FRM Part I · FRM Exam Part I · Simulation and Bootstrapping
In a simulation, the standard error of a plain Monte Carlo estimate with 10,000 independent draws is 0.50. The analyst wants the standard error cut to 0.10 using plain Monte Carlo, with no variance reduction. Approximately how many draws are required?
About 250,000 draws are needed. Standard error falls with the square root of the number of draws, so cutting it by a factor of five requires 25 times as many draws: 10,000 × 25 = 250,000.
- A50,000
- B100,000
- C250,000Correct
- D500,000
Explanation
Standard error scales with 1/√N. To reduce it by a factor of 0.50/0.10 = 5, N must rise by 5² = 25. So N = 10,000 × 25 = 250,000. Choosing 50,000 wrongly scales linearly with the factor of 5.
Did you get it right without looking?
One question tells you little. A timed set on Simulation and Bootstrapping shows your real accuracy, how long you take and where you lose marks.
More Simulation and Bootstrapping questions
- A simulation of a portfolio's one-day loss uses 10,000 independent draws and estimates the mean loss as 2.0 million with sample standard dev…
- A Monte Carlo simulation of 400 independent trials estimates the mean payoff of an option as 6.20 with a sample standard deviation of 5.00. …
- An analyst uses the standard (iid) bootstrap to estimate the 99% one-day VaR of a trading book from 500 historical daily P&L observations. W…
- An analyst has a sample of 50 independent daily returns and wants to estimate the standard error of the sample median using the bootstrap. W…
- Two simulation designs estimate the same expected value with the same 10,000 draws. Design A uses plain independent draws, whose payoff vari…
- A risk manager bootstraps daily returns of an equity portfolio by drawing individual days at random with replacement from five years of data…