Transitions between equilibria in bilingual games under logit choice

Journal of Mathematical Economics 86 (2020) 24–34

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Journal of Mathematical Economics journal homepage: www.elsevier.com/locate/jmateco

Transitions between equilibria in bilingual games under logit choice✩ Srinivas Arigapudi Department of Economics, University of Wisconsin, 1180 Observatory Drive, Madison, WI 53706, United States of America

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Article history: Received 19 March 2019 Received in revised form 17 October 2019 Accepted 22 October 2019 Available online 4 November 2019 Keywords: Evolutionary game theory Stochastic stability Logit choice Bilingual games

a b s t r a c t We study the effect of introducing a bilingual option on the long run equilibrium outcome in a class of two-strategy coordination games with distinct payoff and risk dominant equilibria under the logit choice rule. Existing results show that in the class of two-strategy games under consideration, the inefficient risk dominant equilibrium is selected in the long run under noisy best response models. We show that if the cost of the bilingual option is sufficiently low then the efficient payoff dominant equilibrium will be selected in the long run under the logit choice rule. © 2019 Elsevier B.V. All rights reserved.

1. Introduction Suppose that we have a large number of anonymous myopic agents who are randomly matched to play a two-strategy symmetric coordination game with payoffs as shown in Table 1. We assume that agents choose strategies using a noisy best response model. In these models, best responses are chosen with a very high probability, and with a small probability, sub-optimal strategies are chosen due to noise in the underlying decision making process. It is these small probability events which will eventually lead to the breakdown of and transitions between equilibria, with some occurring more readily than others. This variation in the difficulties of transitions ensures that a single equilibrium, the stochastically stable equilibrium, will be selected in the long run. Assume that c ≤ d < a < c + d. Strategy 1 is then the payoff dominant equilibrium, while strategy 2 is the risk dominant equilibrium.1 , 2 Existing results show that the risk dominant equilibrium (inefficient equilibrium, strategy 2) is selected in the long run under the noisy best response models (see Kandori et al. (1993) and Blume (2003)). This is because in the above class of games, it is relatively easier to disturb the efficient equilibrium compared to disturbing the inefficient equilibrium as the cost ✩ I wish to thank Professor Bill Sandholm, an anonymous referee and the editor of this journal for many helpful comments and suggestions. Financial support from Army Research Office, United States of America Grant MSN201957 is gratefully acknowledged. E-mail address: [email protected]. URL: http://sites.google.com/view/srinivasarigapudi/home. 1 If a player has a uniform prior over other players’ strategies, then the risk dominant strategy is the one which maximizes his expected payoff (Harsanyi et al., 1988). 2 The payoffs in Table 1 are chosen to simplify later computations without losing any major insights. https://doi.org/10.1016/j.jmateco.2019.10.004 0304-4068/© 2019 Elsevier B.V. All rights reserved.

of transition from the risk dominant equilibrium to the payoff dominant equilibrium is greater than the cost of transition from payoff dominant equilibrium to the risk dominant equilibrium (see Section 6, mainly Eq. (6.9)). The work proposed here suggests a feasible way of selecting the efficient equilibrium. This is done by introducing a new strategy called a bilingual option as discussed below. If we allow an extra option of adopting both strategies at an additional cost e > 0 in the game from Table 1, then we have a new game whose payoffs will be as shown in Table 2. These games are known as bilingual games and strategy 3 is called as the bilingual option. For low adoption costs, the bilingual option can be thought of as a hedging strategy in the face of uncertainty about what other players choose. Once sufficient number of agents choose the bilingual option, playing the payoff dominant equilibrium becomes the unique best response. Our results confirm this intuition and show that the efficient equilibrium will be chosen in the long run for low adoption costs. The seminal papers of Foster and Young (1990), Kandori et al. (1993), Young (1993) and Ellison (1993) model the noise using best response with mutations (BRM) in which the probability of a suboptimal choice is independent of its payoff consequences. This model eases the analysis of equilibrium breakdown, as the difficulty of transiting between two equilibria can be determined by counting the number of mutations needed for the transition to occur. In some applications however, it is more realistic to assume that costly mistakes are less likely to happen, i.e., it is more reasonable to assume that the errors in agents choices are payoff dependent as in the logit model of Blume (1993, 2003). When mistake probabilities are payoff-dependent, the probability of transitions between equilibria becomes more difficult to assess, depending not only on the number of suboptimal choices required, but also on the unlikelihood of each such choice. As

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34 Table 1 Coordination game. 1 2

1

2

a, a c, 0

0, c d, d

Table 2 Bilingual game. 1 2 3

1

2

3

a, a c, 0 a − e, a

0, c d, d d − e, d

a, a − e d, d − e a − e, a − e

a consequence, general results on equilibrium breakdown and transitions are only available for two-strategy games in the small noise limit, where only the noise level in agents decisions is taken to zero. In the case of two-strategy coordination games, a noisy best response model induces a birth–death Markov chain on the state space whose stationary distribution can be computed explicitly by standard methods. However for three-strategy games we no longer have this simplification. In order to be able to derive analytical results on equilibrium breakdown and transitions between equilibria in three-strategy games with payoff dependent error structure, we study behavior in the small noise double limit,3 first taking the noise level in agents’ decisions to zero, and then taking the population size to infinity. Sandholm and Staudigl (2016) (henceforth SS16) show that in this double limit, transitions between equilibria can be described in terms of solutions to continuous optimal control problems. We formulate the problem of transitions between equilibria in bilingual games under the logit choice rule in the small noise double limit as an optimal control problem using the results of SS16. We solve this control problem using the results of Sandholm et al. (2018). The solution to the transition problem is then used to compute the set of stochastically stable states. The previous literature on bilingual games mainly focused on cases where the agents either use deterministic best responses in a network setting (Immorlica et al., 2007; Oyama and Takahashi, 2015) or best responses with mutations (Galesloot and Goyal, 1997; Goyal and Janssen, 1997). To the best of my knowledge, this is the first attempt to study the long run equilibrium outcomes of bilingual games under a noisy best response model with payoff dependent error structure in a large population game framework.4 The work closest to ours is by Galesloot and Goyal (1997). They show that for a sufficiently large population size, the payoff dominant equilibrium in the bilingual game will be stochastically stable if and only if e ≤ eBRM in the small noise limit under the BRM model. They characterize the threshold adoption cost of the bilingual option, eBRM in terms of the entries of the payoff matrix of the bilingual game. Our main result on the other hand shows that the payoff dominant equilibrium in the bilingual game will be stochastically stable if e ≤ eL in the small noise double limit under the logit choice rule (see Remark 6.4 for comparison of results under the logit and BRM choice rules). The paper proceeds as follows: Section 2 introduces our class of stochastic evolutionary processes. We formally define the unlikelihood function and describe it under the BRM and logit choice 3 For earlier work on small noise double limit, see Sandholm (2010b) and Staudigl (2012). 4 Alós-Ferrer and Netzer (2010), Marden and Shamma (2012) and Okada and Tercieux (2012) consider logit choice rule but use a normal game framework to compute stochastically stable states.

