(\alpha ,k)$. Then, we obtain

$$\begin{aligned} \frac{\partial g^{{\text{V}}}_{\oplus }(\alpha , k)}{\partial \alpha }&= -\frac{ k(k+1)\alpha ^{\alpha }}{(1+k\alpha )^2} +\frac{(k+1)\alpha ^{\alpha }(1+\ln \alpha )}{1+k\alpha }; \\ \frac{\partial ^2 g^{{\text{V}}}_{\oplus }(\alpha , k) }{\partial \alpha ^2}&= \frac{(k+1)\alpha ^{\alpha }}{1+k\alpha } \left\{ \left[ \frac{k}{1+k\alpha } -(\ln \alpha + 1)\right] ^2 + \frac{k^2}{(1+k\alpha )^2}+ \frac{1}{\alpha } \right\} ; \\ \frac{\partial g^{{\text{V}}}_{\ominus }(\alpha , k) }{\partial \alpha }&= -\left[ \frac{k}{1+(k-1)\alpha }\right] ^{1-\alpha } \left[ \ln \frac{k}{1+(k-1)\alpha } + \frac{(k-1)(1-\alpha )}{1+(k-1)\alpha } \right] ;\\ \frac{\partial ^2 g^{{\text{V}}}_{\ominus }(\alpha , k) }{\partial \alpha ^2}&= \left\{ \frac{2(k-1)}{1+(k-1)\alpha }+ \frac{(1-\alpha )(k-1)^2}{[1+(k-1)\alpha ]^2}\right\} \left[ \frac{k}{1+(k-1)\alpha }\right] ^{1-\alpha } \\&\quad + \left[ \ln \frac{k}{1+(k-1)\alpha } + \frac{(k-1)(\alpha -1)}{1+(k-1)\alpha }\right] ^2 \left[ \frac{k}{1+(k-1)\alpha }\right] ^{1-\alpha }. \end{aligned}$$

The derivations of these equations are delegated to Online Appendix F.

There are two cases.

• Case 1: ${{\varvec{k}}} {\varvec{\ge }} {{\varvec{2}}}$. Then, it is easily checked that (i) $\lim _{\alpha \downarrow 0} g^{{\text{V}}}_{\ominus } (\alpha , k) =k > 1$, (ii) $\lim _{\alpha \uparrow 1} g^{{\text{V}}}_{\oplus } (\alpha , k)= \lim _{\alpha \uparrow 1} g^{{\text{V}}}_{\ominus } (\alpha , k)=1$, (iii) $\lim _{\alpha \uparrow 1}\frac{\partial g^{{\tiny \text{V}}}_{\oplus }(\alpha , k) }{\partial \alpha }= - \frac{k}{1+k} + 1> 0$, (iv) $\lim _{\alpha \downarrow 0}\frac{\partial g^{{\tiny \text{V}}}_{\ominus }(\alpha , k) }{\partial \alpha } = -k(\ln k + k-1)< 0$, (v) $\lim _{\alpha \uparrow 1}\frac{\partial g^{{\tiny \text{V}}}_{\ominus }(\alpha , k)}{\partial \alpha } = 0$, (vi) $\frac{\partial ^2 g^{{\tiny \text{V}}}_{\oplus }(\alpha , k) }{\partial \alpha ^2} > 0$, and (vii) $\frac{\partial ^2 g^{{\tiny \text{V}}}_{\ominus }(\alpha , k) }{\partial \alpha ^2} > 0$. Moreover, by using the l’Hôpital’s rule, we obtain that $\lim _{\alpha \downarrow 0}g^{{\text{V}}}_{\oplus }(\alpha ,k) = k + 1$ ($> k = \lim _{\alpha \downarrow 0} g^{{\text{V}}}_{\ominus }(\alpha ,k)$) and $\lim _{\alpha \downarrow 0}\frac{\partial g^{{\tiny \text{V}}}_{\oplus }(\alpha , k) }{\partial \alpha } = -\infty < 0$.¹⁴ Since both $g_{\oplus }^{{\text{V}}}$ and $g_{\ominus }^{{\text{V}}}$ are continuous in $\alpha$, these facts imply that there is a unique value $\bar{\alpha } \in \ ]0,1[$ such that $g^{{\text{V}}}_{\oplus }(\bar{\alpha },k) = g^{{\text{V}}}_{\ominus }(\bar{\alpha },k)$, that is, $g^{{\text{V}}}(\bar{\alpha },k) = 0$.

