A dynamic game analysis of Internet services with network externalities

Abstract

Internet services, such as review sites, FAQ sites, online auction sites, online flea markets, and social networking services, are essential to our daily lives. Each Internet service aims to promote information exchange among people who share common interests, activities, or goods. Internet service providers aim to have users of their services actively communicate through their services. Without active interaction, the service falls into disuse. In this study, we consider that an Internet service has a network externality as its main feature, and we model user behavior in the Internet service with network externality (ISNE) as a dynamic game. In particular, we model the diffusion process of users of an ISNE as an infinite-horizon extensive-form game of complete information in which: (1) each user can choose whether or not to use the ISNE in her/his turn and (2) the network effect of the ISNE depends on the history of each player’s actions. We then apply Markov perfect equilibrium to analyze how to increase the number of active users. We derive the necessary and sufficient condition under which the state in which every player is an active user is the unique Markov perfect equilibrium outcome. Moreover, we propose an incentive mechanism that enables the number of active users to increase steadily.

Appendix

This appendix contains lemmas and proofs of theorems.

1.1 Appendix A: Lemmas for Section 3

First, we derive the following properties from these basic assumptions (A1)–(A6).

Lemma 1

For the instant payoff function $u(\cdot )$, properties (L1)–(L4) hold:

• (L1) For any finite period $t < \infty$ and $h^t_\alpha \in H^t_\alpha$:

  $$\begin{aligned} u(h^t_\alpha ) = u(n: k, h^t_\alpha ) \quad \text {for any } 0 \le k < +\infty . \end{aligned}$$

• (L2) For any finite periods $t , \tau < \infty$:

  $$\begin{aligned} u(h^t_\alpha , {\tilde{h}}^\tau _\alpha ) \ge u({\tilde{h}}^\tau _\alpha ) \quad \text {for all } h^t_\alpha \in H^t_\alpha \text { and } {\tilde{h}}^\tau _\alpha \in H^\tau _\alpha . \end{aligned}$$

• (L3) u(a : k) is nondecreasing in k.

• (L4) There exists $m^*$, such that $u(a: m^* - 1) \le 0 < u(a: m^*)$.

Proof

We prove each item using the corresponding assumption.

• (L1) From (A1), it immediately follows that $u(h^t_\alpha ) = u(n, h^t_\alpha )=u(n: 2, h^t_\alpha )=\dots =u(n: k, h^t_\alpha )$.

• (L2) In each period from (A2), replacement of any action by “no-action” reduces the instant payoff $u(h^t_\alpha , {\tilde{h}}^\tau _\alpha ) \ge u(n: k, {\tilde{h}}^\tau _\alpha )$. Given $u(n: k, {\tilde{h}}^\tau _\alpha )=u({\tilde{h}}^\tau _\alpha )$ by (L1), $u(h^t_\alpha , {\tilde{h}}^\tau _\alpha ) \ge u({\tilde{h}}^\tau _\alpha )$.

• (L3) From (L2), we have $u(a: k+1) = u(a, a: k) \ge u(a: k)$ for all k.

• (L4) There exists $h^t_\alpha$, such that $u(h^t_\alpha )>0$, by (A5). If we replace the action by “active” in every period from (A2), then the instant payoff increases: $u(a: t) \ge u(h^t_\alpha )$. Thus, $u(a: t) > 0$. Moreover, there exists $m^*$, such that $u(a: m^* - 1) \le 0 < u(a: m^*)$, by assumption (A4) and (L3).$\square$

Lemma 2

If the instant payoff function $u(\cdot )$ satisfies assumptions (A1)–(A6), the following properties of expected payoff (L5)–(L7) hold:

• (L5) Expected payoff function $\varPi _i$ has a positive externality. Likewise, $\varPi _i^a$ and $\varPi _i^n$ have a positive externality.

• (L6) For any finite $h^t_\alpha \in H^t_\alpha$, $\varPi ^a( \overline{\sigma } \mid h^t_\alpha ) = (1-w) u(h^t_\alpha , a) + \varPi ( \overline{\sigma } \mid h^t_\alpha ).$

• (L7) $\varPi ^a(\overline{\sigma } \mid h^0_\alpha ) > \varPi ^n( \overline{\sigma } \mid h^0_\alpha )$ implies that $\varPi ( \overline{\sigma } \mid h^0_\alpha )> \varPi ^n( \overline{\sigma } \mid h^0_\alpha ) > 0$.

Proof

(L5) follows straightforwardly from (A2) and the definition of $\varPi _i$. Clearly, (L6) holds by the definitions of $\varPi$ and $\varPi ^a$.

