MBA 299 – Section Notes

4/11/03

Haas School of Business, UC Berkeley

Rawley

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

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

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

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 . . .

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 smallTherefore “defecting” is always the rule until P=C

Part c.K1=100, K2=50 => there is no pure strategy NEWhy?

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 . . .

COURNOT DUOPOLYMath

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

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

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

BACKWARDS INDUCTION (I)Monk’s Cerecloth

32

L R

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

68

l r

BACKWARDS INDUCTION (II)Shoved Environment

31

46

27

22

13

31

L1 R1

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)

L2 R2 l2 r2

r1

l3 r3

12

A MAJOR MEDIA COMPANY’S ACQUISITION OF A P2P FILE SHARING COMPANY

A Simplified Model of How the Acquisition Was Analyzed

Buy

Join

Join

Join

Join

Abstain

Abstain

Abstain

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

6,6,6,6,61=B2=U3=T4=S5=E

What do you think happened?What are the limits of BI?