Community repeated interaction and strategic delegation

Abstract

A large population of fixed-type agents engage in exclusive pairwise relationships in a decentralized setting. At the onset, agents randomly meet in pairs under private information of individual time-invariant types. They play a voluntary contribution game. At the end of the first period, members of each pair either stay together in the second period, in which case reported information is common knowledge, or quit and meet randomly new partners, under private information of individual types. Thus, either long-term or short-term relationships may arise. We show that there are values of the parameters such that information extracted in the first period has a positive effect on social efficiency. We give an interpretation of our results in terms of advantageous delegation of decisions to uninformed agents. Finally, we consider several extensions of the model in which our results still hold.

Appendices

Appendix A: Proof of Lemma 1

For notational ease, here we omit the dependence on the parameters when confusion will not arise; so, for instance, we will write $U^{d} ( \tau _{j};\tau _{j},M^{d},\bar{\sigma } )$ as $U^{d}( \tau _{j} ;\tau _{j})$, and $\Phi ^{d} (M^{d};\bar{\sigma })$ as $\Phi ^{d}.$

First we state and prove the following facts.

Fact 1

For the i th case of interaction, where $i\in \{ I,II,d)\}$, the following holds: (a) in all feasible and incentive compatible mechanisms, $r_{0}^{i}=0$; (b) in all optimal mechanisms, $r_{2}^{i}=1$.

Proof

(a) We know that low-type agents cannot be charged any positive contribution. Then, their participation constraint coupled with the budget-balance constraint makes the provision of the public good infeasible when both pair members are of the low type. (b) Increasing $r_{2}^{i}$ increases the joint surplus (providing the public good is efficient when both agents are of the high type) and furthermore relaxes the incentive constraint of high-type agents, as $r_{2}^{i}$ only appears on the l.h.s. and with a positive multiplier. $\square$

Fact 2

In all feasible mechanisms, for the i th case of interaction, where $i\in \{ I,II,d)\}$, an optimal vector of contributions exists where the low type pays zero and the high type pays all and only the cost of provision, split in half when there are two high-type agents. So for the j th agent, $c_{j} ^{i}(1,0)=C$, $c_{j}^{i}(1,1)=C/2$ and $c_{j}^{i} (0,1)=0$. All other optimal vectors of contributions are equivalent to this as to the social surplus.

Proof

Assumption 2 is a necessary condition for high-type agents to have an incentive to lie.²⁰ Since there is no reason for granting a declared low type any surplus, this is better allocated to the declared high type, in order to trim the incentives to lie. The proposed vector of contributions complies with this logic, plus that of equal treatment of equals (as is required by anonymity). Other optimal vectors of contributions can only differ from the one proposed above in that the contribution due from the two high-type agents is determined by a lottery, which does not affect the joint surplus. $\square$

Given Fact 2 we are justified in limiting ourselves to consider this vector of contributions in the study of equilibria.

Lemma 1. The bad equilibrium always exists

Proof

When in all upstream pairs the vector of probabilities of continuation $\bar{\sigma }$ $=[ 1,1,1]$ is chosen, a deviant principal choosing $\sigma \ne [ 1,1,1]$ could not have his agents re-matched. Thus the deviant would find himself in exactly the same situation as if he had chosen $\sigma$ $=[ 1,1,1]$, which is enough to conclude that $\sigma$ $=[ 1,1,1]$ is a best reply to itself.

Let us now compare the bad equilibrium with the equilibrium of the one-period static version of the game.

Consider the problem of an upstream principal who sets $\sigma =[1,1,1].$ Facts 1 and 2 allow us to write the incentive constraint of agent j when he is of high type as

$$\begin{aligned} p\left[ b\left( 1,1\right) -C/2\right] (1+1)+(1-p)\left[ b\left( 1,0\right) -C\right] \left( r_{1}^{I}+r_{1}^{II}\right) \ge pb\left( 1,1\right) \left( r_{1}^{I}+r_{1}^{II}\right) \end{aligned}$$

(17)

