The role of aggregate information in a binary threshold game

Abstract

We analyze the problem of coordination failure in the presence of imperfect information in the context of a binary-action sequential game with a tipping point. An information structure summarizes what each agent can observe before making her decision. Focusing on information structures where only “aggregate information” from past history can be observed, we characterize information structures that can lead to various (efficient and inefficient) Nash equilibria. When individual decision making can be rationalized using a process of iterative dominance (Moulin, Econometrica 47:1337–1351, 1979), we derive a necessary and sufficient condition on information structures under which a unique and efficient dominance solvable equilibrium outcome is obtained. Our results suggest that if sufficient (and not necessarily perfect) information is available, coordination failure can be overcome without centralized intervention.

Appendix

Proof of Theorem 1

($\mathbf {\Longrightarrow }$) Let $\tau \in \left\{ 0,\ldots ,(n-\lambda -1)\right\} .$ Suppose there is a PSNE a with exactly $(\lambda +\tau )$ agents participating: $\sum _{i\in N}b_{i}(a)=\lambda +\tau$. Since the set $N_{1}=\{\left( n-\lambda -\tau \right)$, $\ldots$, $n\}$ has $\left( \lambda +\tau +1\right)$ agents, $\sum _{i\in N}b_{i}(a)=\lambda +\tau$ implies that there exists $j\in N_{1}$ such that $b_{j}(a)=0$. Let $\tilde{j}$ be the “last” agent in $N_{1}$ such that $b_{j}(a)=0$, i.e., let $\tilde{j}=\max \{j:j\in N_{1}$ and $b_{j}(a)=0\}$. Given such $\tilde{j}$, we have

$$\begin{aligned} b_{j}(a)= & {} 1\text { for all }j>\tilde{j}\text {, and} \end{aligned}$$

(2)

$$\begin{aligned} \tilde{j}\ge & {} \left( n-\lambda -\tau \right) . \end{aligned}$$

(3)

Consider a profile $\widetilde{a}=(\widetilde{a}_{\widetilde{j}},a_{-\widetilde{j}})$ that differs from a only in $\tilde{j}$’s strategy such that the relevant coordinate of $\tilde{j}$ under a is changed from $b_{\widetilde{j}}(a)=0$ to $b_{\widetilde{j}}(\widetilde{a})=1$. Since a is a PSNE, this unilateral deviation must decrease $\tilde{j}$’s utility to less than zero. In other words at least $\left( \tau +2\right)$ agents in $\left\{ (\tilde{j}+1),\ldots ,n\right\}$ who join under a (with $b_{j}(a)=1$) must no longer join under $\widetilde{a}$ (with $b_{j}(\widetilde{a})=0)$. Define $J=\left\{ j_{1},\ldots ,j_{\tau +2}\right\}$ with $j_{i}<j_{i+1}$ for all $i\le \tau +1$ be the set of the first $\left( \tau +2\right)$ agents such that $b_{j}(a)=1$ and $b_{j}(\widetilde{a})=0$ for $j\in J$ (i.e., the first $\left( \tau +2\right)$ agents switching from joining to not joining after $\tilde{j}$’s deviation). This implies that there are at least $\tau +2$ agents moving after $\tilde{j}$ and hence

$$\begin{aligned} \tilde{j}\le n-\tau -2. \end{aligned}$$

(4)

Inequalities (3) and (4) together imply that

$$\begin{aligned} \left( n-\lambda -\tau +1\right) \le j_{1}\le (n-\tau -1). \end{aligned}$$

(5)

Since $\tilde{j}$ has made a unilateral deviation and $j_{1}\in J$ is such that $b_{j_{1}}(a)=1$ and $b_{j_{1}}(\widetilde{a})=0$, it must be that $j_{1}$’s information covers $\tilde{j}$, i.e., $\kappa (j_{1})\ge \tilde{j}$. By the choice of J, since $j_{1}$ and $j_{2}$ are the first two agents switching from joining under a to not joining under $\widetilde{a}$, and that all agents j with $j_{1}<j<j_{2}$ (if any) have $b_{j}(a)=b_{j}( \widetilde{a})=1$ by definition of $j_{1}$ and $j_{2}$, it follows that $j_{2}$’s information does not cover $j_{1}$, i.e., $\kappa (j_{2})<j_{1}$.²¹ Since $j_{2}\ge j_{1}+1$, Assumption 3 and $\kappa (j_{2})<j_{1}$ then jointly imply that

$$\begin{aligned} \left( j_{1}+1\right) \text {'s information does not cover }j_{1}\text {, i.e., }\kappa (j_{1}+1)<j_{1}. \end{aligned}$$

(6)

Finally, by our choice of J, since $j_{1}$ is the first agent switching after $\tilde{j}$’s deviation, the number of agents who move after agent $\left( j_{1}+1\right)$ and who cover $j_{1}$ but not $j_{2}$ can be at most $\left( n-\left( j_{1}+1\right) \right) -\tau =n-j_{1}-\tau -1$, i.e., $\left| \{j:\kappa (j)=j_{1}\}\right| \le n-j_{1}-\tau -1$. The reason is that the agents in $\{j:\kappa (j)=j_{1}\}$, when this set is non-empty, receive the same information signal under a and $\widetilde{a}$ and will have the same move under both profiles and cannot belong to J. Since $\lambda \ge 2$, the number of agents in $\{j:\kappa (j)=j_{1}\}$ cannot be too large for there to be at least $\left( \tau +2\right)$ agents in J. It is easy to check that if $|\{j:\kappa (j)=j_{1}\}|>n-j_{1}-\tau -1$ then the number of agents who remain in the set $\{j_{1},j_{1}+1,j_{1}+2,\ldots ,n\}\backslash \{j:\kappa (j)=j_{1}\}$ is strictly less than $n-j_{1}+1-(n-j_{1}-\tau -1)=\tau +2.$ Thus,

$$\begin{aligned} |\{j:\kappa (j)=j_{1}\}|\le n-j_{1}-\tau -1. \end{aligned}$$

(7)

Using (5), (6), (7) and letting $j_{1}$ be the $j^{*}$ in Theorem 1 completes the proof.²²

($\Longleftarrow$) Let $\tau \in \{0,\ldots ,n-\lambda -1\}$. By our hypothesis we have $j^{*}\ge n-\lambda -\tau +1$ with $\kappa (j^{*})\ge n-\lambda -\tau$, $\kappa (j^{*}+1)<j^{*}$, and $n-j^{*}-\tau -1\ge |\{j:\kappa (j)=j^{*}\}|$. We will construct a PSNE a such that exactly the last $\left( \lambda +\tau \right)$ agents participate on the path of play.

Fig. 3

Equilibrium strategy profile a with exactly $\left( \lambda +\tau \right)$ participations

Consider a strategy profile a (see Fig. 3) such that:

• For all $j\in J_{1}=\{1,\ldots ,n-\lambda -\tau \}$, $a_{j}\left( l\right) =0$ for all $l\in \{1,\ldots ,\kappa (j)+1\}$, i.e., all of the first $\left( n-\lambda -\tau \right)$ agents never participate regardless of what they observe.

• For all $j\in J_{2}=\{n-\lambda -\tau +1,\ldots ,j^{*}-1\}$, $a_{j}\left( l\right) =1$ for all $l\in \{1,\ldots ,\kappa (j)+1\}$, i.e., all agents moving after the agents in $J_{1}$ but before agent $j^{*}$ choose to participate regardless of what they observe.

