The idea

The price is the discounted expected payoff under the risk-neutral measure. An expectation is an average, and an average can be estimated by sampling. Simulate \(M\) terminal prices, compute the payoff of each, average and discount:

$$ S_T^{(i)} = S \exp\!\left[\left(r - q - \tfrac{1}{2}\sigma^2\right)T + \sigma\sqrt{T}\, Z_i\right], \qquad \hat{V} = \frac{e^{-rT}}{M}\sum_{i=1}^{M} \text{payoff}\!\left(S_T^{(i)}\right) $$

where the \(Z_i\) are independent standard normals.

Error

The estimate has a standard error of \(s/\sqrt{M}\), where \(s\) is the sample standard deviation of the discounted payoffs. Halving the error needs four times the paths. Every Monte Carlo price should be quoted with its standard error or a confidence interval; a price alone says nothing.

To see it happen, watch the simulation on the Monte Carlo page: the paths draw one by one, and the running estimate and its 95% band close in on the Black-Scholes price.

Variance reduction

• Antithetic variates. For every \(Z\) also use \(-Z\). The two payoffs are negatively correlated, so their average has lower variance than two independent draws.
• Control variates. Simulate something with a known expectation alongside the payoff, here the discounted stock price with expectation \(S e^{-qT}\). Adjust the estimate by \(\beta\) times the control's sampling error: $$ \hat{V}_{cv} = \hat{V} - \beta\left(\bar{X} - \mathbb{E}[X]\right), \qquad \beta = \frac{\mathrm{Cov}(V, X)}{\mathrm{Var}(X)} $$ The stronger the correlation, the larger the reduction. In-the-money options correlate strongly with the stock.

American exercise: Longstaff-Schwartz

Early exercise needs the value of continuing at every point, which simulation does not give directly. Longstaff and Schwartz estimate it by regression: step backwards through time, and at each step regress the realised future cash flows on simple functions of the current spot, using only the paths that are in the money. The fitted curve is the continuation value; exercise where the payoff beats it. The result is biased slightly low because the regression is imperfect, and the error is larger than for a European option.

Greeks

Bumping the spot and rerunning with fresh random numbers gives a Delta drowned in noise. Two fixes: reuse the same random numbers for both runs (common random numbers), or differentiate the payoff along each path, the pathwise method. For a call, the pathwise Delta is \(e^{-rT}\, \mathbf{1}\{S_T > K\}\, S_T / S\), averaged over paths.