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Set out what the VaR risk measure is, the assumptions it makes about return distributions and holding periods, and where VaR falls short.

FRM® Part I · Valuation and Risk Models · Measures of Financial Risk · FRM1.VR.1.c

The idea

Value at Risk is a quantile of the loss distribution: the loss that will not be exceeded over a stated holding period with a stated confidence. A one-day 99 percent VaR of USD 5 million means a 1 percent chance of losing more than USD 5 million in a day, under the model's assumptions, and exceedances are not scheduled evenly. Parametric VaR assumes normally distributed returns. VaR = (z × σ − μ) × portfolio value. Here μ and σ are the one-period mean and standard deviation of returns in decimals, so 1.5 percent is 0.015, and z = 1.645 at 95 percent and 2.326 at 99 percent. Longer horizons use the square-root-of-time rule. VaR over T days = VaR over 1 day × √T. This assumes independent returns, constant volatility and an unchanged portfolio. VaR has weaknesses. It says nothing about how bad losses beyond the quantile are, it is unreliable with fat tails, illiquid positions or non-linear payoffs, and it is not guaranteed to be subadditive, so portfolio VaR can exceed the sum of the component VaRs. Exam point: VaR is a threshold, not a worst case.

Formulas

Parametric VaR of a normal portfolio

VaR = (z × σ − μ) × V

z
the standard normal quantile, 1.645 at 95% and 2.326 at 99%
σ
the one-period standard deviation of returns, in decimals
μ
the one-period mean return, in decimals
V
the portfolio value

Assumes normally distributed returns. A positive mean lowers the VaR, so subtract μ from z × σ rather than adding it.

Square-root-of-time scaling of VaR

VaR(T days) = VaR(1 day) × √T

VaR(1 day)
the one-day VaR
T
the horizon in days

Holds only for independent returns, constant volatility and an unchanged portfolio. Multiplying by T instead of √T is the classic error.

A practice question

From the question bank, with the reason each wrong option is wrong.

Which of the following is a recognised limitation of VaR as a risk measure?

  1. It cannot be calculated for a portfolio with more than one position.Why not: VaR is routinely computed for multi-asset portfolios; the limitation lies in tail information and aggregation properties, not in the number of positions.
  2. It can only be estimated for a holding period of exactly one day.Why not: The holding period is a parameter the user chooses; longer horizons are obtained directly or by scaling the one-day VaR.
  3. It always overstates risk when returns are normally distributed.Why not: Under normality VaR is the exact quantile; problems arise when the distribution is fat-tailed, not when it is normal.
  4. It gives no information on the size of losses beyond the VaR quantile. - correct

VaR reports only where the tail begins. Two portfolios with the same VaR can have very different losses beyond it, and VaR is also not subadditive in general.

A harder one

A portfolio is worth USD 50 million. Daily returns are independent and normally distributed with mean zero and standard deviation 1.2%. The 10-day 99% VaR, using the square-root-of-time rule, is closest to:

  1. USD 4.41 million - correct
  2. USD 13.96 millionWhy not: Scales the one-day VaR by 10 instead of by the square root of 10, assuming losses accumulate linearly.
  3. USD 1.40 millionWhy not: Reports the one-day VaR and never scales it to the 10-day horizon.
  4. USD 3.12 millionWhy not: Uses the 95% quantile of 1.645 rather than the 99% quantile of 2.326.

One-day VaR = 2.326 x 0.012 x 50 = USD 1.40 million. Scaling by sqrt(10) = 3.162 gives USD 4.41 million. The rule relies on independent returns and constant volatility.

The hardest tier

A bank's one-day 99% VaR model is correctly calibrated, and exceptions on different days are independent. Over 250 trading days, the probability of observing two or more exceptions is closest to:

  1. 92%Why not: Computes the probability of at least one exception, 1 - 0.99^250, instead of at least two.
  2. 26%Why not: Computes the probability of exactly two exceptions rather than two or more.
  3. 8%Why not: Reports the probability of no exceptions at all, 0.99^250.
  4. 71% - correct

Exceptions are binomial(250, 0.01). P(0) = 0.99^250 = 0.0811; P(1) = 250 x 0.01 x 0.99^249 = 0.2047. So P(2 or more) = 1 - 0.0811 - 0.2047 = 0.714. Even a perfect 99% VaR is exceeded 2.5 times a year on average.

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