Assortative matching in two-sided continuum economies

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 finitenumber 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 characteristicsof the matched trading partners: education, wealth or beauty in householdformation, education or productivity in firms. For this reason market equilibrianeed 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 betweenthe partners and if the marginal product of a type is increasing in the type of thepartner (there are complementarities in types), then necessarily the matchingwill be assortative: higher types match with higher types.

This is a very useful result since it implies that the search for an equilibriumcan 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 acontinuum of agents on each side, when the type assignment is continuous andwhen the production function is differentiable.

For instance consider marriage market with a continuum [0, 1] of men anda continuum [0, 1] of women. Assume that the characteristic of agent i is i; ifthe man i forms a household with the woman j, the production is y(i, j) = ij.Since y12 = 1 > 0, the unique matching pattern is such that the match of theman i is the woman i. Existence of an equilibrium implies that there exists apayoff structure (u, v) where u(i) is the payoff to man i and v(i) is the payoffto woman i such that assortative matching is stable. It is therefore enough toverify that given that a woman j gets v(j), the payoff maximizing match of mani is the woman i, that is that i ∈ arg max ij − v(j), and similarly that the bestmatch for the woman i is the man i, or i ∈ arg max ij − u(j). Because stabilityimplies that u and v are increasing, they are differentiable almost everywhere

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and therefore for almost all i, we have v′(i) = u′(i) = i : the marginal payoffs areequal to the marginal productivities. It follows that by choosing an arbitraryfeasible payoff u(0) ∈ [0, 1] for the man 0, we have for each i,

u(i) = u(0) +∫ i

1

xdx (1)

= u(0) +i2

2− 1

2.

and the women have payoff

v(i) =i2

2+

12− u(0).

Note that these computations are facilitated because total production is in-dependent of how it is allocated. This is no longer the case when there arenontransferabilities, that is when the frontier is not linear with slope −1. InLegros and Newman (2006), we consider situations where the frontier of thefeasible set of payoffs that two agents can achieve is a strictly decreasing fron-tier on the positive orthan. This is typically the case when risk averse agentsshare risk in partnerships, whether or not incentive problems are present. Inan economy with finitely many agents we provide a condition (Generalized In-creasing Differences or GID) that insures that all equilibria of the economy arepayoff equivalent to an equilibrium with assortative matching.

The purpose of this note is to show that this result extends to the continuumeconomy. The extension is non trivial because the argument for the proof is dif-ferent in the continuum and in the finite case. This extension is useful because,as for the transferable case, the derivation of equilibrium payoffs is facilitatedwhen it is known that matching is assortative. We show that the traditionalresult that agents are paid at their marginal productivity continues to hold,which facilitates the search for equilibrium payoffs. However, while in the caseof transferability the marginal productivity is well defined, it is in the generalcase a nontrivial function of the payoff of the partner. Hence while each partnergets paid at the margin at their marginal productivity, one cannot use directlya construction like in (1) but rather we need to solve a system of differentialequations 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 themarket. On one side of the market, we have principals indexed by i ∈ [0, 1] ; onthe 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 compactinterval A and a type assignment specifies the type pi ∈ P of the i-th principal ,

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and specifies a type aj of the j-th agent . We assume that types are taken froma compact interval and that the orders on the indexes are chosen in such a waythat pi and aj are non-decreasing in i ∈ [0, 1] . We will abuse notation and callalternatively pi the principal i and his type, and aj the agent j and his type.We will also write pi < pj and ai < aj when i < j.

Since pi and aj are bounded and monotone, their set of discontinuity pointshas measure zero. In most applications, pi and aj are either continuous or aresimple functions.

Feasible payoffs are described by a frontier φ (p, a, v) specifying the maximumpayoff 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” functionψ (a, p, u) satisfies φ (p, a, ψ (a, p, u)) = u when u ≤ φ (p, a, 0) . We assume thatφ is continuous, strictly decreasing in v on [0, ψ (a, p, 0)], and that φ (p, a, 0) > 0for all (p, a) ∈ P ×A.

A pair (u, v) ∈ R2+ is feasible for (i, j) ∈ [0, 1]2 when u ≤ φ (pi, aj , v) and

v ≤ ψ (aj , pi, 0) .

Definition 1 Payoffs (u, v) are “on the frontier” for (i, j) when u = φ (pi, aj , v)and v ≤ ψ (aj , pi, 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 measureconsistency: 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 aremeasurable payoff functions for principals and agents, satisfying(i) For all i ∈ I,

(πi, ωm(i)

)∈ Φ

(pi, am(i)

)(ii) For all (i, j) ∈ I2, πi ≥ φ (pi, aj , ωj) .

