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Unformatted text preview: 9 y2
9 y3
The second and third equations imply that 2y1 = y0 and 2y3 = y2 . The ﬁrst
equation implies
2
2
1
4
+
=
+
y0
y1
y2
y3
or 4/y1 = 2/y3 , i.e., 2y3 = y1 . Thus if y3 = c, y1 = y2 = 2c, and y3 = 4c.
Adding the equations for the W2 we have
9c = y0 + y1 + y2 + y3 = 25
W0
4 and it follows that
y3 = 4 25
· W0
94 y2 = y1 = 2 25
· W0
94 y0 = 1 25
· W0
94 y2 y3 = 15 (6.19) Once we have the yi ’s we have
y0 y1 = 12 1 y2 y3 = 3 2 y0 + y1 0 194 CHAPTER 6. MATHEMATICAL FINANCE Turing to the two=period case for a general binomial model, the previous
calculation suggests that what we really want to solve for are the terminal
values yi . We cannot achieve any yi . Theorem 6.5 implies that our discounted
wealth Wn /(1 + r)n is a martingale under P ⇤ , so letting p⇤ be the risk neutral
m
probabilities of outcomes m = 0, 1, 2, 3, we must have
3
X
1
p⇤ y m = W 0
(1 + r)2 m=0 m Conversely, given a vector that satisﬁes the last condition, the option with these
payo↵s has value W0 and there is a trading strategy that allows us to replicate
the option. Letting pm be the physical probabilities of out...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
 Spring '10
 DURRETT
 The Land

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