When partner knows best: asymmetric expertise in partnerships

Abstract

This paper analyzes the problem of a principal (she) designing a contract for an agent (he) to form a short-lived partnership to exploit an asset before reselling, as in asset flipping. The agent possesses higher expertise than the principal in the sense that he can form a more-accurate assessment of the resale value of the asset before negotiating the dissolution of the partnership. By dissolving the partnership through a Texas shootout with the agent as proposer, the principal can “neutralize” her partner’s informational advantage and have him disclose the resale value for free. Thus, in the optimal contract, the agent’s superior expertise does not distort the structure of the partnership (i.e., the allocation of shares). The partners attain a higher aggregate surplus ex-ante if the principal commits to giving the asset away to the agent upon dissolving the partnership: She earns a lower revenue but lets all types of the agent enjoy a higher surplus. Thus, at the ex-ante stage, the agent could “bribe” the principal to implement this alternative. However, the additional surplus for lower types is insufficient to compensate the principal, so this higher ex-ante aggregate surplus is unattainable at the interim stage.

Appendix

1.1 Main proofs

Proof of Proposition 1

In a dissolution mechanism, the principal’s first-best payoff is bounded above by $\kappa _{P}w+\left( \kappa _{A}-\theta \right) w=(1-\theta )w.$ This upper bound is attained in the second-best environment in the Texas-shootout equilibrium with the agent as the proposer. The result is established in Proposition 1 of Brooks et al. (2010); a proof is included here for the sake of completeness.

Principal. Let $p\in \left[ 0,\overline{w}\right]$ be the price the agent calls, and let w(p) be the principal’s updated belief about w based on p. If she sells her shares, her expected payoff is $p(1-\theta )$; if she buys the agent’s shares, she makes $w(p)-p\theta .$ Thus, her best response is to sell if $p>w(p),$ buy if $p<w(p),$ and she is indifferent between buying and selling if $p=w(p).$

Agent. Let $\alpha \in [0,1]$ be the probability that the principal buys the agent’s $\theta$ shares. If the agent calls a price $p\in \left[ 0,\overline{w}\right]$ when the dissolution value is w,  his expected payoff is:

$$\begin{aligned} U(p,\alpha ;\theta ,w)&=\alpha p\theta +(1-\alpha )[w-p(1-\theta )] \\ &=\alpha p\theta +(1-\alpha )w-p(1-\alpha )(1-\theta ) \\ &=(1-\alpha )w-p[1-\alpha -\theta ]. \end{aligned}$$

If the principal chooses $\alpha =1-\theta ,$ the agent’s payoff becomes constant in p:

$$\begin{aligned} U(p,1-\theta ;\theta ,w)=(1-\alpha )w. \end{aligned}$$

Thus, $p^{*}=w$ is a best response. Moreover, $w(p^{*})=w,$ so the principal is willing to randomize. Equilibrium payoffs are $\theta w$ for the agent and $(1-\theta )w$ for the principal, the first-best payoffs. $\square$

Proof of Proposition 2

In any incentive-compatible mechanism, the feasible payoff for the principal is:

$$\begin{aligned} u_{P}(\theta ,t)=\frac{1+\delta \alpha }{2}\int _{0}^{\overline{v}} \left[ 2\theta (v)v-\left( v+\frac{1}{\lambda (v)}\right) \theta (v)^{2}\right] f(v){\mathrm {d}}v-U_{A}(0). \end{aligned}$$

In the optimal contract, $U_{A}(0)=0.$ Ignoring the monotonicity constraint, we search for the optimal share allocation rule by pointwise maximization. We look for the maximizer of the following strictly-concave parametric function:

$$\begin{aligned} J(\theta ;v)\,{:}{=}\,2v\theta -\left( v+\frac{1}{\lambda (v)}\right) \theta ^{2}=2v\theta -\left( \frac{v\lambda (v)+1}{\lambda (v)}\right) \theta ^{2} \end{aligned}$$

(5)

subject to the constraint $0\le \theta \le 1.$ For each $v\in [0,\overline{v}],$ the maximizer of (5) is $\theta ^{*}(v)$ in (3). Under Assumption 3, $\theta ^{*}(v)$ is non-decreasing. Therefore, the transfer rule $t^{*}(v)$ ensures incentive compatibility and individual rationality. $\square$

Proof of Proposition 3

In any incentive-compatible mechanism with $U_{A}(0)=0,$ the feasible payoff for the principal is now:

$$\begin{aligned} u_{P}(q,\theta ,t)&=\beta +\int _{0}^{\overline{v}}q(v) \left\{ \frac{1+\delta \alpha }{2}\left[ 2\theta (v)v -\left( v+\frac{1}{\lambda (v)}\right) \theta (v)^{2}\right] -(1-\delta )\beta \right\} f(v){\mathrm {d}}v\\ &=\beta +\int _{0}^{\overline{v}}q(v)\left\{ \frac{1+\delta \alpha }{2} J(\theta (v);v)-(1-\delta )\beta \right\} f(v){\mathrm {d}}v. \end{aligned}$$

Taking $\theta ^{*}(v)$ from (3), we have:

$$\begin{aligned} u_{P}(q,\theta ,t)\le \beta +\int _{0}^{\overline{v}}\max \left\{ \frac{1+\delta \alpha }{2}v\theta ^{*}(v) -(1-\delta )\beta ,0\right\} f(v){\mathrm {d}}v. \end{aligned}$$

