Complementary Monopolies with asymmetric information

Abstract

We investigate how asymmetric information on final demand affects strategic interaction between a downstream monopolist and a set of upstream monopolists, who independently produce complementary inputs. We study an intrinsic private common agency game in which each supplier i independently proposes a pricing schedule contract to the assembler, specifying the supplier’s payment as a function of the assembler’s purchase of input i. We provide a necessary and sufficient equilibrium condition. A lot of equilibria satisfy this condition but there is a unique Pareto-undominated Nash equilibrium from the suppliers’ point of view. In this equilibrium, there are unavoidable efficiency losses due to excessively low sales of the good. However, suppliers may be able to limit these distortions by implicitly coordinating on an equilibrium with a rigid (positive) output in bad demand circumstances.

Appendices

APPENDIX

1.1 Proof of Lemma 1

• (a) suppose that $\theta ^{\prime }>\theta ;$ from revealed preference it must be that

  $$\begin{aligned} (\theta -Q(\theta ))Q(\theta )-\sum _{i=1}^{n}T_{i}(q_{i}(\theta ))\ge (\theta -Q(\theta ^{\prime }))Q(\theta ^{\prime })-\sum _{i=1}^{n}T_{i}(q_{i} (\theta ^{\prime })) \end{aligned}$$

  and

  $$\begin{aligned} (\theta ^{\prime }-Q(\theta ^{\prime }))Q(\theta ^{\prime })-\sum _{i=1}^{n} T_{i}(q_{i}(\theta ^{\prime }))\ge (\theta ^{\prime }-Q(\theta ))Q(\theta ) -\sum _{i=1}^{n}T_{i}(q_{i}(\theta )), \end{aligned}$$

  implying $\left( \theta -\theta ^{\prime }\right) (Q\left( \theta \right) )-Q\left( \theta ^{\prime }\right) )\ge 0$.

• (b) $\Pi ^{A}$ is convex as a supremum of convex (linear here) functions. It is also continuous at the endpoints of the domain $\left[ \underline{\theta },\overline{\theta }\right]$. It is hence absolutely continuous⁴¹ in $\theta .$ Suppose that $\theta ^{\prime }>\theta ;$ from the above “revealed preference” inequalities we obtain $\left( \theta ^{\prime }-\theta \right) Q\left( \theta ^{\prime }\right) \ge \Pi ^{A}(\theta ^{\prime })-\Pi ^{A}(\theta )\ge \left( \theta ^{\prime } -\theta \right) Q\left( \theta \right) )$, so that, dividing throughout by $\left( \theta ^{\prime }-\theta \right)$ and letting $\theta ^{\prime }$ tend toward $\theta$, we obtain $\frac{\partial \Pi ^{A}(\theta )}{\partial \theta } =Q(\theta ) \ge 0$. $\square$

1.2 Proof of Lemma 2

Suppose on the contrary that $\overline{Q}(\widetilde{q}_{i}^{*} (\theta ,\mathbf {T}_{-i}),\theta ,\mathbf {T}_{-i})<\widetilde{q}_{i}^{*}(\theta ,\mathbf {T}_{-i}),$ where $\widetilde{q}_{i}^{*}(\theta ,\mathbf {T}_{-i})$ is defined by (10), i.e. maximizes supplier i’s adjusted profit.

Let $\{Q^{A}(\theta ),\mathbf {q}_{-i}^{A}(\theta )\}$ be solution of the assembler’s problem (3a) in which the constraint $Q\le q_{i}$ has been deleted.⁴² Since, at the solution of (3a), this constraint does not bind, $\overline{Q}(\widetilde{q}_{i}^{*} (\theta ,\mathbf {T}_{-i}),\theta ,\mathbf {T}_{-i})=Q^{A}(\theta )$ and $\overline{\mathbf {q}}_{-i}(\widetilde{q}_{i}^{*}(\theta ,\mathbf {T} _{-i}),\theta ,\mathbf {T}_{-i})=\mathbf {q}_{-i}^{A}(\theta )$ do not depend on $q_{i}$ and the corresponding supplier’s adjusted profit equals

$$\begin{aligned} \left( \theta -Q^{A}(\theta )-h(\theta )\right) Q^{A}(\theta )-\sum _{j=1,j\ne i}^{n}T_{j}(q_{j}^{A}(\theta )). \end{aligned}$$

(33)

Let us on the other hand consider the solution to the problem of maximizing supplier i’s adjusted profit when i is able to select directly $Q(\theta )$ and $\mathbf {q}_{-i}(\theta )$:⁴³

$$\begin{aligned} \{Q^{S}(\theta ),\mathbf {q}_{-i}^{S}(\theta )\}&=\underset{Q,\mathbf {q}_{-i}}{\arg \max } \left( \theta -Q-h(\theta )\right) Q-\sum _{j=1,j\ne i}^{n} T_{j}(q_{j}),\nonumber \\ s.t.\ \ Q&\le q_{j},\ \forall j\ne i. \end{aligned}$$

(34)

