Synchronized matching with incomplete information

Abstract

This paper considers two-sided matching in continuous time without transferable utility or costly effort signaling. Two continua of impatient agents signal their types by delaying before proposing or accepting a match. I use a mechanism design approach to study what matchings and schedules of match times are implementable when there is private information on one or both sides. When only one side has private information, a sufficient condition to implement assortative matching is for the uninformed side to value the log gain in partner quality from waiting more than the informed side, so that it pays to wait. When information is incomplete on both sides, the sufficient conditions to implement assortative matching are much more restrictive: even when coarse matching is considered, a simple sufficient condition is provided under which only simple random matching is implementable.

Proofs

1.1 Proof of Proposition 1

Proof

Incentive compatibility for the X side requires

$$\begin{aligned} U^X(x,x) \ge U^X(x',x), \quad x' \in [0,1]. \end{aligned}$$

Note that taking logs of the direct utility function yields

$$\begin{aligned} \log \left( U^X\left( x',x\right) \right) = \log \left( v\left( x,\mu \left( x'\right) \right) \right) - \rho \tau \left( x'\right) - \log (\rho ), \end{aligned}$$

which is quasi-linear in the log-surplus and the dropout time. Let the indirect utility function for the X side be

$$\begin{aligned} V^X(x) = \max _{x' \in [0,1]} \log \left( U^X\left( x',x\right) \right) . \end{aligned}$$

Then the following, standard characterization of incentive compatibility for a direct mechanism obtains: $\{\mu (x),\tau (x)\}$ is incentive compatible iff $\mu (x)$ is (i) non-decreasing and (ii) $(V^X)'(x) = v_1(x,x)/v(x,x)$.

• To show necessity, first take logs of the direct utility function,

  $$\begin{aligned} \log \left( U^X\left( x',x\right) \right) = \log (v(x,\mu (x')) - \rho \tau (x') - \log (\rho ). \end{aligned}$$

  The envelope condition follows by differentiation of the indirect utility function, which is derived by taking the partial derivative of $\log (U^X(x',x))$ with respect to x and evaluating at $x'=x$, yielding (ii). To show (i), rearrange two incentive compatibility constraints

  $$\begin{aligned} \log (v(x,\mu (x))) - \rho \tau (x) \ge \log (v(x,\mu (x'))) - \rho \tau (x') \end{aligned}$$

  and

  $$\begin{aligned} \log (v(x',\mu (x'))) - \rho \tau (x') \ge \log (v(x',\mu (x))) - \rho \tau (x) \end{aligned}$$

  to get

  $$\begin{aligned} \log (v(x,\mu (x))) + \log (v(x',\mu (x'))) \ge \log (v(x,\mu (x'))) + \log (v(x',\mu (x))). \end{aligned}$$

  But if v(x, y) is strictly log-supermodular and $x \ge x'$, then for the inequality to hold, $\mu (x) \ge \mu (x')$, establishing that the allocation must be monotone. But if the market clears and the match is measure-preserving, that implies $x = y$ for all x and y who match, which must be positive assortative matching. If, instead, v(x, y) were strictly log-submodular and $x \ge x'$, the reverse inequality would have to hold, namely $\mu (x') \ge \mu (x)$.

• To show sufficiency, suppose (i) and (ii) hold. Then

  $$\begin{aligned} \log \left( U^X\left( x,x\right) \right) - \log \left( U^X\left( x',x\right) \right)= & {} \log (U^X(x,x)) -\log (U^X(x',x'))\\&+\log \left( U^X\left( x',x'\right) \right) - \log \left( U^X\left( x',x\right) \right) \\= & {} \log (v(x,\mu (x))) -\log (v(x',\mu (x')))\\&+\log (v(x',\mu (x'))) - \log (v(x',\mu (x)))\\= & {} \int _{x'}^{x} \dfrac{v_1(z,\mu (z))}{v(z,\mu (z))} - \dfrac{v_1(z,\mu (x'))}{v(z,\mu (x'))}\hbox {d}z\\= & {} \int _{x'}^{x}\int _{\mu (x')}^{\mu (z)} \dfrac{v_{12}(z,w)-v_1(z,w)v_2(z,w)}{v(z,2)^2} \hbox {d}w \hbox {d}z. \end{aligned}$$

