Collusive Price Leadership Among Firms with Different Discount Factors

Shuaicheng Liu

doi:10.1515/bejte-2023-0112

Article

Collusive Price Leadership Among Firms with Different Discount Factors

Shuaicheng Liu

Published/Copyright: April 8, 2024

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From the journal The B.E. Journal of Theoretical Economics Volume 24 Issue 2

Abstract

This paper analyzes the effect of price leadership on collusion among firms with different discount factors. We first find that price leadership relaxes the incentive constraints for collusion. We then derive a dynamic collusion path in which the firms with lower discount factors initially occupy the largest market share and then gradually cede it to the firms with higher discount factors. This collusion path is shaped by the conflicting forces of fairness and efficiency. Additionally, price leadership can restore the efficiency implied by differentiated time preferences in repeated games.

Keywords: collusive price leadership; different discount factors; dynamic collusion path

JEL Classification: D43; L13; L41

Corresponding author: Shuaicheng Liu, College of Business, Shanghai University of Finance and Economics, No. 777 Guoding Road, Shanghai, China, E-mail: shuaicheng-liu@163.sufe.edu.cn

Acknowledgments

I would like to thank the editor, Ronald Peeters, and two anonymous referees for their valuable comments. I would also like to thank Wenzhang Zhang, Rongzhu Ke, Xianwen Shi and other participants in 2023 International Conference on Industrial Economics (Zhejiang University).

Appendix

Proof of (1) and (3) in Lemma 1

We first prove the (1) in Lemma 1. Consider such a strategic profile in which all firms publish prices in the first stage and all set (p _i, r _i) = (c, 1/n). In this equilibrium, each firm earns zero profits. Clearly, there is no profitable deviation.

We then prove the (3) in Lemma 1. Suppose not, then we can find at least two firms, i and j, that make positive profits. Suppose the lowest price is p* ∈ (c, p ^m] in equilibrium. Then, we must have p _i = p _j = p*, and for the firm h ≠ i, j, we must have p _h ≥ p*. If firms i and j publish prices at same stage, Bertrand competition implies that the two firms will undercut each other to marginal cost. If the two firms publish prices at different stages, then the follower will undercut slightly to deviate. □

Proof of Theorem 1

By our definition of U _i,t, the NBS is equivalently defined as a sequence of equilibrium profits { π i , t } i ∈ I , t ∈ N * that maximize ∏ i = 1 n Σ t = 1 ∞ δ i t − 1 π i , t . We first prove a series of lemmata.

Lemma A.1.

For the NBS equilibrium, π t = ∑ i = 1 n π i , t = π m for every t.

Proof.

Suppose not, then there exists one period t′ with π _t′ < π ^m. There must be two cases in this equilibrium. In the first case, all firms publish prices at the same stage in period t′. Then we have U _i,t′ ≥ π _t′ for every i to satisfy incentive constraints. Then we can change this equilibrium only in period t′. We pick one firm, say j, to be the follower and the rest to be the leaders. Leaders get the same profits as before, and they won’t deviate. And the profit of firm j changes to be π j , t ′ ′ = π j , t ′ + π m − π t ′ . By this way, the discounted profit of firm j in t′ changes to be U j , t ′ ′ = U j , t ′ + π m − π t ′ , and the joint profit in t′ changes to be π t ′ ′ = π t ′ + π m − π t ′ = π m . As U _j,t′ ≥ π _t′, we have U j , t ′ ′ ≥ π j , t ′ d = π t ′ ′ . So firm j won’t deviate either. But this new equilibrium strictly increases U _j,1, holding the rest U _i,1 constant, a contradiction.

In the second case, at least one firm is the follower in period t′. Without loss of generality, we can assume that firm j is one of the followers. We could construct a new equilibrium using the method in the first case to derive a contradiction. We could keep firm j as the follower and increase its profits by π ^m − π _t′, while the rest of the firms become the leaders with constant profits. □

Lemma A.2.

For the NBS equilibrium, as long as t is large enough, π _i,t will always be zero for every i < n.

Proof.

Suppose not, then we have π _i,t > 0 for some large t + 1. We then construct a new equilibrium in which firm i transfers profit ɛ > 0 to firm n in the period t + 1, and becomes the leader. In the new equilibrium (where we use superscript ′ to denote it), we have:

(A1) U i , 1 ′ U n , 1 ′ = U i , 1 − δ i t ε ( U n , 1 + δ n t ε ) = U i , 1 U n , 1 + ε U i , 1 δ n t − U n , 1 δ i t − δ i t δ n t ε

As long as t > ln(U _n,1/U _i,1)/ln(δ _n/δ _i), we have U i , 1 δ n t − U n , 1 δ i t > 0 . And then we can find a ɛ small enough that U i , 1 δ n t − U n , 1 δ i t − δ i t δ n t ε > 0 and U i , 1 ′ U n , 1 ′ > U i , 1 U n , 1 . Then, such a transfer is feasible and strictly increases ∏ i = 1 n U i , 1 . □

Lemma A.3.

