A simple dynamic contest with a parameterized strength of competition

García-Martínez, José A. Article A simple dynamic contest with a parameterized strength of competition SERIEs - Journal of the Spanish Economic Association Provided in Cooperation with: Spanish Economic Association Suggested Citation: García-Martínez, José A. (2018) : A simple dynamic contest with a parameterized strength of competition, SERIEs - Journal of the Spanish Economic Association, ISSN 1869-4195, Springer, Heidelberg, Vol. 9, Iss. 3, pp. 305-332, https://doi.org/10.1007/s13209-018-0180-6 This Version is available at: https://hdl.handle.net/10419/195277 Standard-Nutzungsbedingungen: Die Dokumente auf EconStor dürfen zu eigenen wissenschaftlichen Zwecken und zum Privatgebrauch gespeichert und kopiert werden. Sie dürfen die Dokumente nicht für öffentliche oder kommerzielle Zwecke vervielfältigen, öffentlich ausstellen, öffentlich zugänglich machen, vertreiben oder anderweitig nutzen. Sofern die Verfasser die Dokumente unter Open-Content-Lizenzen (insbesondere CC-Lizenzen) zur Verfügung gestellt haben sollten, gelten abweichend von diesen Nutzungsbedingungen die in der dort genannten Lizenz gewährten Nutzungsrechte. Terms of use: Documents in EconStor may be saved and copied for your personal and scholarly purposes. You are not to copy documents for public or commercial purposes, to exhibit the documents publicly, to make them publicly available on the internet, or to distribute or otherwise use the documents in public. If the documents have been made available under an Open Content Licence (especially Creative Commons Licences), you may exercise further usage rights as specified in the indicated licence. https://creativecommons.org/licenses/by/4.0/ SERIEs (2018) 9:305–332 https://doi.org/10.1007/s13209-018-0180-6 ORIGINAL ARTICLE A simple dynamic contest with a parameterized strength of competition José A. García-Martínez1 Received: 28 April 2017 / Accepted: 7 July 2018 / Published online: 18 July 2018 © The Author(s) 2018 Abstract This paper analyzes the effect of competition in a dynamic contest in which agents of two types (Aand B) differ in their expected performances; environments where type Aoutperforms type Bare more frequent than those where Boutperforms A. In each period, the population of agents is randomly matched in groups of nmembers (each group faces a particular environment), with the top k<nperforming agents from each group being the winners of the prizes. Hence, the ratio k ndetermines the proportion of winning agents in each group. This ratio also describes the strength of competition in the group: the lower k nis, the higher the level of competition is. Our results show that type Aeventually dominates the entire population with moderate competition, but type Bsurvives in the long run for high levels of competition. Hence, we obtain that no matter how low the expected success rate of a type is, if the strength of competition is high enough those agents with the lowest expected success rate survive in the long run. Keywords Contest ·Competition ·Strength of competition ·Selection process JEL Classification D00 ·D23 ·C73 I wish to express my gratitude to Fernando Vega-Redondo, Ascensión Andina-Díaz, Carlos Alós-Ferrer, Ana Ania, Pablo Beker, Joseph Harrington, Francisco Marhuenda, Dilip Mookherjee, Frédéric Palomino, and Giovanni Ponti for their helpful comments and suggestions. I also thank the Editor Juan D. Moreno-Ternero and two anonymous referees for their insightful comments. I also acknowledge contributions by participants at seminars and conferences at Alicante, Bilbao, Boston, Marseilles, and Ankara. Financial support from the Ministerio de Economía y Competitividad through Project MTM2014-54199-P and the Junta de Andalucía through Project SEJ2011-8065 is gratefully acknowledged. BJosé A. García-Martínez [email protected] 1Departamento de Estudios Económicos y Financieros, Universidad Miguel Hernández, Elche, Alicante, Spain 123 306 SERIEs (2018) 9:305–332 1 Introduction The most common use of contests is as mechanisms to create incentives to work harder. However, they can also be used as selection mechanisms. For example, a contest can select agents that differ in their expected efficiency levels. This can be relevant if the institution or principal cannot either impose the strategy to be followed by an agent or observe the type of agent. We seek to study the role of the number of prizes (k) and the number of contestants (n) in the outcome of the selection process.1These institutional parameters define a ratio k n, which can be seen as a measure of the strength of competition. For example, if k=1 and n=3, three agents compete for only one prize in every group. Similarly, if k=1 and n=10 then 10 agents compete for only one prize. Note that in the second case competition is higher (ceteris paribus). Hence, the strength of competition increases as ratio k ndecreases. We focus on a specific kind of selection, in particular this “strength of competition” in a contest, so we take an evolutionary approach. To that end, we consider a large population with two possible types of behavioral agents: A-agents and B-agents, where environments where type