FRM Exam Part II · Parametric Approaches (II): Extreme Value

EVT-Based VaR and Expected Shortfall Calculation

Q: What is the EVT VaR formula for the generalized Pareto distribution?

VaR(p) = u + (β ÷ ξ) × { [(n ÷ Nu)(1 − p)]^(−ξ) − 1 }. Here u is the threshold, β the scale, ξ the tail index, n the sample size and Nu the number of exceedances. It applies for ξ ≠ 0 and when VaR lies above u.

Q: How do you calculate expected shortfall using extreme value theory?

First compute the EVT VaR. Then ES = (VaR + β − ξu) ÷ (1 − ξ), which requires ξ < 1. The result is always above VaR when ξ is positive.

Q: Why is EVT VaR higher than normal VaR at 99.9%?

Financial losses usually have fat tails, so extreme losses are more likely than the normal predicts. EVT models the tail directly with a positive ξ, so its quantile grows faster at very high confidence levels.

Q: What is the limitation of EVT-based VaR?

The tail parameters are estimated from few data points, so estimates are sensitive to the threshold choice and carry wide confidence intervals. Extrapolating far beyond the data magnifies this uncertainty.

Updated 11 October 2026 · Fact-checked

EVT-based VaR uses a fitted tail model, not the whole distribution, to estimate loss at very high confidence levels. With peaks over threshold, you fit a generalized Pareto distribution above a threshold u, then apply a closed-form VaR formula. Expected shortfall follows from VaR and the tail parameters.

Understand EVT-Based VaR and Expected Shortfall Calculation

Normal VaR fits the centre of the data well but often understates losses at 99.9% because real returns have fat tails. Extreme value theory (EVT) models only the tail, so it is built for rare, large losses.

There are two routes. The block maxima route fits a generalized extreme value (GEV) distribution to the largest loss in each block, such as each month or year. The peaks over threshold (POT) route picks a high threshold u and fits a generalized Pareto distribution (GPD) to the excesses over u. POT uses data more efficiently and is the usual basis for VaR and ES formulas.

The GPD has a scale parameter β > 0 and a tail index ξ. If ξ > 0, the tail is heavy, like a power law, and most financial losses fall here. If ξ = 0, the tail is exponential. A larger ξ means a fatter tail and a faster growth of VaR as confidence rises.

Once you have u, β, ξ, the number of observations n and the number of exceedances Nu, you can extrapolate beyond the sample. This is the key advantage: you can estimate a 99.9% VaR from data that holds few or no observations that extreme. The cost is parameter uncertainty, which grows with the extrapolation.

Expected shortfall is the average loss given that loss exceeds VaR. For a GPD tail with ξ < 1, it has a simple formula. If ξ ≥ 1, the mean is infinite and ES does not exist.

Key formulas to remember

GPD tail estimator (POT): P(X > x) ≈ (Nu ÷ n) × [1 + ξ(x − u) ÷ β]^(−1/ξ), for x > u; Nu is the number of losses above u; n is the total sample size. Valid for ξ ≠ 0.
EVT VaR (ξ ≠ 0): VaR(p) = u + (β ÷ ξ) × { [(n ÷ Nu) × (1 − p)]^(−ξ) − 1 }; p is the confidence level, e.g. 0.999. Use only when VaR(p) lies above u.
EVT VaR (ξ = 0): VaR(p) = u − β × ln[(n ÷ Nu) × (1 − p)]; Exponential tail limit.
EVT expected shortfall: ES(p) = VaR(p) ÷ (1 − ξ) + (β − ξu) ÷ (1 − ξ), for ξ < 1; Equivalent to ES = [VaR + β − ξu] ÷ (1 − ξ). If ξ ≥ 1, ES is infinite.
Normal VaR for comparison: VaR(p) = μ + σ × zp; z0.99 = 2.326; z0.999 = 3.090 (loss with μ and σ for the loss distribution).
Normal expected shortfall: ES(p) = μ + σ × φ(zp) ÷ (1 − p); φ is the standard normal density.

How to solve EVT-Based VaR and Expected Shortfall Calculation questions

Use this order for any EVT VaR or ES question. Work in losses expressed as positive numbers.

1Identify the method: POT with GPD parameters (u, β, ξ) or block maxima with GEV. Most VaR/ES calculations use GPD.
2Write down n, Nu, u, β, ξ and the confidence level p. Compute 1 − p and n ÷ Nu.
3Compute the term (n ÷ Nu) × (1 − p). This is the tail probability scaled to the exceedance distribution.
4Raise it to the power −ξ, subtract 1, multiply by β ÷ ξ and add u. This gives VaR.
5Check that VaR is above u. If not, the formula does not apply.
6For ES, check ξ < 1. Then compute ES = (VaR + β − ξu) ÷ (1 − ξ).
7If asked to compare with normal, compute μ + σz and the normal ES. Explain that EVT is larger when ξ > 0 because of the fat tail.

