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gnoii says

Dear Sir,

Calculations are all about deviation, confidence and then look up table, not any concerned from the annual return. So these returns (2,4M or 14,4M) are just for reference and comment, aren’t they?

Thank you very much.

John Moffat says

But you need those figures to then be able to calculate the value at risk for the particular confidence level.

tashu123 says

Hello Sir,

I did try this chapter from the study texts but didn’t understand it at all…they use a sort of formula for VAR which is horrible.

Thanks a lot for such understandable lecture.

May god bless you sir.

John Moffat says

Thank you for your comment ðŸ™‚

chantsv says

Thank you for this lecture, it’s the first time I’m doing SD. Please may I ask when calculating the VAR over the project’s life, why is the average 6 x $2.4m = $14.4m? As soon as I see “over the life” I think annuity and discounting. Although I realise there is no discount factor in the question, I’m thinking what if part of a bigger question? Many thanks

John Moffat says

I don’t understand why you think discounting – Value at Risk has nothing to do with discounting, and the question will specifically ask for the VaR if it is required.

We multiply by 6 because the question states that the project has a life of 6 years (I assume that you have downloaded the free lecture notes).

chantsv says

Many thanks.

chantsv says

Please let me know if we will ever need to calculate the standard deviation or will it always be given. I have seen in some textbook that there is a very long way to calculate it.

zhixiang85 says

Dear John,

Example 3, I know how to read the distribution table, but how did you get the 1.50 and 2.58? Thank you

John Moffat says

As I say in the lecture, I am only using 1.50 and 2.58 to illustrate how to use the tables. I could have used any figures – all I am doing is explaining how to use the tables!

claudia1 says

Ok Sir, you are perfectly correct of course. I forgot the part where 50 is deducted from 99. Thank you so much for the lecture. Understood.

claudia1 says

Hello Sir, half of 99 is 49.5, so should we be looking at .495 in the table?…just making sure.

bucheeri88 says

Hi Sir, I have a question relating to the normal distribution. Back in my earlier studies of normal distribution, when there is a 95% confidence level, we used to split the 5% (2.5% on each side). Same with a 97.5% confidence level (1.25% on each side). How come we only take the whole 5% on the negative side?

John Moffat says

Because we are only concerned about the amount being lower, not higher.

(You are referring to what is called a two-tail test, where the variable may be higher or lower. This is a one-tailed test.)

adlin says

Understood thank you

John Moffat says

You are welcome ðŸ™‚

adlin says

I m confused as to what table he is referring to…?how did he get 1.645 value ?

John Moffat says

The normal distribution tables that are given to you in the exam (and that are printed in our free lecture notes). I do explain this in the lecture!

3403263nya says

Very understandable. Thank you.

said1988 says

Very happy with this lecture! Thanks a lot

John Moffat says

Thank you for your comment ðŸ™‚

SamTallroth says

I just want to say thankyou, After watching the BPP lectures and reading the books I was confused still on Std Deviations. Your lecture has made it so much clearer.

John Moffat says

Thank you for for your comment ðŸ™‚

herafatima says

Hi John, the concept has been well explained in the lecture, the assumption is the spread is a normal distribution and hence the graph is symmetrical and hence there is a 50% chance of the return being higher or lower than the average return. Our solution of the distance is based on this assumption.

In the exam, can they ask us to calculate the VAR, where the spread is asymmetrical?

John Moffat says

No, that is not possible. VaR always (in real life as well as in exams) assumes a normal distribution (which is symmetrical).

(That is one of the scary things about banks using it in real life (which they do!). Partly it is not necessarily normally distributed, but also even if it is and there is only a 1% chance of there being problems, that is 1 year in 100 and that 1 year could be next year ðŸ™‚ )

herafatima says

Thanks, it is huge relief to know that. ðŸ™‚

John Moffat says

You are welcome ðŸ™‚