Proof of Proposition 2.1.

By construction, a∗ satisfies IR.

Let a′ be an IR and ex-post efficient allocation. Let θ be a profile such that

Let S={i:ai′(θ)=1} be a set of agents that are assigned to contribute under m^' when the type profile is θ, with |S|=n.

Suppose n = 0. Then a∗(θ)=a′(θ) otherwise by (3) the contributing agents in m^* are better-off, violating ex-post efficiency of a′. Suppose n > 0. Then by IR and (3),

Hence S is a subset of the maximal subset of agents in profile θ that are willing to contribute given others contribute. If a∗(θ)≠a′(θ), it implies that mi∗(θ)=1 whenever i ∈ S and there are some agents j∉S but aj∗(θ)=1. Therefore, agents in S are strictly better-off under m^* compared to m^', and agents outside S are weakly better-off because m^* satisfies IR. This contradicts that a′ is ex-post efficient.

Hence we conclude that a∗(θ)=a′(θ). This is true for all but a F-measure zero set of θ.   □

Proof of Lemma 3.1.

1. Fix a symmetric monotone PBE {m1,m2}. Let S = {θ: m¹(θ) = 1} be the event an agent contributes in round 1 and T(n) = {θ: m²(θ, 0, n) = 1} be the event an agent that doesn’t contribute in round 1 contributes in round 2 when there are n round 1 contributors. Let p(n)=P(T(n)|Sc) be the probability of T(n) conditional on the complement of S. Then β(·; 0, n)~ Bin(N – 1 – n, p(n)).^[8] Likewise, β(⋅;1,n)∼Bin(N−1−n,p(n+1)).

   Since the strategies are monotone in h_– i, T(n) ⊂ T (n + 1), which implies p(n + 1) ≥ p(n).

2. For strict FOSD, we need to show that p(N – 1) > p(N – 2).

   Since Assumption 2 and continuity of g with respect to θ imply p(N – 2) < 1 in equilibrium,^[9]I:=T(N−1)∖T(N−2) will be a non-degenerate interval. Suppose p(N−1)=p(N−2), then P(I|Sc)=0, so types in I all contribute in the first round. Consider the type infI, that is, the type such that g(1, N – 1, inf I) = 1. Since p(N – 2) < 1, he strictly prefers to contribute in the second period than to contribute in the first period. By continuity of expected utility with respect to θ and that I is non-degenerate, some types in I also strictly prefers to delay contribution, contradicting that the proposed strategy profile is a PBE. Hence p(N – 1) > p(N – 2).

Proof of Theorem 1.

We will show that the best-response to cut-off strategies is also a cut-off strategy and use a fixed point argument to show the existence of equilibrium cut-offs. Define

Observe that S is a compact and convex set. Define a correspondence T=(T1,T2):S→2S as follows. For 0 ≤ n ≤ N – 1 and s=(s1,s02,...,sN−12)∈S, Tn2(s) is defined to be the solution to the following equation (with variable x):

where Kn∼Bin(N−n−1,pn)^[10] and pn=P(θ≥sn2|θ<s1). The first (second) term in the left-hand side (LHS) of (4) is the continuation utility of contributing (not contributing) in round 2.

Tn2(s) is well defined since the LHS of (4) is continuous, and by Assumption 2 when the LHS of (4) is evaluated at x = 0 it is negative, when evaluated at x = 1 it is positive. Additionally, Assumption 1 implies the LHS of (4) is strictly increasing in x, this ensures the second round best-response is in cut-offs.

Furthermore, since (4) is continuous in s as well as x, T² is upper-hemicontinuous. We claim that T² satisfies Tn2(s)>Tm2(s) whenever n < m. To see this, fix s ∈ S. Since sm2≤sn2,

where the strict inequality follows because m + K_m strictly FOSDs n + K_n and that g satisfies increasing differences in (mi,m−i). Hence Tn2(s)>Tm2(s). This ensures T(s) ⊂ S.

Finally, T¹(s) is defined to be the set of solutions to the following equation (with variable x):

where α(·)~ Bin(N – 1, 1 – F(s¹)). Since s ∈ S, β(·; 1, N – 2) FOSDs β(·; 0, N – 2). The first (second) term is the expected utility of contributing (not contributing) in round 1.

By Assumption 2, when (5) is evaluated at x = 0, it is negative, when evaluated at x = 1, it is non-negative. Furthermore, the LHS of (5) is non-decreasing with respect to x since g(1, n + k, x) is increasing in x and that due to increasing differences, the increase is more pronounced for the summation before the minus signs of (5). Hence the solutions to (5) exist and is an interval. This shows that the first-round best-response to (s1,s02,...,sN−12) can be chosen to be cutoffs. Summarizing what we have so far, T:S→2S is a non-empty, upper-hemicontinuous, convex-valued correspondence on the compact and convex set S. The Kakutani Fixed Point Theorem implies that T has a fixed point. The fixed points of T completely characterizes the set of monotone symmetric PBEs.

