{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw8-1 - 3 Exotic Options 137 1 rT A i.e(1 D0 T)S(1 rT)A < 0...

This preview shows pages 1–4. Sign up to view the full content.

3 Exotic Options 137 implies S > 1 + rT 1 + D 0 αT A , i.e., - (1 + D 0 αT ) S + (1 + rT ) A < 0. Thus if 1 + rT 1 + D 0 αT < 1 α , then when S > A α , the PDE cannot be used. If 1 + rT 1 + D 0 αT > 1 α , then when S > 1 + rT 1 + D 0 αT A , the PDE cannot be used; while A α < S < 1 + rT 1 + D 0 αT A , we have - (1 + D 0 αT ) S + (1 + rT ) A > 0 and the PDE can be used; Putting those cases together, we have that when S < max A α , 1 + rT 1 + D 0 αT A , the PDE can be used, and otherwise the PDE cannot be used. Consequently S f ( A, T ) = A max 1 α , 1 + rT 1 + D 0 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 + rTE - (1 + rT ) A ] . The last expression is greater than zero if S + rTE 1 + rT > A , and less than zero if S + rTE 1 + rT < A . If S + rTE 1 + rT < E , then A > E implies A > S + rTE 1 + rT , i.e., S + rTE - (1+ rT ) A < 0. Thus if S + rTE 1 + rT < E , then when A > E , the PDE cannot be used. If S + rTE 1 + rT > E , then when A > S + rTE 1 + rT , we have S + rTE - (1+ rT ) A < 0 and the PDE cannot be used; while E < A < S + rTE 1 + rT , we have S + rTE - (1 + rT ) A > 0 and the PDE can be used. Putting those cases together, we have that when A < max E, S + rTE 1 + rT , the PDE can be used, and otherwise the PDE cannot be used. Consequently A f ( S, T ) = max E, S + rTE 1 + rT . 9. Suppose that sampling is done discretely at t = t 1 , t 2 , · · · , t K , where 0 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 0 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 0 H ( t - ε ) with ε > 0, using Itˆ o’s lemma, we have dV = ∂V ∂S dS + ∂V ∂H dH + ∂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 = ∂V ∂S dS + ∂V ∂H dH + ∂V ∂t + 1 2 σ 2 S 2 2 V ∂S 2 dt - Δ ( dS + D 0 Sdt ) = ∂V ∂S - Δ dS + ∂V ∂H dH + ∂V ∂t + 1 2 σ 2 S 2 2 V ∂S 2 - ΔD 0 S dt. Let us choose ∂V ∂S = Δ, then this portfolio is risk-free and the return rate is r : ∂V ∂H dH + ∂V ∂t + 1 2 σ 2 S 2 2 V ∂S 2 - ∂V ∂S D 0 S dt = r V ( S, H, t ) - ∂V ∂S S dt. Noticing the expression for dH ( t ) dt , this relation can be rewritten as ∂V ∂t + 1 2 σ 2 S 2 2 V ∂S 2 + ( r - D 0 ) S ∂V ∂S + K X i =1 £ max ( S ( t ) , H ( t - )) - H ( t - )/ δ ( t - t i ) ∂V ∂H - rV = 0 .
3 Exotic Options 139 This means that at t 6 = t i , i = 1 , 2 , · · · , K , V fulfills ∂V ∂t + 1 2 σ 2 S 2 2 V ∂S 2 + ( r - D 0 ) S ∂V ∂S - rV = 0 , 0 S, 0 H and at t = t i , i = 1 , 2 , · · · , or K , the equation ∂V ∂t + K X i =1 £ max ( S ( t ) , H ( t - )) - H ( t - )/ δ ( t - t i ) ∂V ∂H = 0 , 0 S, 0 H holds. It is a hyperbolic equation, and the characteristic relation is dH dt = K X i =1 £ max ( S ( t ) , H ( t - )) - H ( t - )/ δ ( t - t i ) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}