1. MBA 299 – Section Notes 4/11/03 Haas School of Business, UC Berkeley Rawley

2. AGENDA Administrative Exercises 1.Finish off Exercises from Introduction to Game Theory & The Bertrand Trap Problem 2 (see last weeks section notes) Problem 5 d. Problem 6 Problem 7 (done on the board) 2.Cournot duopoly 3.Backwards induction problems

3. ADMINISTRATIVE In response to your feedback –Slides in section –More math CSG entries due Tuesday and Friday at midnight each week Contact info: –rawley@haas.berkeley.edurawley@haas.berkeley.edu Office hours Room F535 –Monday 1-2pm –Friday 2-3pm

4. PROOF THAT ALL IDSDS ARE NE (PROBLEM #5D) Proof by contraction: 1. Assume not => a NE strategy is eliminated by IDSDS 2. Suppose in a two player game strategies s 1, s 2 are a NE 3. WOLOG Let s1 be the first of the strategies to be eliminated by IDSDS 4. Then there must exist a strategy s i that has not yet been eliminated from the strategy set that strictly dominates s 1 5. Therefore U(s 1,s 2 ) < U(s i,s 2 ) 6. A contradiction of the definition of NE since s 1 must be a best response to s 2 (Q.E.D.) Source: Robert Gibbons, “Game Theory for Applied Economists” (1992) p. 13

5. BERTRAND TRAP PROBLEM 6 (I) Parts a and b Part a.) K1=K2=50 0 if p n >$5 d n =50 if p n =$5, n=1,2 50 if p n <$5 profit = 50*(p n -1) max profit by choosing p n =$5 (no game) Part b.) K1=K2=100 0 if p n > p min d n =50 if p n = p min 100 if p n < p min profit = X*(p n -1) max profit by choosing p n =C=$1...

6. BERTRAND TRAP PROBLEM 6 Part b continued and Part c Part b. Why does P=C in party b, where K1=K2=100? Because 50*(P-delta)+50*(P-delta) > 50*P if delta is small Therefore “defecting” is always the rule until P=C Part c. K1=100, K2=50 => there is no pure strategy NE Why? If player 2 charges P 2 =C (and earns zero), player 1 can charge C<P 1 <=$5 and earn 50*(P 1 -C) But if player 1 charges P 1 >C then player 2 will want to increase his price to P 2 = P 1 –e earning 50*(P 2 -C)... But now, if P 2 >C, player 1 will want to charge P 1 =P 2 –e earning 100*(P 1 -C) And so and on...

7. COURNOT DUOPOLY Math Set-up P(Q) = a – Q (inverse demand) Q = q 1 + q 2 C i (q i ) = cq i (no fixed costs) Assume c < a Firms choose their q simultaneously Solution Profit i (q 1,q 2 ) = q i [P(q i +q j )-c] =q i [a-(q i +q j )-c] Recall NE => max profit for i given j’s best play So F.O.C. for q i, assuming q j <a-c q i * =1/2(a-q j * -c) Solving the pair of equations q 1 =q 2 =(a-c)/3 Note that q j < a – c as we assumed

8. COURNOT DUOPOLY Intuition Observe that the monopoly outcome is q m =(a-c)/2 profit m = (a-c) 2 /4 The optimal outcome for the two firms would be to divide the market at the monopoly output level (for example q i =q j =q m /2) But each firm has a strong incentive to deviate at this q m –Check: q m /2 is not firm 2’s best response to q m /2 by firm 1

9. BACKWARDS INDUCTION (I) Monk’s Cerecloth 3232 LR 1 2 What are the BI outcome when a=4? R = (6,8) What is the BI outcome when a = 8? R = (6,8) a -6 6868 lr

10. BACKWARDS INDUCTION (II) Shoved Environment 3131 4646 2727 2222 1313 3131 L1R1 1 2 2 2 1 What are the BI strategies for each player? {L1, L2} {r1, l2, r3} What is the BI outcome? L1, r1 = (2,2) L2R2 l2r2 r1 l3r3 1 2

11. A MAJOR MEDIA COMPANY’S ACQUISITION OF A P2P FILE SHARING COMPANY A Simplified Model of How the Acquisition Was Analyzed Buy Join Abstain Don’t buy 0,0,0,0,0 1 2 3 4 5 -10,0,0,0,0 0,0,0,0,0 2,2,2,0,0 4,4,4,4,0 6,6,6,6,6 1=B 2=U 3=T 4=S 5=E What do you think happened? What are the limits of BI?