018 - 18 Exotic Options Answers to Questions and Problems...

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177 Answers to Questions and Problems 1. Using the following parameter values, find the price of a forward-start put: S 5 100; X 5 100; T – t 5 1 year; s5 0.2; r 5 0.1; d5 0.05; tg 5 0.5. The first step is to value a plain vanilla put on the grant date, which is in one-half year. At that time, the put will have a half year remaining until expiration. Therefore, Forward-Start Put 5 e d ( tg t ) p tg M 5 e –.05(.5) 4.3106 5 4.2042 2. Price all four types of compound options assuming the following parameter values: S 5 100; s5 0.4; r 5 0.1; d5 0.05; X 5 100; x 5 8; T 5 1 year; te 5 0.25 years. The first, and most difficult, step is to find the critical prices for the underlying calls and puts. For underly- ing calls, we have: For underlying puts, the critical stock price satisfies the following relationship: where: z 5 ln 1 S * X 2 1 ( r 2 d 1 0.5 s 2 )( T 2 te ) s Ï T 2 te S * e 2d ( T 2 te ) N ( 2 z ) 1 Xe 2 r ( T 2 te ) N ( 2 z 1 s Ï T 2 te ) 2 x 5 0 S * e 2d ( T 2 te ) N ( z ) 2 Xe 2 r ( T 2 te ) N ( z 2 s Ï T 2 te ) 2 x 5 0 5 4.3106 5 100 e 2 .1(0.5) 0.457765 2 e 2 0.05(0.5) 100(0.402266) 5 100 e 2 .1(0.5) N ( 2 0.106066) 2 e 2 0.05(0.5) 100 N ( 2 0.247487) p tg M 5 Xe 2 r ( T 2 t ) N ( 2 d 2 M ) 2 e 2d ( T 2 t ) S t N ( 2 d 1 M ) d 2 M 5 0.247487 2 0.141421 5 0.106066 d 1 M 5 ln 1 100 100 2 1 [0.1 2 0.05 1 0.5(0.2)(0.2)](0.5) 0.2 Ï 0.5 5 0.247487 18 Exotic Options
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These critical stock prices must be found by an iterative search. The critical stock price for an underlying call is 86.6162, and for an underlying put the critical stock price is 110.1995.These values can be verified as follows, first for underlying calls: Therefore, Thus, we verify that 86.6152 is the correct critical stock price for underlying calls: The same verification can be performed for underlying puts. Before computing the actual option values, we must compute other intermediate results: With our sample values, w 1 5 0.880974 for underlying calls, and w 1 5 –0.323111 for underlying puts. The other intermediate variables are invariant across all option types: w 2 5 0.325000; r5 0.5. We also need the following bivariate cumulative normal probabilities for compound options on underlying calls: For convenience, we also note: xe 2 r ( te 2 t ) 5 8 e 2 0.1 (0.25) 5 7.802479 Xe 2 r ( T 2 t ) 5 100 e 2 0.1 (1) 5 90.483742 Se 2d ( T 2 t ) 5 100 e 2 0.05 (1) 5 95.122942 N ( 2 w 1 1 s Ï te 2 t ) 5 N ( 2 0.680974) 5 0.247944 N ( w 1 2 s Ï te 2 t ) 5 N (0.680974) 5 0.752056 5 0.053158 N 2 ( 2 w 1 1 s Ï te 2 t ; w 2 2 s Ï T 2 t ; 2r ) 5 N 2 ( 2 0.680974; 2 0.075; 2 0.5) 5 0.062904 N 2 ( 2 w 1 ; w 2 ; 2r ) 5 N 2 ( 2 0.880974; 0.325; 2 0.5) 5 0.416949 N 2 ( w 1 2 s Ï te 2 t ; w 2 2 s Ï T 2 t ; r ) 5 N 2 (0.680974; 2 0.075; 0.5) 5 0.564506 N 2 ( w 1 ; w 2 ; r ) 5 N 2 (0.880974; 0.325; 0.5) w 2 5 ln 1 S X 2 1 ( r 2 d 1 0.5 s 2 )( T 2 t ) s Ï T 2 t w 1 5 ln 1 S S * 2 1 ( r 2 d 1 0.5 s 2 )( te 2 t ) s Ï te 2 t 86.6152 e 2 0.05 (0.75) 0.446957 2 100 e 2 0.1 (0.75) 0.315698
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This homework help was uploaded on 04/07/2008 for the course BA 4825 taught by Professor Danısoglu during the Spring '08 term at Middle East Technical University.

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018 - 18 Exotic Options Answers to Questions and Problems...

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