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The even split rule for (concave) symmetric supermodular functions
Economics Letters 186 (2020) 108783
Contents lists available at ScienceDirect
Economics Letters journal homepage: www.elsevier.com/locate/ecolet
The even split rule for (concave) symmetric supermodular functions✩ Hao Jia Department of Economics, Deakin University, 1 Gheringhap Street, Geelong, VIC 3220, Australia
article
info
Article history: Received 29 April 2019 Received in revised form 19 October 2019 Accepted 20 October 2019 Available online 31 October 2019 JEL classification: C78
a b s t r a c t This paper complements Jia (2019) by proving that the even split rule is the only Pareto efficient allocation that breaks down any concave symmetric supermodular function into two supermodular functions. It further provides an alternative proof for Theorem 1 of Jia (2019), which confirms that the even split rule is necessary to ensure any symmetric supermodular function, regardless its convexity or concavity, could be divided into two supermodular functions. © 2019 Elsevier B.V. All rights reserved.
Keywords: Supermodular functions Pareto efficient allocations Even split rule
Supermodularity, as a convenient and increasingly popular property, has been featured in many different areas of economics. For instance, in parametric optimization, when the feasible set has a lattice structure and the objective function is supermodular, the tools of monotone comparative statics (Topkis, 1978; Milgrom and Shannon, 1994; Topkis, 1998) provide a convenient means for making predictions about the parametric dependence of the solution set and of the value function. Another example includes the supermodular games (also known as games with strategic complementarities, as in Topkis, 1979; Vives, 1990; Milgrom and Roberts, 1990, among others), where the strategy spaces have a lattice structure and payoff functions feature certain supermodular properties. It has been widely recognized that many two-sided matching markets feature supermodular production technology (i.e. joint surplus) of some sort. For example, in the job market, recruiting skilled workers for skill-intensive positions usually yields higher productivity; in academia, publishing high-quality works in top-tier journals typically creates a greater impact to the scientific community. This feature has important implications. As shown in Atakan (2006), a positive assortative matching equilibrium (PAM for short1 ) can be achieved in a two-sided matching ✩ I am grateful to an anonymous reviewer for the insightful comments and suggestions. I would like to thank numerous colleagues and seminar participants at Deakin University for their helpful feedback. This paper greatly benefited from the productive comments by Professor John K.-H. Quah on my earlier paper. All remaining errors are my own. E-mail address: [email protected]. 1 In a marriage market, PAM refers to a positive correlation in sorting between the values of the traits of couples (matching of like types): the high type individual matches with an individual of the same high type, and the low type agent couples with an individual of the same low type. https://doi.org/10.1016/j.econlet.2019.108783 0165-1765/© 2019 Elsevier B.V. All rights reserved.
game that features symmetric supermodular production technology with explicit search costs, given the joint surplus is divided between matched couples according to the Nash bargaining solution, which boils down to an even split rule within the transferable utility framework. In other words, the even split rule breaks down any symmetric supermodular joint surplus into two supermodular individual benefit functions and leads to a PAM equilibrium. It is natural to wonder, besides the even split rule, if there are other Pareto efficient allocations that break down any symmetric supermodular function into two supermodular functions. A policy implication of this question arises from the marriage market. When a man and a woman tie the knot and become a couple, they create a joint surplus, also known as marital property. An allocation rule determines the extent of which each spouse can claim for their marital property, which becomes particularly relevant upon divorce or annulment. In most jurisdictions, an even split division of marital property (Rule VI) is strictly mandated by statute. However, there are other allocation rules that have been adopted in practice, such as the equitable distribution statute in Florida (Florida State §61.075). Given the joint surplus features supermodularity, the question becomes whether these systems lead to the socially desired PAM equilibria. A recent paper by Jia (2019) answers this question by showing that the even split rule is the only allocation that features this property, which rationalizes the use of the Nash bargaining solution in various two-sided matching models within the transferable utility framework.2 It is worth noting that, however, 2 A recent thorough survey of economic studies on search and matching models is given in Chade et al. (2017).