• Case 2: ${{\varvec{k}}} = {{\varvec{1}}}$. Then $\lim _{\alpha \downarrow 0}\frac{\partial g^{{\tiny \text{V}}}(\alpha , 1)}{\partial \alpha } = -\infty < 0$ and $\lim _{\alpha \uparrow 1}\frac{\partial g^{{\tiny \text{V}}}(\alpha , 1)}{\partial \alpha }= \frac{1}{2}> 0$.¹⁵ Moreover, we obtain $\frac{\partial ^2 g^{{\tiny \text{V}}}(\alpha , 1)}{\partial \alpha ^2}> 0$. Since $\lim _{\alpha \downarrow 0}g^{{\text{V}}}(\alpha , k) = 1 > 0$, $\lim _{\alpha \uparrow 1}g^{{\text{V}}}(\alpha , k) = 0$, and $g^{{\text{V}}}$ is continuous in $\alpha$, these imply that there is a unique value $\bar{\alpha } \in \ ]0,1[$ with $g^{{\text{V}}}(\bar{\alpha }, k) = 0$.

Moreover, $g^{{\text{V}}}(\bar{\alpha }, k) = 0$ implies $\frac{\partial g^{{\tiny \text{V}}}(\bar{\alpha }, k) }{\partial \alpha }< 0$ because $\lim _{\alpha \downarrow 0} g^{{\text{V}}}(\alpha ,k) > 0$ and $g^{{\text{V}}}$ is continuous in $\alpha$. $\square$

Lemma 4

For each $\alpha \in \ ]0,1[$ and each $k \ge 0$, if $g^{{\text{V}}}(\alpha , k+1)=0$, then $g^{{\text{V}}} (\alpha , k)>0.$

Proof

Let $\alpha \in \ ]0,1[$ and $k \ge 0$ be such that $g^{{\text{V}}}(\alpha , k+1)=0$. Then,

$$\begin{aligned} g^{{\text{V}}}(\alpha , k+1) = \frac{\alpha ^{\alpha } (k+2)}{1 + (k + 1)\alpha } - \left( \frac{k + 1}{1 + k \alpha } \right) ^{1 - \alpha } = 0, \end{aligned}$$

that is,

$$\begin{aligned} \alpha ^{\alpha } (k+1) = \left( \frac{k + 1}{1 + k \alpha } \right) ^{1 - \alpha } \left[ 1 + (k + 1)\alpha \right] - \alpha ^{\alpha }. \end{aligned}$$

(5)

By using (5), $g^{{\text{V}}}(\alpha , k)$ can be rewritten as

$$\begin{aligned}g^{{\text{V}}}(\alpha , k) \nonumber &= \frac{\alpha ^{\alpha } (k+1)}{1 + k \alpha } - \left[ \frac{k}{1 + (k-1) \alpha } \right] ^{1 - \alpha } \nonumber \\&= \frac{1}{1 + k \alpha } \left\{ \left( \frac{k + 1}{1 + k\alpha } \right) ^{1 - \alpha } \left[ 1 + (k + 1)\alpha \right] - \alpha ^{\alpha } \right\} - \left[ \frac{k}{1 + (k-1) \alpha } \right] ^{1 - \alpha } \nonumber \\&= \frac{(1 + k\alpha ) + \alpha }{1 + k\alpha } \left( \frac{k + 1}{1 + k\alpha } \right) ^{1 - \alpha } - \frac{ \alpha ^{\alpha }}{1 + k \alpha } - \left[ \frac{k}{1 + (k-1) \alpha } \right] ^{1 - \alpha } \nonumber \\&= \left( \frac{k + 1}{1 + k \alpha } \right) ^{1 - \alpha } + \frac{\alpha }{1 + k \alpha } \left( \frac{k + 1}{1 + k \alpha } \right) ^{1 - \alpha } - \frac{ \alpha ^{\alpha }}{1 + k\alpha } - \left[ \frac{k}{1 + (k-1) \alpha } \right] ^{1 - \alpha } \nonumber \\ &= \left( \frac{k + 1}{1 + k \alpha } \right) ^{1 - \alpha } - \left[ \frac{k}{1 + (k-1) \alpha } \right] ^{1 - \alpha } - \frac{\alpha }{1 + k \alpha } \left[ \left( \frac{1}{\alpha }\right) ^{1-\alpha } - \left( \frac{k + 1}{1 + k \alpha } \right) ^{1 - \alpha } \right] . \end{aligned}$$

(6)

Let $a \equiv \frac{1}{\alpha }$, $b \equiv \frac{k+1}{1 + k \alpha }$, and $c \equiv \frac{k}{1 + (k-1) \alpha }$. Note that $a> b > c$ and

$$\begin{aligned} b-c&= \frac{k + 1}{1 + k \alpha } - \frac{k}{1 + (k-1) \alpha } \nonumber \\&= \frac{1-\alpha }{(1+k\alpha )[1+(k-1)\alpha ]} \nonumber \\ \ {}&= \frac{\alpha }{1+(k-1)\alpha } \left[ \frac{1-\alpha + k\alpha -k\alpha }{ \alpha (1+k\alpha )}\right] \nonumber \\ \ {}&= \frac{\alpha }{1+(k-1)\alpha } \left( \frac{1}{\alpha } - \frac{k+1}{ 1+k\alpha } \right) \nonumber \\ \ {}&= \frac{\alpha }{1+(k-1)\alpha } \left( a - b \right) . \end{aligned}$$