Finally, we prove the last statement. Let (θ1,{θ2(n)}n=0N−1) be a fixed point of T.

Suppose an agent sees N – 1 contributors in the first round. Since contributions are committed, he contributes whenever his type is above θ˜(N−1), where θ˜(n) is defined in (2). Therefore θ2(N−1)=θ˜(N−1). Part 2 of Lemma 3.1 implies that θ2(N−1)<θ1.

Suppose to the contrary that θ2(0)≤θ1, then for every n, β(·; 1, n) strictly FOSD β(·; 0, n) (we have just shown that θ2(m)<θ2(n) for m > n). Hence when (5) is evaluated at θ¹ the value will be positive instead of zero, a contradiction to θ¹ being its solution.

Since θ1<θ2(0), every other agent who doesn’t contribute in round 1 has a type below θ¹. Therefore if an agent who didn’t contribute in round 1 sees no other contributors, he knows that no one else will contribute in round 2. Hence θ2(0)=θ˜(0). (note that θ²(0) is used off the equilibrium path.)   □

Proof of Lemma 3.2.

Fix a symmetric PBE m with cutoffs (θ1,{a(n),b(n)}n=0N−1) with θ¹ ∈ (0, 1). Note that monotonicity regarding one’s own history hi1 implies a(n) ≤ b(n). Let

be the probability (evaluated by an observer) a first-round contributor would contribute in the second round if there are n other first-round contributors. Similarly, let

Observe that β(k; n, 1) is the probability that X1+Y1=k, where X₁~ Bin(n, p(1, n)), Y1∼Bin(N−n−1,p(0,n+1))^[11] Similarly, β(k; n, 0) is the probability that X2+Y2=k, where X₂~ Bin(n, p(1, n – 1)), Y₂~ Bin(N – n – 1, p(0, n)).

Because the equilibrium is assumed to be monotone,

To prove the lemma, we need to show that at least one of the inequalities in (8) holds strictly when n = 1,..., N – 2.

We first show that an agent who signals to contribute in the first round is more likely to contribute in the second round:

Claim 1p(1, n) > p(0, n) for n = 0,..., N – 1.

We need to show that

which is verified by observing that when a(n) > θ¹ the right hand side is zero, and when a(n) ≤ θ¹ the left hand side is one, and by Assumption 2, in equilibrium a(n) ≠ 1 and b(n) ≠ 0.

Next, we show that seeing more first-round contributor makes everyone more likely to contribute in the second round:

Claim 2a(n + 1) < a(n), b(n + 1) < b(n) for n = 0,..., N – 2.

To see the first inequality, fix a first-round contributor with type θ. His continuation payoff of contributing in the second round under histories h = (1, n + 1) and h = (1, n) are, respectively,

where Z₁~ Bin(n + 1, p(1, n + 1)), W1∼Bin(N−2−n,p(0,n+2)), and Z₂~ Bin(n, p(1, n)), W2∼Bin(N−n−1,p(0,n+1)).

Monotonicity of the equilibrium implies that p(1, n + 1) ≥ p(1, n) and p(0, n + 2) ≥ p(0, n + 1). Together with Claim 1 it follows that Z1+W1 strictly FOSDs Z2+W2.^[12] Since {θ:m2(θ,1,n+1)=1}=[a(n+1),1] is the set of θ on which (10) is positive, and {θ, m²(θ, 1, n) = 1} = [a(n), 1] is the set of θ on which (11) is positive, we have thus proved a(n + 1) < a(n). That b(n + 1) < b(n) follows with an identical argument.

Assume to the contrary that strict FOSD does not hold for some n ∈ {1,..., N – 2}. Then Claim 2, (6), (7) and (8) imply that

A first-round contributor who sees n – 1 other contributors knows that they will contribute in the second round (a(n) < θ¹), and the non-contributors will not contribute in the second round (b(n) > θ¹). That is, there will be exactly n – 1 other final contributors. Sequential rationality then implies that a(n−1)=θ˜(n−1). ^[13] Similarly, an agent who doesn’t contribute in the first round in the history of h−i=n+1 knows that there will be exactly n + 1 contributors. Sequential rationality then implies that b(n+1)=θ˜(n+1). But this produces a contradiction to a(n) ≤ b(n) since θ˜(n+1)<θ˜(n−1).   □

Let m be a symmetric monotone PBE of the contribution game with cheap talk. Then for n = 0,..., N – 1, θ¹ ≤ min{a(n), b(n)} whenever β(·; 1, n) strictly FOSDs β(·; 0, n). In addition, when m is informative:

