Q3) Answer from Question 3:
d1 = [ln(Pa/Pe) + (r + 0·5s2)t]/(st1/2) (from formulae sheet given in examination)
d1 =[ln(340/350) + ((0·04 + 0·5 x 0·42) x 1/6)]/(0·4 x 1/60·5) = –0·055
–d1 = 0·055
N(–d1) = 0·5 + (0·0199 + (0·0239 – 0·0199)/2) = 0·5219[/u]
Put contracts to be bought for a delta hedge = 200,000/(0·5219 x 1,000) = 383·2 rounding to 383 contracts.
Can someone assist how to get the N(-d1)? What formula is this, I am self studying so there is something i am figuring out. Thanks for your help.
d1 = [ln(Pa/Pe) + (r + 0·5s2)t]/(st1/2) (from formulae sheet given in examination)
d1 =[ln(340/350) + ((0·04 + 0·5 x 0·42) x 1/6)]/(0·4 x 1/60·5) = –0·055
–d1 = 0·055
N(–d1) = 0·5 + (0·0199 + (0·0239 – 0·0199)/2) = 0·5219[/u]
Put contracts to be bought for a delta hedge = 200,000/(0·5219 x 1,000) = 383·2 rounding to 383 contracts.
Can someone assist how to get the N(-d1)? What formula is this, I am self studying so there is something i am figuring out. Thanks for your help.
