# Up x xp p with p 0 for x 0 and u x 1 for x 0 1

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Unformatted text preview: +r d 5/4 1/2 1 = = ud 2 1/2 2 and by (6.12) the option prices follow the recursion: Vn (a) = .4[Vn+1 (aH ) + Vn+1 (aT )] (6.17) Example 6.2. Callback options. In this option you can buy the stock at time 3 at its current price and then sell it at the highest price seen in the past for a proﬁt of V3 = max Sm 0m3 S3 Our goal is to compute the value Vn (a) and the replicating strategy n (a) for this option in the binomial model given in (6.16) with S0 = 4. Here the numbers above the nodes are the stock price, while those below are the values of Vn (a) and n (a). Starting at the right edge, S3 (HT T ) = 2 but the maximum in the past is 8 = S1 (H ) so V3 (HT T ) = 8 2 = 6. 188 CHAPTER 6. MATHEMATICAL FINANCE 32 ⇠ 16 ⇠⇠⇠⇠ 0 ⇠ XX XX XX8 3.2 8 .25 8 H H 8 2.24 HH ⇠⇠0 H 4 ⇠ ⇠⇠ .0666 H⇠X X XX X 2 2.4 X 6 1 4 1.376@ @ .1733 @ @ 8 ⇠⇠0 4 ⇠ ⇠⇠ ⇠X XX XX 2 0.8 X 2 .333 @2 @ H 2 1.2 HH H ⇠⇠2 .466 HH⇠⇠⇠⇠ 1 XX XX XX.5 2.2 3.5 1 On the tree, stock prices are above the nodes and option prices below. To explain the computation of the option price note that by (6.17). V2 (HH ) = 0.4(V3 (HHH ) + V3 (HHT )) = 0.4(0 + 8) = 3.2 V2 (HT ) = 0.4(V3 (HT H ) + V3 (HT T )) = 0.4(0 + 6) = 2.4 V1 (H ) = 0.4(V2 (HH ) + V2 (HT )) = 0.4(3.2 + 2.4) = 2.24 If one only wants the option price then Theorem 6.5 which says that V0 = E ⇤ (VN /(1 + r)N ) is much quicker: V0 = (4/5)3 · 1 · [0 + 8 + 0 + 6 + 0 + 2 + 2 + 2 + 3.5] = 1.376 8 Example 6.3. Put option. We wil...
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