(7)

Let $f \colon \mathbb {R}_{+} \rightarrow \mathbb {R}_{+}$ be a function defined by $f(x) = x^{1-\alpha }$. Then, by (6),

$$\begin{aligned} g^{{\text{V}}}(\alpha , k) = [f(b) - f(c)] - \frac{\alpha }{1 + k \alpha }[f(a) - f(b)]. \end{aligned}$$

(8)

Since f is strictly concave,

$$\begin{aligned} \frac{f(b) - f(c )}{b-c} > \frac{f(a) - f(b )}{a-b}. \end{aligned}$$

By (7), this can be rewritten as

$$\begin{aligned} \frac{f(b) - f(c )}{\frac{\alpha \left( a - b \right) }{1+(k-1)\alpha }} > \frac{f(a) - f(b )}{a-b}, \end{aligned}$$

which implies that

$$\begin{aligned} f(b) - f(c ) > \frac{\alpha \left( a - b \right) }{1+(k-1)\alpha } \cdot \frac{f(a) - f(b )}{a-b}. \end{aligned}$$

Since $a-b > 0$ and $\frac{\alpha }{1+(k-1)\alpha } > \frac{\alpha }{1+k\alpha }$,

$$\begin{aligned} f(b) - f(c)> \frac{\alpha }{1+(k-1)\alpha } [f(a) - f(b)] > \frac{\alpha }{1+k\alpha } [f(a) - f(b)], \end{aligned}$$

which implies

$$\begin{aligned}{}[f(b) - f(c)] - \frac{\alpha }{1+k\alpha } [f(a) - f(b)] > 0. \end{aligned}$$

(9)

By (8) and (9), we obtain $g^{{\text{V}}}(\alpha , k) > 0$. $\square$

1.3 B.3 Proof of statement (i)

Note that $h^{{\text{V}}}(0,\alpha ,n) = \binom{n-1}{0}g^{{\text{V}}}(\alpha , 0) = \alpha ^{\alpha } > 0$. Then, we have either

• (a) for each $p \in \ ]0,1[$, $h^{{\text{V}}}(p,\alpha ,n) \ne 0$ or

• (b) there is $\bar{p} \in \ ]0,1[$ such that $h^{{\text{V}}}(\bar{p},\alpha ,n) = 0$.

   

Suppose (a) holds. Then, since $h^{{\text{V}}}$ is continuous in p and $h^{{\text{V}}}(0,\alpha ,n) > 0$, we obtain that for each $p \in [0, 1[$, $h^{{\text{V}}}(p, \alpha , n) > 0$ and $h^{{\text{V}}}(1, \alpha , n) \ge 0$. This implies $p^{{\text{V}}}(\alpha , n) = 1$.

Suppose (b) holds. Then, $\bar{p}$ is a symmetric mixed strategy Nash equilibrium. In order to show the uniqueness, it is sufficient to show $\frac{\partial h^{{\tiny \text{V}}}(\bar{p},\alpha ,n)}{\partial p} < 0$. This is because this fact, together with the facts that $h^{{\text{V}}}(0,\alpha ,n) >0$ and $h^{{\text{V}}}$ is continuous in p, implies that for each $p \in [0, \bar{p}[$, $h^{{\text{V}}}(p,\alpha ,n) > 0$ and for each $p \in \ ]\bar{p},1 ]$, $h^{{\text{V}}}(p,\alpha ,n) < 0$ and hence, $\bar{p}$ is the unique symmetric mixed strategy Nash equilibrium.

By differentiating $h^{{\text{V}}}$ with respect to p and evaluating this at $\bar{p}$, we obtain the following:¹⁶

$$\begin{aligned} \frac{\partial h^{{\text{V}}}(\bar{p},\alpha ,n)}{\partial p} =\sum _{k=0}^{n-2} (n-1) \left( {\begin{array}{c}n-2\\ k\end{array}}\right) \bar{p}^{k} (1-\bar{p})^{n-2-k} \left[ g^{{\text{V}}}(\alpha , k+1) -g^{{\text{V}}}(\alpha ,k) \right] . \end{aligned}$$

(10)

After some manipulation (see Online Appendix F), (10) can be rewritten as follows:

$$\begin{aligned} \frac{\partial h^{{\text{V}}}(\bar{p},\alpha ,n)}{\partial p} = -\frac{n-1}{\bar{p}} \cdot h^{{\text{V}}}(\bar{p},\alpha ,n-1). \end{aligned}$$