Let us show (L7). From property (L6), we have:

$$\begin{aligned} \varPi ^a( \overline{\sigma } \mid h^0_\alpha ) - \varPi ^n( \overline{\sigma } \mid h^0_\alpha )&= (1-w) u(a) + \varPi ( \overline{\sigma } \mid h^0_\alpha ) - \delta \varPi ( \overline{\sigma } \mid h^0_\alpha ) \\&= (1-w) u(a) + (1-\delta )\varPi ( \overline{\sigma } \mid h^0_\alpha ). \end{aligned}$$

If $\varPi (\overline{\sigma } \mid h^0_\alpha ) \le 0$, by (A4), $\varPi ^a( \overline{\sigma } \mid h^0_\alpha ) - \varPi ^n( \overline{\sigma } \mid h^0_\alpha ) < 0$. Thus, if $\varPi ^a( \overline{\sigma } \mid h^0_\alpha ) - \varPi ^n( \overline{\sigma } \mid h^0_\alpha ) > 0$, we have $\varPi ( \overline{\sigma } \mid h^0_\alpha ) > 0$. Moreover, as $\varPi ^n( \overline{\sigma } \mid h^0_\alpha ) = \delta \varPi ( \overline{\sigma } \mid h^0_\alpha )$, $\varPi ( \overline{\sigma } \mid h^0_\alpha )> \varPi ^n( \overline{\sigma } \mid h^0_\alpha ) > 0$. $\square$

1.2 Appendix B: Proofs and Lemmas for Section 4

Proof of Theorem 1

Through this proof, let $h=(h_x, h_\alpha )$ be an infinite history that takes place when every player plays according to $\sigma$. We prove the theorem in three steps. We first show (i) in Steps 1 and 2.

Step 1: First, we show that the total utility of all players $\sum _i P_i(h)$ satisfies $w \sum _i P_i(h) \le \varPi (\overline{\sigma } \mid h^0_\alpha )$ for any $h \in H$. In the following, we fix an infinite history $h=(h_x, h_\alpha )$. For $h_\alpha$, consider the set of periods when “active” is selected, $L=\{ \tau \mid \alpha ^{\tau } = \alpha \}$. Let us denote the smallest element of L by $\tau _1$ and $\ell$th smallest element of L by $\tau _\ell$.

The finite action history of period t taken out from $h_\alpha$ is denoted by $h^t_\alpha =(\alpha ^1,\alpha ^2,\dots , \alpha ^t)$. If $u(h^{\tau _\ell }_\alpha ) \le 0$ for all $\tau _{\ell } \in L$, then, obviously, we have $w \sum _i P_i(h) \le \varPi (\overline{\sigma } \mid h^0_\alpha )$. On the other hand, $u(h^{\tau _1}) < 0$ from (A4). Thus, we assume that there exists $\tau _{\ell ^*}$, such that $u(h^{\tau _{\ell ^*}}_\alpha ) > 0$ and $u(h^{\tau _{\ell }}_\alpha ) \le 0$ for any $\ell < \ell ^*$.

Using the notations, we have:

$$\begin{aligned} \sum _i P_i(h)&= \sum _{\tau =1}^\infty {\mathbf {1}}_a(\alpha ^\tau ) u(h^\tau _\alpha ) \delta ^{\tau -1} \\&= \sum _{\ell =1}^{\ell ^*-1} u(h^{\tau _\ell }_\alpha ) \delta ^{\tau _\ell -1}+\sum _{\ell = \ell ^*}^\infty u(h^{\tau _\ell }_\alpha ) \delta ^{\tau _\ell -1} . \end{aligned}$$

Now, we consider an action history $h'_\alpha$, such that “no-action” is selected until $\tau _{\ell ^*} -\ell ^*$ period and “active” is selected after that. For example, if $h_\alpha = (n,a,n,a,a,n,a\dots )$ and $\tau _{\ell ^*} = 4$, then $h'_\alpha = (n,n,a,a,a,a,a, \dots )$. The sum of total payoff of $h'_\alpha$ is as follows:

$$\begin{aligned} \sum _i P_i(h'_\alpha ) = \delta ^{\tau _{\ell ^*} - \ell ^*} \sum _{k=1}^{\infty } u(a:k) \delta ^{k -1}. \end{aligned}$$

We now show that $\sum _i P_i(h_\alpha ) \le \sum _i P_i(h'_\alpha )$. For $k < \ell ^*$, we have $u(h^{\tau _k}_\alpha ) \le 0$, $\tau _k \le \tau _{\ell ^*} - \ell ^* + k$ and $u(h^{\tau _k}_\alpha ) \le u(a:k)$ from (A3) . Thus, we have $u(h^{\tau _k}_\alpha ) \delta ^{\tau _k - 1} \le \delta ^{\tau _{\ell ^*} - \ell ^*} u(a:k) \delta ^{k-1}$. On the other hand, for $k \ge \ell ^*$, we have $u(\tau _k) \le u(a:k)$, $u(a:k) >0$ and $\tau _k \ge \tau _{\ell ^*} - \ell ^* + k$. Thus, we have $u(h^{\tau _k}_\alpha ) \delta ^{\tau _k - 1} \le \delta ^{\tau _{\ell ^*} - \ell ^*} u(a:k) \delta ^{k-1}$. Therefore, we have $\sum _i P_i(h_\alpha ) \le \sum _i P_i(h'_\alpha )$.