  Where the second equality follows from (ii). Now, assuming log-supermodularity, the monotonicity of positive assortative matching as in (i), and $x\ge x'$ implies the integrand is positive for all $z \in [x',x]$; similarly, log-supermodularity, monotonicity of the allocation, and $x \le x'$ implies the integrand is negative for all $z \in [x,x']$. Either way, the integral is positive, and $\log (U^X(x,x)) \ge \log (U^X(x',x))$, establishing incentive compatibility. Assuming log-submodularity and a decreasing match function yields the analogous conclusions.

To derive the schedule of matching times, note the envelope condition, $(V^X)'(x) = v_1(x,\mu (x))/v(x,\mu (x))$, implies

$$\begin{aligned} V^X(x) = V^X(0) + \int _0^{x} \dfrac{v_1(z,z)}{v(z,z)}\hbox {d}z, \quad V^X(0) = \log (v(0,0))- \log (\rho ) - \rho \tau (0) \end{aligned}$$

in the positive assortative matching case, and

$$\begin{aligned} V^X(x) = V^X(1) - \int _{q}^1 \dfrac{v_1(z,1-z)}{v(z,1-z)}\hbox {d}z, \quad V^X(1) = \log (v(1,0))- \log (\rho ) -\rho \tau (1) \end{aligned}$$

in the negative assortative matching case. Note that $\tau (0)=0$ in the positive assortative matching case, but $\tau (1) = 0$ in the negative assortative matching case, since matching times and partner qualities must move in opposite directions over time. Combining these with the direct utility function evaluated at truth-telling yields

$$\begin{aligned} \rho \tau (x)= & {} \log \left( \dfrac{v(x,x)}{v(0,0)} \right) - \int _0^x \dfrac{v_1(z,z)}{v(z,z)}\hbox {d}z = \int _{0}^{x} \dfrac{v_1(z,z)+v_2(z,z)}{v(z,z)} \hbox {d}z \\&- \int _0^x \dfrac{v_1(z,z)}{v(z,z)}\hbox {d}z = \int _{0}^x \dfrac{v_2(z,z)}{v(z,z)}\hbox {d}z \end{aligned}$$

in the positive assortative matching case and similarly

$$\begin{aligned} \rho \tau (x)= & {} \log \left( \dfrac{v(x,1-x)}{v(1,0)} \right) + \int _x^1 \dfrac{v_1(z,1-z)}{v(z,1-z)}\hbox {d}z \\= & {} -\int _{x}^{1} \dfrac{v_1(z,1-z)-v_2(z,1-z)}{v(z,1-z)}\hbox {d}z + \int _{x}^1\dfrac{v_1(z,1-z)}{v(z,1-z)}\hbox {d}z\\= & {} \int _{x}^1 \dfrac{v_2(z,1-z)}{v(z,1-z)}\hbox {d}z \end{aligned}$$

in the negative assortative matching case. These resemble the bid functions at an all-pay auction, where the true log gains are shaded by an integral term that reflects the agent’s contribution to surplus. This yields the incentive compatible stopping times $\tau (x)$ in (1).

Finally, note that since the quantity $\rho \tau (x)$ enters agent X’s utility directly, $\rho$ vanishes from the payoffs as soon as the strategies are substituted in and the indirect utility function computed. Therefore, equilibrium payoffs in any incentive compatible are independent of $\rho$. $\square$

1.2 Proof of Theorem 2

Proof

• i. To ensure that the mechanism is incentive compatible for the Y side, each y must want to wait for her prescribed partner, so that for all $y = x$ and $y' < y$,

  $$\begin{aligned} e^{-\rho \tau (y)}u(y,y) \ge e^{-\rho \tau (y')}u(y,y'). \end{aligned}$$

  Recall, the reason is that once an X-side agent has stayed in up to time $\tau (x)$, he has signaled that his type is at least x. He could potentially profitably deviate by approaching an agent of type $y > x$, and propose matching. This agent y must find such a proposal unprofitable in order for incentive compatibility to hold.

