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Unformatted text preview: ge pi = 1/2 until somebody deviates; they both charge 0 there
after. (You need to ﬁnd the range of δ .)
ANSWER: If nobody has deviated before:
Payoﬀ to not deviate: 1/8, 1/8, ...⇒ 1/8(1 − δ )
Payoﬀ to deviate: Notice that the only proﬁtable deviation occurs by undercutting the price
and the most proﬁtable undercutting is just to charge inﬁnitesmall less than your competition.
You will get something very close to 1/4 by doing this so 1/4,0,0,0,...⇒ 1/4
3 So we need 1/8(1 − δ ) > 1/4 ⇒ 1 > 1 − δ ⇒ δ > 1 2
We don’t get any meaningful restrictions from the histories with previous deviations. b) (15 points) There are n + 1 modes: Collusion, the ﬁrst day of war (W1 ), the second day
of war (W2 ), ..., and the nth day of war (Wn ). The game starts in the Collusion mode. They
both charge pi = 1/2 in the Collusion mode and pi = p∗ in the war modes (W1 ,. . . , Wn ), where
p∗ < 1/2. If both players charge what they are supposed to charge, then the Collusion mode leads
to the Collusion...
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This document was uploaded on 03/21/2014 for the course ECON 14.12 at MIT.
- Fall '04
- Game Theory