Interacting information cascades: on the movement of conventions between groups

Abstract

When a decision maker is a member of multiple social groups, her actions may cause information to “spill over” from one group to another. We study the nature of these spillovers in an observational learning game where two groups interact via a common player, and where conventions emerge when players follow the decisions of the members of their own groups rather than their own private information. We show that: (i) if a convention develops in one group but not the other group, then the convention spills over via the common player; (ii) when conventions disagree, then the common player’s decision breaks the convention in one group; and (iii) when no convention has developed, then the common player’s decision triggers the same convention in both groups. We also show that information spillovers may reduce welfare, and we investigate the surplus-maximizing timing of spillovers.

Appendix: Equilibrium characterization and proofs

We begin with a few definitions, present three lemmas which completely characterize equilibrium play, and then prove our results. Before we begin, we need some notation. A strategy for player $i^{g}$ is a function $\sigma _{i}^{g}:\{a,r\}^{i-1}\times \{H,L\} \rightarrow \{a,r\}$ that maps the profile of decisions of prior players and her own private signal into a decision, and a strategy for player k is a function $\sigma _{k}:\{a,r\}^{2(k-1)}\times \{H,L\} \rightarrow \{a,r\}$. (Our characterization omits beliefs since these can be easily recovered with Bayes’ Rule.)

A history $\bar{d}_{i}=(d_{1},\ldots ,d_{i})$ is balanced if, for every odd integer $j<i$, we have that $d_{j}\ne d_{j+1}$, and is unbalanced otherwise. A balanced history is one where the cumulative number of a (r) decisions does not exceed the cumulative number of r (a) decisions by more than one as of any player $1,\ldots ,i$. The empty history and any singleton history are trivially balanced. A history $\bar{d} _{i}=(d_{1},\ldots ,d_{i})$ is unbalanced on $\varvec{a}$ if at some point in the profile it switches from a balanced profile to a profile of only a, i.e., if there is an odd $j<i$ such that (i) $(d_{1},\ldots ,d_{j})$ is balanced and (ii) $d_{j}=d_{j+1}=\cdots =d_{i}=a$. Let $D_{i}^{a}$ be the set of i-length histories that are unbalanced on a. Likewise, a history $\bar{d}_{i}=(d_{1},\ldots ,d_{i})$ is unbalanced on $\varvec{r}$ if there is an odd $j<i$ such that (i) $(d_{1},\ldots ,d_{j})$ is balanced and (ii) $d_{j}=d_{j+1}=\cdots =d_{i}=r$ . Let $D_{i}^{r}$ be the set of i-length histories that are unbalanced on r.

Lemma 1

Equilibrium play for predecessors of the common player.

Let $i<k$ and $g\in \{A,B\}$. In equilibrium, $\bar{d}_{i-1}^{g}$ belongs to a row in Table 2(a) and player $i^{g}$’s equilibrium strategy $\sigma _{i}^{g*}(\bar{d}_{i-1}^{g},x_{i}^{g})$ is given by the last two columns.

Table 2 Equilibrium strategies of players 1 through k

Proof

The proof is due to Bikhchandani et al. (1992). $\square$

For instance, if player i observes a history that is unbalanced on a, then she chooses a, ignoring her own signal (i.e., if $\bar{d}_{i}^{g}\in D_{i-1}^{a}$, then Table 2(a) shows that $\sigma _{i}^{g*}(\bar{d} _{i-1}^{g},H)=\sigma _{i}^{g*}(\bar{d}_{i-1}^{g},L)=a$). If player i observes a balanced history, then she follows her own signal.

Lemma 2

Equilibrium play for the Common Player.

In equilibrium, $(\bar{d}_{k-1}^{A},\bar{d}_{k-1}^{B})$ belongs to a row in Table 2(b) and the common player’s equilibrium strategy $\sigma _{k}^{*}(\bar{d}_{k-1}^{A},\bar{d} _{k-1}^{B},x_{k})$ is given by the last two columns.

Proof

The proof is computational. Successive application of Lemma A1 lets us enumerate the set of equilibrium histories and compute the equilibrium probability of each history. We then employ Bayes Rule to compute the common player’s belief and write the table. For details, see the Online Appendix. $\square$

The next result identifies the behavior of players moving after the common player.

Lemma 3

Equilibrium after the common player.

Let $g\in \{A,B\}$.

LA3.1 : :

(Player $k+1^{g}$.) In equilibrium, $\bar{d}_{k{}}^{g}=(\bar{d}_{k-1{}}^{g},d_{k})$ belongs to a row in Table 3(a) and player $k+1^{g}$’s equilibrium strategy $\sigma _{k+1}^{g*}(\bar{d}_{k}^{g},x_{k+1}^{g})$ is given by the last two columns.

