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Dear tutor,
my question is concerned with Black-Scholes formula.
Please tell me how we should calculate N(1.998)?
In the exam we are given the Standard normal distribution table where d=1.99 and than follows d=2.00.Thank you
It is good enough in the exam just to take the nearest.
If you want to be extra clever than apportion between the result for 1.99 and 2.00
(i.e. the value for 1.99 + (80% of the difference between the results for 1.99 and 2.00)Marengo 12/10
In this qs part a) the d1 value is negative so we subtract its N(d1) value from 0.5 right?but in the bpp kit they have added 0.5 to N(d1)It is because the question says:
“Note: You may assume that the delta of a put option is equivalent to N(–d1)”
So although d1 is negative, -d1 is therefore positive.
Is what acceptable?
If you are referring to the first post here, then (as I wrote) – it is good enough in the exam to take the nearest.
There is only one way!!
It is not a question of whether d1 is negative or not – it is whether the number in the ( ) is positive or negative.For a call option we take the N of d1.
For a put option the question tells you to take the N of (-d1)
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I need explanation on how to use the standard normal distribution table if d1 is .8956 d2 is 5421 and if d1 is 1.8394 and d2 is 1.8247
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