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ASSORTATIVE MATCHING AND SEARCH
Robert Shinier
Lones Smith
No.
97-2
March, 1997
massachusetts
institute of
technology
50 memorial drive
Cambridge, mass. 02139
WORKING PAPER
DEPARTMENT
OF ECONOMICS
ASSORTATIVE MATCHING AND SEARCH
Robert Shinier
Lones Smith
No.
97-2
March, 1997
MASSACHUSETTS
INSTITUTE OF
TECHNOLOGY
50 MEMORIAL DRIVE
CAMBRIDGE, MASS. 021329
i
.««
(I
Assortative Matching and Search*^
Robert Shimer*
Department
of
Lones Smith§
Department
Economics
Princeton University
of
Economics
M.I.T.
February, 1996
Abstract
This paper reprises Becker's (1973) neoclassical marriage market
model, assuming search
agents with types
and
frictions:
xGK; match
utility is transferable;
We
There
is
a continuum of heterogeneous
(x, y) yields flow
foregone output
is
output f(x, y)
=
f(y, x),
the only cost of search.
characterize equilibrium and constrained efficient matchings, and
prove equilibrium existence.
Becker's
benchmark
We
then compare matching patterns with
frictionless allocation,
where positively or negatively
complements or substi-
assortative matching arises as agents' types are
tutes in production.
We
formalize a notion of assortative- matching with
search frictions in the spirit of affiliation, and demonstrate that Becker's
2
f(x,y) = (x + y) produces highly
nonassortative matching. Fortunately, we prove that assortative match-
condition no longer suffices --
e.g.
ing does extend to both search settings
type distributions
-
-
- - for all
search frictions and
under stronger assumptions:
supermodularity of
not just /, but also log/x and \ogfxy Examples illustrate the necessity of these conditions. We show that our assortative matching notion
.
in fact implies that
everyone matches with a convex set of types; as a
biproduct, this paper also provides a theory of convex matching
*We
sets.
Chang (M.I.T. Math Dept.) for detecting a serious flaw in a key
spending the time to outline a fix for it. We wish to thank Daron Acemoglu, Susan
Athey, Peter Diamond, Bart Lipman, Andreu MasColell, Andrew McLennan, and Charles Wilson
provided useful comments on this paper or its prequel, as well as participants at the MIT Theory
proof,
are grateful to Sheldon
and
for
Lunch, and seminars at NYU and Princeton. We have also benefited from frequent help received
via sci. math. research. Marcos Chamon provided valuable simulations assistance.
tThis paper answers questions originally posed in a 1994 version of our mimeo "Matching,
Search, and Heterogeneity." That unfinished work now focuses on a host of macro topics, such as
the cross-sectional nature of search inefficiency, and has a more general model
than this one, which is honed for a micro audience.
*
e-mail address:
§e-mail address:
shimer@princeton.edu
lones@lones.mit.edu
in
some respects
INTRODUCTION
1.
we
In this paper,
revisit
matching literature
frictions.
We
what
is
arguably the classic insight of the neoclassical
— when assortative matching
geneous agents with characteristics x
flow return of a
in
[0, 1],
There
who can
match between agents x and y
production function -- for instance, f{x,y)
must search
in a setting
with search
consider a simple model of synergistic matching inspired by Becker's
(1973) seminal analysis of the marriage market.
ric
—
arises
for partners,
is
=
is
a continuum of hetero-
The
only produce in pairs:
f(x,y), where /
Everyone
xy.
is
is
a symmet-
impatient and
with potential matches arriving at a Poisson hazard rate.
Matching precludes further search. Everyone seeks to maximize the present value of
her future payoffs, and
we ask who agrees
match with
to
whom
in a
Such matching decisions turn on how the match output
state.
dynamic steady
Smith
shared.
is
(1996) considers exogenous sharing rules, as he studies the model where utility
not transferable (NTU). This paper instead posits transferable
We
put
first
study search equilibrium
(TU).
— the decentralized solution when match out-
shared according to the Nash bargaining
is
utility
is
rule.
We
provide what we believe
to be the first general proof of existence of such an equilibrium with heterogeneous
types.
1
It is
of independent value, as
point recipe for application elsewhere
strategies to measures of
since search
is
costly,
matching with
maximized.
As a
- -
especially the key continuous
agents.
and the decision
either.
We
unmatched
parses the logical argument into a three
it
of
There are externalities
map from
in equilibrium,
two agents to match precludes others from
result, the present value of
output in the economy
not
is
therefore next examine the constrained efficient outcome. Here,
matching decisions are made by a
social planner seeking to
all
maximize the present
value of output, but unable to circumvent the search frictions faced by individuals.
The
central
theme of this paper, and our most innovative contribution,
is
that the
equilibrium and constrained optimal matching outcomes are united by important
cross-sectional structure.
To introduce
core allocation of a frictionless market:
output (namely, fxy
>
assortative matching.
0),
As
this, recall
If
But
main
insight into the
types are complementary to the match
then other things equal, agents should engage in positively
usual, search frictions create temporal
thus an acceptable match need not be an ideal one.
l
Becker's
Still,
see the text for a discussion of other special case proofs.
matching
rents,
and
the ease with which Becker
derives his result suggests that there might be a simple extension of this neoclassical
insight to a
with
model with search
who
others
all
Namely, agents might be willing to match
frictions:
are not too different than themselves.
we
In this paper,
we demonstrate the negative
dispel any hopes of such a free lunch. Instead,
result
that for an open class of complementarity production functions, individuals
match with types who are
only be willing to
example
is
Fortunately,
We
it
all is
find a simple
arises for
matching
is
=
afforded by the production function f(x,y)
matching pattern, depicted
not
in
An
quite different.
Figure 2 (page 16),
is
(x
+
first
may
especially salient
2
y)
The
.
resulting
no way assortative.
in
Assortative matching does extend to a search setting.
lost:
open subclass of complementary production functions
any search frictions or type
also quite natural,
and
Our
distributions.
essentially asks that
for
which
definition of assortative
matching
sets
be
affiliated.
For types in R, this asks that any two agreeable matches can be severed, and then the
As testimony
greater two and lesser two types agreeably rematched.
of this notion,
we show that
it
to the richness
matches with a convex
in fact implies that everyone
set
of types in R.
This paper therefore also simultaneously provides a theory of convex
matching
We
sets.
research program,
consider this an interesting application of the supermodularity
2
insofar as one
production function (as
and
must combine super/sub-modularity of
in Becker)
(Hi) log cross partial derivative
and
also of
(22) its
logfxy The
.
(2)
the
log marginal product log /a,
latter condition implies a key single
crossing property of matching preferences.
We
use these assumptions to prove that any individual z has a quasiconcave
match surplus function s(z,y)
unmatched option
it is
=
f[z,y)
—
w(z)
—
w(y), where w(y)
value. This immediately produces convex
matching
is
the type y's
sets,
and then
simply a matter of orienting whether matching sets are increasing or decreasing
by Becker's condition
In fact, since one's current value
(2).
on future surplus, or w(z)
=
k
J max(0,
mass of unmatched type-y agents u(y)
our descriptive theory
is
that
it
is
f(z, y)
—
w(z)
—
is
essentially
an option
w(y))u(y)dy, where the
unaffected by a single agent, one view of
exclusively develops
and exploits the properties of
such implicit integral equations.
For an intuitive overview of the quasiconcavity
search frictions so severe that any
entails
2
A
match
is
logic,
think of an economy with
acceptable.
Surplus quasiconcavity
comparing derivatives fy (z,y) and w'(y). The time cost of search renders
good reference
will
be Athey, Milgrom, and Roberts (1996).
7 <
w'(y) a multiple
Thus, the slope of
=
positive for z
>
/y(l>y)
1,
and thus
>
fy( x ,y)
surplus function
z's
unmatched agents x
of the expectation over
1
l's
— jEx fy (x,y). Under
fy (z,y)
is
surplus function
is
this
(i),
is
quasiconcave, simply because
So by continuity, the
lIy{ x iV) f° r a ^ £•
of fy (x,y).
frictionless logic of
Becker's result carries through for 'high' types with search frictions.
at the opposite extreme 2
But
are incomparable under
jfy (x,y)
it
as 1/7
is
this
0,
argument
alone. Here
(i)
guarantees that fy (x,y)/fy (0,y)
jEx fy (x,y) >
=
is
nondecreasing
> Ex (fy (x, y)/fy (0,
y)),
we
where
in y.
given
= Ex fy (x,y),
Loosely, this guarantees that everyone
function
is
quasiconcave
implies s y (z,
y')
<
i.e.
rewriting fy (0,y)
quasiconcave.
>
for y'
y.
falling after
is
by
x fy {x,y).
=
s y (z,y)
by
In other words, the marginal product fy (z, y) adjusts
the SCP, this holds at
(if
everyone matches),
5,
we
all
we
TU
z
as
z,
given the definition of
are aware that considers matching in a
is
>
<
z
z.
search
That paper does not touch
Sattinger (1995).
in the traditional frictionless
literature, as well as existence of a search equilibrium.
In section
and
true at
0,
on our most striking findings -- the link with models
erature.
is
=
w'(y')/w'(y)
=
model with ex ante heterogeneity
matching
and so
z,
<
fy (z,y)-"/Ex fy (x,y) > fy (z,y)-Ex fy (x,y)
that
y.
=
s y (z,y)
false for types z
The only other paper
>
for all y'
is
(i),
Again,
(SCP): For a
our quasiconcavity premise
Finally,
(ii).
By
is
(ii).
any extremum:
than the expected marginal product, or fy (z,y')/fy (z,y)
Ex fy (x,y')/
—
either a 'high' or 'low' type. For z's surplus
is
her surplus
if
< Ex fy (x,y')
then fy (z,y')
proportionately less than the marginal value w'(y), and so
less
into play, for
(Hi) to prove a crucial single-crossing property
z solves fy (z,y)
y, if
Then
have quasiconcave surplus functions under
'low' types
we use
Finally,
comes
(ii)
see that 0's surplus function
increasing for low partners y and decreasing for large y,
by continuity,
because fy (0,y) and
fails,
2,
we summarize the
Our model with
define
relevant results of the frictionless matching
search frictions
and characterize search
we ask who matches with whom.
We
conditions with
some
described in section
equilibria
first
vex equilibrium and optimal matching
is
and
3.
lit-
In sections 4
social optima. In section 6,
derive the conditions that guarantee con-
sets.
illustrative examples.
We
establish the necessity of our
new
This sets the stage for us to define and
explore (both positively and negatively) assortative matching. Section 7 establishes
the existence of a steady state equilibrium.
We appendicize the less intuitive proofs.
THE FRICTIONLESS MATCHING MODEL
2.
Consider a frictionless matching model, with an atomless continuum of agents.
our economy
Everyone
in
x 6
An
[0, 1].
indexed by her publicly observable type, a number
is
agent's type
is
fixed,
and determines her productivity while matched.
Normalize the mass of agents to unity, and
i->
[0,1]
The fraction/mass
[0,1].
assume throughout that L
is
let
most y
of agents with type at
differentiable,
L
the distribution of types be
We
L(y).
is
:
with Borel measurable type density
I.
existence proof alone also requires that I be boundedly finite and positive:
The
<
L
<
<
£{ x )
^
<
oo for
all x.
One view
of our set-up
is
that there
is
implicitly a
continuum of every type of agent, with individuals belonging to the graph
[0, 1],
<i<
When
£{x)},
where
i is
two individuals of types x and y
shall later
AO
need to
is
— agents x and y — are matched together,
purely a function of their types, /
refer collectively to a set of
and twice
when x, y >
0.
It is also
is strictly
=
symmetric, or f(x, y)
differentiable with a uniformly
bounded
:
[0, l]
2
i->
R.
assumptions:
(Regularity Conditions). The -production function f
strictly positive
G
an index number of the type x agent.
they produce a flow output that
We
{(x, i)\x
increasing and
f(y, x), continuous,
first partial derivative.
3
Thus, everyone prefers to be matched with someone rather than to be unmatched,
and, everything else equal, prefers to be matched with higher types of agents.
In the core allocation, wages allocate the scarce resource,
tivity agents.
When
namely high produc-
does positively assortative matching or self-preference obtain,
where everyone matches exclusively with others of the same type?
dition
is
Al-Sup
f
sufficient con-
that characteristics be complementary inputs in the production function:
(Strict
Supermodularity). The own marginal product
increasing in her partner's type.
is strictly
supermodular in the positive quadrant: fxy {x,y)
Al-Sup
of any agent x
>
Equivalently, the production function
is strictly
If
A
>
when x,y >
0.
obtains, then high productivity agents enjoy the highest marginal
product when they are matched with other high productivity agents. In the core
allocation,
matching must be (almost surely) positively assortative.
To
see this,
simply note that any allocation in which some positive measure of agents of type x
match with agents
3
More
of type y
exactly, this follows
^
x admits a simple Pareto-improvement.
from continuity of the
first partials,
If
we
and compactness of
reassign
[0, 1].
such agents to another of her
all
2f(x,y) whenever x
own
+
type, output rises, since f(x,x)
>
f(y,y)
y given Al-Sup. So the unique value-maximizing allocation
^
entails assortative matching, with everyone just
matching with
own
his
type.
Constructing an equilibrium with positively assortative matching provides a useful
benchmark
symmetry
every type x,
=
her match: w°(x)
=
s°(x,y)
f(x,y)
f(x,x)/2.
sum
4( x
since /
an increase
is
>
y)
=
fy( x y)
_
>
symmetric.
like
So
y.
it is
f*(y> y)/ 2
is
Al-Sub
(Strict
Under Al-Sub, the
+
L(y(x))
=
is
social
= A( x
'
y)
-
fy (x,y) ^ fy {y,y)
with
maximum
all
fv(y> y)
as
x
^
y, i.e. for
a
when matched with
value
and coincides with the
matching model when the marginal
own marginal product
agent's
optimum and unique
is
< y2
.
It
follows that
+
if
strictly
is
decreasing
submodular: fxy
<
0.
core allocation entails negatively
Each agent x matches with her
y\
socially optimal allocation.
a decreasing function of her partner's type.
For by Al-Sub, f(x 1 ,y2 )
1.
whenever x\ < x 2 and
Z4,
fv(y> y)l 2
Or, the production function f
assortative matching:
<
negative for
increasing or decreasing in her partner's type y as
Submodularity). An
in her partner's type.
zj,
is
type yields marginal surplus equal to
result obtains in the frictionless
productivity of an agent
L(x)
the output of a partnership
Hence, under Al-Sup, the unique equilibrium of this model has
partner x.
symmetric
~
obtains, then
strictly quasi-concave,
positively assortative matching,
A
define for later use the surplus function for x:
in her partner's
Al-Sup
If
given x, the surplus function
a
We
of the wages of the two partners. If this
x,
For a given
^
4
then we have constructed market wages to decentralize this allocation.
j^ y,
x
'wage' w°(x) for
requires that each agent receive half of the output from
— w°(x) — w°(y) measures how much
with y exceeds the
x
With a market
remainder of this paper.
for the
'opposite' type y(x),
f(x 2 ,yi) > f(xi,yi)
+
there are four agents, z\
where
/(ar 2 ,I/a)
<
z2
<
the allocation in which z\ and z$ are matched and z2 and z$ are matched,
Pareto dominates the two other possible allocations in which these four agents match
in pairs.