For any two periods t ̌ < t ̂ and two firms i < j , π i , t ̂ > 0 and π j , t ̌ > 0 cannot hold at the same time in the NBS equilibrium.

Proof.

Suppose not, then we have π i , t ̂ > 0 and π j , t ̌ > 0 . Then we can intertemporally exchange the profits of firms i and j as follows (where we use superscript ′ to denote the new equilibrium):

(A2) π i , t ̌ ′ = π i , t ̌ + ε ′ , π i , t ̂ ′ = π i , t ̂ − ε π j , t ̌ ′ = π j , t ̌ − ε ′ , π j , t ̂ ′ = π j , t ̂ + ε

By construction, the joint profits in these two periods are unchanged. And by letting firm i become the leader in period t ̂ and firm j become the leader in period t ̌ , their incentive constraints can also be satisfied. The changes in discounted profits of the two firms in period t ̌ are U i , t ̌ ′ − U i , t ̌ = ε ′ − δ i t ̂ − t ̌ ε and U j , t ̌ ′ − U j , t ̌ = δ j t ̂ − t ̌ ε − ε ′ , respectively. As long as ɛ > 0 and ɛ′ > 0 satisfy max { ε , ε ′ } < min { π i , t ̂ , π j , t ̌ } and δ i t ̂ − t ̌ < ε ′ / ε < δ j t ̂ − t ̌ , U i , t ̌ and U j , t ̌ can strictly increase. Since δ _i < δ _j, we can find such ɛ and ɛ′, which implies this new equilibrium strictly increases U _i,1 and U _j,1, holding the rest constant, a contradiction. □

Lemma A.4.

For the NBS equilibrium, if π _i,t = 0 from t′, then π _i−1,t = 0 from t′, for every 1 < i < n.

Proof.

Suppose not, then for some period t ̂ ≥ t ′ , π i − 1 , t ̂ > 0 . As π _i,t = 0 for t ≥ t′, there is one period t ̌ < t ′ that π i , t ̌ > 0 . Otherwise, we have U _i,1 = 0 and ∏ j = 1 n U j , 1 = 0 , which is never an NBS equilibrium. Then we have t ̌ < t ̂ , π i − 1 , t ̂ > 0 and π i , t ̌ > 0 . By Lemma A.3, this is impossible. □

We next complete the proof of Theorem 1. Lemma A.1 implies that the lowest price in each period is the monopoly price p ^m. By Lemma A.2, we define t _i as the minimum t after which π _i,t = 0, thus it is unique. By construction, π i , t i > 0 . Lemma A.2 implies property (2). Lemma A.2 and incentive constraints for collusion imply property (1). Lemma A.4 implies that t _i is bounded and t ₁ ≤ t ₂ ≤ … ≤ t _n−1. Next, we just need to prove properties (3)–(5).

Proof of Property (3)

We prove property (3) by induction. We first consider i = 1. If t ₁ = 1, then we must have π _1,1 > 0. Otherwise, we have U _i,1 = π _1,1 = 0 and ∏ j = 1 n U j , 1 = 0 , which is never an NBS equilibrium. If t ₁ > 1, there must be π _1,t > 0 for 0 < t < t ₁. Otherwise, if π 1 , t 1 ′ = 0 for some t 1 ′ ∈ ( 0 , t 1 ) , then there must be one j > 1 with π j , t 1 ′ > 0 , and by property (2), we have π 1 , t 1 > 0 . Then we can exchange the profits of firms 1 and j between periods t 1 ′ and t ₁ like Lemma A.3. We then show that π 1 , t > max { π j ∣ j > 1 a n d j ∈ I } for 0 < t < t ₁. Otherwise, there is π _j,t ≥ π _1,t > 0 in some period, then we can continue to exchange the profits of firms 1 and j between this period and t ₁ like Lemma A.3. At last, we show that π _i,t ≥ π _i+1,t for i ∈ [2, n − 1] and t ∈ (0, t ₁). Otherwise, we have π _i+1,t > π _i,t ≥ 0 for some period t′ < t ₁, then we can exchange the profits of firms 1 and i + 1 between periods t′ and t ₁ like Lemma A.3. Thus, property (3) holds for i = 1.