Since $\sum _i P_i(h'_\alpha ) = N \delta ^{\tau _{\ell ^*} - \ell ^*} \varPi (\overline{\sigma } \mid h^0_\alpha ) \le N \varPi (\overline{\sigma } \mid h^0_\alpha )$, we finally have $w \sum _i P_i(h_\alpha ) \le w \sum _i P_i(h'_\alpha ) \le \varPi (\overline{\sigma } \mid h^0_\alpha )$.

Step 2: For a given strategy profile $\sigma$, let us denote the probability measure that history h occurs by $\mu _\sigma (h)$. Then:

$$\begin{aligned} \varPi _i(\sigma \mid h^0_\alpha ) = \int _h P_i(h) {\mathrm{{d}}}\mu _\sigma (h). \end{aligned}$$

Thus, by $w \sum _i P_i(h) \le \varPi (\overline{\sigma } \mid h^0_\alpha )$ for all h, we obtain: $w \sum _i \varPi _i (\sigma \mid h^0_\alpha ) \le \varPi (\overline{\sigma } \mid h^0_\alpha ).$ From Steps 1 and 2, we have proven part (i).

Step 3: We finally prove (ii). By the proof of (i), for given $h_\alpha$, (1) if there does not exit $\ell ^*$, such that $u(h^{\tau _{\ell ^*}}_\alpha ) > 0$ then $\sum _i P_i(h) \le 0$, and (2) if there exists $\ell ^*$, such that $u(h^{\tau _{\ell ^*}}_\alpha ) > 0$, then we have $\sum _i P_i(h) \le \varPi ( \overline{\sigma } \mid h^0_\alpha ) < 0$.

In both cases (1) and (2), $\sum _i P_i(h) \le 0$ holds. Thus, we have $\sum _i \varPi _i(\sigma \mid h_\alpha ) \le 0$ for each history h. Therefore, $\sum _i \varPi _i(\sigma \mid h^0_\alpha ) \le 0$ for any strategy profile $\sigma$, which implies that no-action equilibria achieves the maximum consumer surplus. $\square$

The proof of Theorem 2 is complex, because the model considers infinite history. Thus, we need the following two lemmas to show the condition for the uniqueness of the active equilibrium of MPE. We first consider the case where “active” is selected $k^*$ times one after the other. We then derive the condition that every player always selects “active” in the equilibrium after the history.

Lemma 3

Suppose that $\varPi ^a(\overline{\sigma } \mid h^t_\alpha ) \ge \varPi ^n(\overline{\sigma } \mid h^t_\alpha )$ for any $h^t_\alpha$. Given an integer $k^*$, suppose that $\varPi ^a(\overline{\sigma } \mid a:k') > \varPi ^n(\overline{\sigma } \mid a:k')$ for $k' \ge k^*$. Then, in all MPE, “active” is selected successively after $h^{k^*}=(h^{k^*}_x, a:k^*)$.

Proof

Case 1 (that $k \ge m^* - 1$): given $u(a: k, a) > 0$ for all k, such that $k \ge m^* - 1$ and (L3), it follows that $u(a:k+\tau ) > 0$ for $\tau =1,2,\dots$. Hence, for any history $h^{k} = (h^{k}_x, a: k)$ in an equilibrium in a subgame after $h^{k}$, every player selects “active” in her/his turn.

Case 2 (that $k^* \le k < m^*-1$): suppose that, for any history $h^{k+1} = (h^{k+1}_x, a: k+1)$, in an equilibrium of a subgame after $h^{k+1}$, every player selects “active” after the ($k+2$)th period. We then show that every player selects “active” after the ($k+1$)th period in an MPE of a subgame after any history $h^{k} = (h^{k}_x, a: k)$.

Case 2A (that “active” is chosen in the ($k+1$)th period): in a subgame after $h^{k} = (h^{k}_x, a: k)$, if a player that is selected in period $k+1$ selects “active,” then $h^{k+1} = (h^{k+1}_x, a: k+1)$. From the assumption, in an equilibrium of a subgame after $h^{k+1}$, players always select “active” after the $k+2$ period. Thus, the expected payoff to a player when s/he selects “active” in the ($k+1$)th period is $\varPi ^a(\overline{\sigma } \mid a:k)$.

Case 2N (that “no-action” is chosen in the $(k+1)$th period):

Let us show that $\varPi ^a_i(\overline{\sigma } \mid a:k) > \varPi ^n_i(\sigma \mid a:k)$ for any $\sigma$, where $\varPi ^n_i(\sigma \mid a:k)$ is the expected payoff when player i is selected in period $k+1$ and chooses “no-action,” i.e., $h^{k+1}_\alpha = (a:k, n)$, in a subgame from $h^{k} = (h^{k}_x, a: k)$.

