Actuarial Statistics · Basic univariate distributions and generating samples

Simulation Accuracy and How Many Simulations You Need

Updated 11 October 2026 · Fact-checked

Simulation accuracy is measured by the standard error of the simulated estimate, which is s ÷ √n for a mean. To get a margin of error E at a given confidence level, you need n ≥ (z × σ ÷ E)². Accuracy improves with √n, so halving the error needs four times as many simulations.

Understand Simulation Accuracy and Number of Simulations

A simulation estimates a quantity by generating n random values and averaging a function of them. For example, you simulate n claim totals and use their mean to estimate the expected claim. The answer is itself random. Run it again with new random numbers and you get a slightly different value. So you need a way to say how far off it might be.

The tool is the standard error. The simulated values are independent and identically distributed, so the sample mean has variance σ² ÷ n. Its standard deviation, σ ÷ √n, is the standard error. You rarely know σ, so you replace it with the sample standard deviation s from the same simulations.

Because n is large in simulation, the Central Limit Theorem makes the sample mean approximately normal. That lets you build a confidence interval: estimate ± z × s ÷ √n. For 95% confidence, z = 1.96. The half-width of this interval is the margin of error.

To choose n in advance, you set the margin of error you can tolerate and solve for n. You need a guess for σ, often from a small pilot run of, say, a few hundred simulations. Note the √n effect. To cut the error to one tenth, you need 100 times as many runs. This is why simulation is simple but can be slow.

The same idea works for a probability. If you estimate p as the proportion of runs where an event happens, each run is a Bernoulli trial. The standard error is √(p̂(1 − p̂) ÷ n). Estimating small probabilities, such as a ruin probability, needs many more runs to get a useful relative accuracy. Simulation is used in actuarial work to value options, find aggregate claims distributions, test reserves and capital under many scenarios, and price products where no closed form exists.

Key rules to remember

Simulated mean: x̄ = (1 ÷ n) Σ xᵢ; Each xᵢ is the value of the quantity of interest from simulation i. Runs must be independent.
Standard error of the mean: SE = σ ÷ √n, estimated by s ÷ √n; s is the sample standard deviation with divisor n − 1. Use s when σ is unknown.
Approximate confidence interval for the mean: x̄ ± z × s ÷ √n; z = 1.645 for 90%, 1.96 for 95%, 2.576 for 99%. Based on the Central Limit Theorem for large n.
Number of simulations for margin of error E: n ≥ (z × σ ÷ E)²; Round up. Use a pilot estimate of σ. E is in the same units as the mean.
Standard error of a simulated probability: SE = √(p̂(1 − p̂) ÷ n); p̂ is the proportion of runs in which the event occurs. Use a pilot p̂ or p̂ = 0.5 for a cautious n.
Number of simulations for a probability: n ≥ z² × p(1 − p) ÷ E²; The worst case is p = 0.5, giving the largest n.

How to solve Simulation Accuracy and Number of Simulations questions

Use this method for any question on the error of a simulated estimate or the number of runs needed.

1Identify what is being estimated: a mean, a probability or another quantity, and the units.
2Write down what is given: n, the sample mean or proportion, and s or σ, and the confidence level.
3Find the z value for the confidence level, such as 1.96 for 95%.
4For error assessment, compute the standard error, then the interval estimate ± z × SE.
5For sample size, set z × σ ÷ E equal to the required margin, square it, and round n up to the next whole number.
6If σ is not given, say you would use a pilot run to estimate it and state that assumption.
7State the conclusion in words with units, and mention that the normal approximation relies on a large n.

Quickest way: Square the ratio, round up

When to use it: Use when the question gives a target margin of error and asks for the number of simulations.

Compute z × σ ÷ E in one line.
Square it and round up.
Check the scaling: if the old error was E₀ with n₀ runs, the new n = n₀ × (E₀ ÷ E)².
For probabilities, replace σ with √(p(1 − p)).