• For all $j\in J_{3}=\{j^{*},\ldots ,n\}$, $a_{j}\left( l\right) =1$ iff the $l{\text {th}}$ coordinate of j is on the path of play, i.e., agents moving after agent $\left( j^{*}-1\right)$ choose to join only in those coordinates that are relevant under a (i.e., such an agent has “1” in the single coordinate of her strategy that is on the path of play and has “0” elsewhere).

By construction, on the path of play only the last $\left( \lambda +\tau \right) =\left| J_{2}\cup J_{3}\right|$ agents choose to join under a and hence we have $\sum _{i\in N}b_{i}(a)=\lambda +\tau$. Given the payoff in (1) and Assumption 1, the strategies for agents in $J_{1},J_{2}$ and $J_{3}$ then imply

$$\begin{aligned} u_{j}\left( a\right) =\left\{ \begin{array}{l} 0\text {, if }j\le n-\lambda -\tau , \\ f_{j}\left( \lambda +\tau \right) >0\text {, if }j\ge n-\lambda -\tau +1. \end{array} \right. \end{aligned}$$

(8)

The remainder of the proof establishes that the above profile a is a PSNE.

If $j\in J_{2}\cup J_{3}$, then $j\ge n-\lambda -\tau +1$ and using (8), such a j’s unilateral deviation either gives exactly $f_{j}\left( \lambda +\tau \right)$ or decreases j’s payoff from $f_{j}\left( \lambda +\tau \right)$ to 0. Hence j is playing a best response at a.

Next, consider a unilateral deviation of a relevant coordinate by $j\in J_{1}$ with a resulting profile $a^{\prime }=(a_{j}^{\prime },-a_{j}^{\prime })=(a_{j}^{\prime },-a_{j})$. Given $\kappa \left( j^{*}+1\right) <j^{*}$, agents $j^{*}$ and $\left( j^{*}+1\right)$ now both observe a signal at $a^{\prime }$ indicating that one more agent (than at a) has joined. According to $a^{\prime }$, since $a_{-j}=a_{-j}^{\prime }$, both $j^{*}$ and $j^{*}+1$ now switch from “join” ($b_{j^{*}}(a)=b_{j^{*}+1}(a)=1$) to “not join” ($b_{j^{*}}(a^{\prime })=b_{j^{*}+1}(a^{\prime })=0$) after j’s deviation.²³ Now consider agents $j^{\prime }\ge j^{*}+2$ (i.e., $j^{\prime }\in J_{3}\backslash \left\{ j^{*},j^{*}+1\right\}$). There are three possible cases:

• If $\kappa (j^{\prime })<j^{*}$, $j^{\prime }$ now (like agents $j^{*}$ and $\left( j^{*}+1\right)$) gets a signal under $a^{\prime }$ indicating that one more agent has joined than under a. Hence, by construction $j^{\prime }$ will switch from joining to not joining.

• If $\kappa (j^{\prime })=j^{*}$, $j^{\prime }$’s signal will not change and agent $j^{\prime }$ will, as before, participate.²⁴

• If $\kappa (j^{\prime })>j^{*}$, then $j^{\prime }$ gets a signal indicating at least one less agent has joined under $a^{\prime }$ than under a.²⁵ According to $a_{j^{\prime }}=a_{j^{\prime }}^{\prime }$, $j^{\prime }$ will switch from joining to not joining.

To summarize, after the unilateral deviation from $j\in J_{1}$, the set of agents who choose to join on the path of play is given by $\left\{ j\right\} \cup J_{2}\cup \{j:\kappa (j)=j^{*}\}$. It follows that

$$\begin{aligned} \sum \limits _{i\in N}b_{i}(a^{\prime })=1+\left( j^{*}-\left( n-\lambda -\tau +1\right) \right) +|\{j:\kappa (j)=j^{*}\}|\le \lambda -1. \end{aligned}$$

This implies that j’s deviation is not profitable: j’s payoff from her deviation on her relevant coordinate in a is at most $f_{j}\left( \lambda -1\right)$ which is less than 0 (Assumption 1). $\square$

1.1 Proof of Theorem 2

We will introduce some notation, define some key concepts, and prove preliminary lemmas that will be used to prove Theorem 2. The first set of lemmas (Lemmas 1 to 4) holds for all $\mathcal {G}_{0}$ whether or not $\mathcal {G}_{0}$ is dominance solvable and is unrelated to the particular information structure of $\mathcal {G}_{0}$. These lemmas provide useful insights into the iterative reduction process of dominated strategies. The second set of lemmas (Lemmas 5 and 6) relates the existence of information chains in $\mathcal {G}_{0}$ to this reduction process and provides results that represent crucial parts of the proof of Theorem 2.

For convenience, we will abuse notation and use $a_{j}\in \mathcal {G}_{h}$ to indicate that $a_{j}$ is possible in the game $\mathcal {G}_{h}$, i.e., $a_{j}$ has not been eliminated and is a strategy of j in game $\mathcal {G}_{h}$. Similarly $a\in \mathcal {G}_{h}$, $a_{-j}\in \mathcal {G}_{h}$, and $a_{j}(l)\in \mathcal {G}_{h}$ will indicate respectively that the profile a, the contingency $a_{-j}$ and the conditional action $a_{j}\left( l\right)$ are possible in $\mathcal {G}_{h}$.

Definition 1

For $\mathcal {G}_{h}\in \{\mathcal {G}_{0},\ldots ,\mathcal {G}_{M}\}$, a conditional action $a_{j}\left( l\right) \in \mathcal {G}_{h}$ is a best response conditional action (BRCA) in $\mathcal {G}_{h}$ iff there is $a_{-j}\in \mathcal {G}_{h}$ such that the $l^{th}$-coordinate of j’s strategy is on the path of play and $a_{j}$ is a best response to the contingency $a_{-j}$ in $\mathcal {G}_{h}$.

For the games in $\left\{ \mathcal {G}_{1},\ldots ,\mathcal {G}_{M}\right\}$, as dominated strategies are iteratively eliminated, certain paths of play occurring in earlier games may not appear in later games. As this happens, some coordinates of an agent’s strategy become irrelevant, i.e., no path of play that is possible in the game passes through these coordinates. It is important to identify which coordinates and which paths persist. The next lemma shows that an undominated strategy is closely related to BRCAs and these coordinates of individual strategies persist from game to game. Lemma 1 (i) shows that if some coordinates of a strategy in a game consist of BRCAs then these BRCAs survive in that some undominated strategy in that game has these BRCAs in its coordinates, while Lemma 1 (ii) shows that for a strategy to be undominated in a game, all coordinates of that strategy must either be a BRCA or irrelevant in that game.²⁶

Lemma 1

(Persistence) Let $\mathcal {G}_{h}\in \{\mathcal {G}_{0},\ldots ,\mathcal {G}_{M-1}\}$ and $j\in N.$ (i) For $a_{j}\in \mathcal {G}_{h}$ and $L\subseteq \{1,\ldots ,\kappa (j)+1\}$, if for all$\ l\in L$, $a_{j}(l)\$is a BRCA, then there exists $a_{j}^{\prime }\in \mathcal {G}_{h+1}$ such that $a_{j}^{\prime }(l)=a_{j}(l)$ for all $l\in L$. (ii) If $a_{j}^{*}$ is undominated in $\mathcal {G}_{h}$, then $a_{j}^{*}$ is such that for all $l\in \{1,\ldots ,\kappa (j)+1\}$, either $a_{j}^{*}(l)$ is a BRCA in $\mathcal {G}_{h}$ or the $l^{th}$ coordinate of j’s strategy is irrelevant in $\mathcal {G}_{h}$.