As
before,
we can prove the existence of the required core allocation by
demonstrating that the surplus function for each agent x
function of her partner's type, achieving
'The
its
is
a strictly quasi-concave
maximum when
superscript denotes 'zero search frictions'.
her partner
is
y(x).
Remark
Throughout the paper, we
1.
shall generally
maintain either Al-Sup
or Al-Sub. This excludes, for instance, the knife-edge case of additivity, where the
of an agent
own marginal product
p( x )
_(-
g(y)
and
:
for
example, f(x,y)
is efficient.
= max(x 2 y, xy
2
is
Also excluded are production
not monotonic in partner's type
--as Kremer and Maskin
)
—
for
5
(1995) exploit.
A MODEL OF MATCHING WITH SEARCH
3.
In this section,
with search
agents
if
which any allocation
own marginal product
functions for which
=
independent of her partner's type, f(x,y)
is
is
we develop a continuous time,
As per
frictions.
time-consuming process.
agent x uses a rule
like 'only
6
unmatched.
that finding and meeting other
Crucially, with an atomless
match with a type y
surely never match. Rather, everyone
* Action Sets.
mean
usual, frictions
horizon model of matching
infinite
At any instant
continuum of agents,
agent', then she will almost
must be willing to match with
in
continuous time, one
Only the unmatched engage
in (costless)
is
sets of agents.
matched
either
new
search for a
or
partner.
When
two unmatched agents meet, their types are perfectly observable to each
other.
Either
may
and remains so
is
Remark
>
0, i.e. after
both approve,
consummated
it is
St
an elapse time oft with chance e~
.
At the moment
destroyed, both agents instantaneously re-enter the pool of searchers.
2.
In steady state, a
match that
sustain; therefore, to simplify our analysis,
Remark
If
until nature destroys the match. This event occurs with a constant
flow probability 5
the match
veto the proposed match.
3.
Match
dissolutions
is
is
profitable to accept
we disallow match
quits.
one way to maintain a steady-state.
instead assume an inflow of entrants, and that matches are eternal.
moot
for the descriptive theory,
* Preferences.
but
profitable to
is
is
standard
in the
TU
matching
One
Our
could
choice
is
literature.
Agents maximize their lifetime expected present discounted
>
payoffs, using the interest rate r
benchmark
specified later), as the
0.
We
assume transferable
frictionless
utility
model Becker (1973)
is
(in
TU
a sense
and not
For such production functions, matching sets are not convex; some agent may be willing to
match either with y x or y 3 yet unwilling to pair up with y 2 G (2/1,2/3)- This provides a foretaste
of section 6, where we establish that convexity and assortative matching are kindred concepts.
5
,
6
See, for instance,
Diamond
(1982),
Mortensen (1982), and Pissarides (1990).
NTU. As
income of x when matched with y comes from an endoge-
such, the flow
nously determined payoff ix(x\y). These payoffs derive from the output of a match,
+
so that ir(x\y)
=
7r(y|x)
f(x,y) for
Unmatched Agents and
*
function,
Our
i.e.
Jx u(x)dx
is
x and
all
Search.
y.
Unmatched agents earn nothing.
<
Let u
be the unmatched density
£
the mass of unmatched agents with types x G
search frictions are captured by the following simple story.
Were
X
it
C
[0, 1].
possible,
an unmatched individual would meet another unmatched or matched agent according to a Poisson process with constant rendezvous rate p
meet someone who
technically infeasible to
the other individual
is
already matched
is
|/e7C
[0, 1]
is
Our
the mass of those unmatched in Y.
Y
with
whom
x
same type employ the same strategy
A
--or
in direct
is
presumed
equivalently,
is
is
proportion to
beyond
in section 8.
steady state strategy A(x) for agent x G
set of agents
state of this model,
it is
cross-sectional results extend well
we underscore
this (quadratic) search technology, as
measurable
But
p fy u(y)dy. Simply put, the
rate at which one meets others with types in any subset
A
0.
already engaged, and so misses any meeting. Thus, the flow
probability that agent x meets any
* Strategies.
>
willing to match.
7
[0, 1]
(That
specifies a Borel
all
agents of the
follows from our later analysis.) In the steady
time-invariant
8
and depends only on the unmatched density
function u, the payoff-relevant state variable. Next define an agent's matching
M(x)
who
=
A(x)D{y x G A(y)}, the
|
match with
are willing to
(and so by symmetry,
iff
match indicator function
* Steady State.
matches
for every
agents x G [0,1]
flow probability
pu(x)
fM
(
x)
is
5.
x.
set of types y
A
match
=
(x, y) is
whom x
also identify each
1 if
y G M.(x) and
type of agent must exactly balance.
—
The
willing to match,
mutually agreeable
In steady state, the flow creation
i(x)
is
We
x G M(y)).
a: a(x, y)
with
matching
iff
y G
set
-
=
M(x)
with
its
and flow destruction of
The
density of matched
u(x); these agents' matches exogenously dissolve with
flow of matches created by
u(x))
and
otherwise.
unmatched agents of type x
u(y) dy. Putting this together, in steady state for any type x G
5{l(x)
set,
pu(x)
/M(x)
u(y) dy
=
pu(x)
/„*
a(x, y)u(y) dy
is
[0, 1],
(1)
We can express strategies as acceptance sets given an atomless type distribution. With type
atoms, equilibrium existence demands mixed strategies: either probabilistic matching or quitting.
7
8
WLOG, we
restrict attention to stationary
sets. Since no single agent can affect
an alternative acceptance set is optimal at
acceptance
any future state of the economy through her choices,
time t it remains optimal at time t + s.
if
THE DECENTRALIZED ECONOMY
4.
We now
characterize the steady state search equilibria (SE) of this model.
equilibrium requires that
Search
everyone maximizes her present discounted payoffs,
(i)
a match
consummated
both parties
taking
all
other strategies as given, and
accept
it.
Yet this criterion has insufficient cutting power. For example,
(ii)
is
iff
if
everyone
chooses to match with no one, then no one has a strict incentive to deviate. This
outcome does not seem
sensible to us, as
So we also
strategy 'trembles' by others.
doesn't survive the possibility of small
it
insist that
everyone also choose a best
response to any possible strategy one might face at any moment. Equivalently,
in a
SE
all
matches with
strictly positive
1
(ii
)
mutual gains are accepted.
4.1 Search Equilibrium
Let w(x) denote the expected average present value for an
• Value Equations.
unmatched agent
x,
9
assuming x maximizes her expected present value. Similarly,
w(x\y) be the expected average present value for x while matched with
let
SE by
solve for a
defining the implicit equations solved by these
While unmatched, x earns nothing, but
at flow rate p
Bellman
M (i) u(y)cfa/,
-
J"
,
x
=
f
p
w(x\y) —
-±-^
she meets
5,
x enjoys a flow payoff
her match
is
7r(a;|y)
n(x\y)
-
S(w{x\y)
can eliminate w(x\y) from equations
for
w(x) as a function of the equilibrium payoffs
/
X
w(x)=p
f
(2)
and
The average unmatched value
is
y.
- w{x))/r
(3),
,
n
.
(2)
With
—
flow probability
w(x)) jr. Hence,
(3)
and obtain a simple expression
tt(x\-).
— W (X)
n( X \y)
'
,
w
,
when matched with
We
Here,
.
—u(y)dy
destroyed, and she suffers a capital loss (w(x\y)
w (x\y) =
9
.
jr.
r
JM(x)
Similarly,
w(x)
We
values.
and matches with some y € M(x), enjoying a capital gain (w(x\y) - w(x))
w{x)
y.
x
I
\
7
u(y)dy
(A\
(4)
of x equals the rate that x finds matches times the
analogous to the frictionless wage schedule w°.
expected discounted present value of the capital gain over her unmatched value -
where discounting incorporates both impatience and match impermanence.
*
Acceptance
any match that increases one's
In a SE, one accepts
Sets.
expected present value.
(
=>
^ w(x)
w(x\y)
I
(5)
(y<£A(x)
In the borderline case w(x\y)
-k
=
Wage Determination.
since
match output
generally positive:
w(x), y G A(x)
is
neither necessary nor precluded.
Search frictions create temporal bilateral monopoly,
the agents' outside options
less
match surplus s(x,y)
=
f(x,y)
their
(i.e.
unmatched
— w(x) — w(y) >
values),
is
This shifts
0.
the question of wage determination into the realm of bargaining theory.
We
abstain from innovating on this front, and simply follow a
more recently Pissarides
(1990), in
number
of authors,
assuming the Nash bargaining solution, 10
i.e.
the
instantaneous match surplus -- the excess of wages over average unmatched values
(w(x\y)
+
w(y\x))
-
(w(x)
+
w(y))
w(x\y)
So by the optimality condition
in the borderline case w(x\y)
M(x)
—
is
- w(x) =
w(y\x)
(5), this yields
— w(x) =
- w{y)
y G A(x)
w(y\x)
satisfies
(6)
iff
- w(y) =
x E A(y), except possibly
0.
By
definition,
A(x) and
coincide, except possibly in this borderline case.
The
definition of w{x\y)
schedule
from asset value equation
Since the payoffs must
somehow
- w(x) =
and the Nash bargaining
n(x\y)
=
ir(y\x)
(7)
w{x)
+
\(f{x,y)
- w(x) is
+ 7r(y|x) =
schedule:
w(y))
(8)
equal to her flow value from being
half the flow surplus produced by the match.
To describe the equilibrium, observe that
• Synthesis.
Tr(x\y)
w(y)
Nash bargaining payoff
In other words, the flow payoff for an agent x
unmatched plus
-
divide the output of the match, Tr(x\y)
f(x,y), and so (7) yields the equivalent
w(x) as
(3)
(6) yield
7r(x\y)
10
The wage schedule
divided equally.
^
w(x), while
(8) says
n(x\y)
^ w(x)
(3)
implies that w(x\y)
as f{x,y)
—
w(x)
—
w(y)
^
^
0.
See Coles and Wright (1994) for the microfoundations of Nash bargaining in dynamic settings.
0.6
0.4
0.2
Figure
f(x,y)
1:
=
Equilibrium Matching Sets. This depicts the equilibrium
— 50r, 5 = r/2, and a uniform distribution of agents.
xy, with p
y e M(x), then the point (x,y)
Combining
this
with
is
shaded in the graph.
mutual optimality condition.
(5) yields the
where we have used the
fact that y
=
f(x,y)
p
A SE may
whom
(the
(10)
density
it);
and
Proposition
(Hi)
M(x)
is
an implicit system
well-defined, even
— w(x) —
w(y)
=
M);
sets
(ii)
the
+
(w,M,u) where: w
unmatched
values:
u(y) dy
(10)
5)
though we have not specified whether
0.
by specifying:
mass of each
how much everyone's time
is
(i)
who
is
matching with
type searching (the
unmatched
worth (the unmatched value w).
(SE Characterization). A SE can
1
for
-w(x) -w(y)
intuitively be fully described
matching
y £
(9)
2(r
M(x)
y G M(x) when f(x, y)
M(x)
e A(x) and x € A(y) implies y e M(x).
(8) into (4) yields
w(x)
Note that equation
y e
-w(x) -w{y) ^
f(x,y)
Next, substituting
matching sets for
x e M(y) and
If
be
solves the implicit system (10), given
represented as a triple
(M,
it);
M meets the opti-
mality condition (9) given w; and u solves the steady state equation (1) given
Example.
Figure
1
M.
graphically depicts the equilibrium matching sets for the
production function f(x,y)
=
xy
as well as a particular choice of search frictions,
impatience, and type distribution. Since this production function satisfies Al-Sup,
in the frictionless
M(x)
=
{x}.
benchmark agents
are only willing to
As one might expect, with search
10
match with
frictions,
their
everyone
is
own
type,
willing to
accept a range of possible partners. In section 6 we find conditions on the production
function that ensure equilibrium matching sets have approximately this shape.
Value Function
4.2 Properties of the
Before proceeding, we must
Our
first
some basic properties
derive
analysis throughout the paper repeatedly applies the following inequality:
Observation. For any agent x and any
> p
•»(*)
If this
(?)
of the value function.
set
MC
[0. 1],
*(v) dy
j^j
/u
(ID
}
were not true, then the implicit equation (10) would yield a y such that either
y e M(x), y £
f(x,y)
—
Lemma
w(x)
1
—
M. and
>
w(y)
-
f(x.y)
0.
-
w(x)
w(y)
<
0,
or
Either possibility contradicts
(Monotonicity). Given A0.
Intuitively, since higher agents are
the value
y i M(x), y G
(tt)
(9).
w>
is
This leads us to
mcreasing in a SE.
always more productive than lower agents, they
can simply imitate the matching decision of lower agents and do better.
optimize, they will do
The
still
logic underlying
The appendicized proof formalizes
better.
M, and
monotonicity also buttresses continuity:
this
If
they
argument.
Anyone can do
almost as well as a slightly more productive agent, simply by imitating her matching
decision. Consequently, the value function cannot
Lemma
2 (Continuity). Given
A0
;
jump, as proven
the value function
w
is
continuous in a SE. U
We often refer to the derivative of the unmatched value function.
Lemma
3 (Differentiability). Given A0, the value function
tiable in a
SE, and
its
derivative
,
M
W{X)
Monotonic functions are
from
Lemma
1.
When
formula for w'(x)
11
is
is a.e.
w
This
is
a.e.
U
x ^)<y) dy
+ S)+pJM{x) u(y)dy
a.e. differentiable,
the matching set
is
and so the
first
justified:
differen-
.
.
[
^
part of the claim follows
(suitably) differentiable in x, the resulting
a straightforward application of the Fundamental
The appendicized proof
is
given by:
pki{ X)
2(r
in the appendix.
actually establishes that
11
w
is
Lipschitz.
Theorem
of
Calculus: surplus vanishes all along the
we can
safely ignore the effect
differentiate (10)
argues that
(i)
on w'(x) of changes
under the integral
this logic
valid
is still
its
own
right,
we
matching
in the
set; therefore,
set,
sets are
merely continuous and
both these conditions hold
{ii)
is
and simply
appendicized proof carefully
difficult
when matching
Using these lemmata, existence of SE
an important in
The
sign.
the value function Lipschitz at x. and
is
boundary of the matching
a.e. in x.
proved. For expositional ease, and as
defer analysis of this
complex
it
issue to section 7.
CONSTRAINED EFFICIENT MATCHING
5.
In this section,
constrained
we
investigate
dynamic steady
states
where matching decisions are
Everyone matches so as to maximize the global present
efficient:
counted value of output, rather than her own personal value.
a hypothetical social planner
who
is
entrusted with
all
It
matching
dis-
helps to introduce
decisions.
True to
our stated goal, we assume that the planner cannot bypass the search frictions that
We
render matching opportunities infrequent.
look for a stationary social optimum
(SO), the steady state of the optimal dynamic program of a constrained planner.