  Simplifying the incentive condition for the y type yields

  $$\begin{aligned} \rho ( \tau (y)-\tau (y') ) \le \log \left( \dfrac{u(y,y)}{u(y,y') }\right) , \end{aligned}$$

  which is the expression in the statement of the theorem. However, using the closed-form expressions for the matching time functions from (1),

  $$\begin{aligned} \rho (\tau (y) - \tau (y')) = \int _{y'}^{y} \dfrac{ v_2( z, z) }{v(z,z)} \hbox {d}z. \end{aligned}$$

  By strict log-supermodularity, $v_2(x,y)/v(x,y)$ is increasing in x, so

  $$\begin{aligned} \int _{y'}^{y} \dfrac{ v_2( z, z) }{v(z,z)} \hbox {d}z \le \int _{y'}^y \dfrac{v_2(y,z)}{v(y,z)}\hbox {d}z = \int _{y'}^y d \left\{ \log (v(y,z))\right\} = \log \left( \dfrac{v(y,y)}{v(y,y')} \right) . \end{aligned}$$

  Combining these results yields the sufficient condition,

  $$\begin{aligned} \dfrac{v(q,q)}{v(q,q')} \le \dfrac{u(q,q)}{u(q,q') }, \end{aligned}$$

  so that the log losses to the Y side of switching partners outweigh the log losses of the X side, which is the sufficient condition in the statement of the theorem.

  Now consider the noncooperative game in which the agents drop out strategically. I will derive payoffs and solve for the perfect Bayesian equilibrium of the game. Now, if (2) holds, no agent on the X side can manipulate the timing of his exit t to guarantee a different type y is willing to accept than $q' = \tau ^{-1}(t)$. A rejected offer has no impact on the strategies of the other agents in this equilibrium, so the only relevant strategic decision is when to drop out. If dropping out at time t yields surplus

  $$\begin{aligned} \dfrac{v(x, \tau ^{-1}(t) )}{\rho } = S(x,t), \end{aligned}$$

  then the Hamilton–Jacobi–Bellman equation satisfies

  $$\begin{aligned} J_{t}(x) = 0 + e^{-\rho \Delta } \max \left\{ J_{t+\Delta }(x), S(x,t) \right\} . \end{aligned}$$

  Suppose the maximum at time $t+\Delta$ is $J_{t+\Delta }(x)$, so that type x prefers to wait at time t rather than match. Then

  $$\begin{aligned} - \dfrac{J_{t+\Delta }(x)-J_t(x)}{\Delta } + \dfrac{ J_{t+\Delta }(1-e^{-\rho \Delta })}{\Delta } = 0 \end{aligned}$$

  and taking the limit as $\Delta \rightarrow 0$ yields

  $$\begin{aligned} - \dot{J}_t(x) + \rho J_t(x) = 0, \end{aligned}$$

  with solution $J_t(x) = e^{\rho t} Z(x)$, where Z(x) is an arbitrary constant. But at the optimal dropout time, $t^*(x)$, we must have $J_{t^*(x)}(x) = S(x,t^*(x))$, or

  $$\begin{aligned} Z(x)=e^{-\rho t^*(x)}\dfrac{v(x, \tau ^{-1}(t^*(x))) )}{\rho }. \end{aligned}$$

  The objective function for type x on the X side is then

  $$\begin{aligned} \max _{t} e^{-\rho t} \dfrac{v(x, \tau ^{-1}(t))}{\rho } \end{aligned}$$

  yielding the first-order condition,

  $$\begin{aligned} - \rho v(x, x) \tau '(x) + v_{2}(x, x)= 0 \end{aligned}$$

  which, in the case of strict log-supermodularity, can solved along with the boundary condition that $\tau (0)=0$ to yield the same matching schedule as in (1), which is exactly the same matching schedule as in the direct mechanism that implements positive assortative matching. Therefore, this dynamic game implements the same allocation as the static direct mechanism, under assumption (2) or (3).

