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Question
James has estimated an annual standard deviation of $750,000 on one of its projects, based on a normal distribution of returns. The average annual return is $2,400,000.
Estimate the value at risk (VAR) at a 95% confidence level for one year and over the project鈥檚 life of six years.
Answer
For 95% confidence, VAR is 1.645 standard deviations from the mean.
i.e. for one year = 1.645 x $750,000 = $1,233,750
This means that James can be 95% certain that the returns will be $1,166,250聽or more every year ($2,400,000 – $1,233,750).
Over six years, the total standard deviation is square root of ( 6 x ($750,000 squared)) = $1,837,117
Therefore the VAR = 1.645 x 1,837,117 = $3,022,057
This means that James can be 95% certain that the returns will be $11,377,943 or more in total over the six year period ($14,400,000 – $3,022,057).
Is it OK to just take (VAR for one year at 95% confidence) and multiply its results by 6? I would have thought that the 95% applies to one year, not six , as there are more ways things can go wrong over a period of six years …..
I can see how this simplification would work but am not sure the simplification is valid over a long term. I suppose that taking the p(tail) <= 0.05 for one year and 'multiplying up' would include the possibility that tail losses in, say, two years might be counteracted by gains above mean in other years, but I still feel this is not as accurate as it might be.
i.e., having a 95% confidence that returns are more than $1,166,250 for *one* year, is not the same as a 95% confidence that they will be more than $1,166,250 *every* year, for a large number of years.
Surely?
Something has gone wrong with the typing – I will have it corrected (although the lecture itself explains it properly).
Over 6 years, the std devn = sq root (6 x (750,000^2)). (We add variances, which is the square of the std deviation)
The arithmetic from then on is correct and since the std devn over 6 years is a lot greater than that for 1 year, it does fit in with what you say.
Nicely explained as usual…Thanks
You are welcome 馃檪
Great work, understood this first time after failing to understand from virtually everything… 馃檪
I am pleased it helped you 馃檪
Thank you. Your explanation helps me understand the questions and answer. 馃榾
You are welcome – I am very pleased that it helped you 馃檪
Thank you Sir for this lecture on VAR. I was struggling to understand VAR. Now I understood.. 馃檪 馃檪
I am pleased that you found it useful 馃檪
Please, where are the lectures on Adjusted Present Value and Type I-III acquisitions and free cash flows.? i noticed the videos start from chapter 7. i really love the lectures.
There are lectures on Adjusted Present Values. There is no lecture on the types of acquisitions because the techniques involved are dealt with in earlier lectures.
Calculating free cash flows is the same as calculating the net cash flows for investment appraisal and there are lectures on this.
The first six chapters of the lecture notes do not involve calculations and are more background reading. The are for you to read yourself and do not warrant lectures.
Oh ok. Thank you very much. I’m very grateful.
Oh, I understand the logic behind now
Great 馃檪
Perfectly explained. Spent hours trying to grasp from BPP. 20 minutes watching this and I get it!
Great 馃檪
Hi sir,
Excellent lecture. My question though is would the result be different if instead of calculating the VaR for one year you used two weeks for example?
Yes it would because you would need to use the 2 week standard deviation.
Thank you.
I am not sure if this is the right place to post- but I would like to thank Sir Moffat for his brilliant way of explaining things. I passed my P4 June 15 sitting in my first attempt and no base of F9 (as I got exemption for that) whatsoever. His lectures cover good base and this is what you need to pass – to really understand the core areas. Sir Moffat is always quick with his responses which surprisingly you won’t find with your paid classes sometimes. I did not take any other classes for this module and used LSBF book for self study. I would also like to give credit to my study buddy whom I found here in opentuition. Having a study buddy really helps. We did our revision over skype (it works). So good job Sir Moffat for this brilliant website and all your hard work.
Many thanks
Dee
P.S. In case you are wondering, my study buddy passed as well.
Thanks a lot for your comments, and congratulations on passing 馃檪
The way you explain sir is really wonderful. You make the topic so easy.
No better explanation, Thanks John!
Thank you 馃檪
Hello Sir!
Please tell me if the following statement is correct.
Given the average return of 2.4 M and the standard deviation of 0,75 M it means that 2.4/0,75 = 3.2 . looking for 3.2 in the standard normal distribution table we will find the probability to have a positive outcome. Is this correct ?
thank you!
your work is highly appreciated by all of us!
Well….yes (if by positive you means greater than zero), but……
1 You would add 0.5 to the table figure because the table figure gives the probability of being less then the average – there is a 0.5 probability of being more than the average
2 No distribution like this will be perfectly normally distributed (which is one of the practical problems with VaR in practice). If it were truly normal then it could be negative but in reality that is probably impossible – in which case the probability of it being positive would be 100%
3 The tables you are given don’t go so far as 3.2 馃檪
4 This is not the way VaR is used in practice and so you would never be asked for the probability of it b being positive in the exam – you will only be asked the sort of thing I do in the lecture (e.g. 95% confidence)
Thank you, Sir!
I will come back to thank you once again when the results will be released.
I love you John 馃檪
Thank you Sir Moffat ,i am enlightened,i hope this third sitting with OT i will be able to make it.
thank you for explaining this, did the past paper question and had no idea how they got the answer. much thanks.
Very well explained. Thanks for teaching us!
Great lecture.thanx John.so the VAR is1,166,250 for one year and 11,377,943 for six years.just seeking clarification
Yes.
great lecture
Hi John,
Great video! Just one small thing, the value is $1,166,250 not $116,625 as mentioned in the description. (The Video has the right figures)
Thanks 馃檪
John, what if the examiner asks VaR for lets say 6.5 years so we’ll just do Square root of 6.5 * standard deviation?
That’s correct 馃檪
(although I am certain he would ask for a while number of years!)
This is a wonderful lecture explaining how to go about VaR. I’ve been having a lot of trouble being able to understand the concept. This lecture has simplified the VaR concept and I’m actually looking forward to having this question in the June 2015 examination.
Thanks OpenTuition!