Next, consider the incentive constraint of a high-type agent in the downstream mechanism (2). Setting $\Pi =p$ gives us the incentive constraint of the one-period static version of the game

$$\begin{aligned} p\left[ b\left( 1,1\right) -C/2\right] +(1-p)\left[ b\left( 1,0\right) -C\right] r_{1}^{d}\ge pb\left( 1,1\right) r_{1}^{d}. \end{aligned}$$

(18)

Comparing (18) and (17) makes it clear that if $\bar{r}_{1}^{d}$ maximizes the objective function of the static version of the game, then the objective function of a pair when $\sigma =[1,1,1]$ is maximized by setting $r_{1} ^{I}=r_{1}^{II}=\bar{r}_{1}^{d}$; and furthermore that the maximum value of the latter is twice as large as the maximum value of the former. $\square$

Appendix B: Proof of Proposition 1

Proof

Part (a) Consider (17). Under Assumption 3, setting $r_{1}^{I}=r_{1} ^{II}=1$ is feasible and hence the bad equilibrium is first-best efficient. Moreover, as a single upstream principal has the option to play the bad strategy, no inferior equilibrium can exist.

Part (b) First, inspection of (12) shows that for an upstream principal it is weakly dominant to play $\sigma _{0}=1$ and $\sigma _{2}=1$. Indeed, setting $\sigma _{2}=1$ ensures a truthful high-type agent the highest possible second-period surplus, and has no impact on a high-type liar, so overall it has a non-negative impact on the incentive constraint. Similarly, setting $\sigma _{0}=1$ has no impact on truthful high-type agents, and condemns a high-type liar to the lowest possible second-period payoff, so overall it has a non-negative impact on the incentive constraint. If all principals play $\bar{\sigma }_{0}=1,$ $\bar{\sigma }_{1}<1,\bar{\sigma }_{2}=1$, then it turns out $\Pi =1/2,$ for any value of $\bar{\sigma }_{1}$ in the interval[0, 1[. This allows us to focus on the maximization problem of individual upstream principals, since they do not affect each other through their choices. Facts 1 and 2 above allow us to limit attention to upstream mechanisms such that $r_{0}^{I}=r_{0}^{II}=0$ and $r_{2}^{I}=r_{2}^{II}=1$ and downstream mechanisms such that $r_{0}^{d}=0$ and $r_{2}^{d}=1$, and focus on the incentive constraints for high-type agents.

Thanks to Facts 1 and 2, among the maximization variables included in $M^{d}=( r^{d},c_{1}^{d},c_{2}^{d})$, only the value of the scalar $r_{1}^{d}$ remains to be determined. The incentive constraint of a high-type agent in a downstream pair can be written as

$$\begin{aligned} \Pi \left[ b\left( 1,1\right) -C/2\right] +(1-\Pi )\left[ b\left( 1,0\right) -C\right] r_{1}^{d}\ge \Pi b\left( 1,1\right) r_{1}^{d}. \end{aligned}$$

(19)

The optimal solution $r_{1}^{d*}$ is obtained by setting $r_{1}^{d}$ at the highest level (within the unit interval) that is compatible with the constraints. It turns out that $r_{1}^{d*}=1$ if $\Pi$ is below a certain threshold, namely $\Pi \le \Pi ^{*}\equiv \frac{b(1,0)-C}{b(1,0)-C/2}.$

In the following, we use the convenient normalization $b( 1,0) -C=1.$ Then $\Pi ^{*}=\frac{1}{1+C/2}.$ Given that at equilibrium $\Pi =1/2,$ efficiency will prevail iff $\Pi ^{*}=\frac{1}{1+C/2}\ge 1/2$, i.e. iff $C\le 2$. Letting $U_{j}^{d*}( 1;1)$ denote the value of $U_{j}^{d}( \tau _{j};\tau _{j},M^{d},\bar{\sigma })$ for $M^{d}$ belonging to the set of maximizers of the program of a downstream principal, then $U_{j}^{d*}( 1;1) =1$ if $C\le 2$, and if instead $C>2$, then $U_{j}^{d*}( 1;1) =\frac{1}{2}+\frac{1}{C}<1$ and $r_{1}^{d*}=\frac{2}{C}$.