Aoutperforms Bare more likely to occur than environments where Boutperforms A. In this sense, Ais a better type because it has a higher expected success rate.2 In this population of agents, we consider that each individual interacts with randomly selected individuals.3Thus, groups are formed by a random matching process. However, a random matching process would generate a very complicated stochastic system. In the economic literature, for large populations (either countable or uncountable) the population dynamic is usually approximated by a deterministic process, where the frequency of different matches is identified with their corresponding expectations. This simplifying assumption is analyzed in several papers from different points of view. Examples include Boylan (1992,1995), Alós-Ferrer (1999), and Duffie and Sun (2012). In Sect. 2, we make some assumptions that guarantee the existence of a matching process, so we can consider the deterministic process presented in this paper as a good approximation of the complex stochastic system. In the literature on evolutionary selection models the matching process is usually made in pairs, which is a particular case of group size, n=2, where one agent is selected, k=1. We must consider a more general matching process in which n≥2, and k≥1 agents are selected. However, the rationale in this setting is the same as in 1Note that the prizes can be very diverse, for example a promotion, a job opportunity, a contract, a sale, etc. 2As a stylized example, the behavioral rules (Aand B) can be thought of as different available technologies: one of them is better more often than the other, and agents are proficient in either technology Aor B. Another stylized example could be a sales company that promotes people according to their success in selling. The company employs men and women and men sell better to men and women sell better to women. If the potential market has more men than women, men could be the A-agents and women the B-agents. It is not easy to find an application that fits all of the model’s elements because the model seeks to represent a family of complex institutions in a very stylized manner to point out a very specific characteristic of a selection process. Obviously, in any real institution the selection process is influenced by many more factors. However, we believe that the properties identified in our model are robust enough to play a role in more complex situations. 3We assume that the institution cannot either impose the strategy to be followed by an agent or observe the type of agent, so we consider that each individual interacts with randomly selected individuals. Thus, the institution can only determine the size of the group nand the number of prizes k, i.e., the ratio k n. 123 SERIEs (2018) 9:305–332 307 matching in pairs. We consider that there is a continuum population of agents, and we work with the proportions of different kinds of groups of agents. To sum up, this paper analyzes the effect of strength of competition on the characteristics of the successful agents. To that end, we consider that at t=0 each agent is given one behavioral rule, either Aor B. They are not strategic, so they become agents of type Aor B. At each t, the population of Aand Bagents is randomly matched in groups of nagents, with members of each group competing with each other, and each group facing a particular environment. The mechanism selects4the top k<n performing agents from each group, with k∈{1,2,3,...}and n∈{2,3,...}.We assume that at t+1 nonwinners imitate the action of winners at t, so the population at t+1 reproduces the distribution of the type of winners at t.5Consequently, the proportion of A-agents in the population is equal to that of the winners. Then they are again randomly matched and the process repeats. We seek to learn how this competition process changes the characteristics of the population. There is an alternative imitative behavior, which adds a different but very interesting point of view of the dynamic process. Nevertheless, the resulting dynamic is the same as in the first imitative assumption. Let xbe the number of A-agents in a group (and (n−x)the number of B-agents). Under this second imitation behavior, regardless of what environment a group is facing, if the number of individuals who successfully match the environment exceeds a threshold kthen the entire group adopts the strategy matching the environment. However, if the number of agents is below the threshold kthen in groups facing environment Aonly a fraction x kof the members adopt the strategy matching the environment, strategy A, and the remaining 1 −x k adopt B.6Similarly, in groups facing environment Bonly a fraction n−x kcopy Band the rest copy A. This basically means that even if one strategy proves more successful with the current environment it will not automatically dominate the group unless it is sufficiently represented. The higher kis, the more easily the members of the group imitate successful behavior. Thus, the ratio k nmeasures the minimum proportion of members of the group matching the environment needed to cause all members to copy the successful action. Therefore, this ratio also measures the level of conformity in this population. The higher k nis, the more successful agents are needed to cause the group to change behavior, so changing behavior becomes more difficult. On the other hand, alower k nmakes success more important relative to conformism and the environment becomes more competitive. Therefore, in this context, conformism and competition are correlated. In addition, notice that, after both imitation rules, the proportion of A-agents among the nonwinners is equal to the proportion of A-agents in the population of winners. Thus, we can focus on the proportion of A-agents among the agents selected and study how the strength of competition changes the distribution of the population. 