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Sir,
In this question delta hedge to be done for this for N(d1) required.
I calculated the value of d1 which is -0.0551
so i rounded to -0.06 and took the table value 0.0239 and then subtract 0.5 as it is lower than 0 so it comes to 0.4761.but in the Kit the published answer is as follows;
-d1 = 0.055
N(-d1) = 0.5 + (0.0199+(0.0239-0.0199)/2) = 0.5219
Sir, I can’t understand the above published solution.
Sir, Kindly explain. what i did wrong?
The question says to assume that the delta of a put option is N(-d1).
d1 = -0.0551
So – d1 = – – 0.0551 = + 0.0551 (two minus’s make a plus)
So what you are looking up is N (+0.0551)
Hope that explains it 🙂
(You actually only need to use the table figure for 0.06 – the answer apportions between the figures for 0.05 and 0.06 which is strictly more accurate, but you don’t really need to do that in the exam).
@johnmoffat said:
The question says to assume that the delta of a put option is N(-d1).d1 = -0.0551
So – d1 = – – 0.0551 = + 0.0551 (two minus’s make a plus)
So what you are looking up is N (+0.0551)
Sir, why do we need to make d1 to be positive insread of negative and vise versa? I see that question says assume N(-d), but what is the reason for that?
It is not in complience whith instruction at the bottom normal distribution table – “If di < 0, subtract the relevant number above from 0·5.”
Hold on – there are two separate things here.
Firstly, because we are valuing a put option (the normal formula is for a call option) the examiner said in the requirements that delate is N(-d1) (and not N(d1) as usual).
d1 calculated as – 0.0551, therefore – d1 = – – 0.0551 = +0.0551
When it comes to actually looking up the value in the tables, you look up the value of +0.0551 following the normal rules. Since you are now looking up > 0, you add the table figure to 0.5.
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