2
H. Jia / Economics Letters 186 (2020) 108783
Theorem 1 of Jia (2019) is established upon choosing a family of convex functions to obtain key characteristics of the potential allocation rules. This additional restriction excludes concave symmetric supermodular functions from consideration, which limits the scope of its applicable scenarios. Indeed, functions that are both concave and supermodular have important implications in matching models and many other economic settings (see, for example, Smith, 2006; Quah, 2007, among others). Jia (2019) also admits that “it would be beneficial to establish a specific result similar to Theorem 1, focusing on the family of concave and supermodular functions.” The current paper bridges this gap by showing that the even split rule is the only Pareto efficient allocation that breaks down any concave symmetric supermodular function into two supermodular functions. The approach can be easily extended to convex symmetric supermodular functions, which serves as an alternative proof for Theorem 1 of Jia (2019). In particular, this paper confirms that the even split rule is necessary to ensure any symmetric supermodular function, regardless its convexity or concavity, can be divided into two supermodular functions. To ensure consistency, the following notations and concepts are directly borrowed from Jia (2019). Definition 1. A class of allocation rules M is Pareto efficient if and only if any element φ ∈ M with φ : R2+ → [0, 1] possesses the following properties3 : 1. (Pareto efficiency) φ (x, y) ≥ 0 and φ (x, y) + φ (y, x) = 1 for any x, y ≥ 0, 2. (Continuity) φ (x, y) is continuous almost surely on its domain. Definition 2. A bivariate function f : R2+ → R is supermodular, if f (x1 , y1 ) + f (x2 , y2 ) ≥ f (x1 , y2 ) + f (x2 , y1 ),
∀x1 > x2 ≥ 0, y1 > y2 ≥ 0.
(1)
In particular, f is strictly supermodular if Inequality (1) is sharp. The first result of the current paper focuses on concave symmetric supermodular functions. Theorem 1. Consider the sublattice X ⊆ R2+ with (0, 0)′ ∈ X. If the multiplicative function φ (x)f (x) is supermodular for any continuous, concave, symmetric, and supermodular function f (x) on X, where φ ∈ M, then φ (x) ≡ 1/2. Proof. Let φ (x, y) be a Pareto efficient allocation with φ (x, y) f (x, y) and [1 − φ (x, y)]f (x, y) both being supermodular, for any concave symmetric supermodular function f : X → R. Without loss of generality, assume (1, 1)′ ∈ X. Consider functions f0 , f1 : X → R with f0 (x, y) = 1 and f1 (x, y) = x + y. It is easy to check that f0 and f1 are continuous, symmetric, and supermodular. It is worth noting that the two functions f0 , f1 are both concave and convex by definition. One therefore concludes that both φ (x, y)f0 (x, y) = φ (x, y) and [1 − φ (x, y)]f0 (x, y) = 1 − φ (x, y) are supermodular. In particular, for any {x, y} ∈ X, one has
φ (0, 0) + φ (x, y) ≥ φ (0, y) + φ (x, 0), 1 − φ (0, 0) + 1 − φ (x, y) ≥ 1 − φ (0, y) + 1 − φ (x, 0). 3 The over line represents the convex hull operator, which indicates that (0, 0)′ ∈ R2+ .
Note that none of the two inequalities can be sharp if they hold simultaneously for all {x, y} ∈ X, otherwise one gets 2 > 2 by adding them together. So one must have
φ (0, 0) + φ (x, y) = φ (0, y) + φ (x, 0), ⇔ φ (x, y) = φ (0, y) + [φ (x, 0) − φ (0, 0)]. Let φ (0, y) = b(y) and φ (x, 0) − φ (0, 0) = a(x). One can then rewrite the last expression as
φ (x, y) = a(x) + b(y).
