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Hi Sir PYQ J2011 Q4a
How do I calculate N(d1) and N(d2)?
Based on the answer
N(d1)=0.5+0.2291+0.7x(0.2324-0.2291)=0.7314
N(d2)=0.5+0.0753+0.3x(0.0793-0.0753)=0.5765From my understanding N(d1)=0.5+0.2324=0.7374 and N(d2)=0.5+0.0753=0.5753
Did i do it wrong?And I am currently using the BPP text from June 2015, is there any changes or addition from the syllabus?Thanks
There are no changes to the syllabus.
With regard to Black Scholes, you should watch our free lecture on this.
Our lectures are a complete course for Paper P4.I have watched the video. And I’m doing it according to the video
N(d1)=0.5+0.2324=0.7374 and N(d2)=0.5+0.0753=0.5753Instead of doing it like in the PYQ as below
N(d1)=0.5+0.2291+0.7x(0.2324-0.2291)=0.7314
N(d2)=0.5+0.0753+0.3x(0.0793-0.0753)=0.5765I don’t understand the method use in PYQ answers.
I don’t know what you mean by “PYQ”
However, what you have done is fine for the exam – you have rounded D1 and D2 to 2 decimal places and then looked up N(D1) and N(D2) in the tables.
What the “PYQ” answer has done has apportioned between the figures from the tables so as to estimate to more decimal places. That is great if you have the time in the exam, but there is no real need to (because the apportioning would only be approximate anyway since the relationship is not linear).
Dear Sir,
Does delta measures the change in the option price as the value of the underlying share moves by 1% or by 1$?
In the Kaplan text book the answer is $, but in the exam kit it is 1%.
Please help.
If delta is (for example) 0.01 (or 1%) then it means that the option price will change by 1% of the change in the share price.
(So if the share price changes by $1 then the option price will change by $0.01; if the share price changes by $2 then the option price changes by $0.02, and so on)
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