Now, we assume that property (3) holds for i = 2, …, h, where h < n − 1. Then we prove it holds for i = h + 1. We first have π _h+1,t > 0 for t _h < t < t _h+1. Otherwise, if π h + 1 , t h + 1 ′ = 0 for some t h + 1 ′ ∈ ( t h , t h + 1 ) , then there must be one j > h + 1 with π j , t h + 1 ′ > 0 , as π i , t h + 1 ′ = 0 for i ∈ [1, h] by property (2). As π h + 1 , t h + 1 > 0 by construction, we can exchange the profits of firms h + 1 and j between periods t h + 1 ′ and t _h+1 like Lemma A.3. We then show that π h + 1 , t > max { π j ∣ j > h + 1 a n d j ∈ I } for t _h < t < t _h+1. Otherwise, there is π _j,t ≥ π _h+1,t > 0 in one period, then we can continue to exchange the profits of firms h + 1 and j between this period and t _h+1. At last, we show that π _i,t ≥ π _i+1,t for i ∈ [h + 2, n − 1] and t ∈ (t _h, t _h+1). Otherwise, we have π _i+1,t > π _i,t ≥ 0 for some period t′ ∈ (t _h, t _h+1), then we can exchange the profits of firms h + 1 and i + 1 between periods t′ and t _h+1. Thus, property (3) holds for i = h + 1, and it holds for all i < n. □

Proof of Property (4)

If property (4) does not hold, then there exists t _i−1 < t′ < t″ ≤ t _i such that π _i,t
^″ > π _i,t′. By Lemma A.1, π _t
^″ = π _t′ = π ^m. Thus, there must be π _i,t′ < π ^m. Otherwise, we will have π _i,t
^″ = π _i,t′ = π ^m. By property (2), π _j,t′ = 0 for j ∈ [1, i − 1]. Therefore, there must exist a h > i such that π _h,t′ > 0. And by construction, π i , t i > 0 . Then we can exchange the profits of firms i and h between periods t′ and t _i like Lemma A.3. Therefore, property (4) must hold. □

Proof of Property (5)

We first prove two lemmas, which will also be useful for later proofs of Lemma 2 and Proposition 2.

Lemma A.5.

For the NBS equilibrium, if t _i = 1 for some i < n , then 1 ≤ U _i+1,1/U _i,1 ≤ δ _i+1/δ _i.

Proof.

If t _i = 1, then there must be U _i,1 = π _i,1 > 0. We first prove U _i+1,1/U _i,1 ≥ 1. Suppose not, then we have U _i+1,1 < U _i,1. Obviously, transferring profits from firm i to firm i + 1 in first period can directly increase U _i,1 U _i+1,1 in this case.

We then prove U _i+1,1/U _i,1 ≤ δ _i+1/δ _i. Suppose not, then we have

(A3) U i + 1,1 > δ i + 1 δ i U i , 1 .

There must be two cases. In the first case, π _i+1,1 > 0. So transferring profits from firm i + 1 to firm i in first period can directly increase U _i,1 U _i+1,1 in this case. In the second case, π _i+1,1 = 0, and we must have π _i+1,2 > 0. Otherwise, there must be another firm j ≠ i and i + 1, with π _j,2 > 0, and another period t′ > 2 with π _i+1,t′ > 0. Therefore, if j < i, we have 1 < 2, π _i,1 > 0, and π _j,2 > 0, which contradicts Lemma A.3. And if i + 1 < j, we have 2 < t′, π _j,2 > 0, and π _i+1,t′ > 0, which contradicts Lemma A.3 again.

As π _i+1,2 > 0, we could construct a new equilibrium in which firm i + 1 transfers profit ɛ > 0 to firm i in period 2. For the new equilibrium, we have:

(A4) U i , 1 ′ U i + 1,1 ′ = ( U i , 1 + δ i ε ) ( U i + 1,1 − δ i + 1 ε ) = U i , 1 U i + 1,1 + ε ( U i + 1,1 δ i − U i , 1 δ i + 1 − δ i δ i + 1 ε )

By Eq. (A3) we have U _i+1,1 δ _i − U _i,1 δ _i+1 > 0. Therefore we can find a ɛ small enough that U i , 1 ′ U i + 1,1 ′ > U i , 1 U i + 1,1 . Then, such a transfer is feasible and strictly increases ∏ i = 1 n U i , 1 . □

Lemma A.6.

For the NBS equilibrium, if t _i > 1 for some i < n, then ( δ i + 1 / δ i ) t i − 1 ≤ U i + 1,1 / U i , 1 ≤ ( δ i + 1 / δ i ) t i .

Proof.

We first prove U i + 1,1 / U i , 1 ≥ ( δ i + 1 / δ i ) t i − 1 . Suppose not, then we have U i + 1,1 / U i , 1 < ( δ i + 1 / δ i ) t i − 1 . We then construct a new equilibrium in which firm i transfers profit to firm i + 1 in the period t _i. For the new equilibrium, we have:

(A5) U i , 1 ′ U i + 1,1 ′ = U i , 1 − δ i t i − 1 ε ( U i + 1,1 + δ i + 1 t i − 1 ε ) = U i , 1 U i + 1,1 + ε U i , 1 δ i + 1 t i − 1 − U i + 1,1 δ i t i − 1 − δ i t i − 1 δ i + 1 t i − 1 ε

As U i + 1,1 / U i , 1 < ( δ i + 1 / δ i ) t i − 1 , we can find a ɛ small enough that U i , 1 ′ U i + 1,1 ′ > U i , 1 U i + 1,1 , a contradiction.