Proof

(i) Let $a_{j}\in \mathcal {G}_{h}$ be such that $a_{j}(l)$ is a BRCA in $\mathcal {G}_{h}$ for all $l\in L$. By the definition of a BRCA, for each such l, there is a strategy profile $a^{l}=\left( a_{j},a_{-j}^{l}\right) \in \mathcal {G}_{h}$ such that the $l{\text {th}}$ coordinate of j is on the path of play and $a_{j}$ is a best response for $a_{-j}^{l}$ in $\mathcal {G}_{h}$. If $a_{j}$ itself is undominated, then $a_{j}\in \mathcal {G}_{h+1}$ and we are done. If $a_{j}$ is dominated, then by finiteness of $A_{j}$ and transitivity of the dominance relation, there is an undominated strategy $a_{j}^{\prime }\in \mathcal {G}_{h}$ that dominates $a_{j}$. This implies that for each of the contingencies $a_{-j}^{l}\in \mathcal {G}_{h}$, the payoff of j from $a_{j}^{\prime }$ is at least as large as that from $a_{j}$. But, since $a_{j}$ is a best response to $a_{-j}^{l}\in \mathcal {G}_{h}$, we must have $a_{j}^{\prime }(l)=a_{j}(l)$ for all $l\in L$. Finally, as $a_{j}^{\prime }$ is undominated in $\mathcal {G}_{h}$, $a_{j}^{\prime }\in \mathcal {G}_{h+1}$.

(ii) Assume to the contrary that there exists an undominated strategy $a_{j}^{*}\in \mathcal {G}_{h}$ and a non-empty set of coordinates $\widetilde{L}$ of $a_{j}^{*}$ where $\widetilde{L}=\{l:l\in \{1,\ldots ,\kappa (j)+1\}$, $a_{j}^{*}(l)$ is relevant and $a_{j}^{*}(l)$ is not a BRCA in $\mathcal {G}_{h}\}$. Replace of all the coordinates of $a_{j}^{*}$ in $\widetilde{L}$ with the corresponding BRCAs in $\mathcal {G}_{h}$ to construct a new strategy $a_{j}^{\prime }$ which agrees with $a_{j}^{*}$ in all coordinates other than those in $\widetilde{L}$. By construction, the coordinates of $a_{j}^{\prime }$ are either irrelevant or are BRCAs in $\mathcal {G}_{h}$ and if this $a_{j}^{\prime }$ exists in $\mathcal {G}_{h}$, it would dominate $a_{j}^{*}$. Since all j’s contingencies in $\mathcal {G}_{h}$ are also contingencies for j in $\mathcal {G}_{h-1}$ and since every relevant coordinate of $a_{j}^{\prime }$ is a BRCA in $\mathcal {G}_{h}$, it must be the case that these coordinates must also be BRCAs in $\mathcal {G}_{h-1}$. Denoting all these relevant coordinates of $a_{j}^{\prime }$ which are BRCAs by L, Lemma 1 (i) (applied to $\mathcal {G}_{h-1}$) implies that there exists $a_{j}^{\prime \prime }\in \mathcal {G}_{h}$ such that each of $a_{j}^{\prime \prime }$’s coordinates either coincides with that of $a_{j}^{\prime }$ or the coordinate is irrelevant in $\mathcal {G}_{h}$. Thus, since $a_{j}^{\prime \prime }$ “generates exactly the same outcomes as” $a_{j}^{\prime }$ in $\mathcal {G}_{h}$, it dominates $a_{j}^{*}$ contradicting our hypothesis that $a_{j}^{*}$ is undominated in $\mathcal {G}_{h}$. $\square$

In our setting, another consequence of eliminating dominated strategies is that certain relevant coordinates of an agent’s strategies become “fixed” in that for these coordinates all the agent’s strategies take on the same value. Once this has happened in a game, these coordinates remain fixed in subsequent games. We will be particularly interested in cases where the possibility of non-participation is eliminated by dominance and some coordinate of an agent’s strategy is reduced to 1 in a game and is fixed at 1 in all subsequent games. Moreover, if all relevant coordinates greater than or equal to some coordinate are also fixed at one, we will say that this particular coordinate of the agent’s strategy has been strictly reduced to 1 and the coordinate is strictly fixed at 1.

Definition 2

Let $\mathcal {G}_{h}\in \{\mathcal {G}_{0},\ldots ,\mathcal {G}_{M-1}\}$, $j\in N$ and $l\in \left\{ 1,\ldots ,\kappa (j)+1\right\}$. The conditional action $a_{j}\left( l\right)$ is fixed at 1 in $\mathcal {G}_{h+1}$ iff for all $a_{j}\in \mathcal {G}_{h+1},$ $a_{j}(l)=1$. The conditional action $a_{j}\left( l\right)$ is strictly fixed at 1 in $\mathcal {G}_{h+1}$ iff (i) $a_{j}\left( l\right) =1$ is fixed at 1 in $\mathcal {G}_{h+1}$ and (ii) for all $s\in \{1,\ldots ,\kappa (j)+1-l\}$ either $a_{j}\left( l+s\right) =1$ or the $\left( l+s\right) {\text {th}}$ coordinate of j’s strategy is irrelevant in $\mathcal {G}_{h+1}$. If $a_{j}\left( l\right)$ is fixed at 1 in $\mathcal {G}_{h+1}$ and $a_{j}\left( l\right)$ is not fixed at 1 in $\mathcal {G}_{h}$ we will say that the $l{\text {th}}$ coordinate of j’s strategy is reduced to 1 in game $\mathcal {G}_{h}$. If $a_{j}\left( l\right)$ is strictly fixed at 1 in $\mathcal {G}_{h+1}$ and $a_{j}\left( l\right)$ is not strictly fixed at 1 in $\mathcal {G}_{h}$ we will say that the $l{\text {th}}$ coordinate of j’s strategy is strictly reduced to 1 in $\mathcal {G}_{h}$.

Remark 3

As per the terminology we have adopted in Definition 2, a coordinate is “not fixed” in the game in which it is “reduced” and only becomes fixed in subsequent games and $a_{j}\left( l\right) =1$ being strictly fixed at 1 in $\mathcal {G}_{h+1}$ implies that $a_{j}\left( l\right) =1$ is fixed at 1 in $\mathcal {G}_{h+1}$. It is also clear that $a_{j}\left( l\right) =1$ being fixed at 1 in $\mathcal {G}_{h}$ implies that $a_{j}\left( l\right)$ is fixed at 1 in $\mathcal {G}_{h^{\prime }}$ for all $h^{\prime }\ge h$. Moreover, since an irrelevant coordinate of a strategy in a game is irrelevant in all subsequent games, it also follows that if $a_{j}\left( l\right)$ is strictly fixed at 1 in $\mathcal {G}_{h}$ then $a_{j}\left( l\right)$ is strictly fixed at 1 in $\mathcal {G} _{h^{\prime }}$ for all $h^{\prime }\ge h$.

