hw4-1 - 60 2 Basic Options ( E ) =- ln E- [ln S + ( r- D )...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 60 2 Basic Options ( E ) =- ln E- [ln S + ( r- D ) ( T- t )]- 2 ( T- t ) / 2 T- t =- ln E- r ( T- t )- [ln S- D ( T- t )]- 2 ( T- t ) / 2 T- t = ln S * 1- ln S * + 2 ( T- t ) / 2 T- t . Thus we have e- r ( T- t ) Z E max( S , E ) G ( S , T ; S, t ) dS = f ( S * 1 , S * , t ; ) . Consequently the conclusion is proved. b) Based on the result in part a), for V ( S, t ) we have V ( S, t ) = e- r ( T- t ) Z max( S , E ) G ( S , T ; S, t ) dS = e- r ( T- t ) Z E max( S , E ) G ( S , T ; S, t ) dS +e- r ( T- t ) Z E max( S , E ) G ( S , T ; S, t ) dS = f ( S * , S * 1 , t ; ) + f ( S * 1 , S * , t ; ) . Therefore, in the expression for V ( S, t ), the positions of S * and S * 1 are symmetric, i.e., exchanging S * and S * 1 in the expression for V ( S, t ) will generate the same expression. 42. Explain why an American option is always worth at least as much as a European option on the same asset with the same strike price and exercise date by financial reasoning and by mathematical tools. Solution : First we use the financial reasoning to have the conclusion. Because an American option can be exercised at any time before expiry, the holder has more rights than a holder of a European option. Therefore a holder of an American option should always pay at least as much as a holder of a European option does, which means that an American option is always worth at least as much as a European option on the same asset with the same strike price and exercise date. Now let us explain this conclusion by using mathematical tools. Let V ( S, t ) and v ( S, t ) denote the prices of the American and European options, re- spectively, and let G v ( S, t ) be the constraint for the American option. We further assume that , r and D are constants. Set t = T/N , where N is a positive integer. Let V ( S, T ) = G v ( S, T ) and for t n = nt , n = N- 1 , N- 2 , , 0, define V ( S, t n ) = max e- rt Z V ( S , t n +1 ) G ( S , t n +1 ; S, t n ) dS , G v ( S, t n ) , 2 Basic Options 61 where G ( S , t n +1 ; S, t n ) is given by G ( S , t n +1 ; S, t n ) = 1 p 2 ( t n +1- t n ) S e- [ ln S- ln S- ( r- D- 2 / 2)( t n +1- t n ) ] 2 / 2 2 ( t n +1- t n ) . Suppose V ( S, t n +1 ) v ( S, t n +1 ). Then we have v ( S, t n ) = e- rt Z v ( S , t n +1 ) G ( S , t n +1 ; S, t n ) dS e- rt Z V ( S , t n +1 ) G ( S , t n +1 ; S, t n ) dS V ( S, t n ) . At t = t N = Nt = T , the condition V ( S, t N ) = V ( S, T ) v ( S, T ) = v ( S, t N ) holds. Therefore using the induction method we can prove V ( S, t n ) v ( S, t n ) for n = N- 1 , N- 2 , , 0 successively. Letting N and noticing that V ( S, t ) generates V ( S, t ) as N , we can have the con- clusion we need....
View Full Document

Page1 / 16

hw4-1 - 60 2 Basic Options ( E ) =- ln E- [ln S + ( r- D )...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online