Regular Equilibria and Negative Welfare Implications in Delegation Games

Proof of Proposition 1

Suppose that a∗ is regularly implementable from φ. By Assumption 1, we can find h(t,⋅)∈Φi such that h(t,⋅)=φi(⋅)+g(t,⋅) for t∈(−1,1). Note that g satisfies g(0,a−i∗)=0 and ∂g(0,a−i∗)∂t=ε for some ε∈R╲{0}. Let at be the Nash equilibrium induced by (h(t,⋅),φ−i). Then, since φi is the optimal reaction function against the others’ reaction function φj,

Since (φ,a∗) is regular, according to the implicit function theorem, at is uniquely determined and differentiable at a neighborhood of t = 0. Then, total differentiation yields

at t = 0. With matrix expression,

where ei is the i-th unit vector.

We now consider the function ui(at). If φi is optimal, from the first-order condition, differentiating this function by t at t = 0 equals 0. That is,

Then, νi:=1ε(∂aj0∂t)j∈N is a normal vector of ∇ui(a−i) and thus,

which implies that (ν1,ν2,…,νn)=J(φ,a∗)−1.   □

Proof of Proposition 2

Before the proof of the proposition, we note the following fact.

Fact 1.   Suppose that for each i ∈ N, there exists αi∈R╲{0} and vector μ∈RN╲{0} such that ∇ui(a−i)=αiμ. Let νi be a normal vector of ∇ui(a−i) for each i ∈ N. Then, {ν1,…,νn} is linearly dependent.

If a is regularly implementable, from Proposition 1, there exists a tuple of normal vectors (ν1,…,νn) of (∇ui(a−i))i∈N such that (ν1,…,νn)=J(φ,a)−1. Since (φ,a) is regular, J(φ,a)−1 is invertible. Therefore, (ν1,…,νn) is also invertible and thus {ν1,…,νn} is linearly independent. However, this contradicts Fact 1.   □

Proof of Theorem 1

Since Ui(a∗) is convex and open, according to the separating hyperplane theorem, there exists ξi∈Rn,γi∈R such that for each a′∈Ui(a∗), ξi⋅a′>γi and ξi⋅a∗=γi. Without loss of generality, we can say that ξi=∇ui(a∗).

From the assumption, let si:=∂ui+1∂ai(a∗)≠0 for each i∈{1,…,n−1} and sn:=∂u1∂an(a∗)≠0. Then, there exists (βij)j∈N∖{i} such that Bi=(−βi1,…,−βii−1,1,−βii+1,…,−βin)=(1/si)∇ui+1(a∗) for each i∈{1,…,n−1} and Bn:=(−βn1,…,−βnn−1,1)=(1/sn)∇u1(a∗).

Let

Since {∇ui(a∗)} is linearly independent, (B1,…,Bn)⊤ is invertible and thus (ν1,…,νn) is well defined. Note that νi⋅Bi=1 and νi⋅Bj=0 for each i ∈ N and j∈N∖{i}.

We can find βi0 that satisfies ai∗=∑j∈N∖{i}βijaj∗+βi0. We define φi(a−i′)=∑j∈N∖{i}βijaj′+βi0 for each a−i′∈A and i ∈ N. Then, clearly, a∗ is the unique intersection of {φi}i∈N, and (φ,a∗) is regular since

Now, consider a unilateral deviation by player i. Without loss of generality, i = 1. Suppose that player 1 chooses a different function h∈Φ1. If (h,φ−1) induces a Nash equilibrium aˆ, it satisfies

By matrix expression,

On the contrary, since

we have

Therefore,

The last equation is derived from the fact that ∇u1(a∗)=snBn and Bn⋅ν1=0.

Then, ∇u1(a∗)⋅aˆ=∇u1(a∗)⋅a∗=γ1; therefore, aˆ∉U1(a∗), which implies that h is an unprofitable deviation.   □

Proof of Proposition 3

From the assumption, si=∂ui∂a−i<0 for each i ∈ N. As in the proof of Theorem 1, let Bi=(−βi,−i,1)=(1/si)∇u−i(a∗) and φ˜i(a−i)=βi,−ia−i+βi0 for each i ∈ N. Let us check whether the restriction of the above reaction function profile, denoted by φ, is an equilibrium. From Theorem 1, we only consider deviations that yield a corner equilibrium. Note that a deviation that yields ai=0 is not profitable since ui(a∗)⩾ui(0,0)⩾ui(0,a−i) for each a−i. Therefore, consider a deviation h such that C(h,φ−i)≠∅. Let aˉi=φ˜−i−1(0). Note also that ui(a∗)⩾ui(aˉi,0) since (aˉi,0) is on the line {(ai,φ˜−i(ai))}. If ∂ui∂ai(a∗)=0, then β−i,i=0. This implies that C(h,φ−i)=∅. If ∂ui∂ai(a∗)>0, then β−i,i>0, and therefore C(h,φ−i)⊆[0,aˉi]×{0} and aˉi<ai∗. If ∂ui∂ai(aˉi,0)<0, there exists ε∈R++ such that ui(aˉi−ε,0)>ui(aˉi,0). On the contrary, since ∂ui∂aj<0, ui(a∗)<ui(ai∗,0). Since aˉi<ai∗, this contradicts the quasi-concavity of ui. Therefore, ∂ui∂ai(aˉi,0)⩾0. Since ∂2ui∂ai2⩽0, ∂ui∂ai(a)>0 for each a∈C(h,φ−i). Thus, ui(aˉi,0)>ui(a) for each a∈C(h,φ−i). Therefore, h is an unprofitable deviation. In the case of ∂ui∂ai(a∗)<0, we have that β−i,i<0. Then, similarly, we can show that no deviation is profitable, which concludes the proof.   □

Proof of Proposition 4

Given (λ1,λ2), the (interior) equilibrium induced by the reaction function is

On the other hand, the necessary condition implies that

Since each φi is linear, from Proposition 3, when the necessary condition is satisfied, the restriction of this reaction function is an equilibrium.

Note that there are two free variables (λo,λs) and one equation to satisfy (equation (4)). Therefore, there can be multiple equilibria. By calculating the condition, the implementable a∗ is written as

Since the reaction function is written as φi(aj)=−λs+λo2aj+λsβ−c2α, if λs+λc=2, the reaction functions have infinite intersections. In this case, to satisfy (4), λs=2β−cβ. Then, the output level is a∗=β−c4α, which maximizes the total profit of these two companies. This is not regularly implementable.

On the contrary, we can easily check that as long as λs≠2β−cβ and (4β−3c)/β>λs>β/c, it yields the unique equilibrium and positive outputs in the execution stage.   □