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Characterise the difficulties an institution may meet in computing confidence intervals around VaR.

FRM® Part II · Market Risk Measurement and Management · Validating Bank Holding Companies’ Value-at-Risk Models for Market Risk · FRM2.MR.6.c

The idea

A VaR figure is a statistic estimated from a finite sample, so it carries estimation error and a confidence interval should ideally accompany it. Let p be the tail probability, n the sample size and f(q) the density at the quantile. For an empirical quantile the standard error is then roughly given by SE = √(p × (1 − p) ÷ n) ÷ f(q). Far in the tail few observations support the estimate, so the interval is wide, and it narrows only with the square root of the sample size. Analytical intervals rely on distributional assumptions. Volatility clustering and overlapping returns shrink the effective sample, and parameters change over time. Bootstrap or Monte Carlo intervals are computationally heavy and still reflect model risk. A higher confidence level gives a wider interval. For the exam, remember that VaR is a point estimate with sampling error.

Formulas

Standard error of an empirical quantile

SE = √(p × (1 − p) ÷ n) ÷ f(q)

p
the tail probability
n
the number of independent observations
f(q)
the density at the quantile

Assumes iid observations; volatility clustering or overlapping returns shrink the effective sample and widen the true interval. In standard units, multiply by σ × V for money.

A practice question

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

What is a main difficulty in computing a confidence interval for a 99% historical-simulation VaR?

  1. The tail contains too many observations, which makes the quantile unstable.Why not: Reverses the problem: far in the tail there are too few observations, not too many.
  2. The interval can be computed only if the portfolio contains no options.Why not: Wrong: the issue is data in the tail, not the instrument mix; intervals can be bootstrapped for any portfolio.
  3. Few observations fall in the tail, so the estimated quantile is imprecise and the interval is wide. - correct
  4. Historical simulation assumes normality, so the interval is exact and requires no judgement.Why not: Historical simulation does not assume normality, and its intervals are not exact.

At 99% only about 1% of observations lie beyond VaR, so the quantile estimate rests on few points and the standard error is large, particularly for short windows.

A harder one

Daily returns are assumed independent and standard normal. A validator estimates the 99% quantile (z = 2.326) from n = 500 observations. The standard error of an empirical quantile is sqrt(p(1 - p)/n) divided by the density at the quantile, and the standard normal density at 2.326 is 0.0267. The standard error, in units of the return standard deviation, is closest to:

  1. 0.0045Why not: Omits the division by the density at the quantile, using only sqrt(p(1 - p)/n).
  2. 0.0112Why not: Divides by the density at the mean (0.399) rather than at the 99% quantile (0.0267).
  3. 0.0007Why not: Omits the square root, using p(1 - p)/n = 0.0000198 and then dividing by the density.
  4. 0.167 - correct

sqrt(0.99 x 0.01 / 500) = 0.00445. Dividing by f(2.326) = 0.02665 gives 0.167. The thin tail density is why the standard error is so large relative to the quantile of 2.326 (about 7%).

The hardest tier

A parametric normal model gives a 99% one-day VaR of USD 10.47 million for a USD 300 million portfolio with daily volatility of 1.5%. Using 500 independent observations, the standard error of the 99% quantile is 0.167 standard deviations. The validator builds an approximate 95% confidence interval (plus or minus 1.96 standard errors) for the VaR. It is closest to:

  1. USD 10.1 million to USD 10.8 millionWhy not: Applies 1.96 x 0.167 to the return units without scaling by sigma x value (USD 4.5 million), so the interval is far too narrow.
  2. USD 9.7 million to USD 11.2 millionWhy not: Uses one standard error instead of 1.96, which gives roughly a 68% interval, not 95%.
  3. USD 9.8 million to USD 11.1 millionWhy not: Uses the standard error of the volatility estimate alone (z x sigma / sqrt(2n)), ignoring quantile estimation error and giving too narrow an interval.
  4. USD 9.0 million to USD 11.9 million - correct

Standard error in USD = 0.167 x 1.5% x 300 = 0.751 million. 1.96 x 0.751 = 1.47 million. Interval = 10.47 plus or minus 1.47 = 9.0 to 11.9 million, roughly 14% of VaR either side.

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