From revealed preferences,

$$\begin{aligned}&\left( \theta -Q^{A}(\theta )\right) Q^{A}(\theta ) -\sum _{j=1,j\ne i}^{n}T_{j}(q_{j}^{A}(\theta ))\\&\quad \ge \left( \theta -Q^{S}(\theta )\right) Q^{S}(\theta )-\sum _{j=1,j\ne i}^{n}T_{j}(q_{j}^{S}(\theta )), \end{aligned}$$

and

$$\begin{aligned}&\left( \theta -h(\theta )-Q^{S}(\theta )\right) Q^{S}(\theta ) -\sum _{j=1,j\ne i}^{n}T_{j}(q_{j}^{S}(\theta ))\\&\quad \ge \left( \theta -h(\theta )-Q^{A} (\theta )\right) Q^{A}(\theta )-\sum _{j=1,j \ne i}^{n}T_{j}(q_{j}^{A} (\theta )). \end{aligned}$$

This implies

$$\begin{aligned} h(\theta )(Q^{A}(\theta )-Q^{S}(\theta ))\ge & {} 0\Leftrightarrow Q^{S}(\theta ) \le Q^{A}(\theta )=\overline{Q}(\widetilde{q}_{i}^{*} (\theta ,\mathbf {T}_{-i}),\theta ,\mathbf {T}_{-i})\\< & {} \widetilde{q}_{i}^{*}(\theta ,\mathbf {T}_{-i}). \end{aligned}$$

Let us show that the supplier i may reach a greater profit than (33) simply by choosing to sell a quantity of input i equal to $Q^{S}(\theta )$ instead of the quantity $\widetilde{q}_{i}^{*} (\theta ,\mathbf {T}_{-i}).$

This is obviously the case when the solution $\overline{Q}(Q^{S} (\theta ),\theta )$ of the assembler’s problem (3a) under the constraint $Q\le Q^{S}(\theta )$ is such that $\overline{Q}(Q^{S} (\theta ),\theta ,\mathbf {T}_{-i})=Q^{S}(\theta )$.⁴⁴ Let us show that it must indeed be the case that $\overline{Q}(Q^S(\theta ),\theta ,\mathbf T _-i) =Q^S(\theta )$. Suppose on the contrary that the solution of the assembler’s problem (3a) under the constraint $Q\le Q^S(\theta )$ implies a sales level $Q^\prime (\theta )<Q^S(\theta )$. It must then be that

$$\begin{aligned}&\left( \theta -Q^{\prime }(\theta )\right) Q^{\prime }(\theta ) -\sum _{j=1,j\ne i}^{n}T_{j}(\overline{q}_{j}(Q^{\prime }(\theta ),\theta ))\\&\quad >\left( \theta -Q^{S}(\theta )\right) Q^{S}(\theta ) -\sum _{j=1,j\ne i}^{n}T_{j}(\overline{q}_{j}(Q^{S}(\theta ),\theta )). \end{aligned}$$

But, since $h(\theta )\ge 0$ and $Q^{\prime }(\theta )<Q^{S}(\theta ),$ were this true, it should be that

$$\begin{aligned}&\left( \theta -h(\theta )-Q^{\prime }(\theta )\right) Q^{\prime }(\theta ) -\sum _{j=1,j\ne i}^{n}T_{j}(\overline{q}_{j}(Q^{\prime }(\theta ),\theta ))\\&\quad >\left( \theta -h(\theta )-Q^{S}(\theta )\right) Q^{S}(\theta ) -\sum _{j=1,j\ne i}^{n}T_{j}(\overline{q}_{j}(Q^{S}(\theta ),\theta )). \end{aligned}$$

This contradicts (34).

To conclude, supplier i is always able to induce the assembler to select the output and input levels $\{Q^{S}(\theta ),\mathbf {q}_{-i}^{S} (\theta )\}$ by choosing to supply a quantity $Q^{S}(\theta )$ of input i. Then, either $Q^{S}(\theta )<Q^{A}(\theta )$, in which case supplier i clearly obtains greater (adjusted) profits than at $\{Q^{A}(\theta ),\mathbf {q}_{-i}^{A} (\theta )\},$ or $Q^{S}(\theta )=Q^{A}(\theta ),$ in which case an infinitesimal cost of producing input i is enough to ensure that supplier i is better off when supplying a quantity of input i equal to $Q^{S}(\theta )$. $\square$

1.3 Proof of Remark 1

Suppose that $Q(\theta )>\frac{\theta -h(\theta )}{2}>0.$ Obviously $q_{j} (\theta )\ge Q(\theta ).$ Suppose now instead that the supplier i chooses a supply level $q_{i}(\theta )=\frac{\theta -h(\theta )}{2\ }$ and show that it is better off. Indeed, under the constraint $Q\le \frac{\theta -h(\theta )}{2},$ the solution $\overline{Q}(\frac{\theta -h(\theta )}{2},\theta ,\mathbf {T}_{-i})$ of the assembler’s problem (3a) is either $=\frac{\theta -h(\theta )}{2}$ or $<\frac{\theta -h(\theta )}{2}$ so that either

• 1. $\overline{Q}(\frac{\theta -h(\theta )}{2},\theta ,\mathbf {T}_{-i}) =\frac{\theta -h(\theta )}{2}$, and then the supplier i’s adjusted profit is greater than the profit obtained at $Q(\theta )>\frac{\theta -h(\theta )}{2}$ since (i) for $\forall Q(\theta )\ne \frac{\theta -h(\theta )}{2}$, $(\frac{\theta -h(\theta )}{2})^{2}>\left( \theta -h(\theta )-Q(\theta )\right) Q(\theta )$ and (ii) in order to produce a quantity of output $\frac{\theta -h(\theta )}{2}<Q(\theta )$, the assembler does not need greater quantities of inputs other than i and pays accordingly at most the same aggregate transfer $\sum _{j=1,j\ne i}^{n}T_{j}(q_{j}(\theta ))$ to suppliers $j\ne i.$

