# Efficient Dissolution of Partnerships

### A summary by Nick Cao

This is a summary of Cramton, P., Gibbons, R., & Klemperer, P., ``Dissolving a Partnership Efficiently'', *Econometrica*, Vol. 55, No. 3, (May, 1987), 615--632.

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A partnership is jointly owned by agents \(1, \dots, n\), each of whom own shares \(\mathbf{r} = (r_1, \dots, r_n)\) respectively. Their valuations \(\mathbf{v} = (v_1,\dots,v_n)\) are independently distributed according to a distribution \(F\) over \([ \underline{v}, \overline{v} ]\). Social welfare under the ex-post efficient outcome is given by \(W^*(\mathbf{v}) = \max \{v_1,\dots,v_n\}\).

The expected ex-post efficient social welfare is \begin{equation}\label{EW} \mathbf{E}[W^*(\mathbf{v})] = \mathbf{E}[\max \{v_1,\dots,v_n\}] = n \int_\underline{v}^\overline{v} v F(v)^{n-1} f(v) dv \end{equation} where the probability density function of the maximum valuation is $ n F(v)^{n-1} f(v)$. On the other hand, we have \[ H_i(v_i) = \mathbf{E}[\max \{v_1,\dots,v_n\} | v_i] - r_i v_i = v_i F(v_i)^{n-1} + (n-1) \int_{v_i}^\overline{v} v F(v)^{n-2} f(v) dv - r_i v_i \] where: (1) the expectation is now conditioned on knowing the value of $v_i$; (2) evaluating the maximum otherwise remains the same; and (3) an allowance must be made for the non-zero payoff of non-participation. By the fundamental theorem of calculus, we have \[ H_i'(v_i) = F(v_i)^{n-1} + (n-1) v_i F(v_i)^{n-2} f(v_i) - (n-1) v_i F(v_i)^{n-2} f(v_i) - r_i = F(v_i)^{n-1} - r_i \] At $v_i = v_i^*$, the first order condition stipulates that $H_i'(v_i^*) = 0$. Therefore, we must have \begin{equation}\label{v_i^*} F(v_i^*)^{n-1} = r_i \iff v_i^* = F^{-1}(\sqrt[n-1]{r_i}) \end{equation} where $F^{-1}$ can be interpreted as the quantile function of the distribution. Then we may evaluate the function $H_i$ at $v_i^*$ to obtain \begin{equation} H_i(v_i^*) = (n-1) \int_{v_i^*}^\overline{v} v F(v)^{n-2} f(v) dv \end{equation} The condition for dominant incentive compatible, ex-post budget balanced, interim individually rational, and ex-post efficient partnership dissolution is that $\sum_{i=1}^n H_i(v_i^*) \geq (n-1) \mathbf{E}[W^*(\mathbf{v})]$, which is equivalent to \begin{equation} (n-1) \sum_{i=1}^n \int_{v_i^*}^\overline{v} v F(v)^{n-2} f(v) dv \geq n (n-1) \int_\underline{v}^\overline{v} v F(v)^{n-1} f(v) dv \end{equation} Integrating by parts, this condition becomes \[ \sum_{i=1}^n \left( \left[ v F(v)^{n-1} \right]_{v_i^*}^\overline{v} - \int_{v_i^*}^\overline{v} F(v)^{n-1} dv \right) \geq (n-1) \left( \left[ v F(v)^{n} \right]_\underline{v}^\overline{v} - \int_\underline{v}^\overline{v} F(v)^{n} dv \right) \] Evaluating the terms (note that $F(\overline{v}) = 1$ and $F(\underline{v}) = 0$), we obtain \[ \sum_{i=1}^n \left( \overline{v} - v_i^* F(v_i^*)^{n-1} - \int_{v_i^*}^\overline{v} F(v)^{n-1} dv \right) \geq (n-1) \left( \overline{v} - \int_\underline{v}^\overline{v} F(v)^{n} dv \right) \] Then recalling that $F(v_i^*)^{n-1} = r_i$, the condition becomes \begin{equation*} n \overline{v} + \sum_{i=1}^n \left( - r_i v_i^* - \int_{v_i^*}^\overline{v} F(v)^{n-1} dv \right) \geq (n-1) \overline{v} - (n-1) \int_\underline{v}^\overline{v} F(v)^{n} dv \end{equation*} which further simplifies to: \begin{equation}\label{cond} \overline{v} + \sum_{i=1}^n \left( - r_i v_i^* - \int_{v_i^*}^\overline{v} F(v)^{n-1} dv \right) + (n-1) \int_\underline{v}^\overline{v} F(v)^{n} dv \geq 0 \end{equation}

### Equal Ownership

Here, $r_i = 1/n$ for all $i = 1,\dots,n$. As ownerships are equal, equation (\ref{v_i^*}) shows that $v_i^*$ must be equal across all partners. The left hand side of (\ref{cond}) thus simplifies to \[ \overline{v} - v_i^* - n \int_{v_i^*}^\overline{v} F(v)^{n-1} dv + (n-1) \int_\underline{v}^\overline{v} F(v)^{n} dv \] This is equal to \begin{equation}\label{eq_own_intermediate} \overline{v} - v_i^* + (n-1) \int_\underline{v}^{v_i^*} F(v)^{n} dv - \int_{v_i^*}^\overline{v} \left( n F(v)^{n-1} - (n-1) F(v)^{n} \right) dv \end{equation} Taking the derivative of $n F(v)^{n-1} - (n-1) F(v)^{n}$, we obtain \[ n(n-1) F(v)^{n-2} f(v) - n(n-1) F(v)^{n-1} f(v) = n(n-1) F(v)^{n-2} f(v) (1-F(v)) \geq 0 \] as $F(v) \leq 1$ by the axioms of probability. The maximum point is exactly at $F(v) = 1$, which is where $v = \overline{v}$. Here, $n F(\overline{v})^{n-1} - (n-1) F(\overline{v})^{n} = 1$. Hence we have shown that $n F(v)^{n-1} - (n-1) F(v)^{n} \leq 1$ for all $v \in [\underline{v}, \overline{v}]$, and thus we can bound the expression (\ref{eq_own_intermediate}) from above by \[ \overline{v} - v_i^* + (n-1) \int_\underline{v}^{v_i^*} F(v)^{n} dv - \int_{v_i^*}^\overline{v} dv \] This is equal to \[ \overline{v} - v_i^* + (n-1) \int_\underline{v}^{v_i^*} F(v)^{n} dv - (\overline{v} - v_i^*) \] and is equal to \[ (n-1) \int_\underline{v}^{v_i^*} F(v)^{n} dv > 0 \] Hence condition (\ref{cond}) is satisfied and efficient surplus division is possible.

### Concentrated Ownership

Say if partner 1 currently owns the entire partnership: $r_1 = 1$ and $r_i = 0$ for $i = 2, \dots, n$. Then according to equation (\ref{v_i^*}), we have that $v_1^* = F^{-1}(1) = \overline{v}$ and $v_i^* = F^{-1}(0) = \underline{v}$ for all $i = 2, \dots, n$. The left hand side of (\ref{cond}) therefore becomes \[ \overline{v} - v_1^* - (n-1) \int_\underline{v}^\overline{v} F(v)^{n-1} dv + (n-1) \int_\underline{v}^\overline{v} F(v)^{n} dv \] which simplifies simply to \[ (n-1) \int_\underline{v}^\overline{v} F(v)^{n-1} (F(v) - 1) dv < 0 \] As the condition fails, it is impossible to dissolve the partnership in a way that is ex-post efficient, ex-post budget balanced, interim individually rational, and Bayesian incentive compatible.