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The evolution of conventions under condition-dependent mistakes

Abstract

We study the long-run conventions emerging in a stag-hunt game when agents are myopic best responders. Our main novel assumption is that errors converge to zero at a rate that is positively related to the payoff earned in the past. To fully explore the implications of this error model, we introduce a further novelty in the way we model the interaction structure, assuming that with positive probability agents remain matched together in the next period. We find that, if interactions are sufficiently persistent over time, then the payoff-dominant convention emerges in the long run, while if interactions are quite volatile, then the maximin convention can emerge even if it is not risk-dominant. We contrast these results with those obtained under two alternative error models: uniform mistakes and payoff-dependent mistakes.

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Fig. 1
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Notes

  1. 1.

    See Young (2015) for an overview of other mechanisms that can sustain conventions.

  2. 2.

    Their main conclusion is that, differently from Kandori et al. (1993), the payoff-dominant convention is selected in the long run when the population is large enough.

  3. 3.

    Maximin plays a role also in the recent work by Sawa and Wu (2018), where a risk-dominant strategy is no longer guaranteed to be stochastically stable if it is not maximin too.

  4. 4.

    Other possible choice rules encompass—among others—imitation, reinforcement learning and better response (see Young 2001, Section 3.1, for a discussion of different varieties of learning behavior).

  5. 5.

    The role of the revision protocol on the speed of convergence in stochastic evolutionary models has been investigated by Norman (2009). We discuss in Sect. 5 the robustness of our results to different revision protocols.

  6. 6.

    For a thorough review of evolutionary models and coordination games, the interested reader can refer to the recent survey in Newton (2018b).

  7. 7.

    See Peski (2010) for a general framework to study local interaction with an exogenous interaction structure, and Newton (2018a) for a recent extension. See also Weidenholzer (2010) for a recent survey on local interaction models focusing on social coordination.

  8. 8.

    Coexistence of conventions can emerge also under global random interaction when agents are heterogeneous, as in the case they belong to distinct cultural groups (see, e.g., Carvalho 2017).

  9. 9.

    See also the extensions in Shaw and Baer (2011) and Cotton (2009). See Sharp and Agrawal (2012) for evidence on bacteria and Agrawal and Wang (2008) for evidence on insects.

  10. 10.

    This justification is possibly related to the existence of an aspiration level, like in Binmore and Samuelson (1997), where the realized payoff is given by the expected payoff plus a random shock, and the individual changes action only if the realized payoff falls below the aspiration level.

  11. 11.

    Recently, Hwang et al. (2016) make use of intentional mistakes to show that unequal conventions may persist over long periods of time despite being inefficient and not supported by formal institutions.

  12. 12.

    There is a growing literature that investigates the impact of error models on stochastic stability results applied in matching and assignment problems, showing that different error models lead to different predictions (see, e.g., Klaus and Newton 2016; Nax and Pradelski 2015; Boncinelli and Pin 2018).

  13. 13.

    Frey et al. (2012) conduct an experiment to study the emergence of conventions in a networked coordination game and find evidence that the behavior dynamics following an initial deviation leads agents’ choices away from the risk-dominant equilibrium more often than it does from the payoff-dominant equilibrium.

  14. 14.

    In a similar setting, see also the analysis and approximation results by Hwang and Newton (2017).

  15. 15.

    The game, in case action A is not risk-dominant, is sometimes called an assurance game.

  16. 16.

    We note that \(0!=1\) and \(0^0=1\). So, when \(\tau =1\) we have that \(\delta = 1 - 1/(n-1)\) which means that, even if partnerships terminate with certainty, there is always a positive probability to interact with the previous partner through re-matching.

  17. 17.

    It is known that only states belonging to recurrent classes of the model without mistakes can receive positive probability in the stochastically stable distribution. Moreover, if one state receives positive probability then all states in the same recurrent class receive positive probability as well.

  18. 18.

    Since the precise expression for the threshold value of n is cumbersome, we opted to omit it from the proposition, aiming at an easier statement which points to the main result.

  19. 19.

