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FRM Exam Part I · Measures of Financial Risk

Estimating VaR and ES Quantiles and Their Errors

Updated 11 October 2026 · Fact-checked

Estimating VaR means reading a quantile from sample data using order statistics. Expected shortfall is the average of losses beyond that quantile, or of VaRs at equally spaced tail levels. The standard error of a quantile is √[p(1−p)/n] ÷ f(q), so it grows deeper in the tail and shrinks with more data. A QQ-plot checks the distribution.

Understand Estimating VaR and ES Quantiles and Errors

VaR is a quantile of the loss distribution. If you have a sample of n losses, you sort them from smallest to largest. These sorted values are the order statistics. The VaR at confidence level p is the loss that sits at position n × p in that sorted list. When n × p is not a whole number, you round or interpolate. Follow whatever convention the question states.

A sample quantile is only an estimate. A different sample gives a different answer. The standard error measures how much the estimate would move from sample to sample. It depends on three things: the tail probability p, the sample size n, and the density f(q) of the losses at the quantile. Where few observations sit near the quantile, f(q) is small and the estimate is noisy. That is why far-tail VaR is estimated badly. More data helps, but only at a rate of 1/√n.

Expected shortfall (ES) is the average loss given that the loss exceeds VaR. You can estimate it from a sample by averaging the losses beyond the VaR quantile. You can also average the VaRs at k equally spaced tail levels, α + (1−α) × i ÷ (k+1) for i = 1 to k. More levels give a better approximation. A separate variant takes the VaR at the midpoint of each of k equal slices, α + (1−α) × (2i−1) ÷ (2k). The two variants give different levels, so follow the one the question uses. ES is also an estimate with error, and it is often noisier than VaR at the same level, especially for fat-tailed data, because it depends on the extreme tail.

A QQ-plot plots the quantiles of your sample against the quantiles of a reference distribution, usually the normal. If the sample has the same shape as the reference, the points fall near a straight line. A line that does not pass through the origin with slope one points to a difference in location (intercept) or scale (slope) between the sample and the reference. If the ends curve away from the line, the tails are fatter or thinner than the reference. This matters because a fat-tailed sample makes a normal-based VaR too low far out in the tail.

Key formulas to remember

Standard error of a quantile estimate
se(q̂) = √[p(1−p) ÷ n] ÷ f(q)
p is the cumulative probability of the quantile (e.g. 0.95 for the 95% loss quantile). f(q) is the pdf of the loss at q. This is an approximation that works for large n.
Confidence interval for a quantile
q̂ ± z × se(q̂)
Use z = 1.96 for 95% confidence and z = 1.645 for a 90% two-sided interval. The interval is approximate.
Normal VaR quantile
q = μ + z_p × σ
z_p is 1.645 at 95% and 2.326 at 99%. Losses are measured as positive numbers.
Normal pdf at z
φ(z) = exp(−z² ÷ 2) ÷ √(2π); f(q) = φ(z) ÷ σ
Divide by σ when converting from the standard normal to a loss with standard deviation σ. φ(1.645) ≈ 0.1031 and φ(2.326) ≈ 0.0267.
ES as average of tail VaRs
ES_α ≈ average of VaR at α + (1−α) × i ÷ (k+1), for i = 1 to k
These are k equally spaced interior tail levels, not slice midpoints. A larger k gives a more accurate ES. The midpoint variant uses the levels α + (1−α) × (2i−1) ÷ (2k) instead; follow the question.
ES from a sample
ES ≈ average of the losses greater than the VaR estimate
With n losses and level α, this is roughly the average of the worst n × (1−α) losses.

How to solve Estimating VaR and ES Quantiles and Errors questions

Use this order for any question on quantile estimation, standard errors, ES slices or QQ-plots.

  1. 1Identify what is asked: a sample quantile, a standard error or confidence interval, an ES estimate, or a QQ-plot interpretation.
  2. 2Fix the probability p of the quantile and the sample size n. Convert confidence levels to tail probabilities with care: 95% VaR means p = 0.95 on the loss distribution.
  3. 3For a sample quantile, sort the data and find position n × p. Apply the rounding or interpolation rule given in the question.
  4. 4For a standard error, find f(q). Under a normal, compute z_p, then φ(z_p) ÷ σ. Then compute √[p(1−p) ÷ n] and divide by f(q).
  5. 5For a confidence interval, multiply the standard error by the z value and add and subtract it from the estimate.
  6. 6For ES, find the VaR at each slice level (or the tail losses) and take the simple average.
  7. 7For a QQ-plot, read the line first (location and scale), then the ends (tails). Curving ends mean fat or thin tails relative to the reference.
  8. 8Sanity check: the standard error should fall with larger n, rise with a deeper tail, and ES should exceed VaR.

Quickest way: Standard error in four moves

When to use it: Use when a question gives a normal distribution (or a density at the quantile) and asks for a standard error or confidence interval.

  1. Write p(1−p) and divide by n. Take the square root. This is the numerator.
  2. Get φ(z_p) from the table or memory (0.1031 at 95%, 0.0267 at 99%) and divide by σ to get f(q).
  3. Divide the numerator by f(q) to get the standard error.
  4. For the interval, use q̂ ± 1.96 × se. For a quick ranking of options, remember that halving the standard error needs four times the data.

Common mistakes in Estimating VaR and ES Quantiles and Errors

  • Using the density at the mean instead of at the quantile

    Students remember a pdf value of 0.3989 for the standard normal and plug it in.

