FRM Exam Part II · Estimating Market Risk Measures: An Introduction and Overview

Standard Error and Confidence Intervals for VaR and ES

Updated 11 October 2026 · Fact-checked

A VaR estimate is a sample quantile, so it carries sampling error. Its standard error is √(p(1−p)) ÷ (f(q) × √n), where p is the tail probability and f(q) is the density at the quantile. Add ± z × standard error for a confidence interval. More data and less extreme confidence levels give tighter estimates.

Understand Estimating Standard Errors and Confidence Intervals for Risk Measures

A VaR figure from historical data is not a fact. It is an estimate of a quantile built from a finite sample. Draw a different sample and you get a different VaR. The standard error measures how much the estimate would typically move from sample to sample.

Three things drive the standard error. First, the sample size n: more observations make the estimate more stable, but only at the rate of √n. Second, the tail probability p (for 99% VaR, p = 1% = 0.01): the further into the tail, the fewer observations support the estimate, so the error rises. Third, the density f(q) at the quantile: where few observations sit near the quantile (a thin, far tail), the quantile is poorly pinned down, and the error is large.

There are two ways to build a confidence interval. The first uses the standard error and a normal approximation: estimate ± z × standard error. The second uses order statistics and needs no distribution. The number of observations falling below the true p-quantile follows a binomial distribution with parameters n and p. This tells you which ranked observations, such as the 4th and 16th largest losses, bracket the true VaR with the chosen confidence.

The standard error formula needs f(q), which you must assume or estimate. Exam questions usually give you a distribution, such as the normal, so you can compute f(q) directly. For far-tail quantiles in heavy-tailed data, the true density at the quantile is often lower than a normal fit implies, so normal-based standard errors can understate the error and the intervals can be too narrow. This depends on the quantile and on how the distribution is fitted, so it is not always true.

Expected shortfall (ES) is an average of tail losses, so it also has sampling error. Its error is generally larger than the VaR error at the same level when tails are heavy. In practice it is often assessed by simulation or bootstrap rather than a simple formula.

Key formulas to remember

Standard error of a quantile estimate: se(q̂) = √(p(1 − p)) ÷ (f(q) × √n); p is the tail probability (1 − confidence level). f(q) is the pdf at the quantile. This is an asymptotic result for large n.
Confidence interval for VaR (normal approximation): q̂ ± z × se(q̂); z = 1.645 for 90%, 1.96 for 95%, 2.576 for 99% two-sided intervals.
Order-statistic count distribution: Number of observations below the true p-quantile ~ Binomial(n, p); This is the basis of distribution-free confidence intervals for a quantile.
Approximate rank limits: Ranks ≈ np ± z × √(np(1 − p)); Count from the worst loss. Round the lower rank up and the upper rank up or down as the question states. Needs np large enough for the normal approximation.
Effect of sample size: se ∝ 1 ÷ √n; Quadrupling n halves the standard error. Doubling n cuts it by about 29%.

How to solve Estimating Standard Errors and Confidence Intervals for Risk Measures questions

Use this sequence for any question on the precision of a VaR or quantile estimate.

1Identify p, the tail probability. For 99% VaR, p = 0.01. For 95%, p = 0.05.
2Identify the method asked for: the standard error formula, or an order-statistic (binomial) interval.
3For the standard error, find the quantile q and the density f(q) from the stated distribution. For a standard normal, use φ(z). For a normal with standard deviation σ, divide φ(z) by σ.
4Compute √(p(1 − p)), then divide by f(q) × √n. Keep units: the result is in the same units as the VaR (for example USD).
5For the confidence interval, take the VaR estimate ± z × the standard error, using the z for the confidence level of the interval, not the VaR level.
6For the order-statistic interval, compute np and √(np(1 − p)). The ranks are np ± z × √(np(1 − p)), counted from the largest loss.
7Interpret the result: state the range, and say what would narrow it (larger n, a less extreme confidence level).

Quickest way: Scaling shortcut for sample size and tail level

When to use it: Use when the question asks how precision changes with n or p, or asks you to compare two cases without full computation.

Remember se ∝ 1 ÷ √n. Multiply n by k and the standard error is divided by √k.
To compare two sample sizes, take the ratio √(n1 ÷ n2) and apply it to the known standard error.
For a larger tail probability (less extreme VaR), the numerator √(p(1 − p)) rises, but the density f(q) at the quantile rises much faster, so the ratio falls and the standard error is smaller. For a normal distribution, going from p = 0.01 to p = 0.05 moves the numerator from about 0.0995 to 0.2179 and the density from about 0.0267 to 0.1031. The ratio falls from about 3.73 to 2.11. A 95% VaR is normally more precise than a 99% VaR for the same n.
Eliminate options that claim precision improves linearly with n, or that ignore the density term.
For binomial intervals, check np first. The point estimate rank is about np, and the interval width in ranks is about 2z × √(np(1 − p)).