(11)

By (11), the proof is done if we can show $h^{{\text{V}}}(\bar{p},\alpha , n-1) > 0$. Thus, we now show $h^{{\text{V}}}(\bar{p},\alpha ,n-1) > 0$. Since $\lim _{\alpha \downarrow 0}g^{{\text{V}}}(\alpha ,k) > 0$ and $g^{{\text{V}}}$ is continuous in $\alpha$, by Lemmas 3 and 4, $g^{{\text{V}}}(\alpha ,k) < 0$ implies $g^{{\text{V}}}(\alpha , k+1) < 0$. This, together with $h^{{\text{V}}}(\bar{p},\alpha ,n) = 0$ and Lemma 2, implies that there exists $k^{*} > 0$ such that

GV1.:

$g^{{\text{V}}} (\alpha , 0) > 0$;

GV2.:

For each $k \in \{1, \dots , k^{*}-1\}$, $g^{{\text{V}}} (\alpha , k) \ge 0$;

GV3.:

For each $k \in \{k^{*}, \dots , n-1\}$, $g^{{\text{V}}} (\alpha , k) < 0$.

Then, $h^{{\text{V}}}(\bar{p}, \alpha , n) = 0$ is equivalent to

$$\begin{aligned} \underbrace{\sum _{k = 0}^{k^{*}-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \bar{p}^{k}(1-\bar{p})^{n-1-k}g^{{\text{V}}}(\alpha , k)}_{{> 0 \text{ by } \text{ GV1 } \text{ and } \text{ GV2 }}} = \underbrace{-\sum _{k = k^{*}}^{n-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) \bar{p}^{k}(1-\bar{p})^{n-1-k}g^{{\text{V}}}(\alpha , k)}_{{ > 0 \text{ by } \text{ GV3 }}} . \end{aligned}$$

Then, by applying similar arguments as those employed in the proof of statement (ii) of Theorem 1, we obtain $h^{{\text{V}}}(\bar{p},\alpha ,n-1) > 0$. $\square$

1.4 B.4 Proof of statement (ii)

Suppose that $p^{{\text{V}}}(\alpha , n+1) <1$. This implies that there is $\hat{p} \in \ ]0,1[$ with $h^{{\text{V}}}(\hat{p}, \alpha , n+1) = 0$. We now show $h^{{\text{V}}}(\hat{p}, \alpha , n) > 0$. To do this, let $k^{*} > 0$ be such that $g^{{\text{V}}}(\alpha , k^{*}) < 0$ and $g^{{\text{V}}}(\alpha , k^{*}-1) \ge 0$. Since $h^{{\text{V}}}(\hat{p},\alpha , n+1) = 0$, Lemmas 2–4 ensure that such $k^{*}$ exists and GV1–GV2 above and the following slightly modified version of GV3 hold: for each $k \in \{k^{*}, \dots , n\}$, $g^{{\text{V}}}(\alpha , k) < 0$. Then, using similar arguments as those employed in the proof of statement (ii) of Theorem 1, we can prove $h^{{\text{V}}}(\hat{p}, \alpha , n) > 0$.

Now we have either

• (a) for each $p \in \ ]0,1[$, $h^{{\text{V}}}(p,\alpha , n) \ne 0$ or

• (b) there is $\bar{p} \in \ ]0,1[$ such that $h^{{\text{V}}}(\bar{p},\alpha ,n) = 0$.

   

Suppose (a) holds. Since $h^{{\text{V}}}(0,\alpha ,n) >0$ and $h^{{\text{V}}}$ is continuous in p, we obtain that for each $p \in [0,1[$, $h^{{\text{V}}}(p, \alpha , n) > 0$ and $h^{{\text{V}}}(1, \alpha , n) \ge 0$. This implies that $p^{{\text{V}}}(\alpha , n) = 1> p^{{\text{V}}}(\alpha , n+1)$.

Suppose (b) holds. Recall that as shown in the proof of statement (i) of Theorem 2, $\frac{\partial h^{{\tiny \text{V}}}(\bar{p}, \alpha , n)}{\partial p}< 0$. Considering that $h^{{\text{V}}}(0,\alpha , n) >0$ and $h^{{\text{V}}}$ is continuous in p, this implies that for each $p \in [0, \bar{p}[$, $h^{{\text{V}}}(p, \alpha , n) > 0$ and for each $p \in \ ]\bar{p}, 1]$, $h^{{\text{V}}}(p, \alpha , n) < 0$. Hence, $h^{{\text{V}}}(\hat{p}, \alpha , n) > 0$ implies $\bar{p} > \hat{p}$, that is, $p^{{\text{V}}}(\alpha , n) > p^{{\text{V}}}(\alpha , n+1)$. $\square$

C. Appendix: Omitted proofs in Section 5

1.1 C.1 Proof of Theorem 3

The proof is in three steps.