    We opted not to deal with the case \(g(c)/g(d)=\beta (\tau )\), which has measure zero in the parameter space and would bring in the analysis annoying complications. Indeed, when \(g(c)/g(d)=\beta (\tau )\), determining which convention is stochastically stable is complicated by the working of the ceiling function, so that either the risk-dominant convention, the maximin convention or both can be stochastically stable, depending on specific values of n. Also, we opted not to provide a precise expression for the threshold value of n, for the same reason as in Footnote 18. We note that, in general, such threshold here is different than the analogous threshold in Proposition 1.

  20. 20.

    We remind that we have assumed \(c \ge a\). In Sect. 5, we comment on the case \(c < a\) when discussing the payoff structure.

  21. 21.

    Alós-Ferrer and Netzer (2010) show how the result that stochastic stability under the logit dynamics of Blume (1993, 1997) selects potential maximizers in exact potential games depends crucially on the assumption of asynchronous learning, i.e., one and only one agent can revise strategy at each time. A recent strengthening of the notion of stochastic stability which is robust to the specification of revision opportunities and tie-breaking assumptions can be found in Alós-Ferrer and Netzer (2015).

  22. 22.

    If asynchronous learning is assumed as revision protocol, we might simulate inertia by giving revision opportunities to agents who are already best responding and hence would keep choosing the same action by best response.

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Appendix

Appendix

Proof of Lemma 1

Proof

We first show that conventions A and B are recurrent classes. Consider two states, \(s = (\alpha , \mu )\) and \(s' = (\alpha ', \mu ')\), that belong to convention A. Starting from s, we note that with positive probability all matches terminate (probability \(\tau ^{\frac{n}{2}}\)) and new pairs are formed exactly as described by \(\mu '\) (probability \(\prod _{k=0}^{n-1} \frac{1}{n-1-2k}\)). Since \(\alpha =\alpha '\), and any revising agent will maintain action A (so that it is irrelevant which agents actually receive a revision opportunity), we can conclude that \(s'\) can be reached from s with positive probability. If s belongs to convention A and \(s'\) does not, then it means that \(\alpha '(i) = B\) for some agent i. Starting from s, we simply observe that an agent who receives a revision opportunity will never change from A to B, and hence, state \(s'\) cannot be reached with positive probability from s. Altogether, convention A is shown to be a recurrent class. An analogous argument shows that convention B is also a recurrent class.

We now show that no other recurrent class exists. Consider a state s that belongs neither to convention A nor to convention B. We proceed to construct a path of states starting from s and ending in either convention A or convention B, with each step in the path having positive probability to occur. At time t we are in state s, and we consider agent i, with \(x \in \{A,B\}\) being a best response action for i. We denote with K the set of agents who are different from i and choose an action different from x in state s. With probability \(\gamma (1-\gamma )^{n-1}\), the only agent who receives a revision opportunity at time t is agent i, who will choose x with positive probability (either 1 if x is the unique best response or 1 / 2 otherwise). If set K is empty, then we have reached a state in convention x. If set K is non-empty, then consider agent \(j \in K\): With probability at least equal to \(\tau ^2 \frac{1}{n-1}\) the matches involving i and j terminate, and i and j are then matched together. At time \(t+1\), with probability \(\gamma (1-\gamma )^{n-1}\) the only agent who receives a revision opportunity is agent j, who will choose x with positive probability, since j is matched with an agent choosing x, and the fraction of other agents choosing x has not decreased with respect to i’s decision at previous time. As a result, the cardinality of set K gets reduced by 1. If set K is now empty, then we have reached a state in convention x; otherwise, we repeat the above procedure. In a finite number of iterations, equal to the cardinality of K, set K is cleared, and convention x is reached. \(\square \)