Case 2N-1 (that a player selects “no-action” with some positive probability at finite times):¹⁷ consider strategy profile $\sigma$, where player i always selects “active” after a finite history/period.

Let $\varDelta (k+Q)$ be the set of strategy profiles in which every player selects “active” after the ($k+Q$)th period,¹⁸ including period $k+Q$. That is, $\sigma \in \varDelta (k+Q)$ is a strategy profile, such that for every player $i \in I$: $\sigma ^{k+Q+\tau }_i(h^{k+Q-1+\tau }_\alpha ) = 1$ for any $h^{k+Q-1+\tau }_\alpha \in H^{k+Q-1+\tau }$, $\tau = 0, 1, 2, 3, \dots$ for any $\sigma \in \varDelta (k+Q)$.

Then, we show $\varPi ^n_i(\overline{\sigma } \mid a:k) \ge \varPi ^n_i(\sigma \mid a:k)$ for all $\sigma \in \varDelta (k+Q)$. First, from the definition, for any $h^{k+Q-2}$, $\varPi ^n_i(\overline{\sigma } \mid h^{k+Q-2}_\alpha ) = \varPi ^n_i(\sigma \mid h^{k+Q-2}_\alpha )$ and $\varPi ^a_i(\overline{\sigma } \mid h^{k+Q-2}_\alpha ) = \varPi ^a_i(\sigma \mid h^{k+Q-2}_\alpha )$ hold. By assumption of this lemma, $\varPi ^a_i(\overline{\sigma } \mid h^{k+Q-2}_\alpha ) \ge \varPi ^n_i(\overline{\sigma } \mid h^{k+Q-2}_\alpha )$ for all $h^{k+Q-2} \in H^{k+Q-2}$. Thus, when we check the maximum payoff to player i, we can assume that s/he selects “active” at time $k+Q-1$. From the positive externality, we can also assume that other players also select “active” in the ($k+Q-1$)th period. Hence, we obtain $\varPi ^n_i(\overline{\sigma } \mid h^{k+Q-3}_\alpha ) \ge \varPi ^n_i(\sigma \mid h^{k+Q-3}_\alpha )$. In a similar way, we can derive $\varPi ^n_i(\overline{\sigma } \mid a:k) \ge \varPi ^n_i(\sigma \mid a:k)$ for all $\sigma \in \varDelta (k+Q)$. We can apply the same argument for any $Q=0,1,2, \dots (<+\infty )$.

Case 2N-2 (that a player selects “no-action” with some positive probability at infinite times): Consider that a strategy profile where player i may choose “no-action” with a positive probability at an infinite time. Given $\delta < 1$, a T exists that satisfies:

$$\begin{aligned} \varPi ^a(\overline{\sigma } \mid a:k) - \varPi ^n(\overline{\sigma } \mid a:k) > \frac{ \delta ^{T} (\overline{U} - \underline{U})}{ 1 - \delta }. \end{aligned}$$

(B.1)

By Case 2N-1, we have the following:

$$\begin{aligned} \varPi ^a(\overline{\sigma } \mid a:k) - \sup _{\sigma \in \varDelta (T)} \varPi ^n_i( \sigma \mid a:k) = \varPi ^a(\overline{\sigma } \mid a:k) - \varPi ^n(\overline{\sigma } \mid a:k). \end{aligned}$$

(B.2)

On the other hand, because differences in the payoffs after the ($T+1$)th period are at most $\overline{U} - \underline{U}$, it follows that:

$$\begin{aligned} \sup _{\sigma \in \varDelta } \varPi ^n_i(\sigma \mid a:k) - \sup _{\sigma \in \varDelta (T)} \varPi ^n_i( \sigma \mid a:k) \le \frac{ \delta ^{T} (\overline{U} - \underline{U})}{ 1 - \delta }. \end{aligned}$$

(B.3)

Thus, we obtain the following:

$$\begin{aligned} \varPi ^a(\overline{\sigma } \mid a:k) - \sup _{\sigma \in \varDelta } \varPi ^n_i(\sigma \mid a:k)= & {} \{ \varPi ^a(\overline{\sigma } \mid a:k) - \sup _{\sigma \in \varDelta (T)} \varPi ^n_i( \sigma \mid a:k) \} \\&- \{\sup _{\sigma \in \varDelta } \varPi ^n_i(\sigma \mid a:k) - \sup _{\sigma \in \varDelta (T)} \varPi ^n_i( \sigma \mid a:k)\} \\\ge & {} \varPi ^a(\overline{\sigma } \mid a:k) - \varPi ^n(\overline{\sigma } \mid a:k)\\&- \frac{ \delta ^{T} (\overline{U} - \underline{U}) }{ 1 - \delta } > 0, \end{aligned}$$

where the first inequality follows from (3) and (4), and the second follows from (2).