• ii. If v(x, y) is strictly log-submodular, then $\mu (x)$ and $\tau (x)$ are both decreasing as a consequence of Proposition 1 in any incentive compatible allocation, so that the highest types match first to the lowest types on the opposite side. But since u(y, x) is increasing in x, type 1 on the Y side prefers to match instantly with any type x instantly, since she is allocated to the lowest type agent under the proposed match. Then any type x could deviate and propose to 1 right away and match instantly, getting a strictly higher payoff than $v(x,x) e^{-\rho \tau (x)}/\rho$. Therefore, when the Y side can decide whether or not to accept an out-of-equilibrium proposal, the matching unravels and negative assortative matching is not implementable.

$\square$

1.3 Proof of Proposition 3:

Proof

Using (1),

$$\begin{aligned} \rho \tau (x) = \int _{0}^{x} \dfrac{ v_2( z, z)}{v(z,z)} \hbox {d}z. \end{aligned}$$

Taking the derivative twice with respect to x and evaluating at positive assortative matching yields

$$\begin{aligned} \rho \tau ''(x) = \dfrac{[v_{12}(x,x)v(x,x)-v_1(x,x)v_2(x,x)] +[v_{22}(x,x)v(x,x)-v_2(x,x)^2]}{v(x,x)^2} \end{aligned}$$

A sufficient condition for the first term to be positive is that v(x, y) be strictly log-supermodular, and a sufficient condition for the second term to be positive is that v(x, y) be log-convex in y. $\square$

1.4 Proof of Theorem 4:

Proof

i. Note that Proposition 1 provides necessary and sufficient conditions for positive assortative matching to be implementable for each side separately, so that in this proof, I only need to check whether the envelope condition can be jointly satisfied on both sides of the market (or, equivalently, that the incentive compatible matching times on the two sides of the market are the same for types that are matched under positive assortative matching).

Suppose u(y, x) and v(x, y) are strictly log-supermodular. The envelope condition from Proposition 1 and strict log-supermodularity implies the agents’ optimal strategies are to report their types honestly. This yields the matching time schedule for the X side,

$$\begin{aligned} \tau _X(q) = \log \left( \dfrac{v(q,q)}{v(0,0)} \right) - \int _{0}^{q} \dfrac{v_1(z,z)}{v(z,z)}\hbox {d}z = \int _{0}^{q} \dfrac{v_2(z,z)}{v(z,z)}\hbox {d}z, \end{aligned}$$

and similarly for the Y side,

$$\begin{aligned} \tau _Y(q) = \log \left( \dfrac{u(q,q)}{u(0,0)} \right) - \int _{0}^{q} \dfrac{u_1(z,z)}{u(z,z)}\hbox {d}z = \int _{0}^{q} \dfrac{u_2(z,z)}{u(z,z)}\hbox {d}z. \end{aligned}$$

But $\tau _X(0) = \tau _Y(0) = 0$, so that $\tau _X(q) = \tau _Y(q)$ only if their derivatives are equal. This is true if

$$\begin{aligned} \tau _X'(q) = \dfrac{v_2(q,q)}{v(q,q)} = \dfrac{u_2(q,q)}{u(q,q)} = \tau _Y'(q), \end{aligned}$$

which is condition (4). Therefore, matching can be synchronized on the two sides of the market if and only if (4) holds.

ii. Suppose the utility functions are strictly log-submodular. Then negative assortative matching prescribes that the highest type on one side arrive at time zero to match to the lowest type on the other with probability one. But the highest type on the opposite side can deviate, arrive at probability zero, and get the highest possible payoff. Therefore, there is a strictly profitable deviation for some type, and negative assortative matching is not implementable. $\square$

1.5 Proof of Theorem 5

Proof

I will prove the following claim: Supposev(x, y) andu(y, x) are strictly log-supermodular. A coarse matching$\mathcal {M}$is incentive compatible iff it is positive assortative and the downwards local incentive constraints bind.

