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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....
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 Spring '10
 Zhu
 Math

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