4“To win” and “to be selected” are used interchangeably in this paper. 5An equivalent assumption is that only the winners at tare considered for competition at t+1, thus, the nonwinners are out of the contest. 6Notice that in this case (x<kand the environment is A), there are only xagents of type Ain a group, and in addition they are all successful. The rest of the agents in the group are type Band some change to Abut others do not. 123 308 SERIEs (2018) 9:305–332 In this model A-agents perform better than B-agents more often, so we should expect an increase in the strength of competition to punish B-agents and the proportion of B-agents to decrease as competition increases. However, our results show that an increase in competition does not always work this way: In particular, we find that for high enough levels of competition B-agents can persist in the long run, despite being expected to perform worse. More precisely, depending on the strength of competition we find three possible cases: Cases L,Mand H. First, case L: If the strength of competition is too low, the selection process is not strong enough to offset the inertia of the initial population. The dynamic thus depends on the initial conditions and the population eventually becomes homogeneous, i.e. with only type Aor type Bpersists in the long run. Second, case M: If the strength of competition increases (intermediate level) the whole population will become A-agents for any initial mixed population. In this case the selection process is strong enough to eventually select the best performers, as expected. Finally, case H: if competition increases far enough, B-agents also survive.7Thus, surprisingly, we show that no matter how low the success rate of a type is, if the strength of competition is high enough agents of that type survive in the long run. In other words, too much competition is always harmful to the best performers. The intuition behind our results is broadly explained in Sect. 3.1. The contribution of this paper is twofold. First, it presents a family of contest selection mechanisms that parameterizes the strength of competition in a simple way. In evolutionary models agents are usually matched in pairs and one of them is selected. This paper generalizes this idea in contests and considers matchings of nagents with k<nagents selected. Second, we show that this generalization is not innocuous but has a surprising result even in a very simple model. As far as we are aware, there are no similar approaches in the literature on evolutionary models. Our approach is concerned with designing a suitable selection mechanism, which depends on the objective function of the institution.8This approach is related to some extent to classic mechanism design, especially principal-agent models. In such models the information that players have about others players and their individual choices has a major role in the design of the mechanism. However, our approach puts the focus on an institutional characteristic, i.e. the strength of competition, and we try to highlight that it can be an important factor to be considered even in a simple model. This paper is related to Harrington (1998,1999a,b,2000,2003) and GarciaMartinez (2010) because our mechanism can be seen as a generalization of theirs. Harrington uses a selection process in a hierarchical structure to compare the performance of rigid behavior with that of flexible behavior. Agents are randomly matched in pairs (n=2) and one of them is selected. Thus, the strength of competition is fixed. Garcia-Martinez (2010) analyzes a promotion system that works in two steps. The first step is like Harrington’s mechanism: Agents are matched in pairs and one of them is selected. In the second step, the agents selected in the first step are pooled together 7We could consider the level of conformity mentioned above instead of the strength of competition. In that case, we obtain that for high levels of conformity case Lpertains, if the conformity requirement is intermediate case Mpertains and for low levels case Hpertains. 8For any objective function of the institution, there will be an optimal proportion of A-agents, which could vary from 0 to 1. The institution should take the appropriate kand nto attain its objective. 