• 2. or $\overline{Q}(\frac{\theta -h(\theta )}{2},\theta ,\mathbf {T}_{-i}) <\frac{\theta -h(\theta )}{2},$ in which case, by the same argument as in Lemma 2, the supplier i’s adjusted profit must be greater at $\overline{Q} (\frac{\theta -h(\theta )}{2},\theta ,\mathbf {T}_{-i})$ than at $\frac{\theta -h(\theta )}{2}$ and thus strictly greater than at any $Q(\theta )>\frac{\theta -h(\theta )}{2}$. $\square$

1.4 Proof of Lemma 3

Let $\tau =\theta -h(\theta ).$ Define now

$$\begin{aligned} \widehat{\pi }_{i}(\tau ,q)=\left( \tau -Q\right) Q -\sum _{j=1,i\ne j}^{n} T_{j}(Q). \end{aligned}$$

and

$$\begin{aligned} \widehat{\Pi }_{i}^{S}(\tau )=\underset{q}{\max } \widehat{\pi }_{i}(\tau ,q). \end{aligned}$$

Since $\widehat{\pi }_{i}$ is linear in $\tau$, $\widehat{\Pi }_{i}^{S}(\tau )$ is convex in $\tau$ as a supremum of convex (linear) functions, hence absolutely continuous⁴⁵ (AC). By assumption 1, $\tau (\theta )$ is invertible so that $\Pi _{i}^{S}(\theta )=\widehat{\Pi }_{i}^{S}(\tau (\theta ))$ is AC and a.e. differentiable. $\square$

1.5 Proof of Proposition 1

• (a) Necessity

From Lemma 2, $Q^{e}(\theta )=q_{i}^{e}(\theta )$, $\forall i=1,2, \ldots ,n.$ Moreover from Remark 1, it must be that $Q^{e}(\theta )\le \frac{\theta -h(\theta )}{2}.$ On the other hand, we already know that a non-decreasing output function $Q^{e}(\theta ) ($and the input functions $q_{i}^{e}(\theta )=Q^{e}(\theta ))$ solve respectively the Assembler’s Problem (4a) and the Suppliers’ Problems (9) only if the payment schedules, $T_{i}$, are defined by equations (13) and (15).

Substituting in (16) $\sum T_{i}(Q^{e} \left( \theta \right) )$ for its value from (8) and rearranging we obtain:

$$\begin{aligned} 0=\left[ V\left( \theta ,Q\left( \theta \right) \right) -\int _{\underline{\theta }}^{\theta }\left( 1-nh^{\prime } \left( s\right) \right) Q\left( s\right) \mathrm{d}s\right] -\sum _{i=1}^{n}\Pi _{i}^{s}\left( \underline{\theta }\right) . \end{aligned}$$

Evaluating the previous condition at $\theta =\underline{\theta },$ we obtain:

$$\begin{aligned} V\left( \underline{\theta },Q\left( \underline{\theta }\right) \right) =\sum _{i=1}^{n}\Pi _{i}^{s}\left( \underline{\theta }\right) . \end{aligned}$$

Given that $V_{s}(Q(s),s)=(1-nh^{\prime }(s))Q(s),$ the result in Proposition 1 must hold.

• (b) Sufficiency

Let us consider the implementation of a non-decreasing sales function $Q^{e}(\theta )\le \frac{\theta -h(\theta )}{2} \le \frac{\theta }{2}$ when the assembler solves the associate problem $\underset{Q}{\max }\Psi (\theta ,Q)=(\theta -Q)Q -\overline{T(}Q).$ Since $\frac{\partial ^{2}\Psi (\theta ,Q)}{\partial Q\partial \theta }=1>0$, $\Psi (\theta ,Q)$ has increasing differences or is supermodular (see Topkis 1998 or Amir 2005). Therefore, the version of the single-crossing property called CS$^{+}$ in (Fudenberg and Tirole 1991, ch. 7.3.1, p. 259) is here satisfied. Then, there exists $\overline{T}( \cdot )$ such that $\left( Q^{e}( \cdot ),\overline{T}( \cdot )\right)$ is incentive-compatible for this associate problem. Notice that condition (13) where $\sum _{i=1}^{n}T_{i}(Q^{e}(\theta ))$ is replaced by $\overline{T}(Q^{e}(\theta ))$ is necessary in the associate problem as well as in the original one. From condition (13) for any $\theta ^{\prime } >\theta ,$ we have:

$$\begin{aligned} \overline{T}(Q(\theta ^{\prime }))-\overline{T}(Q^{e}(\theta )) =\int _{\theta }^{\theta ^{\prime }}(s-2Q(s))Q^{\prime }(s)\mathrm{d}s, \end{aligned}$$

from what we deduce that $Q^{e}(\theta )\le \frac{\theta -h(\theta )}{2} \le \frac{\theta }{2}$ implies that any transfer function $\overline{T}$ which implements a non-decreasing sales function $Q^{e}(\theta )\le \frac{\theta -h(\theta )}{2}$ in the associate problem must be non-decreasing in Q.

