FRM Exam Part I · Simulation and Bootstrapping

Sampling Error and Number of Simulations in Monte Carlo

Updated 11 October 2026 · Fact-checked

Sampling error is the random gap between a simulation estimate and the true value. The standard error of a Monte Carlo mean is σ ÷ √N, where σ is the standard deviation of the simulated outcomes and N is the number of replications. To halve the error, you need four times as many runs.

Understand Sampling Error and Number of Simulations

A Monte Carlo simulation estimates a quantity, such as an expected payoff or a price, by averaging results from many random scenarios. Each run uses random numbers, so the estimate is itself random. Run the simulation again with new random numbers and you get a slightly different answer. That difference is sampling error, also called simulation error.

You measure sampling error with the standard error of the estimate. The simulated outcomes have a standard deviation σ. The average of N independent outcomes has standard deviation σ ÷ √N. This follows from the Central Limit Theorem and the rule that the variance of an average of independent draws is σ² ÷ N.

The key point is the square root. Accuracy improves slowly. To cut the standard error by a factor of 2 you need 4 times the replications. To cut it by a factor of 10 you need 100 times the replications. Each extra bit of precision costs more than the last.

This creates a trade-off between accuracy and computational cost. Run time grows roughly in proportion to N, but the error falls only with √N. For complex products, such as path-dependent options or large portfolios that need full revaluation, each run is expensive. So analysts choose N to hit a target error, and they often use variance reduction techniques to lower σ instead of just raising N.

In practice σ is unknown, so you estimate it with the sample standard deviation of the simulated results. A confidence interval is then the estimate ± z × standard error. For 95% confidence, z is about 1.96. Note that sampling error is different from model error. A large N removes sampling error but does not fix a wrong model or wrong inputs.

Key formulas to remember

Standard error of a simulated mean: SE = σ ÷ √N; σ is the standard deviation of the simulated outcomes (use the sample standard deviation s if σ is unknown). N is the number of independent replications.
Confidence interval for the estimate: x̄ ± z × σ ÷ √N; z is 1.645 for 90%, 1.96 for 95% and 2.576 for 99% two-sided confidence, based on the normal approximation.
Required number of simulations: N = (z × σ ÷ E)²; E is the target half-width of the confidence interval, the maximum acceptable error at the chosen confidence level.
Scaling rule: SE₂ ÷ SE₁ = √(N₁ ÷ N₂); To cut error by a factor k, multiply N by k². Holds when σ is unchanged.

How to solve Sampling Error and Number of Simulations questions

Most questions give you two of the three items (σ, N, error) and ask for the third, or ask how error changes when N changes. Work through this method.

1Identify what is being estimated and which standard deviation is given. Check that it is the standard deviation of a single simulated outcome, not of the average.
2Write SE = σ ÷ √N and note whether the question asks for standard error, a confidence interval or a required N.
3For a confidence interval, choose the z value for the stated confidence level (1.645, 1.96 or 2.576) and multiply it by the SE.
4For a required N, set the half-width E equal to z × σ ÷ √N and solve N = (z × σ ÷ E)². Round up to the next whole number.
5For a scaling question, use the square-root rule: error falls by k when N rises by k². Do not recompute σ unless the method changes.
6Check units and sense. Error must shrink as N grows. If the answer for N is smaller than expected, you may have forgotten to square.
7If the question mentions variance reduction, remember it lowers σ, so the same N gives a smaller SE.

Quickest way: Square-root scaling shortcut

When to use it: Use when a question gives an existing standard error or N and asks what happens after changing N, or what N is needed for a smaller error.

Find the ratio of the target error to the current error, for example 0.5 ÷ 2 = 0.25 of the original.
Invert and square it: N must grow by (1 ÷ ratio)². Here 4² = 16 times.
Multiply the current N by that factor.
Sanity check: halving error needs 4 times N, a tenth needs 100 times N.