123 SERIEs (2018) 9:305–332 309 and the top fraction θof best-performing agents is eventually selected; this is referred to as “global selection”. In Vega-Redondo (2000) a hierarchical structure is used to select agents, who play in pairs (n=2) a 2 ×2 coordination game, where there is only global selection. The present paper is also related to the literature on tournaments produced since the seminal paper by Lazear and Rosen (1981), in particular to those papers that focus on the selection role of contests, e.g., Rosen (1986), Section V, Hvide and Kristiansen (2002), Tsoulouhas et al. (2007), Azmat and Möller (2009), and Groh et al. (2012). The rest of this paper is organized as follows: Sect. 2describes the model and the dynamic equation; Sect. 3analyzes the dynamics, discusses the results, and provides some intuitions; Sect. 4analyzes the convergence time; and Sect. 5concludes. 2Themodel At time tthere is a continuous population of Aand Bagents. Let at∈[0,1]denote the proportion of A-agents at time t, and 1 −atthe proportion of B-agents. The dynamic function at+1=f(at)describes the evolution of the proportion of A-agents at time t+1 as a function of the proportion of A-agents at t. First, we derive this function. At t, agents are randomly matched in groups of n≥2. We assume that the random matching process has the following properties: First, the probability with which a given agent is matched with agents of given types equals the product of the proportions of agents of the respective types in the population. Second, the proportion of a given class of grouping is equal to the probability (ex-ante) of such a grouping. The existence of a random matching process with these properties is proved in Alós-Ferrer (1999).9 Thus, the proportion of groups containing a number xof A-agents (and (n−x) B-agents) is equal to the probability of such a group, i.e. n xax t(1−at)n−x,letb(at,x) stand for n xax t(1−at)n−x.10 This is also the proportion of agents in groups with xAagents with regard to the initial population (level t) because the groups are composed of equal numbers of agents. Agents face a stochastic environment that is the same for all members of a particular group. However, the environment of each group is stochastically independent of that of other groups. We categorize all the different possible environments into two types. In a type Aenvironment A-agents outperform B-agents. In a type Benvironment B-agents outperform A-agents.11 The probability of an environment of type Ais p>1 2, and that of type Bis (1−p).12 Therefore, each agent faces an uncertain future environment, but there is no aggregate uncertainty because of our assumptions. Therefore, at each level after the random matching, a proportion p((1−p)) of the groups has a type A(B) environment. This is assumed to be i.i.d. across levels, so that the probability 9Alós-Ferrer (1999) gives a constructive existence proof for the case n=2. The generalization to groups of nagents is straightforward. 10 The xis distributed as a binomial distribution, x∼B(n,at). 11 One further kind of environment can be considered in which both rules perform equally. This environment adds no new insights to the analysis, so we do not consider it. 12 This probability pand (1−p)can also be seen as the expected success rates of an agent of type Aand Brespectively. 123 310 SERIEs (2018) 9:305–332 of an agent facing a given environment is independent of the environment that he/she has faced in the past. Therefore, the proportion of agents in groups with a number xof A-agents under a type Aenvironment is b(at,x)p. In such groups A-agents outperform B-agents. The system selects the ktop-performing agents from each group, where k≤n. The agents selected from each group are the winners of that group. The proportion of winning agents is k nwith regard to the initial population (time t). Thus, if a group in a type Aenvironment has more A-agents than vacancies available (i.e. x≥k) then all the agents selected from that group are A-agents, and the proportion of A-agents selected is k nb(at,x)p. However, if x<kthen only a number xof A-agents are selected and some B-agents have to be randomly chosen to fill the k−xvacancies, so the proportion of A-agents selected is x nb(at,x)p. Consequently, in this case, the total proportion of A-agents selected will be: EA a t=k−1 x=0 x nb(at,x)p+n x=k k nb(at,x)p= n x=0min[x,k]1 nb(at,x)p. Analogously, a fraction (1−p)of groups will face a type Benvironment and similar reasoning applies. In that case, the proportion of A-agents selected comprises the A-agents selected from the groups under the type Benvironment that do not have enough B-agents to fill all the kvacancies, i.e. x−(n−k) A-agents: EBa t=n x=n−(k−1)(x−(n−k)) 1 nb(at,x)(1−p). The proportion of A-agents selected will be EA a t+EBa twith regard to the initial population (at t). Finally, the proportion of A-agents is 1 k nEA a t+EBa twith regard to the population of agents selected. We assume that nonwinners imitate the winning type, so this proportion of A-agents will also be the proportion of the whole population at t+1, i.e., at+1. Therefore, the dynamic equation has this form:13 at+1=f(at)=1 k nEA a t+EBa t= n  x=0 min[x,k] kb(at,x)p + n  x=n−(k−1) (x−(n−k)) kb(at,x)(1−p) = n  x=0 (Min[x,k]p+Max[x−(n−k), 0](1−p)) kb(at,x) = n  x=0 (x,k,p,n)b(at,x)(1) We use S[n,k]to denote the system that selects kagents from groups of nagents. We consider the following equilibrium concepts: The point a∗∈[0,1]is said to be a steady state of Eq. (1) if it is a fixed point of f(.), i.e. f(a∗)=a∗. It is obvious that f(0)=0 and f(1)=1. Consequently, a=0 and a=1 are always steady states. The point a∗∈[0,1]is a globally stable equilibrium of (1)ifforalla0∈(0,1), limt→∞ at=a∗. 