As a fictional example, you value an option by the Monte Carlo method with 10,000 paths, and the standard error is 0.20. Assume that the standard deviation of the discounted payoffs on the paths barely changes as you add paths. About how many paths do you need to bring the standard error down to 0.05?
Quiz: Numerical Methods
Last reviewed: October 4, 2026
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5 questions to check your understanding of the chapter “Numerical Methods”. Choose an answer and select “Check answer” to see the answer and explanation. You can also open “Answer and explanation” to see the answer first.
Question 1
Answer and explanation
Answer: C
You need about 160,000 paths. The standard error is the standard deviation of the discounted payoffs on the paths divided by the square root of the number of paths. Bringing 0.20 down to 0.05 means cutting the standard error to a quarter, so the square root of the number of paths must be four times larger. The number of paths must therefore be 4 × 4 = 16 times larger. That gives 10,000 × 16 = 160,000.
40,000 paths is what you get if you assume the standard error is inversely proportional to the number of paths. In fact, quadrupling the paths to 40,000 only halves the standard error, to 0.10. 20,000 paths comes from taking the square root of 4 and doubling the number of paths. With 20,000 paths, the standard error is 0.20 ÷ 1.414 (the square root of 2), or about 0.14. By the same reasoning, gaining one more digit of accuracy (cutting the standard error to a tenth) takes 100 times as many paths.
Learn more: Monte Carlo Simulation
Question 2
Which of the following correctly describes variance reduction methods in the Monte Carlo method?
Answer and explanation
Answer: D
Variance reduction methods use random numbers in ways that reduce the spread of the values on the paths. They leave unchanged the rate at which the standard error shrinks as paths are added, and lower the standard error for the same number of paths. A common example is the antithetic variates method, which pairs a path generated with normal random numbers with a path generated with the same numbers of the opposite sign. Another is the control variates method. On the same paths, it calculates a second value that moves together with the target value and whose expected value is known from a formula. It then uses the error in the estimate of that second value to correct the estimate of the target value.
Variance reduction methods do not cut the computational effort per path. Some methods increase the effort per path, so when choosing a method you compare both the variance and the effort. Random numbers with a long period are used to avoid reusing the same numbers, which would not improve accuracy. That is a separate matter from variance reduction. The standard error still shrinks in inverse proportion to the square root of the number of paths; only the proportionality constant changes.
Learn more: Monte Carlo Simulation
Question 3
As a fictional example, you solve the Black-Scholes equation, rewritten in the logarithm of the price, by the explicit method of the finite difference method. The volatility is constant at 20%, the grid spacing in the price direction is 0.02, and the one year to expiry is divided into time steps of equal length. By the stability guideline that looks at the diffusion term only, what is the smallest number of time steps that avoids instability?
Answer and explanation
Answer: A
The smallest number of time steps is 100. The guideline that looks at the diffusion term only is that the time step is at most the square of the grid spacing divided by the square of the volatility. The square of 0.02 is 0.0004 and the square of 0.20 is 0.04, so the time step is at most 0.0004 ÷ 0.04 = 0.01. Dividing one year into steps of 0.01 gives 1 ÷ 0.01 = 100 steps.
Forty steps give a time step of 1 ÷ 40 = 0.025, 2.5 times the guideline. Under this condition the explicit method diverges. Two hundred steps come from taking the diffusion coefficient as the squared volatility itself (0.04), which gives a guideline of 0.0004 ÷ (2 × 0.04) = 0.005. The coefficient of the second derivative in the log-price equation is half the squared volatility (0.02). Two hundred steps do not cause instability either, but they are not the smallest number that avoids it.
Learn more: Finite Difference Methods
Question 4
When the initial values are smooth, the error from the grid spacing is second order in all three methods of the finite difference method (explicit, implicit, and Crank-Nicolson). Which description of the Crank-Nicolson method is correct?
Answer and explanation
Answer: C
The Crank-Nicolson method averages the values at the previous and next times and solves tridiagonal simultaneous equations for the values at the next time. For smooth initial values, its error from the time step is second order. The explicit and implicit methods have a first-order error from the time step, so the Crank-Nicolson error becomes small faster.
Getting the next values directly from three values at the previous time is the explicit method. The implicit method and the Crank-Nicolson method solve simultaneous equations. Among the three methods, only the Crank-Nicolson method has a second-order error from the time step.
Learn more: Finite Difference Methods
Question 5
As a fictional example, one evaluation of a portfolio takes 3 minutes, and you want 200 sensitivities. The output is a single portfolio value. Which pair gives the time for one-sided bumping (moving one input at a time) and the theoretical time for AAD (adjoint algorithmic differentiation) in reverse (adjoint) mode?
Answer and explanation
Answer: A
Bumping takes 603 minutes and reverse mode under 15 minutes. One-sided bumping is the original evaluation plus 200 evaluations with an input moved, 201 in total. At 3 minutes each, the 201 evaluations take 201 × 3 = 603 minutes. Theoretically, reverse mode costs under 5 times the original evaluation, whatever the number of sensitivities. At 3 minutes per evaluation, that is under 3 × 5 = 15 minutes.
The 600 minutes for bumping comes from forgetting the original evaluation and calculating 200 × 3. Taking reverse mode as 600 minutes confuses it with forward mode automatic differentiation, whose cost grows in proportion to the number of sensitivities. Forward mode needs one calculation for each input direction. The under-15-minutes figure is a theoretical upper bound for a single output, and memory and other costs come separately.
Learn more: Adjoint Algorithmic Differentiation (AAD)