(i) is the feasibility condition, (ii) is the stability condition.2. Existence ofan 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 con-struction (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-equilibriumfor a full measure set I ⊆ [0, 1] , there exists a full measure set I ′ ⊆ I and anI ′-equilibrium (m∗, π, ω) satisfying PAM.

1This avoids matches of the type m (i) = i2 since for i = 1/2, λ (i) = 1/2 > λ (m (i)) = 1/4which means that half the men are allocated to a fourth of the women.

2Feasibility would require only that π (pi) ≤ φ (pi, aj , ω (aj)) and ω (aj) ≤ ψ (aj , pi, 0) .However if π (pi) < φ (pi, aj , ω (aj)) , there would exist ε > 0 such that π (pi) + ε ≤φ (pi, aj , ω (aj) + ε) , contradicting the stability condition. Hence there is no loss of gener-ality in assuming that feasible payoffs for equilibrium matches are on the the frontier.

3They establish nonemptiness of the f -core, where this allows for the possible existence ofa null set of individuals who receive infeasible payoffs. The restriction of this equilibrium tothe set of individuals with feasible payoffs is an equilibrium in our sense.

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Note that any “a.e. identity” map, m(i) = i almost everywhere, satisfiesPAM. In general, the exceptional set will be finite, and for continuous typeassignments, the PAM condition can hold everywhere.

When there are finitely many agents, Legros and Newman (2002) show thatthe following condition is necessary and sufficient for an economy to satisfyPAM.

Definition 5 φ satisfies GID if for all i′ < i, j′ < j, u ≤ φ (pi′ , aj , 0) ,

φ (pi, aj , ψ (aj , pi′ , u)) ≥ φ (pi, aj′ , ψ (aj′ , pi′ , u)) .

3 GID implies PAM

To show that the economy satisfies PAM, it is necessary to show that all equi-libria can be made payoff equivalent to an equilibrium satisfying PAM almosteverywhere. 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: ifthere exist i > i′ with i, i′ ∈ I such that m (pi) < m(pi′). Then,

φ(pi, am(i′), ω (m(i′))

)= φ

(pi, am(i′), ψ

(am(i′), pi′ , πi′

))> φ

(pi, am(i), ψ

(am(i), pi′ , πi′

))≥ φ

(pi, am(i), ωm(i)

),

the strict inequality is GID, the weak inequality follows payoff monotonicityand ωm(pi) ≥ ψ

(am(i), pi′ , πi′

). But then, there exists v > ωm(i′) such that

φ(pi, am(i′), v

)> π (i) , contradicting stability.

When the GID condition is satisfied weakly, the proof of Proposition 6 isprovided below.

We first invoke a key lemma: if i > i′, j > j′,and (pi, aj′) , (pi′ , aj) aresuch that equilibrium payoffs are on the frontier of these matches, then theequilibrium payoffs are also on the frontiers of (pi, aj) , (pi′ , aj′).

Lemma 7 Suppose that GID holds. Let (m,π, ω) be an equilibrium. If i > i′

and j > j′, are such that (πi, ωj′) ∈ Φ (i, j′) and (πi′ , ωj) ∈ Φ (i′, j) , then(πi, ωj) ∈ Φ (i, j) and (πi′ , ωj′) ∈ Φ (i′, j′) .

Proof. Legros and Newman (2006) prove this lemma when j = m (i′) andj′ = 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 alwaysfeasible for pairs (pi, ai) to match and generate their equilibrium payoffs.

Lemma 8 Suppose GID and consider an equilibrium (m,π, ω) . If a or p iscontinuous at i, then (πi, ωi) ∈ Φ (i, i) .

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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(πi, ωi) /∈ Φ (i, i) , we must have either ωi < ψ (ai, pi, πi) or ωi > ψ (ai, pi, πi) .By stability, the first inequality is not possible since it would imply that thereis a beneficial deviation by (i, i) . Therefore for all i′ < i

ωi > ψ (ai, pi, πi) (2)

= ψ(ai, pi, φ

(pi, am(i), ωm(i)

))(3)

≥ ψ(ai, pi′ , φ

(pi′ , am(i), ωm(i)

))≥ ψ (ai, pi′ , πi′) .