Proof of Proposition 1

Proof

Without loss of generality, we set \(h=1\). We start by deriving the stochastic potential \(\rho _A^u\) of convention A under uniform mistakes. Starting from a state in convention B, one mistake is required to turn one agent’s action from B to A. Suppose that k agents choose A, with \(k \ge 1\). The agents who have the highest incentive to switch from B to A are those who are matched to an agent choosing A. We note that we always have a positive probability that an agent playing B is matched with someone who already changed to A. The decision of one such agent involves the comparison between

$$\begin{aligned} (1-\delta ) a + \delta \frac{n-k-1}{n-2}c+\delta \frac{k-1}{n-2}a, \end{aligned}$$
(3)

which is the expected utility of choosing A, and

$$\begin{aligned} (1-\delta ) d + \delta \frac{n-k-1}{n-2}b+\delta \frac{k-1}{n-2}d, \end{aligned}$$
(4)

which is the expected utility of choosing B.

If (3) is larger than or equal to (4), then A can be chosen by best response, increasing the number of agents choosing A from k to \(k+1\). We observe that, if (3) is larger than or equal to (4) when k agents choose A, then this holds a fortiori when the number of such agents is \(k+1\). Hence, the stochastic potential \(\rho _A^u\) is given by the minimum k such that (3) is larger than or equal to (4).

By means of simple algebra we obtain:

$$\begin{aligned} \rho _A^u = {\left\{ \begin{array}{ll} 1 &{} \text {if }\,\, \delta \le \frac{a-d}{a-d+b-c},\\ 1 + \left\lceil \left( 1 - \frac{a-d}{\delta (a-d+b-c)}\right) (n-2)\right\rceil &{} \text {if }\,\,\delta > \frac{a-d}{a-d+b-c}, \end{array}\right. } \end{aligned}$$
(5)

where \(\lceil \cdot \rceil \) denotes the ceiling function.

With an analogous reasoning, we can derive an expression for the stochastic potential of convention B under uniform mistakes, which we denote by \(\rho _B^u\):

$$\begin{aligned} \rho _B^u = {\left\{ \begin{array}{ll} 1 &{} \text {if }\,\, \delta \le \frac{b-c}{a-d+b-c},\\ 1 + \left\lceil \left( 1 - \frac{b-c}{\delta (a-d+b-c)}\right) (n-2)\right\rceil &{} \text {if }\,\,\delta > \frac{b-c}{a-d+b-c}. \end{array}\right. } \end{aligned}$$
(6)

We now compare \(\rho _A^u\) with \(\rho _B^u\) to determine which conventions are stochastically stable. If \(0 < \tau \le {\tilde{\tau }}_m\), then \(0< \delta < {\tilde{\tau }}_m\) since \(\delta < \tau \). Hence \(\rho _A^u = \rho _B^u = 1\), which means that convention A and convention B are both stochastically stable. If \({\tilde{\tau }}_m < \tau \le {\tilde{\tau }}_M\), then \({\tilde{\tau }}_m< \delta < {\tilde{\tau }}_M\) for n sufficiently large, since \(\delta < \tau \) and \(\lim _{n \rightarrow +\infty } \delta = \tau \). Suppose that A is the risk-dominant action and, hence, \((b-c)/(a-d+b-c)< \delta < (a-d)/(a-d+b-c)\). Therefore \(\rho _A^u = 1 < \rho _B^u\), and so only convention A is stochastically stable. By the same token, if B is the risk-dominant action then it is the only stochastically stable convention. Finally, consider \({\tilde{\tau }}_M < \tau \le 1\), which implies \({\tilde{\tau }}_M< \delta < 1\) if n is large enough (for reasons analogous to those given above). We observe that, for n sufficiently large, the following difference,

$$\begin{aligned} \rho _B^u - \rho _A^u = \left( \frac{(a-d)-(b-c)}{\delta (a-d+b-c)}\right) (n-2), \end{aligned}$$
(7)

is strictly greater than 1 if A is the risk-dominant action, implying that convention A is the only stochastically stable convention. By the same token, if B is the risk-dominant action then it is the only stochastically stable convention. \(\square \)

