Reorganizing a partnership efficiently

Abstract

We modify the partnership dissolution model pioneered by Cramton et al. (Econometrica 55:615–632, 1987) to consider the possibility that each partner has an optimal scale and hence values only a fraction of an object, called a block. To achieve efficiency, a partnership should be reorganized so that multiple blocks are allocated one-to-one to the partners who have the highest valuations. The set of initial ownership distributions under which efficient reorganization can be achieved is non-convex. A condition reveals the relationship between the possibility of efficient reorganization for any given partnership and three characteristics that it entails: the number of blocks available (K), the total number of partners (N), and the number of partners who own up to a block (S). Given that K and N are fixed, efficiency can be achieved if and only if S is sufficiently low. In addition, given that N and S are fixed, efficiency can be achieved if and only if K is sufficiently high.

Appendix

Proof of Lemma 1

We first prove that R(q) is the highest revenue that can be generated by any mechanism that implements $q\left( v\right)$ and satisfies IC and IR. Suppose that (q(v), t(v)) satisfies incentive compatibility. This means that

$$\begin{aligned} U_{i}\left( v_{i}\right) =v_{i}Q_{i}(v_{i})-T_{i}\left( v_{i}\right) \ge v_{i}Q_{i}(v_{i}^{\prime })-T_{i}\left( v_{i}^{\prime }\right) , \end{aligned}$$

or, equivalently,

$$\begin{aligned} U_{i}\left( v_{i}\right) \ge U_{i}\left( v_{i}^{\prime }\right) +(v_{i}-v_{i}^{\prime })Q_{i}(v_{i}), \end{aligned}$$

which implies that $U_{i}$ has a supporting hyperplane at $v_{i}^{\prime }$ with the slope $Q_{i}(v_{i}^{\prime })\ge 0$ , and hence $U_{i}(v_{i})$ must be convex and has derivatives $U_{i}^{\prime }(v_{i})=Q_{i}(v_{i})$ almost everywhere. This also means that $Q_{i}(v_{i})$ must be increasing in $v_{i}$ and that

$$\begin{aligned} U\left( v_{i}\right) =U\left( v_{i}^{*}\right) +\int _{v_{i}^{*}}^{v_{i}}Q_{i}(u)du, \end{aligned}$$

where $v_{i}^{*}$ is the type that gives player i the lowest payoff under (q(v), t(v)), and

$$\begin{aligned} T_{i}\left( v_{i}\right) =-v_{i}^{*}Q_{i}(v_{i}^{*})+v_{i}Q_{i}(v_{i})+T_{i}\left( v_{i}^{*}\right) -\int _{v_{i}^{*}}^{v_{i}}Q_{i}(u)du. \end{aligned}$$

Suppose that player $i$ is a buyer (i.e., $r_{i}<\alpha$). In this case, $v_{i}^{*}=0$, and the highest revenue can be extracted without violating the IR constraint by setting $U_{i}\left( 0\right) =0.$ This means that $T_{i}\left( 0\right) =0$ and $T_{i}\left( v_{i}\right) =v_{i}Q_{i}(v_{i})-\int _{0}^{v_{i}}Q_{i}(u)du$. Therefore, the expected revenue from the buyer is

$$\begin{aligned} \int _{0}^{1}T_{i}\left( v_{i}\right) f(v_{i})dv_{i}= & {} \int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}-\int _{0}^{1} \int _{0}^{v_{i}}Q_{i}(u)duf(v_{i})dv_{i} \\= & {} \int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}-\int _{0}^{1} \int _{u}^{1}Q_{i}(u)f(v_{i})dv_{i}du \\= & {} \int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}-\int _{0}^{1}Q_{i}(u) \int _{u}^{1}f(v_{i})dv_{i}du \\= & {} \int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}-\int _{0}^{1}Q_{i}(v_{i})\left[ 1-F\left( v_{i}\right) \right] dv_{i} \\= & {} \int _{0}^{1}[v_{i}-\frac{1-F\left( v_{i}\right) }{f\left( v_{i}\right) } ]Q_{i}(v_{i})f(v_{i})dv_{i}. \end{aligned}$$

