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Unformatted text preview: ⇤
= p⇤ (a)Vn (aH ) + qn Vn (aT ) + qn (a)[Vn (aH )
n Vn+1 (aT )] = Vn+1 (aH )
The proof that Wn+1 (aT ) = Vn+1 (aT ) is almost identical.
Our next goal is to prove that the value of the option is its expected value
under the risk neutral probability discounted by the interest rate (Theorem
6.5). The ﬁrst step is:
Theorem 6.4. In the binomial model, under the risk neutral probability measure Mn = Sn /(1 + r)n is a martingale with respect to Sn .
Proof. Let p⇤ and 1
tails of length n p⇤ be deﬁned by (6.5). Given a string a of heads and
P ⇤ (a) = (p⇤ )H (a) (1 p⇤ )T (a) where H (a) and T (a) are the number of heads and tails in a. To check the
martingale property we need to show that
Sn = sn , . . . S0 = s0 =
(1 + r)
(1 + r)n
where E ⇤ indicates expected value with respect to P ⇤ . Letting Xn+1 = Sn+1 /Sn
which is independent of Sn and is u with probability p⇤ and d with probability
1 p⇤ we have
Sn = sn , . . . S0 = s0
(1 + r)n+1
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- Spring '10
- The Land