Turning to the upstream principal problem, after normalization, the incentive constraint (12) becomes

$$\begin{aligned} U_{j}^{u}(1;1)\equiv & {} p(1+1)+(1-p)\left[ \left( r_{1}^{I}+\sigma _{1} r_{1}^{II}\right) +\left( 1-\sigma _{1}\right) U_{j}^{d*}\left( 1;1\right) \right] \nonumber \\\ge & {} p\left[ (1+C/2)\left( r_{1}^{I}+\sigma _{1}r_{1}^{II}\right) +\left( 1-\sigma _{1}\right) U_{j}^{d*}\left( 1;1\right) \right] \equiv U_{j}^{u}(0;1).\qquad \quad \end{aligned}$$

(20)

The set of feasible triplets $( r_{1}^{I},r_{1}^{II},\sigma _{1})$ that satisfy constraint (20) is certainly non-empty, since it includes all triplets of type $( 0,r_{1}^{II},0)$, as can be immediately checked.

We only have to consider the case $p>\Pi ^{*}$. We show that a symmetric equilibrium exists that improves upon the bad equilibrium. Observe that in the bad equilibrium, which will be denoted $\hat{M}^{u}$, in order to satisfy the incentive constraint of a high-type agent, there must hold either $\hat{r} _{1}^{I}<1$ or $\hat{r}_{1}^{II}<1$, or both. Note also that the constraint needs to be saturated, since any possible slack would penalize the objective function. Let us consider the perturbation $\check{M}^{u}$ of $\hat{M}^{u}$ in which $\check{r}_{1}^{II}=1\$and $\check{r}_{1}^{I}\$ is such that the incentive constraint is saturated (we cannot exclude that $\check{r}_{1} ^{I}<0$), while the remaining components are unchanged. Notice that the value of the objective function when $\check{M}^{u}$ is played is the same as with $\hat{M}^{u}$. In fact, excluding $\sigma _{1}=0$, which would render $r_{1}^{II}$ irrelevant, it is

$$\begin{aligned}&-\frac{\partial (U_{j}^{u}(1;1)-U_{j}^{u}(0;1))}{\partial r_{1}^{II}} \Bigg {/}\frac{\partial (2pU_{j}^{u}(1;1))}{\partial r_{1}^{II}}\\&\quad =-\frac{\partial (U_{j}^{u}(1;1)-U_{j}^{u}(0;1))}{\partial r_{1}^{I}}\Bigg {/}\frac{\partial (2pU_{j}^{u}(1;1))}{\partial r_{1}^{I}}\\&\quad =\frac{1-p-p(1+C/2)}{2p(1-p)}. \end{aligned}$$

In plain words, the increase in the constraint associated with a unit loss in the maximand is the same for both $r_{1}^{I}$ and $r_{1}^{II}$. The equilibrium we are looking for is characterized by the strategy $\tilde{M} ^{u}$ in which either $\tilde{r}_{1}^{I}=\tilde{r}_{1}^{II}=1$ and $\tilde{\sigma }_{1}\ge 0$, or $\tilde{r}_{1}^{I}<1,$ $\tilde{r}_{1}^{II}=1$ and $\tilde{\sigma }_{1}=0$, depending on whether the incentive constraint is satisfied at $r_{1}^{I}=r_{1}^{II}=1$ and $\sigma _{1}=0$. To show this point, we first note that the putative equilibrium $\tilde{M}^{u}$ improves upon $\check{M}^{u}$ (which is equivalent to the bad equilibrium $\hat{M}^{u}$ by construction), since the trade-off between $\sigma _{1}$ and $r_{1}^{I}$ is always favorable to an upstream principal that has set $r_{1}^{II}=1$. Observe that

$$\begin{aligned} \frac{\partial (U_{j}^{u}(1;1)-U_{j}^{u}(0;1))}{\partial \sigma _{1} }=(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] -p\left[ (1+C/2)-U_{j}^{d*}\left( 1;1\right) \right] \text { } \end{aligned}$$

and

$$\begin{aligned} \frac{\partial (2pU_{j}^{u}(1;1))}{\partial \sigma _{1}}=2p(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] . \end{aligned}$$

