Rivalry and Professional Network Formation: The Struggle for Access James C. D. Fisher∗ This Draft: 2.25.2016 (Initial Draft: 7.2014)

Abstract We develop a network formation game where two principals (e.g., partners at a consulting firm) employ agents (e.g., consultants) from their professional networks to help them complete projects. Since agents only work for one principal at a time, the principal’s use of agents is rivalrous. We establish that there is a (pure strategy) equilibrium and we characterize how this rivalry influences equilibrium network structure, as well as the principals’ welfare. We find, for instance, that the principals always hold “minimally overlapping” networks, that the principals’ equilibrium interests are opposed (i.e., in an equilibrium where one of them does best, the other does worst), and that as a principal’s costs of training an agent fall, her payoff and the size of her network increase while the payoff and network of the other partner both decrease. Keywords: comparative statics, network formation, opposition of interests, rivalry, and submodular games. JEL Codes: C7, D2, and D8.

∗

University of Technology Sydney, [email protected]. My thanks to Andreas Blume, Matthew Elliott, Tim Flannery, Willemien Kets, Asaf Plan, Marek Pycia, Mark Walker, John Wooders, Leeat Yariv, and to seminar participants at the University of Arizona, the University of California Los Angeles, the 2014 International Conference on Game Theory, and the 2014 Summer Meeting of the Econometric Society for helpful discussions, comments, and suggestions. All remaining errors are my own.

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1

Introduction

Professional networks play a key role at Deloitte’s consulting practice:1 they are the primary means by which partners obtain consultants to help them complete their projects. Deloitte hires consultants into a pool – any partner may request the help of a consultant and each consultant is expected to provide help, unless she is currently working on another project. When a partner gets a project (from a client), she must decide which consultants in this pool to employ. Since consulting usually requires close collaboration, a partner usually employs consultants with whom she has built a good, productive working relationship, i.e., a partner usually employs consultants from her professional network. Once a consultant takes on a partner’s project, she is usually unavailable to help other partners complete their projects – projects typically involve daylong meetings and other activities at clients’ offices. Thus, the partners’ employment of consultants is rivalrous (at least in the short term), implying the partners may exert a negative externality on each other. To illustrate, consider two partners A and B. If A employs a large number of consultants who are part of her network and part of B’s network, then B’s pool of available, in-network consultants is smaller. This reduces B’s ability to complete certain projects, and diminishes her earnings since, at many consulting practices, a partner’s earnings depends on the revenue she brings in from her completed projects. In contrast, A is in a better position to complete her project and do well. Since partners choose their professional networks, our goal is to understand how this rivalry shapes their network formation decisions and, ultimately, their payoffs. To these ends, we build a stylized, two-stage game of network formation and rivalry.2 In our game, there are two partners, A and B, as well as a finite number of consultants, indexed 1 to N . In the first stage, both partners form their professional networks. It is costly for a partner to include a consultant in her network since she must invest effort (and money) to develop a good, productive working relationship with the consultant. For instance, she often needs to mentor the consultant in her production techniques/technology so that the consultant can be a capable assistant. In the second stage, the partners sequentially receive projects to complete. The projects and order in which they are awarded are determined by nature. Both projects are received in a short succession, so the first is not completed before the second is received. Projects are 1

I am grateful to a Deloitte consultant for helpful conversations the workings of her firm in 2013. While our game is inspired by consulting practices, it is not a complete model of how these firms work. Instead, it is a general model of firms building relationships with small suppliers (e.g., TV news shows and pundits or cities and water sources) because the principals are choosing which suppliers they would like to access and, once accessed by one principal, the resource is (temporarily) unavailable to the other principals. 2

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of heterogeneous difficulty and each requires a different number of consultants to complete – some, like an evaluation of a small call-center’s effectiveness may only require one or two consultants, while others, like an efficiency review of a complex supply chain, may require a dozen or more consultants. Once a partner receives her project, she employs consultants in her network to help her complete it. She may employ any subset of in-network consultants she wants (including the empty set), save that she cannot employ consultants who are already working on an ongoing project. If the partner manages to employ enough consultants, then she completes her project and earns a positive reward, from which she pays her labor and networking costs. If a partner does not employ enough consultants, she fails to complete the project and earns nothing, but is still liable for her labor and networking costs, if any. (For simplicity, the consultants are passive: (i) they always agree to be in a partner’s network and (ii) they always agree to be employed on a partner’s project, provided they are not working on another project. These assumptions reflect the expectations consulting companies frequently have of their employees.) Our solution concept is a pure strategy Nash equilibrium. We begin by establishing that an equilibrium exists (Proposition 1). This non-trivial result illuminates a deep connection between rivalry and submodularity. To prove it, we introduce “simple strategies.” A partner’s simple strategy of size n recommends a network consisting of n consultants and employment strategies. When partners follow simple strategies, their payoffs are submodular in the sizes of their networks/simple strategies. Intuitively, as B’s network grows, the probability A can use consultants she shares with B falls (because of rivalry) and her payoff decreases. Since this happens more quickly when A’s network is larger, we have submodularity. We exploit this submodularity by constructing an auxiliary game where partners use simple strategies, A chooses her network’s size, and B chooses the “negative” of her network’s size. Since the order of B’s choice is inverted, the auxiliary game is a two-player supermodular game and so has a pure strategy Nash equilibrium. Subsequently, we show that each equilibrium of the auxiliary game induces an equilibrium of the full game; the key insight is that a simple strategy is a best response to a simple strategy. We call these induced equilibria “simple equilibria.” In equilibrium, we find that both partners’ networks are “minimally overlapping” (Proposition 2). That is, the partners try to have networks that share as few consultants as possible. The intuition is that by networking with different sets of consultants, the partners minimize the chance that their access to their consultants will be blocked. Not only does this proposition help us understand why networks can be small and disconnected, it also suggests that a partner can improve her payoff by swapping shared consultants for exclusive consul3

tants, perhaps by networking with new hires. A qualitative implication of the proposition, which finds support at Deloitte, is that most consultants are in a small number of partner’s networks. Two properties play a key role in our subsequent analysis: “employment lists” and “employment efficiency.” An equilibrium has employment lists if the first partner to gets a project always uses a ranked list to determine the identities of the consultants she employs: if she employs l consultants, then she employs the first l consultants on her list. This property captures the intuitive idea that each partner uses an address book or other (mental) list. An equilibrium is employment efficient if the first partner to get a project always employs consultants who are exclusively in her network before she employs consultants who are in multiple networks. This ensures that the second partner to get a project has a larger pool of available, in-network consultants than she would if the first partner employed consultants in an arbitrary manner. Thus, the second partner is better able to complete projects and has higher earnings. We focus on equilibria that have employment lists and are employment efficient because these properties are intuitive and these equilibria are “robust” (see Section 4). We call these equilibria “ELEE equilibria.” Since simple equilibria are ELEE equilibria, the existence of ELEE equilibria is assured. We then establish that each ELEE equilibrium is payoff equivalent to a simple equilibrium (Proposition 3). The intuition is that, when employment lists and employment efficiency hold, then the equilibrium behavior of the partners is analogous to their behavior in a permutation of a simple equilibrium. We establish that the partners’ equilibrium interests are opposed. That is, there is an ELEE equilibrium where A does best and B does worst and vice versa (Proposition 4). The intuition for these results is that rivalry (i) causes the auxiliary game to be supermodular and (ii) ensures that a partner’s optimal payoff is always weakly decreasing in the size of the other partner’s network. Since (i), there is a there is a maximal (minimal) equilibrium where A holds her largest (smallest) equilibrium network and B holds her smallest (largest) equilibrium network. Since (ii), the maximal (minimal) equilibrium is most (least) preferred by A and least (most) preferred by B. The desired result then follows from Proposition 3. We also establish that if each partner’s optimal payoff is strictly decreasing in the size of the other partner’s network, then A does best in any equilibrium where B does worst and vice versa (Proposition 4). We develop welfare comparative statics for the ELEE equilibria where the partners’ interests are opposed. For concreteness, we suppose that A’s reward to completing a project weakly increases and that her cost of networking weakly decreases, while B’s reward and cost are held constant. We then compare A and B’s payoffs and network sizes before and 4

after the shift in equilibria where their interests are opposed. We show that A does better and holds a larger network, while B does worse and holds a smaller network (Proposition 5). This result follow naturally from Proposition 3 and weak monotonicity as the changes in A’s reward and cost cause her best response in the auxiliary game to increase. Stepping back, the proposition provides guidance on how a partner can improve her payoff – e.g., by taking classes to become a better mentor/teacher and reducing her networking costs. In light of the opposition of interests, it is natural to wonder if one partner actually does better than the other. We establish that A earns more and has a larger network than B in any ELEE equilibria where she does best and B does worst, provided (i) A and B have the same chance of getting a project, (ii) A has a weakly lower networking cost than B, and (iii) A receives weakly more for completing a project than B (Proposition 6). This result follows naturally from Proposition 5 and the insight that, when A has B’s reward and cost functions, then she holds a larger network and does weakly better than B in the maximal equilibrium of the auxiliary game. The next subsection surveys the related literature. Section 2 presents our model and examples. Section 3 gives our general results. Section 4 focuses on ELEE equilibria and develops the associated results. Section 5 concludes by discussing extensions and robustness. Most proofs are given in the Appendix, and the Online Appendix collects additional results. Related Literature Our work contributes to two literatures – the “trading-on-networks” literature and the “multiple commons” literature. The trading-on-networks literature – e.g., Kranton and Minehart [15], Blume et al. [4], Condorelli et al. [5], and Corominas-Bosch [6] – examine the importance of relationship-specific investments for trading outcomes.3 In Kranton and Minehart, traders first make investments with one another and then traders who have made mutual investments buy or sell a single item, while in Blume et al., Condorelli et al., and Corominas-Bosch traders bargain given an endowment of investments. These papers characterize how investments shape the nature of trade and welfare, as well as how traders make their investments (when the investments are endogenous). We complement this literature by focusing on a non-market environment, by allowing partners to consume multiple units of labor from different suppliers, by examining how the partners’ rivalrous use of consultants affects their networks and welfare, and by employing a different technical approach. The trading-on-networks literature intersects with the network formation literature.4 While other studies in this literature also examine with rivalry, e.g., Jackson and Wolin3

Experimental studies have demonstrated the empirical relevance of relationship-specific investments on market outcomes; see, for instance, Dogan et al. [7]. 4 For overviews of this extensive literature, see Granovetter [10] and Jackson [13], as well as Anshelevich et al. [1], Galeotti et al. [9], and Watts [19], among others.

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sky [14] and Page and Wooders [16], they focus on different forms of it than we do. In particular, Jackson and Wolinsky consider a “Co-Author Game” where there is a rivalry for a player’s limited time and Page and Wooders consider a club-formation game there is a within-club rivalry for a club’s resources. Differences in the natures of the rivalries considered drive differences in the results, especially as concerns network structure. For example, Jackson and Wolinsky show that players’ networks are either entirely overlapping or disjoint, which contrasts with our minimal overlap result. The multiple commons literature – e.g., Ilkilic [12] – examines a generalization of the tragedy of the commons where there are multiple commons. Specifically, Ilkilic considers a game where players are endowed with access to different resources (e.g., cities have aqueducts to different water sources) and each player decides how much to extract from each of her accessible resources. He characterizes the equilibrium extraction rates in terms of the access endowment.5 We complement this literature by allowing cities to choose which resources they have access to and by focusing how rivalry shapes their access decisions. Our work also connects to the matching with externalities literature – e.g., Fisher and Hafalir [8] and Pycia and Yenmez [17] – which explores, from a cooperative standpoint, how workers and firms come together to form productive partnerships. We complement this literature by considering an externality these models cannot capture: rivalry. In general, rivalry depends not only on the map between partners and consultants, but also on the partners’ projects, the order in which they move, and their decisions about which consultants to employ.

2

The Game

This section describes our environment, our solution concept, and gives an example. Environment We consider a two-stage game between two partners, A and B, who both have access to a set of consultants C = {1, . . . , N }, where N > 1. For simplicity, the consultants are passive; that is, (i) they always agree to be in a partner’s network and (ii) they always agree to be work on a partner’s project, provided they are not working on another project. In the first stage, both partners simultaneously form professional networks. That is, they pick subsets of C with whom to build friendly and productive working relationships.6 Let 5

Ilkilic, along with Blume et al., Condorelli et al., and Corominas-Bosch, write in the “network games” literature, which examines how network structure affects the equilibrium play of a game; for overviews of this broad literature, see Granovetter [10] and Jackson [13], as well as Belhaj et al. [3] and Hernandez et al. [11], among others. 6 In the language of Bala and Goyal [2], the partners form “directed links” to consultants. Also, since

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Ni ⊂ C denote partner i’s selection. It is costly for i to form a network as she must invest effort (and money) to develop a good, productive working relationship with each consultant in her network. Let ci (|Ni |) denote i’s cost of holding network Ni , where ci : N → R+ such that ci (0) = 0 and | · | denotes the cardinality of a set and |∅| = 0.7 In the second stage, the partners sequentially get their projects and employ consultants. Initially, none of the consultants are employed. At the start of the stage, nature picks a partner to get her project first – both partners have a 1/2 chance of getting their project first. Once the identity of the first partner is realized, nature draws a project for her. Then this partner employs consultants from her network. Subsequently, nature draws a project for the second partner. Then the second partner employs consultants from her network who are not already employed by the first partner. We imagine that both projects are received in quick succession (e.g., in the same week), so the first is not completed by the time the second arrives. The stage ends with both partners receiving their payoffs. Let X denote the non-empty and finite set of projects that either partner may receive. P For, each partner i, let Pi : X → [0, 1] such that x∈X Pi (x) = 1 be the probability that nature draws project x ∈ X for i. These draws are independent of all other randomness. Let d : X → N++ give the difficulty of a project, i.e., the minimum number of consultants needed to complete it – some projects are more complicated than others and require more help. Observe that every project requires at least one consultant. For each partner i, let ri : X → R+ give i’s reward to completing the project. Thus, if i gets project x, she earns ri (x) if she employs at least d(x) consultants and she earns 0 if she employs less than d(x) consultants. This production technology reflects two assumptions. The first is that clients only pay for completed projects, i.e., there is no residual value to partially-completed projects. The second is that most consultants are of high ability (e.g., Deloitte often hires college graduates in the top of their classes). Thus, they are usually capable assistants who can typically help a partner in whatever ways the partner needs, provided they have developed a good, working relationship with the partner. Hence, in the abstract, in-network consultants are homogenous from a partner’s perspective. (We discuss residual value and consultant homogeneity in the Conclusion.) When i gets her project first, she observes her project xi and then decides which innetwork consultants to employ, if any. (Note that i may choose to forgo a project by not employing any consultants.) A behavioral strategy for i is a σ1i : P(C) × X → P(C) such that σ1i (N , x) ⊂ N for all (N , x) ∈ P(C) × X. Thus, σ1i takes i’s history (Ni , xi ) and returns a partner often needs to train a consultant in her production techniques/technology as part of building a productive working relationship, we can interpret networking costs as training costs. 7 Research in the networking literature (e.g., Bala and Goyal [2] and Jackson [13]) frequently assumes a constant marginal cost of networking, i.e., ci (|N |) = κ|N | for some κ > 0.