Quickest way: Plug-in shortcut for GPD VaR and ES

When to use it: Use when all five inputs are given and the question wants only numbers or a ranking of results.

Compute q = (n ÷ Nu) × (1 − p) first. Note whether it is much less than 1.
Compute q^(−ξ). For ξ = 0.2 and q = 0.02, q^(−0.2) = e^(0.2 × 3.912) ≈ 2.19.
VaR = u + (β ÷ ξ)(q^(−ξ) − 1).
ES = VaR ÷ (1 − ξ) + (β − ξu) ÷ (1 − ξ).
Sanity check: ES must exceed VaR, and VaR must exceed u. Eliminate options that violate this.

Common mistakes in EVT-Based VaR and Expected Shortfall Calculation

Using 1 − p alone inside the bracket instead of (n ÷ Nu) × (1 − p).
The GPD fits only the excesses, so students forget the scaling for the share of data in the tail.
Fix: Always compute n ÷ Nu first and multiply by 1 − p.
Using a positive exponent, q^(+ξ), or forgetting the minus sign.
The formula is remembered loosely.
Fix: The exponent is −ξ. Since q < 1, q^(−ξ) > 1, so VaR rises above u.
Computing ES as VaR ÷ (1 − ξ) only.
The second term (β − ξu) ÷ (1 − ξ) is dropped.
Fix: Include it. It equals zero only when β = ξu.
Applying the ES formula when ξ ≥ 1.
Students plug in without checking the condition.
Fix: State that the tail mean is infinite for ξ ≥ 1, so ES does not exist.
Treating the threshold u as the VaR, or computing excess VaR and not adding u back.
The GPD models excesses over u, not total losses.
Fix: Add u to the excess quantile to get VaR in loss terms.
Assuming EVT VaR is always above normal VaR.
Fat tails are the usual story.
Fix: For ξ > 0 and very high p, EVT is typically larger. At moderate levels such as 95% the two can be close or reversed, so compute both.

Worked examples

Example 1

Daily losses: n = 1,000 observations, threshold u = 2.0% (loss), Nu = 50 exceedances. A GPD fit gives ξ = 0.25 and β = 0.5%. Compute the 99.9% EVT VaR and expected shortfall.

Show the solution

n ÷ Nu = 1,000 ÷ 50 = 20. 1 − p = 0.001. So q = 20 × 0.001 = 0.02.
q^(−ξ) = 0.02^(−0.25). ln(0.02) = −3.912, so the exponent gives e^(0.25 × 3.912) = e^0.978 ≈ 2.659.
VaR = 2.0 + (0.5 ÷ 0.25) × (2.659 − 1) = 2.0 + 2 × 1.659 = 2.0 + 3.318 = 5.32%.
ES = [VaR + β − ξu] ÷ (1 − ξ) = [5.32 + 0.5 − 0.25 × 2.0] ÷ 0.75.
Numerator = 5.32 + 0.5 − 0.5 = 5.32. ES = 5.32 ÷ 0.75 = 7.09%.

Answer: 99.9% EVT VaR ≈ 5.32% and expected shortfall ≈ 7.09%.

Example 2

Daily portfolio losses have mean 0 and standard deviation 1.2%. Using a normal model, find the 99.9% VaR and ES. Compare with the EVT VaR of 5.32% and ES of 7.09% from the first example. Comment.

Show the solution

Normal VaR = 0 + 1.2 × 3.090 = 3.708%.
Normal ES = 1.2 × φ(3.090) ÷ 0.001. φ(3.090) = (1 ÷ √(2π)) × e^(−3.090² ÷ 2) = 0.3989 × e^(−4.774) = 0.3989 × 0.008455 ≈ 0.003373.
Normal ES = 1.2 × 0.003373 ÷ 0.001 = 1.2 × 3.373 = 4.05%.
Compare: EVT VaR 5.32% against normal 3.71%, a difference of about 1.61 percentage points. EVT ES 7.09% against normal 4.05%.
The gap is wider for ES than for VaR because ES averages the tail beyond VaR, and a heavy tail (ξ = 0.25) has far larger extreme losses than the normal.

Answer: Normal 99.9% VaR ≈ 3.71% and ES ≈ 4.05%. EVT gives higher figures (5.32% and 7.09%), showing that the normal model understates tail risk when losses are heavy-tailed.

Exam tips

Check whether the question gives n and Nu. If it gives only the tail fraction Nu ÷ n, use its inverse in the bracket.
Always test the sign and size: VaR > u and ES > VaR. Many wrong options fail this check.
Know the interpretation of ξ: positive means heavy tail, zero means exponential, negative means a finite upper bound.
When a question asks why EVT differs from normal at 99.9%, answer with fat tails, tail-only fitting and extrapolation, plus the caveat of parameter uncertainty.
Remember that ES needs ξ < 1. A question with ξ ≥ 1 is testing this condition.

Practice questions from Parametric Approaches (II): Extreme Value