1. If β(·; 1, N – 1) strictly FOSDs β(·; 0, N – 1), then either max{θ1,θ˜(N−1)}<a(N−1) or θ1=a(N−1)=θ˜(N−1).

2. If β(⋅;1,N−1)=β(⋅;0,N−1), then θ˜(N−1)=b(N−1)<θ1=a(N−2)<θ˜(N−2).

Fix a symmetric monotone PBE m. For any type θ, the expected payoffs for (signalling to) contribute in the first round and not sending the signal are, respectively,

Suppose β(·; 1, n) strictly FOSDs β(·; 0, n). For any θ > a(n), sequential rationality and that g is increasing in θ imply

Since sFOSD holds for n, EU1(θ)>EU2(θ). Hence it is better-off for the type θ to contribute in the first round. This implies that θ ≥ θ¹. This holds for every θ > a(n), hence θ¹ ≤ a(n). For any θ > b(n), sequential rationality and that g is increasing in θ imply

sFOSD then implies EU1(θ)>EU2(θ). Hence it is better-off for this type to contribute in the first round as well, so θ ≥ θ¹. So b(n) ≥ θ¹.

Suppose sFOSD holds for n = N – 1. Then the first part of this lemma implies θ¹ ≤ min{a(N – 1), b(N – 1)}. This means that on the equilibrium path, any agent who ever contributes in the end would do so in the first round. If θ¹ = a(N – 1), then an agent in the history h = (1, N – 1) knows that there will be exactly N – 1 other contributors in the end. Sequential rationality then requires a(N−1)=θ˜(N−1). If θ¹ < a(N – 1), then an agent in the history h = (1,N – 1) thinks that with positive probability some of the other contributors are going to withdraw. Hence θ˜(N−1)<a(N−1).

Suppose sFOSD fails for n = N – 1. Then (6) implies p(1,N−1)=p(1,N−2)=1, which implies a(N – 1) < a(N – 2) ≤ θ¹. Together with the first part of this lemma and Lemma 3.2, we have a(N – 2) = θ¹. Since p(1, N – 2) = 1, sequential rationality requires b(N−1)=θ˜(N−1)<θ1. Therefore (7) implies p(0, N – 1) > 0. Consequently, in a history (h = 1, N – 2) there will be at least N – 2 final contributors, and there can actually be N – 1 final contributors with a positive probability. Hence a(N−2)<θ˜(N−2).

Proof of Theorem 2.

1. (Club Good)

   Suppose g is a club good. We will prove that there exist informative equilibria whose cut-offs are in category (b) and in (c).

   1. (Category b) Let θ1=θ˜(N−1). We need to construct equilibrium cutoff a(n) ∈ [θ¹, 1) for n = 0,..., N – 1. Given that there are n other contributors in the first round and that non first-round contributors would never contribute,^[14]a(n) has to solve the following equation (with variable θ)

      (14)∑k=0nnk1−F(θ)1−F(θ1)kF(θ)−F(θ1)1−F(θ1)n−kg(1,k,θ)−1=0.

      When (14) is evaluated at θ = 1 we get g(1,0,1) – 1 > 0, when evaluated at θ = θ¹ we get g(1, n, θ¹) – 1 ≤ 0 since θ1=θ˜(N−1). Intermediate Value Theorem guarantees at least one solution to (14) in the interval [θ¹,1). In particular, when n = N – 1, a(N−1)=θ˜(N−1) is a solution (category (b)).

      To constitute a monotone equilibrium, we show {a(n)} can be chosen to be decreasing in n. For any n > 0, let a(n) be a solution to (14). Then

      ∑k=0n−1n−1k1−F(a(n))1−F(θ1)kF(a(n))−F(θ1)1−F(θ1)n−1−kg(1,k,a(n))−1<0

      Hence a solution a(n – 1) ∈ (a(n), 1) exists to (14). This deals with the actions on the equilibrium path. The off-path cutoffs {b(n)}’s are the solutions to

      (15)∑k=0nnk1−F(a(n−1))1−F(θ1)kF(a(n−1))−F(θ1)1−F(θ1)n−kg(1,k,θ)−1=0,

      which again can be chosen to be decreasing by a similar argument.

   2. (Category c) Take b(N−1)=θ˜(N−1) (which is on-path). Then θ¹ = a(N – 2) solves the following equation (with variable θ) in the interval (θ˜(N−1),θ˜(N−2)):

      (16)F(θ)−F(θ˜(N−1))F(θ)g(1,N−1,θ)+F(θ˜(N−1))F(θ)g(1,N−2,θ)−1=0

      By the definition of the θ˜(n), (16) is negative when evaluated at θ=θ˜(N−1) and is positive when evaluated at θ=θ˜(N−2). Hence a solution in the said interval exists.