13 We do not actually need to consider the sum from x=0; it suffices to start at x=1. This is because groups with x=0 contain no type A; in those groups only type Bcan survive, independently of k.This also applies to f(a)and f (a). It seems convenient to include this in the expressions only for symmetry with the term where x=n. In addition, (x,k,p,n)stand for (Min[x,k]p+Max[x−(n−k),0](1−p)) k. 123 SERIEs (2018) 9:305–332 311 The point a∗∈[0,1]is a locally stable equilibrium if only for a0∈B(a∗,ε)∩(0,1), limt→∞ at=a∗, where B(a∗,ε) ={a∈(0,1)/ |a−a∗|<ε}with ε>0.Finally, denote a∗[n,k]as an inner steady state for the system S[n,k], i.e. a∗[n,k]belongs to the open interval (0,1). In the following section, the dynamics is analyzed and the intuition behind the result is provided. 3 Results Let a∗be an inner root of the equation f(at)−at=0 that belongs to the open interval (0,1). This root exists and is unique if either k n<(1−p)or k n>p(see the proof of the result below in the “Appendix”). By definition, this root is a steady state. The following result characterizes the dynamic for the selection process specified by Eq. (1). Proposition 1 Assume p >(1−p),k n<1and consider the dynamic equation (1): (1) If k n<(1−p)there is only one inner steady state a∗and it is globally stable. The steady states a =0and a =1are unstable. B-agents survive. (2) If k n∈[(1−p), p]there are no inner steady states. The steady state a =0is unstable and a =1is globally stable. A-agents are eventually the only survivors. (3) If k n>p there is only one inner steady state a∗, which is unstable and divides the interval a ∈(0,1)into two subintervals. The subinterval (0,a∗)is the basin of attraction of the steady state a =0and the subinterval (a∗,1)that of a =1. Both steady states are locally stable. Thus, initial conditions determine whether either A-agent or B-agents are the only survivors. Proposition 1shows that the equilibrium behavior can be characterized according to rate k n, which measures the strength of competition in the contest. Note that for low levels of competition ( k n>p), the selection process is not strong enough to overcome the inertia of the initial population. The dynamic depends on the initial conditions, and the population eventually becomes homogeneous, i.e. either type Aor type B: we refer to this case as case L (Low competition). If the strength of competition increases enough ( k n∈[(1−p), p]), the whole population become type Afor any initial population. In that case the selection process is strong enough to eventually select only A-agents: we refer to this case as case M (Midrange competition). Finally, if the strength of competition is high enough ( k n<(1−p)), B-agents also survive: we refer to this case as case H (High competition). Figure 1shows a phase diagram of each case. Therefore, no matter how low the expected success rate of a type of agent is, if the strength of competition is high enough agents of that type survive in the long run. 123 312 SERIEs (2018) 9:305–332 Fig. 1 Three phase diagrams are plotted with n=20 and PA=0.7. The parameter kvaries: with k=17 in case L, with k=10 in Case M;andk=3inCase H 3.1 Discussion and intuition of the main result To understand why this happens, it must first be observed that the dynamic of the system depends on the probability of each type of agent winning.14 For example, if the system is in a period tand the probability of an A-agent being selected is greater than that of a B-agent, then the proportion of A-agents in period t+1 is greater than in t, i.e. the proportion of A-agents increases and the proportion of B-agents decreases. Now, focus on a particular type of agent who faces one of the two following extreme scenarios: •If agents of this particular type are scarce (say close to extinction) they will generally be matched with agents of the other type.15 Thus, in general, there will only be one agent of this particular type in a group, who will only be selected if he/she outperforms the other type of agents so that he/she is the top performer. In such a context, the probability of this particular type of agent winning is not influenced by an increase in the strength of competition. His/her probability of winning depends almost entirely on his/her probability of outperforming the other type, i.e. it is p if the agent is type Aand (1−p)otherwise. •However, when agents of this particular type abound (say the other type is close to extinction), an agent of this particular type will generally be matched with agents of his/her own type (see footnote 15). Thus, if all the agents in a group are of the same type, they respond in the same way to the same environment. They all perform equally. The competitors of a particular agent in his/her own group are as successful (or unsuccessful) as he/she is. Thus, selection does not depend at all on the performance of this particular type of agent: the probability of winning depends almost