We next introduce notation for a particular set of “participating” strategies $\left( \mathcal {P}\right)$ and a set of “non-participating” strategies $\left( \mathcal {NP}\right)$. These are simply book-keeping devices that we shall use to describe the process of coordinates being “reduced” to 1. Here, $\mathcal {P}$ is the set of strategy profiles where all the agents choose to participate whenever their information signal shows maximal previous participations. The largest coordinate of each agent’s strategy hence takes the value 1 in $\mathcal {P}$.

$$\begin{aligned} \mathcal {P}=\left\{ a\in \mathcal {G}_{0}:a_{j}\left( \kappa (j)+1\right) =1 \text { for all }j\in N\right\} \text {.} \end{aligned}$$

On the other hand, $\mathcal {NP}\left( r\right)$ is the set of strategy profiles where agents decide not to participate upon observing less than ($r-1$) participations. Here, all the coordinates of all the agents’ strategies less than or equal to the $r{\text {th}}$-coordinate are zero.

$$\begin{aligned} \mathcal {NP}\left( r\right) =\left\{ a\in \mathcal {G}_{0}:\text {For all } j\in N\text {, }a_{j}\left( l\right) =0\text { for all }l\le \min \left\{ \kappa (j)+1,r\right\} ,r\in \mathbb {N} \right\} \text {.} \end{aligned}$$

By varying r, we obtain a nested set of subsets of $\mathcal {NP}\left( r\right)$ with $\mathcal {NP}\left( r-1\right) \supseteq \mathcal {NP}\left( r\right)$. We will use the terminology $\mathcal {P}$ in $\mathcal {G}_{h}$ ($\mathcal {NP}\left( r\right) \$in $\mathcal {G}_{h}$) to indicate the strategy profiles in $\mathcal {P}$ (the strategy profiles in $\mathcal {NP}\left( r\right)$) that survive the elimination process from $\mathcal {G}_{0}$ to $\mathcal {G}_{h}$.²⁷ We will also use $a_{j}\in \mathcal {P}$ ($a_{j}\in \mathcal {NP}\left( r\right) )$ to indicate a strategy of j with $a_{j}\left( \kappa (j)+1\right) =1$ (with $a_{j}\left( l\right) =0$ for all $l\le \min \left\{ \kappa (j)+1,r\right\} )$.

Lemma 2

(Unanimous Participation) Let $\mathcal {G}_{h}\in \{\mathcal {G}_{0},\ldots ,\mathcal {G}_{M}\}$. Then $\mathcal {P\ne \varnothing }$ in $\mathcal {G}_{h}$ and every strategy profile $a\in \mathcal {P}$ is a PSNE in $\mathcal {G}_{h}$ with $\sum _{N}b_{j}(a)=n$.

Proof

In $\mathcal {G}_{0}$ since no strategies have been eliminated, $\mathcal {P\ne \varnothing }$. In addition, for all $j\in N$, $a_{j}$ is a best response to $a_{-j}$ for $\left( a_{j},a_{-j}\right) \in \mathcal {P}$ in $\mathcal {G}_{0}$ since any unilateral deviation on the path of play with $a_{j}(\kappa (j)+1)=0$ will reduce agent j’s payoff from positive to zero. Thus, by the persistence Lemma 1, there is then an undominated strategy $a_{j}$ with $a_{j}\left( \kappa (j)+1\right) =1$ in $\mathcal {G}_{0}$. Since this is true for all j, we have $\mathcal {P\ne \varnothing }$ in $\mathcal {G}_{1}$. A similar argument establishes the result inductively. $\square$

Lemma 2 shows that in all games $\mathcal {G}_{h}$ there always exists an (efficient) PSNEO in which each agent receives a payoff of $f_{j}(n)$ —hence the efficient PSNEO will never be eliminated in the reduction process.²⁸ An immediate consequence of Lemma 2 is Corollary 6, which shows that for dominance solvability it is necessary that after the process of iterative dominance, the irreducible game $\mathcal {G}_{M}$ should satisfy $\mathcal {NP}\left( 1\right) =\varnothing .$

Corollary 6

If $\mathcal {NP}\left( 1\right) \ne \varnothing$ in $\mathcal {G}_{M}$ then $\mathcal {G}_{M}$ has at least two outcomes, one in which $\sum _{N}b_{j}=n$ and another in which $\sum _{N}b_{j}=0$ and hence $\mathcal {G}_{0}$ is not dominance solvable.

The following special type of a pre-requisite will play an important role in the sequel:

Definition 3

The pre-requisite of the $l{\text {th}}$ coordinate of j’s strategy is satisfiedexactly for a contingency $a_{-j}$ in the profile $(a_{j},a_{-j})$ if $\sum _{i=1}^{l-1}b_{i}(a_{j},a_{-j})=l-1$, where $b(a_{j},a_{-j})$ is the action profile induced by $(a_{j},a_{-j})$. The contingency $a_{-j}$ is called an exact contingency for the $l{\text {th}}$ coordinate of j’s strategy.

Under an exact contingency $a_{-j}$, the information that j receives represents a full and complete aggregate report of what actually occurs and this report is generated by exactly the first $(l-1)$ individuals participating. Our next lemma shows that for all $r\le \kappa (n)+1$, if $\mathcal {NP}\left( r\right) \ne \varnothing$ in $\mathcal {G}_{h}$, then for each agent, for every (possible) coordinate in the agent’s strategy that is no larger than r,  there exists an exact contingency in $\mathcal {G}_{h}$ for such a coordinate of the agent to be on the path of play.

Lemma 3

(Exact Contingency) Let $\mathcal {G}_{h}\in \{\mathcal {G}_{0},\ldots ,\mathcal {G}_{M}\}$ be such that $\mathcal {NP}\left( r\right) \ne \varnothing$ in $\mathcal {G}_{h}$ for some $r\in \{1,\ldots ,\kappa (n)+1\}$. Then for $j\in N,l\in \{1,\ldots ,\min \left\{ \kappa (j)+1,r\right\} \}$, there is a strategy profile $a^{*}\in \mathcal {G}_{h}$, $a^{*}$ depending on l, such that the $l{\text {th}}$ coordinate of j’s strategy is on the path of play under $a^{*}$, and $\sum _{i=1}^{l-1}b_{i}(a^{*})=\sum _{N}b_{i}(a^{*})=l-1.$

Proof

The proof is done by construction. First, $\mathcal {P\ne \varnothing }$ in $\mathcal {G}_{h}$ (Lemma 2) implies that for each $i\in N$, there is a strategy $\hat{a}_{i}\in \mathcal {G}_{h}$ with $\hat{a}_{i}\left( \kappa (i)+1\right) =1$. Since by the hypothesis $\mathcal {NP}\left( r\right) \ne \varnothing$ in $\mathcal {G}_{h}$, there is also a strategy $\check{a}_{i}\in \mathcal {NP}\left( r\right)$ for all $i\in N$ such that $\check{a}_{i}\left( 1\right) =\cdots =\check{a}_{i}\left( \min \left\{ \kappa (i)+1,r\right\} \right) =0$. Let $j\in N$ and $l\in \left\{ 1,\ldots ,\kappa (j)+1\right\}$ with $1\le l\le r$ and consider a strategy profile $a^{*}$ where (see Fig. 4)

$$\begin{aligned} \text {for }i\in & {} \left\{ 1,\ldots ,l-1\right\} \text {, }a_{i}^{*}=\hat{ a}_{i}\text {,} \nonumber \\ \text {for }i\in & {} \left\{ l,\ldots ,n\right\} \text {, }a_{i}^{*}=\check{ a}_{i}\text {.} \end{aligned}$$

(9)

Hence, under the profile $a^{*}\in \mathcal {G}_{h}$, $a_{j}(l)$ is on the path of play, and $\sum _{1}^{l-1}b_{i}(a^{*})=\sum _{N}b(a^{*})=l-1$, where $a_{i}^{*}\in \mathcal {NP}\left( r\right)$ for all $i\ge l$, establishing the result. $\square$

Fig. 4

The strategy profile $a^{*}$ for the $l{\text {th}}$ coordinate of j ’s strategy (an agent $i\le l-1$ chooses $a_{i}^{*}\in \mathcal {P}$, while an agent $i\ge l$ chooses $a_{i}^{*}\in \mathcal {NP}\left( r\right)$)