Consider now in the original problem the tariff functions $T_{i}(q_{i} )=\frac{1}{n}\overline{T}(q_{i})$, $i=1,2, \ldots ,n.$ Let us now show that they implement the non-decreasing sales and input functions $q_{i}^{e} (\theta )=Q^{e}(\theta )\le \frac{\theta -h(\theta )}{2}\le \frac{\theta }{2}$, $i=1,2, \ldots ,n.$ Suppose on the contrary that there exist $q_{i}(\theta ^{\prime })=q(\theta ^{\prime })\le \frac{\theta ^{\prime }-h(\theta ^{\prime })}{2},i=1,2, \ldots ,n,$ which gives a strictly greater profit to the assembler, i.e.

$$\begin{aligned} (\theta -\widetilde{q})\widetilde{q}-\overline{T}(q(\theta ^{\prime })) >(\theta -Q^{e}(\theta ))Q^{e}(\theta )-\overline{T}(Q^{e}(\theta )), \end{aligned}$$

where $\widetilde{q}\in [0,q(\theta ^{\prime })]$.⁴⁶ Since $\overline{T}$ is non-decreasing, it follows that

$$\begin{aligned} (\theta -\widetilde{q})\widetilde{q}-\overline{T}(\widetilde{q}) >(\theta -Q^{e}(\theta ))Q^{e}(\theta )-\overline{T}(Q^{e}(\theta )), \end{aligned}$$

contradicting the fact that $\overline{T}$ implements $Q^{e}(\theta )$ in the associate problem.

Using now condition (13), where $\sum _{i=1}^{n}T_{i}(Q^{e}(\theta ))$ is replaced by $\overline{T}(Q^{e}(\theta )),$ together with condition (19), we obtain

$$\begin{aligned} \frac{n-1}{n}\overline{T}(Q^{e}(\theta ))= & {} \left[ \theta -h(\theta )-Q^{e} (\theta )\right] Q^{e}(\theta )\nonumber \\&-\int _{\underline{\theta }}^{\theta }(1-h^{\prime }(s))Q^{e}(s)\mathrm{d}s -\frac{1}{n}V(Q^{e}(\underline{\theta }),\underline{\theta }). \end{aligned}$$

(35)

Notice that the definition of the virtual profits implies that $\sum _{i=1} ^{n}\Pi _{i}^{s}\left( \underline{\theta }\right) =V(Q^{e}(\underline{\theta }),\underline{\theta }).$ Given that the tariff functions are identical for all, the $\Pi _{i}^{s} \left( \underline{\theta }\right)$ are identical as well $\forall i=1,2, \ldots ,n.$ It follows that (35) implies

$$\begin{aligned} \frac{n-1}{n}\overline{T}(Q^{e}(\theta ))= & {} \left[ \theta -h(\theta )-Q^{e} (\theta )\right] Q^{e}(\theta )\nonumber \\&-\int _{\underline{\theta }}^{\theta }(1-h^{\prime }(s))Q^{e}(s)\mathrm{d}s -\Pi _{i} ^{s}\left( \underline{\theta }\right) ,\ i=1,2, \ldots ,n, \end{aligned}$$

(36)

which is the necessary (first-order) condition(15) for each principal i. It remains to verify the global second order conditions for each principal i’s problem. We can apply the same proof strategy as in the agent’s case. Consider the associate problem for principal $i: \underset{Q}{\max }\Gamma _{i}(Q,\theta ) =\left( \theta -Q-h(\theta )\right) Q-\frac{n-1}{n}\overline{T}(Q).$ Since the single-crossing property ($\hbox {CS}^{+}$) is here obviously satisfied (i.e. $\frac{\partial ^{2}\Gamma _{i}(\theta ,Q)}{\partial Q\partial \theta }=1-h^{\prime }(\theta )>0$), not only the local second order conditions (35) are satisfied for the associate problem but also the global ones.

Consider now in the original problem the tariff functions $T_{j}(q_{j} )=\frac{1}{n}\overline{T}(q_{j})$, $j\ne i.$ We now show that $\sum _{j\ne i}\frac{1}{n}\overline{T}(q_{j})$ implements the non-decreasing sales and input functions $q_{i}^{e}(\theta )=Q^{e}(\theta )\le \frac{\theta -h(\theta )}{2}\le \frac{\theta }{2}.$ Suppose on the contrary that there exists $q_{i}(\theta ^{\prime })\le \frac{\theta ^{\prime }-h(\theta ^{\prime })}{2},$ which gives a strictly greater profit to supplier i and remembering that the function $\overline{T}$ is increasing, we must have

$$\begin{aligned} (\theta -h(\theta )-\widehat{q})\widehat{q}-\frac{n-1}{n}\overline{T} (\widehat{q})>(\theta -h(\theta )-Q^{e}(\theta ))Q^{e}(\theta )-\frac{n-1}{n}\overline{T}(Q^{e}(\theta )), \end{aligned}$$

where $\widehat{q}\in [0,q_{i}(\theta ^{\prime })]$.⁴⁷ But this contradicts the optimality of $Q^e(\theta )$ for supplier i in the corresponding associate problem. $\square$