Under Assumption 3, $\theta ^{*}(v)$ is continuous and strictly increasing, and so is $\psi (v)\,{:}{=}\,v\theta ^{*}(v).$ Thus, $\psi (v)$ is invertible. Moreover, under the assumption on $\beta ,$ $\frac{2(1-\delta )\beta }{1+\delta \alpha }$ is in the range of $\psi (v),$ so we can define $\underline{v}\,{:}{=}\,\psi ^{-1}\left( 2(1-\delta )\beta /(1+\delta \alpha )\right) .$ This gives:

$$\begin{aligned} u_{P}(q,\theta ,t)\le \beta +\int _{\underline{v}}^{\overline{v}} \left[ \frac{1+\delta \alpha }{2}v\theta ^{*}(v) -(1-\delta )\beta \right] f(v){\mathrm {d}}v. \end{aligned}$$

This upper bound is attained by $q^{*}(v)=I(v\ge \underline{v}),$ which is non-decreasing. Thus, $\hat{\theta }^{*}(v)=q^{*}(v)\theta ^{*}(v)$ is also non-decreasing under Assumption 3. The transfer rule $t^{*}(v)$ from Proposition 2 for $v\ge \underline{v}$ guarantees implementability. $\square$

Proof of Proposition 4

In an incentive-compatible mechanism with $U_{A}(0)=0,$

$$\begin{aligned} u_{P}(\theta ,t)=E\left[ (\theta (v)+\delta \alpha )v-\frac{1}{1+\delta \alpha }\left( v+\frac{1}{\lambda (v)}\right) \frac{(\theta (v)+\delta \alpha )^2}{2}\right] . \end{aligned}$$

We now look for the maximizer of the following strictly-concave parametric function:

$$\begin{aligned} K(\theta ;v)=(\theta +\delta \alpha ) v-\frac{1}{1+\delta \alpha } \left( v+\frac{1}{\lambda (v)}\right) \frac{(\theta +\delta \alpha )^{2}}{2}. \end{aligned}$$

(6)

This function is maximized at $\theta ^{*}_0(v)$ in (4), which is continuous and non-decreasing under Assumption 3. The corresponding transfer rule in this contract is ${\tau }^{*}(v)\,{:}{=}\,\frac{1}{2(1+\delta \alpha )}\left[ (\theta _0^{*}(v)+\delta \alpha )^{2}v -\int _{0}^{v}(\theta _0^{*}(\epsilon )+\delta \alpha )^{2}{\mathrm {d}}\epsilon \right] .$ For all v such that $\hat{\lambda }(v)\le \delta \alpha ,$ we have $\theta _0^{*}(v)=0$ and $\tau ^{*}(v)=\frac{1}{2(1+\delta \alpha )}\left[ (\delta \alpha )^{2}v-\int _{0}^{v}(\delta \alpha )^{2}{\mathrm {d}}\epsilon \right] =0.$ For all $v\ge \underline{v}\,{:}{=}\,\hat{\lambda }^{-1}(\delta \alpha ),$ if any, we have:

$$\begin{aligned} \tau ^{*}(v)&=\frac{1}{2(1+\delta \alpha )}\left[ (\theta _0^{*}(v)+\delta \alpha )^{2}v-\int _{0}^{\underline{v}} (\theta _0^{*}(\epsilon )+\delta \alpha )^{2}{\mathrm {d}}\epsilon -\int _{\underline{v}}^{v}(\theta _0^{*}(\epsilon ) +\delta \alpha )^{2}{\mathrm {d}}\epsilon \right] \\ &=\frac{1}{2(1+\delta \alpha )}\left[ (\theta _0^{*}(v)+\delta \alpha )^{2}v-(\delta \alpha )^2\underline{v} -\int _{\underline{v}}^{v}(\theta _0^{*}(\epsilon )+\delta \alpha )^{2}{\mathrm {d}}\epsilon \right] =:t^{*}_0(v). \end{aligned}$$

This establishes the proposition. $\square$

Proof of Corollary 1

For the principal,

$$\begin{aligned} u_{P}(\theta ^{*}_0,t^{*}_0)&=\frac{1+\delta \alpha }{2}E\left[ I(v\le \underline{v})J \left( \frac{\delta \alpha }{\delta \alpha +1};v\right) +I(v>\underline{v})J \left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1};v\right) \right] \\ &\le \frac{1+\delta \alpha }{2}E\left[ J\left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1};v\right) \right] =u_{P}(\theta ^{*},t^{*}); \end{aligned}$$

the inequality follows from the fact that $\frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1}$ maximizes $J(\theta ;v).$ For the agent of type $v\le \underline{v},$ we have:

$$\begin{aligned} U^*_{A0}(v)=\frac{1+\delta \alpha }{2}\int _{0}^{v} \left( \frac{\delta \alpha }{1+\delta \alpha }\right) ^{2}{\mathrm {d}}\epsilon \ge \frac{1+\delta \alpha }{2}\int _{0}^{v}\left( \frac{\hat{\lambda }(\epsilon )}{\hat{\lambda } (\epsilon )+1}\right) ^{2}{\mathrm {d}}\epsilon =U_{A}^{*}(v). \end{aligned}$$