entirely on how many people are selected. Therefore, 14 Let P(A) t(P(B) t) be the probability of an A(B)-agent being selected in period t. Notice that at+1= A-agents selected agents selected =atP(A) t atP(A) t+btP(B) t =atP(A) t k n ⇔P(A) t=at+1 at k n, analogously P(B) t=bt+1 bt k n. Therefore, P(A) t>P(B) t⇔at+1 at>bt+1 bt⇔at+1 at>(1−at+1) (1−at)>0⇔at+1>at. 15 This happens with a probability close to one. 123 SERIEs (2018) 9:305–332 319 n x+1=n x+2x+2 n−x−1(8) n n−x+1=n (n−(x−1)=n x−1=n x+1(x+1)x (n−x)(n−x+1) (9) The following results are also used in the proofs of the propositions. Claim 1 For any k and n, k x=0(k−x)n x=n x=n−(k−1)(x−(n−k))n x Proof By the symmetry of the binomial coefficient, n x=n n−x. Therefore, k  x=0n x= n  x=n−k+1n x ⇔k  x=0 (k−x)n x= n  x=n−k+1 (k−(n−x))n x ⇔⎛ ⎝ k  x=0 (k−x)n x= n  x=n−(k−1) (x−(n−k))n x⎞ ⎠ Claim 2 For any k and n, n x=0xb(a,x)=an Proof Using (7), (5), and (2), n  x=0 xn xax(1−a)n−x= n  x=0 nn−1 x−1ax(1−a)n−x = n  x=0 nn x−n−1 xax(1−a)n−x =n n  x=0n xax(1−a)n−x−n n  x=0n−1 xax(1−a)n−x =n−n n  x=0n−1 xax(1−a)n−1−x(1−a) Since n−1 n=0, the above expression is equal to n−nn−1 x=0n−1 xax(1− a)n−1−x(1−a)=n−n(1−a)=an  Let z(.) stand for b(a,x)((an −2x)a(n−1)+x(x−1)) Lemma 1 Let h be an integer and 0<h≤n, then, h x=0z(.) =(1−a)n−hah+1(a(n−1)−h)(h+1)n h+1123 320 SERIEs (2018) 9:305–332 Proof This is proved by induction on h.Ifh=1: 1  x=0n xax(1−a)n−x((an −2x)a(n−1)+x(x−1)) =(1−a)na2n(n−1)+na(1−a)n−1a2n(n−1)−2a(n−1) =(1−a)n−1a2n(n−1)((1−a)+na −2) =(1−a)n−1a2n(n−1)(a(n−1)−1) =(1−a)n−hah+1(a(n−1)−h)(h+1)n h+1h=1 It is proved that it holds for h+1 if it holds for h: h+1  x=0n xax(1−a)n−x((an −2x)a(n−1)+x(x−1)) = h  x=0n xax(1−a)n−x((an −2x)a(n−1)+x(x−1)) + n h+1ah+1(1−a)n−h−1a2n(n−1)−2a(n−1)(h+1)+(h+1)(h) =(1−a)n−hah+1(a(n−1)−h)(h+1)n h+1 + n h+1ah+1(1−a)n−h−1a2n(n−1)−2a(n−1)(h+1)+(h+1)(h) =(1−a)n−h−1ah+2n h+1 (1−a)(a(n−1)−h)(h+1) a+a2n(n−1)−2a(n−1)(h+1)+(h+1)(h) a [using expression (8)] =(1−a)n−h−1ah+2 n h+2h+2 n−h−1 ×(n(n−1)−(h+1)( n−1)) a2+((h+1)( h+n−1)−2(h+1)( n−1)) a a =(1−a)n−h−1ah+2 n h+2h+2 n−h−1 ×(n−1)( n−h−1)a2−((h+1)( n−h−1)) a a =(1−a)n−h−1ah+2n h+2(h+2)(a(n−1)−(h+1)) The last expression is equal to (1−a)n−hah+1(a(n−1)−h)(h+1)n h+1but in (h+1)instead of h. 123 SERIEs (2018) 9:305–332 321 Lemma 2 Let h be an integer and 0<h≤n, then, h x=0xz(.) =(1− a)n−hah+1(an −h−1)(h+1)hn h+1 Proof The proof is analogous to the lemma above, and it is also proved by induction on h. If h=1: 1  x=0 xn xax(1−a)n−x((an −2x)a(n−1)+x(x−1)) =0+na(1−a)n−1a2n(n−1)−2a(n−1) =n(1−a)n−1a2(n−1)(an −2)) =(1−a)n−hah+1(an −h−1)(h+1)hn h+1h=1 It is proved that it holds for h+1 if it holds for h: h+1  x=0 xn xax(1−a)n−x((an −2x)a(n−1)+x(x−1)) = h  x=0 xn xax(1−a)n−x((an −2x)a(n−1)+x(x−1)) +(h+1)n h+1ah+1(1−a)n−h−1 ×a2n(n−1)−2a(n−1)(h+1)+(h+1)(h) =(1−a)n−hah+1(an −h−1)(h+1)hn h+1 +(h+1)n h+1ah+1(1−a)n−h−1 ×a2n(n−1)−2a(n−1)(h+1)+(h+1)h [using expression (9)] =(1−a)n−h−1ah+2(h+1) n h+2h+2 n−h−1 (1−a)h(an −h−1) a+a2n(n−1)−2a(n−1)(h+1)+(h+1)h a =(1−a)n−h−1ah+2(an −h−2)(h+1)(h+2)n h+2 123 322 SERIEs (2018) 9:305–332 The last expression is equal to (1−a)n−hah+1(an −h−1)(h+1)hn h+1but in (h+1)instead of h. Lemma 3 The function g(a)in a =1 2is greater than zero, i.e., g(1 2)=f(1 2)−1 2>0 Proof The Eq. (1) evaluated in a=1 2and minus 1 2is g(1 2)=f(1 2)− 1 2=n x=0 min[x,k] kn xp+n x=n−(k−1) (x−(n−k)) kn x(1−p)1 2n−1 2>0⇔ n x=0min[x,k]n xp+n x=n−(k−1)(x−(n−k))n x(1−p)−1 2k2n>0 (using Claim 1) ⇔ n  x=0 min[x,k]n xp+ k  x=0 (k−x)n x(1−p)−1 2k2n>0 ⇔ k  x=0 xn xp+k n  x=k+1n xp+ k  x=0 (k−x)n x(1−p)−1 2k2n>0 ⇔ k  x=0 xn xp+kn  x=0n x− k  x=0n xp+ k  x=0 (k−x)n x(1−p)−1 2k2n>0 [using (3)] ⇔ k  x=0 xn xp+k2n− k  x=0n xp+ k  x=0 (k−x)n x(1−p)−1 2k2n>0 ⇔ k  x=0 xn xp−k k  x=0n xp+ k  x=0 (k−x)n x(1−p)+kp2n−1 2k2n>0 ⇔− k  x=0 (k−x)pn x+ k  x=0 (k−x)n x(1−p)+kp2n−1 2k2n(p+(1−p)) > 0 ⇔− k  x=0 (k−x)n xp+ k  x=0 (k−x)n x(1−p)+1 2k2np−1 2k2n(1−p)>0 ⇔1 2k2n− k  x=0 (k−x)n x(p−(1−p))>0⇔1 2k2n− k  x=0 (k−x)n x>0 [using (3)] ⇔1 2k n  x=0n x−k  x=0 (k−x)n x>0 ⇔1 2k k  x=0n x+1 2k n  x=k+1n x−k  x=0 (k−x)n x>0 ⇔k  x=01 2k−k+xn x+1 2k n  x=k+1n x>0 ⇔k  x=0x−k 2n x+k 2 n  x=k+1n x>0 123 SERIEs (2018) 9:305–332 323 Obviously, k 2n x=k+1n x>0. The expression k x=0x−k 2n xis also positive. First note that x−k 2is negative17 for 0 ≤x≤k 2and positive for k 2<x≤k. In addition, the value of x−k 2for x=iis equal to x=k−iin absolute value; the values are symmetrical. Second, n xtakes increasing values from x=0tox=n 2, and then it decreases symmetrically. As k 2<n 2, the sum of the positive terms has to be greater than the sum of the negative terms in absolute value: n  x=0x−k 2n x>0⇔⎛ ⎜ ⎝ k 2  x=0x−k 2n x+ n  x=k 2+1x−k 2n x>0⎞ ⎟ ⎠ ⇔⎛ ⎜ ⎝ n  x=k 2+1x−k 2n x>−k 2  x=0x−k 2n x⎞ ⎟ ⎠ Claim 3 For any n and a, n  x=0 b(a,x)(x−an)=0 Proof By Claim 2and using (2) n  x=0 b(a,x)(x−an)= n  x=0 xb(a,x)−an n  x=0 b(a,x)= an −an =0 A.2 Proof of Proposition 1. The dynamic of S[n,k]is given by Eq. (1): at+1=f(at)= n  x=0 (x,k,p,n)b(at,x) It is useful to start by presenting an outline of the proof. First it is shown that the function g(a)=f(a)−ais continuous in a=[0,1]with g(0)=g(1)=0. Second, local stability in the steady states a=0 and a=1 is studied by means of the first derivative of g(a). Thus, it is shown that if k n<(1−p)the states a=0 and a=1are unstable because g(0)>0 and g(1)>0; if k n∈[(1−p), p], then a=0 is unstable and a=1 is locally stable because g(0)>0 and g(1)<0; if k n>p, then a=0 and a=1 are locally stable because g(0)<0 and g(1)<0. Third, it is proved that g(a)has no more than one inner root. Fourth, it is proven that there are no periodic points.18 With all these results, the proposition can be easily proved in the following way. First, the function g(a)is continuous with no more than one inner root in a=(0,1), 17 It can be also zero if x=k 2=k 2. However the rationale is the same. Where k 2gives the highest integer less than or equal to k 2. 