25

rules. In Section 3, we describe the formal modeling procedure of revision opportunities. We then set up the transition cost problem in the small noise double limit and define stochastically stable states in this setting. Section 4 provides definitions and introduces matrix notation which will be used in our main analysis. In Section 5.1, we state the verification theorem from Sandholm et al. (2018) which provides a sufficient condition for a value function to be a solution in a certain class of optimal control problems. In Section 5.2, we formulate the transition cost problem in bilingual games in the small noise double limit as an optimal control problem and give a brief outline of how the verification theorem presented in Section 5.1 is used to solve the transition cost problem. In Section 5.3, we compute the costs of certain direct paths in the small noise double limit under the logit choice rule. These costs are used to present the solution of the transition cost problem and also compute the set of stochastically stable states in Section 6. Section 7 concludes. 2. Model 2.1. Population games We consider games in which agents from a population of size N choose strategies from the common finite strategy set S = {1, 2, . . . , n}. The population’s aggregate behavior is described by a state x, an element of the simplex X = {x ∈ Rn+ : ∑population n i=1 xi = 1}. Our main analysis later will focus on cases with three strategies i.e., n = 3. The standard basis vector ei ∈ X ⊂ Rn represents the pure population state at which all agents play strategy i. At population state x = (x1 , x2 , . . . , xn )′ , xi represents the fraction of agents playing strategy i. We suppose that agents are randomly matched to play the symmetric normal form game A ∈ Rn×n . The (i, j)th entry Aij is the payoff a player obtains when he chooses strategy i and his opponent chooses strategy j. The expected to strategy i at population state x is described by ∑payoff n 5 Fi (x) = j=1 Aij xj . In matrix notation, we have F (x) = Ax. 2.2. Noisy best response protocols and unlikelihood function In our model of stochastic evolution, agents occasionally receive opportunities to switch strategies. The formal modeling procedure for revision opportunities and the induced Markov chain on the state space is described in Section 3.1. Upon receiving a revision opportunity, an agent selects a strategy by employing a noisy best response protocol σ η : Rn → int(X ) with noise level η > 0, a function that maps vectors of payoffs to vectors of probabilities of choosing each strategy. For a noisy best response protocol σ η , the unlikelihood function Υj represents the exponential rate of decay of the probability that strategy j is chosen as η approaches zero i.e., η

Υj (π ) = − lim η log σj (π ) η→0

(2.1)

Example 2.1 (Best Response with Mutations). The best response with mutations (BRM) choice protocol is the most commonly used noise model in evolutionary ) theory. Denote the noise level ( game by η and let ϵ = exp − η1 . Under the BRM model, a best

response is chosen with probability 1 − ϵ . With probability ϵ player’s choice is uniformly random over all sub-optimal actions. More formally, we define as follows. Let Sopt (π ) = {k ∈ S : πk ≥ πi for all i ∈ S } 5 We will interchangeably use π (x) to denote the payoff vector i.e., π (x) = F (x).

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S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

denote the set of optimal strategies when the payoff vector is π . Then, the BRM model is defined by6

{

η

σ j (π ) =

if j ∈ Sopt (π ),

1−ϵ |Sopt (π )| ϵ |S ∖Sopt (π )|

(2.2)

otherwise

It is easy to verify that the unlikelihood function for the BRM model is given by

Υj (π ) =

0

if j ∈ Sopt (π ),

1

otherwise

{

(2.3)

Example 2.2 (Logit Choice). The logit choice protocol with noise level η, introduced to evolutionary game theory by Blume (1993), is defined by η

σ j (π ) = ∑

exp(η−1 πj )

k∈S

exp(η−1 πk )

.

(2.4)

It is easy to verify that this protocol has piecewise linear unlikelihood function given by

Υj (π ) = max πk − πj ♦

(2.5)

j from strategy i is given by the unlikelihood of choosing strategy j at the new population state. cxN,y = Υj (F N (y)) for y = x +

1 N

(ej − ei ), j ̸ = i

(3.3)

Let ⟨·, ·⟩ denote the standard inner product on Rn . Let [·]+ denote the positive part of vector i.e., for z ∈ Rn we define [z ]+ ∈ Rn+ by ([z ]+ )i = [zi ]+ .7 Let φ : [0, T ] → X be absolutely continuous and non-pausing, meaning that |φ˙ t | ̸ = 0 for almost all t ∈ [0, T ]. The cost of the continuous path φt in the small noise double limit is derived in SS16 and is given by c(φ ) =

T

∫

⟨Υ (F (φt )), [φ˙ t ]+ ⟩dt .

(3.4)

0

The intuition for Eq. (3.4) is as follows: [φ˙ t ]+ identifies the strategy agents are switching to. Suppose φ˙ t = ej − ei , then we have [φ˙ t ]+ = ej . In this case, the integrand ⟨Υ (F (φt )), [φ˙ t ]+ ⟩ simplifies to Υj (F (φt )) which is the cost of the step. The cost of a discrete path is given by the sum of the costs of its steps. As the number of steps increases, in the limit we have a continuous path whose cost will be given by an integral as in Eq. (3.4).

k∈S

3.3. Transition problem and stochastic stability 3. The small noise double limit 3.1. Induced Markov chain A population game F , a noisy best response protocol σ η and a population size N generate a Markov chain XN ,η on the set of population states X . The process runs in discrete time, with each period taking 1/N units of clock time. During each period, a single agent is chosen at random from the population. This agent updates his strategy by applying the noisy best-response protocol σ η . This procedure generates a N ,η Markov chain XN ,η = {Xt }∞ t =0 on the state space X . The index k denotes the number of revision opportunities that have occurred to date and corresponds to k/N units of clock time. The transition probabilities for the Markov chain XN ,η is given by:

Since revising agents are chosen at random and play each strategy in S with positive probability, the Markov chain XN ,η is irreducible and aperiodic, and so admits a unique stationary distribution µN ,η . Stochastically stable states are the points in the simplex where the mass in the stationary distribution µN ,η , becomes concentrated as the parameters of the process approach their limiting values. Following SS16, formally we have the following definition. Definition 1. Let O(X , x) denote the set of open subsets of X containing x. State x∗ ∈ X is stochastically stable in the small noise double limit if for all δ > 0, O ∈ O(X , x∗ ), there exists N0 ∈ N, such that for all N > N0 , there exists η0 > 0, such that for all η < η0 , we have µN ,η (O) > exp(−ηδ ).

Over short to medium time scales, XN ,η typically converges to an equilibrium of the underlying game F . Over longer periods, runs of suboptimal choices occasionally occur, leading to transitions between the equilibria. In the small noise double limit (first η → 0, then N → ∞), SS16 characterize these transitions as solutions to control problems whose running costs are obtained by composing the unlikelihood function with the payoff function, as we describe next.

