Synthetic financial Products

"DATE "1 week "1 Month "3 Months "6 months "1 year "2 year "15-Jan "0 .848836 "0 .732553 "0 .1045 "0 .33691 "0 .452894 "0 .75465 "where "P (s ,L ,t unknown function of 3 variables "s Log (S "S spot price "t time , measured from today to expiration "L Maximum value of spot price S during the life of the option "The price of a six month calls option will be "S (t "X "r "r (6-month (6-Month "Annualized "1 .9584 "1 .9753 "0 .0897 "0 .0918 "0 .2873 "Solution : in to apply equation ONE need to estimate B9t

, 0 .5 ) and B (t , 0 .5 ) since T 0 .5 ( months equals 0 .5 years ) given the annualized interest rates on six months bound of 8 .97 and 9 .18^ the six moth U .S . and UK interest rates are 1 (2 /2 ) and 2 (4 /2 , respectively . The associated prices are "B (t ,0 .5 1 0 .9901 "1 .035156 "B (t ,0 .5 1 0 .9804 "1 .035156 "Substitution in the values for B and B along with for S (0 .68 , X (0 .70 , and (0 .2873 , we calculate "d1 In (SB /XB 0 .5 2T In (0 .68x 0 .9901 /0 .1 .9753x0 .9804 0 .5 (0 .2873 )2 (0 .5 -0 .0578 "T "The easiest way to compute the values of N (-0 .05786 ) and N (-0 .26101 ) is to use spreadsheet faction such as NORMDIST in Excel . This Excel function yields computed...