The following lemmas allow us to characterize the NE when (qNeq, qN oNeq )∈F0, i.e. zeq= 0. Lemmas 8 and 9 are useful in proving Theorem 10.
Lemma 8. No pN oN and pN such that ∆p = pN oN −pN ≤ −tN oN can be equilibrium strategies.
Proof. Proof: First, we rule out the existence of an NE when ∆p < −tN oN, and then when ∆p=−tN oN.
First, considerpN oN andpN such that ∆p <−tN oN. In this case,pN oN < pN−tN oN. Note that the payoff of the non-neutral ISP when ∆p≤ −tN oN ispN oN−c(by (2.1) and
nN oN = 1, using case 3 of Theorem 1), and is strictly increasing with respect to pN oN. Thus, every pricepN oN < pN−tN oN yields a strictly lower payoff for the non-neutral ISP in comparison with the payoff of the this ISP whenpN oN =pN−tN oN. Thus, there exist a profitable deviation for the non-neutral ISP for strategies such that pN oN−pN <−tN oN. Therefore, no pN oN and pN such thatpN oN −pN <−tN oN can be NE strategies.
Now consider strategiespN oN andpN such that ∆p=−tN oN. In this case, using case 3 of Theorem 1, (qNeq, qN oNeq ) = (0,q˜f) ∈ F0L. Thus, nN = 0 and πN(pN) = 0, i.e. the payoff of the neutral ISP is zero. Consider > 0 such that p0N = pN − > c. In this case, pN oN −p0N >−tN oN. Thus, by Theorem 1, (qeqN, qN oNeq ) ∈F0I or (q
eq N, q
eq
Thus, nN > 0, and πN(p0N) >0. Thus, p
0
N is a profitable deviation for the neutral ISP. Therefore, as long as such a deviation existpN oN and pN such that ∆p=−tN oN cannot be NE. Now, we prove that such deviation always exist. This complete the proof. Note that this deviation does not exist if and only if pN −≤c for all >0. Therefore, this deviation does not exist if only ifpN ≤c. Thus,pN oN ≤c−tN oN < c, which contradicts the fact that if z = 0, peqN oN ≥ c (as mentioned in the beginning of the section). The lemma follows.
Lemma 9. No pN oN and pN such that ∆p≥tN can be equilibrium strategies.
Proof. Proof: First, we rule out the existence of an NE when ∆p > tN, and then when ∆p = tN. Consider pN oN and pN such that ∆p > tN. In this case, pN < pN oN −tN. Note that the payoff of the neutral ISP when ∆p≥tN is pN −c (by (2.1) and nN = 1, using case 2 of Theorem 1), and is strictly increasing with respect to pN. Thus, every price pN < pN oN −tN yields a strictly lower payoff for the neutral ISP in comparison with the payoff of the this ISP whenpN =pN oN−tN. Thus, nopN oN and pN such that
pN oN−pN > tN can be Ne strategies.
Now consider strategiespN oN andpN such that ∆p=tN. In this case, using case 2 of Theorem 1, (qNeq, qN oNeq ) = (˜qf,0)∈F0U. Thus, nN oN = 0 andπN oN(pN oN, z= 0) = 0, i.e. the payoff of the non-neutral ISP is zero. Consider >0 such thatp0N oN =pN oN− > c. In this case, p0N oN−pN < tN. Thus, by Theorem 1, (qNeq, qN oNeq )∈F0I or (q
eq N, q
eq
N oN)∈F0L. Thus, nN oN > 0, and πN oN(p0N oN, z = 0) > 0. Thus, p0N oN is a profitable deviation for the non-neutral ISP. Therefore, as long as such a deviation exist pN oN and pN such that ∆p = tN cannot be NE. Now, we prove that such deviation always exist. This
complete the proof. Note that this deviation does not exist if and only if pN oN −≤c for all > 0. Therefore, this deviation does not exist if only if pN oN ≤ c. Therefore,
pN ≤ c−tN < c, which contradicts the fact that peqN ≥ c (as mentioned after at the beginning of the section.). The lemma follows.
Now, we proceed to prove Theorem 10:
Proof. Proof of Theorem 10: First, in Part 1, we characterize the candidate equilibrium strategies by applying the first order condition on the payoffs. Then, in Part 2, we prove that no unilateral deviation is profitable for ISPs. Thus, the strategies characterized in Part 1 are NE.
Part 1: Note that zeq = 0. First note that by Lemmas 8 and 9, no p
N and pN oN such that ∆p≤ −tN oN or ∆p≥tN can be Nash equilibrium. Thus, we consider−tN oN < ∆p < tN. Note that in this region, 0 < xN < 1, and an NE strategy for ISPs should satisfy the first order optimality conditions. Thus, using (2.8) and (2.9), and item 1 of Theorem 1: peqN =c+1 3(2tN oN+tN) peqN oN =c+1 3(2tN +tN oN) (2.47)
which is unique. Note that peqN ≥c and peqN oN ≥c. In order to prove that this is an NE, it is enough to prove that (i)−tN oN <∆peq=peqN oN−peqN < tN, (ii) a deviation of one of the ISPs by which ∆p is shifted to the region ∆p≤ −tN oN or ∆p ≥tN is not profitable for that ISP.