Proof of Proposition 2

Proof

The determination of the stochastic potential of convention A under payoff-dependent mistakes, which we denote by \(\rho _A^p\), proceeds as for the case with uniform mistakes considered in the proof of Proposition 1, with a few adjustments. We note that the path with minimum total resistance to move from convention B to convention A requires that mistakes are made, if possible, by agents who are matched to someone already choosing A (which is always possible but for the first mistake) and also that mistakes are made sequentially; the reason is that the expected loss of choosing A is lower for an agent linked to someone choosing A, and it decreases in the number of agents choosing A. Hence, the first mistake is weighted by \(f(b-c)\), since \(b-c\) is the payoff loss when A is chosen instead of B. Moreover, each of the following mistakes, until the threshold of \(\rho _A^u\) agents choosing A is reached, is weighted by \(f\left( \frac{\delta (n-k-1)}{n-2} (b-c) - \left[ 1 - \frac{\delta (n-k-1)}{n-2}\right] (a-d) \right) \), where the argument of function f is the expected loss of choosing A instead of B when k agents are choosing A, and the current partner is choosing A. Therefore:

$$\begin{aligned} \rho _A^p = {\left\{ \begin{array}{ll} f(b-c) &{} \text {if } \,\,\delta \le \frac{a-d}{a-d+b-c},\\ f(b-c) \!+ \!\displaystyle {\sum _{k=1}^{\rho _A^u-1}} f\left( \frac{\delta (n-k-1)}{n-2} (b-c) - \left[ 1 - \frac{\delta (n-k-1)}{n-2}\right] (a-d) \right) \!\!\!&{} \!\!\text {if }\,\,\delta > \frac{a-d}{a-d+b-c}. \end{array}\right. } \end{aligned}$$

The stochastic potential of convention B under payoff-dependent mistakes, which we denote by \(\rho _B^p\), is obtained similarly, considering what done for \(\rho _B^u\) and adjusting the weights of mistakes analogously:

$$\begin{aligned} \rho _B^p = {\left\{ \begin{array}{ll} f(a-d) &{} \text {if } \,\,\delta \le \frac{b-c}{a-d+b-c},\\ f(a-d) \!+\! \displaystyle {\sum _{k=1}^{\rho _B^u-1}} f\left( \frac{\delta (n-k-1)}{n-2} (a-d) - \left[ 1 - \frac{\delta (n-k-1)}{n-2}\right] (b-c) \right) \!\!\!&{} \!\! \text {if }\,\,\delta > \frac{b-c}{a-d+b-c}. \end{array}\right. } \end{aligned}$$

We compare \(\rho _A^p\) with \(\rho _B^p\) to determine which conventions are stochastically stable. Suppose that A is the risk-dominant action, \(a-d > b-c\). Since f is a strictly increasing function, and noting that \(\rho _A^u \le \rho _B^u\) always, we can conclude that \(\rho _A^p < \rho _B^p\), which means that convention A is the unique stochastically stable convention irrespectively of \(\delta \), and hence, for every \(\tau \in (0,1]\). If, instead, B is assumed to be the risk-dominant action, then \(a-d < b-c\), and we obtain that convention B is the unique stochastically stable convention for every \(\tau \in (0,1]\). \(\square \)

Proof of Proposition 3

Proof

The determination of the stochastic potential of convention A when mistakes are condition-dependent, which we denote by \(\rho _A^c\), proceeds as for the case with uniform mistakes considered in the proof of Proposition 1, with a few adjustments. We note that the path with minimum total resistance to move from convention B to convention A requires that mistakes are made, if possible, by agents who are matched to someone already choosing A (which is always possible but for the first mistake); the reason is that the resistance to mistake of an agent decreases with decreasing payoff currently earned by the agent, and a miscoordinating agent earns the lowest payoff. Hence, the first mistake is weighted by g(b), since b is the payoff earned by every agent when all agents choose B. Moreover, each of the following mistakes, until the threshold of \(\rho _A^u\) agents choosing A is reached, is weighted by g(d), since d is the payoff earned by an agent playing B who is matched with an agent playing A. Therefore:

$$\begin{aligned} \rho _A^c = {\left\{ \begin{array}{ll} g(b) &{} \text {if } \,\,\delta \le \frac{a-d}{a-d+b-c},\\ g(b) + g(d) (\rho _A^u-1) &{} \text {if }\,\,\delta > \frac{a-d}{a-d+b-c}. \end{array}\right. } \end{aligned}$$
(8)

