Assortative Matching in Two-sided Continuum Economies

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1 Assortative Matching in Two-sided Continuum Economies Patrick Legros and Andrew Newman February 2006 (revised March 2007) Abstract We consider two-sided markets with a continuum of agents and a finite number or continuum of types. We shown that a single crossing condition (GID) implies that any economy satisfies positive assortative matching. 1 Introduction On most markets, the utility from transacting depends on the characteristics of the matched trading partners: education, wealth or beauty in household formation, education or productivity in firms. For this reason market equilibria need to specify not only the payoffs but also the way agents match. When agents are characterized by a one dimensional type, Becker (1973) observed that if the product from the match is perfectly transferable between the partners and if the marginal product of a type is increasing in the type of the partner (there are complementarities in types), then necessarily the matching will be assortative: higher types match with higher types. This is a very useful result since it implies that the search for an equilibrium can be limited to the search for payoffs that make the assortative match stable. Moreover, the search for the equilibrium payoff is facilitated when there is a continuum of agents on each side, when the type assignment is continuous and when the production function is differentiable. For instance consider marriage market with a continuum [0, 1] of men and a continuum [0, 1] of women. Assume that the characteristic of agent i is i; if the man i forms a household with the woman j, the production is y(i, j) = ij. Since y 12 = 1 > 0, the unique matching pattern is such that the match of the man i is the woman i. Existence of an equilibrium implies that there exists a payoff structure (u, v) where u(i) is the payoff to man i and v(i) is the payoff to woman i such that assortative matching is stable. It is therefore enough to verify that given that a woman j gets v(j), the payoff maximizing match of man i is the woman i, that is that i arg max ij v(j), and similarly that the best match for the woman i is the man i, or i arg max ij u(j). Because stability implies that u and v are increasing, they are differentiable almost everywhere 1

2 and therefore for almost all i, we have v (i) = u (i) = i : the marginal payoffs are equal to the marginal productivities. It follows that by choosing an arbitrary feasible payoff u(0) [0, 1] for the man 0, we have for each i, u(i) = u(0) + i = u(0) + i and the women have payoff 1 xdx (1) v(i) = i u(0). Note that these computations are facilitated because total production is independent of how it is allocated. This is no longer the case when there are nontransferabilities, that is when the frontier is not linear with slope 1. In Legros and Newman (2006), we consider situations where the frontier of the feasible set of payoffs that two agents can achieve is a strictly decreasing frontier on the positive orthan. This is typically the case when risk averse agents share risk in partnerships, whether or not incentive problems are present. In an economy with finitely many agents we provide a condition (Generalized Increasing Differences or GID) that insures that all equilibria of the economy are payoff equivalent to an equilibrium with assortative matching. The purpose of this note is to show that this result extends to the continuum economy. The extension is non trivial because the argument for the proof is different in the continuum and in the finite case. This extension is useful because, as for the transferable case, the derivation of equilibrium payoffs is facilitated when it is known that matching is assortative. We show that the traditional result that agents are paid at their marginal productivity continues to hold, which facilitates the search for equilibrium payoffs. However, while in the case of transferability the marginal productivity is well defined, it is in the general case a nontrivial function of the payoff of the partner. Hence while each partner gets paid at the margin at their marginal productivity, one cannot use directly a construction like in (1) but rather we need to solve a system of differential equations to find the equilibrium payoffs. We illustrate this in section 4. 2 Preliminaries We consider a two-sided market with a continuum of agents on each side of the market. On one side of the market, we have principals indexed by i [0, 1] ; on the other side of the market, we have agents indexed by j [0, 1], where [0, 1] is equipped with Lebesgue measure λ. Principals have types in a compact interval P and agents in a compact interval A and a type assignment specifies the type p i P of the i-th principal, 2