(2)
As φ is a Pareto efficient allocation, by definition, φ (t , t) = 1/2, ∀t ≥ 0. Substituting t in (2) gives a(t) + b(t) = 1/2, or b(t) = 1/2 − a(t). Hence, (2) can be further simplified as 1
+ a(x) − a(y). (3) 2 Meanwhile, φ (x, y)f1 (x, y) and [1 − φ (x, y)]f1 (x, y) are both supermodular. I.e., for any 0 ≤ x1 < x2 and 0 ≤ y1 < y2 , one has
φ (x, y) =
φ (x1 , y1 )(x1 + y1 ) + φ (x2 , y2 )(x2 + y2 ) ≥ φ (x1 , y2 )(x1 + y2 ) + φ (x2 , y1 )(x2 + y1 ), [1 − φ (x1 , y1 )](x1 + y1 ) + [1 − φ (x2 , y2 )](x2 + y2 ) ≥ [1 − φ (x1 , y2 )](x1 + y2 ) + [1 − φ (x2 , y1 )](x2 + y1 ). Once again, note that none of the two inequalities can be sharp if they hold simultaneously for all x1 , x2 , y1 , y2 ≥ 0, otherwise one gets x1 + y1 + x2 + y2 > x1 + y2 + x2 + y1 by adding them together. So, one must have
φ (x1 , y1 )(x1 + y1 ) + φ (x2 , y2 )(x2 + y2 ) = φ (x1 , y2 )(x1 + y2 ) + φ (x2 , y1 )(x2 + y1 ). Substituting (3) into the last equality yields
[
] [ ] 1 + a(x1 ) − a(y1 ) (x1 + y1 ) + + a(x2 ) − a(y2 ) (x2 + y2 ) 2 2 [ ] [ ] 1 1 = + a(x1 ) − a(y2 ) (x1 + y2 ) + + a(x2 ) − a(y1 ) (x2 + y1 ), 1
2
2
[a(x1 ) − a(y1 )](x1 + y1 ) + [a(x2 ) − a(y2 )](x2 + y2 ) [a(x1 ) − a(y2 )](x1 + y2 ) + [a(x2 ) − a(y1 )](x2 + y1 ), a(x1 )y1 − a(y1 )x1 + a(x2 )y2 − a(y2 )x2 a(x1 )y2 − a(y2 )x1 + a(x2 )y1 − a(y1 )x2 , [a(y1 ) − a(y2 )][x2 − x1 ] = [a(x1 ) − a(x2 )][y2 − y1 ], a(x2 ) − a(x1 ) a(y2 ) − a(y1 ) ⇒ = . x2 − x1 y2 − y1 Since the last equality holds for all x1 < x2 , y1 < y2 , by choosing y1 = 0, y2 = 1 and defining c ≡ a(1) − a(0), one ⇒ = ⇒ = ⇒
concludes that a(x2 ) − a(x1 )
= c,
x2 − x1
∀x2 > x1 ≥ 0.
(4)
Fixing z = x1 and letting x2 approach x1 from above, which is admissible due to the fact that φ is almost surely continuous on X, the left hand side of (4) becomes a one-sided derivative at z, i.e., a′+ (z) = c. At the same time, by fixing z = x2 and letting x1 approach x2 , one also obtains a′− (z) = c. Hence a′ (z) = c , ∀z ≥ 0, which further suggests that a(z) = cz, or
φ (x, y) =
1
+ cx − cy (5) 2 for some constant c. The last step is to prove that c = 0. To see this, pick yet another symmetric supermodular function f2 (x, y) = ln x + ln y. It is easy to see that f2 is strictly concave on (0, 1] as it is the sum of two concave functions.
H. Jia / Economics Letters 186 (2020) 108783
Since φ (x, y)f2 (x, y) and [1 − φ (x, y)]f2 (x, y) are both supermodular, i.e., for any 0 ≤ x1 < x2 and 0 ≤ y1 < y2 , one has
φ (x1 , y1 )(ln x1 + ln y1 ) + φ (x2 , y2 )(ln x2 + ln y2 ) ≥ φ (x1 , y2 )(ln x1 + ln y2 ) + φ (x2 , y1 )(ln x2 + ln y1 ), [1 − φ (x1 , y1 )](ln x1 + ln y1 ) + [1 − φ (x2 , y2 )](ln x2 + ln y2 ) ≥ [1 − φ (x1 , y2 )](ln x1 + ln y2 ) + [1 − φ (x2 , y1 )](ln x2 + ln y1 ). Once again, note that none of the two inequalities can be sharp if they hold simultaneously, otherwise one gets ln x1 + ln y1 + ln x2 + ln y2 > ln x1 + ln y2 + ln x2 + ln y1 by adding them together, a contradiction. So one must have
φ (x1 , y1 )(ln x1 + ln y1 ) + φ (x2 , y2 )(ln x2 + ln y2 ) = φ (x1 , y2 )(ln x1 + ln y2 ) + φ (x2 , y1 )(ln x2 + ln y1 ).