We
characterize necessary conditions for a SO,
deviations from steady states. That
isting matches, or
(ii)
to
is.
fathom, that for any
we do not allow the planner
model and strategy space, and
But they sidestep a very deep
spirit.
by considering only stationary
employ nonstationary match acceptance
provisos are required by our
initial condition,
issue:
It is
12
ii)
to destroy ex-
strategies.
reflect
our steady-state
theoretically possible,
a nonstationary policy
may
These
if
hard to
beat any station-
ary one the planner could devise. To establish a steady state SE. we safely ignored
nonstationary deviations, as no single agent could affect the future course of the
decentralized economy; since a social planner most certainly can affect the future,
this
is
no longer
WLOG,
and stationary SO existence
Let M*(x) denote the set of
By
agent x.
construction, y
optimal matching
12
set.
and
let
all
agents
G M*(x)
a* be
its
iff
whom
problematic. 13
much more
is
the social planner lets match with
x e M*(y).
We
will call
indicator function: a* (or, y)
M*(-) an agent's
=
1 iff a:
G M*(y).
analogous to the modified golden rule in the economic growth literature. The golden
is the planner's best steady state if she can, at an initial time, costlessly
jump to her prefered steady state, at which she must remain. The two notions coincide for an
infinitely patient planner (r = 0), since the transition path is costless.
This
rule,
13
is
by way of contrast,
Shimer and Smith (1996)
will address
SO
existence allowing for
12
all
dynamic
strategies.
Let
m
matches
2
M+
h->
[0, l]
:
the mass of
is
y)
S. This satisfies a steady state constraint:
£
for (x, y)
J5 m(x,
be the density of matches, so that
6m(x,y)
=
pa*{x,y)u(x)u(y)
(13)
must equal the flow creation
In steady state, the flow dissolution of matches (x, y)
of such matches. Then, being careful not to double-count output, the steady state
average present value of output in the economy
Thus a
pair (M*,
value of output
m; and
M*
m)
among
a
is
SO
only
if
is
JQ JQ
the matching sets
To
relationship (13).
economy remains
solve for a
with respect to
and reach
e,
the arbitrary nature of
^=-1
m = m + efh,
(a la calculus of variations) a pointwise conclusion given
We
fa.
instead recast this as a current-valued Hamiltonian:
is
(
x ,y)
describe the
and
3im(x,y)-
condition, a short-cut to placing multipliers on the constraint that
the Kuhn-Tucker complementary slackness requirements.
p(x, y)
^M
^
In this 'bang-bang' control,
a(s y)
,
when the shadow value
the planner will match them;
(steady state)
(15)
As the density
of
-
5p(x, y)
Optimality only requires that the
=
match of x with y
is
-
is
u(x)
=
£(x)
p J* (a(x, z)p(x,
FOC
z)
— JQ m(x,
+
positive,
,
z) dz,
y)
=
2
r P( x v)-
We
'
we have
a(y, z)p{y, z)) u{z) dz
hold almost everywhere.
13
is
negative, she won't.
order condition determines p: 3i m ( x
unmatched agents x
f(x, y)
of a
when the shadow value
first
14
~
^0^1
[a(x,y)
14
We
match (x,y) (again not double-counting).
[0,1], reflects
2Mm (x,y) =
(14)
the multiplier on the steady state relationship (13), half the
order conditions using the notational shortcuts 3i a
first
The other
differentiate
E
f(x,y)m(x,y)+p(x,y)(pa(x,y)u(x)u(y)-6m(x,y))dxdy
I
planner's value of a
a(x,y) €
can think
Jo
where p(x,y)/2
first
We
m.
at
SO, we could write the planner's problem as a
Lagrangian, insert e-variations on the state variables
The
the present
maximizing the present value of output, subject to the steady state
of the planner
necessary
M* maximize
stationary matching rules, given the initial matched rates
implies that the state of the
^ Jo
\f(x, y)m(x, y) dxdy.
assume they always
(16)
hold.
Let v(x)
= p fM ,( x) p(x, z)u(z)dz = p f
ner's value of a
x
match
we
(x, z),
As
a(x, z)p(x, z)u(z)dz.
p(x, z)
is
unmatched value of agent
interpret v(x) as the social
— namely, the flow rate at which she creates new social present value.
this
with 2IK m
(
x ,y)
=
we can
Combining
rp(x,y) and (16) yields:
p(x, y)
Therefore,
the plan-
=
-v{x)-
(f{x, y)
v(y))/(r
+
(17)
6)
rewrite condition (15) as
(xeM*{y)
f(x,y)-v{x)-v(y)%0=>{
[x$M*{y)
,
with the definition of v yields an implicit equation for the social
Combining
(17)
unmatched
value:
v{x)
=
—
p
r
Jm*(i)
—
-:
+
o
u{y)dy
Proposition 2 (SO Characterization). A SO can
(19)
M*
meets the optimality
condition (18) given v; and u solves the steady state equation (1) given
It
4.
A
may be dominated by
maximum
M*,
triple (v,
nonstationary paths; or
5.
(l).
15
Then
4 (Value Properties). The
strictly increasing, continuous,
and
v'{x)
r
SE
v
not necessarily a SO.
a local, but not a global,
must share the key properties of w:
this
may admit
f
<
0,
SE
unmatched value v
social
+ S + pfM
(Proposition
in fact
,
{x)
5,
nonnegative,
u{y)dy
page 27).
Remark
and
(19).
+
5
This
However, this does not
4 above.
only requires that f
14
is
with derivative a.e. given by
of a triple solving (1), (18),
establish existence of a SO, as explained in
While
may be
a.e. differentiable,
we can prove the existence
follows from existence of a
15
is
(19)
Replacing r by 2f + <5 leaves (18)-(19) identical to (9)-(10), with the
same steady-state equation
Also,
it
and
M*
of the planner's problem.
Remark
Lemma
u) solving (1), (18),
(v,M*,u)
be represented as
where: v solves the implicit system (19), given (M*,u);
Remark
x
(18)
>
0,
which
is still
true.
6.
As seen
in section 2,
DESCRIPTIVE THEORY
supermodularity (Al-Sup) alone ensures that there
positively
is
assortative matching without search frictions in the core allocation/social
- agent
x always matches with another agent
not matching with one's ideal partner
x.
Moreover, the surplus
optimum
loss
<
increasing in the mismatch: If x
is
then x strictly prefers a match with y over
z.
from
y
In the frictional setting, everyone
<
z,
still
has an ideal partner, but since individuals are willing to match with sets of agents,
mismatch
is
the rule in equilibrium or at a constrained optimum.
form that this mismatch takes
match with
z rather
is
quite unpredictable. For example, x
than with y G
z than with another x\
Curiously, the
may
prefer to
— and sometimes would rather match with
(x, z)
This significant violation of positive assortative matching
motivates our new conditions on the production function.
Remark
sets
6.
Throughout
we simply
this section
refer to properties of
matching
— as they are equally true of any SE or SO. For indeed, both SE and SO have
identical value properties,
clarity,
which drive
But
results reported here.
all
our language and notation pertain
to
for brevity
and
SE. Note that while search equilibrium
depends on the exact specification of how surplus
is
divided within a match,
SO
does
not suffer from this ambiguity. This section nonetheless unites these two disparate
concepts, which serves as a further robustness check on our descriptive theory.
6.1
Convexity
First
on our agenda
also agree to
is
convexity: If x
match with y 2 6
x's surplus function s(x,y)
=
match with
Optimality condition
(2/1,2/3)?
f(x,y)
then the answer to this question
willing to
is
— w(x) — w(y)
is 'yes'.
16
is
y\
and y 3
,
(9) tells us
strictly
Section 2 proved that
will she
that
quasiconcave in
Al-Sup
or
if
y,
Al-Sub
ensures a quasiconcave surplus function in the frictionless benchmark model.
Example.
tions.
there
With
is
frictions,
Surprisingly, neither
Al-Sup nor Al-Sub
the supermodular production function f(x,y)
positively assortative
matching
suffices
=
(x
in the frictionless case.
equilibrium matching sets are not convex.
+
with search
2
y)
fric-
in Figure 2,
But with search
Indeed, any x £ (0.20, 0.21)
won't match with her own type, but will match both with higher and lower types.
16
see
Of course, convexity may obtain even if the surplus function is not quasiconcave; however, we
no general proof of convexity that does not rely on quasiconcavity.
15
0.06
0.04
0.02
-0.02
-0.04
0.4
0.2
Figure 2: Non-Convex Matching and Nonquasiconcave Surplus Function.
The left panel depicts the equilibrium matching sets for f(x, y) = (x + y) 2 with p = 40r, S = r/2,
and a uniform distribution of types. All points (x,y) with y £ M(x) are shaded in the graph.
The equilibrium matching sets of any x £ (0.07, 0.21) are not convex. The right panel depicts the
,
non-quasiconcave flow surplus
Also,
=
when f(x,y)
s(0.1,j/) for all
(x
2
+
y)
,
matches with agent
everyone has a strictly (quasi-)concave surplus
maximum
function in the frictionless benchmark, with
matches with another
— (x —
2
y)
.
up.
Indeed, the wage
One might imagine
function in such a
it
x.
way
The example
w°(x)
is
=
when each type x
surplus
2x
2
and surplus s°(x,y)
=
that search frictions simply reduce everyone's value
as to preserve the shape of the surplus function, but shifting
disproves this conjecture.
are clearly not quasiconcave
6.1.1
0.1.
—
for instance,
The
x
surplus functions of
= 0.1
some agents
has two local surplus maxima.
Conditions for Convex Matching Sets
Despite this example, there are restrictions on the production function / that ensure
a quasiconcave surplus function, and therefore convex matching
this section,
we impose
A2-Sup. The
for
all xi
<x 2
A3-Sup. The
for allxi
for allxi
or
Throughout
A2-Sub and A3-Sub below:
production function
log-supermodular:
is
andyi < y 2 fx (xi,yi)fx {x 2 ,y 2 ) > fx {x u y 2 )fx {x 2 ,yi).
,
cross partial derivative of the production function
x
y2
,
y1
<
y2
,
is
of the production function
fx {x 1 ,y l )fx (x 2} y 2 )
x
y2
,
fxy {x u yi)fxy {x 2 ,y 2 )
16
is
log-submodular:
< fx (x u y 2 )fx (x 2 ,yi).
cross partial derivative of the production function
< x 2 andy <
log-supermodular:
fxy {xi,yi)fxy (x 2 ,y2 ) > fxy (xi,y 2 )fxy (x 2 ,yi).
first partial derivative
< x 2 and
A3-Sub. The
A2-Sup and A3-Sup,
first partial derivative of the
< x 2 andy <
A2-Sub. The
for allx x
either
sets.
<
is
log-submodular:
fxy {xi,y 2 )fxy {x 2 ,yi).
A2-Sup* (A2-Sub*) impose
Finally, let
< x 2 and
xi
<
y\
The above assumptions
j/2-
fx (xi,y)/fx (x2,y)
ratio
independent of y
is
C3g(x)g(y), for some constants
some constants
is
,
true
ci,
c2
and
,
c3
fxy {x\,y)/ fX y{%2,y)
iff
c3
=
+
C\
is
C\
+
c 2 (g(x)
and function g
,
and functions g
,
=
form f(x,y)
and
weaker condition asks that f(x, y)
slightly
This
c2
C\,
For example, the
are interrelated.
— and so A2-Sup and A2-Sub both hold
precisely for production functions of the
-
A2-Sup (A2-Sub) when
strict inequality in
:
independent of
[0,1]
+ g(y)) +
R and
c 2 (g(x)
[0, 1]
:
i->-
+ g(y)) +
i—
R.
A
>
c 3 h(x)h(y) for
/i
[0,1]
:
R.
i-»
that A3-Sup and A3-Sub
y, so
both obtain. Thus A2-Sup and A2-Sub jointly imply A3-Sup and A3-Sub, but not
Some
conversely.
of the
most obvious functions that one would write down, such
=
the Cobb-Douglas f{x,y)
{xy)
a
,
satisfy all four
as
weak assumptions.
Proposition 3 (Convex Matching). Posit A0.
Given Al-Sup, A2-Sup, and A3-Sup
(a)
z
G
(0, 1],
and M(0) can
be
;
;
Observe Proposition 3
Proposition 3
That
=
f(L(x),L(y))
let
iff
/ does. This
'units don't matter'
is
M(z)
is
convex for
all
is
convex Vz G
[0,1].
wholly independent of (monotonic transformations of)
is
the type distribution: For instance,
L(x), and
set
chosen convex. With A2-Sup*, M(0) must be convex.
Given Al-Sub A2-Sub, A3-Sub, M(z)
(b)
matching
the
if
we
label each agent
x by her type's percentile
f(x,y), then / satisfies any of the assumptions in
is
comforting, as convexity in
R
is
scale-independent.
a robustness check, suggesting that there are no weaker
conditions on the production function that ensure convex matching sets.
Remark
7.
Extrapolating this
not simply convexity,
types in
6.1.2
We
Rn
.
Since
it
is
(convex coordinate
slices),
and
the scale-independent extension of our theory for productive
adds
little
to our theory,
we have not pursued
this complication.
Proof of Convexity
establish convexity of
strongly quasiconcave.
it is
logic, biconvexity
strictly so except
Theorem
1
matching
As seen
perhaps
sets
by proving that surplus functions are
in figure 3, a function
for a flat global
Given A1,A2,A3-Sup or Al,A2,A3-Sub,
(b)
Given also A2-Sup* or A2-Sub*, s(z,y)
17
strongly quasiconcave
maximum.
(Quasiconcavity). Posit A0 and
(a)
is
fix z.
s(z, y) is strongly quasiconcave in y.
is strictly
quasiconcave in
y.
if
Figure
D
C
B
Quasiconcavity. The surplus function in panel A is not quasiconcave, while those
B and C are quasiconcave, but have flat portions that aren't global maxima. Neither is
3:
in panels
strongly quasiconcave.
That Al-Sup
The
or
surplus function in panel
Al-Sub
D
is
strongly but not strictly quasiconcave.
a key ingredient ensuring a quasiconcave surplus
is
function follows from the analysis of the frictionless model in section
under Al-Sup, the slope of x's
positive
>
x
iff
We
y.
frictionless surplus function,
illustrate the 'importance' of our other
2.
For example,
fy (x,y) — fy (y,y),
assumptions
Example.
f(x,y)
ure
=
(x
+ y)
2, satisfies
The production
(Importance of A2-Sup or A2-Sub)
2
that
(i.e.
they are sometimes necessary conditions) through two examples now, and one
is
later.
function
which generated the non quasiconcave surplus functions
in Fig-
Al-Sup, A2-Sub, and both A3-Sup and A3-Sub. Similarly,
for the
,
production function f(x, y)
=
x
2
+y +x+y—
2
A3-Sup and A3-Sub, the surplus function
Example. (Importance
of
is
xy satisfying Al-Sub, A2-Sup, and
sometimes not quasiconcave.
A2-Sup* or A2-Sub*) Consider the
class of pro-
= xy+c(x+y), which satisfy Al-Sup, A2-Sup, and A3-Sup,
but not A2-Sup*. Putting c = x(^/o~ 2 + A8p - S)/A(r + 5), where x = JQ z t(z)dz is
the mean population type, neatly yields w(x) = ex in equilibrium; thus s(x, y) = xy
duction functions f(x, y)
—
all
i.e.
surplus in
all
Lemma
sure
But type x
matches are weakly agreeable.
matches, and
5 asserts: If the
match with
z as for a
may
elect
=
produces exactly zero
an arbitrary nonconvex matching
own-marginal product of a given type
random match from some
set
M,
is
Choose any
z solve
fy (z,
Then
z
is
Vl )
[0,1]
and subset
L (x,yi)u(x) dx
= Jm fv 7;;
J M u(x)dx
7
uniquely defined, and for
fy {z,y 2 )
G
y\
all
y2
>
^
M
C
Let
[0,1].