Consider first the case $U_{j}^{d*}\left( 1;1\right) =1$. This happens iff $\frac{1}{2}>\Pi ^{*}$. In this case, lowering $\sigma _{1}$ entails no payoff loss, but relaxes the incentive constraint. This is enough to conclude that $\tilde{M}^{u}$ improves upon $\check{M}^{u}$. If instead $U_{j}^{d*}\left( 1;1\right) <1$, the same conclusion follows from the inequality below

$$\begin{aligned} -\frac{\frac{\partial (U_{j}^{u}(1;1)-U_{j}^{u}(0;1))}{\partial \sigma _{1}} }{\frac{\partial (2pU_{j}^{u}(1;1))}{\partial \sigma _{1}}}= & {} -\frac{(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] -p\left[ (1+C/2)-U_{j}^{d*}\left( 1;1\right) \right] }{2p(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] }\\= & {} -\frac{(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] -p(1+C/2)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] }{2p(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] }\\&+\, \frac{pU_{j}^{d*}\left( 1;1\right) C/2}{2p(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] }\\= & {} -\frac{\frac{\partial (U_{j}^{u}(1;1)-U_{j}^{u}(0;1))}{\partial r_{1}^{I}} }{\frac{\partial (2pU_{j}^{u}(1;1))}{\partial r_{1}^{I}}}+\frac{pU_{j}^{d*}\left( 1;1\right) C/2}{2(1-p)\left[ 1-U_{j}^{d*}\left( 1;1\right) \right] }\\> & {} -\frac{\frac{\partial (U_{j}^{u}(1;1)-U_{j}^{u}(0;1))}{\partial r_{1}^{I}}}{\frac{\partial (U_{j}^{u}(1;1)}{\partial r_{1}^{I}}}\text {.} \end{aligned}$$

Secondly, observe that in the putative equilibrium such a favorable trade-off has been exploited as much as feasible. Third, no other favorable trade-off exists. $\square$

Appendix C: Proof of Proposition 2

Case (I) When agents live forever.

Proof

We show that a symmetric equilibrium exists where each principal plays $\sigma _{0}^{t}=1,\sigma _{2}^{t}=1,\sigma _{1}^{t}=\bar{\sigma }_{1}^{t}<1$ for $t=1,\ldots ,n,\ldots$ whenever the optimal one-period static mechanism is inefficient. Setting $\sigma _{0}^{t}=1$ and $\sigma _{2}^{t}=1$ is still dominant as favorable effect on truthtelling obtains and agents cannot obtain more if the pair splits up. As a consequence, as long as $\bar{\sigma }_{1}^{t}<1,$ the maximization problem of a principal is not affected by changes in$\ \bar{\sigma }_{1}^{t},$ as the probability that a member of a dissolved pair will be matched with a high-type agent stays at $\frac{1}{2}$. What there remains to show is that an individual principal does not prefer to deviate to $\sigma _{1}^{t}=1.$ To this end, it is enough to show that the cost of $\sigma _{1}^{t}$ as an instrument to encourage truthtelling is lower than the cost of $r_{1}^{t,k}.$ Let $p^{\left( t\right) },R^{\left( t\right) }$ and $D^{\left( t\right) }$ denote, respectively, the proportion of high-type agents in the population of agents to be re-matched at time t, the continuation payoffs at time t of a truthful and an untruthful high-type agent when the pair is managed by a principal starting in period t and the pair survives. Moreover, let $Q^{\left( t\right) }$ denote the analogous continuation payoff of a high-type agent when the pair breaks up. (Note that it is independent of the current type report.) The cost of $\sigma _{1}^{t}$ is $-\frac{2p^{\left( t\right) }(1-p^{\left( t\right) })(R^{\left( t\right) }-Q^{\left( t\right) })}{(1-p^{\left( t\right) })(R^{\left( t\right) }-Q^{\left( t\right) })-p^{\left( t\right) }\left( D^{\left( t\right) }-Q^{\left( t\right) }\right) }$ and those of $r_{1}^{t,k}$ are $-\frac{2p^{\left( t\right) }(1-p^{\left( t\right) })\left( b\left( 1,0\right) -C\right) }{(1-p^{\left( t\right) })\left( b\left( 1,0\right) -C\right) -p^{\left( t\right) }b\left( 1,1\right) },$ $k=1,2,\ldots$ We can limit attention w.l.o.g. to strategies where $r_{1} ^{t,k}=r,k=1,\ldots ,n,\ldots$. Next we notice that $R^{\left( t\right) }=$ $r\left[ b\left( 1,0\right) -C\right] \frac{\delta }{1-\delta }$ and $D^{\left( t\right) }=rb\ \left( 1,1\right) \frac{\delta }{1-\delta }$, where $\delta$ is the per-period discount factor and $\delta <1.$ Lastly we note that the cost of $\sigma _{1}^{t}$ is equal to the cost of $r_{1}^{t,k}$ when $Q^{\left( t\right) }=0$ and is lower when $Q^{\left( t\right) }>0.$ $\square$

