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Unformatted text preview: 3 Exotic Options 137 implies S > 1 + rT 1 + D αT A , i.e., (1 + D αT ) S + (1 + rT ) A < 0. Thus if 1 + rT 1 + D αT < 1 α , then when S > A α , the PDE cannot be used. If 1 + rT 1 + D αT > 1 α , then when S > 1 + rT 1 + D αT A , the PDE cannot be used; while A α < S < 1 + rT 1 + D αT A , we have (1 + D αT ) S + (1 + rT ) A > 0 and the PDE can be used; Putting those cases together, we have that when S < max µ A α , 1 + rT 1 + D αT A ¶ , the PDE can be used, and otherwise the PDE cannot be used. Consequently S f ( A, T ) = A max µ 1 α , 1 + rT 1 + D αT ¶ . b) When A < E , max( A E, 0) = 0, so the PDE always can be used. When A > E , max( A E ) = A E . Thus ∂ ( A E ) ∂t + L SAT ( A E ) = S A T · 1 r ( A E ) = 1 T [ S + rT E (1 + rT ) A ] . The last expression is greater than zero if S + rT E 1 + rT > A , and less than zero if S + rT E 1 + rT < A . If S + rT E 1 + rT < E , then A > E implies A > S + rT E 1 + rT , i.e., S + rT E (1 + rT ) A < 0. Thus if S + rT E 1 + rT < E , then when A > E , the PDE cannot be used. If S + rT E 1 + rT > E , then when A > S + rT E 1 + rT , we have S + rT E (1+ rT ) A < 0 and the PDE cannot be used; while E < A < S + rT E 1 + rT , we have S + rT E (1 + rT ) A > and the PDE can be used. Putting those cases together, we have that when A < max µ E, S + rT E 1 + rT ¶ , the PDE can be used, and otherwise the PDE cannot be used. Consequently A f ( S, T ) = max µ E, S + rT E 1 + rT ¶ . 9. Suppose that sampling is done discretely at t = t 1 , t 2 , ··· , t K , where ≤ t 1 < t 2 < ··· < t K ≤ T . Let H ( t ) = max ( S ( t 1 ) , ··· , S ( t i * ( t ) )) , where i * ( t ) is the number of samplings before time t. Assume dS = μSdt + σSdX and the dividends are paid continuously with dividend yield D . 138 3 Exotic Options Let V ( S, H, t ) be the value of a lookback option with discrete sampling. Derive the PDE and the jump condition for such a lookback option by using a portfolio Π = V ( S, H, t ) ΔS (without using the general PDE for derivative securities). Solution : Because dH ( t ) = ( 0, if t 6 = t i , i = 1 , 2 , ··· , K, max ( S ( t i ) , H ( t i )) H ( t i ) , if t = t i , i = 1 , 2 , ··· , or K and dH ( t ) dt = K X i =1 £ max ( S ( t ) , H ( t )) H ( t )/ δ ( t t i ) , where H ( t ) = lim ε → H ( t ε ) with ε > 0, using Itˆ o’s lemma, we have dV = ∂V ∂S dS + ∂V ∂H dH + µ ∂V ∂t + 1 2 σ 2 S 2 ∂ 2 V ∂S 2 ¶ dt. Thus if we set Π = V ( S, H, t ) ΔS, then we have dΠ = ∂V ∂S dS + ∂V ∂H dH + µ ∂V ∂t + 1 2 σ 2 S 2 ∂ 2 V ∂S 2 ¶ dt Δ ( dS + D Sdt ) = µ ∂V ∂S Δ ¶ dS + ∂V ∂H dH + µ ∂V ∂t + 1 2 σ 2 S 2 ∂ 2 V ∂S 2 ΔD S ¶ dt....
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This note was uploaded on 02/11/2010 for the course MATH 6203 taught by Professor Zhu during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 Zhu

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