1. Chapter 28 Oligopoly It is the case that lies between two extremes (pure competition and pure monopoly). Often there are a number of competitors but not so many as to regard each of them as having a negligible effect on price. Unreasonable to expect one grand model for many behaviors observed in real life, hence we will introduce some models.

2. We will restrict to the case of two firms or duopoly. The first model is the Stackelberg model, for instance, IBM and its follower. It is a quantity setting game and 1 is the leader who chooses to produce y 1 while after seeing that, the follower, 2 decides to produce y 2. The total output of the market is therefore y 1 +y 2 and the price is p(y 1 +y 2 ). (sequential)

3. Solving backwards, follower’s problem max y 2 p(y 1 +y 2 )y 2 -c 2 (y 2 ). FOC: p(y 1 +y 2 )+p’(y 1 +y 2 )y 2 =c 2 ’(y 2 ). So basically, from FOC, we can derive the reaction function of 2. That is, given y 1, there is an optimal level of y 2. In the following we will work with the case where p(Y)=a-bY and c 2 (y 2 )=0 for all y 2.  2 =(a-b(y 1 +y 2 ))y 2 =ay 2 -by 1 y 2 -by 2 2 and FOC: a-by 1 -2by 2 =0 or y 2 =(a-by 1 )/2b (reaction function).

4. Two points worth mentioning. When y 1 =a/b (1 has flooded the market), then y 2 =0. On the other hand, when y 1 =0, it is as if 2 is the monopolist, so y 2 =a/2b. Let us suppose c 1 (y 1 )=0 for all y 1 and work out the leader’s problem: max y 1 p(y 1 +y 2 )y 1 -c 1 (y 1 ). Now, we can plug in 2’s reaction curve. So  1 =(a-b(y 1 +(a- by 1 )/2b))y 1 =(a-by 1 )y 1 /2. FOC: a/2-by 1 =0 So y 1 =a/2b.

5. Fig. 27.1

6. Plugging this into 2’s reaction curve, we get y 2 =(a-b(a/2b))/2b=a/4b. Show this graphically. Now turn to the Cournot model. Two firms simultaneously decide output levels. We look for the case where 1’s output is a best response to 2’s and vice versa 2’s is a best response to 1’s. Graphically, it is the intersection of the two reaction curves.

7. Since y 2 =(a-by 1 )/2b and symmetrically y 1 =(a-by 2 )/2b, solving these two together, we get y 1 =y 2 =a/3b. Suppose the two quantity setting firms get together and attempt to set outputs so that their joint profit is maximized, i.e., they collude or form a cartel, what will the output levels be? Their problem becomes max y 1,y 2 (a- b(y 1 +y 2 ))(y 1 +y 2 ). FOC: y 1 +y 2 =a/2b.

8. Fig. 27.2

9. Fig. 27.4

10. Together they produce the monopoly output. But there is the temptation to cheat. Look at FOC again. Profit is (a-bY)Y (Y=y 1 +y 2 ), or PY so FOC P+(dP/dY)Y=0 or P+(dP/dY)(y 1 +y 2 )=0. From 1’s perspective, increasing output (if 2’s output is fixed), its profit increases by P+(dP/dY)y 1 =-(dP/dY)y 2 >0. Hence it is a big issue for the cartel to deter members from cheating.

11. Fig. 27.5

12. Often some repeated interactions would help. Consider the punishment strategies. Each firm produces half of the monopoly output and gets profit  m. If there is any cheating in the past, switch to Cournot competition forever and each gets  c. So if a firm deviates, it can at best get  d for one shot and then forever it gets  c. Note that  c  d +  c /r. As long as r<(  m -  c )/(  d -  m ) or future important enough, it will not deviate.

13. Lastly, we talk a bit about Bertrand competition. Two firms simultaneously announce prices p i and p j. There is a market demand x(p). If p i p j, then x i (p i,p j )=0. Suppose both firms have marginal cost c. There is a unique pure Nash equilibrium.

14. Suppose p i ≤p j and p i <c, then i can deviate to p i  c. Suppose p i =c and p j >c, i can increase the price a bit to earn positive profit. Suppose c<p i ≤p j, j earns at most (p i - c)x(p i )/2, undercutting i a bit, it could earn close to (p i -c)x(p i ). We are left with the case where p i =p j =c.