3 and specifies a type a j of the j-th agent. We assume that types are taken from a compact interval and that the orders on the indexes are chosen in such a way that p i and a j are non-decreasing in i [0, 1]. We will abuse notation and call alternatively p i the principal i and his type, and a j the agent j and his type. We will also write p i < p j and a i < a j when i < j. Since p i and a j are bounded and monotone, their set of discontinuity points has measure zero. In most applications, p i and a j are either continuous or are simple functions. Feasible payoffs are described by a frontier φ (p, a, v) specifying the maximum payoff to type p principal matched to a type-a agent who requires a payoff of v. The maximum payoff to this principal is then φ (p, a, 0). The inverse function ψ (a, p, u) satisfies φ (p, a, ψ (a, p, u)) = u when u φ (p, a, 0). We assume that φ is continuous, strictly decreasing in v on [0, ψ (a, p, 0)], and that φ (p, a, 0) > 0 for all (p, a) P A. A pair (u, v) R 2 + is feasible for (i, j) [0, 1] 2 when u φ (p i, a j, v) and v ψ (a j, p i, 0). Definition 1 Payoffs (u, v) are on the frontier for (i, j) when u = φ (p i, a j, v) and v ψ (a j, p i, 0). We write in this case (u, v) Φ (i, j). A match is a 1-1 measurable map m : [0, 1] [0, 1] satisfying measure consistency: for every open interval K I, λ (K) = λ (m (K)). 1 Definition 2 Let I [0, 1] be a set of full measure. An I-equilibrium is a triple (m, π, ω), where m : I I is the match function, π : I R and ω : I R are measurable payoff functions for principals and agents, satisfying (i) For all i I, ( π i, ω m(i) ) Φ ( pi, a m(i) ) (ii) For all (i, j) I 2, π i φ (p i, a j, ω j ). (i) is the feasibility condition, (ii) is the stability condition. 2. Existence of an I-equilibrium for some set I [0, 1] of full measure is implied by Kaneko- Wooders (1996). Note however that with GID one can prove existence by construction (as in section 4). 3 Definition 3 An I-equilibrium (m, π, ω) satisfies positive assortative matching (PAM) if for all i I,, i I, i < i implies that m (i ) m (i). Definition 4 An economy satisfies PAM if whenever (m, π, ω) is an I-equilibrium for a full measure set I [0, 1], there exists a full measure set I I and an I -equilibrium (m, π, ω) satisfying PAM. 1 This avoids matches of the type m (i) = i 2 since for i = 1/2, λ (i) = 1/2 > λ (m (i)) = 1/4 which means that half the men are allocated to a fourth of the women. 2 Feasibility would require only that π (p i ) φ (p i, a j, ω (a j )) and ω (a j ) ψ (a j, p i, 0). However if π (p i ) < φ (p i, a j, ω (a j )), there would exist ε > 0 such that π (p i ) + ε φ (p i, a j, ω (a j ) + ε), contradicting the stability condition. Hence there is no loss of generality in assuming that feasible payoffs for equilibrium matches are on the the frontier. 3 They establish nonemptiness of the f-core, where this allows for the possible existence of a null set of individuals who receive infeasible payoffs. The restriction of this equilibrium to the set of individuals with feasible payoffs is an equilibrium in our sense. 3