It is clear that the only possible scenario for (6) to hold is c = 0, which completes the proof. Two remarks are worth making here. First, the result of Theorem 1 can be further generalized. For instance, by adopting the same approach and choosing a symmetric function f2 (x, y) = g(x) + g(y), one can verify that the conclusion holds as long as function g has a non-zero second order derivative almost surely. In particular, by choosing a strictly convex function g,4 one obtains the following parallel result. Corollary 1. Consider the sublattice X ⊆ R2+ with (0, 0)′ ∈ X. For (x, y) ∈ X and φ ∈ M, if the maps φ (x, y)f (x, y) and (1 − φ (x, y))f (x, y) are supermodular for all convex and supermodular functions f , then φ ≡ 1/2. Note that Corollary 1 has already been established in Jia (2019), and hence the current paper provides an alternative proof for Theorem 1 of Jia (2019). Second, as the result of Theorem 1 only requires one being able to pick a function f2 (x, y) = g(x) + g(y) with g ′′ ̸ = 0, a.s., it does not rely on the concavity or convexity of function f . In other words, the even split rule is the only Pareto allocation to break down any supermodular function, regardless its concavity or convexity, into two supermodular functions.
Substituting (5) into this equation yields
] + cx1 − cy1 (ln x1 + ln y1 ) 2 ] [ 1 + cx2 − cy2 (ln x2 + ln y2 ) + 2 [ ] 1 = + cx1 − cy2 (ln x1 + ln y2 ) 2 [ ] 1 + + cx2 − cy1 (ln x2 + ln y1 ), [
3
1
References
2
⇒ = ⇒ = ⇒
c(x1 − y1 )(ln x1 + ln y1 ) + c(x2 − y2 )(ln x2 + ln y2 ) c(x1 − y2 )(ln x1 + ln y2 ) + c(x2 − y1 )(ln x2 + ln y1 ), c(x1 ln y1 − y1 ln x1 + x2 ln y2 − y2 ln x2 ) c(x1 ln y2 − y2 ln x1 + x2 ln y1 − y1 ln x2 ),
c(ln y1 − ln y2 )(x2 − x1 ) = c(ln x1 − ln x2 )(y2 − y1 ), [ ] ln x2 − ln x1 ln y2 − ln y1 ⇒c − = 0. x2 − x1 y2 − y1 Since the last equality holds for all x1 < x2 , y1 < y2 , by fixing x1 and y1 and letting x2 and y2 approach x1 and y1 from above respectively, the expression within the square brackets in the last equation becomes the derivatives of the logarithmic function at x1 and y1 respectively. The last equality therefore can be rewritten as
( c
1 x1
−
1 y1
)
= 0,
∀x1 , y 1 ∈ X .
(6)
Atakan, A.E., 2006. Assortative matching with explicit search costs. Econometrica 74 (3), 667–680. Chade, H., Eeckhout, J., Smith, L., 2017. Sorting through search and matching models in economics. J. Econ. Lit. 55 (2), 493–544. Jia, H., 2019. The even split rule in positive assortative matching. J. Math. Econom. 81, 57–61. Milgrom, P., Roberts, J., 1990. Rationalizability, learning, and equilibrium in games with strategic complementarities. Econometrica 58 (6), 1255–1277. Milgrom, P., Shannon, C., 1994. Monotone comparative statics. Econometrica 62 (1), 157–180. Quah, J.K.-H., 2007. The comparative statics of constrained optimization problems. Econometrica 75 (2), 401–431. Smith, L., 2006. The marriage model with search frictions. J. Political Econ. 114 (6), 1124–1144. Topkis, D.M., 1978. Minimizing a submodular function on a lattice. Oper. Res. 26, 305–321. Topkis, D.M., 1979. Equilibrium points in nonzero-sum n-person submodular games. SIAM J. Control Optim. 17 (6), 773–787. Topkis, D.M., 1998. Supermodularity and Complementarity. Frontiers of Economic Research, Princeton University Press, Princeton, New Jersey. Vives, X., 1990. Nash equilibrium with strategic complementarities. J. Math. Econom. 19 (3), 305–321.