Case (III) When agents cannot commit intertemporally.

Proof

We adopt the normalization $b( 1,0) -C=$ $b( 1,1) -C/2=1$. Moreover, let us assume that agents obey the upstream principal’s strategy concerning group continuation as long as this does not reduce their expected payoff. We study the conditions for incentive compatibility given that: a lying high-type agent who was initially paired with a low-type mate will certainly quit, so each upstream principal is indifferent as to his own choice of $\sigma _{0}$; truthful high-type agents only accept to obey a strategy entailing $\sigma _{1}<1$ if efficiency prevails in downstream mechanisms (i.e. $U_{{}}^{d*}( 1;1) =1$), a case in which each upstream principal is indifferent as to his own choice of $\sigma _{2}$. As to $\sigma _{1}$, instead, given downstream efficiency, each upstream principal is led to choose the smallest possible value, as a high-type agent’s expected benefit of lying decreases with $\sigma _{1}$. We now show that for some parameter values a good equilibrium exists despite imperfect commitment. Let $\bar{\sigma }_{0}=$ $\bar{\sigma }_{1}=0$ and $\bar{\sigma }_{2}=1$, so $\Pi =\frac{0+2p( 1-p) }{2( ( 1-p) ^{2}+2p( 1-p) ) }=\frac{p}{p+1}$. Efficiency will occur in downstream mechanisms iff $\frac{p}{p+1}\le \Pi ^{*} =\frac{b( 1,0) -C}{b( 1,0) -C/2}=\frac{2}{2+C}$, or $p\le \frac{2}{C}.$ The incentive constraint of high-type agent j writes

$$\begin{aligned} U_{j}^{u}(1;1)\equiv & {} p(1+1)+(1-p)\left( r_{1}^{I}+U_{j}^{d*}\left( 1;1\right) \right) \\\ge & {} p\left( (1+C/2)r_{1}^{I}+U_{j}^{d*}\left( 1;1\right) \right) +(1-p)U_{j}^{d*}\left( 1;1\right) \equiv U_{j}^{u}(0;1) \end{aligned}$$

which, given efficiency in downstream mechanisms, simplifies to

$$\begin{aligned} 2p+(1-p)\left( r_{1}^{I}+1\right) \ge p\left( (1+C/2)r_{1}^{I}+1\right) +(1-p) \end{aligned}$$

whence

$$\begin{aligned} r_{1}^{I*}=\frac{2p}{4p+Cp-2} \end{aligned}$$

(in plain words, the principal maximizes the objective function by choosing the highest value of $r_{1}^{I}$ that satisfies the constraints).