The construction in (9) yields two key implications: First, since all strategies are possible in $\mathcal {G}_{0}$, it follows that $\mathcal {G}_{0}$ satisfies $\mathcal {NP}\left( \lambda \right) \ne \varnothing .$ Second, for the case where $2\le$ $r\le \lambda -1$ as one proceeds along the sequence $\mathcal {G}_{1},\ldots ,\mathcal {G}_{M-1}$, except possibly for the first round (from $\mathcal {G}_{0}$ to $\mathcal {G}_{1})$, the maximum possible ‘reduction’ per round of elimination is ‘one’ in the following sense: If $r\le \lambda -1$ and one has $\mathcal {NP}\left( r\right) \ne \varnothing$ in $\mathcal {G}_{h}$ (i.e., every agent has a strategy in $\mathcal {G}_{h}$ with zeros in all coordinates no larger than the $r{\text {th}}$ coordinate), then it holds that $\mathcal {NP}\left( r-1\right) \ne \varnothing$ in $\mathcal {G}_{h+1}$ (i.e., every agent has a strategy in $\mathcal {G}_{h+1}$ with zeros in all coordinates no larger than the ($r-1){\text {th}}$ coordinate). We summarize the above in Lemma 4. In particular, the maximal reduction Lemma 4 implies that $\mathcal {G}_{1}$ satisfies $\mathcal {NP}\left( \lambda -1\right) \ne \varnothing$ whether or not the game $\mathcal {G}_{0}$ is dominance solvable.

Lemma 4

(Maximal Reduction) (i) $\mathcal {G}_{0}$ satisfies $\mathcal {NP}\left( \lambda \right) \ne \varnothing$. (ii) Let $j\in N$ and $\mathcal {G}_{h}\in \{\mathcal {G}_{1},\ldots ,\mathcal {G}_{M-1}\}$ satisfy $\mathcal {NP}\left( r\right) \ne \varnothing$ in $\mathcal {G}_{h}$ for some $r\in \{2,\ldots ,\lambda -1\}$. Then for all $l\in \{1,\ldots ,\min \{r-1,\kappa (j)+1\}\},a_{j}(l)=0$ is BRCA in $\mathcal {G}_{h}$ and $\mathcal {NP}\left( r-1\right) \ne \varnothing$ in $\mathcal {G}_{h+1}$.

We now present two lemmas that represent the key steps in the proof of Theorem 2.

Lemma 5

(Sufficiency) Consider the canonical sequence of agents given by the ordered set $(i_{1},i_{2},\ldots ,i_{m^{*}})$ where $i_{1}=n$ and $i_{s+1}=\kappa (i_{s})$ and $\kappa (i_{m^{*}})=\kappa ^{m^{*}-1}(n)=0$. If $m^{*}\ge \lambda$ then $\mathcal {G}_{0}$ is dominance solvable.

Proof

Consider the game $\mathcal {G}_{0}$. From $m^{*}\ge \lambda$ and Assumption 3 ($\mathcal {I}$-Monotonicity) we know that $i_{1}$ covers at least ($m^{*}-1)$ agents, i.e., agents $i_{2},\ldots ,i_{m^{*}}$, and hence $\kappa (i_{1})=\kappa (n)\ge m^{*}-1\ge \lambda -1$. Observe that there exists a contingency such that the path of play passes through the $\lambda {\text {th}}$ coordinate of $i_{1}$’s strategy in $\mathcal {G}_{0}$ where all strategies are possible. Moreover, for every contingency in $\mathcal {G}_{0}$ with a path of play passing through the $\lambda {\text {th}}$ or higher coordinate of $i_{1}$, $i_{1}$ knows that at least $(\lambda -1)$ individuals have participated before she moves and hence $a_{n}(\lambda )=a_{i_{1}}(\lambda +s)=1$ is BRCA in $\mathcal {G}_{0}$.²⁹ Thus, using the persistence Lemma 1, we can conclude that for $i_{1}=n$ the $\lambda {\text {th}}$ coordinate is reduced to 1 in $\mathcal {G}_{1}$ and all coordinates $a_{i_{1}}(\lambda )$ and $a_{i_{1}}(\lambda +s)$ are strictly fixed at 1 for all $s\in \{1,\ldots ,\kappa (n)+1-\lambda \}$ in $\mathcal {G}_{1+t}$ for all $t\in \{0,\ldots ,M-1\}$. This implies $\mathcal {G}_{1}$ satisfies $\mathcal {NP}\left( \lambda \right) =\varnothing$. By the maximal reduction Lemma 4, we know that $\mathcal {G}_{1}$ satisfies $\mathcal {NP}\left( \lambda -1\right) \ne \varnothing$.

Similarly, the existence of a chain of length $\lambda$ implies that $\kappa (i_{2})\ge \lambda -2$. By the exact contingency Lemma 3, there is an exact contingency in $\mathcal {G}_{1}$ passing through the $(\lambda -1)^{{th}}$ coordinate of $i_{2}$. Since $\kappa (i_{1})=i_{2}$, if $a_{i_{2}}(\lambda -1)=1$, this path of play must pass through the $\lambda ^{{th}}$ coordinate of $i_{1}$’s strategy. Since $a_{i_{1}}(\lambda )$ is strictly fixed at 1 in $\mathcal {G}_{1+t}$ for all $t\in \{0,\ldots ,M-1\}$, for any contingency in $\mathcal {G}_{1}$ where the path of play passes through the $(\lambda -1){\text {th}}$ coordinate or higher of $i_{2}$, we have that $a_{i_{2}}(\lambda -1)=1$ is a BRCA. By the persistence Lemma 1, $a_{i_{2}}(\lambda -1)$ is strictly fixed at 1 in $\mathcal {G}_{2+t}$ for all $t\in \{0,\ldots ,M-2\}$. In addition, by the maximal reduction Lemma 4, $\mathcal {NP}\left( \lambda -2\right) \ne \varnothing$ in $\mathcal {G}_{2}$.

Using a similar argument repeatedly for $\mathcal {G}_{3},\ldots ,\mathcal {G} _{\lambda -1}$ and $i_{3},\ldots ,i_{\lambda }$, we have that $a_{i_{\lambda }}(1)$ is strictly fixed at 1 in $\mathcal {G}_{\lambda }$ and that for all $a\in \mathcal {G}_{\lambda }$, $\sum _{N}b_{i}(a)\ge \lambda$. This implies that in $\mathcal {G}_{\lambda +1}$ for every $a\in \mathcal {G}_{\lambda +1}$ , for every agent j, and for every relevant coordinate l of j’s strategies on the path of play, $a_{j}(l)=1$ is a BRCA. Using the persistence Lemma 1, it follows that for all $a\in \mathcal {G}_{M}$, $\sum _{N}b_{i}(a)=n$. Thus, $\mathcal {G}_{0}$ is dominance solvable. $\square$

