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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online