In words, the above definition says that state x∗ ∈ X is stochastically stable in the small noise double limit if for any open set O ⊂ X containing x∗ , probability mass of the stationary distribution µN ,η (O) does not vanish at an exponential rate in η once N is sufficiently large. An interpretation of the limiting stationary distribution in the double limit is that it approximates the limiting stationary distribution in η when N is large enough i.e., limN limη µN ,η ≈ limη µN0 ,η for large enough N0 .8 It can be shown that the stochastically stable states are a subset of the recurrent classes of the unperturbed Markov Chain XN ,0 . For bilingual games, there are only two recurrent classes of the process XN ,0 which correspond to the strict Nash equilibria e1 and e2 . To evaluate stochastic stability in the small noise double limit, we must assess the costs of transitions between strict equilibria ei and ej (here i, j = {1, 2} and j ̸ = i):

3.2. Step costs and path costs

C ({ej }, {ei }) = inf{c(φ ) : φ ∈ Φ ({ej }, {ei })}

(

N ,η

N ,η

P Xt +1 = y|Xt

=x

)

⎧ η ⎪ ⎨xi σj (Ax) ∑n η = ⎪ i=1 xi σi (Ax) ⎩ 0

if y = x +

1 (ej N

− ei ),

if y = x,

(3.1)

otherwise

Let x, y ∈ X . Cost of a step from x to y is defined by: cxN,y = − lim η logPNx,,η y η→0

(3.2)

cxN,y is the exponential rate of decay of the probability of a step from x to y as η approaches zero. The cost of switching to strategy 6 |A| denotes the cardinality of the set A and A ∖ B denotes the set difference between the sets A and B.

(3.5)

where for K , L ⊂ X , Φ (K , L), denotes the set of all absolutely continuous paths of arbitrary duration through X from K to L and c(φ ) is the cost of the continuous path φ (see Eq. (3.4)) in the { 7 For a ∈ R, [a] = +

a

if a ≥ 0

otherwise. 8 In cases where the double limit puts all mass on a particular strategy, it is a reasonable guess that we get the same distribution for all large N0 , so the result could hold for a single limit η → 0. This is an interesting open problem which however cannot be solved by the techniques used in the current paper. 0

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

27

small noise double limit. Eq. (3.5) is the transition cost problem9 whose solution will give us the set of stochastically stable states. Adapting Theorem 10 of SS16 to our specific problem, we have the following proposition. Proposition 3.1. For a bilingual game with payoffs as shown in Table 2, strict equilibrium e1 will be stochastically stable in the small noise double limit if and only if C ({e1 }, {e2 }) ≥ C ({e2 }, {e1 }). Similarly, strict equilibrium e2 will be stochastically stable if and only if C ({e2 }, {e1 }) ≥ C ({e1 }, {e2 }). 4. Preliminary analysis 4.1. Notation We follow the notation introduced by SS16 for working with symmetric games A ∈ Rn×n . We use superscripts to refer to rows of A and subscripts to refer to columns. Thus Ai is the ith row of A, Aj is the jth column of A, and Aij is the (i, j)th entry. These objects can be obtained by pre- and post-multiplying A by standard basis vectors: Ai = e′i A,

Aj = Aej ,

Fig. 1. Best response regions for low adoption costs.

Aij = e′i Aej .

In a similar fashion, we use super- and subscripts of the form i − j to denote certain differences obtained from A. Ai−j = Ai − Aj = (ei − ej )′ A, i−j

j

j

Ak−l = Aik − Ail − Ak + Al = (ei − ej )′ A(ek − el ). In this notation, the best response region for strategy i is described by Bi = {x ∈ X : Ai−l x ≥ 0 for all l ∈ S }.

(4.1)

The set Bij = Bi ∩ Bj is the boundary between the best response regions for strategies i and j. We denote the lines as follows: lij = {sei + (1 − s)ej : 0 ≤ s ≤ 1} for i, j ∈ {1, 2, 3} and i ̸ = j. (4.2)

Fig. 2. Best response regions for moderate adoption costs.

4.2. Best response regions The bilingual game with payoffs as given in Table 2 can be represented by the following matrix A.

( A=

a c

0 d

a d

a−e

d−e

a−e

) (4.3)

c ≤ d < a < c + d implies that we have c > 0. d ≥ c reflects the assumption that failing to coordinate is never better than coordinating. Depending on the magnitude of the adoption costs, the best response regions of the bilingual game can be divided into the following three cases: d(a−d) , we have the best Case 1) Low adoption costs: For e ≤ a response regions as shown in Fig. 1. (a−c) Case 2) Moderate adoption costs: For da (a − d) < e < d (a+d−c) , we have the best response regions as shown in Fig. 2. (a−c) Case 3) High adoption costs: For e ≥ d (a+d−c) , we have the best response regions as shown in Fig. 3. 9 To be more precise, Eq. (3.5) is the transition cost problem to the equilibrium ei starting from equilibrium ej . In general, we can define the transition cost problem to the equilibrium ei starting from any state x ∈ X (see Eq. (5.3)). In this paper, we solve the general version of the transition cost problem using optimal control theory results from Sandholm et al. (2018). See Section 5.2 for details.

Fig. 3. Best response regions for high adoption costs.

From Figs. 1–3, it is clear that the best response region of the bilingual option shrinks in size as the cost of the bilingual option increases from low to moderate values and for sufficiently high values of the adoption costs, the best response region of the bilingual option is empty. In this paper, we only consider the effect of the bilingual option on the set of stochastically stable states when the adoption

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S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

to Y ◦ as the interior of Y . Likewise, derivatives of functions f : Y → Rn will be understood as maps Df : Y → L(TY , Rn ) from Y to linear functions from TY to Rn . Let TY (y) = {α (z − y) : for some z ∈ Y and α ≥ 0} ⊆ TY be the tangent cone of Y at y. TY (y) is the set of feasible controls at state y. Assume that the running cost function L : Y × TY → [0, +∞] takes a semi-linear form. Specifically, we assume that for some Lipschitz continuous function Ψ : Y → Rn+ , we have L(y, v ) =

{∑n

i=1

Ψi (y)[vi ]+

if v ∈ TY (y) otherwise

+∞

The constraint that the state remains in Y is built into the definition of running costs. Let Z ⊂ TY be compact and convex. The control problem and its value function V ∗ : Y → R+ are defined as follows: V ∗ (x) = min

T

∫

L(φt , νt )dt

(5.1)

0

costs are low. Subsequent analysis will assume that the adoption d(a−d) costs are low i.e., e ≤ a .

over T ∈ [0, ∞), ν : [0, T ] → Z measurable subject to φ : [0, T ] → Y absolutely continuous, φ0 = x, φT ∈ Ω , φ˙ t = νt for almost every t ∈ [0, T ]. The results proved in Sandholm et al. (2018) provide a sufficient condition for a function V : Y → R+ to be the value function of the above problem. The key requirement is that the Hamilton–Jacobi–Bellman(HJB) equation

4.3. Description of important states for low adoption costs

min(L(x, u) + DV (x)u) = 0

We define a few states (see Fig. 4) here which will later be used to present the solution of the transition problem in Section 6. [ ]′ e e x13 = 1 − 0 d d [ e e ]′ 13 y = 0 1− d [ d ]′ e a − c−e x23 = 0 a−c a−c

hold at almost every x ∈ Y ◦ . Adapting Theorems 3.3 and 3.4 from Sandholm et al. (2018) to our specific problem gives the following version of the verification theorem.