Observe also that $r_{1}^{0*}=2\frac{p}{4p+Cp-2}$, that is, at equilibrium in the static one-period game, the same provision probabilities are played as in the first period of the dynamic mechanism under examination. Thus we know that when the former equilibrium is inefficient ($2\frac{p}{4p+Cp-2}<1$, i.e. $p>\frac{2}{2+C}$) and at the same time the equilibrium of downstream mechanisms having $\Pi =\frac{p}{p+1}$ is efficient (i.e. $p\le \frac{2}{C}$), therefore when $p\in ]\frac{2}{2+C},\frac{2}{C}]$, there exists a dynamic equilibrium strategy that is incentive compatible and improves upon the repetition of the one-period static game (in the first period it ensures the same payoff as the static one, while in the second period efficiency obtains). $\square$

Case (IV) Constructing a good equilibrium when the group size is three.

Proof

We simplify notation and let $b_{k}$ denote the benefit each high-type member derives when their number in the group is k. We continue to assume that low-type members have zero benefit, so they cannot pay any share of the provision cost and at the same time satisfy the participation constraint. This implies once more that they have no incentive to lie. Since high-type members will share the cost of provision equally, the incentive constraint of the one period static game is

$$\begin{aligned}&p^{2}\left( b_{3}-C/3\right) r_{3}^{0}+2p(1-p)\left( b_{2}-C/2\right) r_{2}^{0}+(1-p)^{2}\left( b_{1}-C\right) \\&\qquad r_{1}^{0}\ge p^{2}b_{3}r_{2} ^{0}+2p(1-p)b_{2}r_{1}^{0}+(1-p)^{2}b_{1}r_{0}^{0} \end{aligned}$$

The incentive constraint of the dynamic game is

$$\begin{aligned} U_{j}^{u}(1;1)\equiv & {} p^{2}\left( b_{3}-C/3\right) \left( r_{3}^{I} +\sigma _{3}r_{3}^{II}+\left( 1-\sigma _{3}\right) U_{j}^{d*}\left( 1;1\right) \right) \\&+\,2p(1-p)\left( b_{2}-C/2\right) \left( r_{2}^{I}+\sigma _{2}r_{2} ^{II}+\left( 1-\sigma _{2}\right) U_{j}^{d*}\left( 1;1\right) \right) \\&+\,(1-p)^{2}\left( \left( b_{1}-C\right) \left( r_{1}^{I}+\sigma _{1} r_{1}^{II}\right) +\left( 1-\sigma _{1}\right) U_{j}^{d*}\left( 1;1\right) \right) \\\ge & {} p^{2}\left( b_{3}\left( r_{2}^{I}+\sigma _{2}r_{2}^{II}\right) +\left( 1-\sigma _{2}\right) U_{j}^{d*}\left( 1;1\right) \right) \\&+\,2p(1-p)\left( b_{2}\left( r_{1}^{I}+\sigma _{1}r_{1}^{II}\right) +\left( 1-\sigma _{1}\right) U_{j}^{d*}\left( 1;1\right) \right) \equiv U_{j}^{u}(0;1) \end{aligned}$$

Let us assume here too that the potential social surplus is independent of how agents are matched, or, equivalently. that the potential social surplus per high-type member is the same irrespective of the composition of the triplet. Moreover, let us assume that its value is equal to 1, so $( b_{1}-C) =( b_{2}-C/2) =( b_{3}-C/3) =1$. An easy way to get an intuition of the above is the following. Consider 18 agents, 12 of them of high type. In all the following three alternatives, the social surplus is equal to 12: six groups with two high types and one low type each; four groups with three high types each and two groups with three low type each; three groups with one high type and two low types each and three groups with three high types. In the numerical example below, let us assume that $p=2/3$.

It is immediate to verify that in the static one-period game, efficiency, i.e. $r_{k}^{0}=1,k=1,2,3$ and $r_{0}^{0}=0$, cannot be obtained as soon as $C>\frac{3}{10}$. In fact, after substituting all the values specified above, the incentive constraint becomes $1\ge \frac{10}{27}C+\frac{8}{9}$.

