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MBA 299 – Section Notes

MBA 299 – Section Notes. 4/11/03 Haas School of Business, UC Berkeley Rawley. AGENDA. Administrative Exercises 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)

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MBA 299 – Section Notes

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

  2. AGENDA • Administrative • Exercises • 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) • Cournot duopoly • 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.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 s1, s2 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 si that has not yet been eliminated from the strategy set that strictly dominates s1 • 5. Therefore U(s1,s2) < U(si,s2) • 6. A contradiction of the definition of NE since s1 must be a best response to s2 (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 pn>$5 • dn = 50 if pn=$5,n=1,2 • 50 if pn<$5 • profit = 50*(pn-1) • max profit by choosing pn=$5 (no game) • Part b.) K1=K2=100 • 0 if pn> pmin • dn = 50 if pn= pmin • 100 if pn< pmin • profit = X*(pn-1) • max profit by choosing pn=C=$1 . . .

  6. BERTRAND TRAP PROBLEM 6Part 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 P2=C (and earns zero), player 1 can charge C<P1<=$5 and earn 50*(P1-C) • But if player 1 charges P1>C then player 2 will want to increase his price to P2 = P1 –e earning 50*(P2-C) . . . • But now, if P2>C, player 1 will want to charge P1=P2 –e earning 100*(P1-C) • And so and on . . .

  7. COURNOT DUOPOLYMath • Solution • Profit i (q1,q2) = qi[P(qi+qj)-c] • =qi[a-(qi+qj)-c] • Recall NE => max profit for i given j’s best play • So F.O.C. for qi, assuming qj<a-c • qi*=1/2(a-qj*-c) • Solving the pair of equations • q1=q2=(a-c)/3 • Note that qj < a – c as we assumed • Set-up • P(Q) = a – Q (inverse demand) • Q = q1 + q2 • Ci(qi) = cqi (no fixed costs) • Assume c < a • Firms choose their q simultaneously

  8. COURNOT DUOPOLYIntuition • Observe that the monopoly outcome is • qm=(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 qi=qj=qm/2) • But each firm has a strong incentive to deviate at this qm • Check: qm/2 is not firm 2’s best response to qm/2 by firm 1

  9. BACKWARDS INDUCTION (I)Monk’s Cerecloth 1 • What are the BI outcome when a=4? • R = (6,8) L R 2 l r • What is the BI outcome when a = 8? • R = (6,8) a -6 3 2 6 8

  10. BACKWARDS INDUCTION (II)Shoved Environment 1 • What are the BI strategies for each player? • {L1, L2} • {r1, l2, r3} L1 R1 2 1 r1 2 1 2 2 L2 R2 2 l2 r2 3 1 2 1 3 3 1 l3 r3 4 6 2 7 • What is the BI outcome? • L1, r1 = (2,2)

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

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