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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).
 Spring '10
 DURRETT
 The Land

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