    Fix: Always evaluate the density at the quantile z_p. At 95% it is φ(1.645) ≈ 0.1031, far below 0.3989. Divide by σ as well.

  • Forgetting to divide φ(z) by σ

    The standard normal density is memorised, and the loss scale is ignored.

    Fix: The density of the loss at q is φ(z) ÷ σ. A larger σ means a smaller density and a larger standard error.

  • Thinking more data makes the standard error fall in proportion to n

    Students forget the square root in √[p(1−p) ÷ n].

    Fix: The standard error falls with 1/√n. Four times the sample size halves it.

  • Using p = 0.05 instead of 0.95 in the wrong place, or mixing the two

    Losses and returns use opposite tails.

    Fix: p(1−p) is the same at 0.05 and 0.95. For a symmetric distribution such as the normal, f(q) is also the same at the two tails, so the standard error comes out the same. The choice matters mainly for asymmetric distributions, where f(q) differs between the tails, and for the sign of the quantile: the 95% loss quantile is the 0.95 point of the loss distribution, or the 0.05 point of returns with the sign flipped. Keep your convention consistent.

  • Reading a QQ-plot that bends at the ends as a bad fit everywhere

    Students look for any deviation rather than where it occurs.

    Fix: A good fit in the centre with the ends curving away means the body is close to the reference but the tails are different. Ends bending away more steeply mean fatter tails than the reference.

  • Treating ES as a single sample point

    Students confuse ES with the worst loss or with VaR.

    Fix: ES is an average of losses beyond VaR. Average the tail losses or the tail slice VaRs. It is always at least as large as VaR.

Worked examples

Example 1

Daily losses are modelled as normal with mean 0 and standard deviation USD 2 million. You estimate the 95% VaR from n = 1,000 observations. Estimate the standard error of the 95% VaR estimate and a 95% confidence interval. Use z = 1.645 and φ(1.645) = 0.1031 only to find the VaR quantile and its density. Use z = 1.96 (two-sided) for the 95% confidence interval.

Show the solution
  1. Quantile: q = 0 + 1.645 × 2 = USD 3.29 million.
  2. Density at the quantile: f(q) = φ(1.645) ÷ σ = 0.1031 ÷ 2 = 0.05155.
  3. Numerator: √[p(1−p) ÷ n] = √[0.95 × 0.05 ÷ 1,000] = √0.0000475 = 0.006892.
  4. Standard error: 0.006892 ÷ 0.05155 = 0.1337, about USD 0.134 million.
  5. Confidence interval: the 1.645 was for the VaR quantile, so use the two-sided 95% value 1.96 here. 3.29 ± 1.96 × 0.1337 = 3.29 ± 0.262, which gives 3.028 to 3.552.

Answer: The standard error is about USD 0.134 million. The approximate 95% confidence interval for the 95% VaR is USD 3.03 million to USD 3.55 million.

Example 2

You want the 95% ES of a loss distribution. You use three equally spaced interior levels in the 5% tail and find the VaR at the 96.25%, 97.5% and 98.75% levels. The values are USD 3.6 million, 3.9 million and 4.5 million. Estimate the ES and say how you could improve the estimate.

Show the solution
  1. The three levels are equally spaced interior points of the tail, not slice midpoints: 95% + 5% × i ÷ 4 for i = 1, 2, 3 gives 96.25%, 97.5% and 98.75%.
  2. Each level gets the same weight, so take the simple average of the three VaRs.
  3. Sum: 3.6 + 3.9 + 4.5 = 12.0.
  4. Average: 12.0 ÷ 3 = 4.0.
  5. Check: the 95% VaR must be below 3.6 (the VaR at 96.25%), so the ES of 4.0 exceeds the 95% VaR. This is consistent.

Answer: The ES estimate is USD 4.0 million. Using more levels would reduce the approximation error, because the tail is then sampled more finely.

Exam tips

  • Know the standard error formula by heart and practise it with φ(1.645) and φ(2.326). Most numeric questions are plug-in calculations.
  • Expect conceptual comparisons: standard error rises as you move into the tail and falls with larger n. Use that to rank options without calculating.
  • For QQ-plot questions, link the picture to the answer: a straight line means the sample has the same shape as the reference, a slope other than one or a non-zero intercept points to a scale or location difference, and curved ends mean fat or thin tails.
  • Check which tail and convention the question uses before you compute positions in sorted data. Follow the rounding rule given.
  • If ES is asked from slices, average the VaRs, not the slice probabilities, and check that the answer exceeds VaR.

Practice questions from Measures of Financial Risk

Estimating VaR and ES Quantiles and Errors: frequently asked questions

How do I estimate a VaR quantile from historical simulation data?

Sort the losses from smallest to largest. The VaR at level p is the loss at about position n × p in the sorted list. If that is not a whole number, round or interpolate as the question states.

Why is the standard error of a VaR estimate larger at 99% than at 95%?

Fewer observations lie in the far tail, so the density at the quantile is much smaller. That makes f(q) small, and dividing by it increases the standard error even though p(1−p) is smaller.

How is expected shortfall estimated from tail VaRs?

Take the VaRs at k equally spaced tail levels, α + (1−α) × i ÷ (k+1), and average them. Some questions use the VaR at the midpoint of each equal slice instead, so check which one is asked. More levels give a more accurate answer.

What does a QQ-plot tell me about my data?

It compares sample quantiles with reference quantiles, usually normal. A straight line means a good fit. Ends that curve away from the line indicate tails that differ from the reference, such as fat tails.