Obtain estimates of coherent risk measures by resampling the data in a bootstrapped historical simulation.

FRM® Part II · Market Risk Measurement and Management · Non-Parametric Approaches · FRM2.MR.2.a

The idea

A bootstrapped historical simulation treats the sample of n historical losses as the population and draws new samples from it. Each resample is formed by drawing n observations at random with replacement, so some losses appear several times and others not at all. For every resample you compute the risk measure, and the bootstrap estimate is the average over many resamples. This works for spectral risk measures, a coherent class: weighted averages of the losses ordered from best to worst, with non-negative weights that sum to 1 and never fall as losses get worse. Expected shortfall, the equally weighted average of the tail losses, is the simplest case. VaR is not coherent because it can violate subadditivity. The spread of the resampled estimates gives a standard error and a confidence interval. Bootstrapping measures and improves precision but adds no information. It cannot generate losses worse than those in the sample.

Formulas

Bootstrapped estimate of a risk measure

bootstrap ES = Σ ESᵢ ÷ B

ESᵢ: the expected shortfall computed on resample i
B: the number of resamples

Each resample draws n observations with replacement. The standard deviation of the ESᵢ, not that divided by √B, is the standard error of the estimator.

Chance that a given observation appears in a bootstrap resample

P(drawn at least once) = 1 − (1 − 1 ÷ n)ⁿ

n: the sample size, equal to the resample size

Sampling with replacement. For 250 observations it is about 63.3%, so each resample omits about a third of the distinct observations.

A practice question

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

In a bootstrapped historical simulation, how is each resampled data set formed?

By drawing observations at random without replacement, so that every original observation appears exactly onceWhy not: Sampling without replacement only reorders the sample and returns the same estimate every time, so there is nothing to average.
By drawing observations from the original sample at random with replacement, up to the original sample size - correct
By simulating new returns from a normal distribution fitted to the sample mean and standard deviationWhy not: That is parametric Monte Carlo simulation, not a bootstrap, which resamples the actual historical observations.
By selecting recent observations with a higher probability using an exponentially declining weightWhy not: That describes age-weighted historical simulation, a separate refinement, not the resampling step of the bootstrap.

The bootstrap draws n observations from the n historical losses with replacement, so some appear repeatedly and others not at all. The risk measure is computed on each resample and the results are averaged.

A harder one

A risk manager bootstraps the 90% expected shortfall from 20 daily losses. In each resample the ES is the average of the two largest losses (USD million). The three resamples have tail losses of 3.2 and 2.6, then 3.2 and 3.2, then 2.8 and 2.2. The bootstrapped ES estimate is closest to:

USD 2.87 million - correct
USD 3.20 millionWhy not: Takes the largest resampled loss rather than averaging the resampled ES estimates.
USD 2.67 millionWhy not: Averages the second-largest loss of each resample, which is a 90% VaR-type quantile rather than the tail average.
USD 2.70 millionWhy not: Discards the second resample because its loss is repeated, but repeated draws are a normal feature of bootstrapping.

ES by resample: (3.2+2.6)/2 = 2.90, (3.2+3.2)/2 = 3.20, (2.8+2.2)/2 = 2.50. The bootstrapped estimate is the mean of these: (2.90+3.20+2.50)/3 = 2.87.

The hardest tier

A risk manager has 500 daily losses (USD million). The five worst, in ascending order, are 3.0, 3.4, 3.9, 4.6 and 5.5. A coherent risk measure is estimated by putting weights of 0.10, 0.10, 0.20, 0.25 and 0.35 on these five losses respectively (same order, summing to 1) and zero weight on all other observations. The estimate is closest to:

USD 3.69 millionWhy not: Applies the weights in reverse order, giving the smallest of the five losses the largest weight.
USD 4.08 millionWhy not: Uses equal weights of 0.20, which is the ordinary expected shortfall, not the risk-averse weighting given.
USD 3.00 millionWhy not: Reports the smallest of the five tail losses (a quantile-type VaR) instead of a weighted average of them.
USD 4.50 million - correct

Weighted sum: 0.10x3.0 + 0.10x3.4 + 0.20x3.9 + 0.25x4.6 + 0.35x5.5 = 0.30 + 0.34 + 0.78 + 1.15 + 1.925 = 4.495, about 4.50. The weights rise with severity, so the result exceeds the equal-weight ES of 4.08.

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