Finally, for $v>\underline{v}$ (if any),

$$\begin{aligned} U^*_{A0}(v)&=\frac{1+\delta \alpha }{2}\int _{0}^{\underline{v}}\left( \frac{\delta \alpha }{1+\delta \alpha }\right) ^{2}{\mathrm {d}}\epsilon +\frac{1+\delta \alpha }{2} \int _{\underline{v}}^{v}\left( \frac{\theta ^{*}_0(\epsilon )+\delta \alpha }{1+\delta \alpha }\right) ^{2}{\mathrm {d}}\epsilon \\ &=\frac{1+\delta \alpha }{2}\int _{0}^{\underline{v}}\left( \frac{\delta \alpha }{1+\delta \alpha }\right) ^{2} {\mathrm {d}}\epsilon +\frac{1+\delta \alpha }{2}\int _{\underline{v}}^{v}\left( \frac{\hat{\lambda } (\epsilon )}{\hat{\lambda }(\epsilon )+1}\right) ^{2}{\mathrm {d}}\epsilon \\ &\ge \frac{1+\delta \alpha }{2}\int _{0}^{v}\left( \frac{\hat{\lambda }(\epsilon )}{\hat{\lambda } (\epsilon )+1}\right) ^{2}{\mathrm {d}}\epsilon =U_{A}^{*}(v). \end{aligned}$$

This establishes the result. $\square$

Proof of Corollary 2

We can write the ex-ante aggregate surplus under a contract with effort-choice $\widetilde{e}(\theta ,v)$ and share allocation rule $\widetilde{\theta }(v),$ $\widetilde{S},$ as:

$$\begin{aligned} \widetilde{{\mathcal {S}}}=\frac{1+\delta \alpha }{2}\int _0^{\overline{v}}\left[ 2\widetilde{e} \left( \widetilde{\theta }(v),v\right) -\widetilde{e} \left( \widetilde{\theta }(v),v\right) ^2\right] vf(v) {\mathrm {d}}v. \end{aligned}$$

(7)

For Proposition 2, (7) is:

$$\begin{aligned} {\mathcal {S}}=\frac{1+\delta \alpha }{2}\int _0^{\overline{v}} \left[ \frac{2\hat{\lambda }(v)}{\hat{\lambda }(v)+1} -\left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1}\right) ^2\right] vf(v){\mathrm {d}}v. \end{aligned}$$

Compare this surplus with the surplus under Proposition 4:

$$\begin{aligned}&\int _0^{\underline{v}}\left[ \left( \frac{2\delta \alpha }{\delta \alpha +1}\right) -\left( \frac{\delta \alpha }{\delta \alpha +1}\right) ^2\right] vf(v){\mathrm {d}}v +\int _{\underline{v}}^{\overline{v}}\left[ \frac{2\hat{\lambda }(v)}{\hat{\lambda }(v)+1} -\left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1}\right) ^{2}\right] vf(v){\mathrm {d}}v\\ &\quad \ge \int _{0}^{\overline{v}}\left[ \frac{2\hat{\lambda }(v)}{\hat{\lambda }(v)+1} -\left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1}\right) ^{2}\right] vf(v){\mathrm {d}}v, \end{aligned}$$

where the inequality follows from the fact that $\hat{\lambda }(v)/(\hat{\lambda }(v)+1)\le \delta \alpha /(\delta \alpha +1)$ for $v\le \underline{v}$ and that the function $h(x)=2x-x^2$ is strictly increasing on [0, 1). It follows that ${\mathcal {S}}_{0}\ge {\mathcal {S}}.$ $\square$

Proof of Proposition 5

We prove this proposition by showing that the effort exerted within the partnership, $e^{**}(v,p)=e^{*}(\theta ^{*}(v,p),v,p),$ is non-increasing in p; the result then follows from the fact the integrand in (7), $2e-e^2,$ is strictly increasing on [0, 1). The proof for part (i) is a straightforward extension of Example 6; details are available from the author upon request. Here, we focus on part (ii).

Since $w\le \overline{w},$ the principal can focus on $p\le \overline{w}$; furthermore, we can rule out $p=\overline{w},$ as it leads to a lower ex-ante aggregate surplus than in Proposition 2. Thus, fix $p\in [0,\overline{w}).$ If the exploitation value is null, the agent might as well exert no effort. Otherwise, for every $v\in (0,\overline{v}),$ the agent chooses effort to maximize the function:

$$\begin{aligned} {\mathrm {u}}_{A}(e,p,\theta ,t,v)=\theta e v+\delta \left[ \overline{w}-(1-\theta ) p -\int _{p}^{\overline{w}}G(w|ev){\mathrm {d}}w\right] -t-(1+\delta \alpha )v\frac{e^{2}}{2}. \end{aligned}$$

Under the additional assumptions, this function is strictly concave (as a function of e), so it has a unique maximizer $e^{*}.$ For each triple $(\theta ,v,p)$ with $0\le p<\overline{w}$ and $0<v<\overline{v},$ $e^{*}$ solves the first-order condition:

$$\begin{aligned} \theta -\delta \int _{p}^{\overline{w}}\frac{\partial G(w|e^{*}v)}{\partial \nu }{\mathrm {d}}w-(1+\delta \alpha )e^{*}=0. \end{aligned}$$