Therefore, after history $h^k = (h^k_x, a:k)$, selecting “active” and earning $\varPi ^a(\overline{\sigma } \mid a:k)$ are strictly better than choosing “no-action,” and hence a player selects “active” in the ($k+1$)th period. Given $h^{k+1} = (h^{k+1}_x, a:k+1)$, the outcome where every player selects “active” after $k+2$ is the unique MPE outcome.

From Cases 1 and 2, the proof is completed by induction on k. $\square$

We prove the following lemma to simplify the condition of the previous lemma.

Lemma 4

For any finite action history $h^t_\alpha$, the following inequality holds: $\varPi ^a(\overline{\sigma } \mid h^t_\alpha ) - \varPi ^n(\overline{\sigma } \mid h^t_\alpha ) \ge \varPi ^a(\overline{\sigma } \mid h^0_\alpha ) - \varPi ^n(\overline{\sigma } \mid h^0_\alpha )$.

Proof

By (A3), $u(h^t_\alpha , a: k) \ge u(h^t_\alpha , n, a: k)$ for any $k \ge 0$. Thus, we obtain the following:

$$\begin{aligned} - w \sum _{k=1}^{\infty } u( h^t_\alpha , n, a: k ) \delta ^k \ge&- w \sum _{k=1}^{\infty } u( h^t_\alpha , a: k ) \delta ^k \\ \ge&- \delta w u(h^t_\alpha , a ) - \delta w \sum _{k=1}^{\infty } u( h^t_\alpha , a: k + 1 ) \delta ^k. \end{aligned}$$

Using the above, we have the following:

$$\begin{aligned} \varPi ^a(\overline{\sigma } \mid h^t_\alpha ) - \varPi ^n(\overline{\sigma } \mid h^t_\alpha )&= u(h^t_\alpha ,a) + \delta \varPi (\overline{\sigma } \mid h^t_\alpha , a) - \delta \varPi (\overline{\sigma } \mid h^t_\alpha , n) \\&= u(h^t_\alpha , a) + w \sum _{k=1}^{\infty } u( h^t_\alpha , a, a: k )\delta ^k \\&\qquad - w \sum _{k=1}^{\infty } u( h^t_\alpha , n, a: k )\delta ^k \\ \ge&u(h^t_\alpha , a) + w \sum _{k=1}^{\infty } u( h^t_\alpha , a, a: k )\delta ^k \\&\qquad - w \delta u(h^t_\alpha , a ) - w \delta \sum _{k=1}^{\infty } u( h^t_\alpha , a: k + 1 ) \delta ^k \\ =&(1-w \delta ) u(h^t_\alpha , a) + (1-\delta ) w \sum _{k=1}^{\infty } u( h^t_\alpha , a: k + 1 )\delta ^k \\ \ge&(1-w \delta ) u(a) + (1-\delta ) w \sum _{k=1}^{\infty } u( a: k + 1 )\delta ^k \\ =&\varPi ^a(\overline{\sigma } \mid h^0_\alpha ) - \varPi ^n(\overline{\sigma } \mid h^0_\alpha ), \end{aligned}$$

where the last inequality follows from (L2). $\square$

From the above two lemmas, we can prove the main theorem.

Proof of Theorem 2

We prove the theorem in four steps.

Step 1: We first show (iii). Suppose that MPE $\sigma$ is an active equilibrium. Given that the active equilibrium results in the action history that is a series of “actives” $(a, a, a, \dots )$, $\varPi ^a_i(\sigma \mid h^0_\alpha ) = \varPi ^a(\overline{\sigma } \mid h^0_\alpha )$.

From the fact that $\sigma$ is a payoff-relevant strategy, we have $\sigma ^{t+1}(h^t_\alpha ) = \sigma ^{t+1}(n, h^t_\alpha )$. That is, even if “no-action” is chosen in the first period, “active” will be chosen successively after the second period, because strategy $\sigma$ assigns the same action to $h^0_\alpha$ and (n). Therefore, $\varPi ^n_i(\sigma \mid h^0_\alpha ) = \varPi ^n(\overline{\sigma } \mid h^0_\alpha )$.

When $\varPi ^a( \overline{\sigma } \mid h^0_\alpha ) < \varPi ^n( \overline{\sigma } \mid h^0_\alpha )$, “no-action” must be the best reply in the first period after $h^0$, which contradicts the assumption that $\sigma$ is MPE.

Step 2: We first show “if” is part of (i). From Lemma 4 and $\varPi ^a(\overline{\sigma } \mid h^0_\alpha ) > \varPi ^n(\overline{\sigma } \mid h^0_\alpha )$, we obtain the assumption of Lemma 3. Thus, all MPE are active equilibria.

Step 3: Next, we show (ii) of Theorem 2 in three substeps, 3-1, 3-2, and 3-3.