• Necessity: Suppose $\mathcal {M}$ is incentive compatible. Then the downwards local incentive constraints follow immediately since they are a subset of the original constraints. Taking two local incentive constraints,

  $$\begin{aligned} e^{-\rho \tau _k} V_k(x_k) \ge e^{-\rho \tau _{k-1}} V_{k-1}(x_k), \quad e^{-\rho \tau _{k-1}} V_{k-1}(x_{k-1}) \ge e^{-\rho \tau _{k}} V_{k}(x_{k-1}). \end{aligned}$$

  Rearranging yields

  $$\begin{aligned} V_k(x_k)V_{k-1}(x_{k-1}) \ge V_{k-1}(x_k)V_{k}(x_{k-1}). \end{aligned}$$

  Suppose the set of partners allocated to sub-market $k-1$ is greater than the set of partners allocated to k in the strong set order. Then integrating over $y \in \alpha (X^k)$ and $y' \in \alpha (X^{k-1})$, $y < y'$, on both sides of the strict log-supermodularity inequality, $v(x,y)v(x',y') < v(x,y')v(y',x)$, since the inequality holds pointwise and $x > x'$ and $y' > y$. Integrating on both sides over $y \in \alpha (X^{k-1})$ and $y' \in \alpha (X^{k})$ over all the pointwise inequalities yields

  $$\begin{aligned} V_k(x_k)V_{k-1}(x_{k-1}) < V_{k-1}(x_k)V_{k}(x_{k-1}), \end{aligned}$$

  which is a contradiction: incentive compatibility thus implies a positive assortative coarse match. Now, imagine a downwards local incentive constraint does not bind strictly, so that

  $$\begin{aligned} e^{-\rho \tau _k} V_k(x^k) >e^{-\rho \tau _{k-1}} V_{k-1}(x^{k}). \end{aligned}$$

  The function $V_\ell (x)$ is continuous in x since v(x, y) is differentiable in x. Then there is a type $x^k - \varepsilon$, $\varepsilon >0$ but sufficiently small, for which

  $$\begin{aligned} e^{-\rho \tau _k} V_k(x^k-\varepsilon ) >e^{-\rho \tau _{k-1}} V_{k-1}(x^k-\varepsilon ). \end{aligned}$$

  But this contradicts incentive compatibility, since $x^k - \varepsilon$ should weakly prefer the $k-1$ sub-market. Therefore, downwards local incentive constraints must bind. Now, I can take two downwards incentive constraints on both sides of the market and rearrange to yield

  $$\begin{aligned} \dfrac{V_k(x^k)}{V_{k-1}(x^k)} = \dfrac{U_k(y^k)}{U_{k-1}(y^k)} = \dfrac{e^{-\rho \tau _k}}{e^{-\rho \tau _{k-1}}}. \end{aligned}$$

  Taking logs and summing then yields a recursive equation that defines the times at which the sub-markets meet, since it is without loss of generality that the first market meets at time 0.

• Sufficiency: Suppose downwards local incentive constraints bind, $\alpha$ is positive assortative, and (5) holds. Then $\mathcal {M}$ is incentive compatible. The proof is written in a series of steps:

  • If the local incentive constraints are satisfied for the boundary types, all the incentive constraints are satisfied for the boundary types. First, if v(x, y) is strictly log-supermodular and $\mathcal {M}$ is a positive assortative coarse matching, then for $x > x'$ and $x^k > x^{k'}$ in the strong set order, $v(x,y)v(x',y') > v(x,y')v(x',y)$ holds pointwise. But integrating over $y \in \alpha (X^k)$ and $y' \in \alpha (X^{k'})$ on both sides yields

