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# VaR for Equally likely outcomes

#### nag_san

##### New Member
Hi, The following is an excerpt from 2020 FRM Level 1 Valuation and Risk Models Chapter 1 Measures of Financial Risks. Can someone explain if there is a formula that can be used to calculate VaR for a given confidence level when there is a range of possible equally likely outcomes (Note: the learning objective of this chapter doesn't necessarily state "Calculate VaR" but the below example is not clear in terms of arriving at the answer). For example, one can also say the probability that the loss will lie between 18.5 and 19 is 0.5/50 = 1%. I guess my question is why in this example, they've chosen 20. Is this because the maximum loss is 20? (Note: I could not find a similar thread but if it's part of an existing thread, feel free to move it to that thread please)

As a second simple example, assume the result of an investment with a uniform distribution where all outcomes between a profit of 30 and a loss of 20 are equally likely. In this case, the VaR with a 99% confidence level is 19.5. This is because the probability that the loss will lie between 19.5 and 20 is 0.5/50 = 1%. (Note that we divide the range of losses we are interested in (0.5) by the total range of losses (50) because all outcomes are equally likely.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @nag_san Yes, it's because the lower bound of the (given) uniform distribution is -20.0, and VaR is the quantile associated with a (cumulative) probability such that we need to "start from the bottom" to include the entire left tail. In the uniform distribution, this exercise is essentially the same as a (special case of a) linear interpolation. Linear because it's a uniform distribution. In 3.10 of GARP's chapter 3, there is a nifty formula:

"The probability that Y falls into an interval with lower bound l and upper bound u (i.e., Pr(l < Y < u)) is: [min(u, b) - max(l, a)] / (b - a)"

So any u and l solve for a fraction of the overall distance from a to b. If we are "solving" for a x% confident VaR, we simply want:

(1 - confidence%) = (u - a)/(b - a). I think that's the formula you want. e.g.,

• 99% VaR solves for: 1.0% = (u - a)/(b-a), or
• 95% VaR solves for: 5.0% = (u - a)/(b-a)
Because it's uniform, this is really just a linear interpolation: we want the upper (u) that is the quantile of the uniform distribution such that (u-a) is 1.0% or 5.0% of the entire distribution, from a to b. We "start at a" because this is VaR and we are including the "entire loss tail" from the "worst" (a = -20) to some quantile (u). We are not looking for (eg)
• 1.0% = (u - l)/(b-a) where l > a ...
... because that's the probability of landing somewhere inside the distribution, be we just want the quantile that demarcates "worst 1%" tail. Here the worst tail starts at the extreme of a = -20 and goes to the quantile u = -19.5, which is 1% of the entire distance form a = -20 to b = +30.

So if α% = (u-a)/(b-a) , then u = α%*(b-a) + a. For example, if α% = 1%, then u = 1%*[30- (-20)] + -20 = -19.5, which is expressed in L/P terms as 19.5 (a mathematical loss of -19.5).

So this is just (a special case of) linear interpolation. It's interesting. What if the same distribution were given in L(+)/P(-) or L/P format? Then the distribution is from profit of a = -30 to loss of b = +20. In which case, it's good to see that now a = -30 and b = +20, but %Confident VaR wants:

α% = (b - l)/(b-a), and
l = b - α%*(b-a); e.g.,
l = 20 - 1%*[20 - (-30)] = 20 - 0.5 = 19.5, but it's in L/P format so the +19.5 requires no "translation" as it already represents a loss of 19.5 and the distance from 19.5 to 20.0 is, again, 1%, of the entire distribution. I do find these applications interesting because they allow to tease out exactly what we mean by VaR. I hope that helps!