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a (possibly empty) list of consultants σ1i (Ni , xi ) in Ni who she employs. Partner i pays an exogenously fixed and finite amount w ≥ 0 for each consultant she employs – at many consulting practices a partner’s compensation is linked to the cost of the consultants she employs in the projects she attempts, w is our way of modeling these costs. Thus, her total labor cost is w |σ1i (Ni , xi )|.8 When partner i gets her project second, she sees the set of consultants employed by the other partner T and her project xi . Since the consultants in T are part of an ongoing project, they are unavailable. Next, i decides which available in-network consultants to employ, if any. A behavioral strategy for i is a σ2i : P(C)2 × X → P(C) such that σ2i (N , N 0 , x) ⊂ N \N 0 for all (N , N 0 , x) ∈ P(C)2 × X.9 Thus, σ2i takes i’s history (Ni , T , xi ) and returns a list of consultants σ2i (Ni , T , xi ) in Ni \T who she employs. As before, i has to pay w for each consultant she employs, so her total labor cost is w |σ2i (Ni , T , xi )|. A strategy for partner i is si = (Ni , σ1i , σ2i ), i.e., is a specification of her network and behavioral strategies. Let S i denote i’s finite set of all possible strategies. Let s = (sA , sB ) denote a vector of strategies denote the vector of the partners’ strategies, and let S = S A ×S B be the joint strategy space. Each partner’s (ex-post) payoff is her reward less her labor and networking costs. To fix ideas, consider partner i and let −i be the other partner. Let s = (Ni , σ1i , σ2i , N−i , σ1−i , σ2−i ) ∈ S and let xi and x−i be their projects. Suppose both partners follow s. Then i’s ex-post payoff when she gets her project first is u1i (s, xi , x−i ) = ri (xi )I(|σ1i (Ni , xi )| ≥ d(xi )) − w |σ1i (Ni , xi )| − ci (|Ni |), where I(·) is an indicator function that is equal to one if |σ1i (Ni , xi )| ≥ d(xi ), i.e., if i employs enough consultants to complete her project, and is equal to zero else. When i gets her project second, −i employs σ1−i (N−i , x−i ). Thus, i’s ex-post payoff is u2i (s, xi , x−i ) = ri (xi )I(|σ2i (Ni , σ1−i (N−i , x−i ), xi )| ≥ d(xi )) − w |σ2i (Ni , σ1−i (N−i , x−i ), xi )| − ci (|Ni |), where I(·) is an indicator function that is equal to one if |σ2i (Ni , σ1−i (N−i , x−i ), xi )| ≥ d(xi ), i.e., if i employs enough consultants to complete her project after seeing −i employ σ1−i (N−i , x−i ), and is equal to zero else. Observe that (i) once i gets her project, her payoff only depends on the size of her network and the number of consultant she employs and (ii) 8

Assuming a constant marginal cost of labor is without loss. Our results continue to hold for any total labor cost function that is increasing in the number of consultants employed. 9 To avoid excessive notation, we define ∅\∅ = ∅.

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i can always get a zero payoff by choosing the empty network. Partner i’s ex-ante payoff is Ui (s) =

X (xi ,x−i

(u1i (s, xi , x−i ) + u2i (s, xi , x−i ))Pi (xi )P−i (x−i )/2.

)∈X 2

We occasionally write Ui (s) as Ui (s, ri , ci ) when we wish to emphasize the dependence of i’s ex-ante payoff on her reward and cost functions. Throughout the rest of the paper, we maintain the following simplifying assumption. Assumption 1. Project Completion is (Weakly) Good. For all x ∈ X, we have ri (x) − w d(x) ≥ 0 for each partner i. The assumption guarantees that both partners have a non-negative payoff to employing the number of consultants that are needed to complete a project. While this simplifies the statements and proofs of our results, it is not essential to them. Solution Concept Our solution concept is a pure strategy Nash equilibrium. Definition. An equilibrium is a (pure) strategy vector s? = (s?A , s?B ) such that each partner maximizes her payoff taking her rival’s strategy given, i.e., such that UA (s? ) ≥ UA (sA , s?B ) for all sA ∈ S A and UB (s? ) ≥ UB (s?A , sB ) for all sB ∈ S B . Let E denote the set of equilibria. If E is non-empty, then it is usually nonsingular as the definition places little restriction on off-path behavior and the consultants identities can be permuted. We focus on (pure strategy) equilibria because typical refinements, like subgame perfection, add no economic insight. The reason is simple: after she selects her network, a partner never learns of the network selected by her rival. Thus, the only subgame is the game as a whole, so the sets of subgame perfect equilibria and pure strategy equilibria coincide. Also, in the Online Appendix, we show that no economic insight is gained by refining to a Perfect Bayesian Equilibrium. An Example It is useful to look at two examples. Example 1. A Simple Symmetric Example. Let C = {1, 2} be the set of consultants and let X = {x1 , x2 } be the set of projects. Let d(x1 ) = 1 and d(x2 ) = 2 be the difficulties of the two projects. Let ri (x1 ) = 2 and ri (x2 ) = 5 for i ∈ {A, B} be the rewards to both projects. Let w = 1 and suppose both partners have a constant marginal networking cost of 1/2. Also, Pi (x1 ) = Pi (x2 ) = 1/2 for i ∈ {A, B}.

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? ? ? ? An equilibrium is s? = (NA? , σ1A , σ2A , NB? , σ1B , σ2B ), where NA? = NB? = {1, 2},

  ?     {1} if N = NA & x = x1 {2} if N = NB? & x = x1     ? ? (N , x) = {1, 2} if N = NA? & x = x2 , σ1B (N , x) = {1, 2} if N = NB? & x = x2 , σ1A       ∅ ∅ else else  {1} if N = N ? , x = x , & N 0 = {2} 1 A ? σ2A (N , N 0 , x) = , and ∅ else  {2} if N = N ? , x = x , & N 0 = {1} 1 B ? , σ2B (N , N 0 , x) = ∅ else where N and N 0 are subsets of C and x is a project. It is easy to see this by considering each partner, in turn, and showing that she does best by following s? when her rival does. We do this now for A by considering different networks: • If A holds network ∅, then her payoff is 0. • If A’s network is {1}, her payoff is −1/8. The computation is straightforward. If she gets her project xA first, it is best for her to employ consultant 1 if xA = x1 and for her not to employ consultant 1 if xA = x2 , so she earns 1/2. If she gets her project second, then it is best for her to employ consultant 1 if xA = x1 and B employs only consultant 2, otherwise it is best for her to employ no consultants. Since B employs consultant 2 if and only if she gets project x1 , A’s expected payoff is 1 1/2 1/2 = 1/4. Thus, her payoff from network {1} is 1/2(1/2 + 1/4) − 1/2 = −1/8. • If A’s network is {2}, then her payoff is −1/4. If she gets her project xA first, it is best for her to employ consultant 2 if xA = x1 and for her not to employ consultant 2 if xA = x2 , so her payoff is 1/2. If she gets her project xA second, then she cannot employ consultant 2 as B always employs 2, so her payoff is zero. Hence, her payoff from network {2} is 1/2(1/2) − 1/2 = −1/4. • If A’s network is {1, 2}, her payoff is 1/4. If she gets project xA first, then it is best for her to employ consultant 1 if xA = x1 and for her to employ both consultants ? if xA = x2 – observe that σ1A makes exactly this recommendation – so her payoff is 1/2 1 + 1/2 3 = 2. If she gets x second, then it is best for her to employ consultant A 1 if xA = x1 and B employs consultant 2, otherwise it is best for her to employ no ? consultants – observe that σ2A makes exactly this recommendation – so her payoff is 1 1/2 1/2 = 1/4. Thus, her payoff from network {1, 2} is 1/2(2 + 1/4) − 1 = 1/8. ? ? It follows that it is best for A to hold network NA? = {1, 2} and follow σ1A and σ2A when B plays according to equilibrium. Since an analogous argument gives it is best for B to follow 10

s? when A does, s? is an equilibrium. 4 Example 2. A Second Symmetric Example. Let C = {1, 2, 3, 4} be the set of consultants and let X = {x1 , x2 } be the set of projects. Let d(x1 ) = 2 and d(x2 ) = 3 be the difficulties of the two projects. Let ri (x1 ) = 6 and ri (x2 ) = 10 for i ∈ {A, B} be the rewards to both projects. Let w = 1 and suppose both partners have a constant marginal networking cost of 1/2. Also, Pi (x1 ) = Pi (x2 ) = 1/2 for i ∈ {A, B}. ? ? ? ? An equilibrium is s? = (NA? , σ1A , σ2A , NB? , σ1B , σ2B ), where NA? = {1, 2, 3}, NB? = {2, 3, 4},   ?     {1, 2} if N = NA & x = x1 {3, 4} if N = NB? & x = x1     ? ? σ1A (N , x) = {1, 2, 3} if N = NA? & x = x2 , σ1B (N , x) = {2, 3, 4} if N = NB? & x = x2 ,       ∅ ∅ else else  {1, 2} if N = N ? , x = x , & N 0 = {3, 4} 1 A ? σ2A (N , N 0 , x) = , and ∅ else  {3, 4} if N = N ? , x = x , & N 0 = {1, 2} 1 B ? 0 σ2B (N , N , x) = , ∅ else where N and N 0 are subsets of C and x is a project. This claim is readily verified using an argument analogous to the one employed in Example 1. 4

3

General Results

In this section, we establish equilibrium existence and we show that equilibrium networks are minimally overlapping. Existence of Equilibria Our goal, in this subsection, is to establish the following result. Proposition 1. Existence of a (Pure Strategy) Equilibrium. The set of equilibria E is non-empty. Existence is non-trivial as there are a multitude of ways to choose networks and employ consultants. Fortunately, the use of consultants is rivalrous. Loosely speaking, this means that when one partner chooses a large network and employs many consultants, then the other does better by choosing a smaller network and employing fewer consultants. Hence, our game is, in a sense, a submodular one. Since every two-player submodular game can be rewritten as a supermodular game, an equilibrium exists. 11

To formalize this intuition, we first introduce “simple” strategies. Informally, a sim˜A (n) = (N˜A , σ ple strategy for A (B) of size n ∈ {0, 1, . . . , N } is a tuple s ˜1A , σ ˜2A ) (˜ sB (n) = (N˜B , σ ˜1B , σ ˜2B )) that tells her (i) include the first (last) n consultants in her network and, when she gets a project x, to employ the first (last) d(x) consultants, provided these consultants are not employed by B (A) and there are d(x) consultants in her network.10 For example, the equilibrium strategies in Examples 1 and 2 are simple strategies. ˜(nA , nB ) = (˜ ˜B (nB )). Let s sA (nA ), s We next simplify the main game by supposing that both partners only pick the sizes of their networks and then follow their simple strategies. Formally, consider an auxiliary game where A chooses zA ∈ {0, . . . , N }, B chooses zB ∈ {0, . . . , N }, and then both play the original ˜(zA , N − zB ). Thus, their respective payoffs are UA (˜ game according to s s(zA , N − zB )) and UB (˜ s(zA , N − zB )). An equilibrium of the auxiliary game is a (zA? , zB? ) ∈ {0, . . . , N }2 such s(zA? , N − s(zA , N − zB? )) for all zA ∈ {0, . . . , N } and UB (˜ that (i) UA (˜ s(zA? , N − zB? )) ≥ UA (˜ zB? )) ≥ UB (˜ s(zA? , N − zB )) for all zB ∈ {0, . . . , N }. Let F denote the set of equilibria of this game. It turns out that payoffs in the auxiliary game are supermodular. Thus, its equilibrium set F is a non-empty, complete lattice by Theorem 4.2.1 of Topkis [18]. Lemma 1. Supermodularity and Equilibrium Existence in the Auxiliary Game. Let (zA , zB ) ∈ {0, . . . , N }2 , we have UA (˜ s(zA , N − zB )) and UB (˜ s(zA , N − zB )) are supermodular in (zA , zB ). It follows that F is a non-empty, complete lattice. Proof. See the Appendix.  To see the intuition for supermodularity, focus on A. When zB decreases, i.e., N − zB increases, B adds consultants to her network. This allows her to take on more difficult projects and so reduces the probability that A may employ any “shared” consultant, i.e., a consultant who’s in both her network and in B’s network. Since A has more shared consultants when zA is larger, the reduction in zB decreases A’s payoff faster when zA is larger; hence, supermodularity because of rivalry. It only remains to show that each of the auxiliary game induces an equilibrium of the original game. The next lemma does this. Lemma 2. Effectuate of Equilibria. ˜(zA? , N − zB? ) is an equilibrium of Let (zA? , zB? ) be an equilibrium of the auxiliary game, then s the original game. Proof. See the Appendix.  10

See the Appendix for the formal definition.