4 Note that any a.e. identity map, m(i) = i almost everywhere, satisfies PAM. In general, the exceptional set will be finite, and for continuous type assignments, the PAM condition can hold everywhere. When there are finitely many agents, Legros and Newman (2002) show that the following condition is necessary and sufficient for an economy to satisfy PAM. Definition 5 φ satisfies GID if for all i < i, j < j, u φ (p i, a j, 0), φ (p i, a j, ψ (a j, p i, u)) φ (p i, a j, ψ (a j, p i, u)). 3 GID implies PAM To show that the economy satisfies PAM, it is necessary to show that all equilibria can be made payoff equivalent to an equilibrium satisfying PAM almost everywhere. This is stated below. Proposition 6 If φ (p, a, v) satisfies GID, the economy satisfies PAM. When the GID condition is satisfied strictly, Proposition 6 is immediate: if there exist i > i with i, i I such that m (p i ) < m(p i ). Then, φ ( p i, a m(i ), ω (m(i )) ) = φ ( p i, a m(i ), ψ ( )) a m(i ), p i, π i > φ ( p i, a m(i), ψ ( )) a m(i), p i, π i φ ( p i, a m(i), ω m(i) ), the strict inequality is GID, the weak inequality follows payoff monotonicity and ω m(pi) ψ ( a m(i), p i, π i ). But then, there exists v > ωm(i ) such that φ ( p i, a m(i ), v ) > π (i), contradicting stability. When the GID condition is satisfied weakly, the proof of Proposition 6 is provided below. We first invoke a key lemma: if i > i, j > j,and (p i, a j ), (p i, a j ) are such that equilibrium payoffs are on the frontier of these matches, then the equilibrium payoffs are also on the frontiers of (p i, a j ), (p i, a j ). Lemma 7 Suppose that GID holds. Let (m, π, ω) be an equilibrium. If i > i and j > j, are such that (π i, ω j ) Φ (i, j ) and (π i, ω j ) Φ (i, j), then (π i, ω j ) Φ (i, j) and (π i, ω j ) Φ (i, j ). Proof. Legros and Newman (2006) prove this lemma when j = m (i ) and j = m (i) ; their argument generalizes straightforwardly to j, j as in the Lemma. We next show that at continuity points i of the type assignments, it is always feasible for pairs (p i, a i ) to match and generate their equilibrium payoffs. Lemma 8 Suppose GID and consider an equilibrium (m, π, ω). If a or p is continuous at i, then (π i, ω i ) Φ (i, i). 4

5 Proof. Consider an I-equilibrium and suppose that there exists i I, i / {0, 1}, such that m (i) < i (the case m (i) > i is similar and is omitted). Since (π i, ω i ) / Φ (i, i), we must have either ω i < ψ (a i, p i, π i ) or ω i > ψ (a i, p i, π i ). By stability, the first inequality is not possible since it would imply that there is a beneficial deviation by (i, i). Therefore for all i < i ω i > ψ (a i, p i, π i ) (2) = ψ ( a i, p i, φ ( )) p i, a m(i), ω m(i) ψ ( a i, p i, φ ( )) p i, a m(i), ω m(i) (3) ψ (a i, p i, π i ). The equality is by feasibility of payoffs in equilibrium, the next inequality is by GID applied to i i, the last inequality is by payoff monotonicity and the fact that by stability, π i φ ( p i, a m(i), ω m(i) ). Hence, for all i i, ω i > ψ (a i, p i, π i ). (4) This implies that the match of a i is p k, where k > i. Suppose that there exists i < i such that m (i ) > i, then (p k, a i ) and ( p i, a m(i )) are in a NAM pattern and by Lemma 7, we must have (π i, ω i ) Φ (i, i) which contradicts (4). Hence, for all i < i, m (i ) < i. But then, by measure consistency, principals in the interval (p 0, p i ) have their match in the interval (a 0, a i ). Let m (i) = j, where a j < a i by assumption. If a i is a continuity point of the agent type assignment, choose a sequence T = {j n } (j, i) converging to i; by continuity, a jn a i. Lemma 7 implies that for all j n, π i = φ (p i, a jn, ω jn ), for all j n T. (5) Since ψ is continuous, ω jn = ψ(a jn, p i, π i ) ψ(a i, p i, π i ) ˆω < ω i, where the inequality follows (4). Remember that the match of a i is p k with k > i. Then, π k = φ (p k, a i, ω i ) < φ(p k, a i, ˆω) where the strict inequality follows (2) and (π k, ω i ) Φ (k, i). Since a jn a i and ω jn ˆω there exists j n T such that π k < φ (p k, a jn, ω jn ),implying that (p k, a jn ) can profitably deviate. This contradicts stability of the (p k, a i ) match, and we therefore must have (p i, a i ) Φ (i, i), as claimed. The case in which i is a point of continuity of p proceeds similarly. By measure consistency, for almost all p l (p i, p k ), m (l) > i. Choose a sequence S = {i n } (i, k) converging to i. Then p in p i, and since (p in,a m(in)) and (p k, a i ) are in a NAM pattern, (π in,ω i ) Φ(i n, i) so that π in = φ(p in, a i, ω i ) φ (p i, a i, ω i ) ˆπ < π i, where the inequality is by 2. Then for j = m(i), ω j = ψ (a j, p i, π i ) < ψ (a j, p i, ˆπ), and thus for n sufficiently large, ω j < ψ (a j, p in, π in ), contradicting stability of (p i, a j ). Thus, (p i, a i ) Φ (i, i) if i is a continuity point of p. 5