We then prove U i + 1,1 / U i , 1 ≤ ( δ i + 1 / δ i ) t i . Suppose not, and let T ^[13] = t i , if π i + 1 , t i > 0 t i + 1 , if π i + 1 , t i = 0 , so we have U i + 1,1 / U i , 1 > ( δ i + 1 / δ i ) t i ≥ ( δ i + 1 / δ i ) T − 1 . We then construct a new equilibrium in which firm i + 1 transfers profit to firm i in period T. For the new equilibrium, we have:

(A6) U i , 1 ′ U i + 1,1 ′ = U i , 1 + δ i T − 1 ε ( U i + 1,1 − δ i + 1 T − 1 ε ) = U i , 1 U i + 1,1 + ε U i + 1,1 δ i T − 1 − U i , 1 δ i + 1 T − 1 − δ i T − 1 δ i + 1 T − 1 ε

Again, we can find a ɛ small enough that U i , 1 ′ U i + 1,1 ′ > U i , 1 U i + 1,1 . □

We next complete the proof of property (5). We have proven t ₁ ≤ t ₂ ≤ … ≤ t _n−1. Therefore, if t _i = 1 for some i, we must have t ₁ = t ₂ = … = t _i = 1, and if t _i > 1 for some i, we must have t _j > 1 for all j > i. Thus by Lemmas A.5 and A.6, property (5) holds. □

Proof of Lemma 2

We first prove t i = 1 ⇒ i δ i + Σ j = i + 1 n δ j / n ≤ ( n − i ) / n .

We prove it by induction (backward). Thus, first for firm n − 1, we need to prove t _n−1 = 1 ⇒ (n − 1)δ _n−1 + δ _n ≤ 1. Suppose not, then (n − 1)δ _n−1 > 1 − δ _n. By Theorem 1, t _n−1 = 1 ⇒ t ₁ = t ₂ = … t _n−2 = 1. There must be two cases. In the first case, π _n,1 = 0. Since those n − 1 firms exist for only one period, such that their discount factors do not enter U _i,1, π _1,1 = π _2,1 = … = π _n−1,1 = U _1,1 = U _2,1 = … = U _n−1,1 = π ^m/(n − 1) by Eq. (5) and Lemma A.1, and U _n,1 = π ^m δ _n/(1 − δ _n). So we have

(A7) U n , 1 U n − 1,1 = δ n ( n − 1 ) 1 − δ n = δ n ( n − 1 ) 1 − δ n δ n − 1 δ n − 1 > δ n δ n − 1

But as we assume t _n−1 = 1, U _n,1/U _n−1,1 ≤ δ _n/δ _n−1 by Lemma A.5, a contradiction. In the second case, π _n,1 > 0, so U n − 1,1 = π m − π n , 1 / ( n − 1 ) and U _n,1 = π _n,1 + π ^m δ _n/(1 − δ _n). And once again we can derive U _n,1/U _n−1,1 > δ _n/δ _n−1, which contradicts Lemma A.5. So this proposition holds for n − 1.

We then suppose it holds for firm h + 1 < n and prove that it also holds for firm h, that is, t h = 1 ⇒ h δ h + Σ j = h + 1 n δ j ≤ n − h . Suppose not, then h δ h + Σ j = h + 1 n δ j > n − h ⇒ h δ h + δ h + 1 + Σ j = h + 2 n δ j > n − h ⇒ ( h + 1 ) δ h + 1 + Σ j = h + 2 n δ j > n − h − 1 ⇒ t h + 1 > 1 . We first consider the case where Σ i = 1 h π i , 1 = π m , so we have π _h,1 = U _h,1 = π ^m/h.

There must be two subcases in this case. In the first subcase, π h + 1 , t h = π h + 2 , t h + 1 = … = π n , t n − 1 = 0 . As t _h = 1, we have U _h+1,1/U _h,1 ≤ δ _h+1/δ _h by Lemma A.5. As U h + 1,1 = π m δ h + 1 − δ h + 1 t h + 1 / ( 1 − δ h + 1 ) in this subcase, we have

(A8) U h + 1,1 U h , 1 ≤ δ h + 1 δ h ⇒ δ h + 1 − δ h + 1 t h + 1 1 − δ h + 1 ≤ 1 h δ h + 1 δ h ⇒ 1 − δ h + 1 ≥ h δ h 1 − δ h + 1 t h + 1 − 1

As t _h+1 > 1, we have t _h+1+g > 1 for all g = 1, 2, …, n − 2 − h, and U h + 1 + g , 1 = π m δ h + 1 + g t h + g − δ h + 1 + g t h + 1 + g / ( 1 − δ h + 1 + g ) . Thus, by Lemmas A.5 and A.6, we have