(3-1) We first show that if $\varPi ^a( \overline{\sigma } \mid h^0_\alpha ) = \varPi ^n( \overline{\sigma } \mid h^0_\alpha )$, then all MPE are active equilibria in any subgame after $h^1 = (h^x, a)$ where “active” is chosen in the first period.

To show this, we prove that $\varPi ^a( \overline{\sigma } \mid a:t) > \varPi ^n( \overline{\sigma } \mid a:t)$ for $t (\ge 1)$ if $\varPi ^a( \overline{\sigma } \mid h^0_\alpha ) = \varPi ^n( \overline{\sigma } \mid h^0_\alpha )$. The following inequality follows from (A3):

$$\begin{aligned}&\varPi ^a( \overline{\sigma } \mid a:t) - \varPi ^n( \overline{\sigma } \mid a:t) \\&\quad = u(a:t+1) + w \sum _{k=1}^\infty u(a:k+t+1) \delta ^k - w \sum _{k=1}^\infty u(a:t,n,a:k) \delta ^k \\&\quad \ge u(a:t+1) + w \sum _{k=1}^\infty u(a:k+t+1) \delta ^k - w \sum _{k=1}^\infty u(a:k+t) \delta ^k\\&\quad = (1-w \delta ) u(a:t+1) + (1-\delta ) w \sum _{k=1}^\infty u(a:k+t+1) \delta ^k. \end{aligned}$$

On the other hand, we have $\varPi ^a( \overline{\sigma } \mid h^0_\alpha ) - \varPi ^n( \overline{\sigma } \mid h^0_\alpha )=(1-w \delta ) u(a) + (1-\delta ) w \sum _{k=1}^\infty u(a:k+1) \delta ^k = 0$. Given $u(a:k+1) \ge u(a:k)$ for all $k \in {\mathbb {N}} \cup \{0 \}$ by (L3), from (A5) and (A6) there exists a $\tau$, such that $u(a:\tau +1) > u(a:\tau )$.

Noting that the above strict inequality holds, we can see that:

$$\begin{aligned} \varPi ^a( \overline{\sigma } \mid a:t) - \varPi ^n( \overline{\sigma } \mid a:t)&\ge (1-w \delta ) u(a:t+1) \\&\quad + (1-\delta ) w \sum _{k=1}^\infty u(a:k+t+1) \delta ^k \\&> (1-w \delta ) u(a) + (1-\delta ) w \sum _{k=1}^\infty u(a:k+1) \delta ^k = 0. \end{aligned}$$

Hence, from Lemma 3, every MPE is an active equilibrium in a subgame after $h^1 = (h^x, a)$.

(3-2) Next, we show that there is no MPE where more than one player selects “no-action” with a positive probability in the first period. That is, we show that $\sigma ^1_j(h^0_\alpha )=1$ for all $j(\ne i)$ for all MPE $\sigma$ if $\sigma ^1_i(h^0_\alpha ) < 1$.

As we have shown in (3-1), if “active” is selected in the first period, then every player selects “active” after the second period. Thus, for any MPE $\sigma$, $\varPi ^a( \sigma \mid h^0_\alpha ) = \varPi ^a( \overline{\sigma } \mid h^0_\alpha )$.

Here, we assume that $\sigma$ is an MPE:

$$\begin{aligned} \varPi _j( \sigma \mid h^0_\alpha )&= w \delta \Big [ \sigma ^1_i(h^0_\alpha ) \varPi _j( \sigma \mid a) + (1-\sigma ^1_i(h^0_\alpha ))\varPi _j( \sigma \mid h^0_\alpha ) \Big ] \\&\quad \,\,+ w \Big [\sigma ^1_j(h^0_\alpha ) \varPi ^a_j( \overline{\sigma } \mid h^0_\alpha ) + (1-\sigma ^1_j(h^0_\alpha )) \delta \varPi _j( \sigma \mid h^0_\alpha ) \Big ]\\&\quad \,\, + (1-2w) \delta \varPi _j( \sigma \mid a). \end{aligned}$$

As $\varPi _j( \sigma \mid a) = \varPi _j(\overline{\sigma } \mid a ) = ( \varPi ^a_j( \overline{\sigma } \mid h^0_\alpha ) - u(a))/ \delta$, we have the following:

$$\begin{aligned}&\varPi _j( \sigma \mid h^0_\alpha ) \\&\quad = \frac{ \varPi ^a_j( \overline{\sigma } \mid h^0_\alpha ) \Big [ w \sigma ^1_i( h^0_\alpha ) + w \sigma ^1_j( h^0_\alpha ) + 1 - 2 w \Big ] - u(a) \Big [ w \sigma ^1_i( h^0_\alpha ) + 1 - 2 w \Big ] }{ 1 - \delta w ( 2 - \sigma ^1_i( h^0_\alpha ) - \sigma ^1_j( h^0_\alpha ))}. \end{aligned}$$

