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sir in question 3 of dec 2010 N(-d1) was used as the value for delta in a put option. d1 in that question was -0.055 and in the solution -d1 was goten as .0.055 ( minus sign is taken to the left). thus normal distribution table figure was instead added to 0.5. i don’t quite understand it because in your lectures you said delta can be taken as N(d1).
can you please clarify me
thanksThe question says to assume that the delta of a put option is N(-d1)
(Usually we are looking at call options and then the delta is N(d1). That is why he added that bit to the question because you are not expected to have learned that for a put option it is N(-d1))
– – = + So N( – – 0.055) = N(+0,055)
thank you John, I can see the exam is really full of tricks that one really needs to spot them
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