Fig. 4. Division of best response regions for low adoption costs.

23

y

[

a−d−e

e

]′

= 0 a−d a−d [ ]′ e a−d−e p1 = 0 a−d a−d [e e ]′ 2 p = 0 1−

d d We divide the best response region of strategies 1 and 3 as follows10 (see Fig. 4): B11 = conv{e1 , x13 , y13 , p2 } B12 = conv{p2 , y13 , e3 } B31 = conv{x23 , p1 , y23 } B32

= conv{p , x , y , y } 1

13

13

23

5. Main analysis 5.1. Verification theorem Let Y be a compact convex subset of Rn . Let TY denote the set of tangent vectors from states in the relative interior Y ◦ of Y . Let the set Ω ⊂ Y be closed relative to Y and have piecewise smooth boundary. In what follows, topological statements are made with respect to the relative topology on Y ; for instance, we will refer 10 conv(X ) denotes the convex hull of the set X .

u∈Z

(5.2)

Theorem 5.1 (Verification Theorem (Sandholm et al., 2018)). Let V : Y → R+ be a continuous function that is continuously differentiable a.e. on Y ◦ . Suppose that (i) For every x ∈ Y , there is a time T ∈ [0, ∞) and a measurable function ν : [0, T ] → TY such that the corresponding controlled φ : [0, T ] → Y with φ0 = x satisfies φT ∈ Ω and ∫trajectory T L(φt , νt )dt = V (x); 0 (ii) The HJB equation (5.2) holds a.e. on Y ◦ . (iii) The boundary condition V (x) = 0 holds at all x ∈ Ω . Then V = V ∗ . In our application, n = 3 and the running cost function is L(x, u) = (Υ (F (x)))′ [u]+ , where Υ (F (x)) is the unlikelihood function under the logit choice rule (see Eq. (2.5)). We next briefly explain how the verification theorem is used to solve the transition cost problem in bilingual games. 5.2. Transition cost problem as an optimal control problem In what follows, we restrict our attention to bilingual games. We will now show that the transition cost problem to equilibrium e1 for a bilingual game can be formulated as an optimal ∑3control problem. The state space is given by X = {x ∈ R3+ : i=1 xi = 1}. The bilingual game (see Eq. (4.3)) has two pure strict Nash equilibria which can be represented by the standard basis vectors e1 and e2 . For x ∈ X , let V ∗ (x) ≡ inf{c(φ ) : φ ∈ Φ ({x}, B1 )}

= inf{c(φ ) : φ ∈ Φ ({x}, {e1 })}

(5.3)

where recall that for K , L ⊂ X , Φ (K , L), denotes the set of all absolutely continuous paths of arbitrary duration through X from K to L and c(φ ) is the cost of the continuous path φ (see Eq. (3.4)) in the small noise double limit. We can replace B1 with e1 in the

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

above definition as there exists a zero path cost from any state in the best response region of strategy 1, B1 to equilibrium e1 .11 We observe that by setting Y = X and Ω = B1 , the transition cost problem to equilibrium e1 (see Eq. (5.3)) reduces to an optimal control problem of the form presented in Section 5.1 (see Eq. (5.1)). The solution to the problem is trivial for states x ∈ B1 as there is a path with zero cost starting from x and ending at e1 . The solution is however non-trivial for states x ∈ B2 ∪ B3 . We guess that the solution to the problem starting from such states is as shown in Fig. 5. Using the verification theorem, we show that this guess is indeed the solution to the problem. The details are provided in Section 6.1. By interchanging the roles of indices 1 and 2, we can formulate the transition cost problem to equilibrium e2 as an optimal control problem of the form presented in Section 5.1. The solution to the problem is trivial for states x ∈ B2 as there is a path with zero cost starting from x and ending at e2 . For the non-trivial part of the problem i.e., for states x ∈ B1 ∪ B3 , we guess that the solution is as shown in Fig. 6. Using the verification theorem, we show that this guess is indeed the solution to the problem. The details are presented in Section 6.2. With the solution of transition cost problems, it is straightforward to compute the costs of transitions between strict equilibria (see Eq. (3.5)) and thereby compute the set of stochastically stable states using Proposition 3.1. These details are provided in Section 6.3. We next compute the costs of direct paths shown in Figs. 5 and 6 which are then used in Section 6 to describe the value function of the transition cost problem.

For x, y ∈ X , let γ (x, y) denote the cost of the direct (straightline) path from x to y in the small noise double limit for a three-strategy symmetric game A under the logit choice rule:

φt = (1 − t)x + ty. y = x + c(ek − ej ) ∈ B for some c > 0

(5.5)

(Aj−k x)2 j−k Aj−k

Let x ∈ B1 and suppose that

y = x + c(e2 − e1 ) ∈ B13 for some c > 0 Then, γ (x, y) = cA Lemma 5.4.

1−2

x−

1 2 1−2 c A1−2 2

and c =

(5.6) A1−3 x 3 . A11− −2

1

Let x ∈ B and suppose that

y = x + c(e2 − e3 ) ∈ B13 for some c > 0 2 Then, γ (x, y) = cA1−2 x − 12 c 2 A13− −2 and c =

(5.7)

y = x + c(e2 − e1 ) ∈ l23 for some c > 0 Then, γ (x, y) = cA

x−

1 2 1−2 c A1−2 2

γ (x, y) =

1

∫

[φ˙ t ]′+ Υ (F (φt ))dt 0 1

∫

ce′2 Υ (F (φt ))dt

= 0

1

∫

(F1 (φt ) − F2 (φt ))dt

=c 0 1

∫

(A1−2 x + tA1−2 (y − x))dt

=c 0

( ) 1 = c A1−2 x + A1−2 (y − x) 2

1

2 = cA1−2 x − c 2 A11− −2 □

2

6. Main results

Theorem 6.1. Let A be a bilingual game (4.3). In this case, the value function V ∗ : B2 ∪ B3 → R+ for the transition cost problem with target set B1 is given by the continuous function 1 (A2−3 x)2 2 A2 − 3 2 −3

(5.8)

and c = x1 .