Recall that dominance solvability of $\mathcal {G}_{0}$ implies $\mathcal {G}_{M}$ must satisfy $\mathcal {NP}\left( 1\right) =\varnothing$ (Corollary 6), i.e., in $\mathcal {G}_{M}$ it should not be possible for every individual to have some strategy with zero in the first coordinate. Furthermore, with dominance solvability, using the maximal reduction Lemma 4, we know that $\mathcal {G}_{1}$ satisfies $\mathcal {NP}\left( \lambda -1\right) \ne \varnothing$. Thus, using Lemma 4 repeatedly we conclude that as one proceeds along the sequence $\{\mathcal {G}_{1},\ldots ,\mathcal {G}_{M}\}$ we will encounter (sequentially) games which satisfy “$\mathcal {NP}\left( \lambda -1\right) =\varnothing$ and $\mathcal {NP}\left( \lambda -2\right) \ne \varnothing$” followed by games satisfying “$\mathcal {NP}\left( \lambda -2\right) =\varnothing$ and $\mathcal {NP}\left( \lambda -3\right) \ne \varnothing$” and so on until we will get to the set of games satisfying “$\mathcal {NP}\left( 2\right) =\varnothing$ and $\mathcal {NP}\left( 1\right) \ne \varnothing$” and then to games satisfying $\mathcal {NP}\left( 1\right) =\varnothing$, which, as we have noted, is a condition necessary for dominance solvability. The above process with coordinates successively becoming fixed at 1 breaks down when $\lambda$ is strictly less than the length of the longest information chain $m^{*}$. Before offering a proof, we illustrate the underlying intuition of such break-down using the example below:

Example 7

Let $N=\left\{ 1,\ldots ,6\right\}$, $\lambda =4$, $\kappa \left( 6\right) =\kappa \left( 5\right) =\kappa \left( 4\right) =3$, $\kappa \left( 3\right) =2$, and $\kappa \left( 2\right) =\kappa \left( 1\right) =0$ (and hence $m^{*}=3)$. In $\mathcal {G}_{1}$, the $4^{th}$ coordinate of agents 4,5, and 6 get reduced to 1 and by Lemma 4 all coordinates of all agents less than or equal to the $3^{rd}$ coordinate have 0 as a BRCA. In particular, the $3^{rd}$ coordinate of agent 3 is not reduced to 1. In $\mathcal {G}_{2}$, the $3^{rd}$ coordinate of agent 3 gets reduced to 1 and again by Lemma 4 all coordinates of all agents less than or equal to the $2^{nd}$ coordinate have 0 as a BRCA. Now by Lemma 4, the $2^{nd}$ coordinate of some agent would have to be reduced to 1 in some game in the subsequent sequence $\mathcal {G}_{3},\mathcal {G}_{4},\ldots$ in order for the game to be dominance solvable. However, this cannot happen: Neither 1 or 2 has a $2^{nd}$ coordinate and 0 remains a BRCA in the $2^{nd}$ coordinate of the remaining agents.

Note that the reduction process described above follows the canonical sequence $n=6$ and $\kappa \left( 6\right) =3$ followed by $\kappa \left( 3\right) =2$, and the process ‘fails’ in the $3^{rd}$ step (recall that $\lambda =4$) because $\kappa \left( 2\right) =0\ne 1$—agent 2 has no $2^{nd}$ coordinate that reduces to 1. Our necessity proof hinges on this crucial link between reduction of single coordinates to 1 in each step of the reduction process through dominance and the next element in the canonical sequence having the next lowest coordinate that can then be reduced.

To establish this relationship, we will develop a notion which partitions the sequence $\left\{ \mathcal {G}_{1},\ldots ,\mathcal {G}_{M}\right\}$ into blocks such that the single coordinate reductions occur in the last element of each block. Thus, for $\lambda -1\ge k\ge 1$, we denote the set of games satisfying “$\mathcal {NP}\left( \lambda -k+1\right) =\varnothing$ and $\mathcal {NP}\left( \lambda -k\right) \ne \varnothing$” by the block of games $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\}$ with the first and last games in $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\}$ (as ranked in the sequence $\mathcal {G}_{1},\ldots ,\mathcal {G}_{M}$) being denoted by $\mathcal {G}_{\min }^{\lambda -k}$ and $\mathcal {G}_{\max }^{\lambda -k}$, respectively.³⁰ Note that in all games in the block $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\}$, all agents have strategies in which all the coordinates less than or equal to the $(\lambda -k){\text {th}}$ coordinate is zero, while in the game $\mathcal {G}_{\max }^{\lambda -k}$ the $(\lambda -k){\text {th}}$ coordinate reduces to 1 in all strategies for some individual and that this coordinate is fixed at 1 in all subsequent games starting with $\mathcal {G}_{\min }^{\lambda -k-1}$. Using the sets $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\}$ with $1\le k\le \lambda -1$ we will establish a property of the canonical sequence of agents $i_{1},i_{2},\ldots ,i_{m^{*}}$ that will be critical to demonstrate that the process with successive coordinates becoming fixed at 1 breaks down when $\lambda \$is strictly less than the length of the longest information chain ($m^{*}$).

In Lemma 6, we show that in each round of reduction of zeros to ones (in the sense of Lemma 4) that occurs between the blocks partitioning $\left\{ \mathcal {G}_{1},\ldots ,\mathcal {G}_{M}\right\}$, the reduction of the coordinate always has to start with an agent less than or equal to a distinguished agent from the canonical sequence $i_{1},i_{2},\ldots ,i_{m^{*}}$ (where the selection of this distinguished agent depends on the block $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\}$ under consideration).

Lemma 6

(Non-Reduction) Let $s\in \left\{ 1,\ldots ,\lambda -1\right\}$. If $\mathcal {G}_{h}\in \left\{ \mathcal {G}_{i}^{\lambda -s}\right\}$ then for all agents j such that $j>i_{s+1}$ and $\kappa (j)\ge (\lambda -s-1)$, $a_{j}(\lambda -s)=0$ is a BRCA in $\mathcal {G}_{h}.$

Proof

We will provide a proof by induction. In each step, we explicitly construct a strategy profile to establish the result.

Basis Step. $s=1.$

We first consider the block $\{\mathcal {G}_{i}^{\lambda -1}\}$. The analysis here leads to the reduction that takes place between $\mathcal {G}_{\max }^{\lambda -1}$ and $\mathcal {G}_{\min }^{\lambda -2}$, where the $(\lambda -1){\text {th}}$ coordinate of some agent becomes fixed at 1. Recall that all games in the block $\{\mathcal {G}_{i}^{\lambda -1}\}$ satisfy $\mathcal {NP}\left( \lambda -1\right) \ne \varnothing$ and $\mathcal {G}_{\min }^{\lambda -1}=\mathcal {G}_{1}$.

Consider an agent j with $j>i_{2}$ and $\kappa (j)\ge (\lambda -2)$ in a game $\mathcal {G}_{h}\in \left\{ \mathcal {G}_{i}^{\lambda -1}\right\}$. Since $\mathcal {NP}\left( \lambda -1\right) \ne \varnothing$ in $\mathcal {G}_{h}$, there exists $a_{j^{\prime }}\in \mathcal {NP}\left( \lambda -1\right)$ in $\mathcal {G}_{h}$ for all $j^{\prime }\ge j$. Moreover, since $\kappa (j)\ge (\lambda -2)$ and $\mathcal {NP}\left( \lambda -1\right) \ne \varnothing$, the exact contingency Lemma 3 implies that there exists an exact contingency $a_{-j}^{*}$ for the ($\lambda -1){\text {th}}$ coordinate of j in $\mathcal {G}_{h}$.

Construct a contingency $a_{-j}\in \mathcal {G}_{h}$ as follows (see Fig. 5):

• Let all agents $j^{\prime \prime }<j$ use their strategies from the exact contingency $a_{-j}^{*}$.

• Let all agents $j^{\prime }>j$ use the strategies $a_{j^{\prime }}\in \mathcal {NP}\left( \lambda -1\right)$.