12

The intuition is that when one partner, say B, plays a simple strategy, then A does best by also playing a simple strategy. This is because the (ex-ante) probability that B employs a consultant is monotone increasing in the index of the consultant: B employs consultant N with the highest probability, consultant N − 1 with the second highest probability, and so on. Since A wants to network with consultants she will be able to employ with high probability, she does best by networking with the lowest indexed consultants. Thus, when ˜B (N − zB? ), then the best A can do is play s ˜A (n) for some n ∈ {0, . . . , N }. Since B follows s (zA? , zB? ) is an equilibrium point of the auxiliary game, we necessarily have that A does best by setting n = zA? . Analogous logic applies for B. Proof of Proposition 1. Immediate from Lemmas 1 and 2.  Let E S denote the set of equilibria that are induced by the equilibria of the auxiliary ˜(zA? , N − zB? ) for some (zA? , zB? ) ∈ F }. We refer to ES as the game, i.e., E S = {s ∈ S|s = s set of simple equilibria. Observe that each element of F corresponds to a unique element in E S : s? = (NA? , . . . , NB? , . . .) ∈ E S if and only if (|NA? |, N − |NB? |) ∈ F . Remark. We discuss the utilitarian efficiency of equilibria in the Online Appendix and show, among other results, that equilibria need not be efficient and that inefficiency is often due to “over-investment” in networks. Minimally Overlapping Networks In this subsection, we establish that the partners’ equilibrium networks share as few consultants “as possible.” The next definition formalizes this idea. Definition. Let NA and NB be networks for A and B. We say that NA and NB are minimally overlapping if there does not exists a NA0 ⊂ C and a NB0 ⊂ C, such that |NA | = |NA0 |, |NB | = |NB0 |, and |NA0 ∩ NB0 | < |NA ∩ NB |. That is, A and B’s networks are minimally overlapping if there does not exist a way to shrink the number of consultants in both networks without changing the size of A’s network or B’s network. For instance, when C = {1, 2, 3, 4}, the networks NA = {1, 2, 3} and NB = {2, 3} are not minimally overlapping, while the networks NA = {1, 2, 3} and NB = {3, 4} are minimally overlapping. It is easily seen that the equilibrium networks in Examples 1 and 2 are minimally overlapping. We show that all equilibrium networks are minimally overlapping when the following technical assumption holds. Assumption 2. Strictly Positive Costs and Rewards and Heterogeneous Project Difficulty. We have w > 0 and, for each x ∈ X, we have Pi (x)(ri (x) − w d(x)) > 0 for each partner i. Additionally, for each n ∈ {1, . . . , N }, there is a project x ∈ X such that d(x) = n. 13

The first part of the assumption gives that labor costs are strictly positive, that the reward to completing a project is strictly positive after accounting for labor costs, and that all projects occur with strictly positive probability. This ensures that each partners’ optimal behavior in the second stage is essentially unique. The second part of the assumption requires that projects are of sufficiently heterogeneous difficulty. This guarantees, for instance, that a partner always finds occasion to employ two-thirds of her network. Proposition 2. Equilibrium Networks are Minimally Overlapping. Let Assumption 2 hold and let s? = (NA? , . . . , NB? , . . .) ∈ E, then NA? and NB? are minimally overlapping. The proposition implies, for instance, that if A and B hold small networks, i.e., if |NA? | + |NB? | < |C|, then their networks are disjoint. The proposition reflects a general preference of the partners to share as few consultants as possible: all else equal, fewer shared consultants means that a partner can employ more of the consultants in her network more often and so do better. Before proceeding, we wish to point out that Proposition 2 depends on the Assumption 2’s heterogeneity. Indeed, as we illustrate via an example in the Online Appendix, the absence of heterogeneity can cause the proposition to fail.

4

ELEE Equilibria and Results

In this section, we introduce and motivate the concept of an employment lists and employment efficient (ELEE) equilibrium, and we develop our welfare and comparative statics results. Employment List and Employment Efficient Equilibria In this subsection, we introduce the concepts of employment lists and employment efficiency, we argue that these properties are intuitive refinements, and we show that every equilibrium with these properties is payoff equivalent to a simple equilibrium. We begin by formally defining these properties. Definition. A s = (NA , σ1A , σ2A , NB , σ1B , σ2B ) ∈ S has an employment list for partner i if Ni 6= ∅ implies that we can write Ni as an ordered list {j1 , j2 , . . . , j|Ni | } such that, for each x ∈ X with σ1i (Ni , x) 6= ∅, we have σ1i (Ni , x) = {j1 , j2 , . . . , j|σ1i (Ni ,x)| }. We say s has employment lists if it has employment lists for both A and B. If a strategy profile has employment lists, then when i gets project x first, then she always employs the first |σ1i (Ni , x)| consultants on her list when she employs anyone. For instance, 14

in the equilibrium in Example 1, B’s employment list is {2, 1} as she employs consultant 2 to carry out project x1 and consultants 1 and 2 to carry out project x2 . Likewise, in Example 2, B’s employment list is {4, 3, 2}. Definition. A s = (NA , σ1A , σ2A , NB , σ1B , σ2B ) ∈ S is employment efficient for partner i if, for each x ∈ X, we have: (i) σ1i (Ni , x) ⊂ Ni \N−i when |σ1i (Ni , x)| ≤ |Ni \N−i |, and (ii) Ni \N−i ⊂ σ1i (Ni , x) when |σ1i (Ni , x)| > |Ni \N−i |. We say s is employment efficient if it is employment efficient for both A and B. That is, i’s strategy is employment efficient if it uses “exclusive” consultants in Ni \N−i before shared consultants in Ni ∩N−i . Since every shared consultant i employs is one less consultant for −i, partner i exerts a negative externality by employing shared consultants. Thus, it is inefficient for i to employ shared consultants before she has employed every one of her exclusive consultants.11 For example, the equilibrium strategies in Examples 1 and 2 are employment efficient. We say that a s? ∈ E is an employment list and employment efficient equilibrium (abbreviated ELEE equilibrium) if it has employment lists and is employment efficient. We use E LE to denote the set of ELEE equilibria. We focus on ELEE equilibria for two reasons. First, employment lists and employment efficiency are natural refinements. Employments lists are reasonable because they capture the natural idea that each partner uses an address book or other (mental) list to determine which consultants to employ. Employment efficiency is also natural because it is costless to implement (since both exclusive and shared consultants both cost a partner w) and it weakly increases efficiency. Second, ELEE equilibria are “robust.” That is, there is an ELEE equilibrium for every parameterization of our game, whereas there are parameterizations for which there are no non-ELEE equilibria; see the Online Appendix for details. It is easy to see that that simple strategies have employment lists and are employment efficient. Lemma 3. Simple Strategies Have Employment Lists and are Employment Efficient. ˜(nA , nB ) has employment lists and is employment efficient The vector of simple strategies s for all (nA , nB ) ∈ {0, . . . , N }2 . Proof. Obvious and omitted.  It follows that simple equilibria are ELEE equilibria; in symbols, E S ⊂ E LE . However, the set of ELEE equilibria is substantially larger than the set of simple equilibria. For 11

We discuss utilitarian efficiency in the Online Appendix and show, among other results, that employment efficiency is a necessary, but not sufficient condition for efficiency.

15

instance, it contains all equilibria that differ from a simple equilibrium off of the equilibrium path, as well as all permutations of simple equilibria. The next proposition shows that there is a close relationship between ELEE and simple equilibria: they are payoff equivalent. This equivalency reduces the study of ELEE equilibria to the study of simple equilibria. Proposition 3. Payoff Equivalency of ELEE Equilibria and Simple Equilibria. Let Assumption 2 hold, then each ELEE equilibrium is payoff equivalent to a simple equilibrium where both partners hold the same sized networks as they do in the ELEE equilibrium. ˜(|NA? |, |NB? |) ∈ E S and (ii) That is, for each s? = (NA? , . . . , NB? , . . .) ∈ E LE , we have (i) s that Ui (s? ) = Ui (˜ s(|NA? |, |NB? |)) for each partner i. Proof. See the Appendix.  The key insight behind the proposition is that whenever partner i moves first in the second stage and uses an employment list and employment efficient behavioral strategy, then her behavior in s? is a permutation of her behavior in the corresponding simple equilibrium. Thus, the second partner to move −i has the same number of in-network consultants available in both equilibria. It follows that both partners get the same payoffs in these equilibria. Opposition of Interests In this subsection, we show that the partners’ interests are opposed. Specifically, we establish that there are equilibria where A does best and B does worst and that, under a strict monotonicity condition, A does best whenever B does worst. To understand the monotonicity condition, it helps to understand how rivalry affects the partners’ payoffs when they play simple strategies. To these ends, write bA (n) = arg maxn0 ∈{0,...,N } UA (˜ s(n0 , n)) and bB (n) = arg maxn0 ∈{0,...,N } UB (˜ s(n, n0 )) for A and B’s best responses when playing simple strategies. Let bi (n) = max≤ bi (n) denote the maximal selection of partner i’s best response. Occasionally, we write bi (n, ri , ci ) when we wish to emphasize the dependence of i’s maximal selection on her reward and cost functions. Lemma 4. Weak Monotonicity. Each partner’s payoff to a simple strategy is weakly decreasing in the size of the other partner’s simple strategy when she responds optimally. That is, UA (˜ s(bA (n), n)) and UB (˜ s(n, bB (n)) are weakly decreasing in n. Proof. See the Appendix.  This result follows from the rivalrous use of consultants. Specifically, as nB grows, B takes on more complex projects and employs a larger number of shared consultants when she

16

moves first. Thus, B exerts a greater negative externality on A, causing UA (˜ s(bA (nB ), nB )) to decrease. The strict monotonicity condition is that the partners’ payoffs to simple strategies evaluated at best responses are strictly decreasing in the size of the other partner’s network, which is formalized as follows. Assumption 3. Strict Monotonicity and Single-Valued Best Replies. We have (i) bA (n) and bB (n) are single-valued for all n ∈ {0, . . . , N } and (ii) UA (˜ s(bA (n), n)) and UB (˜ s(n, bB (n)) are strictly decreasing in n. Part (i) is guaranteed if the partners’ payoffs to (simple strategies) are strictly quasi-concave or if they commit to a selection of their best replies before the game begins. Part (ii) holds whenever both partners find it desirable to hold large networks (e.g., the networking cost is low and the returns to difficult projects are high) as then an increase in the size of her rival’s network always increases the number of shared consultants. It is readily verified that the payoffs in Examples 1 satisfy both parts of this assumption, while those in Example 2 do not. We can now give this subsection’s main result. Let W (i) = {s? ∈ E LE |Ui (s? ) ≥ Ui (s) for all s ∈ E LE } be the set of ELEE equilibria where partner i does best. Also, let W (i) = {s? ∈ E LE |Ui (s? ) ≤ Ui (s) for all s ∈ E LE } be the set of ELEE equilibria where i does worst. We occasionally write W (i, ri , ci ) and W (i, ri , ci ) when we wish to emphasize the dependence of these sets on i’s reward and cost functions. Proposition 4. Opposing Interests. Let Assumption 2 hold, then there is an ELEE equilibrium where A does best and B does worst and vice-versa, i.e., W (A)∩W (B) and W (B)∩W (A) are non-empty. When Assumption 3 also holds, then in A does best in any ELEE equilibrium where B does worst and vice versa, i.e., W (A) = W (B) and W (B) = W (A). Proof. See the Appendix.  The proposition gives that there are equilibria where A and B’s interests are opposed, i.e., where A does best and B does worst and vice versa. It also tells us that A does best precisely when B does worst when the strict monotonicity condition holds. To see the intuition, it helps to focus on the auxiliary game. Since this game is supermodular, it has a largest equilibrium, which A weakly most-prefers and B weakly least-prefers (per Lemma 4), and a smallest equilibrium, which B weakly most-prefers and A weakly least-prefers. When Assumption 3 holds, the partners’ preferences over equilibria become strict. Thus, A does best (worst) in only the largest (smallest) equilibrium and B does

17

worst (worst) only in the largest (smallest) equilibrium. The proposition then follows from Proposition 3. Comparative Statics In this subsection, we characterize how the partners’ payoffs and network sizes shift in ELEE equilibria where their interests are opposed as their reward and cost functions shift. We consider what happens in the event of the following shift. Assumption 4. Changes in Rewards and Costs. The cost and reward functions change as follows: 0 0 (i) A’s reward function rA increases to rA , i.e., rA (x) ≤ rA (x) for each project x. (ii) A’s cost function cA decreases to c0A , i.e., c0A (n) ≤ cA (n) for all n. (iii) For simplicity, B’s cost and reward functions do not change.12 Also, cA (n) − c0A (n) is weakly increasing in n. The last part of the assumption is a technical requirement. It is satisfied, for instance, when A has constant marginal costs of networking. Proposition 5. Comparative Statics in Rewards and Costs. Let Assumptions 2 and 4 hold. As A’s reward and cost functions shift, then A’s payoff increases and B’s payoff decreases in the ELEE equilibria that are best for A and worst for 0 B. That is, UA (s0 , rA , c0A ) ≥ UA (s? , rA , cA ) and UB (s0 , rB , cB ) ≤ UB (s? , rB , cB ) for s? = 0 , c0A ) ∩ (NA? , . . . , NB? ) ∈ W (A, rA , cA ) ∩ W (B, rB , cB ) and s0 = (NA0 , . . . , NB0 ) ∈ W (A, rA W (B, rB , cB ). Also, if Assumption 3 holds both before and after the shifts in rewards and costs, then the size of A’s network increases and the size of B’s network decreases, i.e., |NA0 | ≥ |NA? | and |NB0 | ≤ |NB? |. An analogous result holds for the ELEE equilibria that are best for B and worst for A. Proof. See the Appendix.  That is, A’s payoff and network grow and B’s payoff and network both shrink as A’s situation improves. The intuition is that as A’s reward and cost functions shift, her best response in the auxiliary game increases. This increases the size of A’s network and payoff (per Lemma 4) and decreases the size of B’s network and payoff in the maximal and minimal equilibria. The proposition now follows Proposition 3. Comparison of Payoffs and Network Sizes While Proposition 5 tells us what happens as A’s situation improves, it provides no insights into the structure of equilibrium when A has a reward and cost advantage over B. 12

It is readily verified that our results continue to obtain when B’s reward function falls and her cost function increases.

18

In this subsection, we prove that this advantage results in A holding a larger network and earning more than B. The next assumption formalizes the idea that A holds. Assumption 5. Partners’ Costs and Rewards. For each x ∈ X, we have PA (x) = PB (x) and rA (x) ≥ rB (x). For all n, we also have (i) cA (n) ≤ cB (n) and (ii) cB (n) − cA (n) is weakly increasing. Proposition 6. Comparison of Payoffs and Network Sizes. Let Assumptions 2 and 5 hold, then A earns more than B in any ELEE equilibrium where she does best and B does worst, i.e., UA (s? ) ≥ UB (s? ) for all s? = (NA? , . . . , NB? , . . .) ∈ W (A) ∩ W (B). If Assumption 3 also holds, then A has a larger network than B, i.e., |NA? | ≥ |NB? |. Proof. See the Appendix.  The intuition is that A has a larger best response than B in counterfactual the auxiliary game where both partners have rewards rB and costs cB . Thus, A holds a larger network and does better than B in the maximal equilibrium of this game and, by standard arguments, in any element of W (A, rB , cB ) ∩ W (B, rB , cB ). Assumption 5 then allows us to apply Proposition 5 to obtain Proposition 6. Before proceeding, we wish to point out that A need not have a higher payoff or a larger network than B outside of equilibria in W (A) ∩ W (B). We give an example in the Online Appendix.