(8)

Thus, it has the following properties:

$$\begin{aligned} \frac{\partial e^{*}(\theta ,v,p)}{\partial \theta }&=\frac{1}{\delta v\int _{p}^{\overline{w}} \frac{\partial ^2 G(w|ev)}{\partial \nu ^2}{\mathrm {d}}w+(1+\delta \alpha )}>0; \end{aligned}$$

(9)

$$\begin{aligned} \frac{\partial e^{*}(\theta ,v,p)}{\partial v}&=-\frac{\delta e^{*}(\theta ,v,p)\int _{p}^{\overline{w}} \frac{\partial ^2 G(w|e^{*}(\theta ,v,p)v)}{\partial \nu ^2}{\mathrm {d}}w}{\delta v\int _{p}^{\overline{w}} \frac{\partial ^2 G(w|ev)}{\partial \nu ^2}{\mathrm {d}}w+(1+\delta \alpha )}\le 0; \end{aligned}$$

(10)

$$\begin{aligned} \frac{\partial e^{*}(\theta ,v,p)}{\partial p}&=\frac{\delta \frac{\partial G(p|e^{*}(\theta ,v,p)v)}{\partial \nu }}{\delta v\int _{p}^{\overline{w}}\frac{\partial ^2 G(w|ev)}{\partial \nu ^2}{\mathrm {d}}w +(1+\delta \alpha )}\le 0. \end{aligned}$$

(11)

We have:

$$\begin{aligned} \frac{\partial \hat{{\mathrm {u}}}_{A}(\theta ,t,v,p)}{\partial v}&=e^{*}(\theta ,v,p)\left[ \theta -\int _{p}^{\overline{w}}\frac{\partial G(w|e^*v)}{\partial \nu }{\mathrm {d}}w -\frac{1+\delta \alpha }{2}e^{*}(\theta ,v,p)\right] \\ &=\frac{1+\delta \alpha }{2} e^{*}(\theta ,v,p)^{2}, \end{aligned}$$

where the last equality follows from (8). Hence, truthful surplus in a contract with $U_{A}(0)=0$ is $U_{A}(v)=\frac{1+\delta \alpha }{2}\int _{0}^{v} e^{*}(\theta (\epsilon ,p),\epsilon ,p)^{2}{\mathrm {d}}\epsilon$; from the point of view of the ex-ante stage, this is:

$$\begin{aligned} E[U_{A}(v)]=\frac{1+\delta \alpha }{2}\int _{0}^{\overline{v}} \frac{e^{*}(\theta (v,p),v,p)^{2}}{\lambda (v)}f(v){\mathrm {d}}v. \end{aligned}$$

The principal’s payoff in an incentive-compatible contract satisfies:

$$\begin{aligned} u_P(\theta ,t,p)\propto E\left\{ \frac{1}{\lambda (v)}\left[ 2e^{*}(\theta (v,p),v,p) \hat{\lambda }(v) -\left( \hat{\lambda }(v)+1\right) e^{*}(\theta (v,p),v,p)^2\right] \right\} . \end{aligned}$$

Consider the function:

$$\begin{aligned} R(\theta ,v,p)\,{:}{=}\,2e^{*}(\theta ,v,p)\hat{\lambda }(v)- \left( \hat{\lambda }(v)+1\right) e^{*}(\theta ,v,p)^2. \end{aligned}$$

Ignoring the monotonicity constraint, the principal determines the allocation rule $\theta ^{*}(v,p)$ by maximizing $R(\theta ,v,p)$ with respect to $\theta .$ Under Assumption 3, (9) and (10) imply that $\theta ^{*}(v,p)$ is non-decreasing in v,  hence it is implementable. Now, recall $e^{**}(v,p)=e^{*}(\theta ^{*}(v,p),v,p).$ For types v such that $\theta (v,p)=0,$ we have that $e^{**}(v,p)=e^{*}(0,v,p),$ which is non-increasing in p. The same is true for all v such that $\theta (v,p)=1.$ Finally, for all v such that $\theta ^{*}(v,p)\in (0,1),$ we have $e^{**}(v,p)=\frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1}$; here, p does not affect the equilibrium effort whatsoever. It follows that $e^{**}(v,p)$ is non-increasing in p for every v. $\square$

Proof of Corollary 3

Being committed to $p=0,$ the agent’s choice of effort is $e^{*}(\theta )=\frac{\theta +\delta \alpha }{1+\delta \alpha },$ and the ex-ante aggregate surplus is:

$$\begin{aligned} {\mathcal {S}}_0(\theta ): =\frac{1+\delta \alpha }{2}\int _0^{\overline{v}}\left[ 2\frac{\theta (v)+\delta }{1+\delta } -\left( \frac{\theta (v)+\delta }{1+\delta }\right) ^2\right] vf(v){\mathrm {d}}v. \end{aligned}$$

The term in square brackets is bounded above by 1. This bound is attained by setting $\theta (v)=1$ for all $v\in [0,\overline{v}].$ To ensure ex-ante participation, the payment must be no grater than $\frac{1+\delta \alpha }{2}E(v).$ $\square$