    $$\begin{aligned}&\int _{y \in \alpha (X^k)} v(x,y) \dfrac{1}{\int _{z \in \alpha (X^k)}\hbox {d}z} \hbox {d}y * \int _{y \in \alpha (X^{k'})} v(x',y) \dfrac{1}{\int _{z \in \alpha (X_{k'})}\hbox {d}z} \hbox {d}y\nonumber \\&\quad > \int _{y \in \alpha (X^k)} v(x',y) \dfrac{1}{\int _{z \in \alpha (X^k)}\hbox {d}z} \hbox {d}y * \int _{y \in \alpha (X^{k'})} v(x,y) \dfrac{1}{\int _{z \in \alpha (X^{k'})}\hbox {d}z} \hbox {d}y.\nonumber \\ \end{aligned}$$

    (6)

    Now, the local downward local constraint for $X^{k+1}$ is

    $$\begin{aligned}&e^{-\rho \tau _{k+1}} \int _{y \in \alpha (X^{k+1})} v(x^{k+1},y) \dfrac{1}{\int _{z\in \alpha (X^{k+1})}\hbox {d}z}\hbox {d}y \ge e^{-\rho \tau _{k}} \int _{y \in \alpha (X^{k})} v(x^{k+1},y)\\&\quad \times \dfrac{1}{\int _{z\in \alpha (X^{k})}\hbox {d}z}\hbox {d}y \end{aligned}$$

    and the downward local constraint for $x^{k}$ is

    $$\begin{aligned}&e^{-\rho \tau _{k}} \int _{y \in \alpha (X^{k})} v(x^{k},y) \dfrac{1}{\int _{z\in \alpha (X^{k})}\hbox {d}z}\hbox {d}y \ge e^{-\rho \tau _{k-1}} \int _{y \in \alpha (X^{k-1})} v(x^{k},y)\\&\quad \times \dfrac{1}{\int _{z \in \alpha (X^{k-1})}\hbox {d}z}\hbox {d}y. \end{aligned}$$

    Now, taking logs in the second constraint and adding and subtracting terms corresponding to having type $x^{k+1}$ but reporting $x^{k-1}$ yields

    $$\begin{aligned}&- \rho \tau _k + \log \left( \int _{y \in \alpha (X^k) } v(x^{k+1},y) \dfrac{1}{\int _{z\in \alpha (X^k)} \hbox {d}z} \hbox {d}y \right) \ge \\&- \rho \tau _{k-1} + \log \left( \int _{y \in \alpha (X^{k-1}) } v(x^{k+1},y) \dfrac{1}{\int _{z \in \alpha (X^{k-1})} \hbox {d}z} \hbox {d}y \right) \\&+\log \left( \int _{y \in \alpha (X^k) } v(x^{k+1},y) \dfrac{1}{\int _{z \in \alpha (X^k)} \hbox {d}z} \hbox {d}y * \int _{y \in \alpha (X^{k-1}) } v(x^{k},y) \dfrac{1}{\int _{z \in \alpha (X^{k-1})} \hbox {d}z} \hbox {d}y \right) \\&- \log \left( \int _{y \in \alpha (X^k) } v(x^{k},y) \dfrac{1}{\int _{z \in \alpha (X^k)} \hbox {d}z} \hbox {d}y * \int _{y \in \alpha (X^{k-1}) } v(x^{k+1},y) \dfrac{1}{\int _{z\in \alpha (X^{k-1})} \hbox {d}z} \hbox {d}y \right) . \end{aligned}$$

    Now, by the first step of the proof, the last two lines are strictly positive, implying

    $$\begin{aligned}&- \rho \tau _k + \log \left( \int _{y \in \alpha (X^k) } v(x^{k+1},y) \dfrac{1}{\int _{z \in \alpha (X^k)} \hbox {d}z} \hbox {d}y \right) \ge \\&- \rho \tau _{k-1} + \log \left( \int _{y \in \alpha (X^{k-1}) } v(x^{k+1},y) \dfrac{1}{\int _{z \in \alpha (X^{k-1})} \hbox {d}z} \hbox {d}y \right) , \end{aligned}$$

    and the downward local incentive constraints for $x^{k+1}$ and $x^k$ imply that $x^{k+1}$ prefers reporting honestly to reporting $x^{k-1}$. Iterating on this argument and exploiting downward local incentive constraints for types $x^{k-\ell }$ shows that downward local incentive compatibility is satisfied for $x^{k+1}$ for all lower reports. A similar argument can be exploited to show that the upward local incentive compatibility constraints are satisfied for all higher reports. Thus, the local incentive constraints imply the global ones for the boundary types.

