Numerical pricing: Monte Carlo, finite differences and when to use which

PRC · Chapter 413 min readAsked at Optiver, SIG, Akuna, Citadel Securities

Assumes Binomial trees, backward induction and early exercise.

After this lesson you should be able to

  • Choose between a lattice, a PDE solver and Monte Carlo from the payoff.
  • State the error rate of each and what improves it.
  • Explain how Longstaff–Schwartz handles early exercise in a simulation.

Three families of numerical method cover almost everything: lattices, finite-difference PDE solvers and Monte Carlo. The choice is not a matter of taste — it falls out of two features of the payoff, whether it is path-dependent and how many underlyings it has.

MethodErrorStrong atFails at
Binomial latticeO(1/n)O(1/n), oscillatingOne asset, early exercisePath dependence, several assets
Finite differencesO(Δt)+O(Δx2)O(\Delta t) + O(\Delta x^2)One or two assets, Greeks for freeDimension beyond about three
Monte CarloO(1/N)O(1/\sqrt{N})Path dependence, many assetsEarly exercise, and Greeks
Table 4.1 · Choosing the method. The last column is the decision procedure. Path-dependent means Monte Carlo; early exercise means a lattice or PDE; both at once means Longstaff–Schwartz and some care.

Why dimension decides it. A grid method needs points in every dimension, so its cost grows like mdm^d — twenty points per axis is fine in one dimension and hopeless in five. Monte Carlo’s error is O(1/N)O(1/\sqrt{N}) *regardless of dimension*, because it is estimating an average and the central limit theorem does not care how many variables went into each draw. So the crossover is sharp: below about three dimensions grids win comfortably, above it they cannot be built at all.

1001600640000.511/√N scaling, relative to 100 pathsSimulated pathsRelative standard error
Figure 4.2 · Four times as many paths halves Monte Carlo noise. For independent finite-variance simulated payoffs, standard error falls as 1/√N. A fourfold path increase buys only a twofold precision improvement, which is why variance reduction can be more valuable than simply running longer.

Proposition 4.3

Monte Carlo for options

Simulate the underlying under the risk-neutral measure, evaluate the payoff on each path, average, and discount. Use the exact lognormal step rather than an Euler discretisation when you can — for geometric Brownian motion the solution is known, so there is no need to introduce a discretisation error.

Holds when

  • Exact step: St+Δ=Stexp⁡[(r−σ2/2)Δ+σΔ z]S_{t+\Delta} = S_t \exp[(r - \sigma^2/2)\Delta + \sigma\sqrt{\Delta}\,z].
  • For a path-independent payoff, simulate straight to expiry in one step.
  • Antithetic variates are free and typically worth a factor of two; a control variate using the vanilla with a known Black–Scholes price is usually worth much more.

Example 4.4

Sizing a simulation

A Monte Carlo price has a payoff standard deviation of $20\$20. How many paths for a standard error of one cent?

Show the worked solutionHide the worked solution

Worked solution

  1. Formula
    se=σN⇒N=(σse)2\mathrm{se} = \frac{\sigma}{\sqrt{N}} \Rightarrow N = \left(\frac{\sigma}{\mathrm{se}}\right)^2
  2. Substitute
    =(200.01)2= \left(\frac{20}{0.01}\right)^2
  3. Solve
    =20002=4×106= 2000^2 = 4 \times 10^{6}
  4. Answer
    four million paths\text{four million paths}

Sanity check. And one more decimal place would cost four hundred million. That arithmetic is why variance reduction is not an optimisation but a necessity — halving σ\sigma saves three quarters of the work.

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