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I am currently using your videos and notes for CIMA P2 but the video is not yet available for Value at Risk. Please could you confirm the answer to the following question provided in the notes;
Example 4
A bank has estimated that the expected value of its portfolio in 2 weeks time will be $100 million,
with a standard deviation of $5 millionEstimate the value at risk at a 95% confidence level.
Thank you very much for this great study resource.
Kind Regards,
Lauren
Hi Lauren
Thanks for your question & glad you are enjoying the Open Tuition resources.So a 95% confidence level will identify the reduced value of the portfolio that has a 5% chance of occurring.
From the normal distribution tables, 1.65 is the normal distribution value for a one-tailed 5% probability level. Since the value is below the mean, – 1.65 will be needed.
z = (x – ?)/?
(x – 100)/5 = –1.65
x = (–1.65 × 5) + 100 = 91.75Thus there is a 5% probability that the portfolio value will fall to 91.75million or below.
Just checking thats all you needed…. happy to help but sometimes can be better to use these forums (by providing the question AND the answer – which is presumably in your text book) THEN to pinpoint to us the part of the answer you don’t understand or follow …. this is how a tutor can add value here the most…
However, I hope the above has helped a little….
Many Thanks
CathHi Cath,
Thank you very much for your help.
The answer appears to be missing for this question.
Kind Regards,
Lauren
Hi Cath,
You say the value of 1.65 from the normal distribution table. Is that a table that would be provided in the exam? How did you get this from a standard z table if that is what you used.
Do you know when the lecture on this topic will be put up?
Many thanks.
Hi – yes apologies -ill ask about the Value at Risk video..
The tables (Z scores) are given in the CIMA exam.
The relate to a normal distribution curve ( upside down bell shape).
The middle point is our MEAN average figure figureEither side of the curve is 50% area.
A 5% confidence level means that on the left of the mean(below the mean) – is 45%If we obtain the Z score tables …. and search the data in the tables to find a figure as close as possible to 45%
Then you will see somewhere in the middle there is a figure of 0.4505 ( is pretty close)….
I then follow along the row & read at the column side which shows 1.65…. approximate number of standard deviations from the mean.Once we know this figure – we add it into the formulae as above
Hope this helps
Cath
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