Common mistakes in Sampling Error and Number of Simulations

Thinking that doubling N halves the standard error.
Students assume error falls in proportion to N.
Fix: Error falls with √N. Doubling N cuts error by about 29% (a factor of 1 ÷ √2). Halving error needs 4 times N.
Using σ ÷ N instead of σ ÷ √N.
Mixing up standard error with the variance of the mean, which is σ² ÷ N.
Fix: Either use the standard deviation σ ÷ √N, or the variance σ² ÷ N. Never σ ÷ N.
Forgetting to square when solving for N.
Rearranging E = zσ ÷ √N and stopping at √N.
Fix: After isolating √N = zσ ÷ E, square both sides to get N. Round up.
Believing a large N makes the simulation correct.
Confusing sampling error with model error.
Fix: More runs reduce only sampling error. A misspecified model, bad inputs or a poor random number generator still give a biased answer.
Using the wrong z value or one-sided value for a two-sided interval.
Memorising 1.645 and 1.96 without noting their confidence levels.
Fix: Two-sided 95% uses 1.96. Two-sided 90% uses 1.645. Two-sided 99% uses 2.576.
Ignoring that a better method can beat more runs.
Treating N as the only lever.
Fix: Variance reduction (antithetic variates, control variates) lowers σ. Reducing σ by half has the same effect as quadrupling N.

Worked examples

Example 1

A Monte Carlo simulation of an option payoff uses N = 10,000 independent runs. The sample standard deviation of the discounted payoffs is $12.00. Find the standard error of the estimated price, and the 95% confidence interval if the estimated price is $8.40.

Show the solution

SE = σ ÷ √N = 12.00 ÷ √10,000.
√10,000 = 100, so SE = 12.00 ÷ 100 = $0.12.
Half-width = 1.96 × 0.12 = $0.2352.
Interval = 8.40 ± 0.2352 = $8.1648 to $8.6352.

Answer: Standard error = $0.12. 95% confidence interval ≈ $8.16 to $8.64.

Example 2

A risk team's simulation with 25,000 runs gives a standard error of 0.40 (in USD millions) for an expected loss estimate. They want the 95% confidence interval half-width to be no more than 0.098. How many runs are needed, assuming σ is unchanged?

Show the solution

First find σ: SE = σ ÷ √N, so σ = 0.40 × √25,000 = 0.40 × 158.11 = 63.25.
Target half-width E = 0.098, z = 1.96.
N = (z × σ ÷ E)² = (1.96 × 63.25 ÷ 0.098)².
1.96 ÷ 0.098 = 20, so 20 × 63.25 = 1,265.
N = 1,265² = 1,600,225 (approximately 1.6 million).
Check with scaling: current half-width is 1.96 × 0.40 = 0.784. Target 0.098 is 1/8 of that. Required factor is 8² = 64. 25,000 × 64 = 1,600,000, consistent.

Answer: About 1.6 million runs (64 times the original), using σ ≈ 63.25.

Exam tips

Memorise the √N rule as a scaling fact: error ÷ 2 means N × 4, error ÷ 10 means N × 100. Many questions are solved in seconds this way.
Read carefully whether the given standard deviation is for one run or for the average. Only divide by √N for a single-run σ.
If a question asks which change reduces error most efficiently, a variance reduction technique that lowers σ is often the better answer than raising N.
Round N up, not to the nearest integer, when finding the minimum number of runs.
Watch for distractors that claim more simulations remove model risk or bias. They do not.

Practice questions from Simulation and Bootstrapping

Sampling Error and Number of Simulations in other exams

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

Sampling Error and Number of Simulations: frequently asked questions

What is the standard error of a Monte Carlo simulation?

It is the standard deviation of the simulated estimate, equal to σ ÷ √N. Here σ is the standard deviation of the individual simulated outcomes and N is the number of runs. It tells you how far a typical estimate lies from the true value.

How many simulations do I need for a given accuracy?

Choose the confidence level and the acceptable error E, then use N = (z × σ ÷ E)². You usually estimate σ from a pilot run. Round the answer up.

Why does Monte Carlo accuracy improve only with the square root of N?

The variance of an average of N independent draws is σ² ÷ N, so its standard deviation is σ ÷ √N. Each added run contributes less new information than the one before it. That is why large gains in accuracy are costly.

Can I reduce sampling error without running more simulations?

Yes. Variance reduction techniques such as antithetic variates and control variates lower σ, so the same N gives a smaller standard error. They add some work per run but often pay off.