hw3-2 - 2 Basic Options 39 2 p + 1 2 (t)S 2 p + [r(t) D...

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2 Basic Options 39 p ∂t + 1 2 σ 2 ( t ) S 2 2 p S 2 + [ r ( t ) - D 0 ( t )] S p S - r ( t ) p = 0 , 0 S, t T, p ( S, T ) = max(1 - S, 0) , 0 S. Therefore p ( S, t ) = Ee - γ ( t ) h N ( - d 2 ) - ( S/E ) e β ( t ) N ( - d 1 ) i = Ee - γ ( t ) N ( - d 2 ) - Se β ( t ) - γ ( t ) N ( - d 1 ) is the value of a put option with time-dependent parameters: ∂p ∂t + 1 2 σ 2 ( t ) S 2 2 p ∂S 2 + ( r ( t ) - D 0 ( t )) S ∂p ∂S - r ( t ) p = 0 , 0 S, t T, p ( S, T ) = max( E - S, 0) , 0 S. If we deFne δ ( t ) = γ ( t ) - β ( t ) = Z T t D 0 ( s ) ds, then the solution p ( S, t ), d 1 and d 2 can be written as p ( S, t ) = Ee - γ ( t ) N ( - d 2 ) - Se - δ ( t ) N ( - d 1 ) and d 1 = µ ln Se - δ ( t ) Ee - γ ( t ) + α ( t ) ¶` (2 α ( t )) 1 / 2 , d 2 = d 1 - (2 α ( t )) 1 / 2 . 25. a) Show that if D 0 = 0, then c ( S, t ) max( S - E, 0), which means that for this case the value of an American call option is the same as the value of a European call option. b) Show that if r = 0, then p ( S, t ) max( E - S, 0), which means that for this case the value of an American put option is the same as the value of a European put option. Solution : a) If D 0 = 0, we have c ( S, t ) = e - r ( T - t ) Z 0 max( S 0 - E, 0) G ( S 0 , T ; S, t ) dS 0 e - r ( T - t ) max µZ 0 ( S 0 - E ) G ( S 0 , T ; S, t ) dS 0 , 0
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40 2 Basic Options = max Se - D 0 ( T - t ) - Ee - r ( T - t ) , 0 · = max S - Ee - r ( T - t ) , 0 · max( S - E, 0) where G ( S 0 , T ; S, t ) = 1 σ p 2 π ( T - t ) S 0 e - [ ln( S 0 /S ) - ( r - D 0 - σ 2 / 2)( T - t ) ] 2 / 2 σ 2 ( T - t ) . Therefore we obtain our conclusion.
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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.

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hw3-2 - 2 Basic Options 39 2 p + 1 2 (t)S 2 p + [r(t) D...

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