# are iid with h p x1 1 p h h p x1 1 1 p h then

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Unformatted text preview: or the European version. It certainly cannot be strictly less since one possibility in the American option is to continue each time and this turns it into a European option. For some of the theoretical results it is useful to notice that ✓ ◆ g⌧ ⇤ V0 = max E ⌧ (1 + r)⌧ (6.21) where the maximum is taken over all stopping times ⌧ with 0 ⌧ N . In Example 6.7 ⌧ (T ) = 1 and ⌧ (H ) = 3, i.e., if the stock goes down on the ﬁrst step we stop. Otherwise we continue until the end. 1 41 V0 = · 6 · + · 6 · 2 58 ✓ ◆3 4 = 2.4 + 0.384 5 Proof. The key to prove the stronger statement ⇤ Vn (a) = max En (g⌧ /(1 + r)⌧ ⌧ n n ) ⇤ where En is the conditional expectation given the events that have occurred up to time n. Let Wn (a) denote the right-hand side. If we condition on the ﬁrst n outcomes being a then P (⌧ = n) is 1 or 0. In the ﬁrst case we get gn (a). In the ⇤ second case Wn (a) = [p⇤ (a)Wn+1 (aH ) + qn (a)Wn+1 (aT )]/(1 + r), so Wn and n Vn satisfy the same r...
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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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