11 One such path involves switching to strategy 1 from strategy 2 until reaching the boundary of the simplex and then switching to strategy 1 from strategy 3 until reaching the equilibrium e1 . As strategy 1 is the best response in B1 , the unlikelihood of choosing it is zero by definition and so the total path cost is zero.

if x ∈ B2 , if x ∈ B3 .

0

(6.1)

The optimal feedback control in this case is (see Fig. 5)

ν ∗ (x) = e3 − e2

(6.2)

Proof. For x ∈ B3 , switching to strategy 3 from strategy 2 i.e., movement along the direction e3 − e2 costs zero as strategy 3 is the best response. Therefore, in this region the optimal control is e3 − e2 . From Lemma 5.2, the cost of switching to strategy 3 from strategy 2 for the states in B2 until reaching a state in B23 is V (x) = 1 (A2−3 x)2 2 2−3 . This implies that for x ∈ int(B ), we have DV (x) = 2 A2 − 3 (A2−3 x) 2−3 . 3 A A22− −3 2

The running cost function L(x, u) = (Υ (F (x)))′ [u]+ for

x ∈ B under logit choice rule is given by

[

Let x ∈ B13 and suppose that

1−2

(from Eq. (2.5))

We now compute as follows:

L(x, u) = A2−1 x

A1−3 x 3 . A13− −2

Lemmas 5.2–5.4 follow from Lemma 3 of SS16. Lemma 5.5.

e′2 Υ (F (φt )) = F1 (φt ) − F2 (φt )

V (x) = (5.4)

jk

Lemma 5.3.

Clearly, φ˙ t = y − x = c(e2 − e1 ) and therefore [φ˙ t ]′+ = ce′2 .

∗

For j, k ∈ {2, 3} and k ̸ = j, let x ∈ Bj . Suppose that

1 2

Fl (φt ) = Al (x + t(y − x)) = Al x + tAl (y − x)

{

γ (x, y) = c(φ ), where φ : [0, 1] → X is defined by

Then γ (x, y) =

Proof. Since y ∈ l23 , we have y1 = 0 and therefore c = x1 from Eq. (5.8). Let π = F (φt ) = Aφt , where φt = x + t(y − x). For l = i, j and k we have

6.1. Transitions to state e1

5.3. Costs of certain direct paths

Lemma 5.2.

29

0

A2−3 x [u]+

]

(6.3)

where u = ea − eb for a, b = {1, 2, 3} and a ̸ = b. Here, the vector u represents the six basic directions of motion. We show that for x ∈ B2 , the optimal control is again e3 − e2 . This entails verifying the HJB equation (5.2) for states x ∈ int(B2 ) by showing that the function H(x, u) = L(x, u) + DV (x)u is 0 for u = e3 − e2 and that it is non-negative for the remaining five basic directions of motion. We verify this in Appendix A. □ 6.2. Transitions to state e2 Theorem 6.2. Let A be a bilingual game (4.3). In this case, the value function V ∗ : B3 ∪ B1 → R+ for the transition cost problem with

30

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

Fig. 5. Transitions to state 1.

Fig. 6. Transitions to state 2.

target set B2 is given by the continuous function

⎧ 0 if x ∈ B31 , ⎪ ⎪ ⎪ ⎪ 1 (A3−2 (x+x1 (e3 −e1 )))2 ⎪ if x ∈ B32 , ⎪ −2 ⎪ A33− ⎪2 2 ⎪ )2 ( ⎨ 1−3 1−3 2 1 (A3−2 y13 )2 V ∗ (x) = A 1−3x A1−2 x − 1 A 1−3x A11− if x ∈ B11 2 −2 + 2 2 A1 − 2 A1 − 2 A33− ⎪ −2 ⎪ ⎪ ⎪ 2 2 1 (A3−2 y13 )2 1−2 ⎪ ⎪ k1 A1−2 x − 21 k21 A11− y − 21 k22 A13− 2 −2 + k2 A −2 + 2 ⎪ A33− ⎪ −2 ⎪ ⎩ 1 if x ∈ B2 (6.4) where k1 = x1 , y − x = k1 (e2 − e1 ) and k2 =

A1−3 y 3 A13− −2

.

for x ∈ B11 is given by

if x ∈ B3 ∖ l23 , if x ∈ B1 ∖ l23 ,

(6.5)

if x ∈ (B1 ∪ B3 ) ∩ l23

Proof. We compute the path costs generated by the controls in Eq. (6.5) (see Fig. 6) and then verify the corresponding HJB equation (5.2) in Appendix B. Consider the states x ∈ B3 i.e., x ∈ B31 ∪ B32 . For x ∈ B31 , switching to strategy 3 from strategy 1 until reaching B23 costs zero as strategy 3 is the best response here. Therefore e3 − e1 is the optimal control in this region. Path Costs for states in B32 : For x ∈ B32 , the total cost of the path can be decomposed into two parts. The first part involves switching to strategy 3 from strategy 1 (this costs zero as strategy 3 is the best response in this region) until reaching the boundary of the simplex l23 (denote this state by y). The second part involves switching to strategy 2 from strategy 3 until reaching B23 . From Lemma 5.2, the total cost of this path is given by 1 (A3−2 y)2 2

2 A33− −2

A1−2

involves switching to strategy 3 from strategy 1 until reaching the state y13 . Since strategy 3 is the best response here, the cost of this path is zero. Finally, the third part involves switching to strategy 2 from strategy 3 until reaching B23 . From Lemma 5.2, the cost of (A3−2 y13 )2 this path is 12 . Adding the three path costs, the total cost 3−2 A3−2

The optimal feedback controls in this case is (see Fig. 6)

⎧ ⎨e3 − e1 ∗ ν (x) = e2 − e1 ⎩ e2 − e3

H(x, u) = L(x, u) + DV (x)u is 0 for u = e3 − e1 and that it is non-negative for the remaining five basic directions of motion in Appendix B. Path Costs for states in B11 : For x ∈ B11 , the total cost can be divided into three parts. The first part involves switching to strategy 2 from strategy 1 until reaching a state on B13 (denote this state by y). From Lemma 5.3, the cost of this path is k1 A1−2 x − 1 −3 1 2 1−2 k A , where k1 = A 1−3x and y − x = k1 (e2 − e1 ). The second part 2 1 1−2

(6.6)

Here y = x + k(e3 − e1 ), for some k > 0. Since y ∈ l23 , we have y1 = 0 which implies that k = x1 . We verify the corresponding HJB equation (5.2) by showing that the function

V (x) = k1 A1−2 x −

1 2

2 k21 A11− −2 +

1 (A3−2 y13 )2 2

(6.7)

