Post on 30-Dec-2015
description
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?