QUANTITY PRECOMMITMENT AND BERTRAND COMPETITION YIELD COURNOT OUTCOMES: A CASE WITH PRODUCT DIFFERENTIATION: REPLY

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1 QUANTITY PRECOMMITMENT AND BERTRAND COMPETITION YIELD COURNOT OUTCOMES: A CASE WITH PRODUCT DIFFERENTIATION: REPLY XIANGKANG YIN La Trobe University YEW-KWANG NG Monash University Since the publication of our paper (Yin and Ng, 1997), we have received some very helpful comments on the paper. In his recent comment, Schulz (2000) indicates that price reaction functions in the second-stage subgame of our model are discontinuous in certain circumstances so that Lemma 1 in the paper is incorrect. The editor, R. Damania, points out a mistake in the proof of Lemma 1 in his letter to us. We sincerely appreciate these criticisms, which are greatly helpful for us to re ne our work. After a careful check, we nd that not only Lemma 1 is invalid but the price reaction function given in Theorem 1 also has to be revised. Fortunately, these errors do not affect the main results of the paper. The erratum to the previous mistakes is given below. Figure 1 below is borrowed from Schulz (2000), which can be obtained by applying demand functions (3)±(7) in our original paper. In the gure, line C i (i ˆ 1, 2) plots the price pairs on which the demand for the rm i's product just hits its capacity while the demand for the rm j's product is not binding by either the upper or lower boundary; i.e. C i : p 1 i ( p j) ˆ (1 ã)á ã p j (1 ã 2 )K i : (1) Z i in the gure (i ˆ 1, 2) is the line on which the demand for the rm i's product just becomes zero while the demand for the rm j's product hits neither the upper nor lower boundary; i.e. Z i : p i ˆ (1 ã)á ã p j : Or inversed, p 2 j ( p i) ˆ [ p i (1 ã)á]=ã: (2) Axy gives the area, where the demand for rms 1 and 2's products has the characteristics of the rst and second subscripts, respectively. x and y in the subscript can be c, u, and z, indicating that the demand is constrained by the production capacity, unconstrained demand and zero demand, respectively. The equations below Axy show the demand functions in that area and the demand in Auu is q i ˆ á=(1 ã) ( p i ã p j )=(1 ã 2 ). It should be noticed that Figure 1 presents a most comprehensive situation with nine areas, which requires small production capacities of both rms. When capacities are large, C 1 and C 2 move to left and down, respectively, so that some areas may disappear. An extreme is no capacity constraints and Figure 1 has only four areas: Auu, Auz, Azu, and Azz. With the help of Figure 1, we can revise the price reaction function as follows. # Blackwell Publishers Ltd, 108 Cowley Road, Oxford OX4 1JF, UK and 350 Main Street, Malden MA 02148, USA and the University of Adelaide and Flinders University of South Australia 2000.

2 114 AUSTRALIAN ECONOMIC PAPERS MARCH Figure 1. Theorem 19 (I) If 2K i ãk j < á, the price reaction function is 1 8 á K i ãk j p j 2 [0, á K j ãk i ] >< R i ( p j ) ˆ p 1 i ( p j) p j 2 (á K j ãk i, á ãk i ) >: minf p 2 i ( p j), á=2g p j 2 [á ãk i, á] (II) If 2K i ãk j. á, the price reaction function is 8 (á ãk j =2 p j 2 [0, á K j ãk i ] (á ãk j )=2 p j 2 (á K j ãk i, á ãk i )\ (á K j ãk i, minf~p 1, ^p 1 g) >< R i ( p j ) ˆ maxf p 1 i ( p j), p 3 i ( p j)g p j 2 (á K j ãk i, á ãk i )\ (minf~p 1, ^p 1 g, á ãk i,) (á ãk j )=2 p j 2 [á ãk i, á] \ (á ãk i, minf p 1, ^p 1 g) >: minfmax[ p 2 i ( p j), p 3 i ( p j)], á=2g p j 2 [á ãk i, á] \ (minf p 1, ^p 1 g, á] 1 An interval (x, y) below should be understood as an empty set if x > y or both x and y are negative. If x, 0 but y. 0, the interval (x, y) should be understood as [0, y).