      Lemma 3.2 and A.1 imply that b(n) > ... > b(N – 2) ≥ θ¹ for n ≤ N – 2. Therefore, a contributor that doesn’t contribute in the first round will contribute in the second round only if the other N – 1 players contribute in the first round. This means that the first-round contributors in any history h = (1, n) with n < N – 2 knows there will be at most n other final contributors. Hence the on-path a(n) for n < N – 2 solves (14) with θ1∈(θ˜(N−1),θ˜(N−2)). Such cutoffs therefore exist by the same argument as before. The remaining off-path cutoffs b(n) are solved similarly.

2. (Public Good)

   Suppose g is a public good and there is an informative monotone symmetric PBE with round 1 contribution cutoff θ¹ ∈ (0, 1). Lemma 3.2 implies that EU1(θ)>EU2(θ) for all θ, where EU1,EU2 are given by (12) and (13) respectively. This contradicts θ¹ ∈ (0, 1) in a PBE.

Proof of Theorem 3.

1. (Commitment)

   Let σ={mit} be a PBE strategy profile of the game with commitment. Consider the set of profile E such that for all (θ1,...,θN)∈E, θi∈(θ˜(N−1),θ˜(N−2)) for all i. Ex-post efficiency requires that under σ every type in E contributes. However, let i be the earliest contributor on the equilibrium path of σwho first contributes in period t with history h^t – 1. In this period, his continuation expected payoff is positive only if he believes in the future everyone is going to contribute with probability one. But this is impossible since by definition the history h^t – 1 is that no one ever contributes before, and the posterior probability that there are some players whose type is below θ˜(N−1) given this history is thus still positive. Hence player i’s action at t given by σ is not sequentially rational, contradicting that σ is an equilibrium.

2. (Cheap Talk)

   We will build a strategy in which types higher than θ˜(N−1) signal to contribute in round 1, but gradually drop out if they don’t see enough contributors in subsequent rounds.

   A history up to date t is ht=(h1t,...,hNt), where hit=(ai1,...,ait). hit is called consistent if ait≥ait′ whenever t > t^'. That is, a contributor has a consistent history at the start of round t if he signals to contribute in all previous rounds (he acts as if he is following a monotone strategy). Hence, the beliefs in any consistent history is the same as that on the equilibrium path, and the beliefs in any inconsistent history can be defined in an arbitrary way. We are going to construct a strategy and a belief such that any deviator is believed to be the low types who will never contribute in the end. To define it formally, fix a history h^t and a player i, let

   nt(ht)=∑j≠i,hjtis consistent ajt.

   That is, nt(ht) is the number of players other than i who always signals to contribute until round t. The strategy for player i is defined as

   mi1(θi)=1if θi≥θ˜(N−1)0else

   mit(θi,ht−1)=1if θi≥θ˜(nt−1(ht−1))0else 

   for each t = 2,..., T. Hence,

   The posterior probability density for θ_j held by player i at each history h^t is given by

   f(θj|ht)=f(θj)1−F(θ˜(nt−1)if ajt−1=1,hjtis consistent f(θj)F(θ˜(nk−1)−F(θ˜(nk−2)))if ajk=1,ajk+1=0for some k⩽t−2,hjtis consistent f(θj)F(θ˜(N−1))if hjtis inconsistent.

   Note that if ht+1=(ht,ht) is an on-path history with nt(ht)=nt+1(ht+1) for some t, then the equality will continue to hold until t = T following this history because agents stop to drop out. It follows that the on-path allocation of the above profile is the a* characterized in Proposition 2.1.

   What remains is to verify that the strategy-belief profile forms a monotone PBE. We do this by checking that there is no incentive to deviate at any given history.

   Suppose hit is consistent. Then there are two possible deviations in round t. One is mit+1(θ,ht)=1 when θ<θ˜(nt). The best outcome of this kind of deviation is to get nT=nt other contributors in the end, but g(1, n^t, θ) – 1 < 0 since θ<θ˜(nt). Hence this is not a profitable deviation. (The argument fails if g is a public good because player i can benefit from others’ contributions even if himself does not contribute) The other deviation is mit+1(θ,ht)=0 when θ>θ˜(nt). The best outcome deviating in this way is that n^T is the same as the on-path one, but the worst outcome is that the deviation makes n^T smaller, which isn’t profitable either.

   Suppose hit is inconsistent. Then by the constructed belief, player i is already believed to contribute zero in the end. Hence any choice of player i will not affect the outcomes following this history. Therefore there is no profitable deviations either.   □