4 The function g being strictly convex indicates g ′′ ̸ = 0 for sure.
Contents lists available at ScienceDirect
Economics Letters journal homepage: www.elsevier.com/locate/ecolet
The even split rule for (concave) symmetric supermodular functions✩ Hao Jia Department of Economics, Deakin University, 1 Gheringhap Street, Geelong, VIC 3220, Australia
article
info
Article history: Received 29 April 2019 Received in revised form 19 October 2019 Accepted 20 October 2019 Available online 31 October 2019 JEL classification: C78
a b s t r a c t This paper complements Jia (2019) by proving that the even split rule is the only Pareto efficient allocation that breaks down any concave symmetric supermodular function into two supermodular functions. It further provides an alternative proof for Theorem 1 of Jia (2019), which confirms that the even split rule is necessary to ensure any symmetric supermodular function, regardless its convexity or concavity, could be divided into two supermodular functions. © 2019 Elsevier B.V. All rights reserved.
Keywords: Supermodular functions Pareto efficient allocations Even split rule
Supermodularity, as a convenient and increasingly popular property, has been featured in many different areas of economics. For instance, in parametric optimization, when the feasible set has a lattice structure and the objective function is supermodular, the tools of monotone comparative statics (Topkis, 1978; Milgrom and Shannon, 1994; Topkis, 1998) provide a convenient means for making predictions about the parametric dependence of the solution set and of the value function. Another example includes the supermodular games (also known as games with strategic complementarities, as in Topkis, 1979; Vives, 1990; Milgrom and Roberts, 1990, among others), where the strategy spaces have a lattice structure and payoff functions feature certain supermodular properties. It has been widely recognized that many two-sided matching markets feature supermodular production technology (i.e. joint surplus) of some sort. For example, in the job market, recruiting skilled workers for skill-intensive positions usually yields higher productivity; in academia, publishing high-quality works in top-tier journals typically creates a greater impact to the scientific community. This feature has important implications. As shown in Atakan (2006), a positive assortative matching equilibrium (PAM for short1 ) can be achieved in a two-sided matching ✩ I am grateful to an anonymous reviewer for the insightful comments and suggestions. I would like to thank numerous colleagues and seminar participants at Deakin University for their helpful feedback. This paper greatly benefited from the productive comments by Professor John K.-H. Quah on my earlier paper. All remaining errors are my own. E-mail address: [email protected]. 1 In a marriage market, PAM refers to a positive correlation in sorting between the values of the traits of couples (matching of like types): the high type individual matches with an individual of the same high type, and the low type agent couples with an individual of the same low type. https://doi.org/10.1016/j.econlet.2019.108783 0165-1765/© 2019 Elsevier B.V. All rights reserved.
game that features symmetric supermodular production technology with explicit search costs, given the joint surplus is divided between matched couples according to the Nash bargaining solution, which boils down to an even split rule within the transferable utility framework. In other words, the even split rule breaks down any symmetric supermodular joint surplus into two supermodular individual benefit functions and leads to a PAM equilibrium. It is natural to wonder, besides the even split rule, if there are other Pareto efficient allocations that break down any symmetric supermodular function into two supermodular functions. A policy implication of this question arises from the marriage market. When a man and a woman tie the knot and become a couple, they create a joint surplus, also known as marital property. An allocation rule determines the extent of which each spouse can claim for their marital property, which becomes particularly relevant upon divorce or annulment. In most jurisdictions, an even split division of marital property (Rule VI) is strictly mandated by statute. However, there are other allocation rules that have been adopted in practice, such as the equitable distribution statute in Florida (Florida State §61.075). Given the joint surplus features supermodularity, the question becomes whether these systems lead to the socially desired PAM equilibria. A recent paper by Jia (2019) answers this question by showing that the even split rule is the only allocation that features this property, which rationalizes the use of the Nash bargaining solution in various two-sided matching models within the transferable utility framework.2 It is worth noting that, however, 2 A recent thorough survey of economic studies on search and matching models is given in Chade et al. (2017).
