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Unformatted text preview: Sn
Sn = sn , . . . S0 = s0 =
(1 + r)n
(1 + r)n
since E ⇤ Xn+1 = 1 + r by (6.10).
Notation. To make it easier to write computations like the last one we will let
En (Y ) = E (Y |Sn = sn , . . . S0 = s0 ) (6.15) or in words, the conditional expectation of Y given the information at time n.
A second important martingale result is
Theorem 6.5. Assume that the holdings n (a) can be determined from the
outcomes of the ﬁrst n stock movements and let Wn be the wealth process deﬁned
by (6.14). Under P ⇤ , Wn /(1 + r)n is a martingale, and hence the value has
V0 = E ⇤ (Vn /(1 + r)n ). 187 6.3. CONCRETE EXAMPLES Proof. The second conclusion follows from the ﬁrst and Theorem 6.3. A little
arithmetic with (6.14) shows that
(1 + r)n+1
(1 + r)n n ✓ Sn+1
(1 + r)n+1 Sn
(1 + r)n ◆ Since n (a) is an admissible gambling strategy and Sn /(1+ r)n is a martingale,
the desired result follows from Theorem 5.12. 6.3 Concrete Examples Turning to examples, we will often use the following binomial model because it
leads to easy arithmetic
u = 2, d = 1/2, r = 1/4 (6.16) The risk neutral probabilities p⇤ = 1...
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- Spring '10
- The Land