,
.
(20)
y\,
(x,y 2 )u(x)dx
< JM fy
IM u x dx
(
18
M.
Assume Al-Sup and A3-
5 (Single-Crossing Property). Posit AO.
Sup or Al-Sub and A3-Sub.
the same for a
then any higher type
has a lower own-marginal product from the match with z than with
Lemma
set.
)
(21)
Remark
=
f(x, y)
8.
One can
+ c2
Ci
verify that (21) binds for production functions of the
+
(g(x)
+
g(y))
obtain, as fxy (xi,y)/fxy (x 2 ,y)
fact, (21) is reversed if
c3
is
h(x)h(y) --
Al-Sup and A3-Sub
Next, for the formal proof of
Theorem
So
y.
Lemma
5
is
quite tight; in
Al-Sub and A3-Sup) both
(or
1,
where both A3-Sup and A3-Sub
i.e.
independent of
we
form
obtain.
give a convenient characterization
of quasiconcave functions that are almost everywhere differentiable.
Q-l.
If o{y)
>
o{x) and o'(x)
Q-2.
If cr(y)
>
a(x) and
A
continuous and almost everywhere differentiable
Lemma
is (a)
a
If
6.
is
<
2/3,
is
with a(y 2 )
Lemma
and
and w'(y)
^
0,
Vx,y.
then y
^x
implies a'(x)
^
0,
Vx,y.
(0,1), this
and
(b) strictly
lemma admits
< min(a(yi),a(y 3 )}.
>
a simple proof.
>
Since a{y )
]
>
a(y 2 ) implies a'(y 2 )
map
cr
[0, 1]
:
— M
i
>-
so under Q-2.
Then
not strictly quasiconcave but Q-2 holds.
is
Step
surplus
is
For fixed
defined for
s y (z, y\) exists
•
1.
Also, s y (z,y)
2.
characterization in
1:
y by
Lemma 6.
and
s(z, y 2 )
by Q-2. This
Suppose, for
<
there exists yi
a(y 2 ) and y 1
N
the surplus function s(z,y)
defined for
Lemma
Lemma
5 with
yi )fy(
>
s(z,
y r ), then
,
j/i
is
is
<
y 2 o'{y 2 )
,
<
a contradiction.
w|
We
M = M.(yi)
M dx
since
/
continuous by
A0
by A0
differentiable
establish quasiconcavity using the
for all z
s y (z, y±)
>
0,
and y x < y 2 (*)
enough
and use
z.
Choose
y\
pJM yi )fy( x iy^ u x dx
2 r + S)+pSmm) u [x)d
{
)
We
show that
with w'(yi) defined.
differentiability
(
holds:
,
under supermodularity. 17
(
>
is
is
Valid For 'High' Types.
for large
x >yi) u ( x ) dx
„
3.
a.e. y,
Namely, we prove that
Conclusion of (*)
JM(
.,_
M'^=
JM
So
a.e.
z,
is
strictly increasing at
Define z as in
Lemma
3:
^ mM
l(
v
(on ,
(22)
< fy (z, y^ - w'(yi) < fy (z, y x ) - to'(j/i) = s y (z, y Vz > z, by Al-Sup.
Step 2: Implication (*) is Valid For 'Low' Types. Fix y 2 > y x From
x )
•
17
implies a'(x)
appendicize the general proof, for which a'(y2 ) need not be defined.
Proof of Theorem
if
^x
cr'(x) is defined,
on
differentiable
by Q-2. Similarly, a(y 3 )
We
defined, then y
strongly quasiconcave under Q-l,
example, that a
2/2
is
.
For the proof of y 2 < yi => s y (z,yi) < 0, one instead shows that the premise s(z,y 2 ) > s(z,yi)
high types in step 1, while the implication is valid for low types in step 2. Proofs under
fails for
submodularity are totally analogous.
19
the implicit equation (10) and inequality (11)
/9
w(y 2 ) - w(yi) >
Divide through by w'(yx) and
w(y 2 ) - w( yi )
the final inequality
is
>
=
fy (z,yi) — w'(yi) >
<
z,
so s(z, y 2 )
is
3:
suppose s(z,y 2 )
=
Every Type
<
z
iff
increasing in z for
s y (z,yi)
Step
M = M(yi)
in the
3.
(i)
integral over y' G [y x y 2 \.
is its
,
- w(y
2)
>
[2/1,2/2])
yi )
Jmyi)
<
as well.
Apply
Lemma
6.
Here, the key ingredient
and divide through by
(20).
{f(x,y2)
-
f(x, yi ))u{x)dx
M {yi )fy{.^y\)u{x)dx
.
<
z then z
<
z also.
So
Under the
basic assumptions, by
If
s(z,y)
>
whom
s(z, y)
>
for all y,
w(z)
=
(10),
could choose a non-convex matching set
yield zero surplus; even that cannot
if
and so
for
is
=
(2/1,2/2),
alone.
1,
some
each agent z
y,
Theorem
convex and
z.
the z
Whether
(ii)
z
1
there
matches
M(z).
from
Lemma
1.
Agent
a positive mass of matches happen to
happen under A2-Sup*,
a strictly quasiconcave surplus function. Finally,
is
z
2/1
Theorem
affect the convexity of
by
any
for
larger than the z associated with
is
the set of agents y for
increasing in y, and so
f{z,yi).
!-
with these boundary types does not
<
-
f{z,y2 )
final inequality in (23)
most two types who produce exactly zero surplus with
s(z,y)
<
1 )
,
has a strongly quasiconcave surplus function.
establishes that
y lt as
Either 'High' or 'Low'.
satisfies inequality (23); if z
Proof of Proposition
yi )
numerator and denominator, and replace y2 by y 2 to get
associated with that pair
If
f(z,y 2 )-f(z,
Under A2-Sup*, the
fy&Vl)
are at
>
appropriately defined. For by A2-
s(z, y{) implies s y (z, y{)
is
f{z,y 2 )-f(z,
So z
3:
fy(z,yi)
& w(y
s(z,2/i)
0.
z,
>
y'
A3-Sup: Integrate inequality (21) over y 2 6
Put
Lemma
f(x, yi ))u(x)dx
true
is
Then
•
f{x,yi))u(x)dx
i
z,
z
-
JM{yi) fy (x,y )u(x)dx
For some such
strict if
from
definition
its
)
Sup, fy (z, y')/fy (z, yi)
is
x '^)
+ 8)+pfU{n) u{x)dx
2(r
Jmyi) (f(x,y 2 -
>
w'{yi)
By A2-Sup,
/m(j/1 )(/(
if
for
then each agent has
A1,A2,A3-Sub obtain, s(0,y)
is
trivially strictly quasiconcave.
Example. (Importance
of
A3-Sup or A3-Sub) Finding examples
of production
functions associated with non-quasiconcave surplus functions obeying Al-Sub and
20
A2-Sub
(or
Al-Sup and A2-Sup)
every type
may be
verify this.
For instance, step
increasing below
is
rather problematic. This difficulty arises because
either 'high' or a 'low', even
1
yielded strict inequality, so the surplus function
Likewise, any failure of convexity
z.
the center of the acceptance set,
One production
+ 260x +
2378 - 248x +
f(x,y)
2378
+
2627
- 746x
Al-Sup and A2-Sup
+ 480xy
756y + U96xy
- 248y + U96xy
- 746y + 4500x|/
756x
must therefore be
260?/
0.5
<
z2
fxy (z 1 ,z l )fxy (z 2 z 2 )
,
,
- fxy (z u
z2
f =
function also does not satisfy A0, because
G
[0,
1/2]
if (x,
y)
G
[0,
1/2] x (1/2,
S
=
r,
equilibrium,
p
=
all
[0,
1/2]
if (x,
y)
G (1/2,
1]
x
if (a;,
y)
G
1]
x (1/2,
480 4500
it is
x
y)
(1/2,
[0,
1]
1/2]
1]
A3-Sup, consider any
-
1496
2
= -78016 <
not differentiable
when x
or y
z\
<
0.
The
is
1/2.
can be restored, however, suitably smoothing out the creases.
Differentiability
If
a hole in
is:
if (x,
see that this production function does not satisfy
To
is
which we have not been able to produce.
function that satisfies
2626
analytically impossible to
if it is
and the type distribution
156r,
is
uniform, one can prove that in
matches are acceptable, and that
w(x)
=
( 1056.62
+
470. 25x
if
x G
1409.63x
if
x G (1/2,1]
[0,
1/2]
<
[586.92
+
Most importantly, the equilibrium surplus functions
minimized when x matches with
0.5,
of types x
G
[0.437, 0.438] are
and hence are not quasiconcave.
Bound Functions
6.1.3
Nonempty, convex matching
bound functions a,b
:
sets are
[0,1] i-> [0,1],
almost fully described by lower and upper
namely a(x)
=
inf{y y G M.(x)} and b(x)
=
|
sup{y y G M(x)}. Bound functions do not encode whether 'boundary' matches are
|
consummated
(e.g.
whether x matches with a(x)). But since the boundaries have
zero measure, value functions and
Also, by continuity of /
Remark
sets are
9.
unmatched
and w, boundary partners are
Under A0, we impose with
nonempty.
If
rates are unaffected by this choice.
ever
1,
or provide zero surplus.
trivial loss of generality that
M(x) — 0, then w(x) =
21
or
by
(10).
For x
matching
>
0,
this
contradicts
0,
Lemma
impossible. Next,
is
i.e.
for
may
=
0,
is
quasiconvex and
is
convex and
are
sets
quasiconcave.
b
some x x < x 2 < x z such that b(x
>
x )
>
y
y,
also true that y
it is
>
/(0,0)
on the equilibrium.
effect
>
not quasiconcave then there exists y with {x\b(x)
choose y large enough that
=
then s(0, 0)
(Matching Set Bounds). Assume matching
2
nonempty. Then a
Proof. If b
w(0)
if
G M(0) with no other
and we can adopt the convention
Theorem
xi
and
1,
y} not convex,
b(x 2 ), and b(x 3 )
G
M(x
x )
and y G
G M(y),x 2 £ M(y), x 3 G M(y), violating convexity. Similarly
>
y.
M(x 3 ).
We
So
for a.
Next, quasiconcave surplus functions imply continuous matching set bounds:
Theorem
(Matching Set Continuity).
3
A1,A2,A3-Sub, a and
on an interval only
Proof.
Suppose a(y)
A0 and Lemma
be
flat at its
So w(a)
=
by
value
a G
or
is
on that
1
(0, 1) for all
y G
G
[yi,
—
2, s(a, y)
maximum
=
continuous on (0,1), and a bound function
b are
if its
for all y
>
0.
Then
[2/1,2/2]-
45° line, a discontinuity in a
who
A
6.2 Assortative
6.2.1
bound function
are indifferent about
Theorem
continuous by
is
at
x G
(0, 1)
x,
1), s(a,
•)
is
nonpositive.
argument establishes that
similar
matching with
of Assortative
Because matching sets
M are not
definition of assortative
matching
either-or
Observation. Given
and y 2 G M(a;i), then
since s
when
b is
reflected across the
corresponds to an interval
which we just ruled out.
Matching
The Meaning
Al-Sub imposes an
constant
y 2 }. Since a's surplus function can only
(by strong quasiconcavity,
(10), contradicting a
is
interval.
not constant. Finally, since matching sets are symmetric
of types
Given Al,A2,A3-Sup or
Posit A0.
Matching
singletons with search frictions, an appropriate
is
not a priori obvious. For example, Al-Sup or
form of assortative matching
are agents x\
either y x
< x2
,
y\
<
£ M(xi), or y 2 G
y x G M(xi) and y 2 G M.(x 2 ), then either y x G
22
y-i-
even by Figure
Under Al-Sup,
M(x 2 ),
M(x 2 ),
satisfied
or both.
if
yx G
2.
M(x 2
Under Al-Sub,
or y 2 G M(xi), or both.
)
if
A
-9—9
2/2
C
B
2/2
—
*"
9
2/2
i
~
Figure
4:
positively
i
2/2
+ -9
Vi
2/1
i
2/i
" Xi
X~2
Assortative Matching.
i
X2
X2
Xi
Panels
A
X3
and B depict graphically the
negatively assortative matching, respectively.
and
-?
i
i
If
definition of
the pairs indicated by
filled
dots
Panel C depicts the proof of
match, then the pairs indicated by hollow dots match
then so must middle types
some
agent,
Lemma 4. If low and high types (x\ and x 3 ) match with
(x 2 ), given positively or negatively assortative matching.
as well.
To
0,
see why, note that y x
respectively.
G
M(x 2 )
and y 2 G M(xi) imply s(x 2
,
2/i)
>
and s(x u y 2 )
>
Summing yields f{x 2 ,yi)+f(x u y 2 ) > w(x )+w(x 2 + w{yi) + w(y 2 ).
)
1
+
Also Al-Sup implies f(x u yi)
>
inequalities yield s(a;i,i/i)
> f{x 2 ,yi) + f(x u y 2 ). Together
f(x 2 ,y 2 )
>
or s(x 2 ,y 2 )
0,
these
or both, as required.
0,
This observation speaks to a weaker notion of assortative matching than what
we
(3
intend.
To formalize our
V J3' the maximum,
idea, let
of vectors
matching indicator function a
given matching pairs
if
(3
=
(3
A
(3
and
(3'
Call matching positively assortative
f3'.
is affiliated:
(x, y)
and
(3'
be the component-wise minimum, and
=
18
a0A(3')a0VJ3') > a0)a0')
(x', y').
Matching
the reverse (negative affiliation) inequality obtains.
Definition. Let x Y
< x 2 and
y\
M(x 2 and
y 2 G M(xi) imply y 1 G
assortative
if
)
yi
G M(xi) and y 2 G
<
y2
Matching
.
M(x 1
M(x 2
)
These natural definitions are depicted graphically
M(x 2 and
)
in panels
A
is
frictionless
(resp. opposite) types.
matching ones, as given
Our new
if iji
E
negatively
y 2 G M(xi).
and B of Figure
Positively (resp. negatively) assortative matching describes a preference for
with similar
any
Equivalently,
M(x 2 ). Matching
imply y x G
for
the
negatively assortative
positively assortative
is
and y 2 G
)
19
is
if
4.
matching
definitions generalize the traditional
in section 2:
The presumed contrary matches
don't exist with singleton matching sets, and so the definitions vacuously hold.
6.2.2 Characterization of Assortative
With
18
Matching
assortative matching, matching sets are convex,
Milgrom and Weber (1982)
is
and bounds
a, b well-defined.
the classic (auction-theory) economic application of this concept.
it has the advantage of easily extending to discrete types: The
is immediate.
probability-of-matching function must be affiliated. Likewise, the extension to
19
This abstract formulation of
23
Theorem 4
(Assortative Matching and Convexity). Given
M(x)
atively assortative matching,
Proof.
The two
Take any X\
M(x 2
)
Figure
G
< x 2 < x 3 and
y2 G
6
M(x 2 ).