  • If the local incentive constraints are satisfied for the boundary types, all the incentive constraints are satisfied for the interior types. Take $x \in (x^{k}, x^{k+1})$. If the downwards incentive constraint holds for the boundary type,

    $$\begin{aligned} e^{-\rho \tau _k} V_k(x^k) \ge e^{-\rho \tau _{k'}} V_{k'}(x^k) \end{aligned}$$

    with $k' < k$, we also have

    $$\begin{aligned} e^{-\rho \tau _k} V_k(x) \ge e^{-\rho \tau _{k'}} \left( \dfrac{V_{k'}(x^k)}{V_k(x^k)} \dfrac{V_{k}(x)}{V_{k'}(x)}\right) V_{k'}(x), \end{aligned}$$

    but the term in parentheses is greater than 1, so that

    $$\begin{aligned} e^{-\rho \tau _k} V_k(x) \ge e^{-\rho \tau _{k'}} V_{k'}(x), \end{aligned}$$

    so that x prefers k to $k'$. The same argument can be applied to show the upwards incentive constraints are satisfied for interior types, taking $k'>k$ and noting the opposite log-supermodularity inequality will be satisfied.

  • The upwards local incentive constraints are automatically satisfied; thus, only the downwards local constraints are relevant. Take the binding downwards local incentive constraint

    $$\begin{aligned} e^{-\rho \tau _{k+1}} V_{k+1}(x^{k+1}) = e^{-\rho \tau _{k}} V_{k}(x^{k+1}). \end{aligned}$$

    This is equivalent to

    $$\begin{aligned} e^{-\rho \tau _{k+1}} V_{k+1}(x^k) = e^{-\rho \tau _{k}} \left( \dfrac{V_{k}(x^{k+1})}{V_{k+1}(x^{k+1})} \dfrac{V_{k+1}(x^k)}{V_{k}(x^k)}\right) V_{k}(x^k), \end{aligned}$$

    but the term in parentheses is strictly less than 1, since $V_\ell (x)$ is strictly log-supermodular. This yields

    $$\begin{aligned} e^{-\rho \tau _{k+1}} V_{k+1}(x^k) < e^{-\rho \tau _{k}}V_{k}(x^k), \end{aligned}$$

    so that the upwards local incentive constraints are satisfied.

  • If (5) is satisfied, then so are the downwards local constraints; thus$\mathcal {M}$is incentive compatible. The binding downwards local incentive constraints imply that for each $k = 1, ..., K$,

    $$\begin{aligned} e^{-\rho \tau _k} V_k(x^k) = e^{-\rho \tau _{k-1}} V_{k-1}(x^k), \quad e^{-\rho \tau _k} U_k(y^k) = e^{-\rho \tau _{k-1}} U_{k-1}(y^k). \end{aligned}$$

    Taking logs and substituting in (5) shows that the downwards local incentive constraints are all satisfied. $\square$

1.6 Proof of Proposition 6

Proof

I will show that (5) cannot hold for any $q^* \in (0,1)$ if v(x, y) exhibits larger log gains than u(y, x). If v(x, y) exhibits larger log gains than u(y, x), then for all x, y, and $y'<y$, we have

$$\begin{aligned} v(x,y)u(x,y') > u(x,y)v(x,y'). \end{aligned}$$

Integrate over $y' < q^* \le y$ to get

$$\begin{aligned} v(x,y)\int _{0}^{q^*} u(x,z)\hbox {d}z > u(x,y) \int _{0}^{q^*} v(x,z)\hbox {d}z, \end{aligned}$$