3 2000 REPLY: QUANTITY PRECOMMITMENT AND BERTRAND COMPETITION 115 Proof. (I) Consider 2K 2 ãk 1 < á. (i) p 1 2 [0, á K 1 ãk 2 ] For a given p 1, rm 2 chooses the optimal price p 2 ˆ á K 2 ãk 1 if q 2 ˆ K 2 or p 2 ˆ (á ãk 1 )=2 ifq 2 ˆ á p 2 ãk 1. The latter is the best price response iff it is in A cu. Since 2K 2 ãk 1 < á, á K 2 ãk 1 > (á ãk 1 )=2, the price reaction function is á K 2 ãk 1. (ii) p 1 2 (á K 1 ãk 2, á ãk 2 ) There are two candidates of optimal price. One is to choose maximum price when the capacity is binding, which is the line C 2, i.e., p 1 2 ( p 1). The other is p 3 2 ( p 1) ˆ [(1 ã)á ã p 1 ]=2, (3) which maximises pro ts for demand q 2 ˆ á=(1 ã) ( p 2 ã p 1 )=(1 ã 2 ). Recalling (1), p 1. á K 1 ãk 2 and 2K 2 ãk 1 < á,wehave p 1 2 ( p 1) ˆ [(1 ã)á ã p 1 ]=2 [(1 ã)á ã p 1 2(1 ã 2 )K 2 ]=2. p 3 2 ( p 1) [á ãk 1 2K 2 ã 2 K 2 ]=2. p 3 2 ( p 1): Since p 3 2 ( p 1) is the best price response iff it falls in A uu, the above inequality implies that the price reaction function in this interval must be p 1 2 ( p 1). (iii) p 1 2 [á ãk 2, á] (I)±(ii) shows that p 3 2 ( p 1), p 1 2 ( p 1)when p 1 2 (á K 1 ãk 2, á ãk 2 ). Thus, p 3 2 ( p 1), á K 2. Since p 3 2 ( p 1) is atter than Z 1,itisinA zc or A zu on this interval and cannot be the best price response. Noting that p 2 ˆ á=2 maximises pro t for demand q 2 ˆ á p 2, the best price response should be Z 1 if p 2 ˆ á=2 is above Z 1 or p 2 ˆ á=2 if it is below Z 1 ; i.e. minf p 2 2 ( p 1), á=2g. (II) Assuming 2K 2 ãk 1. á (i) p 1 2 [0, á K 1 ãk 2 ] As (I)±(i) shows the price reaction function on this interval is (á ãk 1 )=2. (ii) p 1 2 (á K 1 ãk 2, á ãk 2 ) In this interval, rm 2 has three possible price choices, i.e., p 2 ˆ (á ãk 1 )=2 or p 1 2 ( p 1)or p 3 2 ( p 1). Their corresponding pro ts are, respectively (á ãk 1 ) 2 =4, K 2 p 1 2 ( p 1) ˆ K 2 [(1 ã)á ã p 1 (1 ã 2 )K 2 ], fá=(1 ã) [ p 3 2 ( p 1) ã p 1 ]=(1 ã 2 )g p 3 2 ( p 1) ˆ [(1 ã)á ã p 1 ] 2 =4(1 ã 2 ) Let ~p 1 be the point where the rst pro t gure is equal to the second, we have ~p 1 ˆ [(á ãk 1 ) 2 =4K 2 (1 ã 2 )K 2 (1 ã)á]=ã: (4) Thus, p 2 ˆ (á ãk 1 )=2 leads to a larger pro t than p 1 2 ( p 1)iff p 1, ~p 1. Let ^p 1 be the point where the rst pro t gure is equal to the third, we have

