## 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.