Proof of Proposition 6

A partnership-constitution mechanism is now given by the tuple $(q,\theta _{1},\theta _{2},t_{1},t_{2}),$ where $q:\left[ 0,\overline{v}\right] ^{2}\rightarrow \{0,1\}$ is the partner-selection function—with $q(v_{1},v_{2})=1$ denoting agent 1 being selected and $q(v_{1},v_{2})=0$ denoting agent 2 making partner—and $(\theta _{i},t_{i})$ is the contract if agent $i=1,2$ is chosen. Once an agent is selected, the problem is the same as before; thus, incentive compatibility for the agents leads to:

$$\begin{aligned} U_{Ai}(v_{i},v_{-i})=U_{Ai}(0,v_{-i})+\frac{1+\delta \alpha }{2} \int _{0}^{v_{i}}\theta (\epsilon ){\mathrm {d}}\epsilon . \end{aligned}$$

In any incentive-compatible mechanism where the lowest-type agents are held to their outside option, denoting the profile of agents’ types by $\mathbf{v}=(v_{1},v_{2}),$ we have:

$$\begin{aligned}&u_{P}(\theta ,t) \\ &\quad =\frac{1+\delta \alpha }{2}\int _{0}^{\overline{v}}\int _{0}^{\overline{v}}\left[ q(\mathbf{v})J (\theta _{1}(\mathbf{v});v_{1})+(1-q(\mathbf{v}))J(\theta _{2}(\mathbf{v});v_{2})\right] f(v_{1})f(v_{2}) {\mathrm {d}}v_{1}{\mathrm {d}}v_{2} \\ &\quad \le \frac{1+\delta \alpha }{2}\int _{0}^{\overline{v}}\int _{0}^{\overline{v}}\left[ q(\mathbf{v}) J(\theta ^{*}(v_{1});v_{1})+(1-q(\mathbf{v}))J(\theta ^{*}(v_{2});v_{2})\right] f(v_{1})f(v_{2}){\mathrm {d}}v_{1}{\mathrm {d}}v_{2}\\ &\quad \le \frac{1+\delta \alpha }{2}\int _{0}^{\overline{v}}\int _{0}^{\overline{v}} \max \left\{ J(\theta ^{*}(v_{1});v_{1}),J(\theta ^{*}(v_{2});v_{2})\right\} f(v_{1})f(v_{2}){\mathrm {d}}v_{1}{\mathrm {d}}v_{2}\\ &\quad =\frac{1+\delta \alpha }{2}\int _{0}^{\overline{v}}\int _{0}^{\overline{v}}\max \left\{ \theta ^{*}(v_{1})v_{1},\theta ^{*}(v_{2})v_{2}\right\} f(v_{1})f(v_{2}){\mathrm {d}}v_{1}{\mathrm {d}}v_{2} \end{aligned}$$

where $\theta ^{*}(v_{i})$ is as in (3). The last upper bound is attained by setting:

$$\begin{aligned} q^{*}(v_{1},v_{2})=I\left( \theta ^{*}(v_{1})v_{1}\ge \theta ^{*}(v_{2})v_{2}\right) , \end{aligned}$$

which—under Assumption 3—equals $q^{*}(v_{1},v_{2})=I\left( v_{1}\ge v_{2}\right) .$ The transfer rules $t_{i}^{*}(\mathbf{v})$ ensure incentive compatibility and individual rationality. $\square$

1.2 Additional computations for Example 4

In Example 4, the cdf and hazard rate for v are:

$$\begin{aligned} F(v)=\left\{ \begin{array}{ll} \frac{3}{8}v &{}\quad 0\le v<2, \\ \frac{2}{3}+\frac{v}{24} &{}\quad 2\le v\le 8; \\ \end{array}\right. \quad \lambda (v)=\left\{ \begin{array}{ll} \frac{3}{8-3v} &{}\quad 0\le v<2, \\ \frac{1}{8-v} &{}\quad 2\le v\le 8; \end{array}\right. \end{aligned}$$

and the function $J(\theta ;v)$ from (5) is:

$$\begin{aligned} J(\theta ;v)=\left\{ \begin{array}{ll} 2v\theta -\frac{8}{3}\theta ^2 &{}\quad 0\le v<2, \\ 2v\theta -8\theta ^2 &{}\quad 2\le v\le 8. \end{array}\right. \end{aligned}$$

Following Toikka (2011), we perform a change of variable on v to obtain a uniformly distributed type. Take the quantile function $v=F^{-1}(p),$ which is given by:

$$\begin{aligned} F^{-1}(p)=\left\{ \begin{array}{ll} \frac{3}{8}p &{}\quad 0\le p<\frac{3}{4}, \\ 24p-16 &{}\quad \frac{3}{4}\le p\le 1, \end{array}\right. \end{aligned}$$

and define $\widetilde{J}(\theta ,p)\,{:}{=}\,J(\theta ,F^{-1}(p))$; we have:

$$\begin{aligned} \widetilde{J}(\theta ,p)=\left\{ \begin{array}{ll} \frac{16}{3}p\theta -\frac{8}{3}\theta ^2 &{}\quad 0\le p<\frac{3}{4}, \\ 16(3p-2)\theta -8\theta ^2 &{}\quad \frac{3}{4}\le p\le 1. \end{array}\right. \end{aligned}$$