5

Conclusion

We study a game where players make resource-specific investments in order to utilize resources at a later date and where subsequent access to these resources can be blocked by another player. We show that there is a pure strategy equilibrium and that all equilibrium networks are minimally overlapping. We also show that there are equilibria where the partners’ interests are opposed and we develop comparative statics for these equilibria. While our simple model lacks the complexity of real world consulting practices, (i) it captures many of their silent features (e.g., consultant-specific investments, rivalry in consultant use, and the partners general self-interest), (ii) it can easily be embedded in more complicated models, thereby allowing our results to serve as benchmarks, and (iii) it is portable to different economic contexts (e.g., TV shows and pundits or cities and water sources). Our model’s strength is its simplicity, which allows us to construct an equilibrium and leverage this construction to examine how rivalry shapes equilibrium networks and payoffs.

19

We conclude with a discussion of extensions and robustness. It is often the case that one partner is a better salesperson than the other and is able to obtain projects more frequently. Thus, she has a higher probability of moving first in the second stage. We consider this extension in the Online Appendix and we show that all of our core results continue to obtain. We also show that that increases in the probability this partner moves first increase her welfare and decrease the other partner’s welfare. The assumption that clients do not pay for half-completed project is not essential. We develop an extension with residual value in the Online Appendix and show that existence and minimal overlap continue to obtain, as do all of our results for simple equilibria. The key insight is that residual value does not mitigate rivalry, so the auxiliary game remains submodular and our arguments go through. The assumption of two partners is not needed for our minimal overlap result, but is essential to other results. While rivalry ensures the auxiliary game is submodular when there are more than two partners, we can no longer invert the order of one of the partner’s strategies to make the auxiliary game supermodular and establish our existence, welfare, and comparative statics results. Nevertheless, because rivalry is insensitive to the number of partners, our minimal overlap result continues to obtain; see the Online Appendix for details. Consultant homogeneity is necessary for our results. While there are good reasons to think that consultants are (approximately) the same from a partner’s perspective, there are times when this is not the case. When consultants are heterogeneous, simple strategies are not best responses to simple strategies because partners favor the more productive consultants. Thus, an equilibria of the auxiliary game may not induce equilibrium of the full game. Nevertheless, our results are compatible with certain kinds of heterogeneity – e.g., it is easily seen that our results continue to obtain when (i) projects come in two varieties, say red and blue, (ii) consultants are either productive on red project or blue projects, and (iii) partners face a constant marginal cost of networking.

A

Proofs Appendix

Before we prove Lemma 1, it helps to formally define simple strategies and give a closed form expression for payoffs under simple strategies. Formally, a simple strategy for A of size ˜A (n) = (N˜A , σ n ∈ {0, . . . , N } is a tuple s ˜1A , σ ˜2A ) such that, for every (N , N 0 , x) ∈ P(C)2 ×X,

20

we have N˜A = {1, . . . , n},  {1, . . . , d(x)} if d(x) ≤ |N | and N = N˜ A , σ ˜1A (N , x) = ∅ else    if d(x) ≤ |N \N 0 |, N 0 ⊂ {d(x) + 1, . . . , N },  {1, . . . , d(x)} σ ˜2A (N , N 0 , x) = and N = N˜A    ∅ else

.

˜B (n) = (N˜B , σ Likewise, a simple strategy for B of size n ∈ {0, . . . , N } is a tuple s ˜1B , σ ˜2B ) 0 2 such that, for every (N , N , x) ∈ P(C) × X, we have N˜B = {N + 1 − n, . . . , N },  {N + 1 − d(x), . . . , N } if d(x) ≤ |N | and N = N˜ B σ ˜1B (N , x) = , ∅ else    if d(x) ≤ |N \N 0 |, N 0 ⊂ {1, . . . , N − d(xB )},  {N + 1 − d(x), . . . , N } σ ˜2B (N , N 0 , x) = and N = N˜B    ∅ else

.

Lemma A1. Payoffs when Partners Play Simple Strategies. Let i ∈ {A, B} and let (ni , n−i ) ∈ {0, . . . , N }2 . Then, Ui (˜ s(ni , n−i )) = +

X

(ri (x) − w d(x))

{x|d(x)≤ni } ni X X

Pi (x) 2

(ri (x) − w d(x))gi (l, n−i )

l=1 {x|d(x)=l}

Pi (x) − ci (ni ) 2

where, for l ∈ {1, . . . , N }, X

gi (l, n−i ) = 1 −

P−i (x).

{x|N +1−l≤d(x)≤n−i }

Proof. The proof is computational. Nevertheless, we give a complete proof for completeness. We perform the computation for A as the it is analogous for B. The claim is true if nA = 0. Thus, we take nA ≥ 1. We begin by computing A’s payoff when she gets her project first. It is useful to think about the difficulty of A’s project xA . If d(xA ) ≤ nA , then A employs σ ˜1A = {1, . . . , d(xA )}, completes her project, and earns rA (xA ) − w d(xA ) (before networking costs). If d(xA ) > nA , 21

then A employs σ ˜1A = ∅, does not complete her project, and earns 0. Thus, before accounting P for network costs, A’s expected payoff from getting her project first are {x|d(x)≤nA } (rA (x) − w d(x))PA (x). Next, we compute A’s payoff when she gets her project second. Let xA and xB be A and B’s projects. We need to consider the possibility that A can not complete her project. Let l = d(xA ). Since B follows her simple strategy there are four sub-cases: (i) d(xB ) > nB and l > nA , (ii) d(xB ) ≤ nB and l > nA , (iii) d(xB ) > nB and l ≤ nA , and (iv) d(xB ) ≤ nB and l ≤ nA . If cases (i) or (ii), then σ ˜2A = ∅ and so A earns nothing. If case (iii), then σ ˜2A = {1, . . . , l}, A completes her project, and earns rA (xA ) − w d(xA ) (before networking costs). Since A follows her simple strategy, she completes her project if and only if B does not employ a consultant with an index at or below l.13 Since B employs no consultants, the desired result follows. If case (iv), then A might not complete her project because B might employ consultants with indices at or below l. Since B follows her simple strategy, she will employ a consultant with an index at or below l if an only if she gets a project xB with d(xB ) ≥ N + 1 − l.14 Thus, if d(xB ) ≥ N + 1 − l, A does not complete her project and earns 0. However, if d(xB ) < N + 1 − l or d(xB ) > nB , A completes her project and earns rA (xA ) − w d(xA ). It follows that, A’s expected payoff from getting project xA second are (1) 0 if d(xA ) > nA P and (2) (rA (xA ) − w d(xA ))(1 − {x|N +1−l≤d(x)≤nB } PB (x)) if d(xA ) ≤ nA . The first follows directly from sub-cases (i) and (ii). The second follows from sub-cases (iii) and (iv), where A gets rA (xA ) − w d(xA ) if either d(xB ) < N + 1 − l or d(xB ) > nB , an event of probability P 1− {x|N +1−l≤d(x)≤nB } PB (x) = gA (l, nB ). Thus, before networking costs, A’s expected payoff P from getting her project second is {x|d(x)≤nA } (rA (x) − w d(x))gA (d(x), nB )PA (x). Since A has a 1/2 chance of getting her project first and a 1/2 chance of getting her project second, her expected payoff inclusive of networking costs are X

(rA (x) − w d(x))

{x|d(x)≤nA }

PA (x) + 2

X

(rA (x) − w d(x))gA (d(xA ), nB )

{x|d(x)≤nA }

PA (x) − cA (nA ). 2

Reordering the second sum shows that this expression is equivalent to the one given in the statement of the lemma.  Proof of Lemma 1. We first show that A’s payoff is supermodular. Let nA , nB , and ˜A = {1, . . . , nA } To see this, suppose B employs a set of consultants T and recall that A’s network N ˜(nA , nB ). If T 6⊂ {l + 1, . . . , N }, then σ under s ˜2A = ∅ (by construction) and A fails to complete her project. ˜A \T | ≥ l, and so σ If, however, T ⊂ {l + 1, . . . , N }, then |N ˜2A = {1, . . . , l}, implying A completes her project. 14 Recall that B employs {N + 1 − d(xB ), . . . , N }. If d(xB ) < N + 1 − l, then B does not employ a with an index at or below l as d(xB ) < N + 1 − l =⇒ N + 1 − d(xB ) > l. If, however, d(xB ) ≥ N + 1 − l, then B employs a consultant with an index at or below l as d(xB ) ≥ N + 1 − l =⇒ N + 1 − d(xB ) ≤ l. 13

22

n0B be in {0, . . . , N } with n0B ≥ nB . We first establish that the difference UA (˜ s(nA , nB )) − UA (˜ s(nA , n0B )) is weakly increasing in nA , for all nA ∈ {0, . . . , N }. Thus, for n0A ≥ nA with n0A ∈ {0, . . . , N }, UA (˜ s(n0A , nB )) − UA (˜ s(n0A , n0B )) ≥ UA (˜ s(nA , nB )) − UA (˜ s(nA , n0B )) UA (˜ s(n0A , nB )) + UA (˜ s(nA , n0B )) ≥ UA (˜ s(n0A , n0B )) + UA (˜ s(nA , nB )). That is, UA (˜ s(nA , nB )) is submodular in (nA , nB ). It follows that UA (˜ s(zA , N − zB )) is 2 supermodular in (zA , zB ) ∈ {0, . . . , N } . It remains to establish that UA (˜ s(nA , nB )) − UA (˜ s(nA , n0B )) is weakly increasing in nA . To these ends, use Lemma A1 to write UA (˜ s(nA , nB )) −

UA (˜ s(nA , n0B ))

=

nA X

X

(rA (x) − w d(x))(gA (l, nB ) − gA (l, n0B ))

l=1 {x|d(x)=l}

PA (x) . 2

Since gA (l, n) is weakly decreasing in n and Assumption 1 holds, the sum on the right hand side is weakly increasing in nA . An analogous argument gives that B’s payoff is also supermodular. Thus, the second part of the lemma follows immediately from Topkis’ Theorem 4.2.1 as the strategy space is finite.  To prove Lemma 2, we need to show that a partner always does best by playing her simple strategy whenever the other plays a simple strategy. The argument is non-trivial given the complexity of the strategy space, so we make it via three lemmas. The first lemma describes an optimal way for a partner to employ her consultants, given her network and given that the other partner plays a simple strategy. The second lemma calculates a partner’s payoff to a network under these optimal employment strategies. The third lemma uses the first two lemmas to show that a simple strategy is a best response to a simple strategy. Before continuing, it is useful to characterize optimal second stage behavior. Remark A1. Optimal Second Stage Behavior. Suppose partner i gets project x and has non-empty network N . (If N is empty, i cannot employ any consultants and so gets 0.) It is easily seen that, if i gets her project first, then it is optimal for her to employ exactly d(x) consultants when d(x) ≤ |N | and to employ zero consultants otherwise. Likewise, if i gets her project second, then it is optimal for her to employ exactly d(x) consultants when d(x) ≤ |N \T | and to employ zero consultants otherwise, where T is the set of consultants initially employed by −i. Under Assumption 2, this is actually the unique optimal behavior because employing more than the needed number 23

of consultants is wasteful and employing less leaves surplus on the table as the project is incomplete. #

Our goal is to develop a pair of second stage strategies that implement a partner’s optimal behavior, given her network and given the other partner plays her simple strategy. To these ends, N be a non-empty network of consultants and let n = |N |. The standard form of N for A is a labeling of the consultants in N such that the lowest indexed consultant receives the label m1 , the second lowest indexed consultant receives the label m2 , and so on until the highest indexed consultant gets the label mn . The standard form of N for B is a labeling of the consultants in N such that the highest indexed consultant receives the label m1 , the second highest indexed consultant gets the label m2 , and so on until the lowest indexed consultant gets the label mn . We now develop a candidate pair of second stage behavioral strategies for A. Let NA be NA NA (N , N 0 , x) = ∅ for all (N , N 0 , x) ∈ (N , x) = σ ˆ2A a network for A. If NA is empty, let σ ˆ1A P(C)2 × X. If NA is non-empty, let {m1 , . . . , mn } be the standard form of NA for A, where n = |NA |. For each (N , N 0 , x) ∈ P(C)2 × X, let  {m , m . . . , m } if d(x) ≤ |N | and N = N 1 2 A d(x) NA σ ˆ1A (N , x) = ∅ else,    if d(x) ≤ |N \N 0 |, N 0 ⊂ {md(x) + 1, . . . , N },  {m1 , m2 , . . . , md(x) } NA (N , N 0 , x) = σ ˆ2A and N = NA    ∅ else. NA NA We refer to σ ˆ1A and σ ˆ2A as A’s hat strategies given NA . Under these strategies, when A gets project x, she employs the d(x) consultants in her network with the lowest indices, when these consultants are available and there are d(x) consultants in her network. Observe that, when NA = {1, . . . , n} for some n ∈ {0, . . . , N }, then m1 = 1, m2 = 2, . . ., and mn = n, so the hat strategies are the same as A’s second stage simple strategies. NB NB Consider B. Let NB be a network for B. If NB is empty, let σ ˆ1B (N , x) = σ ˆ2A (N , N 0 , x) = ∅ for all (N , N 0 , x) ∈ P(C)2 × X. If NB is non-empty, let {m1 , . . . , mn } be the standard form

24

of NB for B, where n = |NB |. For each (N , N 0 , x) ∈ P(C)2 × X, let  {m , m . . . , m } if d(x) ≤ |N | and N = N 1 2 B d(x) NB σ ˆ1B (N , x) = ∅ else,    if d(x) ≤ |N \N 0 |, N 0 ⊂ {1, . . . , md(x) − 1},  {m1 , m2 , . . . , md(x) } NB σ ˆ2B (N , N 0 , x) = and N = NB     ∅ else. NB NB We refer to σ ˆ1B and σ ˆ2B as B’s hat strategies given NB . Under these strategies, when B gets project x, she employs the d(x) consultants in her network with the highest indices, when these consultants are available there are d(x) consultants in her network. Observe that, when NB = {N + 1 − n, . . . , N } for some n ∈ {0, . . . , N }, then m1 = N , m2 = N − 1, . . ., and mn = N + 1 − n, so the hat strategies are the same as B’s second stage simple strategies.