2 A33− −2

We verify the corresponding HJB equation (5.2) by showing that the function H(x, u) = L(x, u) + DV (x)u is 0 for u = e2 − e1 and that it is non-negative for the remaining five basic directions of motion in Appendix B. Path Costs for states in B12 : For x ∈ B12 , the total cost can be divided into three parts. The first part involves switching to strategy 2 from strategy 1 until reaching a state on l23 (denote this state 2 by y). From Lemma 5.5, the cost of this path is k1 A1−2 x − 21 k21 A11− −2 , where k1 = x1 and y − x = k1 (e2 − e1 ). The second part involves switching to strategy 2 from strategy 3 until reaching the state 2 y13 . From Lemma 5.4, the cost of this path is k2 A1−2 y − 12 k22 A13− −2 , where k2 =

A1−3 y 3 . A13− −2

Finally, the third part involves switching to

strategy 2 from strategy 3 until reaching B23 . From Lemma 5.2, (A3−2 y13 )2 the cost of this path is 12 . Adding the three path costs, 3 −2 A3−2

we get V (x) = k1 A1−2 x −

1 2

2 1−2 k21 A11− y− −2 + k2 A

1 2

2 k22 A13− −2 +

1 (A3−2 y13 )2 2

2 A33− −2

(6.8)

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

where y − x = k1 (e2 − e1 ), k1 = x1 = e′1 x and k2 =

A1−3 y 3 A13− −2

. We

verify the corresponding HJB equation (5.2) by showing that the function H(x, u) = L(x, u) + DV (x)u is 0 for u = e2 − e1 and that it is non-negative for the remaining five basic directions of motion in Appendix B. □

first compare the cutoff values eL , eBRM under the two choice rules for the case when c = d. d For c = d, eBRM = a− . Since 1 < da < 2, it is straightforward 2 to verify that

(

√

d 1− 6.3. Stochastically stable states In the absence of the bilingual option, the costs of transitions between the equilibria for the coordination game (payoffs in Table 1) under logit choice are given by Sandholm (2010a): C

old

C

old

({e1 }, {e2 }) = ({e2 }, {e1 }) =

1 2 1

(a − c)2

(

) and

a+d−c d2

(

2

(6.9)

)

a+d−c

Since by assumption c + d > a, we have C old ({e2 }, {e1 }) > C old ({e1 }, {e2 }). In words, this means that the cost of transition from the risk dominant equilibrium to the payoff dominant equilibrium is higher than the reverse transition. Therefore, the risk dominant equilibrium (which in our case is inefficient) is selected in the long run. The new transition costs in the presence of the bilingual option from Theorems 6.1 and 6.2 are given by: C new ({e1 }, {e2 }) = W ∗ (e1 )

= +

e

(a − c) −

d(

1

(

1 ( e )2 2

d

(

(a − d) 1 −

2

(a + d − c)

) e d

)2 ) −e

a−d

(from Eq. (6.4))

C new ({e2 }, {e1 }) = W ∗ (e2 )

=

1

(

2

e2

)

a−d

(from Eq. (6.1))

From the above expressions it is clear that for small values of e, the cost of transition from the risk dominant equilibrium to the payoff dominant equilibrium (C new ({e2 }, {e1 })) will be lower than the reverse transition (C new ({e1 }, {e2 })) and so the efficient equilibrium will be selected in the long run. Formally, we have the following: Proposition 6.3. In a bilingual game (4.3) with low adoption costs, the payoff dominant equilibrium stable ) } if and only { ( e1 is√stochastically if e ≤ eL where eL = min d 1 −

c +d−a c

, da (a − d) , under the

31

2d − a

)

d

=

a−d

√ > 1 + 2dd−a

a−d 2

and

d a

(a − d) >

a−d 2

.

d From Proposition 6.3, we therefore have that eL > eBRM = a− . 2 Therefore, for adoption costs e such that eBRM < e < eL , the efficient equilibrium is selected in the long run under the logit choice rule but not under the BRM choice rule. In words, this means that it is easier to select the efficient equilibrium under logit choice compared to the BRM model for c = d. There can however be cases where it is easier to select the efficient equilibrium under BRM model compared to the logit choice rule. For example, with a = 5, c = 3 and d = 4, we have eL = 0.734 and eBRM = 0.8. Therefore, for adoption costs e such that 0.734 = eL < e < eBRM = 0.8, the efficient equilibrium is selected in the long run under the BRM choice rule but not under the logit choice rule.

7. Conclusion In this paper we considered the transition cost problem, a control problem associated with large deviations properties, which is used to assess the probable time until a transition between a given pair of stable equilibria, and to determine the most likely path that this transition will follow. Solving the transition problem in a class of bilingual games, we have shown that if the cost of the bilingual option is sufficiently low then the efficient equilibrium will be stochastically stable in the small noise double limit under the logit choice rule. The current paper provides a necessary and sufficient condition for the payoff dominant equilibria to be stochastically stable in a class of bilingual games for low adoption costs. It is an interesting open problem to provide necessary and sufficient conditions for characterizing stochastic stability in bilingual games for moderate and high adoption costs by solving the transition cost problem. For high adoption costs, the best response region of the bilingual option is empty (see Fig. 3). This suggests that the analysis of the transition problem in bilingual games for such costs will be similar to the analysis in a two-strategy coordination game with distinct payoff and risk dominant equilibria (see Eq. (6.9)). The analysis of the transition problem in bilingual games for moderate adoption costs (see Fig. 2) is however more challenging. We leave this for future research.

logit choice rule in the small noise double limit. Appendix A. Proof of Theorem 6.1 Proof. See Appendix C. □ Remark 6.4 (Galesloot and Goyal (1997)). show that for a sufficiently large population size, the payoff dominant equilibrium e1 in a bilingual game (4.3) is stochastically stable under the BRM model if and only if e ≤ eBRM , where

= max{min{s1 , s2 }, min{s1 , s3 }} a−c s1 =

Proof. For x ∈ int(B2 ), we have H(x, e3 − e2 ) = L(x, e3 − e2 ) + DV (x)(e3 − e2 )

= A2−3 x +

BRM

e

2

s2 = s3 =

(a − c)(a − d)d (a − c)d + (a − d)(c + d − a) (a − c)(a − d)d a(c + d − a)

We now compare the stochastic stability results for a bilingual game (4.3) under the BRM model and the logit choice rule. We

A2−3 x 3 A22− −3

3 A23− −2

=0 H(x, e3 − e1 ) = L(x, e3 − e1 ) + DV (x)(e3 − e1 )

= A2−3 x + = A2−3 x + ≥0

A2−3 x 3 A22− −3

3 A23− −1

A2−3 x (a − d)

(d − c) (since x ∈ B2 and a > d ≥ c)

32

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

H(x, e2 − e3 ) = L(x, e2 − e3 ) + DV (x)(e2 − e3 ) 2−3

A

=0+ =A

x

3 A22− −3

2−3

Appendix B. Proof of Theorem 6.2

3 A22− −3

Proof. For x ∈ int(B32 ), differentiating Eq. (6.6) we have

(

x DV (x) =

(since x ∈ B2 )