and over $y>q^*$ to get

$$\begin{aligned} \int _{q^*}^{1}v(x,z)\hbox {d}z \int _{0}^{q^*} u(x,z)\hbox {d}z > \int _{q^*}^{1} u(x,z)\hbox {d}z \int _{0}^{q^*} v(x,z)\hbox {d}z, \end{aligned}$$

and rearrange to get

$$\begin{aligned} \dfrac{\int _{q^*}^{1}v(x,z)\hbox {d}z}{\int _{0}^{q^*} v(x,z)\hbox {d}z} > \dfrac{\int _{q^*}^{1} u(x,z)\hbox {d}z}{\int _{0}^{q^*} u(x,z)\hbox {d}z}. \end{aligned}$$

This implies that for all $q^* \in (0,1)$, (5) fails, so that there can be no non-trivial coarse matching with two sub-markets. But note also that the same calculations apply when considering two sub-markets in any coarse matching, $[q^{\ell -1},q^{\ell })$ and $[q^{\ell },q^{\ell +1})$, $q^{\ell -1}<q^\ell <q^{\ell +1}$:

$$\begin{aligned} \int _{q^{\ell }}^{q^{\ell +1}}v(x,z)\hbox {d}z \int _{q^{\ell -1}}^{q^{\ell +1}} u(x,z)\hbox {d}z > \int _{q^\ell }^{q^{\ell +1}} u(x,z)\hbox {d}z \int _{q^{\ell -1}}^{q^\ell } v(x,z)\hbox {d}z, \end{aligned}$$

so that (5) must fail in general.$\square$

1.7 Proof of Proposition 7

Proof

Define the function

$$\begin{aligned} g(q) = \int _{0}^{q} u(q,z)\hbox {d}z \int _{q}^{1}v(q,z)\hbox {d}z - \int _{0}^{q}v(q,z)\hbox {d}z \int _{q}^{1}u(q,z)\hbox {d}z. \end{aligned}$$

If q is a zero of g(q), then (5) is an equality, and q partitions [0, 1] an implementable coarse matching (Fig. 2).

Fig. 2

Intuition for existence argument

Note that $q=0$ and $q=1$ are always solutions of g(q), so that simple random matching is always implementable. To provide a condition under which there are is at least one interior solution, note first that g(q) is continuous in q, $g(0)=0$, and $g(1)=0$: to ensure another solution exists, a sufficient condition is

$$\begin{aligned} g'(0)g'(1) <0. \end{aligned}$$

This implies there is some quantile $q'$ for which $g(q')>0$, and some quantile $q''$ for which $g(q'')<0$, and by the intermediate value theorem, there exists an interior point at which $g(q^*)=0$, $q^* \in (0,1)$. Computation of these derivatives yields

$$\begin{aligned} g'(0) = u(0,0)\int _{0}^{1}v(0,z)\hbox {d}z - v(0,0) \int _{0}^{1} u(0,z)\hbox {d}z \end{aligned}$$

and

$$\begin{aligned} g'(1) = u(1,1) \int _{0}^{1} v(1,z) \hbox {d}z - v(1,1) \int _{0}^{1} u(1,z)\hbox {d}z \end{aligned}$$

Then $g'(0)g'(1)$ is a quantity proportional to

$$\begin{aligned} \int _{0}^{1} \dfrac{v(0,z)}{v(0,0)} - \dfrac{u(0,z)}{u(0,0)} \hbox {d}z \times \int _{0}^{1} \dfrac{v(1,z)}{v(1,1)} - \dfrac{u(1,z)}{u(1,1)}\hbox {d}z. \end{aligned}$$

Note that the integrands now must take the opposite signs by the alternatingly larger log gains property, so that the overall value of this expression is negative, implying $g'(0)g'(1)<0$, and at least one interior zero $q^*$ exists. Then there exists a partition $\mathcal {Q} = \{ [0,q^*),[q^*,1]\}$ for which positive assortative matching is implementable. $\square$