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Explain the POT technique, which models only the observations that exceed a high cutoff.

FRM® Part II · Market Risk Measurement and Management · Parametric Approaches (II): Extreme Value · FRM2.MR.3.c

The idea

The peaks-over-threshold (POT) technique chooses a high threshold u and keeps only the observations that exceed it, working with the excesses x − u. By the Pickands-Balkema-de Haan theorem, for a high enough threshold the excesses follow a generalized Pareto (GP) distribution. The tail probability is then estimated as the fraction of observations above u, written Nu ÷ n, times the GP survival function of the excess. Choosing u is a bias-variance trade-off. Too low and the GP approximation is poor, giving bias. Too high and few exceedances remain, giving noisy estimates. A mean excess plot, the average excess against the threshold, should be roughly linear above a good threshold. An upward slope signals a fat tail with ξ > 0. The usual mistake is treating the threshold as harmless, when it drives the answer.

Formulas

Mean excess function of a generalised Pareto tail

e(u) = (β + ξ × u) ÷ (1 − ξ)

β
the GP scale parameter
ξ
the GP shape parameter, below 1
u
the threshold

Linear in u, with slope ξ ÷ (1 − ξ), so an upward-sloping mean excess plot signals ξ > 0. The expected loss beyond u is u + e(u).

Sample mean excess over a threshold

e(u) = Σ(xᵢ − u) ÷ Nu

xᵢ
each observation above the threshold
u
the threshold
Nu
the number of exceedances

Averages the excesses, not the raw losses. Plotted against u, it should be roughly linear above a good threshold.

Approximate standard error of the POT shape estimate

SE(ξ) = (1 + ξ) ÷ √Nu

ξ
the shape parameter
Nu
the number of exceedances

Maximum likelihood fit. It falls with √Nu, so a very high threshold leaving few exceedances gives an imprecise shape.

A practice question

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

Which description of the peaks-over-threshold (POT) technique is correct?

  1. Split the sample into blocks and fit a distribution to the maximum of each block.Why not: Describes the block maxima (GEV) approach, not POT.
  2. Choose a high threshold and fit a distribution to the amounts by which the observations exceed it. - correct
  3. Fit a normal distribution to all observations above the threshold using their mean and standard deviation.Why not: POT uses the generalized Pareto distribution for excesses, not a normal fit.
  4. Delete observations above the threshold as outliers and fit the distribution to the rest.Why not: Reverses the logic: POT keeps the extreme observations and discards the rest.

POT keeps only the exceedances, forms the excesses x - u, and fits a generalized Pareto distribution to them. Everything below the threshold is used only to estimate the exceedance probability.

A harder one

A risk manager plots the sample mean excess against the threshold for a loss series. Above a threshold of 2 percent the plot is roughly a straight line sloping upward. What does this suggest?

  1. The excesses are exponential, with xi = 0, so the tail is thin.Why not: An exponential tail gives a flat mean excess plot, not an upward slope.
  2. The tail is bounded, with xi < 0, because the mean excess is increasing.Why not: A bounded tail (xi < 0) gives a downward-sloping line.
  3. The threshold is too high and should be lowered until the plot becomes horizontal.Why not: Linearity above u is the desired sign of a GP tail; a horizontal line would mean xi = 0, not a better threshold.
  4. A GP fit above 2 percent is reasonable, with xi > 0 (heavy tail). - correct

For a GP tail the mean excess is linear in the threshold, e(u) = (beta + xi u) / (1 - xi), so a roughly linear plot supports a GP fit, and the sign of the slope is the sign of xi. An upward slope means xi > 0.

The hardest tier

With 2,000 observations and a true shape xi = 0.2, the standard error of the maximum likelihood estimate of xi in a POT fit is approximately (1 + xi) / sqrt(Nu), where Nu is the number of exceedances. A 90th-percentile threshold gives Nu = 200 and a 99th-percentile threshold gives Nu = 20. What is the approximate standard error at the 99th-percentile threshold, and what does it illustrate?

  1. 0.085, showing that a very high threshold gives a precise shape estimateWhy not: Uses the 200 exceedances of the 90th-percentile threshold instead of the 20 at the 99th.
  2. 0.27, showing that a very high threshold leaves too few exceedances - correct
  3. 0.060, showing that the error falls in proportion to the number of exceedancesWhy not: Divides by Nu rather than sqrt(Nu); the standard error falls with the square root of the count.
  4. 0.072, showing that a very high threshold leaves too few exceedancesWhy not: Reports the variance, (1 + xi)² / Nu, rather than the standard error, which is its square root.

Standard error = 1.2 / sqrt(20) = 0.268, against 0.085 with 200 exceedances: with the threshold at 99 percent the shape is estimated roughly three times less precisely. This is the variance side of the bias-variance trade-off.

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