LA3.2 : :

(Player $k+2^{g}$.) In equilibrium, $\bar{d}_{k+1{}}^{g}=(\bar{d}_{k-1}^{g},d_{k},d_{k+1}^{g})$ belongs to a row in Table 3(b) and player $k+2^{g}$’s equilibrium strategy $\sigma _{k+2}^{g*}(\bar{d}_{k+1}^{g},x_{k+2}^{g})$ is given by the last two columns.

LA3.3 : :

(Subsequent players.) Let $i>k+2$. In equilibrium, $\bar{d}_{i-1{}}^{g}=(\bar{d}_{k-1}^{g},d_{k},d_{k+1}^{g},d_{k+2}^{g},\ldots ,d_{i-1}^{g})$ belongs to a row in Table 3(c) and player $i^{g}$’s equilibrium strategy $\sigma _{i}^{g*}(\bar{d}_{i-1}^{g},x_{i}^{g})$ is given by the last two columns.

Table 3 Equilibrium strategies of players $k+1$ through N

For instance, if $\bar{d}_{k-1}^{g}\in D_{k-1}^{a}$ and $d_{k}=a$, then the top row of Table 3(a) shows that player $k+1^{g}$ is in a cascade on a.

Proof

The proof is a computational exercise that mirrors the Proof of Lemma 2. The details are in the Online Appendix. $\square$

Proof of Proposition 1

Lemma 1 shows that, in equilibrium, player i observes a history $\bar{d}_{i-1}$ in either $D_{i-1}^{a}$, $D_{i-1}^{b}$, or $D_{i-1}^{r}$. For histories where the number of a decisions exceeds the number of r decisions by two or more, we have $\bar{d}_{i-1}\in D_{i-1}^{a}$ and player i chooses a by Table  2. Likewise, for histories $\bar{d}_{i-1}\in D_{i-1}^{r}$, the number of r decisions exceeds the number of a decisions by two or more, and she chooses r. Otherwise, player i observes a history in $D_{i-1}^{b}$ in which case she follows her own signal. $\square$

Proof of Proposition 2

Follows directly from Lemma 2 since a cascade on a (r) occurs when a group’s history is in $D_{k-1}^{a}$ ($D_{k-1}^{r}$) by Lemma 1. $\square$

Proof of Proposition 3

Follows from Lemmas 2 and 3. We illustrate with P3.1. Since both groups are in cascades on a, we have that $\bar{d}_{k-1}^{A}$ and $\bar{d}_{k-1}^{B}$ are in $D_{k-1}^{a}$. Thus, Lemma 2 gives that the common player chooses a. Hence, successive application of Lemma 3 gives that every subsequent player in group g chooses a. The remaining cases are analogous.

Proof of Proposition 4

The proof is a computation exercise and omitted. The details are in the Online Appendix. $\square$

Proof of Proposition 5

In a basic cascade, the players’ asymptotic payoff is

$$\begin{aligned} \lim _{i\rightarrow \infty }w^{B}(i)=\lim _{i\rightarrow \infty }\frac{2p-1}{2} \Lambda (i)=\frac{2p-1}{2(1-\lambda )}. \end{aligned}$$

In an interacting cascade, the payoff of the common player is

$$\begin{aligned} w(k,k)=\frac{2p-1}{2(1-\lambda )}\left[ 1-\lambda ^{k}+\frac{1}{2(1-\lambda ) }[\lambda (1-\lambda ^{\frac{k-1}{2}})]^{2}\right] . \end{aligned}$$

Hence, $w(k,k)\ge \lim _{i\rightarrow \infty }w^{B}(i)$ if and only if

$$\begin{aligned} 1-\lambda ^{k}+\frac{1}{2(1-\lambda )}[\lambda (1-\lambda ^{\frac{k-1}{2} })]^{2}\ge 1, \end{aligned}$$

equivalently,

$$\begin{aligned} (1-\lambda )(1-3\lambda ^{k-2})+\lambda ^{-2}(\lambda ^{\frac{3}{2}}-\lambda ^{\frac{k}{2}})^{2}\ge 0. \end{aligned}$$

Since $p\in (1/2,1),$ then $\lambda =2p(1-p)\in (0,1/2)$. If $k\ge 5,$ then $1-3\lambda ^{k-2}>0$ and $\lambda ^{\frac{3}{2}}-\lambda ^{\frac{k}{2}}>0$, and hence $w(k,k)>\lim _{i\rightarrow \infty }w^{B}(i)$. While if $k=3,$ then $\lambda ^{\frac{3}{2}}-\lambda ^{\frac{k}{2}}=0$ and $1-3\lambda \ge 0$ if $\lambda \le 1/3$ (i.e., $p\ge \frac{1}{6}\sqrt{3}+\frac{1}{2}$), and hence $w(k,k)\ge \lim _{i\rightarrow \infty }w^{B}(i)$. Since $w^{B}(j)<\lim _{i\rightarrow \infty }w^{B}(i)$ for each j, then $w(k,k)>w^{B}(i)$ for all i if either $k\ge 5$ or $k=3$ and $p\ge \frac{1 }{6}\sqrt{3}+\frac{1}{2}$.