• Step 1: For each ${\varvec{\alpha }} {\varvec{\in }}\ ]{{\varvec{0}}}, {{\varvec{1}}}[$, $\frac{{\varvec{\partial }} {{\varvec{h}}}^{{\tiny{\textbf{V}}}}({{\varvec{p}}},{\varvec{\alpha }}, {{\varvec{2}}})}{ {\varvec{\partial }} {{\varvec{p}}}} < {{\varvec{0}}}$. Note that

  $$\begin{aligned} \frac{\partial h^{{\text{V}}}(p,\alpha , 2)}{ \partial p} = g^{{\text{V}}}(\alpha , 1) - g^{{\text{V}}}(\alpha , 0) = \frac{2\alpha ^{\alpha }}{1+\alpha } -1 - \alpha ^{\alpha } = \frac{(\alpha ^{\alpha } - 1)-\alpha (1+\alpha ^{\alpha })}{1+\alpha }. \end{aligned}$$

  Since $\alpha \in \ ]0,1 [$, we obtain $\alpha ^{\alpha } -1 <0$, which implies $\frac{\partial h^{{\tiny \text{V}}}(p,\alpha ,2)}{ \partial p} < 0$.

• Step 2: For each ${\varvec{\alpha }} {\varvec{\in }}\ ]{{\varvec{0}}}, {{\varvec{1}}}[$, ${{\varvec{1}}} + {\varvec{\alpha }} > {{\varvec{2}}}^{{\varvec{\alpha }}}$. Let $\varphi (\alpha ) \equiv 1 + \alpha$ and $\psi (\alpha ) \equiv 2^{\alpha }$. Note that $\lim _{\alpha \downarrow 0} \varphi (\alpha )= \lim _{\alpha \downarrow 0} \psi (\alpha )= 1$ and $\lim _{\alpha \uparrow 1} \varphi (\alpha )= \lim _{\alpha \uparrow 1} \psi (\alpha )= 2$. Then, $\varphi '(\alpha ) = 1 > 0$, $\varphi ''(\alpha ) = 0$, $\psi '(\alpha ) = 2^{\alpha } \ln 2 >0$, and $\psi ''(\alpha ) = 2^{\alpha } \left( \ln 2\right) ^2>0$. Since both $\varphi$ and $\psi$ are continuous, for each $\alpha \in \ ]0, 1[$, $\varphi (\alpha ) > \psi (\alpha )$, that is, $1 + \alpha > 2^{\alpha }$.

• Step 3: Conclusion. Let $\alpha \in \ ]0, 1[$. As we have shown in the proof of statement (i) of Theorem 1, $\frac{\partial h^{{\tiny \text{PE}}}(p,\alpha ,2)}{\partial p} < 0$. Moreover, by Step 1, $\frac{\partial h^{{\tiny \text{V}}}(p,\alpha ,2)}{ \partial p}< 0$. Thus, $h^{{\text{PE}}}(p, \alpha , 2)- h^{{\text{V}}}(p, \alpha , 2) > 0$ implies that if $p^{{\text{V}}}(\alpha , 2) = 1$, then $p^{{\text{PE}}}(\alpha , 2) = 1$ (i.e., $\textit{PP}(\alpha ,2) = 1$); otherwise, $p^{{\text{PE}}}(\alpha , 2) > p^{{\text{V}}}(\alpha , 2)$ (i.e., $\textit{PP}(\alpha ,2) < 1$). We now show that $h^{{\text{PE}}}(p, \alpha , 2)- h^{{\text{V}}}(p, \alpha , 2) > 0$. Then,

  $$\begin{aligned} h^{{\text{PE}}}(p, \alpha , 2)- h^{{\text{V}}}(p, \alpha , 2)&= (1-p)[g^{{\text{PE}}}(\alpha , 0) - g^{{\text{V}}}(\alpha , 0)] +p[g^{{\text{PE}}}(\alpha , 1) - g^{{\text{V}}}(\alpha , 1)]\\&=(1-p)(\alpha ^{\alpha }-\alpha ^{\alpha })+p\left[ \frac{2\alpha ^{\alpha }}{2^{\alpha }} - 1 - \left( \frac{2\alpha ^{\alpha }}{1+\alpha } - 1\right) \right] \\&= p 2\alpha ^{\alpha } \left[ \frac{(1+\alpha ) - 2^{\alpha }}{2^{\alpha }(1+\alpha )} \right] \\&>0, \end{aligned}$$

where the last inequality follows from $(1+\alpha ) - 2^{\alpha } > 0$ (Step 2). Hence, $\textit{PP}(\alpha ,2) \le 1$. Moreover, we can find $\bar{\alpha } \in \ ]0, 1[$ with $\textit{PP}(\bar{\alpha },2) < 1$ (e.g., $\textit{PP}(0.5,2) < 1$). $\square$