4 116 AUSTRALIAN ECONOMIC PAPERS MARCH ^p 1 ˆ [(1 ã 2 ) 1=2 (á ãk 1 ) (1 ã)á]=ã, (5) which implies that p 2 ˆ (á ãk 1 )=2 brings more pro t than p 3 2 ( p 1)iff p 1, ^p 1. Let p 1 be the intersection of p 1 2 ( p 1) and p 3 2 ( p 1); i.e., p 1 ˆ [2(1 ã2 )K 1 (1 ã)á]=ã. Since p 1 2 ( p 1) is steeper than p 3 2 ( p 1), we can see that p 3 2 ( p 1). p 1 2 ( p 1) on the left of p 1 and p 1 2 ( p 1) > p 3 2 ( p 1) on the right. Moreover, for any p 1,if p 3 2 ( p 1) falls in A uu, it results in more pro t than p 1 2 ( p 1). It can also be shown that ~p 1. á K 1 ãk 2 and p 3 2 (^p 1) is below the line C 1 (or its extension). Thus, the price reaction function shifts from p 2 ˆ (á ãk 1 )=2 to p 3 2 ( p 1)at p 1 ˆ ^p 1 if ^p 1, ~p 1 and ^p 1 2 (á K 1 ãk 2, p 2 ),2 and then switches to p 1 2 ( p 1)at p 1 ˆ p 1 if p 1, á ãk 2. But if ~p 1 < ^p 1 or ~p 1 < á ãk 2 and ^p 1 =2 (á K 1 ãk 2, p 2 ), the price reaction function shifts from p 2 ˆ (á ãk 1 )=2 to p 1 2 ( p 1)at p 1 ˆ ~p 1 and remains on it until the end of the interval. If ~p 1. á ãk 2 and ^p 1. á ãk 2 or ~p 1. á ãk 2 and ^p 1 =2 (á K 1 ãk 2, p 2 ), the reaction function remains on p 2 ˆ (á ãk 1 )=2 for the whole interval. De ne ( ^p 1 ˆ ^p 1 ^p 1 2 (á K 1 ãk 2, p 2 ) : (6) 1 otherwise The price reaction function in this interval is (á ãk 1 )=2 when p 1 < minf~p 1, ^p 1 g maxf p 2 2 ( p 1), p 3 2 ( p 1)g when p 1 > minf~p 1, ^p 1 g : (iii) p 1 2 [á ãk 2, á] Let p 1 be determined by p 2 2 ( p 1) ˆ á=2 so that p 1 ˆ (1 ã=2)á. As (I)±(iii) shows, the price reaction function cannot be p 2 ˆ á=2 on the left of p 1. Hence, on the left of p 1 there are three possible price choices, i.e., p 2 ˆ (á ãk 1 )=2 or p 2 2 ( p 1) (i.e., line Z 1 )or p 3 2 ( p 1). The pro t of adopting p 2 2 ( p 1) is equal to fá [ p 1 (1 ã)á]=ãg[ p 1 (1 ã)á]=ã ˆ (á p 1 )[ p 1 (1 ã)á]=ã 2. Let p 1 be the point where this pro t is equal to (á ãk 1 ) 2 =4, i.e., p 1 is the smaller solution of the equation p 2 1 (2 ã)á p 1 (1 ã)á 2 ã 2 (á ãk 1 ]=4 ˆ 0: (7) It can be shown that the intersection of p 2 2 ( p 1) and p 3 2 ( p 1) is on the left of p 1 and p 2 2 ( p 1)is steeper than p 3 2 ( p 1). Then, if the price reaction function is p 2 ˆ (á ãk 1 )=2 at p 1 ˆ á ãk 2, it shifts to p 3 2 ( p 1) at the p 1 ˆ ^p 1 when ^p 1, p 1. Afterwards, it switches to p 2 2 ( p 1) at the intersection of p 2 2 ( p 1) and p 3 2 ( p 1) to reach p 2 ˆ á=2. When p 1, ^p 1 it shifts to p 2 2 ( p 1)at p 1 ˆ p 1 and moves toward á=2. If the price reaction function at p 1 ˆ á ãk 2 is p 2 2 ( p 1)or p 3 2 ( p 1), the development is similar, except that there is no shift. Thus, the price reaction function in this interval is (á ãk 1 )=2 when p 1 < minf p 1, ^p 1 g minfmax[ p 2 2 ( p 1), p 3 2 ( p 1)], á=2g when p 1 > minf p 1, ^p 1 g: j With this new Theorem 19, Lemma 1 in our original paper should be changed to 2 The second condition is to ensure that p 3 2 (^p 1) falls in A uu.

5 2000 REPLY: QUANTITY PRECOMMITMENT AND BERTRAND COMPETITION 117 Lemma 19. Firm i's price reaction function has one and only one discontinuity point at p j ˆ p j or ~p j or ^p j when 2K i ãk j. á and ãk j, á. Otherwise, it is conninuous. Proof. Since maxf:, :g and minf:, :g are continuous operators, the discontinuity occurs at the point where the price reaction function shifts from p i ˆ (á ãk j )=2 to p 1 i ( p j)or p 2 i ( p j) or p 3 i ( p j). But this can happens only once when 2K i ãk j. á and ãk j, á as Theorem 19 illustrated. j The bold lines in Figure 2 illustrates a typical price reaction curve of rm 2 when the discontinuity conditions in Lemma 19 are satis ed. Thus, there is a shift at p 1 ˆ ^p 1. If the line p 3 2 ( p 1) is lower, the shift may occur at p 1 ˆ ~p 1. p 1. A rare situation is that K 1 is very small but K 2 is suf ciently large so that that the line C 1 is long but C 2 is short. In turn, the shift occurs on the right of p 1 ˆ á ãk 2. Because of the discontinuity of the price reaction curves, the price subgame in the second stage does not guarantee a unique pure strategy equilibrium as indicated in Theorem 2 in the original paper. However, this theorem is not a necessity for Theorem 3. Hence, Theorem 3 holds for the discontinuous price reaction functions but the proof should be revised. Theorem 3. In the equilibrium of the entire game, there are no idle capacities. Proof. Suppose ( p1 e, pe 2, K 1, K 2 ) is the equilibrium of the entire game and at least one rm, say, rm 2, has certain idle production capacity. If the equilibrium leads to a pure price Figure 2.

