# hw4-1 - 60 2 Basic Options ξ E = ln E[ln S r D T t σ 2 T...

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Unformatted text preview: 60 2 Basic Options ξ ( E ) =- ln E- [ln S + ( r- D ) ( T- t )]- σ 2 ( T- t ) / 2 σ √ T- t =- ln E- r ( T- t )- [ln S- D ( T- t )]- σ 2 ( T- t ) / 2 σ √ T- t = ln S * 1- ln S * + σ 2 ( T- t ) / 2 σ √ 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 ; σ ) . 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 ; σ ) + f ( S * 1 , S * , t ; σ ) . 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 σ , r and D are constants. Set Δt = T/N , where N is a positive integer. Let ˜ V ( S, T ) = G v ( S, T ) and for t n = nΔt , n = N- 1 , N- 2 , ··· , 0, define ˜ V ( S, t n ) = max µ e- rΔt 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 σ p 2 π ( t n +1- t n ) S e- [ ln S- ln S- ( r- D- σ 2 / 2)( t n +1- t n ) ] 2 / 2 σ 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- rΔt Z ∞ v ( S , t n +1 ) G ( S , t n +1 ; S, t n ) dS ≤ e- rΔt Z ∞ ˜ V ( S , t n +1 ) G ( S , t n +1 ; S, t n ) dS ≤ ˜ V ( S, t n ) . At t = t N = NΔt = 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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hw4-1 - 60 2 Basic Options ξ E = ln E[ln S r D T t σ 2 T...

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