Suppose that player i is a seller (i.e., $r_{i}\ge \alpha$ ). In this case, $v_{i}^{*}=1$, and the highest revenue can be extracted without violating the IR constraint by setting $U_{i}\left( 1\right) =1.$ This means that $T_{i}\left( 0\right) =0$ and $T_{i}\left( v_{i}\right) =-1+v_{i}Q_{i}(v_{i})-\int _{1}^{v_{i}}Q_{i}(u)du.$ Therefore, the expected revenue from the seller is

$$\begin{aligned} \int _{0}^{1}T_{i}\left( v_{i}\right) f(v_{i})dv_{i}= & {} -1+\int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}-\int _{0}^{1} \int _{1}^{v_{i}}Q_{i}(u)duf(v_{i})dv_{i} \\= & {} -1+\int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}+\int _{0}^{1} \int _{0}^{u}Q_{i}(u)f(v_{i})dv_{i}du \\= & {} -1+\int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}+\int _{0}^{1}Q_{i}(u) \int _{0}^{u}f(v_{i})dv_{i}du \\= & {} -1+\int _{0}^{1}v_{i}Q_{i}(v_{i})f(v_{i})dv_{i}+\int _{0}^{1}Q_{i}(v_{i})F \left( v_{i}\right) dv_{i} \\= & {} -1+\int _{0}^{1}[v_{i}+\frac{F\left( v_{i}\right) }{f\left( v_{i}\right) } ]Q_{i}(v_{i})f(v_{i})dv_{i}. \end{aligned}$$

Thus, the expected revenue of a mechanism (q(v), t(v)) that satisfies IC and IR is

$$\begin{aligned} \sum \limits _{i=1}^{N}E(t_{i}(v))= & {} \sum \limits _{i=1}^{N}E(T_{i}\left( v_{i}\right) ) \\= & {} \sum \limits _{i=1}^{N}E[v_{i}q_{i}(v)-\int _{v_{i}^{*}}^{v_{i}}q_{i}(u,v_{-i})du]-\sum \limits _{i=1}^{N}U_{i}\left( v_{i}^{*}\right) \\= & {} \sum \limits _{i=1}^{N}E[v_{i}q_{i}(v)-(v_{i}-\phi _{i}(v_{i}))q_{i}(v)]-\sum \limits _{i=1}^{N}U_{i}\left( v_{i}^{*}\right) \\= & {} \sum \limits _{i=1}^{N}E[\phi _{i}(v_{i})q_{i}(v)]-S=R(q), \end{aligned}$$

where the last equality is due to the fact that

$$\begin{aligned} S=\sum \limits _{\left\{ i|r_{i}\ge \alpha \right\} }E[\phi _{i}(v_{i})] \text{, } \end{aligned}$$

which can be seen with integration by parts. This completes the proof of the first point of the Lemma.

The “only if” part of the second point: It follows immediately from the first point that BB is satisfied if $R(q)\ge 0.$

The “if” part of the second point: Suppose that q(v) satisfies $R(q)\ge 0.$ Let

$$\begin{aligned} h_{i}(v_{i})\equiv & {} E_{-i}[v_{i}q_{i}(v_{i},v_{-i})-\int _{v_{i}^{*}}^{v_{i}}q_{i}(u,v_{-i})du]{{\ and}} \nonumber \\ c_{i}\equiv & {} \frac{R(q)}{N}+v_{i}^{*}-\frac{1}{N-1}\sum _{j\ne i}E[\phi _{j}(v_{j}|v_{j}^{*})q_{j}(v)]. \end{aligned}$$

(13)

Also let

$$\begin{aligned} t_{i}(v)=h_{i}(v_{i})-\frac{1}{N-1}\sum _{j\ne i}q_{j}(v_{j})-c_{i}, \forall v. \end{aligned}$$