Fig. 5

An illustration of the contingency $a_{-j}$ for Basis Step

By the construction of $a_{-j}$, the first $\left( \lambda -2\right)$ agents participate on the path of play, while the other agents playing before j choose to not participate. Since, $j>i_{2}=\kappa (n)=\kappa (i_{1}),$ j is not covered by any individual. It follows that all agents moving after j do not participate along the path of play since all such agents after j are using their strategies from $\mathcal {NP}\left( \lambda -1\right)$. Hence, irrespective of whether $a_{j}(\lambda -1)=1$ is possible or not in $\mathcal {G}_{h}$, $a_{j}(\lambda -1)=0$ is a BRCA for j in $\mathcal {G}_{h}$, establishing Basis Step.³¹

Applying the above to the game $\mathcal {G}_{\max }^{\lambda -1}\in \left\{ \mathcal {G}_{i}^{\lambda -1}\right\}$ and using the persistence Lemma 1, if $\mathcal {G}_{\min }^{\lambda -2}$, the next game after $\mathcal {G}_{\max }^{\lambda -1}$ in the sequence $\mathcal {G}_{1},\ldots ,\mathcal {G}_{M}$, exists, we then have:

$$\begin{aligned} \text {If }j>i_{2}\text { and }\kappa (j)\ge (\lambda -2)\text {, then there is }a_{j}\in \mathcal {G}_{\min }^{\lambda -2}\text { with }a_{j}(\lambda -1)=0. \end{aligned}$$

(10)

In other words, a reduction of the $\left( \lambda -1\right) {\text {th}}$ coordinate to 1 of any agent moving after $i_{2}$ cannot take place in the block $\left\{ \mathcal {G}_{i}^{\lambda -1}\right\}$ and therefore any such reduction (if any) has to necessarily take place for an agent moving before $i_{2}$.

Inductive Step. $s\ge 2.$

Consider the inductive hypothesis: If $\mathcal {G}_{h}\in \left\{ \mathcal {G}_{i}^{\lambda -s+1}\right\}$ then for all agents $j^{\prime }$ such that $j^{\prime }>i_{s}$ and $\kappa (j^{\prime })\ge (\lambda -s)$, $a_{j^{\prime }}(\lambda -s+1)=0$ is a BRCA in $\mathcal {G}_{h}$.

Using this inductive hypothesis, we need to show that if $\mathcal {G}_{h}\in \left\{ \mathcal {G}_{i}^{\lambda -s}\right\}$ then for all agents j such that $j>i_{s+1}$ and $\kappa (j)\ge (\lambda -s-1)$, $a_{j}(\lambda -s)=0$ is a BRCA in $\mathcal {G}_{h}.$

Analogous to (10), use the inductive hypothesis and Lemma 1 to obtain:

$$\begin{aligned} \text {If }j^{\prime }>i_{s}\text { and }\kappa (j^{\prime })\ge (\lambda -s) \text {, then there is }a_{j^{\prime }}\in \mathcal {G}_{\min }^{\lambda -s} \text { with }a_{j^{\prime }}(\lambda -s+1)=0. \end{aligned}$$

(11)

Consider the first game in the block $\{\mathcal {G}_{i}^{\lambda -s}\}$: $\mathcal {G}_{\min }^{\lambda -s}$.

Let j be such that $j>i_{s+1}$ and $\kappa (j)\ge (\lambda -s-1)$. We will consider two cases: Case 1. $a_{j}\left( \lambda -s\right) =0$ for all $a_{j}\in \mathcal {G}_{\min }^{\lambda -s}$. Case 2. There exists $a_{j}\in \mathcal {G}_{\min }^{\lambda -2}$ with $a_{j}\left( \lambda -s\right) =1$.

Case 1. Since $\mathcal {NP}\left( \lambda -s\right) \ne \varnothing$ in $\mathcal {G}_{\min }^{\lambda -s}$, $a_{j}\left( \lambda -s\right) =0$ is possible and the $\left( \lambda -s\right) {\text {th}}$ coordinate of j is relevant (Lemma 3) in $\mathcal {G}_{\min }^{\lambda -s}$. In addition, as $a_{j}\left( \lambda -s\right) =0$ for all $a_{j}\in \mathcal {G} _{\min }^{\lambda -s}$, $a_{j}(\lambda -s)=0$ is a BRCA in $\mathcal {G}_{h}$ , completing the proof in this case.

Case 2. In this case we have

$$\begin{aligned} \text {there exists }a_{j}\in \mathcal {G}_{\min }^{\lambda -s}\text { with } a_{j}\left( \lambda -s\right) =1. \end{aligned}$$

(12)

Let $j^{\prime }$ be any agent that covers j, i.e., $\kappa (j^{\prime })\ge j$. Since $\kappa (j)\ge (\lambda -s-1)$ and $\kappa (j^{\prime })\ge j,$ we have $\kappa (j^{\prime })\ge \left( \lambda -s\right)$. In addition, we know that since $j^{\prime }$ covers j and $j>i_{s}$ it must be the case that agent $j^{\prime }$ moves after $i_{s}$.³² Statement (11) shows that for any such $j^{\prime }$, there is $a_{j^{\prime }}\in \mathcal {G}_{\min }^{\lambda -s}$ with $a_{j^{\prime }}(\lambda -s+1)=0$. This allows us to construct the following strategy profile $a\in \mathcal {G}_{\min }^{\lambda -s}$ (and hence a contingency $a_{-j}\in \mathcal {G}_{\min }^{\lambda -s}$) to show that $a_{j}\left( \lambda -s\right) =0$ is a BRCA for j in $\mathcal {G}_{\min }^{\lambda -s}$ (see Fig. 6):

Fig. 6

An illustration of the strategy profile a for Inductive Step

• Let agent j use $a_{j}\in \mathcal {G}_{\min }^{\lambda -s}$ with $a_{j}\left( \lambda -s\right) =1$ (see (12)).

• Let all agents $j^{\prime \prime }$, who do not cover j (i.e., $\kappa (j^{\prime \prime })<j$), use the strategies in the corresponding exact contingency $a_{-j}^{*}$ so that the $\left( \lambda -s\right) {\text {th}}$ coordinate of j’s strategy is on the path of play. (This includes all agents moving before j and possibly some agents moving after j. Notice that $\mathcal {NP}\left( \lambda -s\right) \ne \varnothing$ in $\mathcal {G}_{\min }^{\lambda -s}$ and the exact contingency Lemma 3 imply that this is possible.)

• Let all agents $j^{\prime }$ with $j^{\prime }>j$ and $\kappa (j^{\prime })\ge j$ use $a_{j^{\prime }}\in \mathcal {G}_{\min }^{\lambda -s}$ such that $a_{j^{\prime }}(\lambda -s+1)=0$. (Recall that our above arguments for $j^{\prime }$ imply that this is possible.)