0, there exists
4,
then x 2 G M(y') and x 3 G
y'
<
imply x 2 G
)
y2
M(y2 ).
,
like
(0, 1)
implies
M(x) convex for
all x.
20
assume positively assortative matching.
M(y2 ) and
with Xi G
[0,1],
7^
M(y2
Vx G
cases are symmetric, so
assortative matching. If y'
xi
/
positively or neg-
If y'
M(y2
>
y2
M(y2 ), assuming
imply x 2 £
)
Since
the point y 3 in panel
like
,
M(y 2 ).
x3 G
C
of
positively
the point y x in panel C, then x 2 G M(y') and
=
if y'
Finally,
y 2 obviously x 2 G
,
D
M(y 2 ).
This very close relationship between our two notions of matching set cohesiveness,
convexity and assortative matching,
further evidence that
is
we have indeed found
the appropriate definition of assortative matching. And, like convexity, assortative
matching
independent of the type distribution.
is
Assortative matching implies convexity (Theorem 4) and thus well-defined
functions (Theorem
tative
Lemma
1
To formalize a
if it is
weakly monotonic, and
7 (Assortative
between the parallel notions of assor-
strictly so
Posit convex matching sets. If a and
decreasing and
1
call
a function into
when valued
[0, 1]
strongly
in (0, 1).
Matching and Monotonic Bound Functions).
G M(l), then matching
(b)
link
matching and monotonic bound functions,
monotonic
(a)
2),
bound
is
b are
positively assortative,
G M(0), then matching
is
G M(0), and
strongly increasing,
while
if
a and
b
are strongly
negatively assortative.
Conversely, under A0, a and b are nondecreasing (nonincreasing) with positively
(negatively) assortative matching.
Part
(a)
captures one interpretation of our research program, which
set convexity plus Becker's condition
Al
6.2.3 Conditions For Assortative
We
yields assortative matching.
Matching
Theorem 4 thus
implies that matching
is
not necessarily assortative in
environment; however, the assumptions that ensure convex matching sets are
almost sufficient for assortative matching.
20
that matching
have shown matching sets are not necessarily convex when there are search
frictions.
this
is
While the matching
From Lemma
1,
set
assumption
is
We
state our
main
descriptive result:
endogenous, this theorem makes almost no assumptions.
we know that under A0, w(x) >
24
and so M(x)
^
when x >
0.
0.8
0.6
0.4
0.2
0.4
0.2
Figure
0.6
Low Types Match with High Types, But Not with Low Types.
5:
This depicts the equilibrium matching sets (with (x, y) shaded iff y £ M(x)) for f(x, y) = x + y+xy,
with p = 50r, S = r/2, and a uniform type distribution. The lower bound a(x) is falling on (0, 0.4).
Proposition 4 (Assortative Matching). Posit A0, and
(a)
Given A1,A2,A3-Sup and f(0,y)
(b)
Given Al,A2,A3-Sub, there
Proof of
(Proposition
3).
when valued
or
(Theorem
1
Then
M(0)
is
since b
=
k for
Lemma
increasing.
Proof of
(b).
all y.
>
=
0.
y implies
1
G M(0).
It
all
>
is
is
1)
implies
2), it is
3),
0, for if
satisfies
>
0.
+
+
xy.
y
6(1)
To prove
decreasing in
y,
and so M(0)
2), it is
search frictions,
is
strongly
w(0)
=
M(0)
is
=
and
0, s(0, 0)
=
k)
>
and that s(0,y)
0;
from
nonempty; s(0,y) increasing
6(0)
=
1.
2),
Lemma
Then
since a
and
b
in
are
and strongly monotonic
7.
Consider a production function that
some agents are
25
is
this follows
=
0,
and /(0,y) >
Figure 5 illustrates the matching sets for one such function, f(x,y)
With
=
we use the assumption
Al-Sup, A2-Sup, and A3-Sup, but that has /(0,0)
for y
x
of f(0,y)
constant only
G M(l), whence
sets are convex,
they are strongly decreasing. Apply
Example. (Importance
b,
are convex
strongly increasing.
quasiconvex and quasiconcave, respectively (Theorem
(Theorem
1
quasiconvex (Theorem
matching
follows that a(l)
is
M(0)
positively assortative.
a contradiction. So
0,
matching.
nonempty. Then s(l,y)
is
strongly increasing,
is
Since a
A1,A2,A3-Sub imply
>
M(l)
Theorem
7 implies matching
for all y
0,
This implies s(0,y)
increasing in y. Quite easily w(0)
s(0,y)
sets except possibly
As w(l) >
3).
G M(0), and a
G M(0), and a(0)
convex,
matching
quasiconcave (Theorem
is
convex,
that f(0,y)
all
of the proof of
1
0.
k, there is positively assortative
Hence they are described bound functions a and
increasing in y (step
1.
>
k
negatively assortative matching.
is
From A1,A2,A3-Sup,
(a).
=
let
willing to
match with
0,
=
since
f(0,y)
>
s(0,0)
<
when y >
By
0.
therefore, (10) implies that w(0)
0;
continuity, very low types won't
which we state here
set boundaries,
for
Corollary. Posit A0. Bounds a and
/(0,
In
=
•)
match with each
/(0,0)/2, and so
other.
of Proposition 4 establishes an important characteristic of
The proof
and
=
>
k,
matching
emphasis.
strongly increasing given
b are
A1,A2,A3-Sup
and strongly decreasing given A1,A2,A3-Sub.
summary, we have found a powerful generalization of negatively and
tively assortative
matching to a
Under
frictional setting.
posi-
the three submodularity
assumptions, matching sets are decreasing, convex, and continuous, while under the
and
three supermodularity assumptions,
agents have a zero
the proviso that type
marginal product, matching sets are increasing, convex, and continuous.
6.3 Ideal Partners
In the frictionless world of section 2, every individual x only matches with her 'ideal
partner', that type y
means that
who maximizes
this ideal partner
s°(x, y). Assortative
graph
is
matching without
By way
increasing.
frictions
of comparison, let's
define the set of ideal partners p*(x) of type x, those partners yielding the greatest
surplus:
y*
G p*(x)
s(x,y*)
iff
>
s(x,y) for
all y.
Continuity of the ideal partner
correspondence and quasiconcavity of the surplus function (and hence convexity of
matching
sets) are closely linked, further reinforcing
Theorem
(a)
5 (Ideal Partners
Given s(x,
•)
our approach.
and Quasiconcavity).
Posit A0.
strongly (strictly) quasiconcave for all x,
21
p*
=4
[0, 1]
:
[0, 1]
is
nonempty, convex-valued and upper hemicontinuous (single-valued and continuous).
Conversely, assume p*(x)
(b)
(single-valued
and /(l,
Proof of
is
•)
=
(a).
is
nonempty, convex-valued, and upper hemicontinuous
and continuous) for
k, s(x,
is
•)
Clearly, p*
compact. Since the
all x.
Given Al-Sup and /(0, )
strongly (strictly) quasiconcave for
^
set of
as s
is
continuous by
A0 and Lemma
Since A1,A2,A3 imply each surplus function
clusions of the first part of
Theorem
26
is
2,
Al-Sub
and
[0,1]
convex, p*
is
For strictly
continuous.
strongly quasiconcave (Theorem
from these primitive assumptions.
is
5 also follow
is
Maximum Theorem.
quasiconcave surplus functions with single-valued maximizers, p*
21
k, or
all x.
maximizers of a quasiconcave function
convex-valued, while u.h.c. follows from Berge's
=
1),
the con-
Proof sketch of
(b).
We
prove in the appendix that everyone
Then we show that under Al,
partner.
- -
violation of quasiconcavity
instance, under Al-Sup,
if
x
y*
a local
is
^
x prefers
ideal partner correspondence
under Al alone, an intuitive generalization of assortative
section
Strong monotonicity of p*
2.
of s(x,y)
—
(strict
y'
^
a
For
O
y* to y*.
weakly monotonic
is
frictionless
monotonicity, except
matching from
when valued
or
however, obtains only under our stronger assumptions Al,A2,A3.
1),
Theorem
(Monotonic Ideal Partners).
6
(a)
Given Al-Sup (Al-Sub), p*
(b)
Given Al,A2,A3-Sup (Al,A2,A3-Sub), p*
Proof of
Let x x
(a).
< x2
,
with y x
>
s{x u y 2 ) and s(x 2 ,y2 )
f{x\,y 2 )
+
f(x 2 ,yi). So y x
We
prove part
strict given
(b) in
<
Posit AO.
nondecreasing (nonincreasing)
is
s(xi,yi)
is
minimum
then y* can never be someone's ideal partner.
prefers y' to y*, then x'
we show that the
Finally,
if
someone's ideal
is
>
<E
is
strongly increasing (decreasing).
s(x 2 ,yi).
definition of p*,
Summing yields f(xi,y 1 )+f(x 2 ,y 2 >
)
y 2 under Al-Sup, and yx
>
y 2 under Al-Sub.
the appendix, showing that one of the above inequalities
A1,A2,A3. Note that a
still
stronger claim, p*
doesn't obtain under any reasonable assumptions:
increasing surplus function under
so do nearby types; hence p*(x)
The
monotonic,
strictly
is
highest (lowest) type has an
A1,A2,A3-Sup (A1,A2,A3-Sub), and by
=
{1} for an open interval of types near
continuity,
1
(near
return to the question of the existence of SE.
Proposition 5 (SE Existence). Given AO and Al-Sup or Al-Sub, a SE
The
0).
SEARCH EQUILIBRIUM EXISTENCE
7.
We now
By
p*(xx), y 2 G p*(x 2 ).
technical difficulties with both a continuum of agents
exists.
and a continuous
strategy space are well-known. For the larger picture for our proof, note that as a
matching-theoretic model, goods are individual-specific, and insignificant (measure
zero) changes in
to
match with
itself.
22
sets
can have drastic consequences:
x, her fortunes are sunk.
players' actions,
into
matching
we look
To embed the
for a fixed point of the
If
requisite
map from
everyone opts not
anonymity into the
players' value functions
Since values here play an analogous role to prices in Walrasian models,
22
While a slight shift in all values may still lead everyone not to match with some agent x, those
matches must have furnished little surplus, and this must only slightly affect the value of x.
27
Even though a search
this corresponds to the usual equilibrium in price space.
equilibrium
encode
all
But
is
a
triple,
our value function program works because values essentially
needed information
in search-theoretic
analogue
for
matching
8).
Much more
present with no
resolution constitutes a major
its definitive
must be continuous
continuously follow from value functions
(Lemma
is
u.
Since matches must be sought out in a time costly
contribution of this paper.
fashion, their availability
M and unmatched densities
matching models, an additional hurdle
Walrasian setting, and
in the
sets
is
subtle, however,
determine unmatched measures
is
(Lemma
That matching
in the values.
sets
not hard to show given the right space
the fact that matching sets continuously
-- our 'fundamental matching lemma'.
9)
This generally proves the key continuity result for the quadratic search technology.
Lemma
w
'""""
Lemma
8
~~~~~^ ~ - "
a
*"
~
~
9
~~ ""
"*
„
u
A
Existence of Search Equilibrium:
Recipe. The two maps: value
and then a to the unmatched density u, are well-defined
and continuous, respectively by lemmata 8 and 9; the composite map w h-» u is thus well-defined
and continuous. Our fixed point proof proceeds using the value function .best-response map T.
Figure
6:
functions
w to match indicator functions a,
Remark
A
10.
plausible alternative approach, taking the limits of finite agent
approximations, would entail studying a largely different model -- for then mixed
strategies
for
would be needed
in equilibrium.
We thus develop the general methodology
continuum agent search-theoretic matching models that can be applied elsewhere.
Remark
There are a few previous existence theorems
11.
agent two-sided matching search models.
Some assume
heterogeneous
for
for simplicity
an exogenous
type distribution, and thus entirely eschew the trickier aspects of search theory. All,
however, take advantage of threshold matching sets to prove existence directly with
them. In this class of existence proofs, Morgan (1995) and Burdett and Coles (1995)
are the best.
It
must be underscored that with such
our fundamental matching
By Lemma
2,
matching
sets
9), as it
we consider value functions
continuous functions on
of
lemma (Lemma
the weak topology:
were,
is
easily verified.
as elements of the space C[0,
with the sup norm:
[0,1]
M, we work
w
easily structured strategy sets,
\\w\\
=
sup l6
[
0)1 ]
|w(x)|.
1]
of
Instead
with the associated match indicator functions a, and
a € ^([O,
2
l]
),
with norm
28
||a:||,ci
=f f
\a(x,y)\dx dy
<
oo.
Lemma
8.
AO and Al-Sup
Posit
a(w) from value functions
This result
Sub
needed
is
is
proven
to
Any
or Al-Sub.
match indicator functions
in the
appendix.
It is
map w
is
the only place where
must be
verified
Next, unmatched densities u are given a weak topology,
9
unmatched density implied by
for the quadratic search technology
The proof of this
result
the steady-state equation
existence of a solution u
To prove continuity
(IFT). For this,
is
most
(1),
—
in fact positive definite,
IFT,
Tu
for
is
—
for
= JQ
||u||,c,i
0.
A
\u(x)\dx.
u(a) from match
h->
and continuous.
nontrivial, but its underlying idea
=
basis.
the steady-state equation (1)
both well-defined
it is
fixed point
is
not. Consider
argument shows the
natural to apply the Implicit Function
derivative r„
its
or Al-
u(a) for each a, while ad hoc reasoning proves uniqueness.
of u(a),
T and
is
say T(a, u(a))
we happen upon the remarkable
is
on an ad hoc
(Fundamental Matching Lemma). The map a
indicator functions to the
Al-Sup
a(w). To apply our proof methodology
i->
elsewhere, this no knife-edge atoms property
Lemma
h-»
continuous.
an atom of zero surplus matches
for existence. Either rules out
that would preclude a continuous
map w
Borel measurable
must be continuous, and F u
fact that
and hence
Theorem
invertible.
Here
with our quadratic search technology, T u
invertible.
As
it
so happens,
never simultaneously invertible and continuous
we
can't apply the
(later, in
Remark
13).
So we instead take inspiration from the IFT contraction-mapping proof, and modify
it
for
Proof of Proposition
Point
•
Given values
5.
Theorem program
Step
glV6n
1:
Y
where vw
9,
is
in §17.4 of
w
in C[0,
the
and u
W
(
\
""P/
1],
we
fixed point of the
u(z)dz
mapping
).
Consider the
map T
:
C[0,
1] i->
C[0,
1]
lmax (/( a: >?/) -w(y),w(x))uw (dy)
2(r
is
2
follow the Schauder Fixed
+ S) + pu
unmatched measure implied by the value function w,
= J
l]
Stokey and Lucas (1989).
The Best Response Value:
t
and
£ 2 ([0,
our context with convex but not open neighborhoods of a in
the mass of
Tw = w
is
unmatched
agents.
By
as in
Lemmas
Proposition
1,
8
a
a SE. 23
23
This conclusion is also true of two much more natural candidate best response mappings:
Tw(x) equal to the R.HS of (10) or the implied value of w{x) from solving (10). But neithermapping is closed on a small enough family S-
29
To
C
bounded convex subset 3
closed,
T(S)
C[0,
1]
such that
Step
The Family
2:
< w{x) <
satisfying
=
where k
clearly
if
w
E
T
y), as in the
sup y f{-,y)], then so
[0,
since
So
x.