If $k=3$ and $p<\frac{1}{6}\sqrt{3}+\frac{1}{2}$, then $w(k,k)<\lim _{i \rightarrow \infty }w^{B}(i)$, so there is some i such that $w(k,k)<w^{B}(i)$. $\square$

Proof of Proposition 6

We prove the result by establishing that the gain from moving the common player exceeds the loss when p is large. Since players in positions 1, $\ldots$, $k-3$ in both groups are unaffected by the move, we only need to show that the (i) minimum gain for players $k-2^{A}$, $k-1^{A}$, $k-2^{B}$, and $k-1^{B}$ exceeds the (ii) maximum loss for the common player and players $k+1^{A}$, $k+1^{B}$, $\ldots$, $N^{A}$, and $N^{B}$. With this in mind, we proceed by constructing a (i$^{\prime }$) lower bound on the gain and (ii$^{\prime }$) an upper bound on the loss. We then establish that (i$^{\prime }$) is strictly greater than (ii$^{\prime }$) when p is sufficiently close to 1. The desired result follows.

Table 4 Gain to players $k-2$ and $k-1$

First, we compute a lower bound on the gain for players $k-2^{A}$, $k-1^{A}$ , $k-2^{B}$, and $k-1^{B}$. Table 4 computes the gain to each of these players. It is clear from the table that the total gain to all four players is at least $G(p)=2(2p-1)\lambda ^{\frac{k-1}{2}}\Lambda (k-3)$.

Table 5 Loss to Common Player and Subsequent Players

Second, we compute an upper bound on the loss for the common player and subsequent players in both groups. Table 5 computes the loss to each of these players. A bit of algebra shows that every row of Table 5 is less than or equal to

$$\begin{aligned} T(p)=\frac{2p-1}{2}\lambda ^{k-2}\Lambda (4)+2(2p-1)\left( \frac{\lambda }{2} \right) ^{2}(\Lambda (k-1)^{2}-\Lambda (k-3)^{2}). \end{aligned}$$

Consequently, the loss to the common player and all subsequent players is at most $L(p)=2(N-k+\frac{1}{2})T(p)$.

We now establish that $G(p)\ge L(p)$ for p sufficiently close to 1, this then implies that the move of the common player is surplus improving. To do this, consider the related inequality

$$\begin{aligned}&\underbrace{2\left( 1-\lambda ^{\frac{k-3}{2}}\right) }_{\text {LHS}(p)}\ge \underbrace{ \left( N-k+\frac{1}{2}\right) \lambda ^{\frac{k-3}{2}}(1-\lambda ^{2})}_{\text {RHS} _{1}(p)}\nonumber \\&\qquad +\underbrace{\left( N-k+\frac{1}{2}\right) (1-\lambda )\lambda ^{\frac{5}{2}}\left( \frac{\Lambda (k-1)^{2}}{\lambda ^{k/2}}-\frac{\Lambda (k-3)^{2}}{\lambda ^{k/2}}\right) }_{\text {RHS}_{2}(p)}. \end{aligned}$$

(1)

Observe that $G(p)=\frac{2p-1}{1-\lambda }\lambda ^{\frac{k-1}{2}}\text {LHS} (p)$ and that $L(p)=\frac{2p-1}{1-\lambda }\lambda ^{\frac{k-1}{2}}(\text {RHS }_{1}(p)+\text {RHS}_{2}(p)).$ Note that (i) $\text {LHS}(1)=2$, (ii) $\text { RHS}_{1}(1)=0$, and (iii) $\text {RHS}_{2}(1)=0$. The first and second equalities are obvious. The third equality follows from the fact that $(1-\lambda )\lambda ^{\frac{5}{2}}(\frac{\Lambda (k-1)^{2}}{\lambda ^{k/2}}- \frac{\Lambda (k-3)^{2}}{\lambda ^{k/2}})=2\lambda -\lambda ^{\frac{k-1}{2} }(1+\lambda )$, which equals 0 when $p=1$ since $k>1$. It follows that (1) is true with strict inequality when $p=1$. Since both sides of (1) are continuous in p, there is a $p_{k}$, with $\frac{1 }{2}<p_{k}<1$, such that (1) is true for all $p\ge p_{k}$. At every $p\in [p_{k},1)$, we have that $\frac{2p-1}{1-\lambda }\lambda ^{\frac{k-1}{2}}>0$. Multiplying both sides of (1) by $\frac{ 2p-1}{1-\lambda }\lambda ^{\frac{k-1}{2}}$ shows that $G(p)\ge L(p)$. $\square$