The equality is by feasibility of payoffs in equilibrium, the next inequality is byGID applied to i′ ≤ i, the last inequality is by payoff monotonicity and the factthat by stability, πi′ ≥ φ

(pi′ , am(i), ωm(i)

). Hence, for all i′ ≤ i,

ωi > ψ (ai, pi′ , πi′) . (4)

This implies that the match of ai is pk, where k > i. Suppose that there existsi′ < i such that m (i′) > i, then (pk, ai) and

(pi′ , am(i′)

)are in a NAM pattern

and by Lemma 7, we must have (πi′ , ωi) ∈ Φ (i′, i) which contradicts (4). Hence,for all i′ < i, m (i′) < i.But then, by measure consistency, principals in the interval (p0, pi) have theirmatch in the interval (a0, ai). Let m (i) = j, where aj < ai by assumption.If ai is a continuity point of the agent type assignment, choose a sequenceT = {jn} ⊂ (j, i) converging to i; by continuity, ajn → ai. Lemma 7 impliesthat for all jn,

πi = φ (pi, ajn , ωjn) , for all jn ∈ T. (5)

Since ψ is continuous, ωjn = ψ(ajn , pi, πi) → ψ(ai, pi, πi) ≡ ω̂ < ωi, where theinequality follows (4).Remember that the match of ai is pk with k > i. Then,

πk = φ (pk, ai, ωi)< φ(pk, ai, ω̂)

where the strict inequality follows (2) and (πk, ωi) ∈ Φ (k, i) . Since ajn → aiand ωjn → ω̂ there exists jn ∈ T such that πk < φ (pk, ajn , ωjn) ,implying that(pk, ajn) can profitably deviate. This contradicts stability of the (pk, ai) match,and we therefore must have (pi, ai) ∈ Φ (i, i) , as claimed.The case in which i is a point of continuity of p proceeds similarly. By measureconsistency, for almost all pl ∈ (pi, pk) , m (l) > i. Choose a sequence S = {in} ⊂(i, k) converging to i. Then pin → pi, and since (pin,am(in)) and (pk, ai) are in aNAM pattern, (πin,ωi) ∈ Φ(in, i) so that πin = φ(pin , ai, ωi) → φ (pi, ai, ωi) ≡π̂ < πi, where the inequality is by 2. Then for j = m(i), ωj = ψ (aj , pi, πi) <ψ (aj , pi, π̂) , and thus for n sufficiently large, ωj < ψ (aj , pin , πin) , contradictingstability of (pi, aj). Thus, (pi, ai) ∈ Φ (i, i) if i is a continuity point of p.

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Suppose that GID holds and that (m,π, ω) is an I-equilibrium. Let Ep bethe set of points of discontinuity of the type assignment p and Ea be the set ofpoints of discontinuity of the type assignment a. Let E = Ep∩Ea. and I ′ = I\E;then I ′ is of full measure. By Lemma 8 for all i ∈ I ′, (πi, ωi) ∈ Φ (i, i) . Definingm∗ (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, wecan restrict attention to matches where the man i is matched with the womani. This can be part of an equilibrium if there exist payoff maps u and v suchthat the following conditions are satisfied almost everywhere:

u(i) = φ (pi, ai, v(i)) (6)u(i) = max

jφ (pi, aj , v(j)) (7)

v(i) = maxjψ (ai, pj , u (j)) (8)

(6) is feasibility: in a match payoffs must be on the frontier; (7) and (8) are theincentive compatibility conditions: a man i must prefer to match with womani given that women ask payoffs v. Note the similarity between the incentiveconditions and those obtained when we consider screening problems. Like inthis literature, the search for equilibrium payoffs is facilitated by looking at localconditions, something that is acceptable as long as a single crossing condition issatisfied: if the local incentive compatibility condition holds, then single crossingimplies that larger types have even less incentives to deviate downward, insuringglobal stability. The GID condition is a single crossing condition. It can beequivalently expressed as: 4

Definition 9 Let i′ < i, j′ < j, and suppose that v and v′ are such thatφ (pi′ , aj′ , v′) = φ (pi′ , aj , v) , then, φ (pi, aj , v) ≥ φ (pi, aj′ , v′) .

However a local incentive condition in the continuum is useful if it is knownthat the payoff is differentiable in the type of the partner. Here since φ is differ-entiable, it is enough to show that the payoffs u and v must be differentiable. Inscreening models, when payoffs are quasi-linear, incentive compatibility indeedimplies that payoffs are monotonically increasing in types.

In our case, we do not have quasi-linarity of payoffs (otherwise the frontierwould 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 compatibilityindeed implies that payoffs must be non-decreasing. For any j the incentivecompatibility of the man i requires u (i) ≥ φ (pi, aj , v (j)) ; since φ is increasing,

4To see this note that if v, v′ satisfy φ`pi′ , aj′ , v′

´= φ (pi′ , aj , v) , then letting u be the

common value we have v = ψ`aj , pj′ , u

´and v′ = ψ

`aj′ , pi′ , u

´, and substituting this into

the terms in the inequality gives the GID condition in Definition 5.

