Step 1: Self-Generating Maximization (SGM) Program

Given contracts \({w_2}, \ldots ,{w_N}\), \({P_1}\) chooses \({w_1}\) to solve the program \[\mathop {\max }\limits_{{w_1} \in W} \left( {{B_1} - {w_1}} \right) \cdot \phi \left( {{e^*}\left( w \right)} \right) = \mathop {\max }\limits_{{w_1} \in W} {u_1}\left( {{w_1},w} \right).\]

That is, \({P_1}\) cares directly about the contract she offers and about the aggregate of the contracts that all parties offer. Instead of thinking of \({P_1}\) as choosing only her own contract, we can think of her as choosing the aggregate contract \(w\) that will be offered to the agent. That is, write \[w = {w_1} + \sum\limits_{i = 2}^N {{w_i}} .\]

Then \({w_1} \in W\) if and only if \(w \in W + \sum\nolimits_{i = 2}^N {{w_i}} .\) \({P_1}\)'s problem can therefore be written as \[\mathop {\max }\limits_{w \in W + \sum\nolimits_{i = 2}^N {{w_i}} } {u_1}\left( {w - \sum\limits_{i = 2}^N {{w_i}} ,w} \right).\]

If an aggregate contract \(\bar w \in W\) is an equilibrium aggregate contract, then in each principal's problem of choosing an aggregate contract, they each optimally offer contract \(\bar w\). That is, for each principal \({P_i}\), \[\bar w \in \mathop {\arg \max }\limits_{w \in W + \bar w - {{\bar w}_i}} {u_i}\left( {w - \sum\limits_{j \ne i} {{{\bar w}_j}} ,w} \right).\]

Since \(\bar w\) solves this problem for each \(i\), it also solves the sum of these problems: \[\bar w \in \mathop {\arg \max }\limits_{w \in \bigcap\nolimits_{i = 1}^N {W + \bar w - {{\bar w}_i}} } \sum\limits_{i = 1}^N {{u_i}\left( {w - \sum\limits_{j \ne i} {{{\bar w}_j}} ,w} \right)} .\] Plugging in the definition of this objective, we have \[\Lambda \left( {w,\bar w} \right) = \sum\limits_{i = 1}^N {{u_i}} \left( {w - \sum\limits_{j \ne i} {{{\bar w}_j}} ,w} \right) = \sum\limits_{i = 1}^N {\left( {{B_i} - \left( {w - \sum\limits_{j \ne i} {{{\bar w}_j}} } \right)} \right)}  \cdot \phi \left( {{e^*}\left( w \right)} \right),\] or \[\Lambda \left( {w,\bar w} \right) = \left( {B - Nw + \left( {N - 1} \right)\bar w} \right) \cdot \phi \left( {{e^*}\left( w \right)} \right),\]

where \(B = \sum\nolimits_{i = 1}^N {{B_i}} \). Therefore, if \(\bar w\) is an equilibrium aggregate, then \[\bar w \in \mathop {\arg \max }\limits_{w \in \bigcap\nolimits_{i = 1}^N {W + \bar w - {{\bar w}_i}} } \Lambda \left( {w,\bar w} \right)\] for some \({\bar w_1}, \ldots ,{\bar w_N}\) such that \({\bar w_1} +  \cdots  + {\bar w_N} = \bar w\). Putting these steps together, we have the following lemma.

Lemma 1: If \(\bar w\) is an equilibrium aggregate contract, then \[\bar w \in \mathop {\arg \max }\limits_{w \in W\left( {\bar w} \right)} \Lambda \left( {w,\bar w} \right),\] where \[W\left( {\bar w} \right) = \bigcup\limits_{{{\bar w}_1} +  \cdots  + {{\bar w}_N},{{\bar w}_i} \in W} {\bigcap\limits_{i = 1}^N {W + \bar w - {{\bar w}_i}} } .\] At this point, we have shown only that if \(\bar w\) is an equilibrium aggregate contract, it solves this SGM program. That is, solving this program is a necessary, but not sufficient, condition for \(\bar w\) to be an equilibrium aggregate contract. In other words, all we have shown at this point is that if \(\bar w\) does not solve this program, it cannot be an equilibrium aggregate contract. Lemma 4 below will derive necessary and sufficient conditions for \(\bar w\) to be an equilibrium aggregate contract when Assumption S  is satisfied.

I now state several lemmas that simplify this problem. Lemmas 2 and 3 simplify the set \(W\left( {\bar w} \right)\).

Lemma 2: Suppose \({X_1}, \ldots ,{X_N}\) are convex cones. Then \[\bigcap\limits_{i = 1}^N {{X_i}}  \subset \sum\limits_{i = 1}^N {\frac{1}{N}{X_i}.} \]Proof of Lemma 2: Take \(x \in \bigcap\nolimits_{i = 1}^N {{X_i}} \). Then \(x \in {X_i}\). Since \({X_i}\) is a convex cone, \(\frac{1}{N}x \in {X_i}\) for each \(i\). Therefore \[x = \sum\limits_{i = 1}^N {\frac{1}{N}x}  \in \sum\limits_{i = 1}^N {\frac{1}{N}{X_i}} ,\]which establishes the claim. Q.E.D.

