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First of all THANKS is advance John for the great work you are doing here.
I am struggling here with deltas N(d1) and N(d2) in the question Marengo. According to the normal calculations we calculate d1 according to the formula and look up the corresponding figures using the tables. If d1 is greater than 0, we add 0.5 and vice versa.
The problem is I can’t figure out what the examiner has done here…
N(d1)=0.5+0.2291+0.7x (0.234 – 0.2291) = 0.7134
N(d2)=0.5+0.0753+0.3x (0.0793 – 0.0753) = 0.5765
This procedure has been done in a number of other questions as well but I can’t figure it. Maybe its something to do with the tables as I am only looking up to two decimal places i.e instead of 0.617 only looking up 0.61.
Plus I would appreciate if you can recommend me some readings for deltas and delta hedges as I can’t find something in detail in bpp or kaplan.
All he has done is approximated between the results from the tables.
I don’t think that the figures you have quoted come from Marengo!
Using Marengo however as an example, we want to look up 0.055 in the tables. However we can only look up 0.05 and 0.06, so what he has done is taken the figure for 0.05 (0.0199) and then added 0.5 of the difference between 0.05 and 0.06 from the tables (0.0199 and 0.0239). In that way he has approximated to 0.055.
With regard to deltas and delta hedges, I am surprised if BPP and Kaplan have no detail. Have you read our Course Notes (and watched the free lecture that goes with them)?
Hi John,
I don’t understand why N(-d1) is plus with 0.5, shouldn’t it be (0.5-(0.0199+(0.0239-0.0199)/2)?
And in BPP text, with the option to abandon, given the d1 is 1.6424 and N(d1) is 0.9495, the N(-d1) is calculated this way “1-0.9495 = 0.0505”. My question would be why using 1 not 0.5?
Many thanks!
D1 is negative, and therefore -D1 is positive (minus a negative number is a positive number).
The question tells you to use N(-D1)
I do not have BPP books and so I cannot really help with your second question without seeing the question itself.
Hi John,
I have been working out Marengo from the ACCA Dec 2010 Past Questions and Answers. I expect that the value of d1 should be -0.149 instead of 0.55.
It appears the formula used in the solving the problem was wrong. ACCA usedd1 = [ln(Pa/Pe) + (r + 0·5s^2)t]/(st^1/2)
Instead of using
d1 = [ln(Pa/Pe) + (r +( 0·5s^2)t]/(st^1/2)
Please could you confirm if i am right?
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