The condition (i) can be proved by (2.47). From this equation, ∆peq = tN−tN oN
3 .
Thus, ∆peq >−t
Part 2: Now, we should prove that condition (ii) holds, i.e. no unilateral deviation is profitable. First, in Case 2-a, we rule out the possibility of a unilateral deviation when
−tN oN <∆p < tN for both neutral and non-neutral ISPs. Then, we consider ∆p≤ −tN oN and ∆p ≥ tN, and in Cases 2-NoN and 2-N, we rule out the possibility of a unilateral deviation in these regions for ISP N and NoN, respectively.
Case 2-a: First, note that by concavity of the payoffs (using (2.8) and (2.9)) as long as−tN oN <∆p < tN, i.e. 0< xN <1, a unilateral deviation by one of the ISPs frompeqN or peqN oN decreases this ISP’s payoff. Thus, we should consider the deviation by ISPs by which ∆p≤ −tN oN or ∆p≥tN.
Case 2-NoN:Now, consider the deviations by the non-neutral ISP. FixpN =peqN, and consider two cases. In Case 2-NoN-i (respectively, Case 2-NoN-ii), we consider deviation by ISP NoN such that ∆p≥tN (respectively, ∆p≤ −tN oN).
Case 2-NoN-i: Suppose the non-neutral ISP increases her price frompeqN oN to make ∆p≥tN. In this case,nN oN = 0, and the payoff of the ISP is zero (by (2.1)). Since in the candidate equilibrium strategy this payoff is non-negative, this deviation is not profitable.
Case 2-NoN-ii: Now, consider the case in which the non-neutral ISP decreases her price to make ∆p≤ −tN oN. In this case, nN oN = 1 and πN oN(p0N oN, z = 0) =p0N oN −c (by (2.1)). Thus, the payoff is a strictly increasing function of p0N oN, and is maximized at
p0N oN =peqN −tN oN. We show that πN oN(p0N oN, z = 0) < πN oN(peqN oN, z= 0). Note that
πN oN(p0N oN, z= 0) = 13(tN −tN oN). In addition, using (2.47), (2.1), 0 ≤xN ≤1, (2.5), and the fact that withpeqN and peqN oN,qeqN oN−qeqN = 0:
πN oN(peqN oN, z= 0) = 1 9
(2tN +tN oN)2
Thus:
πN oN(p0N oN, z= 0)< πN oN(peqN oN, z= 0) ⇐⇒ 3(tN2 −t2N oN)<4tN2 +t2N oN + 4tNtN oN
⇐⇒ t2N + 4t2N oN+ 4tNtN oN >0
where the last inequality is always true. Thus, this deviation is not profitable for ISP NoN.
These cases prove that no deviation form (2.47) is profitable for ISP NoN.
Case 2-N: Now, consider a deviation by the neutral ISP from (2.47). Similar argu- ment can be done for the neutral ISP. Fix,pN oN =peqN oN, and consider two cases. In Case 2-N-i (respectively, Case 2-N-ii), we consider deviation by ISP N such that ∆p≤ −tN oN (respectively, ∆p≥tN).
Case 2-N-i: Suppose the neutral ISP increases her price frompeqN to get ∆p≤ −tN oN. In this case,nN = 0, and the payoff of this ISP is zero. Since in the candidate equilibrium strategy the payoff is non-negative, this deviation is not profitable.
Case 2-N-ii: Now, consider the case in which the non-neutral ISP decreases her price such that ∆p ≥ tN. In this case, nN = 1 and πN(p0N) = p0N −c. Thus, the payoff is a strictly increasing function of p0N, and is maximized at p0N =peqN oN −tN. We show that
πN(p0N) < πN(peqN). Note that πN(p0N) = 13(tN oN −tN) (by (2.1)). In addition, using (2.47), (2.1), 0≤xN ≤1, (2.5), and the fact that with peqN and p
eq N oN,q eq N oN −q eq N = 0: πN(peqN) = 1 9 (2tN oN +tN)2 tN oN+tN Thus: πN(p0N)< πN(peqN) ⇐⇒ 3(t 2 N oN−t2N)<4tN oN2 +t2N+ 4tNtN oN ⇐⇒ t2 N oN+ 4t2N + 4tNtN oN >0
where the last inequality is always true. Thus, this deviation is not profitable for ISP N. Thus, there is no profitable deviation for ISP N. This completes the proof, and the lemma follows.