It follows that

$$\begin{aligned} E_{i}[h_{i}(v_{i})]=E[\phi _{i}(v_{i})q_{i}(v)] \end{aligned}$$

and hence

$$\begin{aligned} \sum _{i=1}^{N}c_{i}=0{ {and }}\sum _{i=1}^{N}t_{i}(v)=0, \end{aligned}$$

that is, BB is satisfied. To see that (q, t) satisfies IR,  note that

$$\begin{aligned} T_{i}\left( v_{i}\right)= & {} h_{i}(v_{i})-\frac{1}{N-1}\sum _{j\ne i}E[\phi _{j}(v_{j})q_{j}(v)]-c_{i} \\= & {} h_{i}(v_{i})-\frac{R(q)}{N}-v_{i}^{*}. \end{aligned}$$

Hence, for all $v_{i},$

$$\begin{aligned} U_{i}\left( v_{i}\right) =E_{-i}[v_{i}q_{i}(v_{i},v_{-i})]-T_{i}\left( v_{i}\right) =\int _{v_{i}^{*}}^{v_{i}}E_{-i}[q_{i}(u,v_{-i})]du+\frac{ R(q)}{N}+v_{i}^{*}, \nonumber \\ \end{aligned}$$

(14)

where the second equality is given by (13). Moreover, by the definition of the worst type $v_{i}^{*},$

$$\begin{aligned} \int _{0}^{v_{i}}E_{-i}[q_{i}(u,v_{-i})]du\ge \int _{0}^{v_{i}^{*}}E_{-i}[q_{i}(u,v_{-i})]du-v_{i}^{*}=0 \text{ for } \text{ the } \text{ buyer, } \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{v_{i}}E_{-i}[q_{i}(u,v_{-i})]du-v_{i}\ge \int _{0}^{1}E_{-i}[q_{i}(u,v_{-i})]du-1 \text{ for } \text{ the } \text{ seller }. \end{aligned}$$

It follows from (14) that

$$\begin{aligned} U_{i}\left( v_{i}\right) =\int _{0}^{v_{i}}E_{-i}[q_{i}(u,v_{-i})]du+\frac{ R(q)}{N}\ge 0, \text{ for } \text{ the } \text{ buyer } \end{aligned}$$

and

$$\begin{aligned} U_{i}\left( v_{i}\right) =\int _{1}^{v_{i}}E_{-i}[q_{i}(u,v_{-i})]du+\frac{ R(q)}{N}+1\ge 0, \text{ for } \text{ the } \text{ seller, } \end{aligned}$$

that is, IR is satisfied. In addition, to see that $(q\left( v\right) ,t\left( v\right) )$ satisfies IC,  consider any $v_{i}$ and $v_{i}^{^{\prime }}.$ It follows that

$$\begin{aligned} U_{i}\left( v_{i}\right) -U_{i}\left( v_{i}^{^{\prime }}\right)= & {} \int _{v_{i}^{^{\prime }}}^{v_{i}}E_{-i}[q_{i}(u,v_{-i})]du \\\ge & {} \int _{v_{i}^{^{\prime }}}^{v_{i}}E_{-i}[q_{i}(v_{i}^{^{\prime }},v_{-i})]du \\= & {} E_{-i}[v_{i}q_{i}(v_{i}^{^{\prime }},v_{-i})-v_{i}^{^{\prime }}q_{i}(v_{i}^{^{\prime }},v_{-i})]. \end{aligned}$$

This implies that

$$\begin{aligned} U_{i}\left( v_{i}\right)\ge & {} E_{-i}[v_{i}q_{i}(v_{i}^{^{\prime }},v_{-i})]-T_{i}\left( v_{i}^{^{\prime }}\right) \\= & {} v_{i}Q_{i}(v_{i}^{^{\prime }})-T_{i}\left( v_{i}^{^{\prime }}\right) , \end{aligned}$$

which means that IC is satisfied. $\square$