Lemma A2. Optimality of Hat Strategies. Two parts: (i) Let NA ⊂ C, then, for all (σ1A , σ2A ) such that (NA , σ1A , σ2A ) ∈ S A and all nB ∈ NA NA {0, . . . , N }, we have UA (NA , σ ˆ1A ,σ ˆ2A , s˜B (nB )) ≥ UA (NA , σ1A , σ2A , s˜B (nB )). (ii) Let NB ⊂ C, then, for all (σ1B , σ2B ) such that (NB , σ1B , σ2B ) ∈ S B and all nA ∈ NB NB {0, . . . , N }, we have UB (˜ sA (nA ), NB , σ ˆ1B ,σ ˆ2B ) ≥ UB (˜ sA (nA ), NB , σ1B , σ2B ). Proof. It is clear that hat strategies implement the behavior described in Remark OA 1. Thus, not other behavioral strategies can do better and the desired result follow from the fact that all projects occur with strictly positive probability. Nevertheless, we give a detailed proof for completeness. We only prove part (i) since an analogous argument gives part (ii). Since UA (s) =

X

(u1A (s, xA , xB ) + u2A (s, xA , xB ))PA (xA )PB (xB )/2,

(xA ,xB )∈X 2

it suffices to show, for each (xA , xB ) ∈ X 2 , that NA NA u1A (NA , σ ˆ1A ,σ ˆ2A , s˜B (nB ), xA , xB ) ≥ u1A (NA , σ1A , σ2A , s˜B (nB ), xA , xB ) NA NA u2A (NA , σ ˆ1A ,σ ˆ2A , s˜B (nB ), xA , xB ))

≥ u2A (NA , σ1A , σ2A , s˜B (nB ), xA , xB )).

(A.1) (A.2)

Since this argument is trivial if NA = ∅ , we take NA to be non-empty. Let {m1 , . . . , mnA } be the standard form of NA , where nA = |NA |. 25

NA Let (xA , xB ) ∈ X 2 . Suppose A gets her project first. If d(xA ) ≤ |NA |, then σ ˆ1A = NA {m1 , . . . , md(x) } and so A completes her project. If, however, d(xA ) > |NA |, then σ ˆ1A =∅ and A does not complete her project. This behavior is optimal per Remark A1, so we necessarily have that equation (A.1) holds. Now, suppose A gets her project second. Let l = d(xA ). Since B follows her simple strategy, there are four cases: (i) d(xB ) > nB and l > nA , (ii) d(xB ) > nB and l ≤ nA , (iii) NA d(xB ) ≤ nB and l > nA , and (iv) d(xB ) ≤ nB and l ≤ nA . If (i) or (iii), then σ ˆ2A = ∅ as l > |NA | and A does not complete her project. This is optimal per Remark A1. Thus, we necessarily have that equation (A.2) holds. If (ii), then B employs no consultants, i.e., σ ˜1B = ∅. Since d(xA ) ≤ nA = |NA \˜ σ1B |, we NA have σ ˆ2A = {m1 , . . . , ml }, so A completes her project. This is also optimal per Remark A1. Thus, we necessarily have that equation (A.2) holds. If (iv), then A may or may not complete her project. Since A follows her hat strategy, she completes her project if and only if B does not employ a consultant with an index at or below ml .15 Since B follows her simple strategy, she will employ a consultant with an index at or below ml if and only if d(xB ) ≥ N + 1 − ml .16 NA = ∅ and A does not complete her project. This is If d(xB ) ≥ N + 1 − ml , then σ ˆ2A optimal by Remark A1. Specifically, B employs all consultants with indices of N + 1 − d(xB ) and above, let T denote this set. Since ml ≥ N + 1 − d(xB ), NA less T is a subset of {m1 , . . . , ml−1 }. Thus, l > |NA \T | and so Remark A1 gives that it is best for A not to complete her project. Thus, we necessarily have that equation (A.2) holds. NA = {m1 , . . . , ml } and A completes her project. This is If d(xB ) < N + 1 − ml , then σ ˆ2A optimal by Remark A1. Simply, B employs all consultants with indices of N + 1 − d(xB ) or above, let T denote this set. Since N + 1 − d(xB ) > ml , NA less T contains {m1 , . . . , ml }. It follows that |NA \T | ≥ l, so Remark A1 gives that it is best for A to complete her project. Thus, equation (A.2) necessarily holds. 

Lemma A3. Payoffs to Different Networks. Let i ∈ {A, B}, let n−i ∈ {0, . . . , N }, and let Ni ⊂ C. Then: Ni Ni ˜−i (n−i )) = 0. (i) If Ni = ∅, then Ui (Ni , σ ˆ1i ,σ ˆ2i ,s NA To see this, suppose B employs a set of consultants T . If T 6⊂ {ml + 1, . . . , N }, then σ ˆ2A = ∅ (by construction) and A fails to complete her project. If, however, T ⊂ {ml + 1, . . . , N }, then |NA \T | ≥ l and NA so σ ˜2A = {m1 , . . . , ml }, implying A completes her project. 16 Recall that B employs {N +1−d(xB ), . . . , N }. If d(xB ) < N +1−ml , then B does not employ a consultant with an index at or below ml as d(xB ) < N +1−ml =⇒ N +1−d(xB ) > ml . If, however, d(xB ) ≥ N +1−ml , then B employs a consultant with an index at or below ml as d(xB ) ≥ N + 1 − ml =⇒ N + 1 − d(xB ) ≤ ml . 15

26

(ii) If Ni 6= ∅, let {m1 , . . . , mn } be the standard form of Ni for partner i with n = |Ni |, then Ni Ni ˜−i (n−i )) = Ui (Ni , σ ˆ1i ,σ ˆ2i ,s

+

X

(ri (x) − w d(x))

{x|d(x)≤n} n X X

Pi (x) , 2

(ri (x) − w d(x))

l=1 {x|d(x)=l}

Pi (x) Ni hi (l, n−i ) − ci (n) 2

where, for l ∈ {1, . . . , N },  1 − P {x|N +1−ml ≤d(x)≤n−i } P−i (x) if i = A Ni hi (l, n−i ) = P 1 − if i = B. {x|ml ≤d(x)≤n−i } P−i (x) Proof. The proof is computational and analogous to the Proof of Lemma A1. The only difference is when a partner moves second. We document this difference for completeness. We focus on A as the argument for B is analogous and, as usual, we take NA 6= ∅ to avoid trivialities. Let xA and xB be A and B’s projects. Let l = d(xA ) and let nA = |NA |. There are four cases: (i) d(xB ) > nB and l > nA , (ii) d(xB ) ≤ nB and l > nA , (iii) d(xB ) > nB and l ≤ nA , NA = ∅ and A earns nothing, while if and (iv) d(xB ) ≤ nB and l ≤ nA . If (i) or (ii), then σ ˆ2A NA (iii) then B employs no consultants, so σ ˆ2A = {m1 , . . . , ml } and A earns rA (xA ) − w l. NA If (iv), then A may or may not complete her project. Since A follows σ ˆ2A , she will complete her project if and only if B employs no consultant with an index at or below ml . Since B follows her simple strategy, she will employ a consultant with an index at or below ml if and only if d(xB ) ≥ N + 1 − ml . Thus, A earns rA (xA ) − w l if d(xB ) < N + 1 − ml and earns 0 if d(xB ) ≥ N + 1 − ml . It follows that, A’s expected payoff from getting project xA second are (i) 0 if d(xA ) > nA P and (ii) (rA (xA ) − w d(xA ))(1 − {x|N +1−ml ≤d(x)≤nB } PB (x)) if d(xA ) ≤ nA . The first is follows directly from cases (i) and (ii). The second follows from cases (iii) and (iv), where A gets rA (xA ) − w d(xA ) if d(xB ) < N + 1 − ml or d(xB ) > nB , an event of probability P 1 − {x|N +1−ml ≤d(x)≤nB } PB (x). Thus, before networking costs, A’s expected payoff from P A getting her project second is {x|d(x)≤nA } (rA (x) − w d(x))hN A (d(xA ), nB )PA (x).  Lemma A4. Optimality of Simple Strategies. Two parts: ˜B (nB )) ≥ (i) Let NA ⊂ C, let nA = |NA |, and let nB ∈ {0, . . . , N }, then UA (˜ sA (nA ), s NA NA ˜B (nB )). UA (NA , σ ˆ1A , σ ˆ2A , s ˜B (nB )) ≥ (ii) Let NB ⊂ C, let nB = |NB |, and let nA ∈ {0, . . . , N }, then UB (˜ sA (nA ), s 27

NB NB UB (˜ sA (nA ), NB , σ ˆ1B ,σ ˆ2B ).

Proof. We prove part (i) as the argument for part (ii) is analogous. If NA is empty, the result is true as A makes zero. Thus, we take NA to be non-empty. ˜A ˜A N N ˜A (nA ) = (N˜A , σ Write s ˜1A , σ ˜2A ) and recall that σ ˆ1A =σ ˜1A and σ ˆ2A =σ ˜2A since N˜A = {1, . . . , nA }, i.e., A’s hat strategies are the same as her second stage strategies under her ˜A ˜A N N ˜B (nB )) = UA (N˜A , σ ˜B (nB )). So we simple strategy of size nA . Thus, UA (˜ sA (nA ), s ˆ1A ,σ ˆ2A ,s ˜A ˜A N N N N ˜B (nB )) − UA (NA , σ ˜B (nB )) ≥ 0. only need to show that UA (N˜A , σ ˆ1A , σ ˆ2A , s ˆ1AA , σ ˆ2AA , s Since both of A’s networks are of size nA , Lemma A3 gives ˜A ˜A N N NA NA ˜B (nB )) − UA (NA , σ ˜B (nB )) UA (N˜A , σ ˆ1A ,σ ˆ2A ,s ˆ1A ,σ ˆ2A ,s nA X X PA (x) N˜A A (hA (l, nB ) − hN = (rA (x) − w d(x)) A (l, nB )). 2 l=1 {x|d(x)=l}

˜

NA A Since Assumption 1 holds and PA ≥ 0, we only need to show that hN A (l, nB )−hA (l, nB ) ≥ 0. Let {m ˜ 1, . . . , m ˜ nA } be the standard from of N˜A for A and let {m1 , . . . , mnA } be the standard form of NA for A. Thus, ˜

NA A hN A (l, nB ) − hA (l, nB ) =

X

X

PB (x) −

{x|N +1−ml ≤d(x)≤nB }

PB (x).

{x|N +1−m ˜ l ≤d(x)≤nB }

For every k ∈ {1, . . . , nA }, we have m ˜ k ≤ mk .17 Thus, N + 1 − ml ≤ N + 1 − m ˜ l . It follows that X ˜A NA (l, n ) = PB (x) ≥ 0, hN (l, n ) − h B B A A {x|N +1−ml ≤d(x)
where the inequality follows from the fact that PB ≥ 0.  Proof of Lemma 2. Let (zA? , zB? ) be an equilibrium of the auxiliary game. We will show ˜(zA? , N −zB? ) = (˜ ˜B (N −zB? )) is an equilibrium by showing that (i) UA (˜ that s sA (zA? ), s s(zA? , N − ˜B (N −zB? )) for all sA ∈ S A and (ii) that UB (˜ zB? )) ≥ UA (sA , s s(zA? , N −zB? )) ≥ UB (˜ sA (zA? ), sB ) for all sB ∈ S B . We first focus on (i). Since (zA? , zB? ) is an equilibrium of the auxiliary game, UA (˜ s(zA? , N − zB? )) ≥ UA (˜ s(nA , N −zB? )) for all nA ∈ {0, . . . , N }. Thus, Lemma A4 implies that UA (˜ s(zA? , N − NA NA ˜B (N −zB? )) NA ⊂ C. Hence, Lemma A2 implies that UA (˜ zB? )) ≥ UA (NA , σ ˆ1A ,σ ˆ2A ,s s(zA? , N − ˜B (N − zB? )) for all (NA , σ1A , σ2A ) ∈ S A . Since the argument for zB? )) ≥ UA (NA , σ1A , σ2A , s (ii) is analogous, the lemma obtains.  17

To see this, note that for every j ∈ {1, . . . , nA }, we have j ≤ mj . Simply, m1 ≥ 1. Since m2 > m1 , we necessarily have m2 ≥ m1 + 1 ≥ 2. Continuing establishes the desired result. Since m ˜ 1 = 1, m ˜ 2 = 2, . . ., and m ˜ nA = nA , we have m ˜ k ≤ mk for every k ∈ {1, . . . , nA }.

28

To prove Proposition 2, we need two lemmas. The first shows that the partners behave according to Remark OA 1 in any equilibrium when Assumption 2 holds. The second lemma shows that if one consultant is in both partners’ networks, then the networks “cover” the set of consultant. Before proceeding, we introduce a notion inspired by sequential rationality. Let s = (NA , σ1A , σ2A , NB , σ1B , σ2B ) ∈ S and let Eis denote the set of sets of consultants partner i employs under s when she gets her project first, i.e., Eis = {T ∈ P(C)|T = σ1i (Ni , x) for some x ∈ X}. Definition. We say that a s = (NA , σ1A , σ2A , NB , σ1B , σ2B ) ∈ S is weakly rational for partner i if she behaves optimally in the second stage when the other partner follows s. s That is, if, for each (T , x) ∈ E−i × X, we have: (i) |σ1i (Ni , x)| = d(x) when d(x) ≤ |Ni | and |σ1i (Ni , x)| = 0 when d(x) > |Ni |; and (ii) |σ2i (Ni , T , x)| = d(x) when d(x) ≤ |Ni \T | and |σ2i (Ni , T , x)| = 0 when d(x) > |Ni \T |. We say that s is weakly rational if it is weakly rational for both partners A and B. Note that simple strategies of any size are always weakly rational. Lemma A5. Every Equilibrium is Weak Rationality. Let Assumption 2 hold, then each s? ∈ E is weakly rational. The lemma follows from the fact that every project occurs with positive probability. Thus, if a partner follows a strategy that is not weakly rational, she can do strictly better in expectation by switching to a strategy that is weakly rational, violating the conjecture of equilibrium. Proof. This is almost obvious; we give the proof for completeness. We argue by contra? ? ? ? diction. Let s? = (NA? , σ1A , σ2A , NB? , σ1B , σ2B ) ∈ E. We suppose, without loss, that A’s ? ? ) is not weakly rational for her; the argument is analogous for B. We , σ2A strategy (NA? , σ1A ? ? ? , σ2A ) is not a best response for A when B plays according to s? , will establish that (NA , σ1A a contradiction. ? ? There are three ways in which (NA? , σ1A , σ2A ) may not be weakly rational: ? ? 1. There is an x0 ∈ X such that |σ1A (NA? , x0 )| = 6 d(x0 ) when d(x) ≤ |NA? | or |σ1A (NA? , x)| = 6 ? 0 when d(x) > |NA |. In this case, upon getting project x first, A does strictly better by employing exactly d(x) consultants when d(x) ≤ |NA? | or employing no consultants when d(x) > |NA? |. ? 2. There is an (T 0 , x0 ) ∈ EBs ×X, such that |σ2A (NA? , T 0 , x0 )| = 6 d(x0 ) when d(x0 ) ≤ |NA? \T 0 | ? or |σ2A (NA? , T 0 , x0 )| = 6 0 when d(x0 ) > |NA? \T 0 |. In this case, upon getting project x0 second and observing B employ consultants in T 0 , A does strictly better by employing exactly d(x0 ) consultants when d(x0 ) ≤ |NA? \T 0 | or employing no consultants when d(x0 ) > |NA? \T 0 |. 29