Proof of Corollary 2

The proof is straightforward. Indeed differentiating (19) with respect to $\theta ,$ one obtains $V_{s}( \cdot )=V_{Q}Q^{e\prime }(\theta )+V_{s}( \cdot )$, implying $V_{Q}Q^{e\prime }(\theta )=0.$ Given that $V_{Q}=\theta -nh(\theta )-2Q^{e}(\theta ),$ and given $Q^{D}\left( \theta \right)$, then $V_{Q}Q^{e\prime }(\theta )=0$ implies

$$\begin{aligned} 2\left[ Q^{D}(\theta )-Q^{e}(\theta )\right] Q^{e\prime }(\theta )=0, \end{aligned}$$

yielding the results in the preceding Corollary. $\square$

1.1 Proof of Lemma 4

Given (17), $V(Q^{e}(\theta ),\theta )$ is obviously continuous at any $\theta$ where $Q^{e}(\theta )$ is continuous. So we only have to consider the values of $\theta$ at which $Q^{e}\left( \theta \right)$ is discontinuous. Let $\widetilde{\theta }$ be such a point. Since from Lemma 1, the equilibrium sales function is a.e. differentiable, points of discontinuity are isolated and it is right and left continuous with $\underset{\theta \downarrow \widetilde{\theta }}{\lim }Q^{e}(\theta )=Q^{+}(\widetilde{\theta })> \underset{\theta \uparrow \widetilde{\theta }}{\lim }Q^{e}(\theta )=Q^{-}(\widetilde{\theta }).$ From (19), $\underset{\theta \downarrow \widetilde{\theta }}{\lim }V(Q^{e}(\theta ),\theta )-V(Q^{e} (\underline{\theta }),\underline{\theta })=V(Q^{+} (\widetilde{\theta }),\widetilde{\theta } )-V(Q^{e}(\underline{\theta }), \underline{\theta })=\underset{\theta \uparrow \widetilde{\theta }}{\lim } V(Q^{e}(\theta ),\theta )-V(Q^{e} (\underline{\theta }),\underline{\theta }) =V(Q^{-}(\widetilde{\theta }),\widetilde{\theta }) -V(Q^{e}(\underline{\theta }),\underline{\theta }).$ It follows that $V(Q^{+}(\widetilde{\theta }),\widetilde{\theta })=V(Q^{-} (\widetilde{\theta }),\widetilde{\theta }).$ Notice in addition that, since $Q^{D}(\widetilde{\theta })=\arg \underset{Q}{\max }V(Q, \widetilde{\theta }),$ it must be that $Q^{-}(\widetilde{\theta })<Q^{D}(\widetilde{\theta })<Q^{+}(\widetilde{\theta })$. $\square$

1.2 Proof of Example 1

An equilibrium output function maximizing the virtual aggregate surplus function has already been shown to be given by $Q^{D}(\theta )$, where here $Q^{D}(\theta )= \frac{(n+1)\theta -n}{2}.$ The assembler’s and subcontractor j’s first-order conditions are respectively over $(\frac{n}{n+1} ,1]:(\theta -2Q\left( \theta \right) )=\sum _{i=1}^{n}T_{i}^{\prime } (Q^{e}(\theta ))$ and $2\theta -1-2Q\left( \theta \right) =\sum _{i=1,i\ne j}^{n}T_{i}^{\prime }(Q^{e}(\theta )).$ Subtracting the second from the first we obtain $1-\theta =T_{j}^{\prime }(Q^{e} (\theta )).$ Using (ii) we can express $\theta$ as a function of Q,  with

$$\begin{aligned} \theta =\frac{2Q+n}{n+1}, \end{aligned}$$

and substitute this value for $\theta .$ We obtain $T_{j}^{\prime } (Q)=\frac{1-2Q}{n+1}$ and then integrating with respect to Q, we obtain $T_{j}(Q)=K_{j}+\frac{Q-Q^{2}}{n+1}$, $\forall Q\in \left[ 0,\frac{1}{2}\right]$, where $K_{j}$ is a constant. From the assembler’s participation constraint and suppliers optimization behavior, $\sum _{j=1}^{n}K_{j}=0.$ It is easy to see that choosing $Q=0$ maximizes the assembler’s profit for $\theta \in \left[ 0,\frac{n}{n+1}\right] .$ It follows that $\sum _{j=1} ^{n}T_{j}(Q)=\frac{n(Q-Q^{2})}{n+1}$. $\square$

1.3 Proof of Remark 2

From equation (7) we obtain

$$\begin{aligned} \Pi ^{A}(\theta )&=\int _{\frac{n}{n+1}}^{\theta }Q^{e}(\theta )\mathrm{d}\theta =\frac{\left[ \theta -n(1-\theta )\right] ^{2}}{4(1+n)}\nonumber \\ \text {for all }\theta&\in \left[ \frac{n}{n+1},1\right] \text { and }\Pi ^{A}(\theta )=0\text { otherwise,} \end{aligned}$$

(37)

so that the ex ante expected profit of the assembler, $E\left[ \Pi ^{A}\right] ,$ is equal to

$$\begin{aligned} E\left[ \Pi ^{A}\right] =\int _{\frac{n}{n+1}}^{1} \frac{\left[ \theta -n(1-\theta )\right] ^{2}}{4(1+n)} \mathrm{d}\theta =\frac{1}{12(n+1)^{2}}. \end{aligned}$$