2
H. Jia / Economics Letters 186 (2020) 108783
Theorem 1 of Jia (2019) is established upon choosing a family of convex functions to obtain key characteristics of the potential allocation rules. This additional restriction excludes concave symmetric supermodular functions from consideration, which limits the scope of its applicable scenarios. Indeed, functions that are both concave and supermodular have important implications in matching models and many other economic settings (see, for example, Smith, 2006; Quah, 2007, among others). Jia (2019) also admits that “it would be beneficial to establish a specific result similar to Theorem 1, focusing on the family of concave and supermodular functions.” The current paper bridges this gap by showing that the even split rule is the only Pareto efficient allocation that breaks down any concave symmetric supermodular function into two supermodular functions. The approach can be easily extended to convex symmetric supermodular functions, which serves as an alternative proof for Theorem 1 of Jia (2019). In particular, this paper confirms that the even split rule is necessary to ensure any symmetric supermodular function, regardless its convexity or concavity, can be divided into two supermodular functions. To ensure consistency, the following notations and concepts are directly borrowed from Jia (2019). Definition 1. A class of allocation rules M is Pareto efficient if and only if any element φ ∈ M with φ : R2+ → [0, 1] possesses the following properties3 : 1. (Pareto efficiency) φ (x, y) ≥ 0 and φ (x, y) + φ (y, x) = 1 for any x, y ≥ 0, 2. (Continuity) φ (x, y) is continuous almost surely on its domain. Definition 2. A bivariate function f : R2+ → R is supermodular, if f (x1 , y1 ) + f (x2 , y2 ) ≥ f (x1 , y2 ) + f (x2 , y1 ),
∀x1 > x2 ≥ 0, y1 > y2 ≥ 0.
(1)
In particular, f is strictly supermodular if Inequality (1) is sharp. The first result of the current paper focuses on concave symmetric supermodular functions. Theorem 1. Consider the sublattice X ⊆ R2+ with (0, 0)′ ∈ X. If the multiplicative function φ (x)f (x) is supermodular for any continuous, concave, symmetric, and supermodular function f (x) on X, where φ ∈ M, then φ (x) ≡ 1/2. Proof. Let φ (x, y) be a Pareto efficient allocation with φ (x, y) f (x, y) and [1 − φ (x, y)]f (x, y) both being supermodular, for any concave symmetric supermodular function f : X → R. Without loss of generality, assume (1, 1)′ ∈ X. Consider functions f0 , f1 : X → R with f0 (x, y) = 1 and f1 (x, y) = x + y. It is easy to check that f0 and f1 are continuous, symmetric, and supermodular. It is worth noting that the two functions f0 , f1 are both concave and convex by definition. One therefore concludes that both φ (x, y)f0 (x, y) = φ (x, y) and [1 − φ (x, y)]f0 (x, y) = 1 − φ (x, y) are supermodular. In particular, for any {x, y} ∈ X, one has
φ (0, 0) + φ (x, y) ≥ φ (0, y) + φ (x, 0), 1 − φ (0, 0) + 1 − φ (x, y) ≥ 1 − φ (0, y) + 1 − φ (x, 0). 3 The over line represents the convex hull operator, which indicates that (0, 0)′ ∈ R2+ .