1.2 C.2 Proof of Theorem 4

Pick any $\alpha \in \ ]0,1[$. Let $y^{{\text{PE}}}_{2}(\alpha )$ (respectively, $y^{{\text{PE}}}_{1}(\alpha )$) be the Nash equilibrium provision level of a public good under any PEM when there are two participants (respectively, one participant). The notation $y^{{\text{V}}}_{2}(\alpha )$ and $y^{{\text{V}}}_{1}(\alpha )$ are similarly defined. Let $\bar{p} \equiv p^{{\text{PE}}}(\alpha , 2)$ and $\tilde{p} \equiv p^{{\text{V}}}(\alpha , 2)$. Note that by Proposition 1, $y^{{\text{PE}}}_{2}(\alpha ) = 2(1-\alpha )> \frac{2(1-\alpha )}{1+\alpha } = y^{{\text{V}}}_{2}(\alpha )$ and $y^{{\text{PE}}}_{1}(\alpha ) =(1-\alpha )= y^{{\text{V}}}_{1}(\alpha )$. Then,

$$\begin{aligned} Ey^{{\text{PE}}}(\alpha , 2)&= \bar{p}^2 y^{{\text{PE}}}_{2}(\alpha ) + 2\bar{p}(1-\bar{p})y^{{\text{PE}}}_{1}(\alpha )\\&= 2(1-\alpha )\bar{p}^2 + 2(1-\alpha )\bar{p}(1-\bar{p})\\&=2(1-\alpha )\bar{p};\\ Ey^{{\text{V}}}(\alpha , 2)&= \tilde{p}^2 y^{{\text{V}}}_{2}(\alpha ) + 2\tilde{p}(1-\tilde{p}) y^{{\text{V}}}_{1}(\alpha )\\&= \frac{2(1-\alpha )}{ 1+\alpha } \cdot \tilde{p}^2 + 2(1-\alpha )\tilde{p}(1-\tilde{p})\\&=2(1-\alpha )\tilde{p} - \left[ 2(1-\alpha ) - \frac{2(1-\alpha )}{1+\alpha }\right] \tilde{p}^2. \end{aligned}$$

Since $\bar{p} \ge \tilde{p}$ (Theorem 3) and $2(1-\alpha )> \frac{2(1-\alpha )}{1+\alpha }$, we obtain

$$\begin{aligned} Ey^{{\text{PE}}}(\alpha , 2) - Ey^{{\text{V}}}(\alpha , 2)&= 2(1-\alpha )\bar{p} - 2(1-\alpha )\tilde{p}+\left[ 2(1-\alpha ) - \frac{2(1-\alpha )}{1+\alpha }\right] \tilde{p}^2\\&= 2(1-\alpha )(\bar{p} - \tilde{p}) + \left[ 2(1-\alpha ) - \frac{2(1-\alpha )}{1+\alpha }\right] \tilde{p}^2\\&>0. \end{aligned}$$

This implies $\textit{PL}(\alpha ,2) < 1$. $\square$

1.3 C.3 Proof of Theorem 5

The proof is in three steps.

• Step 1: For each ${\varvec{\alpha }} {\varvec{\in }}\ ]{{\varvec{0}}}, {{\varvec{1}}}[$, ${{\varvec{2}}}^{{{\varvec{1}}}-{\varvec{\alpha }}} > \frac{{{\varvec{2}}}}{{{\varvec{1}}}+{\varvec{\alpha }}}$. Let $\zeta (\alpha ) \equiv 2^{1-\alpha }$ and $\xi (\alpha ) \equiv \frac{2}{1+\alpha }$. Note that $\lim _{\alpha \downarrow 0} \zeta (\alpha )= \lim _{\alpha \downarrow 0} \xi (\alpha )= 2$ and $\lim _{\alpha \uparrow 1} \zeta (\alpha )= \lim _{\alpha \uparrow 1} \xi (\alpha )= 1$. Then, $\zeta '(\alpha ) = -2^{1-\alpha }\ln 2 < 0$, $\zeta ''(\alpha ) = 2^{1-\alpha }(\ln 2)^2 > 0$, $\xi '(\alpha ) = -\frac{2}{(1+\alpha )^2} < 0$, and $\xi ''(\alpha ) = \frac{4}{(1+\alpha )^3} >0$. Moreover, $\lim _{\alpha \downarrow 0} \zeta '(\alpha )=-2\ln 2 > -2 = \lim _{\alpha \downarrow 0} \xi '(\alpha )$ and $\lim _{\alpha \uparrow 1} \zeta '(\alpha )=-\ln 2 < -0.5 = \lim _{\alpha \uparrow 1} \xi '(\alpha )$. Since both $\zeta$ and $\xi$ are continuous, we obtain that for each $\alpha \in \ ]0, 1[$, $\zeta (\alpha ) > \xi (\alpha )$, that is, $2^{1-\alpha } > \frac{2}{1+\alpha }$.