≥0

H(x, e2 − e1 ) = L(x, e2 − e1 ) + DV (x)(e2 − e1 )

=0+ =A

x

L(x, u) = A3−1 x

[

(a − d)

H(x, e1 − e3 ) = L(x, e1 − e3 ) + DV (x)(e1 − e3 ) A2−3 x 3 A22− −3

x+

DV (x)(e3 − e1 ) =

3 A21− −3

3 A21− −2 3 A22− −3

H(x, e1 − e3 ) − H(x, e1 − e2 ) =

=

A2−3 x 3 A22− −3

3 2−3 (A21− −3 − A1−2 )

3 A22− −3 3 A22− −3

(since x ∈ B2 )

From the above series of inequalities, we can conclude that H(x, e1 − e3 ) ≥ H(x, e1 − e2 ) for all x ∈ int(B2 ). We next show that H(x, e1 − e2 ) ≥ 0 for all x ∈ int(B2 ) which will complete the proof. The function H(x, e1 − e2 ) is linear in x and the set B2 is a closed convex polyhedral set with extreme points e2 , x23 and y23 . By the Fundamental Theorem of Linear Programming, we know that H(x, e1 − e2 ) attains its minima at one of these three extreme points. By definition, x23 , y23 ∈ B2 , therefore we have A2−3 x23 = A2−3 y23 = 0. We compute as follows12 :

(since y23 ∈ B2 ) A22−3

A2−3 3 1−2 A22− −3 e =d+ (c − a) (a − d)

( since e ≤

d a

)

(a − d) and a > d ≥ c > 0

Since H(x, e1 − e2 ) is non-negative at all the extreme points of the convex set B2 , it will be non-negative at all the points in B2 . By Theorem 5.1, we can now conclude that V ∗ (x) = V (x) and so the proof is complete. □

2 3−2 (A33− −2 + A3−1 (0))

2 A33− −2

A3−2 y

DV (x)(e1 − e2 ) =

= (since x23 ∈ B2 )

H(y23 , e1 − e2 ) = A2−1 y23

≥0

A3−2 y

2 3−2 (A31− −2 + A3−1 (1))

2 A33− −2

(

≥0

2 3−2 (A33− −1 + A3−1 (−1))

= A3−2 y H(x, e3 − e2 ) = L(x, e3 − e2 ) + DV (x)(e3 − e2 ) = 0 + A3−2 y ≥0 (since y ∈ B3 ) H(x, e2 − e3 ) = L(x, e2 − e3 ) + DV (x)(e2 − e3 ) = A3−2 x − DV (x)(e3 − e2 ) = A3−2 (x − y) 2 = x1 A31− (since x − y = x1 (e1 − e3 )) −3 = x1 (d − c) ≥0 (since d ≥ c) ) (

H(x23 , e1 − e2 ) = A2−1 x23

≥0

)

2 A33− −2

DV (x)(e3 − e2 ) =

A2−3 x

≥0

A3−2 y

= A3−1 x − DV (x)(e3 − e1 ) = A3−1 x ≥0 (since x ∈ B3 ) ( )

A2−3 x

H(e2 , e1 − e2 ) = A22−1 +

]

=0 H(x, e3 − e1 ) = L(x, e3 − e1 ) + DV (x)(e3 − e1 ) =0 H(x, e1 − e3 ) = L(x, e1 − e3 ) + DV (x)(e1 − e3 )

H(x, e1 − e2 ) = L(x, e1 − e2 ) + DV (x)(e1 − e2 )

=A

0 [u]+ .

A3−2 x

( (since x ∈ B2 and a > d ≥ c)

= A2−1 x +

2 ′ (A3−2 + A33− −1 e1 )

HJB equation verification for states in B32 : We now compute as follows:

(a − c)

≥0

2−1

2 A33− −2

)

The running cost for x ∈ B3 is

3 A22− −1 A2−3 x 2−3 A2−3

2−3

A3−2 y

A3−2 y

)

2 A33− −2

= A3−2 y H(x, e1 − e2 ) = L(x, e1 − e2 ) + DV (x)(e1 − e2 ) = A3−1 x + A3−2 y ≥0 (since x, y ∈ B3 ) H(x, e2 − e1 ) = L(x, e2 − e1 ) + DV (x)(e2 − e1 ) = A3−2 x − DV (x)(e1 − e2 ) = A3−2 (x − y) 2 = x1 A31− (since x − y = x1 (e1 − e3 )) −3 = x1 (d − c) ≥0 (since d ≥ c) HJB equation verification for states in B11 differentiating Eq. (6.7) we have DV (x) = k1 A1−2 + (A1−2 x)

12 For DV (x) to make sense on the boundary of the set B2 , we modify the definition in Eq. (4.1) as follows: Bi = {x ∈ aff(X ) : Ai−l x ≥∑0 for all l ∈ S }. n Here aff(X ) is the affine hull of the set X , aff(X ) = {x ∈ Rn : i=1 xi = 1}. The derivative DV (x) will then be defined on the boundary of the set B2 ∩ X .

2 A33− −2

A1−3 3 A11− −2

2 − k1 A11− −2

The running cost for x ∈ B1 is L(x, u) = 0

[

A1−2 x

A1−3 x [u]+ .

]

: For x ∈ int(B11 ),

A1−3 3 A11− −2

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34 2 2 1−2 1−2 1−2 DV (x)(e2 − e1 ) = k1 A12− x + k1 A11− −1 − A −2 + k2 (A1−2 − A2−1 )

We now compute for x ∈ int(B11 ) as follows: 2 2 1−2 x + k1 A11− DV (x)(e2 − e1 ) = k1 A12− −2 −1 − A

( 1−2

+A

= −A1−2 x H(x, e2 − e1 ) = L(x, e2 − e1 ) + DV (x)(e2 − e1 )

=A

−

−2 1−2 DV (x)(e2 − e3 ) = k1 A21− x 3 +A

1−3

2 = A1−2 (x − y) + k1 A12− −3

1−2

=A

x−

( = A1−3 x 1 − = A1−3 x

(c ) d

1−2

+A )

2 A11− −3

−2 = A1−3 x + A1−2 x + k1 A31− 1

(since c , d > 0 and x ∈ B1 ) A1−3 2 1−3 k1 A11− −2 1−3 A1−2 1−3 (since A1−3 0)

1−2

+A

−

=0+ = k1 (d − c) ≥0 (since k1 ≥ 0 and d ≥ c) H(x, e3 − e1 ) = L(x, e3 − e1 ) + DV (x)(e3 − e1 ) 2 = A1−3 x − k1 A11− −3 ( )

B12

HJB equation verification for states in differentiating Eq. (6.8) we have DV (x) = k1 A1−2 + A1−2 xe′1 −