(A9) U h + 1 + g , 1 U h , 1 = U h + 1 + g , 1 U h + g , 1 U h + g , 1 U h + g − 1,1 … U h + 1,1 U h , 1 ≤ δ h + 1 + g t h + g δ h + g t h + g δ h + g t h + g − 1 δ h + g − 1 t h + g − 1 … δ h + 1 δ h ⇒ 1 − δ h + 1 + g ≥ h δ h δ h + 1 t h + 1 − 1 δ h + 2 t h + 2 − t h + 1 … δ h + g t h + g − t h + g − 1 − δ h + 1 t h + 1 − 1 δ h + 2 t h + 2 − t h + 1 … δ h + g + 1 t h + g + 1 − t h + g

For firm n, U n , 1 = π m δ n t n − 1 / ( 1 − δ n ) , so by the same argument, we could get

(A10) 1 − δ n ≥ h δ h δ h + 1 t h + 1 − 1 δ h + 2 t h + 2 − t h + 1 … δ n − 1 t h − 1 − t n − 2

Summing Eqs. (A8)–(A10) across all g, we get

(A11) ∑ j = h + 1 n ( 1 − δ j ) ≥ h δ h 1 − δ h + 1 t h + 1 − 1 + δ h + 1 t h + 1 − 1 − δ h + 1 t h + 1 − 1 δ h + 2 t h + 2 − t h + 1 + δ h + 1 t h + 1 − 1 δ h + 2 t h + 2 − t h + 1 − ⋯ = h δ h

And Σ j = h + 1 n ( 1 − δ j ) ≥ h δ h ⇔ h δ h + Σ j = h + 1 n δ j ≤ n − h , which contradicts the assumption that h δ h + Σ j = h + 1 n δ j > n − h by law of indirect proof.

For the second subcase, π h + 1 + g , t h + g > 0 for some g. Since δ _h+1+g > δ _h+g, this increases Σ j = h + 1 n U j , 1 compared to the U _j,1 in Eqs. (A8)–(A10). Thus, in this subcase, we would obtain a stricter condition than Eq. (A11), that is, ∑ j = h + 1 n ( 1 − δ j ) > h δ h , which is a contradiction again. At last, for the second case, Σ i = 1 h π i , 1 < π m , so we have U _h,1 < π ^m/h. Thus, we would obtain ∑ j = h + 1 n ( 1 − δ j ) > h δ h again, a contradiction. Therefore, t i = 1 ⇒ i δ i + Σ j = i + 1 n δ j / n ≤ ( n − i ) / n holds for i = h, and it holds for all i < n. □

We then prove i δ i + Σ j = i + 1 n δ j / n ≤ ( n − i ) / n ⇒ t i = 1 .

We prove it by induction. Thus, first for i = 1, we need to prove δ 1 + Σ j = 2 n δ j ≤ n − 1 ⇒ t 1 = 1 . Suppose not, then t ₁ > 1. Thus, U _1,1 > π ^m. There must be two cases. In the first case, π 2 , t 1 = π 3 , t 2 = … = π n , t n − 1 = 0 . So we have U g , 1 = π m δ g t g − 1 − δ g t g / ( 1 − δ g ) , where g = 2, 3, …, n − 1. By Lemma A.6, for firm g > 2, we have:

(A12) U g , 1 U 1,1 = U g , 1 U g − 1,1 U g − 1,1 U g − 2,1 … U 2,1 U 1,1 ≥ δ g t g − 1 − 1 δ g − 1 t g − 1 − 1 δ g − 1 t g − 2 − 1 δ g − 2 t g − 2 − 1 … δ 2 t 1 − 1 δ 1 t 1 − 1 ⇒ 1 − δ g ≤ δ 1 t 1 − 1 δ g δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ g − 1 t g − 1 − t g − 2 − δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ g t g − t g − 1 ⇒ 1 − δ g < δ 1 δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ g − 1 t g − 1 − t g − 2 − δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ g t g − t g − 1

And for firm g = 2, we also have:

(A13) 1 − δ 2 ≤ δ 1 t 1 − 1 δ 2 1 − δ 2 t 2 − t 1 < δ 1 1 − δ 2 t 2 − t 1

For firm n, U n , 1 = π m δ n t n − 1 / ( 1 − δ n ) . Thus, by the same argument, we have

(A14) 1 − δ n ≤ δ 1 t 1 − 1 δ n δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ n − 1 t n − 1 − t n − 2 < δ 1 δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ n − 1 t n − 1 − t n − 2

Summing Eqs. (A12)–(A14) across all g, we get

(A15) ∑ j = 2 n ( 1 − δ j ) < δ 1 1 − δ 2 t 2 − t 1 + δ 2 t 2 − t 1 − δ 2 t 2 − t 1 δ 3 t 3 − t 2 + δ 2 t 2 − t 1 δ 3 t 3 − t 2 − ⋯ = δ 1