•
T]
since f{x 2 y)
,
if
Step
>
To
3,
then
f
p
Tw.
so that for
< f{x 2 ,y)-w{y)
G
all
x 6
on
)n for x 2
[0,1]
> x1
any
for
y,
,
is
1]
that
< x 2 and
xi
,
max(f{x,y)-w{y),w{x))
weak monotonicity
for
x2 >
of
X\.
max(/(x 2 ,y) - f{x u y),w{x 2 - w{x x ))uw {dy)
)
- f(x u y) =
Tw
x
w{x x ) < w{x 2 ) when
integral over y, establishing
Tw
f** fx (x, y)dx
+ 5)+pu
<
shares the Lipschitz
{x 2
-
xi)k, and
bound k
as well.
w{x 2 ) - w{x x ) <
This establishes
3-
[0, 1]
and
w
see this, first note that |max(,4,
|r
Also, since
EQUICONTINUITY of T(S): We prove
3:
-x
w
Lemma 2. This subset of C[0,
convex. We can also immediately verify
2{r
by assumption,
w €
{x 2
max(A, B) - max(C, D) < max(A - C,B - D),
Tw{x 2 ) — Tw{x\) <
— Xi)k
is
is its
1
that
3; {ii)
proof of
nonempty, closed, bounded, and
Tw. Now,
{x 2
€
continuous as an operator.
is
< w{x 2 ) - w{xi) <
sup y f{x, y) and
snp xy fx {x,
nondecreasing in
Then
w
6 3 when
Let 3 be the space of Lipschitz functions
9:
by assumption f{x 1 ,y)-w{y)
is
Tw
(i)
an equicontinuous family (which by the Arzela-Ascoli Theorem, yields the
is
compactness of T(S) needed by Schauder); {Hi)
•
mapping T, we need a nonempty,
establish the existence of a fixed point in the
mW - t«*)\ <
? f°
£
-
3, if \x
that for
<
x'\
77,
all e
then \Tw{x)
B) - max(C, D)\ < \A-C\
mx
-s
y)
'
<sup|/(s,y)
-
W
™ll
f{x',y)\
+
>
0,
-
there
is
Tw{x')\
+ \B-D\
an
<
implies
e.
24
~ mWI) vMy)
\w{x) -w{x')\
y
Now, /
is
uniformly continuous, so we
sup y \f{x,y)
\w{x)
•
-
Step
24
Here
—
w{x')\
4:
is
f{x',y)\
<
sn\o
y
is
may
fx {x,y)\x
-
x'\.
Continuity of T: This
is
implies that
equicontinuous.
an appendicized algebraic exercise.
a proof of this claim: Assume without
the result
is
immediate.
If
C<
=A-D
A<D - B =
30
A > B. Then if C > D,
D, we consider two more
< A - C = \A - C\. Second
\D - B\.
loss of generality that
assume A > D. Then |max(^,B> - ma,x(C,D)\
< D. Then |max(A, B) - max(C, D)\= D -
cases. First
A
< sup y fx {x,y),
This proves that T(S)
is
enough to x so that
x' close
arbitrarily small. Also, w'{x)
|max(i4, B) - max(C, D)\ = \A-C\, and
assume
choose
CONCLUSION
8.
The model
• Robustness.
such,
to investigate
we have focused on the quadratic search
establish existence of a search equilibrium.
to
any anonymous search technology: By
meets others
as
we can rigorously
our core descriptive theory applies
we mean that the
= fQ
=
rate that a searcher
u(y)dy of unmatched agents, as
fully
is
symmetric). Just
expect that our descriptive results obtain in a model with
like
essentially double the notation,
Summary.
it
of agents in the economy, with
'class'
types within this class (and hence the production function
do Becker's, we
may
p/u).
have assumed for simplicity only one
two distinct classes of agents
•
Still,
this
on the mass u
does in the linear technology (where p
many
technology, where
proportional to their presence in the unmatched pool. This rate
is
well depend, for instance,
We
25
the one that can be fully solved. As
is
men/ women. But
workers/firms or
and
is
therefore
left
this
would
as a future robustness exercise.
This paper has pushed Becker's matching insights into a quite
Exploiting the implicit integral equation (10), we gave
plausible search setting.
a thorough characterization of cross-sectional matching patterns in the equilibrium
and constrained
social
optimum
of a frictional matching model. In so doing,
we have
developed important new techniques for establishing existence, as well as extended
the reach of the supermodularity research program into a wide
new
arena.
APPENDICES: OMITTED PROOFS
VALUE FUNCTION PROPERTIES
A.
• Monotonicity:
From equation
Lemma
Proof of
(10)
and inequality
First
1.
(11), for
w{x 2 ) - w{xi) > p
Im(x
)
prove strict monotonicity.
=
by
(24).
+
2 (r
)
)
u (y)dy
< x2
any x\
w(x 2 — w(xi). As f(x 2 ,y) > f{x\,y)
Now we
establish
But then w(x 2 )
=
25
u{y)dy
=
w(xi)
(24)
6)
for all y
w(x 2 )
If
weak monotonicity.
,
—.
Jm( Xi
Solve for
we
>
0,
w(x 2 ) — w(xi) >
w(xi) for some x\
=
by
(10),
< x2
whence w(x)
,
—
0.
then
at
0ur Fundamental Matching Lemma in particular exploits a positive definite operator, and thus
need not pursue a very difficult and messy analysis using perturbation theory for linear operators.
31
x G
all
for
•k
[0,
by weak monotonicity. Inequality (11) with matching
£2],
defined since fx
=
Q
.
.
— W(Xi)
<
W(X
2J
{
K
,
pJy
l
z3
D(B,C) be
Let
D(B, C)
=
such that
(x)
s(x,x)
1:
M+
is
e (B, C),
M
Lemma
-
Xi,
we have
x{),
M+
0,
<
s(x,y)
a.e. x.
<
and so
s(x, y)
and
We now
—>
>
M+
First,
for all y,
a contradiction.
D
X\).
M+
d and \c-c'\
there
is
<
neighorhood, D(M(x), M(x'))
continuous at
for all n,
>
for all e
(x, y),
so
,
:
= {y\s(x, y) > 0}.
B and C: namely,
(x)
d) G (C, B), with \b-b'\
if
:
.
Define a correspondence
3.
compact-valued. Take any sequence (x n ,y n )
<
d}.
a neighorhood
e.
nonempty- valued:
is
w(x)
—
by
M+
G M + (x n
prove that
with y n
as well since s
is
then
(10);
u.h.c.
is
)
if
and
for all n.
A0
continuous by
M
M
+ (x)
+ (x), establishing u.h.c. Fixing x = x for all n,
Thus, y G
n
+ is compact- valued.
+ (x) C
closed too. Also,
[0, 1] is bounded, whence
and
is
x2 >
(11),
w(x 2 ) — w(xi) < k(x 2 —
Lipschitz:
3(6',
in that
= for some x, then
= f(x,x) — 2w(x) > 0,
,
well-
is
7:
continuous at x
is
if x' is
is
Then s(x n y n ) >
k(x 2
This
))u(y)dy
1
7.
Proof of
inf{rf|V(6, c)
Step
•
M+
x,
<
f(xi,y)
the Hausdorff distance between sets
Call a correspondence
around
)
0.
for all
and
Also, by (10)
the set of nonnegative surplus matches,
[0, 1],
Then
set.
>
2]
k{x 2 - x x ) p JM{x2) u{y)dy
{f(x2,y)-f{x u y))u{y)dy
-—
< —7
T7
2(r + 6) + p J
u(y)dy
2(r + 5) + pJM{x2) u(y)dy
M{x2)
—.
• Differentiability:
[0, 1]
8))
M = [0,x
X2)
t
not only continuous, but
is
1 ).
-
w(xi), and using f(x 2 y)
'-
)
w
-
-x
fz {z,y)dz < k(x 2
+
(2 (r
set
= max^ fx (x, y).
Define k
2.
^^
w(x 2 ) - w{ Xl ) <
Solving for w(x 2 )
,
pfM{X2) (f(x2,y)- f(xi,y)-w{x 2 + w(x
.
,
Lemma
p J^ /(x 2 y) u(y) dy/
a continuous function on a compact
is
f{x 2i y)-f{xi,y)
So
>
x 2 yields a contradiction: w(x 2 )
Continuity: Proof of
2
Lemma
2.
M
We
e w (e),
call
if
M
x an e-continuity point of a correspondence M, and say x belongs
for all x' sufficiently close to x,
and compact valued, Theorem
our nonatomic density u) for
for all
n
contains
=
1,
a.e.
2
,
x as
a.e.
well.
£>(M + (x),M + (x')) <
•
Step
2:
D(M(x),M(x')) <
e.
Since
M+
is
to
u.h.c.
I-B-III-4 in Hildenbrand (1974) implies that (given
all
e
>
0, a.e.
x
is
an £-continuity point of
M+
.
Then
x G C M +(l/n), and the countable intersection n n C M +(l/n)
That
1/n. So
is,
for a.e. x, for all n, if x'
M+
is a.e.
is
sufficiently close to x,
continuous.
Decomposition of the Value Function's Slope. The
32
sets
M + (x)
and M(x)
=
only at points where s(x,y)
differ
0.
Thus we can
define a
function that everywhere coincides with w:
w
-w(x) -w{y)
f{x,y)
=
(x)
p
—>
Take any sequence x n
Add and
Then
x.
for each n, use (25) to
and divide through by x n —
+ 5)
2{r
Xr
expression for w'(x)
-
- {w{x n ) -
f{x, y))
definition,
Lemma
2,
—
f(x,y)\
we have
is
closed-
liminf
M + (x„)
=
hm
for
M + (x n
As
n\x n
)
)
)
—
to the desired
way (Step
a significant
M + (x)
for
> f(x,y)-w(x)-w(y)ify G M
— x\ and \w(x n — w(x)\ < k\x u —
)
—
x\
4).
a.e. x.
By
n
)\M + (.x).
x\
from the
)
< 2k
u(y)dy
u(y)dy
M+(x n )-M+(i)
D(M + (x n ),M + (x)) —
I-B-II-1 in
= limsupM + (x„).
M + (x)
26
must vanish
Since
0.
M+
(-)
is
Hildenbrand (1974) then yields
Next, the integral of the atomless
as well
M+
Integrating over M(x) or
first
term
by the sequential continuity
f(x n y)
,
p
flf
>m M( and
~
equivalent for
a.e. x.
limsupj.
33
- w(x n + w(x n
u(y)dy
+ S){x n -x)
fix, y)
2(r
M+(x)
Mk = U m
(x) is
in the brackets in (26) vanishes,
-w + {x)
— = hm
usual, liminffc
in brackets vanishes if
> f(x n ,y) — w(x n - w(y) > 0. Similarly,
—w(y) < for y G M + {x) \M + (x n ). Thus for all n,
2K,\x n
x
26
>
)
(x)
(26)
Hildenbrand (1974).
which the
w + (x n
-
4:
M+
u{y)dy
+ (:r
and nonempty- valued, Theorem
Step
differ in
djj
result I-D-I-(9) in
At x
do not
continuous at x, then lim Xn _>x
density u over
•
M + (x)
-w(x n -w(y)
M+(x n )-M+(x)
M+
<
-w(x n
f(x n ,ij)
u(y)dy
w(x))
and that the remaining terms reduce
f(x n ,y)-w(x n )-w(y)
-2K\x n -x\ < f{x n ,y)
If
3),
term
first
Surplus vanishes at the boundary of
3:
Since also \f(x n ,y)
proof of
P rove ^hat the
e
M(x) and
if
the resulting
X
^
= Ja\b~ §b\a~
continuous at x (Step
Step
)
X
.
M+(x)
•
w + (x n — w + (x).
)
M+(x n )-M+(x)
(/fan, y)
is
expand
-w(x n -w(y)
f(x n ,y)
)
M+
(25)
x:
w + (x n — w + (x)
where fA _ B
u(y)dy
5)
fM+ ,Af (x, y) - w(x) - w(y)) u(y)dy /2(r + 5) from
subtract p
expression,
+
2(r
M+(x)
Mk = H m
U( >m Me.
)
)
J^
i
Z
^sii--
8^
<<f
Figure
w+
Replace
w = w+
by
(2(r
We claim that
,
Illustration of
7:
solve for lim 2;n _>x (K;(x n )
M+
(x) \
for a.e. x, (27) simplifies to (12).
M(x) with fx (x,
y)
M+
(x)
s(x, y)
=
a.e. x:
for a.e. y either s(x,y)
(x, y)
whenever y G
with s(x, y)
any fixed
y, if
such x
isolated,
is
=
and
G
(x,y)
and the
any
set of
is
x'
we conclude that
so by Fubini again,
^
such x
not in S.
simplify:
fx (x,y) u{y)dy
(27)
a positive measure of
is
27
Since
0.
claim by showing that for
S denote
Let
S
/
s x (x, y)
is
the set of pairs
(Borel) measurable. For
/
satisfies s(x' y)
:
countable and has measure
is
Fubini's Theorem,
5
0.
0;
So every
therefore,
has measure
for a.e. x, for a.e. y, (x, y)
Lemma
• Single Crossing Property: Proof of
1
0.
x close to x
By
and
<£
0,
S.
Lemma
5,
we
5.
establish the following useful input result:
(Density- Free Integral Comparison). Let
cf),ip
:
[0,1]
i—>
E
both be
increasing (decreasing) functions with a nondecreasing (nonincreasing) ratio
on
[0, 1].
measure,
If for
some
12. Figure
JM ?/)(x)u(x) dx =
But then
27
We
density u
fM ip{x)u{x) dx
Remark
0,
also (p(x)
and
DESCRIPTIVE THEORY
B.
Before proving
=
First note that
0.
x),
This equivalence obtains, and can be
will prove the
or s x (x,y)
/
-
w(x))/(x n
-- or equivalently with
\M(x), we
s x (x, y)
1.
of (27), unless there
/ w'(x)
/
S, then
for all y, for a.e. x, (x, y)
Claim
—
+ 5)+pJM+{x) u{y)dy)w'(x) = p JM+{x)
shown through algebraic manipulation
y G
Claim
=
0,
:
[0, 1] (->
then
,
fM (f>(x)u(x) dx >
set
MC
[0, 1]
of positive
0.
As
7 depicts the graphical inspiration for the proof.
a threshold z G
^
M+ and for some
(fr/ip
as x
^
z
[0, 1]
exists such that ip(x)
if cf)/ip is
for all
monotonic nondecreasing
thank Randall Dougherty (Ohio State Univ. Math Dept.)
34
^
for the clean
x
(loosely,
proof below.
^
<j>
z.
is
more convex than
>
function u
Assume
Proof.
[0, 1]
1(0)
0,
So
ip).
the positive area of
WLOG
while 1(1)
0,
and so I(x)
<
that
=
by defining u(x)
=
the positive and negative areas of
if
=
<f>
for all
must outweigh the negative
<f>
and
are increasing.
tp
M.
x £
x e
is
M
If
=
Define I(x)
by assumption. Then /
for all
$*
^
M <f>(x)u(x) dx =
Lemma
Proof of
We
5.
quasiconvex, as
is
<
0, dq(-)
=
ip is
increasing,
>
in the last expression are zero,
So JM
0.