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when j < i, we have φ (pi, aj , v (j)) ≥ φ (pj , aj , v (j)) , but the right hand sideis by (6) equal to u (j) proving that payoffs are non-decreasing. A similarlogic shows that v is increasing in i. Since increasing functions are differentiablealmost everywhere, the global incentive compatibility conditions can be replacedalmost everywhere by the first order conditions

dφ (pi, aj , v (j))dj

∣∣∣∣j=i

= 0

dψ (ai, pj , u (j))dj

∣∣∣∣j=i

= 0.

Proposition 10 Suppose that the type assignments p and a are continuouslydifferentiable and that GID holds. Then there exist payoffs u, v that stabilize theassortative matching in which man i is matched with woman i and these payoffssolve the system

u′ (i) = p′iφ1 = −p′iψ2

ψ3(9)

v′(i) = a′iψ1 = −a′iφ2

φ3(10)

Proof. If the type assignment maps p and a are differentiable, the first orderconditions can be written

a′iφ2 + v′ (i)φ3 = 0 (11)p′iψ2 + u′(i)ψ3 = 0, (12)

where for k = 2, 3, φk = φk (pi, ai, v (i)) and ψk = ψk (ai, pi, u (i)) . Henceincentive compatibility requires solving a system of differentiable equations inu, v. However, contrary to the screening problem there is an additional feasibilityconstraint to take into account. Since (6) is an identity, we can differentiate withrespect to i and get

u′(i) = p′iφ1 + a′iφ2 + v′ (i)φ3. (13)

Hence, (11) and (13) imply that

u′ (i) = p′iφ1.

Men i’s marginal payoff is indeed equal to his marginal product when he ismatched with woman i. Using the incentive compatibility condition of thewoman (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-payofftransferability. Having type-type transferability is not enough for PAM if highertypes find it also more difficult to transfer surplus to their partner. Note thatthe rate at which a man can transfer to a woman is given by the ratio p′ψ2/ψ3,

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and incentive compatibility requires that this equates the marginal productivityof the man.

Consider first a situation where there is perfect transferability, that is whenφ(p, a, v) = y(p, a)− v. Then, φ3 = ψ3 = −1 and φ2 = ψ1 and ψ2 = φ1 and thesecond equality in conditions (9) and (10) are trivial; we have u′ (i) = p′iy1(pi, ai)and v′ (i) = a′iy2 (pi, ai) .

With non-transferabilities φ2 is not equal to ψ1 in general, and the secondequality 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 − u2. This frontier satisfies

GID.5

Note that φ1 = a2√pa−v , ψ2 = a, ψ3 = −2u; since u =

√pa− v, we have

indeed p′φ1 = −p′ψ2ψ3

. The same is true in (10) and therefore we need to solvethe system

u′(i) = p′iai

2√piai − v(i)

v′ (i) = a′ipi

u(i)2 + v (i) = piai.

Suppose for instance that pi = ai = i for all i ∈ [0, 1]. Then, the conditions are

u′ (i) =i

2√i2 − v(i)

v′ (i) = i

u (i)2 + v (i) = i2.

At i = 0, the unique feasible payoff consistent with individual rationality isu (0) = v(0) = 0. Then v (́ı) = i implies that v (i) =

∫ i0xdx = i2

2 . The feasibilitycondition implies that u (i) = i√

2. We verify that u′ (i) = i

2√i2−v(i)

= 1√2.

We represent below a typical frontier for a match {i, i} and the equilbriumpayoffs of men and women.

If one is interested in total surplus maximization, the payoffs should beu(i) = i and v(i) = 0 when i < 1/2 since the slope of the frontier at v = 0is smaller than −1. This obviously illustrates the fact that non-transferabilityimplies that surplus division and surplus maximization are conflicting with eachother. Since the payoff to men is linear while the payoff to women is quadraticin type, women ”catch up” with men when i is large enough; moreover, whilethe 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. Obviouslythese results are specific to the example.

5In fact this economy admits a TU representation: use the change of variable F (u) = u2

and consider the new frontier φ∗(p, a, v) = φ (p, a, v)2 . Then F (u) = pa − v and there istransferability; clearly φ∗ satisfied GID. Legros and Newman (2006) show that in this case φalso satisfies GID.

8

i

i2 v

u

12i

−

i

,u v

22

v(i)

u(i)

Frontier for a match {i,i} Equilibrium payoffs

Figure 1: Frontier and Equilibrium Payoffs when φ (pi, ai, v) =√i2 − v.

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References

[1] Becker, G. (1973), “A Theory of Marriage, Part I, Journal of Political Econ-omy, 81: 813-846.”

[2] Kaneko, M., and M. Wooders (1996), “The Nonemptiness of the f -Core ofa Game Without Side Payments,” International Journal of Game Theory,25: 245-258.

[3] Legros, P. and A.F. Newman (2006), “Beauty is a Beast, Frog is a Prince: As-sortative Matching with Nontransferabilities,” forthcoming, Econometrica.

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