From the above equality, we obtain the following:

$$\begin{aligned}&\varPi _j( \sigma _j,\sigma _{-j} \mid h^0_\alpha )-\varPi _j( {\tilde{\sigma }}_j,\sigma _{-j} \mid h^0_\alpha )\\&\quad = \frac{ (\sigma ^1_j( h^0_\alpha ) - {\tilde{\sigma }}^1_j( h^0_\alpha )) w \left[ (1-\delta ) \varPi ^a_j( \overline{\sigma } \mid h^0_\alpha ) + \delta u(a) ( 1 - 2 w + \sigma ^1_i( h^0_\alpha )w )\right] }{ [ 1 - \delta ( 2 - \sigma ^1_i( h^0_\alpha ) - \sigma ^1_j( h^0_\alpha ))w ][1 - \delta ( 2 - \sigma ^1_i( h^0_\alpha ) - {\tilde{\sigma }}^1_j( h^0_\alpha ))w ]}, \end{aligned}$$

where ${\tilde{\sigma }}^t_j( h^{t-1}_\alpha ))=1$ for all $t>1$. Given $\varPi ^n(\overline{\sigma } \mid h^0_\alpha ) = w \delta \varPi ^a(\overline{\sigma } \mid h^0_\alpha ) + (1-w )\delta [ \varPi ^a(\overline{\sigma } \mid h^0_\alpha ) - u(a) ]$, if $\varPi ^a(\overline{\sigma } \mid h^0_\alpha ) = \varPi ^n(\overline{\sigma } \mid h^0_\alpha )$, then $(1-\delta ) \varPi ^a(\overline{\sigma } \mid h^0_\alpha ) + u(a) ( 1 - w ) \delta = 0$ holds. If $\sigma ^1_i( h^0_\alpha ) < 1$ and $\sigma ^1_j( h^0_\alpha ) > {\tilde{\sigma }}^1_j( h^0_\alpha )$, we have the following:

$$\begin{aligned}&\varPi _j( \sigma _j,\sigma _{-j} \mid h^0_\alpha )-\varPi _j( {\tilde{\sigma }}_j,\sigma _{-j} \mid h^0_\alpha )\nonumber \\&\quad > \frac{ (\sigma ^1_j( h^0_\alpha ) - {\tilde{\sigma }}^1_j( h^0_\alpha )) w \left[ (1-\delta ) \varPi ^a_j( \overline{\sigma } \mid h^0_\alpha ) + \delta u(a) ( 1 - w )\right] }{ [ 1 - \delta ( 2 - \sigma ^1_i( h^0_\alpha ) - \sigma ^1_j( h^0_\alpha ))w ][1 - \delta ( 2 - \sigma ^1_i( h^0_\alpha ) - {\tilde{\sigma }}^1_j( h^0_\alpha ))w ]} = 0. \end{aligned}$$

(B.4)

Therefore, $\varPi _j( \sigma \mid h^0_\alpha )$ is strictly increasing in $\sigma ^1_j( h^0_\alpha )$. Thus, player j can be made better off by selecting “active” in the first period, which states that if $\sigma ^1_i(h^0_\alpha ) < 1$ then $\sigma ^1_j(h^0_\alpha ) = 1$ for all $j (\ne i)$ when $\sigma$ is an MPE.

(3-3) From (3-2), we can see that if a symmetric MPE exists, then it is an active equilibrium. Next, to show the existence of an active equilibrium, we prove that $\overline{\sigma }$ is an equilibrium.

In a similar manner to the above, we can derive the following:

$$\begin{aligned}&\varPi _i( \sigma _i , \overline{\sigma }_{-i} \mid h^0_\alpha ) = \frac{ \varPi ^a( \overline{\sigma } \mid h^0_\alpha ) \Big [ w \sigma ^1_i( h^0_\alpha ) + 1 - w \Big ] - u(a) ( 1 - w ) }{ 1 - \delta w ( 1 - \sigma ^1_i( h^0_\alpha ))};\nonumber \\&\varPi _i( \sigma _i,\overline{\sigma }_{-i} \mid h^0_\alpha )-\varPi _i( {\tilde{\sigma }}_i,\overline{\sigma }_{-i} \mid h^0_\alpha ) \nonumber \\&\quad = \frac{ (\sigma ^1_i( h^0_\alpha ) - {\tilde{\sigma }}^1_i( h^0_\alpha )) w \left[ (1-\delta ) \varPi ^a_i( \overline{\sigma } \mid h^0_\alpha ) + \delta u(a) ( 1 - w )\right] }{ [ 1 - \delta ( 1 - \sigma ^1_i( h^0_\alpha ))w ][1 - \delta ( 1 - {\tilde{\sigma }}^1_i( h^0_\alpha ) )w ]} = 0, \end{aligned}$$

(B.5)

where $\sigma ^t_i(h^{t-1}) = 1$ for any $t>1$. The last equality comes from $(1-\delta ) \varPi ^a(\overline{\sigma } \mid h^0_\alpha ) + u(a) ( 1 - w ) \delta = 0$ . Therefore, the player cannot be made better off by varying $\sigma ^1_i(h^0_\alpha )$. Thus, $\overline{\sigma }$ is an equilibrium.