<f>(x)u(x)
dx
>
D
0.
under Al-Sup
i.e.
analogous.
uniquely defined by Al-Sup, as the marginal product fy (z,y{) of a
supermodular production function
strictly
Clearly
cj)(x)/ip(x) yields
supermodular case --
just establish the
is
two terms
first
/(•)
and A3-Sup. The submodular proof
First, z
extend u to
£ q(x)I'(x) dx = g (l)7(l) - q (0)1(0) - ft I(x) dq(x)
term nonnegative, as
final
[0, 1],
[0, 1].
where we have integrated by parts. The
and the
area.
ip(x')u(x') dx'.
Substituting with / and the nondecreasing quotient q(x)
f
balance for some
ip
is
strictly increasing in z.
Next, integrating the A3-Sup inequality over x\ G [z,x 2
]
and then x 2 £
[a?i,;z],
we discover that
for all y 2
fxy(x, y 2 ) (fy (x, yi)
- fy (z, yi))> fxy (x,
>
>
?/i,
and
for all
x
y x ) (fv (x, y2 )
z and crucially also
d fy{x,V2) - fy {z,y2 )
dx fy (x,yi) - fy (z,yi)
Let
=
(j)(x)
fy (x,y 2 )
nondecreasing,
Claim
4>,tp
- fy (z,y 2 ) and
tp(x)
are increasing (by Al-Sup),
(a)
mm(a(y
•
x
<
z.
This
equivalent to
is
>0
fy (x,y l )
- fy (z,y
x
Then
).
and fM ip(x)u(x)dx
—
by
(f>/i/j
l
[Strong
D
Version]:
),a(y 3 )) and o(y 2 )
Step
1:
continuous and
If
<
a
Proof of
Lemma
6.
not strongly quasiconcave, then a(y 2 )
is
max(a(t/i),a(y 3 )) for some
Inequality Case.
a.e.
is
(20); so
implies (21).
1
• Characterization of Quasiconcavity:
Part
=
all
- fy (z, y 2 ))
If
differentiate, there
is
<
yx
< y%<
< mm{a(yi),a(y 3 )),
Vz
<
<
1-
then as a
is
y 2 e (yi,y3 ) near enough y 2 that a(y 2 )
<
a(y 2 )
35
mm{a(yi),a(y3)) and a'(y 2 )
> a(y 2 and
given a(yi)
•
Step
=
cr(y 2 )
some
for
,
Now,
cr(y 3 ).
If
either
>
(m), replace
by
yi
a
If
is
V e
(yi,
a'(x)
<
If («), <t'(x)
y3 )-
by Q-2.
Step
1:
by Q-l.
=
Ofor
=
o(y 2 )
all
Then
cr(y 3 ).
some y e (yi,y 3 ),
for
)
=
or
we admit
or (m), replace y 3 or y x (as y
either
a
(i)
constant
is
(m) a(y 2 < a(y 2 )
for
)
=
e (yi,y 3 ), but a(yi)
a;
Matching and Bounds: Proof of
Monotonic bound functions
imply positively assortative matching. Choose x\
and y 2 € M(xi). Then we immediately have
strongly increasing implies 6(x 2 )
is
j/2
<
b(x 2 ), and so y 2 €
M(x 2 ),
—
or y 2
only possible
=
Then
l-
if
Xi
equivalently y 2 G
ensures that yi G
•
Step
^
< x
a(x) and yi
y 2 ) by
some
implies
or replace y 2 by
y,
Lemma
7.
=> assortative
matching. As
M(x)
x
=
7^
if
y2
G
M(x 2 ).
M^),
x >
0.
By Theorem
0.
M(y 2 ),
>
b(x x )
y2
as claimed.
=
That
1.
o^) >
If
.
y\
<
y2 G
y 2 with y x £
)
whence matching
<
y 2 and a(x 2 )
6(x 2 )
>
y2
,
=
b(x\)
=
=
)
But
b
<
y2
4,
matching
repetition
=
if
,
.
Then
M(x 2 ). Then
>
y2
,
and so
M(x) = 0, contrary
G
M(y 2 ),
or
to
M(x 2
)
Lemma
1
First,
unless
a, b well-defined.
we simply prove that
If not,
>
b(x r )
b(x 2 ).
>
b(x 2 ) for
b is
some
Also choose any
by the definition of positively
assortative matching. This contradicts the definition 6(x 2 )
• Ideal Partners and Quasiconcavity: Proof of
36
which
bound function a
and thus bounds
four similar cases,
yi
<
so suppose
0,
convex, implies x 2 G
there exists y : G M(xi), with y x
y:
yi
positively assortative.
is
sets are convex,
among
a, b
M(x 2
y\.
then a(x 2 )
y 2 precludes y 2
nondecreasing with positively assortative matching.
< x2
<
Similarly, a strongly increasing lower
For by (10), w(x)
To avoid tedious
pair x x
y\
Otherwise, 6(x 2 )
M(y 2 ), and M(y 2
G
1
>
< x 2 and
Assortative matching => monotonic bound functions.
2:
or
If (n)
the two cases are analogous, we simply prove that strongly increasing bounds
y2
)
case', yielding a contradiction.
,
k Assortative
=
If (ii)
y 2 in the 'equality
•
> a(y 2
a(y)
(ii)
>
x, so o'(x)
.
> a(y 2
If (i), ct'(x)
strongly but not strictly quasiconcave,
the additional possibility that cr(yi)
on [yi,y3 ], or
and y3 >
y2
WLOG
assume
some y 2 G (yi,y2 ).
for
)
>
y3
)
constant on [yi,y 2 ], or (u) a(yi)
is
< a{y 2
cr(x)
> a(y 2 and
min(o-(t/ 1 ),cr(y 3 )),
<
a'(y 2 )
or y 2 by y2 in the 'inequality case', yielding a contradiction.
y"i
Version]:
(b) [Strict
given a(y 3 )
=
a{y 2 )
a
(i)
G (yi,y 2 ), or (Hi) a(y 2 )
yi
x G (yi,y2), but u(y 3 )
for all
Pari
<
>
y 2 while a'(y 2 )
Equality Case.
2:
o-(yi)
<
yx
)
Q-l then yields a contradiction:
exists.
=
sup{y|y G
Theorem
5(b).
M(x 2 )}.
We just
prove the supermodular case; the submodular case in analogous.
Step
•
=
p*(l)
{1}, since
Theorem
=
p*(0)
Everyone
1:
1
has an increasing surplus function by step
Given f{0,y)
1.
=
correspondences, because p*(x)
Step
•
Then
Surplus
2:
<
there exists
>
s(x,y)
y\
<
is
€ V*{U2) only
x'
<
>
if x'
Then p*{y 2
=
)
(for p* single-valued)
<
y2
Vz
M
whenever
<
s(x',y 2 ) only
> s(x,y 2 ),
Since
w
is
with a unique maximizer
is
^
differentiable, so
implies
y'
y as
^
maximize
to
which
x,
is s,
Z
s
(x)
1:
=
measure on
Let x
^
x'
:
Z
s
(x)
Assume
28
w
and y
Z
+
s
and thus
x,
p*(y 2 ) only
if
Observe that
true for
is
all
(0, 1)
is
Lemma
3.
only
=
y
is
3 establishes
x by Theorem
if s
w
(x, y)
0.
Under
increasing (decreasing) at y:
s y (x',y')
1,
=
Higher (lower) types must choose higher (lower) partners
a(w):
i->
=
We
[0, 1].
Pi
>
G
p*{x') implies y'
^
^
y as x'
D
x.
EQUILIBRIUM EXISTENCE
s(x,y)
Al-Sub, f(x,y)
Thus,
x:
Proof of
Surplus Function
{y
<E
quasiconcave by Theorem
is
their surplus function, or y'
• Continuity of
Step
x'
6(b).
For the proof of
and y € p*(x) D
Since s(x, )
x.
^
x'
C.
•
^
as x'
x'
strictly quasiconcave.
Al-Sup, the surplus function of higher (lower) types
s y (x',y)
if
0, a contradiction. Finally, a strongly quasiconcave function
continuous at
is
).
)
everywhere differentiable under A1,A2,A3.
this
not.
)
Similarly, since s(x,y 3 )
x.
Suppose
in y.
< 1, with s(x,y 2 < mm(s(x,yi),s(x,y 3 ))
WLOG, assume s(x,y 3 > s(x,y 2 Since
Monotonic Ideal Partners: Proof of Theorem
*
decreasing, and so
is
strongly quasiconcave
s(x,y 2 ), Al-Sup implies that s(x',yi)
%'
x.
- w(0) - w(y)
k
nonempty, convex- valued, and u.h.c.
is
and s(x,y 2 ) < max(s(x,yi),s(x,y 3 )).
s{x,y\)
=
s(0,y)
k,
of the proof of
1
desired result follows from the intermediate value theorem for
The
{0}.
someone's ideal partner. Given Al-Sup, we have
is
/
0} and
/j,(Zs (x))
8.
Rarely Constant
Z =
s
{(x,y)
:
s(x,y)
=
in
one Variable.
and
0},
let
y',
with s(x,y)
^
f(x',y)
=
+
s(x',y)
=
s(x,y')
=
for
an uncountable number of
Toby Gifford (Math Dept., Washington U.
=
x.
28
^
for a.e. x.
Under Al-Sup
f(x,y'), from which s(x',y')
contains at most one point whenever x
>
0.
+
or
follows.
x'
Then
for
some
in St. Louis) tightened the logic of this
37
Define
be Lebesgue
\i
claim that under Al-Sup or Al-Sub, fi(Zs (x))
f(x',y')
(x')
is
Lemma
k,
there
paragraph.
are infinitely
many
(x n ) with (i(Zs (x n ))
whereupon Y^=i
1/k,
and so
s
t
Z
s {
oo
oo
= Y, MZs(Xn) \N)=Y, KZs(Xn))
\ N
n=l
Xn )
n=l
n(Zs (x)) =
Step
for a.e. x,
many x with
countably
and by Fubini's theorem
x jj){Zs )
(fi
Close Surplus Functions Rarely Differ
2:
=
OO
n=l
/
this contradiction, there are only
Given
i
s
)
\
|J
=
(0)
lim(/iX/i)(E s (77))
in Sign.
As
r\
-»
0,
the
s
=
(l/A;)
=
=
lim(MX/x)(E,(l/fc))
(/.x
M )(n~ ^.(l/fc)) =
(//x
M )(Zs
77/2,
and
)
=
fc—>00
Finally, let
i^ and
w2
be value functions, with
corresponding match indicator functions.
then
s 2 {x,y)
=
<
Si(x,y)
f{x,y)
-77,
{(x,y)\ai(x,y)
then
<
s 2 {x,y)
/ a 2 (x,y)} C E
||wi(a:)
0,
Si (t?),
so that 5
Claim
2.
=
1.
We
first
f{x,y)
-
Wi(x)
-
lirri|[
l(;i
we prove
«i,
>
wi(y)
a2
r),
= a 2 (x,y) = 1, while
= a 2 (x,y) = 0. Consequently,
whose Lebesgue measure vanishes
—>
77
0.
1
,
that the
as
— o^lk =0.
_ u 2 ||^.o ll^i
Fundamental Matching Lemma: Proof of
WLOG
— zu 2 (a:)|| <
and so ai(x,y)
and ai(x,y)
0,
=
S\(x,y)
If
- w 2 (x) - w 2 (y) >
This implies the desired continuity
•
So
.
T)—>0
if
0.
0.
= {{x,y) with \s(x,y)\ € [0, 77]} shrinks monotonically to n^=1 S
= Z By the countable intersection property of measures,
s
>
fj,(Z s (x))
set £5(77)
_1
oo.
rr i:/
s
s
(oo
s
=
{Zs {x n ))
JL
l
= Z (xi) C\ Z (xj) contains at most one point, TV = U°° =1
is countable,
fi(N) = 0. Also, Z (x \ N and Z {xj) \ N are disjoint for all ^ j. Thus
Since Xij
•
>
Lemma
map a
For each match indicator function a, there
Renormalize time
9.
h->
u(a)
exists a
well-defined.
is
unique unmatched
density u(a) that satisfies steady-state condition (1).
•
Step
1:
Proof of Existence of
Theorem, the
compact, since
set 5" of
£ is integrable.
easy to check that
=
-
\& Q is
+ pj
Here,
we
are speaking of
its
f
[0, 1]
< u <
with
is
of
weak-*
1 into
itself:
(28)
a(x,y)u{y)dy
in
u 6
a fixed point
general version, such as
38
^Q
£ is
]
weak continuous
29
there
Schauder's Fixed Point Theorem,
29
a suitable version of Alaoglu's
So consider the following mapping
1
it is
By
measurable functions u on
*««(*)
Then
u(a):
Theorem
"3
given
«G?,
a G
i.e.
[0, 1].
So by
(1) holds.
5.1.3 in Istratescu (1981).
Step
•
Proof of Uniqueness of
2:
We now
u(a):
Suppose that there are two essentially distinct solutions
supremum 30
differ
i*i
(x)
on a
>
set of positive
>
#2
RHS <
Once
again,
Claim
Step
•
Yet
if
=
6>i
LHS > RHS
The map a
3.
<
i-»
92
,
=
Defining G(a,u)(x)
a
iff
G(a, u)
G u (a,u)(g) =
w(a)
H
is
—
u(x)(l
The
0.
lim t ^
=
Write
then
6>
>
2
is
1,
9
+
P [9 Jo
G u (a,u) = I + pH.
self-adjoint,
and
Since u
is
=
=
=
=
2
since
and u 2
11 1
since (28) easily precludes
RHS <
and so
9 2 fa(x,y)ui(y)dy,
measure of
measure of
x,
>
0.
9 2 in (29), as \/p
a contradiction.
x,
continuous.
2:
i.e.
£(x)
For fixed a Q
—
derivative
+
tg)
G u (a, u)
=
uo(x)
is
pu
(x)
we
£{x),
u
let
,
be the unique
J ao(x,y)uo(y)dy.
see that
a linear operator on
-G(a,u))/t
=
solves (1)
it
£ 2 [0,
au J + pu
31
l]:
ag
g (l
+
p f^
symmetric (a(x,y)
—
a(x,y)), the operator
fa
au + u /o ap
Since
a
is
also neatly positive-definite
on the space
functions square-integrable with respect to density l/u, because
2(g,Hg)
1
9 2 for a positive
+ p J a(x,y)u(y)dy) —
(G(a,u
1
>
a positive measure of x.
> ui{x)/u 2 {x) >
A Steady-State Operator.
1:
>
61
fa(x,y)u 2 (y)dy <
in (29) for a positive
unmatched density from Claim
given
weakly greater. Then
9 2 u\ to get
8 2 in (29). Since #1
this violates (29).
is
strict inequality for
Integrate u 2
-
Then
of U\{x)/u 2 (x) (resp. u 2 {x)/ui{x)) over
measure. Furthermore, 9 2
with
1*2(2) for all x,
Posit #1
and so
WLOG that 9\
and assume
[0, 1],
to (28) in 7.
l/p
u 2 (x)
x G
1*1,1*2
+ Ja(x,y)u 2 {y)dy
l/p + J a(x,y)ui{y)dy
Mx) =
Let 61 (resp. 9 2 ) be the essential
pursue a different tack.