6 Suppose that GID holds and that (m, π, ω) is an I-equilibrium. Let E p be the set of points of discontinuity of the type assignment p and E a be the set of points of discontinuity of the type assignment a. Let E = E p E a. and I = I\E; then I is of full measure. By Lemma 8 for all i I, (π i, ω i ) Φ (i, i). Defining m (i) = i, i I, we have that (m, π, ω) is an I -equilibrium. 4 Computing Equilibrium Payoffs Since under GID, all equilibria are essentially equivalent to one with PAM, we can restrict attention to matches where the man i is matched with the woman i. This can be part of an equilibrium if there exist payoff maps u and v such that the following conditions are satisfied almost everywhere: u(i) = φ (p i, a i, v(i)) (6) u(i) = max φ (p i, a j, v(j)) (7) j v(i) = max ψ (a i, p j, u (j)) (8) j (6) is feasibility: in a match payoffs must be on the frontier; (7) and (8) are the incentive compatibility conditions: a man i must prefer to match with woman i given that women ask payoffs v. Note the similarity between the incentive conditions and those obtained when we consider screening problems. Like in this literature, the search for equilibrium payoffs is facilitated by looking at local conditions, something that is acceptable as long as a single crossing condition is satisfied: if the local incentive compatibility condition holds, then single crossing implies that larger types have even less incentives to deviate downward, insuring global stability. The GID condition is a single crossing condition. It can be equivalently expressed as: 4 Definition 9 Let i < i, j < j, and suppose that v and v are such that φ (p i, a j, v ) = φ (p i, a j, v), then, φ (p i, a j, v) φ (p i, a j, v ). However a local incentive condition in the continuum is useful if it is known that the payoff is differentiable in the type of the partner. Here since φ is differentiable, it is enough to show that the payoffs u and v must be differentiable. In screening models, when payoffs are quasi-linear, incentive compatibility indeed implies that payoffs are monotonically increasing in types. In our case, we do not have quasi-linarity of payoffs (otherwise the frontier would have slope 1!) but as long as φ (p, a, v) is increasing in p and ψ (a, p, v) is increasing in a (higher types are more productive ), incentive compatibility indeed implies that payoffs must be non-decreasing. For any j the incentive compatibility of the man i requires u (i) φ (p i, a j, v (j)) ; since φ is increasing, 4 To see this note that if v, v satisfy φ `p i, a j, v = φ (p i, a j, v), then letting u be the common value we have v = ψ `a j, p j, u and v = ψ `a j, p i, u, and substituting this into the terms in the inequality gives the GID condition in Definition 5. 6

7 when j < i, we have φ (p i, a j, v (j)) φ (p j, a j, v (j)), but the right hand side is by (6) equal to u (j) proving that payoffs are non-decreasing. A similar logic shows that v is increasing in i. Since increasing functions are differentiable almost everywhere, the global incentive compatibility conditions can be replaced almost everywhere by the first order conditions dφ (p i, a j, v (j)) dj = 0 j=i dψ (a i, p j, u (j)) dj = 0. j=i Proposition 10 Suppose that the type assignments p and a are continuously differentiable and that GID holds. Then there exist payoffs u, v that stabilize the assortative matching in which man i is matched with woman i and these payoffs solve the system u (i) = p iφ 1 = p i ψ 2 ψ 3 (9) v (i) = a iψ 1 = a i φ 2 φ 3 (10) Proof. If the type assignment maps p and a are differentiable, the first order conditions can be written a iφ 2 + v (i) φ 3 = 0 (11) p iψ 2 + u (i)ψ 3 = 0, (12) where for k = 2, 3, φ k = φ k (p i, a i, v (i)) and ψ k = ψ k (a i, p i, u (i)). Hence incentive compatibility requires solving a system of differentiable equations in u, v. However, contrary to the screening problem there is an additional feasibility constraint to take into account. Since (6) is an identity, we can differentiate with respect to i and get u (i) = p iφ 1 + a iφ 2 + v (i) φ 3. (13) Hence, (11) and (13) imply that u (i) = p iφ 1. Men i s marginal payoff is indeed equal to his marginal product when he is matched with woman i. Using the incentive compatibility condition of the woman (12) we obtain (9); (10) is obtained in a similar fashion. These equalities translate two effects present in models with non-transferability: the usual type-type transferability in transferable utility cases and a type-payoff transferability. Having type-type transferability is not enough for PAM if higher types find it also more difficult to transfer surplus to their partner. Note that the rate at which a man can transfer to a woman is given by the ratio p ψ 2 /ψ 3, 7