+A

1−2

y

3 A13− −2

−

+

A1−3 3 A13− −2

3 A12− −1 3 A13− −2

+

: For x ∈

2 ′ k1 A11− −2 e1

+

)

3 A13− −2

−2 2 2 1−2 1−2 x + k2 A21− y + k2 A13− = k1 A12− −3 + A 3 −A −2 2 1−2 = k1 A12− (x − y) −3 + A 2 1−2 = k1 A12− −3 + k1 A1−2

(since d ≥ c)

int(B12 ),

H(x, e3 − e1 ) = A1−3 x − k1 (d − c) From the above series of inequalities, we have H(x, e3 − e2 ) − H(x, e3 − e1 ) = A1−2 x

≥0

2 ′ A12− −1 e1 )

A1−3

( −2 k2 A31− 2

−3 A31− 2

3 A12− −1

≥0 (since c , d > 0 and x ∈ B1 )

+

−3 A11− 3

3 A13− −2

= k1 (d − c)

d

(

2 k2 A13− −2

+

)

2 H(x, e1 − e3 ) = k1 A11− −3

3 A11− −2

≥0

+ k2 (A

3 A13− −2

3 A12− −1

2 = k1 A11− −3

2 A11− −3

(c )

1−2

y

3 A11− −3

(

2 k1 A11− −3

= A1−3 x

2 1−2 1−2 2 1−2 x − k1 A11− DV (x)(e1 − e3 ) = k1 A11− −3 + A −2 + k2 (A1−3 + A2−1 )

(

−

= = H(x, e1 − e3 ) = L(x, e1 − e3 ) + DV (x)(e1 − e3 )

= A1−3 x 1 −

(since d ≥ c)

= A1−3 x + A1−2 x − k1 (d − c)

3 A11− −3 A1−2 x 1−3 A1−2

2 k1 A11− −3

= k1 (d − c)

2 −2 1−2 = A1−3 x − k1 A12− x + k1 A21− −3 + A 1

+ A1−2 x

≥0 −2 DV (x)(e1 − e3 ) = k1 A11− 3 +

2 = k1 A11− −3

2 1−2 H(x, e3 − e2 ) = A1−3 x − k1 A12− y −3 + A

x

+ A1−2 x

3 A11− −2

2 1−2 = k1 A11− −2 + k1 A2−3

≥0

x − DV (x)(e2 − e3 ) 2 k1 A11− −3

2 1−2 = k1 A12− y −3 − A 2 1−2 H(x, e2 − e3 ) = A1−2 x + k1 A12− y −3 − A

=A x+ −A x = k1 (d − c) ≥0 (since k1 ≥ 0 and d ≥ c) H(x, e3 − e2 ) = L(x, e3 − e2 ) + DV (x)(e3 − e2 ) =A

3 A13− −2

2 1−2 2 1−2 DV (x)(e2 − e3 ) = k1 A12− y + k2 A13− −3 + k2 A2−3 − A −2

2 1−2 = k1 A11− x −3 − A H(x, e2 − e3 ) = L(x, e2 − e3 ) + DV (x)(e2 − e3 )

1−3

−

)

≥0

3 1−3 (since A11− −2 = d = −A2−3 )

2 k1 A11− −3

−3 A31− 2

3 A12− −1

H(x, e1 − e2 ) = 0 + A1−2 x

2 2 1−2 x + k1 A11− = k1 A12− −2 −3 − A

1−2

−3 A21− 1

3 A13− −2

=0

A1−3 2 2−3 − k1 A11− −2 1−3 A1−2

3 A11− −2

2 k2 A13− −2

−

)

H(x, e2 − e1 ) = A1−2 x − A1−2 x

(since x ∈ B1 ) 3 A12− −3

3 A13− −2

3 A12− −1

= −A1−2 x

x

≥0

y

3 A12− −1

(

= A1−2 x − A1−2 x =0 H(x, e1 − e2 ) = L(x, e1 − e2 ) + DV (x)(e1 − e2 ) = 0 − DV (x)(e2 − e1 ) 1−2

33

) ′

e1

3 A12− −1 3 A13− −2

) ′

e1

since x ∈ B1

(

)

Therefore, H(x, e3 − e2 ) ≥ H(x, e3 − e1 ) for all x ∈ B12 . We now show that H(x, e3 − e1 ) ≥ 0 for all x ∈ int(B12 ) which will complete the proof. The function H(x, e3 − e1 ) is linear in x and the constraint set B12 is a closed convex set with extreme points e3 , p2 and y13 . By the Fundamental Theorem of Linear Programming, we know that

34

S. Arigapudi / Journal of Mathematical Economics 86 (2020) 24–34

H(x, e3 − e1 ) attains its minima at one of these three extreme points. We now compute as follows13 : H(e3 , e3 − e1 ) = A13−3 − k1 (d − c)

= e − 0(d − c) ≥0

(since k1 = x1 = (e3 )1 = 0)

H(y13 , e3 − e1 ) = A1−3 y13 − k1 (d − c)

= A1−3 y13 − 0(d − c)

(since k1 = x1 = y13 1 = 0) (since y13 ∈ B1 )

≥0

References

H(p , e3 − e1 ) = A 2

The leading coefficient of h(e) i.e., the coefficient of e2 of h(e) is 2dc 2 > 0. Therefore, we have that h(e) ≥ 0 for 0 ≤ e ≤ e or e ≥ e and h(e) < 0 for e < e < e. Recall that our computations are for the case with low adopd(a−d) tion {costs i.e., e} ≤ (see Section 4.2). Therefore, for e ≤ a d min e, a (a − d) , we have h(e) ≥ 0. By definition for such values of e, C new ({e1 }, {e2 }) ≥ C new ({e2 }, {e1 }), which implies that state e1 is stochastically stable from Proposition 3.1. □

1−3 2

p − k1 (d − c) ( e) e =e +e 1− − (d − c) d d ( d e

since k1 = x1 = p21 =

=

ec

e) d

d

≥0 Since H(x, e1 − e2 ) is non-negative at all the extreme points of the convex set B12 , it will be non-negative on B12 . □ Appendix C. Proof of Proposition 6.3

Proof. For e ≥ 0, let e

f (e) ≡ C new ({e1 }, {e2 }) =

+

1

((

(

(a − d) 1 −

2

(a − c) −

d

) e d

1 ( e )2 2

d

(a + d − c)

)2 )

−e

a−d

g(e) ≡ C new ({e2 }, {e1 }) =

1

(

e2

)

2 a−d Clearly f (e) and g(e) are both quadratic in e and therefore their difference h(e) = f (e) − g(e) is also quadratic in e. Let e < e denote the roots of the quadratic function h(e). Straightforward computations yield

(

√

e=d 1−

c+d−a c

)

(

√

and e = d 1 +

c+d−a c

) .

Since c ≤ d < a < c + d, we have e, e > 0.

13 We have a minor technical difficulty of defining the derivatives on the boundary as before. See Footnote 12.

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