First, we take the (piecewise) partial derivative of $\widetilde{J}(\theta ,p)$ with respect to $\theta$:

$$\begin{aligned} \widetilde{J}'_{\theta }(\theta ,p)=\left\{ \begin{array}{ll} \frac{16}{3}(p-\theta ) &{}\quad 0<p<\frac{3}{4}, \\ 48p-32-16\theta &{}\quad \frac{3}{4}<p\le 1. \end{array}\right. \end{aligned}$$

Next, compute $H(\theta ,p)\,{:}{=}\,\int _{0}^{p}\widetilde{J}'_{\theta }(\theta ,r){\mathrm {d}}r$:

$$\begin{aligned} H(\theta ,p)=\left\{ \begin{array}{ll} \frac{8}{3}p^{2}-\frac{16}{3}\theta p &{}\quad 0\le p<\frac{3}{4}, \\ 12+8\theta -16(2+\theta )p+24p^2 &{}\quad \frac{3}{4}\le p\le 1. \end{array}\right. \end{aligned}$$

For each fixed $\theta ,$ both pieces of $H(\theta ,p)$ are parabolas in p. Thus, its convex hull, $\overline{H}(\theta ,p),$ is obtained by “patching” $H(\theta ,p)$ with a linear (possibly flat) function; there are two values $p_{1}<\frac{3}{4}$ and $p_{2}\ge \frac{3}{4}$ for p and two parameters $a,b\in {\mathbb {R}}$ such that:

$$\begin{aligned} \overline{H}(\theta ,p)=\left\{ \begin{array}{ll} \frac{8}{3}p^{2}-\frac{16}{3}\theta p &{}\quad 0\le p<p_{1}, \\ ap+b &{}\quad p_{1}\le p<p_{2}, \\ 12+8\theta -16(2+\theta )p+24p^2 &{}\quad p_{2}\le p\le 1. \end{array}\right. \end{aligned}$$

Now, take the (piecewise) partial derivative of $\overline{H}(\theta ,p)$ with respect to p; call it $\overline{h}(\theta ,p)$:

$$\begin{aligned} \overline{h}(\theta ,p)\,{:}{=}\,\overline{H}'_{p}(\theta ,p)=\left\{ \begin{array}{ll} \frac{16}{3}(p-\theta ) &{}\quad 0<p<p_{1}, \\ a &{}\quad p_{1}\le p\le p_{2}, \\ 24p-16(2+\theta ) &{}\quad p_{2}<p\le 1. \end{array}\right. \end{aligned}$$

Integrate $\overline{h}(\theta ,p)$ with respect to $\theta$ to obtained the “ironed” objective function for the principal, $\overline{J}(\theta ,p)$:

$$\begin{aligned} \overline{J}(\theta ,p)\,{:}{=}\,\int _{0}^{\theta }\overline{h}(s,p) {\mathrm {d}}s=\left\{ \begin{array}{ll} \frac{16}{3}p\theta -\frac{8}{3}\theta ^2 &{}\quad 0\le p<p_{1}, \\ a\theta &{}\quad p_{1}\le p<p_{2}, \\ 16(3p-2)\theta -8\theta ^2 &{}\quad p_{2}\le p\le 1. \end{array}\right. \end{aligned}$$

In order to identify $a, b, p_{1},$ and $p_{2},$ notice that the linear “patch” on $H(\theta ,p)$ must paste smoothly with its two pieces:

$$\begin{aligned}&\frac{16}{3}(p_{1}-\theta )=a=48p_{2}-16(2+\theta );\quad (\text{smooth pasting}) \end{aligned}$$

(12)

$$\begin{aligned}&\frac{8}{3}p^2_{1}-\frac{16}{3}\theta p_{1}=ap_{1}+b; \quad (\text{value matching at}\ p_{1}) \end{aligned}$$

(13)

$$\begin{aligned}&24p_{2}^{2}-16(2+\theta )p_{2}+4(3+2\theta )=ap_{2}+b. \quad (\text{value matching at}\ p_{2}). \end{aligned}$$

(14)

From (1), we get $p_{1}=9p_{2}-6-2\theta .$ From (1) and (3), $b=-24p_{2}^{2}+4(3+2\theta ).$ Thus, (2) becomes $\frac{16}{3}p^2_{1}=-24p_{2}^{2}+4(3+2\theta ).$ Combining the latter equation with (1) yields the quadratic equation $144 p_{2}^2-72(3+\theta )p_{2}+81+27\theta +8\theta ^2=0.$ At the same time, we want to maximize the third piece of $\overline{J}(\theta ,p)$ with respect to $\theta .$ (Notice that the boundaries for each piece, $p_{1}$ and $p_{2},$ may in principle depend on $\theta$ itself.) For $p\in (0,p_{1}$), the maximizer is $\theta (p)=p$; for $p\in [p_{2},1),$ we have $\theta (p)=3p-2$—recall that $p_{2}\ge \frac{3}{4}>\frac{2}{3}.$ Plugging the latter expression evaluated at $p_{2}$ in the quadratic equation above yields:

$$\begin{aligned} 144 p_{2}^2-72(3p_{2}+1)p_{2}+81+27(3p_{2}-2)+8(3p_{2}-2)^2=0. \end{aligned}$$