The stochastic potential of convention B under condition-dependent mistakes, which we denote by \(\rho _B^c\), is obtained similarly, considering what done for \(\rho _B^u\) and adjusting the weights of mistakes analogously:

$$\begin{aligned} \rho _B^c = {\left\{ \begin{array}{ll} g(a) &{} \text {if } \,\,\delta \le \frac{b-c}{a-d+b-c},\\ g(a) + g(c) (\rho _B^u-1) &{} \text {if }\,\,\delta > \frac{b-c}{a-d+b-c}. \end{array}\right. } \end{aligned}$$
(9)

We now compare \(\rho _A^c\) with \(\rho _B^c\) to determine which conventions are stochastically stable. If \(0 < \tau \le {\tilde{\tau }}_m\), then \(0< \delta < {\tilde{\tau }}_m\) since \(\delta < \tau \); hence, \(\rho _A^c = g(b) > g(a) = \rho _B^c\), since g is a strictly increasing function and \(b>a\), and so convention B is the unique stochastically stable convention.

We now consider the case where \({\tilde{\tau }}_m < \tau \le {\tilde{\tau }}_M\), which implies \({\tilde{\tau }}_m< \delta < {\tilde{\tau }}_M\) for n sufficiently large, since \(\delta < \tau \) and \(\lim _{n \rightarrow +\infty } \delta = \tau \). Suppose that A is the risk-dominant action and, hence, \((b-c)/(a-d+b-c)< \delta < (a-d)/(a-d+b-c)\). Therefore, we have that \(\rho _A^c = g(b)\) and \(\rho _B^c = g(a) + g(c)(\rho ^u_B -1)\). Since \(\rho ^u_B\) grows unboundedly in n, we can conclude that convention A is the unique stochastically stable convention when n is large enough. If, instead, B is assumed to be the risk-dominant action, then by the same token \(\rho _B^c = g(a)\) and \(\rho _A^c = g(b) + g(d)(\rho ^u_A -1)\), and we obtain that convention B is the unique stochastically stable convention. Finally, we consider the case where \({\tilde{\tau }}_M < \tau \le 1\), which implies \({\tilde{\tau }}_M< \delta < 1\) if n is large enough (for reasons analogous to those given above). Taking the difference between (8) and (9), and considering n sufficiently large, we can write

$$\begin{aligned} \rho _A^c - \rho _B^c< (>) \; 0 \Longleftrightarrow \frac{g(c)}{g(d)} > (<) \; \frac{\rho _A^u}{\rho _B^u}, \end{aligned}$$
(10)

since both \(\rho _A^u\) and \(\rho _B^u\) grow unboundedly when \(n \rightarrow \infty \). By using (5) and (6) to substitute for \(\rho _A^u\) and \(\rho _B^u\), and noting that \(\lim _{n \rightarrow +\infty } \delta = \tau \), we obtain

$$\begin{aligned} \rho _A^c - \rho _B^c< (>) \; 0 \Longleftrightarrow \frac{g(c)}{g(d)} > (<) \; \frac{\tau (a-d+b-c)-(a-d)}{\tau (a-d+b-c)-(b-c)} = \beta (\tau ). \quad \end{aligned}$$
(11)

Therefore, \(g(c)/g(d) > (<) \beta (\tau )\) implies that \(\rho _A^c < (>) \; \rho _B^c\), and hence, convention A (B) is the unique stochastically stable convention. \(\square \)

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Bilancini, E., Boncinelli, L. The evolution of conventions under condition-dependent mistakes. Econ Theory 69, 497–521 (2020). https://doi.org/10.1007/s00199-019-01174-y

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Keywords

  • Risk-dominant
  • Payoff-dominant
  • Maximin
  • Stag hunt
  • Stochastic stability

JEL Classification

  • C72
  • C73