MBA 299 Section Notes

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

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ADMINISTRATIVE

• In response to your feedback
• Slides in section
• More math
• CSG entries due Tuesday and Friday at midnight
  each week
• Contact info
• rawley_at_haas.berkeley.edu
• Office hours Room F535
• Monday 1-2pm
• Friday 2-3pm

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PROOF THAT ALL IDSDS ARE NE (PROBLEM 5D)

• Proof by contraction
• 1. Assume not gt 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) lt 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

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BERTRAND TRAP PROBLEM 6 (I)Parts a and b

• Part a.) K1K250
• 0 if pngt5
• dn 50 if pn5, n1,2
• 50 if pnlt5
• profit 50(pn-1)
• max profit by choosing pn5 (no game)

Part b.) K1K2100 0 if pngt pmin dn 50 if
pn pmin 100 if pnlt pmin profit
X(pn-1) max profit by choosing pnC1 . . .
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BERTRAND TRAP PROBLEM 6Part b continued and Part
c

• Part b.
• Why does PC in party b, where K1K2100?
• Because 50(P-delta)50(P-delta) gt 50P if delta
  is small
• Therefore defecting is always the rule until
  PC
• Part c.
• K1100, K250 gt there is no pure strategy NE
• Why?
• If player 2 charges P2C (and earns zero), player
  1 can charge CltP1lt5 and earn 50(P1-C)
• But if player 1 charges P1gtC then player 2 will
  want to increase his price to P2 P1 e earning
  50(P2-C) . . .
• But now, if P2gtC, player 1 will want to charge
  P1P2 e earning 100(P1-C)
• And so and on . . .

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COURNOT DUOPOLYMath

• Set-up
• P(Q) a Q (inverse demand)
• Q q1 q2
• Ci(qi) cqi (no fixed costs)
• Assume c lt a
• Firms choose their q simultaneously

Solution Profit i (q1,q2) qiP(qiqj)-c qi
a-(qiqj)-c Recall NE gt max profit for i
given js best play So F.O.C. for qi, assuming
qjlta-c qi1/2(a-qj-c) Solving the pair of
equations q1q2(a-c)/3 Note that qj lt a c as
we assumed
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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 qiqjqm/2)
• But each firm has a strong incentive to deviate
  at this qm
• Check qm/2 is not firm 2s best response to
  qm/2 by firm 1

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BACKWARDS INDUCTION (I)Monks Cerecloth
1

• What are the BI outcome when a4?
• R (6,8)

L
R
2
l
r

• What is the BI outcome when a 8?
• R (6,8)

3 2
a -6
6 8
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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
2
L2
R2
l2
r2
3 1
2
1 3
3 1
l3
r3
4 6
2 7

• What is the BI outcome?
• L1, r1 (2,2)

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A MAJOR MEDIA COMPANYS ACQUISITION OF A P2P FILE
SHARING COMPANYA Simplified Model of How the
Acquisition Was Analyzed
6,6,6,6,6
Join
1B 2U 3T 4S 5E
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
Dont buy
What do you think happened? What are the limits
of BI?
0,0,0,0,0