Note that none of the two inequalities can be sharp if they hold simultaneously for all {x, y} ∈ X, otherwise one gets 2 > 2 by adding them together. So one must have
φ (0, 0) + φ (x, y) = φ (0, y) + φ (x, 0), ⇔ φ (x, y) = φ (0, y) + [φ (x, 0) − φ (0, 0)]. Let φ (0, y) = b(y) and φ (x, 0) − φ (0, 0) = a(x). One can then rewrite the last expression as
φ (x, y) = a(x) + b(y).
(2)
As φ is a Pareto efficient allocation, by definition, φ (t , t) = 1/2, ∀t ≥ 0. Substituting t in (2) gives a(t) + b(t) = 1/2, or b(t) = 1/2 − a(t). Hence, (2) can be further simplified as 1
+ a(x) − a(y). (3) 2 Meanwhile, φ (x, y)f1 (x, y) and [1 − φ (x, y)]f1 (x, y) are both supermodular. I.e., for any 0 ≤ x1 < x2 and 0 ≤ y1 < y2 , one has
φ (x, y) =
φ (x1 , y1 )(x1 + y1 ) + φ (x2 , y2 )(x2 + y2 ) ≥ φ (x1 , y2 )(x1 + y2 ) + φ (x2 , y1 )(x2 + y1 ), [1 − φ (x1 , y1 )](x1 + y1 ) + [1 − φ (x2 , y2 )](x2 + y2 ) ≥ [1 − φ (x1 , y2 )](x1 + y2 ) + [1 − φ (x2 , y1 )](x2 + y1 ). Once again, note that none of the two inequalities can be sharp if they hold simultaneously for all x1 , x2 , y1 , y2 ≥ 0, otherwise one gets x1 + y1 + x2 + y2 > x1 + y2 + x2 + y1 by adding them together. So, one must have
φ (x1 , y1 )(x1 + y1 ) + φ (x2 , y2 )(x2 + y2 ) = φ (x1 , y2 )(x1 + y2 ) + φ (x2 , y1 )(x2 + y1 ). Substituting (3) into the last equality yields
[
] [ ] 1 + a(x1 ) − a(y1 ) (x1 + y1 ) + + a(x2 ) − a(y2 ) (x2 + y2 ) 2 2 [ ] [ ] 1 1 = + a(x1 ) − a(y2 ) (x1 + y2 ) + + a(x2 ) − a(y1 ) (x2 + y1 ), 1
2
2
[a(x1 ) − a(y1 )](x1 + y1 ) + [a(x2 ) − a(y2 )](x2 + y2 ) [a(x1 ) − a(y2 )](x1 + y2 ) + [a(x2 ) − a(y1 )](x2 + y1 ), a(x1 )y1 − a(y1 )x1 + a(x2 )y2 − a(y2 )x2 a(x1 )y2 − a(y2 )x1 + a(x2 )y1 − a(y1 )x2 , [a(y1 ) − a(y2 )][x2 − x1 ] = [a(x1 ) − a(x2 )][y2 − y1 ], a(x2 ) − a(x1 ) a(y2 ) − a(y1 ) ⇒ = . x2 − x1 y2 − y1 Since the last equality holds for all x1 < x2 , y1 < y2 , by choosing y1 = 0, y2 = 1 and defining c ≡ a(1) − a(0), one ⇒ = ⇒ = ⇒
concludes that a(x2 ) − a(x1 )
= c,
x2 − x1
∀x2 > x1 ≥ 0.
(4)
Fixing z = x1 and letting x2 approach x1 from above, which is admissible due to the fact that φ is almost surely continuous on X, the left hand side of (4) becomes a one-sided derivative at z, i.e., a′+ (z) = c. At the same time, by fixing z = x2 and letting x1 approach x2 , one also obtains a′− (z) = c. Hence a′ (z) = c , ∀z ≥ 0, which further suggests that a(z) = cz, or
φ (x, y) =
1
+ cx − cy (5) 2 for some constant c. The last step is to prove that c = 0. To see this, pick yet another symmetric supermodular function f2 (x, y) = ln x + ln y. It is easy to see that f2 is strictly concave on (0, 1] as it is the sum of two concave functions.