Observe that the incentive constraint of the dynamic game is efficiently satisfied by setting $r_{k}^{h}=1,k=1,2,3$ and $r_{0}^{h}=0$, for $h=I,II$, and $\sigma _{0}=\sigma _{2}=\sigma _{3}=1,\sigma _{1}=0$ when $C=\frac{3}{7}>\frac{3}{10}$. Notice that, in this way, efficiency can not only be obtained in the first period and in surviving groups but also in the downstream mechanism (in fact, here it is $\Pi =\frac{1}{3}$, so efficiency can be obtained as long as $C\le \frac{12}{7}$: in this case, the static incentive constraint becomes $1\ge \frac{7}{27}C+\frac{5}{9}$). After substituting all the values specified above, the dynamic incentive constraint is satisfied and the value of both sides, which represents the overall payoff to a high-type member, is equal to 2 (1 per period, which corresponds to efficiency). As to the one-period static mechanism, when $C=\frac{3}{7}$, the incentive constraint is optimally satisfied by setting $r_{0}^{0}=0,r_{2}^{0}=r_{3} ^{0}=1,r_{1}^{0}=\frac{8}{9}.$ In fact, lowering $r_{1}^{0}$ is the most cost-effective way to relax the incentive constraint (this corresponds to intuition and can be easily proved), and furthermore the incentive constraint is satisfied as an equality. The equilibrium expected payoff of a high type in the static one-period mechanism (corresponding to the lhs of the incentive constraint) is $\frac{80}{81}<1$. This is enough to state that also with more than two agents per group there exists an equilibrium with selective group stabilization that improves upon the repetition of the static one-period game. $\square$

Case (VI). When the potential social surplus depends on how the types are matched in pairs.

Under the assumption that $0<2b( 1,1) -C<2[ b( 1,0) -C]$, we prove the following statement:

• (a) there is a non-negligible set of parameter values in which the bad equilibrium is the only equilibrium;

• (b) there is a non-negligible set of parameter values in which a good equilibrium exists and the social surplus is strictly greater than that associated with the bad equilibrium.

Proof. Note that in our problem all choice variables enter linearly in both the objective function and the constraints. Moreover, we only need to consider the incentive constraint of a high-type agent. Therefore, we can conveniently use the notion of the cost of a choice variable, which in our case is independent of the variable’s value. Recall that the marginal cost of the choice variable $x_{i}$ is equal to the marginal change in the objective when the constraint is marginally relaxed by using variable $x_{i}$. This cost can be negative if relaxing the constraint increases the objective function. To ease notation, let us denote a high-type agent’s net benefit from the public good when the mate is of the high type by h under truthtelling and by H if he lies and his mate reports truthfully. Therefore $h\equiv b\left( 1,1\right) -C/2$ and $H\equiv h+C/2$. Moreover, a high-type agent’s net benefit from the public good when he reports truthfully and his mate is of the low type is denoted by z. Therefore $z\equiv b(1,0)-C.$

Part (a) We will prove the existence of a non-negligible set of parameter values for which the statement holds. Normalize the payoffs so that z $=1,$ without loss of generality, and consider the case $0<h<H<1$. Using Facts 1 and 2 (see Appendix A), we see that for an upstream principal, setting $\sigma _{0}=1$ and $\sigma _{2}=0$ is dominant, since it encourages truthtelling and is efficient (it is easily proved that $U_{{}}^{d*}( 1;1) \ge h$ at equilibrium). Now suppose, for the sake of argument, that a good equilibrium always exists, that is, $\bar{\sigma }_{1}<1$ for any set of values of the parameter. The proof consists in showing that a single principal can profitably deviate by setting $\sigma _{1}=1$. Let us consider the cost of the variables for encouraging truthtelling. The cost of variable $r_{1}^{I}$ is $-\frac{1-p}{1-p-pH},$ that of $r_{1}^{II}$ is $-\frac{(1-p)\sigma _{1}}{( (1-p)-pH) \sigma _{1}},$ and that of $\sigma _{1}$ is $-\frac{(1-p)(r_{1}^{II}-U_{{}}^{d*}( 1;1) )}{(1-p)(r_{1}^{II}-U_{{}}^{d*}( 1;1) )-p( H-U_{{} }^{d*}( 1;1) ) }$.