(38)

The profit of subcontractor i,  for all $\theta \in \left[ \frac{n}{n+1},1\right] ,$ is easily obtained from $T_{i}(Q)=\frac{Q-Q^{2} }{n+1},$ by setting $Q=\frac{(n+1)\theta -n}{2},$ implying:

$$\begin{aligned} \Pi _{i}^{S}(\theta )=\frac{\left[ 2-\theta +n(1-\theta )\right] \left[ \theta -n(1-\theta )\right] }{4(n+1)}, \end{aligned}$$

for all $\theta \in \left[ \frac{n}{n+1},1\right]$ and $\Pi _{i}^{S} (\theta )=0$ otherwise. The expected profit of subcontractor i,  denoted $E\left[ \Pi _{i}^{S}\right]$, is

$$\begin{aligned} E\left[ \Pi _{i}^{S}\right] =\int _{\frac{n}{n+1}}^{1}\frac{(2-\theta +n(1-\theta ))(\theta -n(1-\theta ))}{4(n+1)}\mathrm{d}\theta =\frac{1}{6(n+1)^{2}}, \end{aligned}$$

(39)

yielding the results in Remark 2. $\square$

1.4 Proof of Corollary 3

Let us write the condition (19) respectively at some $\theta \in (x_{i},x_{i+1}]$ and at $x_{i}:$

$$\begin{aligned} \int _{\underline{\theta }}^{\theta }V_{s}^{\prime }(Q(s),s)\mathrm{d}s =V(Q_{i}, \theta )-V(Q(\underline{\theta }),\underline{\theta }), \end{aligned}$$

(40)

and

$$\begin{aligned} \int _{\underline{\theta }}^{x_{i}}V_{s}^{\prime }(Q(s),s)\mathrm{d}s =V(Q_{i-1}, x_{i})-V(Q(\underline{\theta }),\underline{\theta }). \end{aligned}$$

(41)

In the equation above, we consider that $\left[ \underline{\theta } ,\overline{\theta }\right]$ is divided in $n+1$ intervals $[x_{j},x_{j+1} )$ such that $Q(\theta )=Q_{j}$, $\forall \theta \in (x_{j},x_{j+1}],$ with $Q_{j+1}>Q_{j}$, $x_{0}=\underline{\theta }$ and $x_{n+1}=\overline{\theta }.$ Subtracting (41) from (40) we obtain

$$\begin{aligned} \int _{x_{i}}^{\theta }V_{s}^{\prime }(Q_{i},s)\mathrm{d}s =V(Q_{i},\theta ) -V(Q_{i-1} ,x_{i}). \end{aligned}$$

The LHS equals $V(Q_{i},\theta )-V(Q_{i},x_{i})$ so that we conclude that $V(Q_{i-1},x_{i})=V(Q_{i},x_{i})$. $\square$

1.5 Proof of Remark 3

• (i) Let us first derive (23). Notice first that substituting in (15) $\sum T_{i}(Q\left( \theta \right) )$ for its value from (8), accounting for $\Pi ^{A} (\underline{\theta })=0,$ and rearranging we obtain an expression which, evaluated at $\theta =\underline{\theta }$, yields

  $$\begin{aligned} {\displaystyle \sum \limits _{i=1}^{i=n}} \Pi _{i}^{S}(\underline{\theta })=\Pi ^{S}(\underline{\theta }) =V(Q^{e} (\underline{\theta }),\underline{\theta }). \end{aligned}$$

  Let us now consider the expected aggregate suppliers’ profit $E[\Pi ^{S}].$ From equation (14) in the paper, and the above result, we obtain

  $$\begin{aligned} E[\Pi ^{S}]= & {} V(Q^{e}(\underline{\theta }),\underline{\theta }) +nE\left[ \int _{\underline{\theta }}^{\theta }(1-h^{\prime } (s))Q^{e}(s)\mathrm{d}s\right] \\= & {} V(Q^{e}(\underline{\theta }),\underline{\theta })+(n-1)E \left[ \int _{\underline{\theta }}^{\theta }Q^{e}(s)\mathrm{d}s\right] \\&+\,E\left[ \int _{\underline{\theta }}^{\theta }(1-nh^{\prime } (s))Q^{e}(s)\mathrm{d}s\right] . \end{aligned}$$

Using now equation (7) and condition (19 ), it turns out that

$$\begin{aligned} E[\Pi ^{S}]=E\left[ V(Q^{e}(\theta ),\theta \right] +(n-1)E\left[ \Pi ^{A}(\theta )\right] . \end{aligned}$$

(42)

Moreover, introducing in (42) $E\left[ \Pi ^{A}\right] =E[(\theta -Q^{e}(\theta ))Q^{e}(\theta )]-E[\Pi ^{S}]$ and using the definition of $V(Q(\theta ),\theta ),$ we obtain:

$$\begin{aligned} E[\Pi ^{S}]=E\left[ (\theta -h(\theta ) -Q^{e}(\theta ))Q^{e}(\theta )\right] . \end{aligned}$$

(43)

• (ii) Let us now derive (24).