H. Jia / Economics Letters 186 (2020) 108783
Since φ (x, y)f2 (x, y) and [1 − φ (x, y)]f2 (x, y) are both supermodular, i.e., for any 0 ≤ x1 < x2 and 0 ≤ y1 < y2 , one has
φ (x1 , y1 )(ln x1 + ln y1 ) + φ (x2 , y2 )(ln x2 + ln y2 ) ≥ φ (x1 , y2 )(ln x1 + ln y2 ) + φ (x2 , y1 )(ln x2 + ln y1 ), [1 − φ (x1 , y1 )](ln x1 + ln y1 ) + [1 − φ (x2 , y2 )](ln x2 + ln y2 ) ≥ [1 − φ (x1 , y2 )](ln x1 + ln y2 ) + [1 − φ (x2 , y1 )](ln x2 + ln y1 ). Once again, note that none of the two inequalities can be sharp if they hold simultaneously, otherwise one gets ln x1 + ln y1 + ln x2 + ln y2 > ln x1 + ln y2 + ln x2 + ln y1 by adding them together, a contradiction. So one must have
φ (x1 , y1 )(ln x1 + ln y1 ) + φ (x2 , y2 )(ln x2 + ln y2 ) = φ (x1 , y2 )(ln x1 + ln y2 ) + φ (x2 , y1 )(ln x2 + ln y1 ).
It is clear that the only possible scenario for (6) to hold is c = 0, which completes the proof. Two remarks are worth making here. First, the result of Theorem 1 can be further generalized. For instance, by adopting the same approach and choosing a symmetric function f2 (x, y) = g(x) + g(y), one can verify that the conclusion holds as long as function g has a non-zero second order derivative almost surely. In particular, by choosing a strictly convex function g,4 one obtains the following parallel result. Corollary 1. Consider the sublattice X ⊆ R2+ with (0, 0)′ ∈ X. For (x, y) ∈ X and φ ∈ M, if the maps φ (x, y)f (x, y) and (1 − φ (x, y))f (x, y) are supermodular for all convex and supermodular functions f , then φ ≡ 1/2. Note that Corollary 1 has already been established in Jia (2019), and hence the current paper provides an alternative proof for Theorem 1 of Jia (2019). Second, as the result of Theorem 1 only requires one being able to pick a function f2 (x, y) = g(x) + g(y) with g ′′ ̸ = 0, a.s., it does not rely on the concavity or convexity of function f . In other words, the even split rule is the only Pareto allocation to break down any supermodular function, regardless its concavity or convexity, into two supermodular functions.
Substituting (5) into this equation yields
] + cx1 − cy1 (ln x1 + ln y1 ) 2 ] [ 1 + cx2 − cy2 (ln x2 + ln y2 ) + 2 [ ] 1 = + cx1 − cy2 (ln x1 + ln y2 ) 2 [ ] 1 + + cx2 − cy1 (ln x2 + ln y1 ), [
3
1
References
2
⇒ = ⇒ = ⇒
c(x1 − y1 )(ln x1 + ln y1 ) + c(x2 − y2 )(ln x2 + ln y2 ) c(x1 − y2 )(ln x1 + ln y2 ) + c(x2 − y1 )(ln x2 + ln y1 ), c(x1 ln y1 − y1 ln x1 + x2 ln y2 − y2 ln x2 ) c(x1 ln y2 − y2 ln x1 + x2 ln y1 − y1 ln x2 ),
c(ln y1 − ln y2 )(x2 − x1 ) = c(ln x1 − ln x2 )(y2 − y1 ), [ ] ln x2 − ln x1 ln y2 − ln y1 ⇒c − = 0. x2 − x1 y2 − y1 Since the last equality holds for all x1 < x2 , y1 < y2 , by fixing x1 and y1 and letting x2 and y2 approach x1 and y1 from above respectively, the expression within the square brackets in the last equation becomes the derivatives of the logarithmic function at x1 and y1 respectively. The last equality therefore can be rewritten as
( c
1 x1
−
1 y1
)
= 0,
∀x1 , y 1 ∈ X .
(6)
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4 The function g being strictly convex indicates g ′′ ̸ = 0 for sure.