And Σ j = 2 n ( 1 − δ j ) < δ 1 ⇔ δ 1 + Σ j = 2 n δ j > n − 1 , which contradicts the assumption that δ 1 + Σ j = 2 n δ j ≤ n − 1 in the beginning. And for the second case, π g , t g − 1 > 0 for some g. Since δ _g > δ _g−1, this increases Σ j = 2 n U j , 1 compared to the U _j,1 in Eqs. (A12)–(A14). Thus, in this case, we would obtain a stricter condition than Eq. (A12), that is, 1 − δ g < δ 1 t 1 − 1 δ g δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ g − 1 t g − 1 − t g − 2 − δ 2 t 2 − t 1 δ 3 t 3 − t 2 … δ g t g − t g − 1 . Thus, we can still derive the contradiction that δ 1 + Σ j = 2 n δ j > n − 1 . Therefore, this proposition holds for i = 1.

We then suppose it holds for firm i = h − 1 < n − 1 and prove that it also holds for firm h, that is, h δ h + Σ j = h + 1 n δ j / n ≤ ( n − h ) / n ⇒ t h = 1 . Suppose not, then we have t _h > 1. And h δ h + Σ j = h + 1 n δ j ≤ n − h ⇒ ( h − 1 ) δ h + Σ j = h n δ j ≤ n − h + 1 ⇒ ( h − 1 ) δ h − 1 + Σ j = h n δ j ≤ n − ( h − 1 ) ⇒ t h − 1 = 1 ⇒ t 1 = t 2 = … = t h − 2 = 1 . Thus, we must have π _1,1 = π _2,1 = … π _h−1,1 = U _1,1 = U _2,1 = … U _h−1,1 by Eq. (5).

There must be two cases for the first period. In the first case, Σ i = 1 h − 1 π i , 1 = π m . Thus, by Lemma A.5, we have U _h,1 ≥ U _h−1,1 = π ^m/(h − 1) > π ^m/h. In the second case, Σ i = 1 h − 1 π i , 1 < π m , and we must have U _h,1 ≥ π ^m/h as well. Otherwise, if U _h,1 < π ^m/h, we will have U h − 1,1 = π m − π h , 1 / ( n − 1 ) > π m / h > U h , 1 , which contradicts Lemma A.5. Thus, there must be U _h,1 ≥ π ^m/h.

As t _h > 1, we have t _h+g > 1 for g = 1, 2, …, n − 1 − h. And there must be two cases. In the first case, π h + 1 , t h = π h + 2 , t h + 1 = … = π n , t n − 1 = 0 . Therefore, using the same method of deriving Eqs. (A12)–(A14), we can get

(A16) 1 − δ h + 1 ≤ h δ h t h − 1 δ h + 1 1 − δ h + 1 t h + 1 − t h < h δ h 1 − δ h + 1 t h + 1 − t h 1 − δ g < h δ h δ h + 1 t h + 1 − t h δ h + 2 t t + 2 − t h + 1 … δ g − 1 t g − 1 − t g − 2 − δ h + 1 t h + 1 − t h δ h + 2 t t + 2 − t h + 1 … δ g t g − t g − 1 × ( g = 2 … , n − 1 ) 1 − δ n < h δ h δ h + 1 t h + 1 − t h δ h + 2 t t + 2 − t h + 1 … δ n − 1 t n − 1 − t n − 2

By summing Eq. (A16), we can get Σ j = h + 1 n ( 1 − δ j ) < h δ h ⇔ h δ h + Σ j = h + 1 n δ j > n − h , a contradiction. And finally for the case where some π g , t g − 1 > 0 , we can still get this contradiction, as we are simply imposing stricter conditions. Therefore, i δ i + Σ j = i + 1 n δ j / n ≤ ( n − i ) / n ⇒ t i = 1 holds for i = h, and it holds for all i < n. □

Proof of Proposition 2

Before proceeding, we establish three claims.

Claim 1.

If t _h−1 = 1 and h δ h + Σ i = h + 1 n δ i > n + 1 − h for h > 1, then t _h > 2.

Proof.

Suppose not, then we have t _h ≤ 2. As h δ h + Σ i = h + 1 n δ i > n + 1 − h > n − h , t _h = 2 by Lemma 2. When π _h+1,2 = 0, π _h,2 is maximized to π ^m. And U _h,1 = π _h,1 + δ _h π _h,2 is larger when π _h,1 > 0 than when π _h,1 = 0. As t _h−1 = 1, we have t ₁ = t ₂ = … t _h−2 = 1 and U _1,1 = U _2,1 = … = U _h−1,1 ≤ π ^m/(h − 1). When π _h,1 > 0, we must have U _h,1 = U _h−1,1. Otherwise, transferring profits from firm h in the first period to firm h − 1 can directly increase U _h−1,1 U _h,1. As δ _j < 1, h δ h + Σ i = h + 1 n δ i > n + 1 − h ⇒ δ h > 1 / h . Thus, we can get 1/h < δ _h < U _h,1/π ^m < 1/(h − 1).