• Step 2: For each ${\varvec{\alpha }} {\varvec{\in }}\ ]{{\varvec{0}}}, {{\varvec{1}}}[$, ${{\varvec{U}}}_{\textsf{P}}^{{\textbf{PE}}}(\bar{{{\varvec{p}}}},{\varvec{\alpha }} ,{{\varvec{2}}}) - {{\varvec{U}}}_{\textsf{P}}^{{\textbf{V}}}(\tilde{{{\varvec{p}}}},{\varvec{\alpha }}, {{\varvec{2}}}) > {{\varvec{0}}}$. Pick any $\alpha \in \ ]0,1[$. Let $\bar{p} \equiv p^{{\text{PE}}}(\alpha , 2)$ and $\tilde{p} \equiv p^{{\text{V}}}(\alpha , 2)$. Note that

  $$\begin{aligned} U_{\textsf{P}}^{{\text{PE}}}(\bar{p},\alpha ,2)&= (1-\alpha )^{1-\alpha }\alpha ^{\alpha }\left[ (1-\bar{p}) + \bar{p} \cdot 2^{1-\alpha } \right] ;\\ U_{\textsf{P}}^{{\text{V}}}(\tilde{p},\alpha ,2)&= (1-\alpha )^{1-\alpha }\alpha ^{\alpha }\left[ (1-\tilde{p}) + \tilde{p} \cdot \frac{2}{1+\alpha } \right] . \end{aligned}$$

It then follows that

$$\begin{aligned}&U_{\textsf{P}}^{{\text{PE}}}(\bar{p},\alpha ,2) - U_{\textsf{P}}^{{\text{V}}}(\tilde{p},\alpha ,2)\\&= (1-\alpha )^{1-\alpha }\alpha ^{\alpha }\left[ (1-\bar{p}) + \bar{p} \cdot 2^{1-\alpha } \right] - (1-\alpha )^{1-\alpha }\alpha ^{\alpha }\left[ (1-\tilde{p}) + \tilde{p} \cdot \frac{2}{1+\alpha } \right] \\&=(1-\alpha )^{1-\alpha }\alpha ^{\alpha }\left\{ (1-\bar{p}) + \bar{p} \cdot 2^{1-\alpha } - \left[ (1-\tilde{p}) - \tilde{p} \cdot \frac{2}{1+\alpha } \right] \right\} \\&> (1-\alpha )^{1-\alpha }\alpha ^{\alpha }\left\{ (1-\bar{p}) + \bar{p} \cdot \frac{2}{1+\alpha } - \left[ (1-\tilde{p}) - \tilde{p} \cdot \frac{2}{1+\alpha }\right] \right\} \\&= (1-\alpha )^{1-\alpha }\alpha ^{\alpha } \left[ -(\bar{p} - \tilde{p}) + (\bar{p} - \tilde{p}) \cdot \frac{2}{1+\alpha } \right] \\&= (1-\alpha )^{1-\alpha }\alpha ^{\alpha } \left[ (\bar{p} - \tilde{p}) \cdot \frac{1 - \alpha }{1+\alpha } \right] \\&\ge 0, \end{aligned}$$

where the first inequality follows from $2^{1-\alpha } > \frac{2}{1+\alpha }$ (Step 1) and the second inequality follows from $\bar{p} \ge \tilde{p}$ (Theorem 3).

• Step 3: Conclusion. By Step 2 and the fact that $U_{\textsf{P}}^{{\text{PE}}}(\bar{p},\alpha ,2) = U_{\textsf{NP}}^{{\text{PE}}}(\bar{p},\alpha ,2)$ and $U_{\textsf{P}}^{{\text{PE}}}(\tilde{p},\alpha ,2) = U_{\textsf{NP}}^{{\text{PE}}}(\tilde{p},\alpha ,2)$, we obtain $U^{{\text{PE}}}(\alpha , 2) - U^{{\text{V}}}(\alpha , 2) > 0$. This implies $\textit{EP}(\alpha ,2) < 1$. $\square$

D. Appendix: Expected participation level and probability of breakdown

In this section, we first consider the expected participation level under any PEM, $n \cdot p^{{\text{PE}}}(\alpha , n)$, and under the VCM, $n \cdot p^{{\text{V}}}(\alpha , n)$.¹⁷ It is not clear whether the expected participation level increases or decreases with the number of agents, n, because one factor (the number of agents) increases while another (participation probability) decreases. Table 4 presents the expected participation level under both any PEM and the VCM when $\alpha$ varies from 0.1 to 0.9 and n from 2 to 500. We then observe the following:

Table 4 Numerical results for the expected participation level

• PEM: If $\alpha = 0.5$, then $n \cdot p^{{\text{PE}}}(\alpha , n)$ monotonically decreases as n increases. If $\alpha = 0.9$, then $n \cdot p^{{\text{PE}}}(\alpha , n)$ monotonically increases as n increases. For the other cases, $n \cdot p^{{\text{PE}}}(\alpha , n)$ is non-monotonic in n. Specifically:

  • ${\varvec{\alpha }} = {\varvec{0.1}}$: In this case, $p^{{\text{PE}}}(0.1 ,n) = 1$ if $n \le 4$; otherwise, $p^{{\text{PE}}}(0.1 ,n) < 1$. Thus, $n \cdot p^{{\text{PE}}}(0.1, n)$ monotonically increases until $n = 4$. When n increases from 4 to 5, $n \cdot p^{{\text{PE}}}(0.1, n)$ also increases while $p^{{\text{PE}}}(0.1 ,n)$ decreases. However, $n \cdot p^{{\text{PE}}}(0.1, n)$ monotonically decreases over $n =5$ as n increases.

  • ${\varvec{\alpha }} = {\varvec{0.2}}$: In this case, $p^{{\text{PE}}}(0.2 ,n) = 1$ if $n \le 3$; otherwise, $p^{{\text{PE}}}(0.2, n) < 1$. Thus, $n \cdot p^{{\text{PE}}}(0.2, n)$ monotonically increases until $n = 3$. However, it monotonically decreases over $n =3$ as n increases.

  • $\varvec{\alpha } \varvec{\in } \{\varvec{0.3, 0.4, 0.6, 0.7, 0.8}\}$: In this case, $p^{{\text{PE}}}(\alpha , n)$ monotonically decreases as n increases. If $\alpha \in \{0.3, 0.4, 0.6, 0.7\}$ (resp. $\alpha = 0.8$), $n \cdot p^{{\text{PE}}}(\alpha , n)$ increases in n until $n=3$ (resp. $n = 6$) and then decreases.

• VCM: If $\alpha \in \{0.3, \dots , 0.9\}$, $n \cdot p^{{\text{V}}}(\alpha , n)$ monotonically increases as n increases. If $\alpha \in \{0.1,0,2\}$, $n \cdot p^{{\text{V}}}(\alpha , n)$ is non-monotonic in n. Specifically:

  • ${\varvec{\alpha }} = {{\varvec{0.1}}}$: In this case, $p^{{\text{V}}}(0.1 ,n) = 1$ if $n \le 6$; otherwise, $p^{{\text{V}}}(0.1 ,n) < 1$. Thus, $n \cdot p^{{\text{V}}}(0.1, n)$ monotonically increases until $n = 6$. When n increases from 6 to 7, $n \cdot p^{{\text{V}}}(0.1, n)$ also increases until $n =7$ while $p^{{\text{V}}}(0.1 ,n)$ decreases. However, $n \cdot p^{{\text{V}}}(0.1, n)$ monotonically decreases over $n =7$ as n increases.

  • ${\varvec{\alpha }} = {{\varvec{0.2}}}$: In this case, $p^{{\text{V}}}(0.2, n) = 1$ if $n \le 3$; otherwise, $p^{{\text{V}}}(0.2, n) < 1$. Thus, $n \cdot p^{{\text{V}}}(0.2, n)$ monotonically increases until $n = 3$. Interestingly, it also increases until $n = 19$ while $p^{{\text{V}}}(0.2 ,n)$ decreases. However, $n \cdot p^{{\text{V}}}(0.2, n)$ monotonically decreases over $n =19$ as n increases.¹⁸

These observations tell us how the expected participation level varies in group size depending on the value of preference parameter $\alpha$ under both any PEM and the VCM.

Next, we consider the probability of breakdown under any PEM, $\left( 1 - p^{{\text{PE}}}(\alpha , n)\right)^n$, and under the VCM, $\left( 1 - p^{{\text{V}}}(\alpha , n)\right)^n$. Table 5 presents the probability of breakdown under both any PEM and the VCM when $\alpha$ varies from 0.1 to 0.9 and n from 2 to 500. This table indicates that under both mechanisms, the probability of breakdown monotonically increases as the number of agents increases.

Table 5 Numerical results for the probability of breakdown

It is, however, hard to obtain analytical results on both the expected participation level and the probability of breakdown. This is because the expected utility functions take complicated forms in our model, and thus, we cannot derive the explicit forms of $p^{{\text{PE}}}(\alpha , n)$ and $p^{{\text{V}}}(\alpha , n)$.¹⁹