3. Both 1 and 2. We only consider the second case, since it is the hardest and the other two are analogous. 0 0 Suppose case 2 occurs. Let σ2A be a new strategy for A such that σ2A (N , N 0 , x) = ? 0 σ2A (N , N 0 , x) for all (N , N 0 , x) ∈ (P(C)2 × X) \{(NA? , T 0 , x0 )} and such that σ2A (NA? , T 0 , x0 ) selects d(x0 ) consultants from NA? \T 0 when d(x0 ) ≤ |NA? \T 0 | or select no consultants when ? 0 ? ? d(x0 ) > |NA? \T 0 |. Let s0 = (NA? , σ1A , σ2A , NB? , σ1B , σ2B ). 0 Consider the difference of A’s payoff under s and s? . Since (i) A has the same network in both strategies, (ii) follows the same behavioral strategy when she gets her project first, and (iii) follows the same behavioral strategy when she gets her project second, unless B gets a project that causes her to employ T 0 and A gets project x0 , UA (s0 ) − UA (s? ) =

X

(u2A (s0 , x0 , xB ) − u2A (s? , x0 , xB ))

? (N ? ,x )=T 0 } {xB |σ1B B B

PA (x0 )PB (xB ) . 2

Since PA > 0 and PB > 0, this difference is strictly positive if u2A (s0 , x0 , xB ) > u2A (s? , x0 , xB ), which is exactly the case since A’s behavior under s0 is optimal and her behavior under s? ? ? is sub-optimal. Thus, (NA? , σ1A , σ2A ) is not a best response for A when B plays according to s? .  Lemma A6. Covering. Let Assumption 2 hold and let s? = (NA? , . . . , NB? , . . .) ∈ E, then NA? ∩ NB? 6= ∅ implies that C ⊂ NA? ∪ NB? . The intuition is that if NA? ∩ NB? 6= ∅ and C is not in NA? ∪ NB? , then there is at least one consultant who is not in either partner’s network. We show that one partner, say A, can always do strictly better by swapping this consultant for one who she shares with B. Thus, s? cannot be an equilibrium, a contradiction. The lemma follows. Proof. While this result is almost obvious, we give a proof for completeness. We argue by contradiction. Suppose that NA? ∩ NB? 6= ∅ and that C 6⊂ NA? ∪ NB? . Then there is a consultant j ∈ C\(NA? ∪ NB? ). We will show that A can do strictly better by swapping a consultant k ∈ NA? ∩ NB? for consultant j, when B follows s? . It follows that s? cannot be an equilibrium. We proceed by constructing a “post-swap” strategy for A and showing that this strategy ? ? ? leads to a strictly higher payoff than A gets in equilibrium. Write s? = (NA? , σ1A , σ2A , NB? , σ1B , ? σ2B ) for the equilibrium given in the statement of this lemma. Let XA ⊂ X such that x ∈ XA implies d(x) = |NA? \NB? | + 1. Also, let XB ⊂ X such that

30

? x ∈ XB implies d(x) = |NB? |.18 By Lemma A5, x ∈ XB if and only if σ1B (NB? , xB ) = NB? .19 0 0 We now write down a post-swap strategy for A, which we denote (NA0 , σ1A , σ2A ). Since 0 ? 0 A swaps k for j, we have NA = {j} ∪ NA \{k}. We choose σ1A so that, when A observes her network is NA0 and observes her project is x, she employs exactly the same consultants ? as she would under σ1A , when she observes her network is NA? and observes her project is x, save she swaps k for j. Formally, for every x ∈ X, let

 ? σ ? (N ? , x) if k ∈ / σ1A (NA? , x) 1A A 0 0 σ1A (NA , x) = {j} ∪ σ ? (N ? , x)\{k} if k ∈ σ ? (N ? , x). 1A A 1A A 0 And for every (N , x) ∈ (P(C)\NA0 ) × X, let σ1A (N , x) = ∅. ? 0 We choose σ2A so that, when A observes her network is NA0 , observes B employ T ∈ EBs , and observes her project is x, she employs exactly the same consultants as she would under ? σ2A , when she observes her network is NA? , observes B employ T , and observes her project ? is x, save she swaps k for j. Formally, for every (T , x) ∈ EBs × X, let

 ? σ ? (N ? , T , x) if k ∈ / σ2A (NA? , T , x) 2A A 0 σ2A (NA0 , T , x) = {j} ∪ σ ? (N ? , T , x)\{k} if k ∈ σ ? (N ? , T , x). A 2A A 2A
? 0 And for every (N , N 0 , x) ∈ (P(C)2 × X) \ {NA0 } × EBs × X , let σ2A (N , N 0 , x) = ∅. 0 We make one modification to σ2A before proceeding: if A observes B employ NB? and then gets a project in XA , she employs all of NA0 \NB? . Formally, for every x ∈ XA , let 0 0 0 ) ∈ S A . Let s0 = , σ2A (NA0 , NB? , x) = NA0 \NB? . It is readily verified that (NA0 , σ1A σ2A 0 0 ? ? (NA0 , σ1A , σ2A , NB? , σ1B , σ2B ). Before we establish that A makes strictly under s0 than s? , we need two preliminary facts. 0 ? (NA0 , x)| = |σ1A (NA? , x)| for all x ∈ X, i.e., A uses the same number of consultants First, |σ1A to complete project x under s0 and s? when she moves first in the second stage. This is a 0 direct consequence the construction of σ1A . Second, for all (xA , xB ) ∈ X 2 \(XA ×XB ), we have 0 ? ? ? |σ2A (NA0 , σ1B (NB? , xB ), xA )| = |σ2A (NA? , σ1B (NB? , xB ), xA )|. That is, given A and B’s projects are in X 2 \(XA × XB ), then A uses the same number of consultants to complete xA under s0 ? and s? when she moves second. Let us establish this. Since σ1B (NB? , xB ) = NB? if and only if 18

Observe that XA and XB are non-empty. Since j is not in both partner’s networks, we have |NA | ≤ N −1 and |NB | ≤ N − 1. Thus, we have 0 ≤ |NA? \NB? | ≤ N − 1. Hence, Assumption 2 gives that there is a x0 and x00 in X such that d(x0 ) = |NA? \NB? | + 1 and d(x00 ) = |NB? |. 19 ? To see this, let x ∈ XB . Since every equilibrium is weakly rational, we must have |σ1B (NB? , x)| = ? ? ? ? ? d(x) = |NB |, which implies σ1B (NB , x) = NB . Conversely, let x ∈ / XB . Then either d(x) < |NB |, implying ? ? ? |σ1B (NB? , x)| < |NB? |, or d(x) > |NB? |, implying |σ1B (NB? , x0 )| = 0. In both cases, σ1B (NB? , x0 ) is a strict ? subset of NB .

31

? xB ∈ XB , we have (σ1B (NB? , xB ), xA ) ∈ {NB? } × XA if and only if (xA , xB ) ∈ XA × XB . Thus,
? ? (σ1B (NB? , xB ), xA ) ∈ EBs × X \ ({NB? } × XA ) if and only if (xA , xB ) ∈ X 2 \ (XA × XB ).
? 0 ? Since |σ2A (NA0 , T , xA )| = |σ2A (NA? , T , xA )| for all (T , xA ) ∈ EBs × X \ ({NB? } × XA ) by construction,20 we have the secondary preliminary fact. Recall that A’s ex-post payoff depends only on the size of her network and the number of consultants she employs. Since A holds the same sized network under s0 and s? , the first preliminary fact implies u1A (s0 , xA , xB ) = u1A (s? , xA , xB ) for all (xA , xB ) ∈ X 2 , while the second preliminary fact implies that u2A (s0 , xA , xB ) = u2A (s? , xA , xB ) for all (xA , xB ) ∈ X 2 \(XA × XB ). Thus,

X

UA (s0 ) − UA (s) =

(u2A (s0 , xA , xB ) − u2A (s? , xA , xB ))

(xA ,xB )∈XA ×XB

PA (xA )PB (xB ) . 2

Under s? , when A moves second, she employs no consultants when her project xA is in XA and B’s project xB is in XB . Since B employs her entire network, A is left with |NA? \NB? | consultants to possibly employ. Since x ∈ XA implies d(x) > |NA? \NB? |, it is optimal for A to employ no consultants per Remark A1. Since s? is an equilibrium, Lemma A5 tells us that this is exactly what A does. Hence, u2A (s? , xA , xB ) = −cA (|NA? |). Under s0 , when A moves second, she employs NA0 \NB? if her project xA is in XA and B’s 0 . Since |NA0 \NB? | = |{j} ∪ NA? \NB? | = 1 + |NA? \NB? | project xB is in XB by construction of σ2A (as j ∈ / NB? and k ∈ NB? ) and d(xA ) = 1 + |NA? \NB? |, A completes her project and gets a payoff of u2A (s? , xA , xB ) = rA (xA ) − w d(xA ) − cA (NA? ). It follows that UA (s0 ) − UA (s) =

X

(rA (xA ) − w d(xA ))PA (xA )PB (xB ) > 0,

(xA ,xB )∈XA ×XB

where the strict inequality follows from Assumption 2.  Proof of Proposition 2. There are two cases NA? ∩ NB? = ∅ and NA? ∩ NB? 6= ∅. If NA? ∩ NB? = ∅, then |NA? ∩ NB? | = 0 and NA? and NB? are (trivially) minimally overlapping. If NA? ∩ NB? 6= ∅, we argue by contradiction. Suppose NA? and NB? are not minimally overlapping, then there are NA0 ⊂ C and NB0 ⊂ C with |NA? | = |NA0 |, |NB? | = |NB0 |, and |NA0 ∩ NB0 | < |NA? ∩ NB? |. Since NA? ∩ NB? 6= ∅, Lemma A6 gives that C ⊂ NA? ∪ NB? , so the sets NA? and NB? \NA? partition C. It follows that N = |C| = |NA? | + |NB? | − |NA? ∩ NB? |, since |NB? \NA? | = |NB? | − |NA? ∩ NB? |. Thus, N < |NA0 | + |NB0 | − |NA0 ∩ NB |. Since |NB0 \NA0 | = |NB0 | − |NA0 ∩ NB0 |, we have N < |NA0 | + |NB0 \NA0 |. Since NA0 and NB0 \NA0 are disjoint subsets of C, we necessarily have that |NA0 | + |NB0 \NA0 | ≤ N . Thus, N < N , a contradiction.  20

This equality does not hold on {NB? } × XA due to our modification.

32

To prove Proposition 3 we need a lemma. We say that a s = (NA , . . . , NB , . . .) ∈ S has the covering property if NA ∩ NB 6= ∅ implies C ⊂ NA ∪ NB . Lemma A7. Payoff Equivalent Strategies. Two parts: (i) Let s = (NA , σ1A , σ2A , NB , σ1B , σ2B ) ∈ S such that: (a) s has the covering property, (b) s is employment efficient for B, (c) s is weakly rational for A, and (d) σ1B fulfills part (i) of the definition of weak rationality for B. Then, UA (s) = UA (˜ s(|NA |, |NB |)). (ii) Let s = (NA , σ1A , σ2A , NB , σ1B , σ2B ) ∈ S such that: (a) s has the covering property, (b) s is employment efficient for A, (c) s is weakly rational for B, and (d) σ1A fulfills part (i) of the definition of weak rationality for A. Then, UB (s) = UB (˜ s(|NA |, |NB |)). Proof. While this result is straightforward, we give a proof for completeness. We prove part ˜ denote s ˜(|NA |, |NB |) = (N˜A , σ (i) as the argument for part (ii) is analogous. Let s ˜1A , σ ˜2A , N˜B , σ ˜1B , σ ˜2B ). We prove the lemma by establishing that (i) u1A (s, xA , xB ) = u1A (˜ s, xA , xB ) 2 and (ii) u2A (s, xA , xB ) = u2A (˜ s, xA , xB ) for every (xA , xB ) ∈ X . It follows that UA (s) = UA (˜ s). Let nA = |NA | and nB = |NA |. We need to establish a preliminary result: |NA ∩ NB | = |N˜A ∩ N˜B |. There are two cases, NA ∩ NB = ∅ and NA ∩ NB 6= ∅. If NA ∩ NB = ∅, then N ≥ nA + nB . Since |N˜A ∩ N˜B | = |{N + 1 − nB , . . . , nA }| =