Notice that by construction, there are (exactly) $\left( \lambda -s\right)$ agents participating on the path of play (i.e., $\sum _{N}b_{i}(a)=\lambda -s)$. In particular, only agent j and the first $\left( \lambda -s-1\right)$ agents moving before j participate and no agent moving after j participates. In addition, for each agent $j^{\prime }$ that covers j, the $\left( \lambda -s+1\right) {\text {th}}$ coordinate of $j^{\prime }$ is on the path of play. Since $\sum _{N}b_{i}(a)=\lambda -s\le \lambda -2$, we have that $a_{j}\left( \lambda -s\right) =0$ is a BRCA for j in $\mathcal {G}_{\min }^{\lambda -s}$. In addition, under exactly the same strategy profile a, $a_{j^{\prime }}\left( \lambda -s+1\right) =0$ is a BRCA for all $j^{\prime }$ with $j^{\prime }>j$ and $\kappa (j^{\prime })\ge j$. If $\mathcal {G}_{\min }^{\lambda -s}=\mathcal {G}_{\max }^{\lambda -s}$, our proof is complete. If not, the persistence Lemma 1 implies that $a_{j^{\prime }}\left( \lambda -s+1\right) =0$ persists and is possible in the next game after $\mathcal {G}_{\min }^{\lambda -s}$ which we denote by $\mathcal {G}_{\min +1}^{\lambda -s}$. Hence, if $\mathcal {G}_{\min +1}^{\lambda -s}$ exists, we have the following:

$$\begin{aligned} \text {if }j^{\prime }>i_{s}\text { and }\kappa (j^{\prime })\ge (\lambda -s) \text {, then there is }a_{j^{\prime }}\in \mathcal {G}_{\min +1}^{\lambda -s} \text { with }a_{j^{\prime }}(\lambda -s+1)=0. \end{aligned}$$

(13)

Using an argument similar to the above used for $\mathcal {G}_{\min }^{\lambda -s}$, we can show that $a_{j}\left( \lambda -s\right) =0$ is a BRCA in $\mathcal {G}_{\min +1}^{\lambda -s}$ for any agent $j>i_{s+1}$. The persistence Lemma 1 and the repeated use of this argument establish Inductive Step for all games in $\left\{ \mathcal {G}_{i}^{\lambda -s}\right\}$. $\square$

Proof of Theorem 2

Consider the canonical sequence of agents given by the ordered set $(i_{1},i_{2},\ldots ,i_{m^{*}})$ where $i_{1}=n$, $i_{2}=\kappa (i_{1})=\kappa ^{1}(n)$, $\cdots$, and $\kappa (i_{m^{*}})=\kappa ^{m^{*}-1}(n)=0$. By Proposition 3, $m^{*}$ is the maximum length of an information chain in $\mathcal {G}_{0}$. The sufficiency Lemma 5 already establishes that $m^{*}\ge \lambda$ implies that $\mathcal {G}_{0}$ is dominance solvable. Hence we only need to show that if $\mathcal {G}_{0}$ is dominance solvable then we have $m^{*}\ge \lambda$.

In $\mathcal {G}_{0}$, choose any $j\in N$ and any coordinate$\ l\le \kappa (j)+1$ in j’s strategy with $l\le \lambda -1.$ Since all strategies are possible in $\mathcal {G}_{0}$, we can construct an exact contingency such that the path of play passes through the $l{\text {th}}$ coordinate of j’s strategy. Hence, $a_{j}(l)=0$ is a BRCA. Using the persistence Lemma 1 (i) it follows that $\mathcal {G}_{1}$ satisfies $\mathcal {NP}\left( \lambda -1\right) \ne \varnothing$.

By the maximal reduction Lemma 4, as we proceed along the sequence $\mathcal {G}_{1},\ldots ,\mathcal {G}_{M},$ we will encounter (sequentially) the ($\lambda -1$) non-empty blocks $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\} _{\lambda -1\ge k\ge 1}$ where in the block $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\}$ all agents have strategies in which all the coordinates less than or equal to the $(\lambda -k){\text {th}}$ coordinate are zero and in the game $\mathcal {G}_{\max }^{\lambda -k}$ the $(\lambda -k){\text {th}}$ coordinate of some individual is reduced to 1 and becomes fixed at 1 in all subsequent games.

Consider the first block $\left\{ \mathcal {G}_{i}^{\lambda -k}\right\}$ with $k=\lambda -1$.³³ By definition, some agent’s ($\lambda -1){\text {th}}$ coordinate is reduced to 1 in game $\mathcal {G}_{\max }^{1}$. There must be some individual $j^{\prime }$ such that $\kappa (j^{\prime })\ge \lambda -1.$ Thus, noticing that the hypothesis of Lemma 6 is non-vacuously satisfied, using the persistence Lemma 1, we can conclude that $j^{\prime }\le i_{2}$.

Notice that the existence of these ($\lambda -2$) more of such blocks implies that the hypothesis of the non-reduction Lemma 6 is non-vacuously satisfied in each step of the reduction process and that in the game $\mathcal {G}_{\max }^{\lambda -s}$ for $\lambda -1\ge s\ge 2$, the $(\lambda -s){\text {th}}$ coordinate can be on the path of play and reduces to 1 in $\mathcal {G}_{\max }^{\lambda -h}$ only for some individual $j\le i_{s}$ where $\kappa (j)\ge \lambda -s-1$.³⁴ Hence, for $\mathcal {NP}\left( 1\right) =\varnothing$ to be true (i.e., the $(\lambda -\lambda +1){\text {th}}=1^{\text {st}}$ coordinate of some agent’s strategy to be reduced to 1), we must have an agent $j\le i_{\lambda }$ with $\kappa (j)\ge 0$. Accordingly, we have that $\mathcal {G}_{0}$ being dominance solvable implies the existence of the canonical sequence $i_{1},\ldots ,i_{m^{*}}$ of length at least equal to $\lambda$. $\square$

Proof of Proposition 4

We proceed with our analysis using a procedure similar to backward induction, focusing mainly on the agents in the canonical information chain $\left\{ i_{1},\ldots ,i_{m}\right\}$ with $i_{1}=n$, ..., $i_{m}=\kappa ^{m-1}\left( n\right)$.

Given that $m\ge \lambda$, we have $\kappa \left( n\right) \ge \lambda -1$, i.e., $i_{1}=n$ has at least $\lambda$ information sets. Agent n’s beliefs in each of her information sets are on the participation history of all agents i such that $i_{1}>i>i_{2}$.³⁵ For n’s information sets $H_{n}\left( l\right)$ with $l\ge \lambda -1$, it is a dominant strategy for n to participate regardless of n’s beliefs in $H_{n}\left( l\right)$. For any of n’s other information sets, n’s optimal strategy depends on n’s beliefs of the participation history of i such that $i_{1}>i>i_{2}$.

Next consider agent $i_{2}=\kappa \left( n\right)$. The canonical information chain implies that $\kappa \left( i_{2}\right) \ge \lambda -2$. Given agent n’s optimal play, for $i_{2}$’s information sets $H_{i_{2}}\left( l\right)$ with $l\ge \lambda -2$, it is a dominant strategy for $i_{2}$ to participate regardless of $i_{2}$’s beliefs. And $i_{2}$’s optimal strategy in her other information sets again depend on her beliefs of the participation history of all agents i such that $i_{2}>i>i_{3}$.

Applying a similar arguments as those for $i_{1}$ and $i_{2}$ repeatedly for the remaining agents in $\left\{ i_{1},\ldots ,i_{m}\right\}$, we have that agent $i_{m}$ who has only one information set—by definition, $\left\{ i_{1},\ldots ,i_{m}\right\}$ is the longest canonical information chain and hence $\kappa \left( i_{m}\right) =0$—participates with probability 1 in any perfect Bayesian equilibrium.

Given the optimal strategies of agents in $\left\{ i_{1},\ldots ,i_{m}\right\}$, we have that all the m agents in the chain participate on the equilibrium path in every perfect Bayesian equilibrium. Given that $m\ge \lambda$, this further implies that in every perfect Bayesian equilibrium, there are at least $\lambda$ agents participating on the equilibrium path (regardless of what the other agents’ strategies are). Given the payoff specifications, it follows that all the other agents participate in every perfect Bayesian equilibrium as well, i.e., there is a unique perfect Bayesian equilibrium outcome where everyone participates. $\square$