Expanding the binomial squared and regrouping terms, this equation reduces to the linear equation $-6p_{2}+5=0,$ so we get $p_{2}=\frac{5}{6}$; thus, $\theta (p_{2})=\frac{1}{2}$ and, from (1), $p_{1}=\frac{1}{2}.$ The optimal ironed allocation rule in terms of p is then:

$$\begin{aligned} \overline{\theta }^{*}(p)=\left\{ \begin{array}{ll} p &{}\quad p<\frac{1}{2}, \\ \frac{1}{2} &{}\quad \frac{1}{2}\le p<\frac{5}{6}, \\ 3p-2 &{}\quad p\ge \frac{5}{6}. \end{array}\right. \end{aligned}$$

Changing variables back to v by setting $p=F(v)$ yields $\overline{\theta }^{*}(v)$ in Example 4.

1.3 Commitment to $p=\overline{w}$

If the principal commits to $p=\overline{w},$ the agent’s production payoff is ${\mathrm {u}}_{A}(e;\theta ,t,v)=\theta e v+\delta \theta \overline{w}-t-(1+\delta \alpha )v\frac{e^{2}}{2},$ which is maximized at $e^{*}(\theta )=\frac{\theta }{1+\delta \alpha }$; since the agent has a guaranteed dissolution payoff, he will exert less effort: $e^{*}(\theta )<\theta$ for all $\theta >0.$ Expected payoffs in a constitution contract are now:

$$\begin{aligned} u_{A}(\widetilde{v};v)&=\frac{\theta (\widetilde{v})^{2}}{2(1+\delta \alpha )}v+ \delta \theta (\widetilde{v})\overline{w}-t(\widetilde{v});\\ u_{P}(\theta ,t)&=E\left\{ (1-\theta (v))e^{*}(\theta (v))v+ \delta \left[ \alpha e^{*}(\theta (v))v -\theta (v)\overline{w}\right] \right\} +E[t(v)]. \end{aligned}$$

Proposition A1

(Price commitment—$p=\overline{w}$) Under Assumption 3, the optimal contract when the principal commits ex-ante to dissolution price $p=\overline{w}$ is as follows. The optimal share allocation rule is:

$$\begin{aligned} \theta ^{*}_1(v)\,{:}{=}\,\min \left\{ (1+\delta \alpha )\frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1},1\right\} . \end{aligned}$$

Define $\underline{v}$ as $\hat{\lambda }^{-1}\left( \frac{1}{\delta \alpha }\right)$ if $\max \{\hat{\lambda }(v):v\in [0,\overline{v}]\}>\frac{1}{\delta \alpha }$ and as $\overline{v}$ otherwise. An agent of type $v\le \underline{v}$ pays $t_1^{*}(v)\,{:}{=}\,\frac{1}{2(1+\delta \alpha )}\left[ \theta _1^{*}(v)^{2}v-\int _{0}^{v}\theta _1^{*}(\epsilon )^{2}{\mathrm {d}}\epsilon \right] +\delta \overline{w}\theta ^{*}_1(v),$ while an agent of type $v>\underline{v},$ if any, is awarded full ownership for $t_{1f}^{*}(v)\,{:}{=}\,\frac{1}{2(1+\delta \alpha )}\left[ \underline{v}-\int _{0}^{\underline{v}}\theta _1^{*}(\epsilon )^{2}{\mathrm {d}}\epsilon \right] +\delta \overline{w}.$ When the partnership is dissolved, the agent sells his shares back to the principal.

Proof

In an incentive-compatible mechanism, we have:

$$\begin{aligned} t(v)=\frac{\theta (v)^{2}}{2(1+\delta \alpha )}v+\delta \overline{w}\theta (v)- \frac{1}{2(1+\delta \alpha )}\int _0^v\theta (\epsilon )^{2}{\mathrm {d}}\epsilon . \end{aligned}$$

The principal’s payoff is:

$$\begin{aligned} u_{P}(\theta ,t)=\frac{1}{1+\delta \alpha }E\left[ (1+\delta \alpha )\theta (v)v -\left( v+\frac{1}{\lambda (v)}\right) \frac{\theta (v)^2}{2}\right] . \end{aligned}$$

So, we look for the maximizer of the strictly-concave parametric function:

$$\begin{aligned} L(\theta ;v)=(1+\delta \alpha ) v\theta -\left( v+\frac{1}{\lambda (v)}\right) \frac{\theta ^{2}}{2}. \end{aligned}$$

The maximizer is $\theta ^{*}_1(v),$ which is non-decreasing under Assumption 3; the transfer function is $\tau ^{*}(v)\,{:}{=}\,\frac{1}{2(1+\delta \alpha )}\left[ \theta _1^{*}(v)^{2}v-\int _{0}^{v}\theta _1^{*} (\epsilon )^{2}{\mathrm {d}}\epsilon \right] +\delta \overline{w}\theta ^{*}_1(v).$ For $v\le \underline{v},$ we have:

$$\begin{aligned} \tau ^{*}(v)=\frac{1}{2(1+\delta \alpha )}\left[ \theta _1^{*}(v)^{2}v-\int _{0}^{v}\theta _1^{*} (\epsilon )^{2}{\mathrm {d}}\epsilon \right] +\delta \overline{w}\theta ^{*}_1(v)=:t^{*}_1(v); \end{aligned}$$