We first consider the case where π h + 1 , t h = π h + 2 , t h + 1 = … = π n , t n − 1 = 0 . As t _h > 1, we have t _h+g > 1 for all g = 1, 2, …, n − 1 − h, and U h + g , 1 = π m δ h + g t h + g − 1 − δ h + g t h + g / ( 1 − δ h + g ) , and U n , 1 = π m δ n t n − 1 / ( 1 − δ n ) . By Lemma A.6, we have

(A17) U h + g , 1 π m / ( h − 1 ) < U h + g , 1 U h , 1 = U h + g , 1 U h + g − 1,1 U h + g − 1,1 U h + g − 2,1 … U h + 1,1 U h , 1 ≤ δ h + g t h + g − 1 δ h + g − 1 t h + g − 1 δ h + g − 1 t h + g − 2 δ h + g − 2 t h + g − 2 … δ h + 1 t h δ h t h ⇒ 1 − δ h + g > ( h − 1 ) δ h 2 δ h + 1 t h + 1 − 2 δ h + 2 t h + 2 − t h + 1 … δ h + g − 1 t h + g − 1 − t h + g − 2 − δ h + 1 t h + 1 − 2 δ h + 2 t h + 2 − t h + 1 … δ h + g t h + g − t h + g − 1 1 − δ n > ( h − 1 ) δ h 2 δ h + 1 t h + 1 − 2 δ h + 2 t h + 2 − t h + 1 … δ n − 1 t n − 1 − t n − 2

By summing Eq. (A17), we can get Σ i = h + 1 n ( 1 − δ i ) > ( h − 1 ) δ h 2 ⇔ ( h − 1 ) δ h 2 + Σ i = h + 1 n δ i < n − h . Let H ( h , δ h ) = ( h − 1 ) δ h 2 + 1 − h δ h . As 1/h < δ _h < 1/(h − 1), H(2, δ _h) > 0 for δ _h ∈ (1/2, 1). And we have ∂ H ( h , δ h ) / ∂ h = δ h 2 − δ h < 0 , and ∂H(h, δ _h)/∂δ _h = h(2δ _h − 1) − 2δ _h < 0 for h > 2. As lim_h→+∞ H(h, 1/(h − 1)) = 0, we have H(h, δ _h) > 0 for h > 2. And we can get h δ h + Σ i = h + 1 n δ i < n + 1 − h , which is a contradiction. At last, for the case where π h + g , t h + g − 1 > 0 for some g. Since δ _h+g > δ _h+g−1, this increases Σ i = h + 1 n U i , 1 , and we can get the same conditions as Eq. (A17). □

Claim 2.

If t _i > 1, then U i , 1 ≥ π m δ i t i − 1 .

Proof.

If t _i > 1, then we have δ i t i − 1 < 1 . We then prove Claim 2 by induction. First for i = 1. If t ₁ > 1, and we have U 1,1 > π 1,1 = π m > π m δ 1 t 1 − 1 . Thus Claim 2 holds for i = 1.

We then suppose it holds for i = h − 1 and prove that it also holds for i = h. Suppose not, then we have U h , 1 < π m δ h t h − 1 . There must be two cases. In the first case, t _h−1 > 1, so we have U h − 1,1 ≥ π m δ h − 1 t h − 1 − 1 by the assumption of induction. As t _h ≥ t _h−1 > 1, we have U h , 1 / U h − 1,1 < δ h t h − 1 / δ h − 1 t h − 1 − 1 ≤ δ h t h − 1 − 1 / δ h − 1 t h − 1 − 1 , which contradicts Lemma A.6.

In the second case, t _h−1 = 1. If t _h > 2, we will have U h , 1 > δ h π h , 2 = δ h π m > π m δ h t h − 1 , a contradiction. So we must have t _h = 2. In the proof of Lemma 2, we have shown that if t _h > 1, there must be U _h,1 ≥ π ^m/h. So we have δ _h > 1/h.^[14] As U _h,1 ≥ π ^m/h, for firm g > h, we have U g , 1 / ( π m / h ) ≥ U g , 1 / U h , 1 ≥ δ g t g − 1 − 1 / δ g − 1 t g − 1 − 1 δ g − 1 t g − 2 − 1 / δ g − 2 t g − 2 − 1 … δ h + 1 t h − 1 / δ h t h − 1 by Lemma A.6. Thus, as in Eq. (A16), we have the following inequalities:

(A18) 1 − δ h + 1 ≤ h δ h δ h + 1 1 − δ h + 1 t h + 1 − t h 1 − δ g ≤ h δ h δ g δ h + 1 t h + 1 − t h δ h + 2 t t + 2 − t h + 1 … δ g − 1 t g − 1 − t g − 2 − δ h + 1 t h + 1 − t h δ h + 2 t t + 2 − t h + 1 … δ g t g − t g − 1 ( g = 2 … , n − 1 ) 1 − δ n ≤ h δ h δ n δ h + 1 t h + 1 − t h δ h + 2 t t + 2 − t h + 1 … δ n − 1 t n − 1 − t n − 2