Now, set $h=1/3,$ $H=4/5,$ and $p=.685.$ Then $U_{j}^{d*}( 1;1)$ is a monotone function of $\bar{\sigma }_{1}$ with range $U_{{} }^{d*}( 1;1) \in [ 1/3,0.418]$.²¹ Moreover, it is straightforward to verify that the cost of $\sigma _{1}$ is increasing in $r_{1}^{II}.$ Next, suppose the other principals set $\bar{\sigma }_{1}<1.$ Then the deviator optimally sets $r_{1}^{I}=1,$ $\sigma _{0}=1,$ $\sigma _{1}=1,$ $\sigma _{2}=0,$ and $r_{1}^{II}=0.96$, since in this way the incentive compatibility constraint is satisfied at the minimum cost. In fact, then $-\frac{(1-p)}{(1-p)-p( H) }=1.352$ and $-\frac{(1-p)(r_{1}^{2}-U_{{}}^{d*}( 1;1) )}{(1-p)(r_{1} ^{2}-U_{{}}^{d*}( 1;1) )-p( H-U_{{}}^{d*}( 1;1) ) }$ > 1.352 for all admissible values of $U_{{}} ^{d}( 1;1) ,$ which implies that a single principal profitably deviates from $\sigma _{1}=\bar{\sigma }_{1}<1$: the cost of the instrument $\sigma _{1}$ is greater than those of $r_{1}^{I}$ and $r_{1}^{II}$. Notice that the choice to set $r_{1}^{I}=1$ and $r_{1}^{II}=0.96$ puts to a minimum the costs of $\sigma _{1}.$ We conclude that the good equilibrium cannot exist. By continuity, this result also holds for quadruples of parameters sufficiently near to that we have considered. QED.

Part (b) Consider the case $z<H,$ and $h<H.$ We normalize the payoffs so that $H=1,$ without loss of generality. For an upstream principal, setting $\sigma _{0}=1$ is dominant since it encourages truthtelling without costs. Suppose there exists a set of parameters for which setting $\sigma _{2}=1$ is also dominant. Then if at equilibrium $\bar{\sigma }_{1}<1$, it is $\Pi =1/2.$ Now, the cost of $\sigma _{1}$ is $-\frac{(1-p)(r_{1}^{II}z-U_{{} }^{d*}( 1;1) )}{(1-p)(r_{1}^{II}z-U_{{}}^{d*}( 1;1) )-p( r_{1}^{II}-U_{{}}^{d*}( 1;1) ) }$, and those of $r_{1}^{I}$ and $r_{1}^{II\text { }}$ are $-\frac{(1-p)z}{(1-p)z-p}$ and $-\frac{(1-p)z\sigma _{1}}{( (1-p)z-p) \sigma _{1}}$, respectively. Finally note that the cost of $\sigma _{1}$ is increasing in $r_{1}^{II}$ and for $r_{1}^{II}=1$ is lower than the cost of the other variables. Therefore, a principal determines the optimal value of $\sigma _{1},$ without considering the behavior of the others, simply believing $\Pi =1/2$. Then, a simple proof that a good equilibrium exists is to show that parameters exist for which at equilibrium $\sigma _{0}=\sigma _{2}=1$. For $\Pi =1/2$, one necessarily has $U_{{}}^{d*}( 1;1) =\Pi r_{1}^{d}=\frac{1}{2}\frac{h}{1-z},$ with $\frac{h}{1-z}<1$: otherwise the assumption $z<H$ is violated. Moreover, setting $\sigma _{2}=1$ is dominant iff $\frac{\partial (U_{{}}^{u}(1;1)-U_{{}}^{u}(0;1))}{\partial \sigma _{2}}>0,$ that is, assuming $\Pi =1/2$, iff $p[ h( r_{1}^{I}+r_{1}^{II}) -\frac{1}{2}\frac{h}{1-z}] >0,$ which is satisfied when $r_{1} ^{I}+r_{1}^{II}=2$ iff $z<\frac{3}{4}.$ Therefore, a mobility equilibrium exists, for instance, when $z=1/2$, $h=1/3$ and $H=1$. The surplus of this equilibrium is larger than twice the surplus of the one-period static version of the game since the cost of satisfying the incentive constraint is lower than in the bad equilibrium. $\square$