Common mistakes in Simulation Accuracy and Number of Simulations

Forgetting to square the ratio and giving n = zσ ÷ E.
You solve for the error formula and stop before isolating n.
Fix: Rearrange SE = σ ÷ √n to get n = (σ ÷ SE)². Check that a smaller E gives a larger n.
Rounding n down.
Standard rounding habit.
Fix: Always round up, since rounding down gives a margin larger than the target.
Thinking doubling the simulations halves the error.
You assume error falls in proportion to n.
Fix: Error falls with √n. To halve it you need four times as many runs.
Using the variance instead of the standard deviation in z × σ ÷ √n.
Questions sometimes give the sample variance.
Fix: Take the square root first. Check whether the given figure is s² or s.
Using a bias-free reading of the standard error, treating it as the error of a single run.
Confusing the spread of individual outcomes with the spread of their mean.
Fix: s describes single runs. s ÷ √n describes the mean of n runs.
Ignoring that more runs do not remove modelling error.
Treating the simulation error as the only error.
Fix: State that the standard error covers sampling noise only. Wrong model assumptions or a poor random number generator are separate sources of error.

Worked examples

Example 1

A simulation of 400 independent annual claim totals gives a sample mean of ₹52,000 and a sample standard deviation of ₹16,000. Find the standard error and an approximate 95% confidence interval for the expected claim total.

Show the solution

SE = s ÷ √n = 16,000 ÷ √400 = 16,000 ÷ 20 = ₹800.
The margin of error is 1.96 × 800 = ₹1,568.
Lower limit = 52,000 − 1,568 = ₹50,432.
Upper limit = 52,000 + 1,568 = ₹53,568.

Answer: The standard error is ₹800 and the approximate 95% confidence interval is ₹50,432 to ₹53,568.

Example 2

A pilot run suggests the standard deviation of a simulated present value is ₹5,000. How many simulations are needed so that the 95% margin of error for the mean is no more than ₹250? What happens to n if the margin is tightened to ₹125?

Show the solution

n ≥ (z × σ ÷ E)² with z = 1.96, σ = 5,000, E = 250.
z × σ ÷ E = 1.96 × 5,000 ÷ 250 = 9,800 ÷ 250 = 39.2.
n ≥ 39.2² = 1,536.64, so round up to n = 1,537.
For E = 125, the ratio is 78.4 and 78.4² = 6,146.56, so n = 6,147.
Halving E multiplies n by 4, which matches 1,536.64 × 4 = 6,146.56.

Answer: You need at least 1,537 simulations for a margin of ₹250, and at least 6,147 for a margin of ₹125.

Exam tips

Write the formula first, then substitute. Method marks are given even if the arithmetic slips.
State that the interval relies on the Central Limit Theorem and a large n.
In sample size questions, always say where σ comes from, such as a pilot run, and round n up.
For written answers on uses of simulation, give concrete actuarial examples such as option pricing or aggregate claims where no closed form exists.
In computer-based papers, report the estimate with its standard error, not the estimate alone.

Practice questions from Basic univariate distributions and generating samples

Simulation Accuracy 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.

FRM Part ISampling Error and Number of Simulations

Simulation Accuracy and Number of Simulations: frequently asked questions

How many simulations are enough?

There is no fixed number. It depends on the variability of the output and the precision you need. Use n ≥ (z × σ ÷ E)² with a pilot estimate of σ.

Why does accuracy improve so slowly with more simulations?

The standard error is σ ÷ √n. Because of the square root, you need four times as many runs to halve the error. Ten times the precision needs a hundred times the runs.

Do I use s or σ in the standard error?

σ is usually unknown in a simulation, so you use the sample standard deviation s from the same runs. With a large n the difference is small and the normal z value is acceptable.

Does a large number of simulations guarantee a correct answer?

No. It only reduces sampling error. If the model, the distribution or the random number generation is wrong, the estimate stays wrong however many runs you do.