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In question number 3(a) of December 2010 question paper on delta hedge, the answer sheet arrived at the solution
–d1 = 0•055 (this one is clear)
N(–d1) = 0•5 + (0•0199 + (0•0239 – 0•0199)/2) = 0•5219 (this is doubt)
I am able to arrive at –d1 = 0•055 and it is right no question on this aspect of answer.
and from the above I have to arrive at d1=-0.055 from standard normal distribution table given in exam for which the value is 0.0199
as sign is negative we have to deduct the same from 0.5, thus 0.5-0.0199= 0.4801
But I am unable to understand the following step given in answer sheet of December 2010 exams
N(–d1) = 0•5 + (0•0199 + (0•0239 – 0•0199)/2) = 0•5219
Can any one clarify the same.
Regards
somayajuluYou are right. Your method is very collect as well. There are two methods of finding N(d1). The one you know is the best for somebody like you and me when rounded to 2 decimals. The other one is that used in the suggested solution.
The method in sugg sol. has got 3 decimals for d1 = -0.055. So first they will find table fig for the first 2 dec which is same 0.0199 that is supposed to be subtracted as you collectly said…but they will find another table fig for the 3rd dec. which is approx nearer to 0.006 (next right on table) and get its table fig as 0.0239. Lastly an avge of the difference of the 2 table figures get to be added to 0.0199 ie 0.0199+((0.0239-.0.0199)/2) to find an exact N(-d1) -Put Option.
Your method is just right as well and you cant be marked wrong I hope.
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