 0

if nA + nB ≤ N

n + n − N A B

if nA + nB > N,

˜A ∩ N ˜B |. (by definition of simple strategies) we have |N˜A ∩ N˜B | = 0. Thus, |NA ∩ NB | = |N If NA ∩ NB 6= ∅, then the covering property implies C ⊂ NA ∪ NB . Thus, we have NA \NB , NB \NA , and NA ∩ NB partition C, so N = |NA \NB | + |NB \NA | + |NA ∩ NB |. Since |Ni \N−i | = ni − |Ni ∩ Ni | for i ∈ {A, B}, we have |NA ∩ NB | = nA + nB − N . Since NA ∩NB 6= ∅, we have |NA ∩NB | > 0, implying nA +nB −N > 0 and |N˜A ∩ N˜B | = nA +nB −N . Thus, |NA ∩ NB | = |N˜A ∩ N˜B |. Suppose A gets project first. Let xA and xB be A and B’s projects. Since s is weakly rational for A, |σ1A (NA , xA )| = d(xA ) when d(xA ) ≤ nA and |σ1A (NA , xA )| = 0 when d(xA ) > nA . Likewise, |˜ σ1A (NA , xA )| = d(xA ) when d(xA ) ≤ nA and |˜ σ1A (NA , xA )| = 0 when d(x) > nA by definition of a simple strategy. Thus, A employs the same number of consultants under both strategies, i.e., |σ1A (NA , xA )| = |˜ σ1A (NA , xA )|. Since A holds the same sized network ˜ and since A’s ex-post earnings are determined by the number of consultants under s and s she employs and the size of her network, we have u1A (s, xA , xB ) = u1A (˜ s, xA , xB ). Since (xA , xB ) were arbitrary, the desired result follows. Suppose A gets her project second. Let xA and xB be A and B’s projects. There are four 33

cases: (i) d(xB ) > nB and d(xA ) > nA , (ii) d(xB ) > nB and d(xA ) ≤ nA , (iii) d(xB ) ≤ nB and d(xA ) > nA , and (iv) d(xB ) ≤ nB and d(xA ) ≤ nA . In each case, we establish that A’s behavioral strategies σ2A and σ ˜2A select the same number of consultants when B follows σ1B and σ ˜1B respectively, i.e., that |σ2A (NA , σ1B (NB , xB ), xA )| = |˜ σ2A (N˜A , σ ˜1B (N˜B , xB ), xA )|. Since A’s ex-post payoff is determined entirely by the size of her network, which is the same under both strategies, and the number of consultants that she employs, we have u2A (s, xA , xB ) = u2A (˜ s, xA , xB ). Since (xA , xB ) were arbitrary, the desired result follows. If (i), then |σ1B (NB , xB )| = 0 by the weak rationality of σ1B and |σ2A (NA , ∅, xA )| = 0 by the weak rationality of s for A. Likewise, |˜ σ1B (N˜B , xB )| = 0 and |˜ σ2A (N˜A , ∅, xA )| = 0 by the construction of the simple strategies. Thus, |σ2A (NA , ∅, xA )| = |˜ σ2A (N˜A , ∅, xA )|. If case (ii), then |σ1B (NB , xB )| = 0 by the weak rationality of σ1B and |σ2A (NA , ∅, xA )| = σ2A (N˜A , ∅, xA )| = d(xA ) by the weak rationality of s for A. Likewise, |˜ σ1B (N˜B , xB )| = 0 and |˜ d(xA ) by the construction of simple strategies. Thus, |σ2A (NA , ∅, xA )| = |˜ σ2A (N˜A , ∅, xA )|. If case (iii), then |σ1B (NB , xB )| = d(xB ) by the weak rationality of σ1B and |σ2A (NA , σ1B (NB , xB ), xA )| = 0 by the weak rationality of s for A. Likewise, |˜ σ1B (N˜B , xB )| = d(xB ) and |˜ σ2A (N˜A , σ ˜1B (N˜B , xB ), xA )| = 0 by the construction of simple strategies. Thus, |σ2A (NA , σ1B (NB , xB ), xA )| = |˜ σ2A (N˜A , σ ˜1B (N˜B , xB ), xA )|. If case (iv), then |σ1B (NB , xB )| = d(xB ) by the weak rationality of σ1B . It is useful to think about the number of consultants left for A after B moves. Since s is employment efficient for B, σ1B (NB , xB ) ⊂ NB \NA when d(xB ) ≤ |NB \NA |, leaving A with nA consultants in NA . If d(xB ) > |NB \NA |, then employment efficiency implies B employs all consultants in NB \NA and so employs d(xB ) − |NB \NA | consultants from NA ∩ NB . This leaves A with |NA ∩ NB | − (d(xB ) − |NB \NA |) consultants in NA ∩ NB and with |NA \NB | consultants in NA \NB . Thus, there are

|NA \σ1B (NB , xB )| =

 n

if d(xB ) ≤ |NB \NA |

A

N − d(x ) if d(x ) > |N \N | B B B A

˜ is also employment efficient and σ consultants left for A after B moves under s. Since s ˜1B satisfies part (i) of the definition of weak rationality for B, an analogous argument gives that there are  n if d(xB ) ≤ |N˜B \N˜A | A |N˜A \˜ σ1B (N˜B , xB )| = N − d(x ) if d(x ) > |N˜ \N˜ | B

B

B

A

consultants left for A after B moves. Since (i) |NB \NA | = nB − |NA ∩ NB |, (ii) |N˜B \N˜A | =

34

nB − |N˜B ∩ N˜A |, and (iii) |NA ∩ NB | = |N˜A ∩ N˜B |, we have |NA \σ1B (NB , xB )| = |N˜A \˜ σ1B (N˜B , xB )|.

(A.3)

˜. That is, A has the same number of consultants left after B moves in both s and s Since s is weakly rational for A, |σ2A (NA , σ1B (NB , xB ), xA )| = d(xA ) when d(xA ) ≤ |NA \σ1B (NB , xB )| and |σ2A (NA , σ1B (NB , xB ), xA )| = 0 when d(xA ) > |NA \σ1B (NB , xB )|. Likewise, |˜ σ2A (N˜A , σ ˜1B (N˜B , xB ), xA )| = d(xA ) when d(xA ) ≤ |N˜A \˜ σ1B (N˜B , xB )| and |˜ σ2A (N˜A , σ ˜1B (N˜B , xB ), xA )| = 0 when d(xA ) > |N˜A \˜ σ1B (N˜B , xB )| by construction of simple strategies.21 Hence, |σ2A (NA , σ1B (NB , xB ), xA )| = |˜ σ2A (N˜A , σ ˜1B (N˜B , xB ), xA )| by equation (A.3). The desired result follows.  Corollary A1. Employment Efficient Equilibria and Simple Strategies. Let Assumption 2 hold and let s? = (NA? , . . . , NB? , . . .) ∈ E be employment efficient, then Ui (s? ) = Ui (˜ s(|NA? |, |NB? |)) for each partner i. Proof. Immediate from Lemmas A5 to A7.  The corollary, by itself, does not imply Proposition 3 because both partners strategies are ˜(|NA? |, |NB? |). Thus, while there are no profitable defections to s? , there different in s? and s ˜(|NA? |, |NB? |). As we show in the next proof, employment can be profitable defections to s lists mitigate this concern. ? ? ? ? ) ∈ E LE , let , σ2B , NB? , σ1B , σ2A Proof of Proposition 3. Let s? = (s?A , s?B ) = (NA? , σ1A ˜ denote s ˜(n?A , n?B ). We prove the proposition by showing n?A = |NA? | and n?B = |NB? |. Let s ˜(n?A , n?B ) ∈ E S . that (n?A , N − n?B ) ∈ F as then Lemma 2 implies s We argue that (n?A , N − n?B ) ∈ F by contradiction. Suppose that (n?A , N − n?B ) ∈ / F, then at least one partner, say A, does strictly better in the auxiliary game by picking a new network size n0 , with n0 ∈ {0, . . . , N } and n0 6= n?A , i.e.,

UA (˜ s(n0 , n?B ))) > UA (˜ s(n?A , n?B )).

(A.4)

The argument is analogous if B is the partner who does better. Construct a network NA0 of size n0 for A as follows. If n0 ≤ |C\NB? |, let NA0 consist of n0 elements of C\NB? . If n0 > |C\NB? |, then NB? 6= 0 as n0 ≤ N and so B has an ˜A \˜ ˜B , xB )|, then the the definition of σ Let us establish these facts. Suppose that d(xA ) > |N σ1B (N ˜2A gives ˜ ˜ ˜A \˜ ˜B , xB )|, then the definition of that |˜ σ2A (NA , σ ˜1B (NB , xB ), xA )| = 0. Now suppose that d(xA ) ≤ |N σ1B (N ˜A , σ ˜B , xB ), xA )| = d(xA ) if σ ˜B , xB ) ⊂ {d(xA )+1, . . . , N }. We next establish σ ˜2A gives that |˜ σ2A (N ˜1B (N ˜1B (N ˜ ˜B , xB ) = {N + 1 − d(xB ), . . . , N } and N ˜A = that σ ˜1B (NB , xB ) ⊂ {d(xA ) + 1, . . . , N }. Recall that σ ˜1B (N ˜ ˜ ˜ ˜ {1, . . . , nA }. Thus, NA \˜ σ1B (NB , xB ) = {1, . . . , min{nA , N − d(xB )}}. Hence, d(xA ) ≤ |NA \˜ σ1B (NB , xB )| ˜B , xB ) ⊂ implies that d(xA ) ≤ N − d(xB ). It follows that N + 1 − d(xB ) ≥ d(xA ) + 1 and so σ ˜1B (N {d(xA ) + 1, . . . , N }. 21

35

employment list {j1 , . . . , jn?B } for s? . Let NA0 consist of C\NB? and the last n0 − |C\NB? | elements of B’s employment list. Thus, NA0 = C\NB? ∪ {jn?B , jn?B −1 , . . . , jψ+1 , jψ }, where 0 0 ψ = n?B + 1 − (n0 − |C\NB? |). Also, construct weakly rational strategies σ1A and σ2A for 0 A when her network is NA0 and B follows s? . For all x ∈ X, let σ1A (NA0 , x) select d(x) 0 consultants from NA0 when d(x) ≤ n0 and let σ1A (NA0 , x) be empty when d(x) > n0 . For all ? 0 0 other (N , x) ∈ P(C) × X, let σ1A (N , x) = ∅. For all (T , x) ∈ EBs × X, let σ2A (NA0 , T , x) select d(x) consultants from NA0 \T when d(x) ≤ |NA0 \T | and select no consultants when ? 0 d(x) > |NA0 \T |. For all other (N , T , x) ∈ EBs × X, let σ2A (N , T , x) = ∅. 0 0 0 0 0 Let sA = (NA , σ1A , σ2A ). By construction, sA ∈ S A and that (s0A , s?B ) is weakly rational for A. We will prove that UA (s0A , s?B ) = UA (˜ s(n0 , n?B )). Given this, we have UA (s?A , s?B ) ≥ UA (s0A , s?B ) = UA (˜ s(n0 , n?B )), where the weak inequality follows from the fact (s?A , s?B ) is an equilibrium. Corollary A1 gives that UA (s?A , s?B ) = UA (˜ s(n?A , n?B )), since s? is employs(n0 , n?B )), a contradiction of our initial ment efficient. Thus, we have UA (˜ s(n?A , n?B )) ≥ UA (˜ supposition (A.4). It remains to establish that UA (s0A , s?B ) = UA (˜ s(n0 , n?B )). This follows from Lemma A7, we just need to verify that the antecedents hold. To these ends, recall that (s0A , s?B ) is weakly ? satisfies rational for A by construction. Since s? is an equilibrium, Lemma A5 gives that σ1B part (i) of the definition of weak rationality for B. Additionally, (s0A , s?B ) has the covering property: if NA0 ∩ NB? 6= ∅, then C\NB? ⊂ NA0 by construction and so C is in NA0 ∪ NB? . Finally, (s0A , s?B ) is employment efficient for B. If NA0 ∩ NB? = ∅, this is trivially the case. If NA0 ∩ NB? 6= ∅, observe that NB? \NA0 = {j1 , . . . , jψ−1 } since we constructed NA0 to contain the last n0 − |C\NB? | elements of B’s employment list. Since B sticks to her employment list ? when she gets a project x, we have σ1B (NB? , x) = {j1 , . . . , jd(x) } when B employs consultants. ? ⊂ NB? \NA0 . If, however, d(x) > ψ − 1, then Thus, if d(x) ≤ |NB? \NA0 | = ψ − 1, we have σ1B ? . Thus, Lemma A7 applies.  NB? \NA0 ⊂ σ1B Proof of Lemma 4. Follows immediately from Lemma A1 since the lemma gives UA (˜ s(nA , nB )) is weakly decreasing in nB for all nA .  The next lemma is useful in the Proof of Proposition 4. Recall that the set of equilibria of the auxiliary game F is a complete lattice and so has unique minimal and maximal elements. s(zA? , N − zB? )) ≥ Ui (˜ s(zA , N − zB )) for all (zA , zB ) ∈ F } Let W F (i) = {(zA? , zB? ) ∈ F |Ui (˜ denote the set of equilibria in the auxiliary game where partner i does best and let W F (i) = {(zA? , zB? ) ∈ F |Ui (˜ s(zA? , N − zB? )) ≤ Ui (˜ s(zA , N − zB )) for all (zA , zB ) ∈ F } denote the set of all equilibria in the auxiliary game where i does worst. Lemma A8. Equilibria with Opposing Interests in the Auxiliary Game. Let (z A , z B ) be the maximal element of F and let (z A , z B ) be its minimal element. We have 36

(i) (z A , z B ) ∈ W F (A) ∩ W F (B) and (ii) (z A , z B ) ∈ W F (B) ∩ W F (A). Also, if Assumption 3 holds, then we have (iii) W F (A) = W F (B) = {(z A , z B )} and (iv) W F (B) = W F (A) = {(z A , z B )}. Proof. Parts (i) and (ii) are obvious, so we only prove (iii); the argument for (iv) is analogous. We establish part (iii) by showing that (z A , z B ) is the unique element of W F (A) and the unique element of W F (B). s(bA (N − z ?B ), N − To begin, let (zA? , zB? ) ∈ W F (A). Then, for every (zA , zB ) ∈ F , UA (˜ z ?B ) = UA (˜ s(zA? , N − z ?B ) ≥ UA (˜ s(zA , N − z B )) = UA (˜ s(bA (N − zB ), N − zB ) , where the equalities are due to equilibrium and the inequality is due to the fact (zA? , zB? ) ∈ W F (A). Since UA (˜ s(bA (n), n) is strictly decreasing by Assumption 3, this implies zB ≤ zB? for all zB played in some equilibrium. Thus, zB? = z B . Since A’s best response is single valued by Assumption 3, we have zA? = bA (N − zB? ) = bA (N − z B ) = z A . Hence, (zA? , zB? ) = (z A , z B ), implying that W F (A) = {(z A , z B )}. Next, consider W F (B). Let (zA? , zB? ) ∈ W F (B). Then, for every (zA , zB ) ∈ F , we s(zA , N − z B )) = UB (˜ s(zA , bB (zA )). Since s(zA? , N − z ?B ) ≤ UB (˜ have UB (˜ s(zA? , bB (zA? )) = UB (˜ ? ? UA (˜ s(n, bB (n)) is strictly decreasing, we have zA = z A . Thus, zB = z B since B’s best response is unique, implying (zA? , zB? ) = (z A , z B ). Hence, W F (B) = {(z A , z B )}.  Proof of Proposition 4. We begin by showing that W (A) ∩ W (B) is non-empty. Subsequently, we show that W (A) = W (B) when Assumption 3 holds. The arguments that (i) W (B) ∩ W (A) 6= ∅ and (ii) that W (B) = W (A) when Assumption 3 holds are analogous. We first establish that W (A) ∩ W (B) is non-empty. Consider (z A , z B ). Lemma A8 and the fact that s0 = (NA0 , . . . , NB0 , . . .) ∈ E S if and only if (|NA0 |, N − |NB0 |) ∈ F (per the s(z A , N −z B )) ≤ construction of simple equilibria) imply UA (˜ s(z A , N −z B )) ≥ UA (s0 ) and UB (˜ 0 0 ˜(z A , N −z B ) ∈ W (A)∩W (B). (To UB (s ) for all s ∈ E S . Thus, Proposition 3 implies that s ? see this, let s ∈ E LE , then Proposition 3 gives that UA (s? ) = UA (s0 ) and UB (s? ) = UB (s0 ) for an s0 ∈ E S . Thus, UA (s? ) ≤ UA (˜ s(z A , N − z B )) and UB (s? ) ≥ UB (˜ s(z A , N − z B )). Since ˜(z A , N − z B ) ∈ W (A) and s ˜(z A , N − z B ) ∈ this holds for all elements of E LE , we have s W (B).) Next, we establish that W (A) = W (B) when Assumption 3 holds. First, we show that W (A) ⊂ W (B). Let s? = (NA? , . . . , NB? , . . .) ∈ W (A), then (|NA? |, N − |NB? |) ∈ W F (A).22 Thus, Lemma A8 gives that (|NA? |, N − |NB? |) ∈ W F (B). The construction of simple equilibria then implies that UB (˜ s(|NA? |, |NB? |)) ≤ UB (s0 ) for all s0 ∈ E S . Thus, 22

To see this, suppose that s = (NA , . . . , NB , . . .) ∈ W (A) and that (|NA |, N − |NB |) 6∈ W F (A). Then, UA (s? ) = UA (˜ s(|NA |, |NB |)) < UA (˜ s(z A , N − z B )) by Lemma A8 and Proposition 3, where (z A , z B ) is the ˜(z A , N − z B ) ∈ E S ⊂ E LE by Lemmas 2 and 3, UA (˜ maximal element of F . Since s s(z A , N − z B )) ≤ UA (s). Thus, UA (s) < UA (s), an impossibility.