6 118 AUSTRALIAN ECONOMIC PAPERS MARCH strategy of rm 2, then ( p1 e, pe 2 ) must be on the line p 2 ˆ (á ãk 1 )=2 or p 2 2 ( p 1)or p 3 2 ( p 1) or p 2 ˆ á=2. Hence, p2 e is independent of rm 2's capacity and it has an incentive to reduce its idle capacity, which implies that ( p1 e, pe 2, K 1, K 2 ) cannot be the equilibrium of the entire game. Consider now that rm 2 adopts a mixed price strategy in equilibrium. Since the mixed price strategy can only occur at the discontinuity point of the price reaction function, it implies that p1 e ˆ ^p 1 or p 1 or ~p 1. For the rst two cases, ( p1 e, pe 2 ) is the intersection of the line C 1 and p 1 ˆ ^p 1 or C 1 and p 1 ˆ p 1. Because C 1, ^p 1 and p 1 are independent of K 2 (see (1), (5)±(7)), the price p2 e is independent of rm 2's capacity. Turning to the case where p1 e ˆ ~p 1. Figure 3 illustrates the most complicated situation in the sense that both rms have discontinuous price reaction functions, resulting in multiequilibria at E 1 and E 2 in the price subgame. 3 Let us focus on E 1. The shift of price reaction curve at ~p 1 from p 2 ˆ (á ãk 1 )=2 to p 1 2 ( p 1) (i.e., line C 2 ) implies that p 1 2 (~p 1). p 3 2 (~p 1). 4 Recalling (1), (3) and (4), this inequality gives (1 ã)á ã~p 1. 2(1 ã 2 )K 2 and (á ãk 1 ) 2 =4K 2. (1 ã 2 )K 2 : (8) Since E 1 is on C 1, we obtain from the reverse of p 1 1 ( p 2) that p e 2 ˆ [~p 1 (1 ã 2 )K 1 (1 ã)á]=ã: (9) The demand for good 2 along line C 1 is q e 2 ˆ á pe 2 ãk 1. Thus, the rm 2's equilibrium pro t in the second-stage subgame is ð ˆ p e 2 qe 2 ˆ pe 2 (á pe 2 ãk 1), which 2 ˆ (á 2 p e 2 ãk 1)@ p e 2 =@K 2 ˆ ã 2 (á 2 p e 2 ãk 1)[(1 ã 2 ) (á ãk 1 ) 2 =4K 2 2 ], 0: Figure 3. 3 In equilibrium E 2, the rm 2's capacity is binding. But actually, we can show that it is not the equilibrium of the entire game following the same arguments in the proof of E 1. If rm 1's price reaction function is continuous, then there exists only one equilibrium, E 1. The intersection of ~p 1 and ~p 2 cannot be an equilibrium point because it falls in the area A uu. 4 Otherwise, it shifts to p 3 2 ( p 1) before ~p 1.

7 2000 REPLY: QUANTITY PRECOMMITMENT AND BERTRAND COMPETITION 119 To get the second equality, (4) and (9) are used. From (8) and q2 e. á 2 pe 2 ãk 1, the inequality is immediate, which implies that rm 2 has an incentive to reduce the capacity and ( p1 e, pe 2, K 1, K 2 ) is not an equilibrium. j When the capacity constraints of both rms are binding, the price reaction function is R i ( p j ) ˆ á K i ãk j, which leads to the price subgame equilibrium (11) in our original paper. Therefore, the main conclusions, Theorem 4 and Proposition 1 there, hold. REFERENCES Schulz, N. 2000, `A Comment on Yin, Xiangkang and Yew-Kwang Ng: Quantity Precommitment and Bertrand Competition Yield Cournot Outcomes: A Case with Product Differentiation' Australian Economic Papers, vol. 39, pp. 108±112. Yin, X. and Ng, Y.-K. 1997, `Quantity Precommitment and Bertrand Competition Yield Cournot Outcomes: A Case with Product Differentiation' Australian Economic Papers, vol. 36, pp. 14± 22.