From the condition (19), we can write

$$\begin{aligned} \left[ (\theta -h(\theta )-Q^{e}(\theta ))Q^{e}(\theta )\right]= & {} V(Q^{e}(\overline{\theta }),\overline{\theta }))\\&-\int _{\theta }^{\overline{\theta }}(1-nh^{\prime }(s)) Q^{e}(s)\mathrm{d}s+(n-1)h(\theta )Q^{e}(\theta ). \end{aligned}$$

Integrating both sides between $\underline{\theta }$ and $\overline{\theta },$ and then integrating the RHS by parts, we obtain (24). $\square$

1.6 Proof of Lemma 5

Let us denote by $Q_{1}\left( \theta \right)$ the first equilibrium output function and by $Q_{2}\left( \theta \right)$ the second equilibrium output function described in Lemma 5.

Given (43), the difference $E\left[ \Pi ^{S}(Q_{2}(\theta ),\theta )\right] -E\left[ \Pi ^{S}(Q_{1}(\theta ),\theta )\right]$ between the expected aggregate suppliers’ profits under equilibrium output functions $Q_{2}(\theta )$ and $Q_{1}(\theta )$ equals

$$\begin{aligned} \int _{x}^{\overline{\theta }}\left( \theta -h(\theta )-Q^{D} \left( \theta \right) \right) Q^{D}\left( \theta \right) f(\theta )\mathrm{d}\theta -\int _{x}^{\overline{\theta }}\left( \theta -h(\theta )-Q^{D} \left( x\right) \right) Q^{D}\left( x\right) f(\theta )\mathrm{d}\theta . \end{aligned}$$

(44)

Since $\frac{\theta -h(\theta )}{2}=\arg \underset{Q}{\max }\Pi ^{S} (Q,\theta )>Q^{D}(\theta )=\frac{\theta -nh(\theta )}{2} >Q^{D} \left( x\right)$, $\forall \theta \in (x,\overline{\theta }],$ the concavity of $\Pi ^{S}(Q,\theta )$ with respect to Q implies that $\Pi ^{S}(Q^{D}(\theta ),\theta )>\Pi ^{S} (Q^{D}(x),\theta )$, $\forall \theta \in (x,\overline{\theta }]$, so that (44) is $>0$. $\square$

1.7 Proof of Lemma 6

Let us consider the difference between the expected aggregate suppliers’ profit under the equilibrium output functions $Q_{C}(\theta )$ and $Q_{H}(\theta )$, which, given (24), is

$$\begin{aligned}&E\left[ \Pi ^{S}(Q_{C}(\theta ),\theta )\right] -E\left[ \Pi ^{S} (Q_{H}(\theta ),\theta )\right] \\&\quad = \int _{x_{0}}^{x_{1}}Q^{D}\left( \theta \right) g(\theta )\mathrm{d}\theta -\left[ \int _{x}^{x_{1}}Q^{D}(x_{1})g(\theta )\mathrm{d}\theta +\int _{x_{0}}^{x}Q^{D}(x_{0})g(\theta )\mathrm{d}\theta \right] \\&\quad = \int _{x_{0}}^{x_{1}}Q^{D}\left( \theta \right) g(\theta ) \mathrm{d}\theta -[Q^{D}(x_{1})(G(x_{1})-G(x))+Q^{D}(x_{0})(G(x)-G(x_{0}))], \end{aligned}$$

where $g(\theta )=(n-1)-n(1-h^{\prime }(\theta ))F(\theta )$ and $G(\theta )=\int _{\underline{\theta }}^{\theta }g(s)\mathrm{d}s.$

Now, integrating by parts, $\int _{x_{0}}^{x_{1}}Q^{D}\left( \theta \right) g(\theta )\mathrm{d}\theta =Q^{D}(x_{1})G(x_{1})-Q^{D}(x_{0})G(x_{0}) -\int _{x_{0} }^{x_{1}}Q^{D\prime }(\theta )G(\theta )\mathrm{d}\theta ,$ so that

$$\begin{aligned}&E\left[ \Pi ^{S}(Q_{C}(\theta ),\theta )\right] -E\left[ \Pi ^{S}(Q_{H} (\theta ),\theta )\right] \\&\quad =(Q^{D}(x_{1})-Q^{D}(x_{0}))G(x) -\int _{x_{0}}^{x_{1}}Q^{D\prime }(\theta )G(\theta )\mathrm{d}\theta \\&\quad =\int _{x_{0}}^{x_{1}}Q^{D\prime }(\theta )(G(x)-G(\theta ))\mathrm{d}\theta . \end{aligned}$$

Let now $z=\frac{\theta -nh(\theta )}{2}.$ From Assumption 1, this is a monotone increasing function so that there is an inverse function $\Theta (z).$ On the other hand, $\mathrm{d}\theta =\frac{2}{1-nh^{\prime } (\theta )}\mathrm{d}z.$ Remembering that $Q^{D\prime }(\theta )=\frac{1}{2} (1-nh^{\prime }(\theta ))$ a simple change of variables leads to

$$\begin{aligned} E\left[ \Pi ^{S}(Q_{C}(\theta ),\theta )\right] -E\left[ \Pi ^{S}(Q_{H}(\theta ),\theta )\right] =\int _{z_{0}}^{z_{1}}\left( G\left( \Theta \left( \frac{z_{0}+z_{1}}{2}\right) \right) -G(\Theta (z))\right) \mathrm{d}z, \end{aligned}$$

(45)

where $z_{i}=\frac{x_{i}-nh(x_{i})}{2}$, $i=0,1$ and $\Theta (\frac{z_{0} +z_{1}}{2})=x$⁴⁸.

