Sir I was doing ‘Q59 UNIGLOW from Kaplan kit sep14 to aug15 edition’ on option pricing. It is not difficult question but I just need to understand one point. In answer it says we can use Delta to construct delta hedge and in order to protect against a fall in Uniglow’s share price, the easiest hegde would be to write (sell) options on Uniglow’s shares. (I am protecting the investment of my company in Uniglow’s shares). So my question whether the option we sell is call or put option?
We sell a call option (you could achieve the same ‘protection’ by buying a put option, but in the exam you always sell a call to create a delta hedge, unless told otherwise).
I do explain the reasoning behind this in the lectures.
i still can’t figure out how you reached the figure of 0.045 while calculating d1 (0.1+0.5*0.4^2)0.25 . My answer is 0.105 i’m really confused please help..
Because that is was comes out of the equations!!!!
The price of the share in 3 months could obviously be anything. However it is more likely that it will be less than 1.80 than more than 1.80, so the people selling the options will want to charge more for a put option because they are more likely to have to pay out on it.
(and don’t ask me why the price is more likely to be less than 1.80 than more!! That is down to the statistics why is how the derived the formulae, and I have certainly no intention of going through their proof 🙂 )
Not all scientific calculators are the same – there are two different ways that they work. If yours does not have a +/- button then it uses a different logic, and you will have to look in the instruction book how to do it.
Sir, In Example % while calculating value of call option on the third step u used e raised to power -.04… aint that wrong as time is .25 and r is .1 when we multiply that it comes to .025.. :O dont know how u came to to .04 :O waiting for ur reply thanks
I assume you were happy in arriving at the figure of -0.6886 for D1 (the workings are at the end of the lecture notes as well).
We look up 0.69 in the tables (we can only look up for 2 decimal places, so 0.6886 becomes 0.69). So do this you look along the 0.6 row and as you move through the columns it gives the figure for 0.60, 0.61, 0.62 and so on. We want 0.69 and so it is the 0.6 row and the 0.09 column, and the figure from the tables is 0.2549.
Because D1 is negative, we subtract then 0.2549 away from 0.5. (Had it been positive, as D2 is, then we would add 0.5). This rule is given at the bottom of the tables.
You should be able to follow the rest of the answer at the back of the Lecture Notes.
(How did you manage to sort out the first four examples, but not example 5? :-))
John i have this question. it concerns calculatind d1 if a questions is given n i calculate d1 and a figure say 1.2812 and i then round it to 1.28 and another candidate calculates his and round it to 1.3. these two answers will give different values for N(d1) which will lead to a different value for a call option. will the two of us have all the marks. I am asking because when calculating the figure for natual log of Pa/Pe due to rounding candidates will have different answers thank you
Firstly most if not all of the marks are for proving you understand what is happening rather than for the final answer. Secondly, when rounding you should really round to the number of decimals needed for the tables, so why round to 1.3 when the tables allow you to look up 1.28 🙂
I am sorry, but you can only download the course notes – the lectures can only be watched online. It is the only way that we can keep this website free of charge.
If it was a call option exercisable immediately, then the option gives you the right to buy a share at a fixed price.
So, for example, if the current share price is $4.00 and you could buy an option giving you the right to buy the share at an exercise price of $3.70, then you could buy the share for $3.70 and immediately sell it for $0.30. Nobody is going to give you that right free! You would be prepared to pay $0.30 for the option. Then you could use it and buy a share for $3.70. You have then spent $4.00 in total and you own a share worth $4.00 🙂
(But of course, that is only if the option were exercisable immediately. In practice the option will be the right to buy a share at a fixed price on a future date, and to get the value of that we need to use all the formulae.)
A very basic question: Value of an option is basically the premium payable right?
Yes 🙂
Thanks alot
Hello sir…
am confused, why do u said the call option is cheaper than the put option in this case?
the call option 5cent n the put option 28cent
The prices of the two options depend on how likely they are to end up being exercised.
@ Sir John Moffat,
Sir I was doing ‘Q59 UNIGLOW from Kaplan kit sep14 to aug15 edition’ on option pricing.
It is not difficult question but I just need to understand one point.
In answer it says we can use Delta to construct delta hedge and in order to protect against a fall in Uniglow’s share price, the easiest hegde would be to write (sell) options on Uniglow’s shares. (I am protecting the investment of my company in Uniglow’s shares).
So my question whether the option we sell is call or put option?
We sell a call option (you could achieve the same ‘protection’ by buying a put option, but in the exam you always sell a call to create a delta hedge, unless told otherwise).
I do explain the reasoning behind this in the lectures.
Thank you very much sir.
You are welcome 🙂
thank you. incredible lecture
Thank you for the comment 🙂
i still can’t figure out how you reached the figure of 0.045 while calculating d1 (0.1+0.5*0.4^2)0.25 . My answer is 0.105 i’m really confused please help..
