22 with sn1 ah 1 1r p a n sn1 ah 1r p

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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).

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