£ 2 ([0,
1],
l/u) of
32
£ g(x)Hg(x)/u(x)dx
2
+ g{x)g{y))a(x,y)dxdy
+ 2g{x)g{y) + (/(y) 2 u(x)/'u(y)] a(ar, t/)dxdy
2 JoJo(9{x) u{y)/u(x)
2
Jo Jo i9(x)
u(y)/u{x)
SiSi\9( x )V u (y)/ u ( x )
+ 9(y)y/u{x)/u{y)] 2 a(x,y)dxdy >
boundedly positive (given p
<
00, £
30
>
£
>
0)
and
finite (u
<
^
<
00),
That is, 9i is the least number such that Ui(:e)/u2(:e) > #1 only on a set of measure 0.
To save on space, hereafter suppress x and (x,y) arguments whenever there is no ambiguity.
32
We thank Robert Israel (UBC Math Dept.) for remarking on this fact.
31
39
£ 2 ([0,
in
1],
£ 2 [0,
G u (a
Step
•
real
)
2:
now
into a,
and
H
So
1].
and hence
1],
w
,
= £ 2 [0,
l/u)
a self-adjoint and positive definite linear operator
is
spectrum
its
and nonnegative, and the spectrum
real
is
and thus excludes
positive,
G u (a
0:
A New Steady-State Operator. 33
with range [0,p]. Let T(a, ) = I - Q
\I/
—
Then G(a,u)
r(a,ii)/o!/i+
+ /mt)r(a,u)
(1
+ Q!u)r
(l
tt
(Q!,u)(/i).
So
r(a
=
(£/u )T u (a ,uo)(h) given
oo, the linear operator
Remark
,'"o)
A = T u (a
of a, u,
G u (a
is
^
Xx
thus, the derivative
To apply the IFT, T and F u must be continuous
=
means that Z
^
if
T~
supremum norm), and
Step
Lemma
Tu
so continuity of
demands that Z
at (a ,u
= L1
,
G u (a
.
Since
|a-a
<
|
p,
\\
< y/p\\a-a
and
is.
)
T u maps
1x^x^2,.
The form T„ =
).
requires that a weaker
L2
u
,
=
are the ambient
while invertibility requires
\\a-a
)(h)
Z
V,
demands a weak norm on
8
Jau).
G u (a,u)(h) =
X
+#
/
(not a
norm on Z than ^
X =
Z = L2
if
^
=
-C
2
,
then
.
Setting up a Triple Induction. Pick a with range
3:
closetoa
But
exists.
+
u )T u (a ,u
since
1],
apply the Holder inequality.) For instance,
(for instance, to
•
l
rule
where X,
Z,
u - £/(l
bounded away from
£ 2 [0,
invertible in
=
+Ja
(1
is
)
£ 2 [0, 1].
invertible in
is
T(a,u)
=
,Uo)(h)
Since (£/u
0.
T map
and the range;
)
For simplicity, embed p
i.e.
,
u
and by the product
,u Q )
13. Let the operator
Banach spaces
continuity
,
of
[0,p]
1
iL -
and
is
small too. Recursively
T(a, u k )
(30)
\\jii
define u^+i by
A(u k+l for k
=
step,
assume
r,
>
e
that
1
i*i
(i)
<
and
—
Wo Ik 2
Step
to (30).
4:
is
We
But any
1
1
\\
<
L2
Part
1
u k+x e
-C
F(a, u
)
by
T
(30). Since
||ck
—
ck
||xl
u
2
-
)
[0, 1] is
well-defined.
- u k _i\\c2 <
e, (li) \\u k
are determined later.
vanishing with
2
-
We
of Induction. The
To
is
As an induction
d k and (Hi) u k
,
> -r, where
get the induction going, observe
continuous in a, and r(ao, "o)
verify part (Hi) with k
first
equality below adds
=
=
0,
1 later.
= r(a
,"o)
In a recurring theme, the next parameterizes the vector from (a ,Uo) to
(a,u k ) by (a s ,u s )
33
<
d
= A(u k -
A is invertible,
-u
\\u k
u\—uq = —A~
1
•
Since
0, 1,
uq)
=
(a ,u
)
+
s(a
-a
,u k
thank Sheldon Chang (M.I.T. Math Dept.)
errors in executing this proof
- u
for
<
s
<
1,
for outlining the rest of the
program are ours
40
),
alone.
and applies
(§):
proof that follows.
-
C(l)
-u
A(u k+l
=
((0)
)
('(s)ds,
Jq
= A(u k -u
the Fundamental
i.e.
-
)
= r u {ao,u )(uk-u Q -
If
we
(J(-)dx)
2 II
On
[(
first
2
<
first
2
we
to (a s ,u ks )
to apply (§); finally use (J(-)d(, s )
=
T a {a,u)(b)
in (31), substitute
-
4///[(r ua (a( S ,w^)K
<x
iIa ' s{Uk ~
4 fff ( 2
(1 +
JJJ V
J(-)
d£, s
<
for
fbu/(l
+ Jau) 2
-
d£ s dxds
){u k
wo)]
2
-u
)}
.
±
.
remaining terms,
moreover, \a^ s
<
£s
—
2
+
u
-u
(u ks
£, s
)
In the second- term
1.
bounded above by
is
*"
°—
(
dxds
)
2
~
~ a ° )Uk ' s - iJ{as a ° ){Uk -?°
r
) dxdZ s ds
2
3
]
+
(1
S)
J assUkts)
J
=
is
u kis ) 3
2
{a
2
J
2X 2 + 2Y 2
on the
first
term
(32)
J
"°>
aQ)Ukfcs*
ff{
x
)dxds
"l ;
J J \{1 + J a s u ks ) J
in (32),
2
and replace
all
inner
< J a Ju by the Cauchy-Schwartz inequality. To bound the
+ (a, u k
recall that £ < £ < oo. Also, as (a s u ks = (1 - s)(a u
2
2
)
,
-
(1
—a
To remove the
\
s
must be bounded.
and H^olU 2 <
(31)
Substituting into the previous expression yields
J«^
Y) 2 <
products (Jau) 2
,u ks )
dx ds
g
U ° )I{as
(I
\
(a(. s
)]
at most:
is
and then parameterize the
Thus, (31)
+
d^ s dxds
2
-a
and u k ^ s
)
W,d« +
7^^ + pL&^
a
Apply (X
and
<
~
(
s
,
dx
}
3
-2£ Jag Jah/(1 + Jau)
JJJ
,
2
))
+ £jag/(l + / au) 2 w.r.t. both a and u to get
= 2£fagfbu/(l + Jau) 3 -£jbg/(l + Jau) 2 and T uu (a,u)(g h) =
T u (a,u)(g)
T ua (a,u){b,g)
+4 fff
=
£
2
-u
J(A(u k+1
2j [T a {a s u ks ){a
s
2
)(u k
+4jJJ[r uu {a^,u k ^){u ks -u
Differentiate
=
)|||2
± Y) 2 < 2X 2 + 2Y 2
a^ = Of + ^ (a; -a
by
)]ds
,
then the Cauchy-Schwartz inequality
dx+
)}
s
)
apply (X
in (31):
)
2
-u
(u k
,
vector from (a ,u
-u
+ T a (a u ks )(a - a
)
- u Q + T a (a s ,u ks )(a - a Q )ds
,u ks ))(u k
see that ||A(it fc+1
- T u {a s u ks ))
term
s
,
J{-) dx,
)
-T u (a
)
,u Q )]
(X±Y) 2 < 2X 2 + 2Y 2 and
apply
r u( a o, u
the
J^(T u (a ,u
of Calculus.
J^[T u (a s ,u s )(u k -u
)
=
-T(a
[T(a,u k )
Theorem
oo,
<
f s )(a ,Uo)
\a s
and £s
Now,
—a
\
+
<
C*(a».«fc*)i a11 the
\a
from
integrals
since
—a
\\u k
-
\,
and
\u k ^ s
(32), all
2
uo\\ c2
<
41
\
<
< a <
satisfy
\u ks
—u
)
\
<
\u k
e
<
p;
—u
denominators and a term / u
we have max(J" u 2k ^ s dy J u 2k ^ s dy) <
,
a terms
—u
)
,
2
^s
\.
dy
oo by induction assumption,
B <
oo.
Because u k >
—r
and
1
<
|a|
- pr >
is small enough that
m'm(J a s u ks J a s u k ^ s ) > -pr, where r >
Finally, we find a new upper bound for ||A(ii fc+ i - t*o) 11^.2
p,
,
1/2.
:
(
+
_
(1
)6
Taking square
-u
\\A(u k+1
)\\
-a
(\\a
= J (||q <
Let
<
d
and choose
1,
<
||wfc-wo||£2
Iki
must
roots, there
< J
L2
u j B JJ (a - a
_
(1
-u
\\
\\
+
-a
other induction assumption,
\\
+
C2
<
d/2 and
oo, such that
- u Q \\l
\\u k
+
2 )
\\
<
L2
-a
C\\a
C2
\\
+ C||" - a olU
«olk»
-u
||ui
dx
)
-
d(l
d)/2.
2
Then
k~l
< d/2 by the triangle inequality and the
- a ||.e2 + \\u k - u ||.c2 < d. Since the inverse
d
||a
i.e.
J <
- wolUO IK -
\\u k
||£,2
+ d-\
L 2(l
-u
L 2\\u k
2
dydx J(u k - u
exist constants C,
aoll^a
||a
2
)
)
_1
-
l
<
aolk*
<
^/ 2 and \A~ \Jd
||a-ao|k 2 an(i d small enough; therefore, ||«fc-wolU 2
<
£ implies IK+i-'Uoll.c 2
operator
•
Step
A'
5:
Subtract
1
has a
Part
(*
fc
)
2
norm, |^
finite
of Induction. Next, (*
- (*
fc
and apply
-i),
= A{u k -
A(u k+ i - u k )
=
in part 1, except
as well as {A'
1
]
-1
•
1
•
X(||a
Step
6:
—
if
aolk 2
Part
3
+
=
+
u k -\
-
fc )
-/
omitting the
1
fe
=
)
t(u k
r(a, u
fc
final
-
- u k -i),
to get
_i)]
fc
-u
fc
_ 1 )dt
u*-i)d*
T Q term
in each expression,
||tifc
-
Wfe-ilka) ||^ -^fc-ilk 2
,
k
L2
\\
>
||^i
+d
1.
—
- uk
k
<
)
For
fc
A
<
dk
,
This latter inequality holds for ||a-a |k 2
d.
=
1,
we want
liolk 2 ) as small as
we
,
)
42
-
last ingredient,
£ja {u k+1 -u k
=
r^—
r
2
(1 + J a w
,.
||u 2
«i||£ 2 /ll u i
~~
u o\\l 2 <
like.
expand (*
£
)
)
e.
-r(a,Mjt) by (30).
r u (a,^i)(u
(«*
<
oo for which:
of Induction. For the
{u k+1
-u
1/2 for
d fc+1 we use our induction assumption Hwfc-Wfc-iH^
K(\\a-a
and d small enough
|
<
u
A(it fc+1
- r u (a,tifct))
< K(||a-ao|U2 +
||A(ufc+i-Ufc)||£,2
||u fc+ i-'Ufc||£2
<
[r(a,
){u k -ujfe_i)
(r u (a ,wo)
K
):
fc
with u kt
fc
1
similarly yields a constant
To prove
(§)
u _x) -
T u (a ,u
= /
Proceeding just as
C||a
•
|
-wjb
+ —-7
1 + J au
fc
fc
)
into
=
Substituting the steady-state identity u
Uk+i
I
=
+
Because u
positive.
>
for k
=
1,
a Cauchy sequence. Since
=
u
close to
T(a,u 00 ). So
-C
2
As per
.
>
for all e
Since
\\u
— u
\\
* Continuity of the
a
>
7]
L
i
<
—
<
Tw2(z)\
- Tw 2 {x)\ <
\\u
=
so that
|Z>i|
\Di\
e.
+
(/mi(x)
JmxCsO
—
— u
<
||.c2
for e small
establishes that (u k )
converges to some limit
whenever
e
We
w2
W\,
e
+
\,
\
\D3
||a
prove that for
—
sup z
if
S,
—
|ii>i(.z)
where
(
+
(2 (r
d y)
+
S)
Jm'(x)
wi(y))
VwA d y)
— w 2 (z)\ <
77;
- /^( X)
>
<
V-,
then
+
pu )D equals
l
l
1
(dy)^
w i{ x vm
)
(
d y)
M^)
-
wifc))
Wl (dy)
i/
M^Ki
- Mv)) VwMy) + /Mc (l)
+
<5)
+
(dtf))
pu x
- My))"w
+
2
5)
(dy)
+
+
/mj(«)
w2(x)^
2 ,(dy)J
pu 2
all
integrands
the arguments of the integrals are uniformly bounded and
all
Lemmas
8
and
P (/mi(x) (f( x ^)
(x)
this
there ex-
^2(^)1
uniformly small by the definition of the weak topology, as
2
small
and where
2 (r
~P (Im
is
by the triangle inequality, the difference
i/
P (/m,(i) (/( x >?/)
continuous (by
unmatched
/
\wi(z)
and
and thus
ao||.c2
e
all
- w i(y)) v™Mv) + IMnx)M x ) w
2 (r
are continuous,
is
enough.
by the Cauchy-Schwartz inequality,
uq\\h2
^^' y) ~ My)) v*n
(My) -
< sup z
\\u
x,
\D 2
'
is
> —r
it
term
1/2, the first
the essentially unique feasible
is
For any
p \Jmux) (/( x y)
which
>
an u(a).
all
p (/m,(x) (/foy)
-p
u
Operator T.
such that for
sup 2 \Tvj\{z)
\Twxix)
pr
)
usual, Uoo satisfies the recursion (30),
0, if
proves the desired continuity of
ists
—
1
a Banach space,
is
[0, 1]
density solving (1) given a, then
enough.
-u
{u k+ i
—(ul/£)e
\\
yields
)
Using the The Cauchy Limit. The induction
is
is
J
i
Step
Moo that
au k
and more generally since
•
7:
(u /e)
> — (ul/£)\\u k+ — u Q >
Thus, u k+ i
ja
2
7-7-7
1
+Ja u
£(1
(/(*. y)
9);
and
- My))
finally,
where
"w 2 (dy)
~ Mv))
+ jM
Vw*(dy)
43
(2 (r
c
+ JK
{x)
c
+ 5) + pu 2 )D 3
w 2 (x)vW2 {dy)j
(x)
w 2 (x)vW2 {dyy)
equals
This term measures the
loss
w2
suffers witii value function
x
from using the matching
<
we have \f(x,y)-w (x)-w
As l^ - w <
M
difference
the symmetric
y 6 (Mi(i) UM
\ (Mi(i) nM
than 2n.
strictly
So the third term
between Mi(x) and M
set
Mi(i) rather than
2t)
for all
2 (x).
2
2
rj,
\
(x))
2
2
(y)\
set
(x)),
less
is
2 (x).
2
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Milgrom, and
P.
S.,
J.
Roberts
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44
MIT mimeo.
08
9
040
Date Due
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DUPL
3 9080 02163 2838