for $v>\underline{v},$ if any such v exists,

$$\begin{aligned} \tau ^{*}(v)&=\frac{1}{2(1+\delta \alpha )}\left[ v-\int _{0}^{\underline{v}}\theta _1^{*} (\epsilon )^{2}{\mathrm {d}}\epsilon -\int _{\underline{v}}^{v}{\mathrm {d}}\epsilon \right] +\delta \overline{w}\\ &=\frac{1}{2(1+\delta \alpha )}\left[ \underline{v}-\int _{0}^{\underline{v}}\theta _1^{*} (\epsilon )^{2}{\mathrm {d}}\epsilon \right] +\delta \overline{w}=:t^{*}_{1f}(v). \end{aligned}$$

This establishes the proposition. $\square$

Corollary A1

(The value of commitment, II) Denote by $u_{P}(\theta ^{*}_1,t^{*}_1)$ and $U^*_{A1}(v)$ the payoffs for the principal and a type-v agent, respectively, under the contract in the proposition above. We have $u_{P}(\theta ^{*},t^{*})\ge u_{P}(\theta ^{*}_1,t^{*}_1)$ and $U^*_{A}(v)\ge U^*_{A1}(v)$ for all v.

Proof

For the principal, we have:

$$\begin{aligned} u_{P}(\theta ^{*}_1,t^{*}_1)&=(1+\delta \alpha )E\left\{ I(v<\underline{v})\left[ \frac{v\hat{\lambda }(v)}{\hat{\lambda }(v)+1}-\left( v+\frac{1}{\lambda (v)}\right) \frac{1}{2}\left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1}\right) ^2\right] \right\} \\ &\quad +\frac{1}{1+\delta \alpha }E\left\{ I(v>\underline{v})\left[ (1+\delta \alpha )v-\left( v+\frac{1}{\lambda (v)}\right) \frac{1}{2}\right] \right\} \\ u_{P}(\theta ^{*}_1,t^{*}_1)&=(1+\delta \alpha )E\left\{ I(v<\underline{v})\left[ \frac{v\hat{\lambda }(v)}{\hat{\lambda }(v)+1}-\left( v+\frac{1}{\lambda (v)}\right) \frac{1}{2}\left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1} \right) ^2\right] \right\} \\ &\quad +(1+\delta \alpha )E\left\{ I(v>\underline{v})\left[ \frac{1}{1+\delta \alpha }v-\left( v+\frac{1}{\lambda (v)}\right) \frac{1}{2}\left( \frac{1}{1+\delta \alpha }\right) ^2\right] \right\} \\ &=\frac{1+\delta \alpha }{2}E\left[ I(v<\underline{v})J\left( \frac{\hat{\lambda }(v)}{\hat{\lambda }(v)+1}; v\right) +I(v>\underline{v})J\left( \frac{1}{1+\delta \alpha };v\right) \right] \\ &\le \frac{1+\delta \alpha }{2}E\left[ J\left( \theta ^{*}(v);v\right) \right] =u_{P}(\theta ^{*},t^{*}), \end{aligned}$$

where $J(\theta ;v)$ is the function in (5). For the agent of type $v\le \underline{v},$

$$\begin{aligned} U^*_{A1}(v)&=\frac{1}{2(1+\delta \alpha )}\int _{0}^{v}\left( (1+\delta \alpha )\frac{\hat{\lambda }(\epsilon )}{\hat{\lambda }(\epsilon )+1}\right) ^2{\mathrm {d}}\epsilon \\ &=\frac{1+\delta \alpha }{2}\int _{0}^{v}\left( \frac{\hat{\lambda }(\epsilon )}{\hat{\lambda }(\epsilon )+1} \right) ^2{\mathrm {d}}\epsilon =U_{A}^{*}(v). \end{aligned}$$

Finally, for the agent of type $v>\underline{v},$ if any,

$$\begin{aligned} U^*_{A1}(v)&=\frac{1}{2(1+\delta \alpha )}\left[ \int _{0}^{\underline{v}}\left( (1+\delta \alpha ) \frac{\hat{\lambda }(\epsilon )}{\hat{\lambda }(\epsilon )+1}\right) ^2{\mathrm {d}}\epsilon +\int _{\underline{v}}^v{\mathrm {d}}\epsilon \right] \\ &=\frac{1+\delta \alpha }{2}\left[ \int _{0}^{\underline{v}}\left( \frac{\hat{\lambda }(\epsilon )}{\hat{\lambda }(\epsilon )+1}\right) ^2{\mathrm {d}}\epsilon +\left( \frac{1}{1+\delta \alpha }\right) ^2(v-\underline{v})\right] \\ &\le \frac{1+\delta \alpha }{2}\left[ \int _{0}^{\underline{v}}\left( \frac{\hat{\lambda }(\epsilon )}{\hat{\lambda }(\epsilon )+1}\right) ^2{\mathrm {d}}\epsilon +\int _{\underline{v}}^{v} \left( \frac{\hat{\lambda }(\epsilon )}{\hat{\lambda }(\epsilon )+1}\right) ^2{\mathrm {d}}\epsilon \right] =U^*_{A}(v). \end{aligned}$$

This establishes the result. $\square$