(0.1 + 0.5 x 0.4 x 0.4) x 0.25 = (0.1 + 0.08) x 0.25 = 0.18 x 0.25 = 0.045
thank you so much looks like my calculator had issues
You are welcome 🙂
on example 5 why is the put option more epensive than call option?
Because that is was comes out of the equations!!!!
The price of the share in 3 months could obviously be anything. However it is more likely that it will be less than 1.80 than more than 1.80, so the people selling the options will want to charge more for a put option because they are more likely to have to pay out on it.
(and don’t ask me why the price is more likely to be less than 1.80 than more!! That is down to the statistics why is how the derived the formulae, and I have certainly no intention of going through their proof 🙂 )
am unable to calculator the figure for e.
I have a scientific one with e and ln.
which key to press for +-
Not all scientific calculators are the same – there are two different ways that they work. If yours does not have a +/- button then it uses a different logic, and you will have to look in the instruction book how to do it.
in mine i just type shift, e^x,-.25 etc. mine doesnt have a +/- button either but it still works by just typing ‘-‘ before any figure
Sir,
In Example % while calculating value of call option on the third step u used e raised to power -.04… aint that wrong as time is .25 and r is .1 when we multiply that it comes to .025.. :O
dont know how u came to to .04 :O
waiting for ur reply
thanks
example 5
You are correct – it should be 0.025 (and the final answer should be 4c).
I will re-record the lecture.
thanks 🙂
Please leave it like it is, makes us think, not just silly writing all down 🙂
Good point Braske77 (and thank you 🙂 )
Im gettign lost in example 5 on the calculation of p on the e to the power of -0.04 what is the e value is it 2.7183
Yes it is – and you must have a calculator with an ‘e’ button on it 🙂
found the e button thanx
Hi Tutor
Please give me an example on how to use the distribution table i got lost in Example 5 on the column 0.09 why that column?
I assume you were happy in arriving at the figure of -0.6886 for D1 (the workings are at the end of the lecture notes as well).
We look up 0.69 in the tables (we can only look up for 2 decimal places, so 0.6886 becomes 0.69). So do this you look along the 0.6 row and as you move through the columns it gives the figure for 0.60, 0.61, 0.62 and so on. We want 0.69 and so it is the 0.6 row and the 0.09 column, and the figure from the tables is 0.2549.
Because D1 is negative, we subtract then 0.2549 away from 0.5.
(Had it been positive, as D2 is, then we would add 0.5). This rule is given at the bottom of the tables.
You should be able to follow the rest of the answer at the back of the Lecture Notes.
(How did you manage to sort out the first four examples, but not example 5? :-))
i only started practicing exmaple 5 then i got lost. Many thanx for your advice
John i have this question. it concerns calculatind d1 if a questions is given n i calculate d1 and a figure say 1.2812 and i then round it to 1.28 and another candidate calculates his and round it to 1.3. these two answers will give different values for N(d1) which will lead to a different value for a call option. will the two of us have all the marks. I am asking because when calculating the figure for natual log of Pa/Pe due to rounding candidates will have different answers
thank you
Two things.
Firstly most if not all of the marks are for proving you understand what is happening rather than for the final answer.
Secondly, when rounding you should really round to the number of decimals needed for the tables, so why round to 1.3 when the tables allow you to look up 1.28 🙂
How can I Have Lectures of Business Valuation
There are no lectures yet on business valuation.
How can i download this?or this is only for watching?
I am sorry, but you can only download the course notes – the lectures can only be watched online.
It is the only way that we can keep this website free of charge.
Can u provide me solution of paper strategic financial management 3.7 of december 2006 of ACCA………?
I don’t think that I have it any longer – I will check later.
Do remember that the examiner (and the syllabus) has changed twice since then.
okk,,,..
sir in example you taken t as 0.4 in formula of call option and put option i guess thats a mistake it should be 0.25 right?
i mean example 5.
I have watched my lecture again, and it seems that I have taken ‘t’ as 0.25 correctly (it is ‘s’ that is 0.4).
You can of course check the answer at the back of the Course Notes. I think it is correct.
Dear John, it seems when calculating ‘-rt’ in example 6 (52,48 minute of the lecture and so on) you’ve multiplied 0,1 (r) by 0,4 instead of 0,25 (t)….
Ooops – you are correct.
Sorry 🙁
Thanks John!
These are very well explained lectures and are a great help. Thank you sir and Open Tuition.
I have a very basic ques.. But its really confusing me. why the value of option is share price – excercise price?
If it was a call option exercisable immediately, then the option gives you the right to buy a share at a fixed price.
So, for example, if the current share price is $4.00 and you could buy an option giving you the right to buy the share at an exercise price of $3.70, then you could buy the share for $3.70 and immediately sell it for $0.30.
Nobody is going to give you that right free! You would be prepared to pay $0.30 for the option. Then you could use it and buy a share for $3.70. You have then spent $4.00 in total and you own a share worth $4.00 🙂
(But of course, that is only if the option were exercisable immediately. In practice the option will be the right to buy a share at a fixed price on a future date, and to get the value of that we need to use all the formulae.)
thank u!
thank you