Step 4: Finally, we show the “only if” part of (i).

From equation (B.4) and (B.5), we see that, if $\varPi ^a(\overline{\sigma } \mid h^0_\alpha ) = \varPi ^n(\overline{\sigma } \mid h^0_\alpha )$, then there exists an asymmetric equilibrium, where $\sigma ^1_i(h^0_\alpha )<1$ and $\sigma ^1_j(h^0_\alpha )=1$ for $j\ne i$. Moreover, there is no active equilibrium if $\varPi ^a(\overline{\sigma } \mid h^0_\alpha ) < \varPi ^n(\overline{\sigma } \mid h^0_\alpha )$ from Step 1. $\square$

1.3 Appendix C: Lemmas for Section 6

In the following, we provide a lemma and its proof for Section 6. For history $h^t$, the difference between the expected payoff of selecting “no-action” at time $t+1$ and that of selecting “active” at time $t+1$ is $\varPi ^a(\overline{\sigma } \mid h^t) - C - \left[ \varPi ^n(\overline{\sigma } \mid h^t) - \delta w C/( 1- (1-w)\delta )\right] = \varPi ^a(\overline{\sigma } \mid h^t)-(1-\delta ) C/( 1- (1-w)\delta ) - \varPi ^n(\overline{\sigma } \mid h^t)$. Using this, we have the following lemma.

Lemma 5

Suppose that $\varPi ^a(\overline{\sigma } \mid h^t) - C(1- \delta )/(1-\delta + w \delta )\ge \varPi ^n(\overline{\sigma } \mid h^t)$ for any $h^t$. Given an integer $k^*$, suppose that $\varPi ^a(\overline{\sigma } \mid (h_x^{k'}, a:k')) - C(1- \delta )/(1-\delta + w \delta ) > \varPi ^n(\overline{\sigma } \mid (h_x^{k'},a:k'))$ for $k' \ge k^*$. Then, in all MPE, “active” is selected successively after $h^{k^*}=(h^{k^*}_x, a:k^*)$.

Proof

As most of the proof encompasses the same argument as the proof of Lemma 3, we only provide an outline of the proof.

Case 1 (that $k \ge m^* - 1$):

Case 1-1 (that the number of members is N): for any history $h^{k} = (h^{k}_x, a: k)$, in an equilibrium in a subgame after $h^{k}$, every player selects “active” in her/his turn as s/he can earn a positive instant payoff.

Case 1-2 (that the number of members is less than N): Suppose that in every equilibrium of a subgame after $h^{k}=(h^{k}_x, a: k)$, every player selects “active” after the ($k+1$)th period if the number of members is $\ell +1$. We then show that every player selects “active” after the ($k+1$)th period in an MPE of a subgame after any history $h^{k} = (h^{k}_x, a: k)$ if the number of members is $\ell$. If the selected player $x^{k+1}$ is a member, s/he selects “active”, because s/he obtains a positive instant payoff. We consider the case that the selected player is a nonmember and show that all of them select “active” in an equilibrium.

Case 1-2A (that the selected player is a nonmember and s/he selects “active” in the ($k+1$)th period): if the selected player is a nonmember and selects “active” in period $k+1$, then the number of members becomes $\ell +1$. From the assumption, in an equilibrium of a subgame after $h^{k+1}$, the player always selects “active” after the ($k+2$)th period.

Case 1-2N (that the selected player is a nonmember and s/he selects “no-action” in the ($k+1$)th period): Using the same logic as in the proof of Lemma 3, Case 2N yields that selecting “active” is strictly better than choosing “no-action” in a subgame from $h^{k} = (h^{k}_x, a: k)$ in which the number of members is $\ell$.

From Cases 1-1 and 1-2, by induction on $\ell$, we have that after history $h^k = (h^k_x, a:k)$, the outcome where every player selects “active” after $k+1$ is the unique MPE outcome.

Case 2 (that $k~* \le k < m^* - 1$): Suppose that for any history $h^{k+1} = (h^{k+1}_x, a: k+1)$, in an equilibrium of a subgame after $h^{k+1}$, every player selects “active” after the ($k+2$)th period. Using the same logic as in Lemma 3, we obtain that every player selects “active” after the ($k+1$)th period in an MPE of a subgame after any history $h^{k} = (h^{k}_x, a: k)$.

From Cases 1 and 2, the proof is completed by induction on k. $\square$

In addition, if $\varPi ^a(\overline{\sigma } \mid h^t) - C(1- \delta )/(1-\delta + w \delta )\ge \varPi ^n(\overline{\sigma } \mid h^t)$ for any $h^t$, then we can see that $\varPi ^a(\overline{\sigma } \mid h^t) - C$ is positive for any history $h^t$.