# 3 a little arithmetic with 614 shows that wn1 wn 1

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Unformatted text preview: he risk neutral probability p⇤ so that 1 (p⇤ S0 u + (1 p⇤ )S0 d) = S0 (6.4) 1+r Solving we have p⇤ = 1+r d ud 1 p⇤ = u (1 + r) ud (6.5) The conditions in (6.1) imply 0 < p⇤ < 1. Taking p⇤ (6.2) + (1 p⇤ )(6.3) and using (6.4) we have V0 = 1 (p⇤ V1 (H ) + (1 1+r p⇤ )V1 (T )) (6.6) i.e., the value is the discounted expected value under the risk neutral probabilities. Taking the di↵erence (6.2) (6.3) we have ✓ ◆ 1 1 (S1 (H ) S1 (T )) = (V1 (H ) V1 (T )) 0 1+r 1+r 184 CHAPTER 6. MATHEMATICAL FINANCE which implies that 0 = V1 (H ) S1 (H ) V1 (T ) S1 (T ) (6.7) To explain the notion of hedging we consider a concrete example. Example 6.1. A stock is selling at \$60 today. A month from now it will either be at \$80 or \$50, i.e., u = 4/3 and d = 5/6. We assume an interest rate of r = 1/18 so the risk neutral probability is 4 19/18 5/6 = 4/3 5/6 9 Consider now a call option (S1 V0 = 65)+ . By 6.6) the value is 18 4 120 · · 15 = = 6.3158 19 9 19 Being a savy businessman you o↵...
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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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