18 It is possible in difference equations for a solution not to be a steady point. Thus, point bis called a periodic point of xt+1=f(xt)if fk(b)=bfor a positive integer k,i.e.bis again reached after kiterations. See Elaydi (1996). 123 324 SERIEs (2018) 9:305–332 and g(0)=0 and g(1)=0. Second, if k n<(1−p), then g(0)>0 and g(1)>0, so g(a)must have at least one root, but as it cannot be possible for it to have more than one g(a)necessarily has a unique inner root ˆa.Asg(ˆa)is necessarily negative,19 that inner root is a stable steady state. As g(a)>0ifa∈0,ˆaand g(a)<0ifa∈ˆa,1, and there are no periodic points, this inner root is a globally stable steady state and the other two steady states (a=0 and a=1) are unstable. Third, if k n∈[(1−p), p] then g(0)>0 and g(1)<0, so the minimum number of roots compatible with this case is two, which is not possible. Therefore there cannot be any inner roots, and g(a)>0 for all a∈(0,1).Fourth,if k n>p, then g(0)<0 and g(1)<0, so g(a)necessarily has a unique inner root ˆa∈(0,1)with g(a)<0ifa∈0,ˆaand g(a)>0ifa∈ˆa,1. This inner root is an unstable stable steady state, and the other two steady states (a=0 and a=1) are locally stable. This completes the proof. It is straightforward to show that g(a)is continuous because f(a)is a polynomial. In addition, it is obvious that f(0)=0 and f(1)=1, so g(0)=g(1)=0. Using Eq. (1), it is straightforward to show that the first derivative of g(a)= f(a)−ais:20 g(a)=f(a)−1= n  x=0 (x,k,p,n)b(a,x)x−na a(1−a)−1(10) If a=0 the terms of the series g(a)are equal to zero except for x=1.21 g(0)=p kn−1≥0⇐⇒ k n≤p If a=1 the terms of the series g(a)are equal to zero except for x=n−1 and x=n. g(1)=kp+(k−1)(1−p) kn(−1)+kp+k(1−p) kn−1=n k(1−p)−1≥0⇐⇒ k n≤(1−p) To prove that g(a)=0 has no more than one inner root in a∈(0,1), it suffices to show that g(a)has no more than one inflection point in a∈(0,1)since g(0)= g(1)=0 and f(a)is continuous in (0,1). The following Claim gives a close form of the second derivative, g(a). Claim 4 The function g(a)=f(a)=−(k+1)n k+1(1−a)−k−1a−k−1(p(1− a)na2k−(1−p)(1−a)2kan) 19 The function g(a)must be decreasing at a=ˆa, because it is positive to the left side, negative to the right, and continuous. 20 Note that (x,k,p,n)is independent of aand ∂b(a,x) ∂a=x−na a(1−a)b(a,x). 21 The first derivative of g(a)is the sequence of the derivatives of each term of g(a). It is necessary to simplify the expression to obtain the properly defined derivative function. 123 SERIEs (2018) 9:305–332 325 Proof It is straightforward to show that the second derivative22 of g(a)is:23 g(a)=f(a)= n  x=0 (x,k,p,n)(x−an)(x−(n−1)a)−x(1−a) a2(1−a)2b(a,x) = n  x=0 (x,k,p,n)(an −2x)a(n−1)+x(x−1) a2(1−a)2b(a,x) Where if z(.) =b(a,x)((an −2x)a(n−1)+x(x−1)), g(a)=n x=0(x,k,p,n)z(.) a2(1−a)2 First, we rewritten this functions depending on h x=0z(.) and h x=0xz(.), thus, we will be able to apply Lemmas 1and 2to simplify the expression. g(a)=f(a) = n  x=0 (x,k,p,n)z(.) a2(1−a)2= n  x=0 (Min[x,k]p+Max[x−(n−k), 0](1−p)) k z(.) a2(1−a)2 = n  x=0 (Min[x,k]z(.)p+Max[x−(n−k), 0]z(.)(1−p)) ka2(1−a)2 =k−1 x=0xz(.) +n x=kkz(.)p+n x=n−(k−1)(x−(n−k))z(.)(1−p) ka2(1−a)2 =k x=0xz(.) +kn x=k+1z(.)p+n x=n−(k−1)((x−(n−k))z(.))(1−p) ka2(1−a)2 =k x=0xz(.)p+kn x=0z(.) −k x=0z(.)p+n x=0(x−(n−k)z(.) −n−k x=0(x−(n−k))z(.)(1−p) ka2(1−a)2 =k x=0xz(.)p+kn x=0z(.) −k x=0z(.)p+n x=0xz(.) −(n−k)n x=0z(.)−n−k x=0xz(.) −(n−k)n−k x=0z(.)(1−p) ka2(1−a)2 Notice that, from Lemmas 1and 2with h=n,n x=0z(.) =n x=0xz(.) =0 because n n+1=0, thus: g(a)=k x=0xz(.)p−kk x=0z(.)p ka2(1−a)2−n−k x=0xz(.)−(n−k)n−k x=0z(.)(1−p) ka2(1−a)2 (11) By Lemmas 1and 2with h=k, after some algebra, the first term of (11) above is k x=0xz(.)p−kk x=0z(.)p ka2(1−a)2=−(1−a)n−k−1ak−1(k+1)n k+1p and by Lemmas 1and 2with h=n−k, after some algebra, the second term of (11)is: n−k x=0xz(.)−(n−k)n−k x=0z(.)(1−p) ka2(1−a)2=− (1−p)(1−a)k−1a(n−k)−1(n−k)((n−k)+1)(n n−k+1) k [using expression (9)] 22 The second derivative of g(a)is the sequence of the second derivatives of each term of g(a).Itis necessary to simplify these terms to obtain the properly defined derivative function. 23 As mentioned above (x,k,p,n)is independent of aand ∂2b(a,x) ∂a2= (x−an)(x−(n−1)a)−x(1−a) a2(1−a)2b(a,x). 123 326 SERIEs (2018) 9:305–332 =− (1−p)(1−a)k−1a(n−k)−1(n−k)((n−k)+1)(n k+1)(k+1)k (n−k)(n−k+1) k=−(1−p)(1−a)k−1 a(n−k)−1n k+1(k+1) Therefore, g(a)=−p(1−a)n−k−1ak−1(k+1)n k+1 −(−(1−p)(1−a)k−1a(n−k)−1n k+1(k+1)) =−(k+1)n k+1(1−a)−k−1a−k−1(p(1−a)na2k−(1−p)(1−a)2kan).  