8 and incentive compatibility requires that this equates the marginal productivity of the man. Consider first a situation where there is perfect transferability, that is when φ(p, a, v) = y(p, a) v. Then, φ 3 = ψ 3 = 1 and φ 2 = ψ 1 and ψ 2 = φ 1 and the second equality in conditions (9) and (10) are trivial; we have u (i) = p i y 1(p i, a i ) and v (i) = a i y 2 (p i, a i ). With non-transferabilities φ 2 is not equal to ψ 1 in general, and the second equality in (9) and (10) is not redundant. As an illustration of these conditions, consider φ (p, a, v) = pa v, then ψ (a, p, u) = pa u 2. This frontier satisfies GID. 5 a Note that φ 1 = 2 pa v, ψ 2 = a, ψ 3 = 2u; since u = pa v, we have indeed p φ 1 = p ψ 2 ψ 3. The same is true in (10) and therefore we need to solve the system a i u (i) = p i 2 p i a i v(i) v (i) = a ip i u(i) 2 + v (i) = p i a i. Suppose for instance that p i = a i = i for all i [0, 1]. Then, the conditions are u i (i) = 2 i 2 v(i) v (i) = i u (i) 2 + v (i) = i 2. At i = 0, the unique feasible payoff consistent with individual rationality is u (0) = v(0) = 0. Then v (í) = i implies that v (i) = i 0 xdx = i2 2. The feasibility condition implies that u (i) = i 2. We verify that u i (i) = = 1 2 i 2 v(i) 2. We represent below a typical frontier for a match {i, i} and the equilbrium payoffs of men and women. If one is interested in total surplus maximization, the payoffs should be u(i) = i and v(i) = 0 when i < 1/2 since the slope of the frontier at v = 0 is smaller than 1. This obviously illustrates the fact that non-transferability implies that surplus division and surplus maximization are conflicting with each other. Since the payoff to men is linear while the payoff to women is quadratic in type, women catch up with men when i is large enough; moreover, while the degree of inequality is an inverted U for i less than 2/ 2, it is increasing for larger values of i : women not only catch up but leave men behind. Obviously these results are specific to the example. 5 In fact this economy admits a TU representation: use the change of variable F (u) = u 2 and consider the new frontier φ (p, a, v) = φ (p, a, v) 2. Then F (u) = pa v and there is transferability; clearly φ satisfied GID. Legros and Newman (2006) show that in this case φ also satisfies GID. 8

9 u uv, v(i) i u(i) 1 2i i 2 v 2 2 i Frontier for a match {i,i} Equilibrium payoffs Figure 1: Frontier and Equilibrium Payoffs when φ (p i, a i, v) = i 2 v. 9

10 References [1] Becker, G. (1973), A Theory of Marriage, Part I, Journal of Political Economy, 81: [2] Kaneko, M., and M. Wooders (1996), The Nonemptiness of the f -Core of a Game Without Side Payments, International Journal of Game Theory, 25: [3] Legros, P. and A.F. Newman (2006), Beauty is a Beast, Frog is a Prince: Assortative Matching with Nontransferabilities, forthcoming, Econometrica. 10