Now, the next step is to prove the concavity of $G(\Theta (z))$ with respect to z. It turns out that:

$$\begin{aligned}&\frac{d^{2}G(\Theta (z))}{\mathrm{d}z^{2}}\\&\quad =\frac{2n}{(1-nh^{\prime }(\theta ))} \left[ -(1-nh^{\prime }(\theta ))(1-h^{\prime }(\theta ))f(\theta ) +h\text {''}(\theta )(n-1)(1-F(\theta ))\right] \\&\quad =\frac{2nf(\theta )}{(1-nh^{\prime }(\theta ))} \left[ -(1-nh^{\prime }(\theta ))(1-h^{\prime }(\theta )) +h\text {''}(\theta )(n-1)h(\theta )\right] . \end{aligned}$$

From Assumptions 1 and 3, this is negative and accordingly $G(\Theta (z))$ is concave with respect to z.

This is enough to show that the RHS of (45) is positive. From the concavity of $G(\Theta (z))$ with respect to z,

$$\begin{aligned} \int _{z_{0}}^{\frac{z_{0}+z_{1}}{2}}\left( G\left( \Theta \left( \frac{z_{0} +z_{1}}{2}\right) \right) -G(\Theta (z))\right) dz \ge \frac{dG\left( \Theta \left( \frac{z_{0}+z_{1}}{2}\right) \right) }{dz} \frac{1}{8}(z_{1}-z_{0})^{2}, \end{aligned}$$

and

$$\begin{aligned} \int _{\frac{z_{0}+z_{1}}{2}}^{z_{1}}\left( G\left( \Theta \left( \frac{z_{0} +z_{1}}{2}\right) \right) -G(\Theta (z))\right) dz \ge -\frac{dG\left( \Theta \left( \frac{z_{0}+z_{1}}{2}\right) \right) }{dz} \frac{1}{8}(z_{1}-z_{0})^{2}, \end{aligned}$$

so that $\int _{z_{0}}^{z_{1}}(G(\Theta (\frac{z_{0}+z_{1}}{2}))-G(\Theta (z)))dz\ge 0$. $\square$

1.8 Proof of Proposition 2

Given (23) and (29), the suppliers’ aggregate expected profits may be written along a semi-regular equilibrium as

$$\begin{aligned}&\int _{\underline{\theta }}^{\theta ^{SR}}(\theta -h(\theta ) -Q^{D}(\theta ^{SR}))Q^{D}(\theta ^{SR})f(\theta )\mathrm{d}\theta \\&\quad +\int _{\theta ^{SR}}^{\overline{\theta }}(\theta -h(\theta ) -Q^{D}(\theta ))Q^{D}(\theta )f(\theta )\mathrm{d}\theta . \end{aligned}$$

Their derivative with respect to $\theta ^{SR}$ is easily obtained as:

$$\begin{aligned} \frac{1-nh^{\prime }(\theta ^{SR})}{2} \int _{\underline{\theta }}^{\theta ^{SR}} (\theta -h(\theta )-2Q^{D}(\theta ^{SR}))f(\theta )\mathrm{d}\theta . \end{aligned}$$

Remember the constraint $Q(\theta )\le \frac{\theta -h(\theta )}{2}$, $\forall \theta \in \left[ \underline{\theta },\overline{\theta }\right]$.

For any $\theta ^{SR}$ such that the previous constraint is satisfied, i.e. $Q^{D}(\theta ^{SR})=\frac{\theta ^{SR} -nh(\theta ^{SR})}{2}\le \frac{\underline{\theta } -h(\underline{\theta })}{2},$ the above derivative is strictly positive. Letting now $\theta ^{c}$ be such that $\frac{\theta ^{c}-nh(\theta ^{c})}{2}=\frac{\underline{\theta }-h(\underline{\theta })}{2}$⁴⁹, the solution of our problem is $\theta ^{SR}=\theta ^c$ whenever the nonnegativity of sales constraint is satisfied, i.e. $\frac{\theta ^c-nh(\theta ^c)}{2}>0.$ Otherwise it has to be such that $\theta ^{SR}=\Theta (0),$ where $\Theta (0)-nh(\Theta (0))=0:$ the optimal sales level is zero for all $\theta \in \left[ \underline{\theta },\Theta (0)\right]$.

The necessity part of the Proposition is straightforward. Suppose that (27) does not hold over some interval $\left[ \theta _{0},\theta _{1}\right] .$ Then the equilibrium sales function which is identical to the semi-regular (or regular) one defined in Proposition 2 except that $Q(\theta )=Q^{D}(\theta _{0})$, $\theta \in \left[ \theta _{0},x\right] ,$ and $Q(\theta )=Q^{D} (\theta _{1})$, $\theta \in \left[ x,\theta _{1}\right] ,$ where $x-nh(x)=\frac{\theta _{0}-nh(\theta _{0})+\theta _{1}-nh(\theta _{1})}{2}$, strongly Pareto-dominates the original one.⁵⁰$\square$