By summing Eq. (A18), we can get Σ i = h + 1 n ( 1 − δ i ) / δ i ≤ h δ h . As δ _h > 1/h, hδ _h > 1. When δ _h+1 = … = δ _n = 1, we have h δ h > 1 ⇔ h δ h + Σ i = h + 1 n ( 1 − δ i ) > 1 ⇔ h δ h + Σ i = h + 1 n δ i > n − h + 1 . As (1 − δ _i)/δ _i decreases as δ _i increases and (1 − δ _i)/δ _i = 1 − δ _i for δ _i = 1, we have h δ h + Σ i = h + 1 n δ i > n − h + 1 for all δ _h < δ _i < 1. Thus, by Claim 1, we have t _h > 2, which is a contradiction. Therefore, Claim 2 holds for i = h, and it holds for all i < n. □

Claim 3.

If h δ h + Σ i = h + 1 n δ i ≤ n + 1 − h for h > 1, then π _h,1 > 0.

Proof.

Suppose not, then we have π _h,1 = 0. And h δ h + Σ i = h + 1 n δ i ≤ n + 1 − h ⇒ ( h − 1 ) δ h + δ h + Σ i = h + 1 n δ i ≤ n + 1 − h ⇒ ( h − 1 ) δ h − 1 + Σ i = h n δ i < n + 1 − h , so we have t _h−1 = 1 by Lemma 2. Thus, we have π ^m/h < U _h−1,1 = π ^m/(h − 1) ≤ U _h,1. Thus, as in the Claim 2, we can also derive h δ h + Σ i = h + 1 n δ i > n − h + 1 , which is a contradiction. □

Proof of (1) in Proposition 2

The proof of t _i = 1 ⇒ t _i−1,1 = t _i = 1 is obvious. Then we prove t _i−1 = t _i ⇒ t _i = 1. Suppose not, then we have t _i−1 = t _i > 1. Then U i , 1 = δ i t i − 1 π i , t i < δ i t i − 1 π m , which contradicts Claim 2. □

Proof of (2) in Proposition 2

We first prove U i − 1,1 = U i , 1 ⇒ i δ i + Σ j = i + 1 n δ j / n ≤ ( n + 1 − i ) / n .

According to Lemma A.5, when U _i−1,1 = U _i,1, there must be t _i−1 = 1. So we must have U _1,1 = U _2,1 = … = U _i−1,1 ≤ π ^m/(i − 1). There must be two cases. In the first case, t _i = 1, so U _i−1,1 = U _i,1 must be true. Otherwise, transferring profits from firm i in the first period to firm i − 1 can directly increase U _i−1,1 U _i,1. And by Lemma 2, t i = 1 ⇒ i δ i + Σ j = i + 1 n δ j / n ≤ ( n − i ) / n < ( n + 1 − i ) / n .

In the second case, t _i > 1. For any firm g > i, by Lemma A.6, we have:

(A19) U i , 1 ≥ δ i t i δ i + 1 t i δ i + 1 t i + 1 δ i + 2 t i + 1 … δ g − 1 t g − 1 δ g t g − 1 U g , 1 = δ i t i δ i + 1 t i δ i + 1 t i + 1 δ i + 2 t i + 1 … δ g − 1 t g − 1 δ g t g − 1 δ g t g − 1 − δ g t g 1 − δ g π m < δ i 1 − δ g π m

Since U _i−1,1 ≤ π ^m/(i − 1), in order to ensure that U _i−1,1 = U _i,1, we should guarantee 1/(i − 1) ≥ δ _i/(1 − δ _g) for g = i + 1, i + 2, …n, which gives us 1 − δ _g ≥ (i − 1)δ _i. Summing this inequality from g = i + 1 to n gives ( i − 1 ) δ i + Σ j = i + 1 n δ j ≤ n − i ⇒ ( i − 1 ) δ i + 1 + Σ j = i + 1 n δ j ≤ n + 1 − i ⇒ i δ i + Σ j = i + 1 n δ j ≤ n + 1 − i . □

We then prove i δ i + Σ j = i + 1 n δ j / n ≤ ( n + 1 − i ) / n ⇒ U i − 1,1 = U i , 1 .

By Claim 3, i δ i + Σ j = i + 1 n δ j / n ≤ ( n + 1 − i ) / n ⇒ π i , 1 > 0 ⇒ U i − 1,1 = U i , 1 . □

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Received: 2023-10-23

Accepted: 2024-03-17

Published Online: 2024-04-08

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Articles in the same Issue

https://doi.org/10.1515/bejte-2023-0112

Keywords for this article

collusive price leadership; different discount factors; dynamic collusion path