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- November 21, 2010 at 4:29 pm #46105
Hi,

Could anyone help please?

1) I was wondering where the examinator took the amount 1.645 from when calculating in Pilot paper Q3 projectVaR element: N(0.95) = 1.645 ?Many thanks and good luck with studying!!!

November 21, 2010 at 7:08 pm #71036I have the same doubt. I have tryong to find it out, but till date in vain. not only this question many others also I did not understand how the value of “N”is derived at

November 22, 2010 at 6:41 am #71037Let me explain with u what i understand about VaR, and pls correct me if I have any mistake

1- Normal distribution of 95% is 1.645, this figure is taken from Standard normal distribution table

2- VaR is the maximum amount that can be expected to lose over a period of time with 95%/99% probability. As my understanding, VaR is the maximum amount which could

**add or less**to a cost or return at a given level of probability over a given period of time. It comes from the symmetric characteristic of the curve which represents probability of profit and loss.For this VaR questions, i think, we just need to remember two important and simple figures i.e 99% and 95% probabilities. For example 95% probability means only 95% to get more money than the certain level and 5% probability to lose money. And the Normal distribution of 95% and 99% probability is 1.645 and 2.335.

Assumption that we have to borrow of $1m in next month for 6 months time, the cost of borrowing will be LIBOR+1%, the volatility of LIBOR is 1.5% per annum. Determine total interest expenses for this contract with 95% confidence if LIBOR

is 5% at current time.Note that Normal distribution of 95% is 1.645, this figure is taken from Standard normal distribution table

Current interest expenses = $1m*(5%+1%)*6/12= $30,000

Maximum loss for 95% per 6 months = $1m*1.645*1.5%*(6/12)^(1/2)=$17,448

Actual interest expenses= 30,000+17,448= $47,448That means there is 5% that actual interest will be higher than $47,448 and 95% that the actual interest will be less than $47,448

November 22, 2010 at 10:52 am #71038Thanks for explaining, however still do not understand how you read these values from Standard normal distribution table (rows/columns)…

November 22, 2010 at 7:55 pm #71039AnonymousInactive- Topics: 0
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Haven’t attempted that question, and need to revise VAR, but good thing is, I remember how to get 1.645 for 95%

We’re used to finding the numbers in the table by using the numbers in rows and columns for option pricing. But this time, you have to find row and column numbers by using the numbers in the table.

Now the numbers in the distribution table are probabilities. But the probabilities are till 50%. As you can see it starts from 0.0000 and ends at 0.4990 which is 49.9%.

So we deduct 95%-50%= 45%(this is something you can find in the table)

Now look where this 45% or 0.45 lies in the table.

As you can see, you can’t find it, but you can find the closest probabilities to it, which are 0.4495 and 0.4505 meaning 44.95% and 45.05%. This means 0.45 is between them.

Now move your eyes up, to the rows and see where 0.4495 and 0.4505 are located. They’re in row 0.04 and 0.05.

Now as you know 45% is between 0.4495 and 0.4505, you can find out its row location too by adding 0.04 and 0.05 and dividing it by 2. You’ll get 0.045.

Now you need 45% column’s location. This is 1.6 since 0.4495 and 0.4505 are also there and 0.45 is located between them.

So 45% column location is 1.6 and its row location is 0.045. Add the column and row’s location together and you get 1.645 🙂

If you use the table to find 1.645 or 1.65 if you round it off, you’ll get 0.4505 and when you add 0.5, like we’re normally used to doing, you get…0.9505 which none other than 95% 🙂

November 23, 2010 at 9:39 am #71041What a detailed answer, couldn´t be more detailed! 🙂 Somehow I didn´t figure out that we have to deduct 50 first, now it´s clear! Thanks a lot Cyma!

November 23, 2010 at 8:04 pm #71042Thanks a lot Cyma.

Well explianed and well understood.

All the very best for your papers - AuthorPosts

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