We now show that there is no more than one inflection point in a∈(0,1). First, note that if ¯ais an inflection point, then g(¯a)=0. Thus, by Claim 4, g(¯a)=0⇔p(1−¯a)n¯a2k−(1−p)(1−¯a)2k¯an=0 ⇔p(1−¯a)n¯a2k (1−p)(1−¯a)2k¯an=1⇔(1−¯a)n−2k ¯an−2k p (1−p)=1 ⇔1−¯a ¯ap (1−p)1 n−2k =1⇔⎛ ⎜ ⎜ ⎝ ¯a=1 1+p (1−p)1 2k−n ⎞ ⎟ ⎟ ⎠ (12) Therefore, the Eq. (12) can have only one real root in the interval a∈(0,1). Consequently, there is no more than one inflection point in a∈(0,1), which means that the function g(a)has no more than one root in a∈(0,1). Before concluding, it is proved that there are no periodic points.24 It has been proved that the function g(at)has either one inner root or none, i.e. there is only one inner steady state in (0,1)or none at all. If it has none, the function g(at)is necessarily positive,25 so there are obviously no periodic points because at<at+1 for all at∈(0,1). If there is one steady state in (0,1)any possibility of there being periodic points completely disappears if the function f(at)is increasing. Note that if f(at)is increasing, then for all atequal to or greater than the inner steady state (ˆa) either at<at+1for all at∈(ˆa,1)or at>at+1for all at∈(ˆa,1), and always at+1≥ˆafor any at∈(ˆa,1). Thus, it suffices to prove that the function f(at)is increasing. The following claim is needed to prove that f(at)>0.  Claim 5 Let the function r(x)=b(a,x)(x−an)and h(x)be a positive function and increasing in x. Then n x=0h(x)r(x)>0 24 See footnote 18. 25 In any event, even if g(at)was negative there would be no periodic points. 123 SERIEs (2018) 9:305–332 327 Proof The expression b(a,x)(x−an)is negative26 if x≤anand positive if x> an. Where angives the highest integer less than or equal to an. By Claim 3, n  x=0 r(x)=0⇐⇒ ⎛ ⎝ an  x=0 r(x)+ n  x=an+1 r(x)=0⎞ ⎠ ⇐⇒ ⎛ ⎝− an  x=0 r(x)= n  x=an+1 r(x)⎞ ⎠ As h(x)>0 is a function increasing in x, − an  x=0 h(x)r(x)< n  x=an+1 h(x)r(x)⇐⇒ n  x=0 h(x)r(x)>0 That f(at)>0 is now proved. From expression (10), f(at)= n  x=0 (x,k,p,n)b(a,x)x−na a(1−a)>0⇐⇒ n  x=0 (x,k,p,n)b(a,x) (x−na)>0 It is straightforward to show that the expression (x,k,p,n)takes values in [0,1] and is increasing in x. Therefore, by Claim 5,f(at)>0, and f(at)is increasing. Periodic points are therefore not possible. It can be concluded that: (1) If k n<(1−p)then g(0)>0 and g(1)>0. Thus, on the one hand, the function g(a)is continuous and g(0)=g(1)=0. On the other hand g(a)is positive around a=0 and negative around a=1. Consequently, by Bolzano’s Theorem there is at least one inner root. It has been proved that there cannot be more than one inner root. Therefore, there is only one inner root a∗and it is globally stable. (2) If k n∈[(1−p), p]then g(0)>0 and g(1)<0. In that case, as g(0)=g(1)=0, the function g(a)is positive around a=0 and around a=1. There cannot be more than one inner root, so the function g(a)must be positive for a∈(0,1).The only possibility of having an inner point is for the inner point to be a minimum of the function, so that the function would be positive for a∈(0,1). However, this is not possible because there is no more than one inflection point. Therefore, there is necessarily no inner root. The steady state a=0 is unstable, and the steady state a=1 is globally stable. (3) If k n>p, then g(0)<0 and g(1)<0. In that case, as g(0)=g(1)=0, the function g(a)is negative around a=0 and positive around a=1. There is necessarily only one inner root a∗for the same reason as in point 1). This unique inner root a∗must be unstable and divides the interval a∈(0,1)into two subintervals. The subinterval (0,a∗)is the basin of attraction of a=0 and (a∗,1)of a=1 which are locally stable. The proof is complete.  26 It can be also zero if x=an=an. However the rationale is the same. 123 328 SERIEs (2018) 9:305–332 First we show the proofs of Proposition 4and 5and then that of Propositions 2and 3. A.3 Proof of Proposition 4. Aclosedformofg(a)is first obtained. The Eq. (1) with k=1 can be rewritten as f(a)= n  x=0 min[x,1]b(a,x)p+ n  x=n (x−(n−1))b(a,x)(1−p) = n  x=1 b(a,x)p+n nan(1−p) =n  x=0 b(a,x)−n 0a0(1−a)n−0p+an(1−p) =1−(1−a)np+an(1−p) Thus, the function g(a)can be written as g(a;n)=(1−(1−a)n)p+an(1−p)−a Let ˜abe the unique inner root of g(a;¯n)=0, where n=¯n, and k n=1 ¯n<(1−p), see Proposition 1, i.e., a∗[¯n,k=1]=˜a. Let ˇabe the unique inner root of g(a;¯n+1)=0, where n=(¯n+1), and k n=1 ¯n+1<(1−p), i.e., a∗[¯n+1,k=1]=ˇa. The proof of Proposition 1 shows, on the one hand, that g(0;¯n+1)=g(1;¯n+1)=0. On the other hand, if k n=1 ¯n+1<(1−p), the function g(a;¯n+1)>0ifa∈(0,ˇa), and g(a;¯n+1)<0 if a∈(ˇa,1). Consequently, if the function g(a;¯n+1)is negative in a=˜a, i.e., g(˜a;¯n+1)<0, then ˜a>ˇabecause g(ˇa;¯n+1)=0, and the proof is complete. Thus, we only need to prove that g(˜a;¯n+1)<0. Before doing this we characterized, the steady state ˜a. As the state ˜ais a steady state with n=¯n, g(˜a;¯n)=1−(1−˜a)¯np+˜a¯n(1−p)−˜a=0 ⇐⇒ 1−(1−˜a)¯np+˜a¯n(1−p)−˜a(p+(1−p)) =0 ⇐⇒ (1−˜a)−(1−˜a)¯np=˜a−˜a¯n(1−p) ⇔p (1−p)=˜a−˜a¯n (1−˜a)−(1−˜a)¯n(13) We now prove that g(˜a;¯n+1)<0, g(˜a;¯n+1)=1−(1−˜a)¯n+1p+˜a¯n+1(1−p)−˜a<0 ⇔p−(1−˜a)¯n+1p+˜a¯n+1(1−p)−˜a(p+(1−p)) < 0⇐⇒ (1−˜a)−(1−˜a)¯n+1 p<˜a−˜a¯n+1(1−p) ⇔˜a−˜a¯n+1 (1−˜a)−(1−˜a)¯n+1>p (1−p) [using (13)] ⇔˜a−˜a¯n+1 (1−˜a)−(1−˜a)¯n+1>p (1−p)=˜a−˜a¯n (1−˜a)−(1−˜a)¯n 123