Monte Carlo pricing

~9 min read

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Why it matters

Once a payoff depends on a whole path (barriers, Asians, cliquets, Bermudans with a twist) or on several underlyings, trees and PDEs struggle; Monte Carlo scales with the number of paths, not the dimension of the state, in the naive sense that matters for interviews. The idea is one line: simulate under the risk-neutral dynamics, average the discounted payoff. Interviewers probe whether you simulate the right drift, whether you discount inside the average, and how error falls with \(N\).

Core concepts

The estimator. For a European-style (path-dependent is fine) claim, \[ \hat V = \frac{1}{N}\sum_{i=1}^N e^{-rT} g(S^{(i)}), \] where \(S^{(i)}\) are i.i.d. paths generated under \(Q\). By the law of large numbers this converges to \(e^{-rT} E^Q[g]\), which is the no-arbitrage price in a complete (or well-hedged) model.
Simulate the risk-neutral SDE, not the real-world one. Under BS, \(dS = (r-q)S\,dt + \sigma S\,dW^Q\). Using \(\mu\) instead of \(r-q\) prices with the wrong measure — a classic fail (see Black–Scholes intuition and TVM / discounting).
Euler (and better) schemes. Discrete GBM has an exact solution: \(S_{t+\Delta} = S_t \exp\bigl((r-q-\sigma^2/2)\Delta + \sigma\sqrt{\Delta}\,Z\bigr)\), \(Z\sim N(0,1)\). For local vol / stochastic vol / rates, you usually Euler-discretize and watch bias from \(\Delta t\).
Error has two pieces. Statistical error \(\sim \sigma_g / \sqrt{N}\) (standard error of the mean). Bias from time-stepping, American approximation, etc. Quote a price with a standard error; doubling accuracy in RMSE costs \(4\times\) paths unless you reduce variance.
Variance reduction (know the names): antithetic variates (\(Z\) and \(-Z\)); control variates (e.g. use the BS European you can price in closed form); importance sampling; stratified sampling; quasi-MC (Sobol). Interviewers love: “what’s your control?”
Greeks from MC. Bump-and-revalue (noisy, especially gamma). Pathwise derivatives when the payoff is smooth enough. Likelihood ratio / score method when it isn’t. Common seed (same \(Z\)s) when bumping.
Early exercise is the Achilles heel. Vanilla MC is forward-only; American needs an exercise decision. Longstaff–Schwartz: regress continuation value on basis functions of the state, using the cross-section of paths. Know that this exists and is biased; don’t claim you “just take max at each step” on a single path (that’s wrong — you’d be using future information).
When not to use MC. One-factor European vanillas: use BS. Low-dimensional Americans: trees/PDE are faster and better for early exercise. MC shines at high-dimensional Europeans and path-dependent payoffs.

Mental model

  under Q:
    draw Z1, Z2, ... -> path S(t)
    payoff g(path)
    discount to t=0

  repeat N times
  price ≈ average of the discounted payoffs
  stderr ≈ stdev / sqrt(N)

  Wrong:
    simulate with mu, then discount at r     (mixed measures)
    average first, discount never            (numeraire error)
    exercise using that path's own future    (lookahead bias)

Monte Carlo is a numerical integral of the discounted payoff against the risk-neutral density (or path measure).

Interview questions

Write the Monte Carlo estimator for a European call in BS, including the SDE you simulate. Answer: \(S_T = S_0\exp\bigl((r-q-\sigma^2/2)T + \sigma\sqrt{T}Z\bigr)\), \(Z\sim N(0,1)\). \(\hat C = e^{-rT} \frac1N \sum \max(S_T^{(i)}-K,0)\). Drift is \(r-q\), not \(\mu\).
You run \(N = 10{,}000\) paths and the stderr is 0.20. How many paths for stderr \(\approx 0.02\)? Answer: Error scales \(1/\sqrt{N}\), so you need \(100\times\) paths: \(N = 1{,}000{,}000\). This is why variance reduction and quasi-MC matter in production.
Why can’t you price an American put by, on each path, taking \(\max(K-S_t, \text{discounted remaining})\) using that path’s future? Answer: That uses information from after \(t\) on the same path — not adapted, upward biased (you’d exercise with perfect foresight). You need an exercise policy that depends only on the current state, e.g. Longstaff–Schwartz regression across paths.
Give one control-variate example for an Asian call. Answer: Use the geometric-average Asian, which has a closed-form BS-like price, as control: \(\hat V + \beta\bigl(\text{analytic geo} - \hat V_{\text{geo}}\bigr)\). Highly correlated with the arithmetic Asian, so variance drops a lot. \(\beta\) is estimated or set to 1.
Pathwise vs. bump-and-revalue for delta. When does pathwise fail? Answer: Pathwise: differentiate the payoff along the path, \(\partial g/\partial S_0\), then average. Fails when \(g\) is not Lipschitz / the derivative doesn’t exist in \(L^1\) — digital, barrier-on-touch, or “max” at a kink if you’re not careful (call is OK almost everywhere; digital is not). Then use likelihood ratio or smoothing.

Watch

6. Monte Carlo Simulation — MIT 6.0002 (John Guttag). What Monte Carlo is: inferential statistics, casinos, and why averaging random samples works.
Session 6B: Monte Carlo Simulations in Finance & Investing — Aswath Damodaran. Distributions instead of point estimates; how to read a simulated output.
Monte Carlo Simulation for Option Pricing with Python (Basic Ideas Explained) — QuantPy. Risk-neutral paths to a discounted expected payoff.