Lemma 3: For any \(\bar x \in X\), \[\bigcup\limits_{{x_1} +  \cdots  + {x_N} = \bar x,{x_i} \in X} {\bigcap\limits_{i = 1}^N {X + {x_i} = X + \frac{1}{N}\bar x.} } \]Proof of Lemma 3: To prove \(\left(  \subset  \right)\), take \({x_1}, \ldots ,{x_N}\) such that \(\sum\nolimits_{i = 1}^N {{x_i} = \bar x} \). From Lemma 2, we know that \[\bigcap\limits_{i = 1}^N {X + {x_i}}  \subset \sum\limits_{i = 1}^N {\frac{1}{N}\left( {X + {x_i}} \right).} \]Since \(X\) is a convex cone, \(\frac{1}{N}X = X\). We therefore have that \[\sum\limits_{i = 1}^N {\frac{1}{N}\left( {X + {x_i}} \right) = X + \sum\limits_{i = 1}^N {\frac{1}{N}{x_i}} }  = X + \frac{1}{N}\bar x,\]since \({x_i}\) is just a scalar. Since \({x_1}, \ldots ,{x_N}\) was arbitrary, this holds for all \({x_1}, \ldots ,{x_N}\) for which \(\sum\nolimits_{i = 1}^N {{x_i} = \bar x} \). This establishes \(\left(  \subset  \right)\).

To prove \(\left(  \supset  \right)\), suppose \(x \in X + \frac{1}{N}\bar x\). Let \({x_i} = \frac{1}{N}x\). Then \(\sum\nolimits_{i = 1}^N {\frac{1}{N}{x_i} = x} \), and \(\frac{1}{N}x \in X + \frac{1}{N}x\), establishing the claim. Q.E.D.

As a corollary to lemmas 2 and 3, we have the following characterization of \(W\left( {\bar w} \right)\).

Corollary 1: For any \(\bar w \in W\), \[W\left( {\bar w} \right) = \bigcup\limits_{{{\bar w}_1} +  \cdots  + {{\bar w}_N} = \bar w,{{\bar w}_i} \in W} {\bigcap\limits_{i = 1}^N {W + \bar w - {{\bar w}_i} = W + \left( {1 - \frac{1}{N}} \right)\bar w.} } \]Taking stock, by Corollary 1 and Lemma 1, if an aggregate contract \(\bar w\) is an equilibrium aggregate contract, then \[\bar w \in \mathop {\arg \max }\limits_{\bar w \in W + \left( {1 - 1/N} \right)\bar w} \Lambda \left( {w,\bar w} \right).\]Again, solving this program is a necessary, but not sufficient condition for \(\bar w\) to be an equilibrium aggregate contract. However, under Assumption S, it is also sufficient, as the following lemma shows.

Lemma 4: Suppose Assumption S holds. Then \(\bar w \in W\) is an equilibrium aggregate contract if and only if \[\bar w \in \mathop {\arg \max }\limits_{\bar w \in W + \left( {1 - 1/N} \right)\bar w} \Lambda \left( {w,\bar w} \right).\]Proof of Lemma 4: The necessary conditions have already been established. Now, suppose \(\bar w\) solves this program. Then let \({\bar w_i} = \frac{1}{N}\bar w\) for \(i = 1, \ldots ,N\). Principal \(i\)'s program is to \[\mathop {\max }\limits_{w \in W + \left( {1 - 1/N} \right)\bar w} \left( {{B_i} - \left( {w - \sum\limits_{j \ne i} {{{\bar w}_j}} } \right)} \right) \cdot \phi \left( {{e^*}\left( w \right)} \right) = \mathop {\max }\limits_{w \in W + \left( {1 - 1/N} \right)\bar w} \left( {{B_i} - \left( {w - \frac{{N - 1}}{N}\bar w} \right)} \right) \cdot \phi \left( {{e^*}\left( w \right)} \right).\]Since \({B_1} =  \cdots  = {B_N}\), \(N{B_i} = B\), so this program is equivalent to \[\mathop {\max }\limits_{w \in W + \left( {1 - 1/N} \right)\bar w} \left( {B - Nw + \left( {N - 1} \right)\bar w} \right)\phi \left( {{e^*}\left( w \right)} \right) = \mathop {\max }\limits_{w \in W + \left( {1 - 1/N} \right)\bar w} \Lambda \left( {w,\bar w} \right).\]Therefore, since \(\bar w\) solves the aggregate program, it also solves each principal's program. Q.E.D.