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Illustrate how a Monte Carlo simulation is used to approximate moments or other quantities, and outline the basic steps of carrying it out.

FRM® Part I · Quantitative Analysis · Simulation and Bootstrapping · FRM1.QA.13.a

The idea

A Monte Carlo simulation approximates an expectation by averaging over random draws. The steps are to specify the data generating process and its parameters, generate N random draws or paths from it, compute the quantity of interest (a payoff, loss or moment) for each draw, and then average across the draws. In symbols, E[g(X)] is estimated by the sample mean of g(xi). So the second moment uses the mean of x², not the square of the mean. Standard error = s ÷ √N. Here s is the sample standard deviation of the draws. Probabilities are the fraction of draws meeting a condition. For derivative pricing, simulate under the risk-neutral drift and discount the average payoff. Exam points: the answer is an estimate with sampling error, and the standard error falls only in proportion to one over the square root of N.

Formulas

Standard error of a Monte Carlo estimate

SE = s ÷ √N

s
the sample standard deviation of the simulated values
N
the number of independent replications

Needs independent draws. Halving the error takes four times the replications; quasi-random sequences break the formula.

A practice question

From the question bank, with the reason each wrong option is wrong.

Which sequence correctly outlines the basic steps of a Monte Carlo simulation used to approximate the expected value of a function of a random variable?

  1. Specify the process, simulate N draws, compute the function for each, then average and compute the standard error. - correct
  2. Collect N historical observations, sort them, discard the outliers, then read the expected value from the middle of the sorted list.Why not: Describes a historical-sample summary, not a simulation from a specified process, and the outlier trimming biases the moments.
  3. Draw one very long random path, compute the function once at its end, then multiply by N to scale up the estimate.Why not: Uses a single draw and scales by N; the estimate is the average over many draws and one draw gives no standard error.
  4. Compute the function at the mean of the inputs, then add a random error term to represent the remaining uncertainty.Why not: Substitutes the function of the mean for the mean of the function; for nonlinear functions these differ (Jensen's inequality).

The steps are: (1) specify the data generating process and its parameters, (2) generate N random draws or paths, (3) compute the quantity of interest for each draw, (4) average over the draws, and (5) estimate the standard error as the sample standard deviation divided by the square root of N.

A harder one

A Monte Carlo simulation of 10,000 independent paths gives discounted option payoffs with a sample mean of 4.20 and a sample standard deviation of 12.0. What is the approximate 95% confidence interval for the option value?

  1. 4.00 to 4.40Why not: Uses 1.645, the one-sided 95% (two-sided 90%) quantile, instead of 1.96 for a two-sided 95% interval.
  2. 3.96 to 4.44 - correct
  3. -19.32 to 27.72Why not: Uses the standard deviation of the payoffs (12.0) as if it were the standard error of the mean.
  4. 4.198 to 4.202Why not: Divides the standard deviation by N instead of by the square root of N, so the standard error is far too small.

Standard error = 12.0 / sqrt(10,000) = 0.12. The interval is 4.20 +/- 1.96 x 0.12 = 3.96 to 4.44. The error of the estimate shrinks with the square root of N, not with N.

The hardest tier

A risk-neutral Monte Carlo valuation of a 1-year European call (S0 = 100, strike 105, risk-free rate 3% continuously compounded, volatility 20%) uses terminal prices S_T = S0 x exp[(r - sigma²/2)T + sigma x sqrt(T) x z]. Five standard normal draws are -0.8, 0.3, 1.1, -1.5 and 0.6. The stock's expected real-world return is 9%. What is the estimated option value, closest to?

  1. 6.40Why not: Averages the payoffs but forgets to discount them back at the risk-free rate.
  2. 10.37Why not: Uses the 9% real-world expected return as the drift; risk-neutral pricing requires the risk-free rate.
  3. 6.21 - correct
  4. 7.57Why not: Drops the -sigma²/2 term from the drift, so the simulated prices are biased upward under risk-neutral lognormal dynamics.

Terminal prices use drift (0.03 - 0.02) = 0.01. The payoffs are averaged to 6.40 and discounted by exp(-0.03) to give 6.21. The drift must be the risk-free rate less half the variance, and the average payoff must be discounted.

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