37

Proposition 3 gives that s? ∈ W (B) since every other ELEE equilibrium maps to a simple equilibrium with a (weakly) higher payoff. Next, we establish that W (B) ⊂ W (A). Let s? = (NA? , . . . , NB? , . . .) ∈ W (B), then (|NA? |, N − |NB? |) ∈ W F (B). Lemma A8 then implies s(|NA? |, |NB? |)) ≥ UA (s0 ) for all s0 ∈ E S . that (|NA? |, N − |NB? |) ∈ W F (A). Then, UA (˜ Proposition 3 then gives that s? ∈ W (A). It follows that that W (A) = W (B).  We need three lemmas to prove Proposition 5. The first lemma shows that A’s best response in the auxiliary game increases with the shift in her reward and cost functions. The second lemma shows that this increase causes the minimal and maximal equilibria of the auxiliary game to increase. The third lemma shows that these increases increase A’s welfare and decrease B’s welfare. For each z ∈ {0, . . . , N }, let φA (z) = arg maxz0 ∈{0,...,N } UA (˜ s(z 0 , N − z)) and φB (z) = arg maxz0 ∈{0,...,N } UA (˜ s(z, N − z 0 )) denote A and B’s best responses in the auxiliary game.23 We occasionally write φA (z, rA , cA ) and φB (z, rB , cB ) when we wish to emphasize the dependence of these best replies on the partners’ reward and cost functions. Lemma A9. Shifts in Best Responses. 0 Let Assumption 4 hold, then φA (z, rA , cA ) is less than φA (z, rA , c0A ) in the strong set order for each z ∈ {0, . . . , N }. Proof. This follows from an application of Topkis’ [18] Monotonicity Theorem. Let θ ∈ {0, 1} and let  U (˜ A s(zA , N − zB ), rA , cA ) if θ = 0 f (zA , zB , θ) = 0 0 U (˜ A s(zA , N − zB ), rA , cA ) if θ = 1. We will show that f (zA , zB , θ) is supermodular in (zA , θ) for each zB ∈ {0, . . . , N }. Given this, Theorem 2.8.1 in Topkis [18] implies that ρ(zB , θ) = arg maxzA ∈{0,...,N } f (zA , zB , θ) is 0 weakly increasing in θ in the strong set order. Since φA (z, rA , cA ) = ρ(z, 0) and φA (z, rA , c0A ) = 0 ρ(z, 1), it follows that φA (z, rA , cA )  φA (z, rA , c0A ) for all z ∈ {0, . . . N }. To show that f (zA , zB , θ) is supermodular in (zA , θ), let θ0 = 1 and θ = 0. By Lemma A1, f (zA , zB , θ0 ) − f (zA , zB , θ) =

+

X

0 (rA (x) − rA (x))

{x|d(x)≤zA } zA X X

PA (x) 2

0 (rA (x) − rA (x))gA (l, N − zB )

l=1 {x|d(x)=l}

PA (x) + (cA (zA ) − c0A (zA )). 2

0 Since rA ≥ rA and cA ≥ c0A (and gA ≥ 0 and PA ≥ 0), this sum is positive. Additionally, the sum is increasing in zA – the first two terms are trivially increasing in zA and the last term 23

Observe that φA (z) = bA (N − z) and that φB (z) = {z 0 ∈ {0, . . . , N }|N − z 0 ∈ bB (z)}.

38

is increasing in zA by Assumption 4. Thus, for zA0 ≥ zA , f (zA0 , zB , θ0 ) − f (zA0 , zB , θ) ≥ f (zA , zB , θ0 ) − f (zA , zB , θ) f (zA0 , zB , θ0 ) + f (zA , zB , θ) ≥ f (zA , zB , θ0 ) + f (zA0 , zB , θ), that is, f is supermodular in (zA , θ).  Lemma A10. Comparisons of Extremal Elements. Let Assumption 4 hold, let F denote the pre-shift equilibrium set of the auxiliary game before the parameter shift, and let F 0 denote post-shift equilibrium set. Let (z A , z B ) and (z A , z B ) be the maximal and minimal elements of F and let (z 0A , z 0B ) and (z 0A , z 0B ) be the maximal and minimal elements of F 0 . Then (z A , z B ) ≤ (z 0A , z 0B ) and (z A , z B ) ≤ (z 0A , z 0B ). Proof. The argument is standard (e.g., Topkis’ [18]) and omitted.  Lemma A11. Comparative Statics of the Auxiliary Game. Let Assumption 4 hold. For all (zA? , zB? ) ∈ W F (A, rA , cA ) ∩ W F (B, rB , cB ) and all (zA0 , zB0 ) ∈ 0 , c0A ) ∩ W F (B, rB , cB ), we have W F (A, rA 0 UA (˜ s(zA0 , N − zB0 ), rA , c0A ) ≥ UA (˜ s(zA? , N − zB? ), rA , cA ) and

(A.5)

UB (˜ s(zA0 , N − zB0 ), rB , cB ) ≤ UB (˜ s(zA? , N − zB? ), rB , cB ).

(A.6)

If Assumption 3 also holds both before and after the shifts in rewards and costs, then (zA? , zB? ) ≤ (zA0 , zB0 ). Analogous results obtain for equilibria were B does best and A does worst. Proof. This result is obvious; we give the proof for completeness. We begin by proving that equations (A.5) and (A.5) hold. The result for network sizes is an immediate consequence of Lemmas A8 and A10. The argument for ELEE equilibria where B does best and A does worse is analogous. Let (z A , z B ) and (z 0A , z 0B ) be as in the statement of Lemma A10. To simplify notation, let Ui (nA , nB ) denote Ui (˜ s(nA , N − nB ), ri , ci ) for each partner i and let UA0 (nA , nB ) 0 denote UA (˜ s(nA , N − nB ), rA , c0A ). Lemma A8 implies that UA (zA? , zB? ) = UA (z A , z B ), that UB (zA? , zB? ) = UB (z A , z B ), that UA0 (zA0 , zB0 ) = UA0 (z 0A , z 0B ), that UB (zA0 , zB0 ) = UB (z 0A , z 0B ). Thus, we only need to show that UA0 (z 0A z 0B ) ≥ UA (z A , z B ) and UB (z 0A z 0B ) ≤ UB (z A , z B ) to 0 establish equations (A.5) and (A.6). To these ends, let bA (n) denote bA (n, rA , cA ), let bA (n) 0 denote bA (n, rA , c0A ), and write UA (z A , z B ) = UA (˜ s(bA (N − z B ), N − z B ), rA , cA ) ≤ UA (˜ s(bA (N − z 0B ), N − z 0B ), rA , cA ) 0

0 0 ≤ UA (˜ s(bA (N − z 0B ), N − z 0B ), rA , c0A ) ≤ UA (˜ s(bA (N − z 0B ), N − z 0B ), rA , c0A ) = UA0 (z 0A , z 0B ).

39

The first and second equalities are standard, the first inequality follows from Lemma 4 since (z A , z B ) ≤ (z 0A , z 0B ), the second inequality follows from the fact A’s costs fall and her rewards increase, and the third inequality is due to optimality. An analogous argument gives gives UB (z 0A z 0B ) ≤ UB (z A , z B ).  Proof of Proposition 5. We only establish the result for ELEE equilibria where A does best and B does worst. The argument for ELEE equilibria where B does best and A does worst is analogous. Let s? = (NA? , . . . , NB? , . . .) ∈ W (A, rA , cA )∩W (B, rB , cB ) and let s0 = (NA0 , . . . , NB0 , . . .) ∈ 0 W (A, rA , c0A ) ∩ W (B, rB , cB ). Since (|NA? |, N − |NB? |) ∈ W F (A, rA , cA ) ∩ W F (B, rB , cB ) 0 and since (|NA0 |, N − |NB0 |) ∈ W F (A, rA , c0A ) ∩ W F (B, rB , cB ), Lemma A11 gives that 0 UA (˜ s(|NA0 |, N − |NB0 |), rA , c0A ) ≥ UA (˜ s(|NA? |, N − |NB? |), rA , cA ) and that UB (˜ s(|NA0 |, N − s(|NA? |, N − |NB? |)) and s(|NA? |, N − |NB? |), rB , cB ). Since Ui (s? ) = Ui (˜ |NB0 |), rB , cB ) ≤ UB (˜ 0 Ui (s0 ) = Ui (˜ s(|NA0 |, N − |NB0 |)) for each partner i by Proposition 3, we have UA (s0 , rA , c0A ) ≥ UA (s? , rA , cA ) and UB (s0 , rB , cB ) ≤ UB (s? , rB , cB ). If Assumption 3 also holds, then we have (|NA? |, N − |NB? |) ≤ (|NA0 |, N − |NB0 |) by Lemma A11.  The next lemma assists with the Proof of Proposition 6. Lemma A12. Symmetric Fundamentals and Equilibrium Structure. Let Assumptions 2 and 5 hold, then, UA (s? , rB , rB ) ≥ UB (s? , rB , rB ) for all s? = (NA? , . . . , NB? , . . .) ∈ W (A, rB , cB ) ∩ W (B, rB , cB ). If Assumption 3 also holds, then |NA? | ≥ |NB? |. Proof. We first establish that A does better than B. Subsequently, we establish that A’s network is weakly larger than B’s network. Let F denote the equilibrium set of the auxiliary game when A has B’s reward and cost function and let (z A , z B ) be it’s maximal element. We need a preliminary fact: N − z B ≤ z A . To see this, recall that φA (z) = bA (N − z) and that φB (z) = {z 0 ∈ {0, . . . , N }|N − z 0 ∈ bB (z)}. Since z A ∈ φA (z B ) and z B ∈ φB (z A ), we have z A ∈ bA (N − z B ) and N − z B ∈ bB (z A ). Since bA = bB , it follows that N − z B ∈ bA (z A ) and z A ∈ bB (N − z B ). Hence, N − z B ∈ φA (N − z A ) and that N − z A ∈ φB (N − z B ). So (N − z B , N − z A ) ∈ F , implying (N − z B , N − z A ) ≤ (z A , z B ), which gives N − z B ≤ z A . We now establish that A does better than B. Since both partners’ have the same costs and rewards, the game is symmetric, i.e., UA (˜ s(nA , nB ), rB , cB ) = UB (˜ s(nB , nA ), rB , cB ), so they have the same best response, i.e., bA (n) = bB (n). Consequently, UA (˜ s(z A , N − z B ), rB , cB ) = UA (˜ s(bA (N − z B ), N − z B ), rB , cB ) ≥ UA (˜ s(bA (z A ), z A ), rB , cB ) = UB (˜ s(z A , bB (z A )), rB , cB ) = UB (˜ s(z A , N − z B ), rB , cB ). The first equality is standard, the first inequality is due to the fact N − z B ≤ z A , second 40

˜(z A , N − z B ) ∈ equality is due to symmetry, and the last equality is standard. Since s W (A, rB , cB ) ∩ W (B, rB , cB ) by Proposition 3 (the argument is analogous to the Proof of Proposition 4), we have the desired result. It remains to show that |NA? | ≥ |NB? |. Since (|NA? |, N − |NB? |) ∈ W F (A, rB , cB ) ∩ W F (B, rB , cB ), Lemma A8 gives (|NA? |, N − |NB? |) = (z A , z B ). The desired result follows from the fact N − z B ≤ z A .  Proof of Proposition 6. Since rA ≥ rB , cA ≤ cB , and cB − cA is weakly increasing, Proposition 5 gives that UA (s0 , rA , cA ) ≥ UA (s? , rB , cB ) and UB (s0 , rB , cB ) ≤ UB (s? , rB , cB ) for all s? ∈ W (A, rB , cB ) ∩ W (B, rB , cB ) and s0 ∈ W (A, rA , cA ) ∩ W (B, rB , cB ). Thus, the first part of Proposition 6 follows directly from Lemma A12. As to the second part, observe that Assumption 3 continues to hold when A has B’s reward and cost functions.24 Thus, Proposition 5 implies that |NA0 | ≥ |NA? | and |NB0 | ≤ |NB? | for all s? = (NA? , . . . , NB? , . . .) ∈ W (A, rB , cB ) ∩ W (B, rB , cB ) and s0 = (NA0 , . . . , NB0 , . . .) ∈ W (A, rA , cA ) ∩ W (B, rB , cB ), so Proposition 6 follows from Lemma A12. 

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To see this, recall that UA (˜ s(nA , nB ), rB , cB ) = UB (˜ s(nB , nA ), rB , cB ). Thus, bA (n, rB , cB ) = bB (n, rB , cB ) and UA (˜ s(bA (n, rB , cB ), n), rB , cB ) = UB (˜ s(n, bB (n, rB , cB )), rB , cB ). Since bB is single-valued and since UB (˜ s(n, bB (n, rB , cB ), n), rB , cB ) is strictly decreasing in n by the hypothesis, we have bA is single valued and that UA (˜ s(bA (n, rB , cB ), n), rB , cB ) is strictly decreasing in n, i.e., Assumption 3 continues to hold.

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