Common mistakes in Estimating Standard Errors and Confidence Intervals for Risk Measures

Using z for the VaR confidence level when building the interval around VaR.
Both use 'confidence level' and 1.645 or 2.326 sit in memory from VaR calculations.
Fix: Keep two separate levels. The VaR level sets p and the quantile. The interval level sets the z multiplier.
Forgetting to divide by f(q), or using the cdf instead of the pdf.
Students remember the √(p(1 − p)) ÷ √n part and drop the density term.
Fix: Write the full formula first. f(q) is the pdf at the quantile, such as φ(1.645) = 0.1031 for the standard normal.
Forgetting to scale the density when σ is not 1.
Tables give φ for a standard normal only.
Fix: For N(μ, σ²), f(q) = φ(z) ÷ σ. A larger σ gives a larger standard error in the same ratio.
Thinking doubling the sample halves the error.
Intuition says error falls in proportion to data.
Fix: Error falls with √n. Halving needs four times the data.
Treating the confidence interval as covering future losses.
Confusing sampling uncertainty with the loss distribution itself.
Fix: The interval is about where the true VaR lies, not about the range of future P&L.
Counting order-statistic ranks from the wrong end.
Losses and returns are ranked in opposite directions.
Fix: State the convention. If working with losses, rank 1 is the largest loss. With returns, VaR sits in the lower tail.

Worked examples

Example 1

Daily P&L is normal with mean 0 and standard deviation USD 1 million. You estimate 95% VaR from 500 observations. Using the quantile standard error, find the standard error and a 95% confidence interval for VaR. Take VaR = USD 1.645 million, φ(1.645) = 0.1031.

Show the solution

p = 0.05, so √(p(1 − p)) = √0.0475 = 0.2179.
Density at the quantile for σ = 1 million: f(q) = 0.1031 ÷ 1 = 0.1031 per USD million.
√n = √500 = 22.36.
se = 0.2179 ÷ (0.1031 × 22.36) = 0.2179 ÷ 2.305 = 0.0945, so about USD 0.0945 million.
95% interval: 1.645 ± 1.96 × 0.0945 = 1.645 ± 0.185.
Lower = 1.460, upper = 1.830.

Answer: Standard error ≈ USD 94,500. The 95% confidence interval for VaR is about USD 1.46 million to USD 1.83 million.

Example 2

You hold 1,000 daily losses and estimate 99% VaR. Using the binomial approach with a normal approximation, find the approximate 95% confidence interval in terms of ranked losses (rank 1 = largest loss).

Show the solution

p = 0.01 and n = 1,000, so np = 10.
np(1 − p) = 1,000 × 0.01 × 0.99 = 9.9, and √9.9 = 3.146.
Half-width in ranks = 1.96 × 3.146 = 6.17.
Rank limits: 10 − 6.17 = 3.83 and 10 + 6.17 = 16.17.
Rounding to whole ranks gives about the 4th and 16th largest losses.
The point estimate sits near the 10th largest loss.

Answer: The true 99% VaR lies between roughly the 4th and 16th largest losses with about 95% confidence. The wide rank range shows how imprecise a 99% estimate from 1,000 observations is.

Exam tips

Write p, n and f(q) at the top before you calculate. Most lost marks come from a missing term.
Watch the units. The standard error should be in the same units as the VaR, such as USD or EUR.
Know the direction of effects: more data narrows the interval, a more extreme confidence level widens it, and heavier tails make the estimate less reliable.
For interpretation questions, say that a wide interval means a point VaR should not be treated as exact. This matters when comparing VaR to limits or exceptions.
If a question asks about ES precision, say it is usually harder to estimate than VaR because it depends on the extreme tail, and that simulation or bootstrap is a common route.

Practice questions from Estimating Market Risk Measures: An Introduction and Overview

Estimating Standard Errors and Confidence Intervals for Risk Measures in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

FRM Part IConfidence Intervals and Standard Error

Estimating Standard Errors and Confidence Intervals for Risk Measures: frequently asked questions

What is the standard error of a VaR estimate?

It measures how much a sample VaR varies from sample to sample. For a quantile it equals √(p(1 − p)) ÷ (f(q) × √n). It rises when p is more extreme, the density at the quantile is low, or n is small.

How do I compute a confidence interval for a quantile estimate?

There are two routes. Take the estimate ± z × standard error, or use order statistics, where the count of observations below the true quantile is binomial with parameters n and p. The second route needs no distributional assumption.

How does sample size affect VaR precision?

Precision improves with the square root of n. You need four times as many observations to halve the standard error. Extreme confidence levels, such as 99%, need much more data than 95% for the same precision.

Is the confidence interval for ES computed the same way?

Not with the simple quantile formula. ES averages the tail beyond VaR, so its error depends on the tail shape. In practice it is usually assessed by simulation or bootstrap. Expect it to be less precise than VaR when tails are heavy.