Diagnosing a Monte Carlo Pricing Experiment

What a Pricing Experiment Needs to Diagnose

A Monte Carlo estimate can disagree with the desired price for two different reasons. A finite number of simulated paths creates sampling error; replacing a continuous-time SDE by a time-stepping scheme creates time-discretization bias. Increasing the number of paths reduces the first but does not, by itself, reduce the second. A useful diagnosis therefore compares the simulation with an analytic benchmark and evaluates both errors on their own scales.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{Q})\) carrying a Brownian motion \(B_t\). The measure \(\mathbb{Q}\) is the pricing, or risk-neutral, measure in this experiment; the target is an expectation under \(\mathbb{Q}\), not under a physical measure \(\mathbb{P}\). For an adapted, progressively measurable process \(H_s\) satisfying \(\mathbb{E}^{\mathbb{Q}}\int_0^T H_s^2\,ds<\infty\), the Ito integral \(\int_0^T H_s\,dB_s\) is defined in the usual \(L^2\) sense. The deterministic integrands below satisfy this condition.

The drift coefficient is \(\kappa(\theta-x)\), and the diffusion coefficient is \(\eta\). Both are globally Lipschitz in \(x\) and satisfy a linear growth bound: for some constant \(C\), their absolute values are at most \(C(1+|x|)\). The standard existence and uniqueness theorem for Lipschitz SDEs gives a unique strong solution (adapted to the given Brownian motion); it also implies uniqueness in law. Here the solution can be written explicitly, so the benchmark can be derived directly.

The Analytic Benchmark

Proof. Multiplying the SDE by the deterministic integrating factor \(e^{\kappa t}\), and applying the ordinary product rule with this finite-variation factor, gives the solution \(X_T=\theta+(x_0-\theta)e^{-\kappa T}+\eta\int_0^T e^{-\kappa(T-s)}\,dB_s\). The integrand is progressively measurable and its squared integral is finite, since \(\int_0^T e^{-2\kappa(T-s)}\,ds=(1-e^{-2\kappa T})/(2\kappa)<\infty\). A deterministic Ito integral is Gaussian with mean zero and variance equal to the integral of the squared integrand. Thus \(X_T\) has the stated mean and variance. For any square-integrable random variable, \(\mathbb{E}^{\mathbb{Q}}[X_T^2]=\operatorname{Var}^{\mathbb{Q}}(X_T)+(\mathbb{E}^{\mathbb{Q}}X_T)^2\). Multiplying this identity by \(e^{-rT}\) proves the price formula. \(\square\)

This benchmark is for the continuous-time model. If paths are generated using a time step, their expected payoff can differ from \(P\), even if the number of simulated paths is arbitrarily large.

What Euler–Maruyama Estimates

For \(N\) equal time steps, let \(h=T/N\) and \(t_n=nh\). Euler–Maruyama simulates \(X_{n+1}^{(h)}=X_n^{(h)}+\kappa(\theta-X_n^{(h)})h+\eta(B_{t_{n+1}}-B_{t_n})\). The Brownian increments are independent \(N(0,h)\) variables under \(\mathbb{Q}\). The simulated discounted payoff is \(Y_h=e^{-rT}(X_N^{(h)})^2\), and its expectation \(P_h=\mathbb{E}^{\mathbb{Q}}[Y_h]\) is the price for this discrete scheme.

Proof. Subtract \(\theta\) from each Euler update. Then \(X_{n+1}^{(h)}-\theta=a(X_n^{(h)}-\theta)+\eta\Delta B_n\), where \(\Delta B_n=B_{t_{n+1}}-B_{t_n}\). Iterating the recurrence gives \(X_N^{(h)}=\theta+a^N(x_0-\theta)+\eta\sum_{j=0}^{N-1}a^{N-1-j}\Delta B_j\). The increments are independent, centered, and each has variance \(h\). The mean is therefore \(\theta+a^N(x_0-\theta)\); the variance is \(\eta^2h\sum_{j=0}^{N-1}a^{2(N-1-j)}\), which is the stated sum after reordering. The terminal state is Gaussian because it is a deterministic linear combination of independent Gaussian increments. Applying its second-moment identity and then discounting proves the formula for \(P_h\). \(\square\)

The difference \(P_h-P\) is the weak time-discretization bias for this payoff. The formulas apply for every \(h>0\); in particular, the variance sum is valid even if \(a\) is negative. For a stable-looking time-stepping sequence one often chooses \(|1-\kappa h|<1\), but this restriction is not needed for the finite-\(N\) formula.

Confidence Intervals and Their Target

Let \(Y_{h,1},\ldots,Y_{h,M}\) be independent copies of the discretized payoff. The Monte Carlo estimator is \(\widehat P_{M,h}=M^{-1}\sum_{i=1}^M Y_{h,i}\). It is unbiased for \(P_h\), because linearity of expectation gives \(\mathbb{E}^{\mathbb{Q}}[\widehat P_{M,h}]=P_h\). It is not generally unbiased for the continuous-time price \(P\).

Proof. Independence gives zero covariance between distinct path payoffs, so the variance of their average is \(M^{-2}\sum_{i=1}^M\operatorname{Var}^{\mathbb{Q}}(Y_{h,i})=\sigma_h^2/M\). To check the sample-variance claim, expand \(s_h^2=(M-1)^{-1}(\sum_iY_{h,i}^2-M\widehat P_{M,h}^2)\). The expectations of the two terms in parentheses are \(M(\sigma_h^2+P_h^2)\) and \(M(\sigma_h^2/M+P_h^2)\). Their difference is \((M-1)\sigma_h^2\); division by \(M-1\) proves the claim. \(\square\)

For large \(M\), the usual approximate 95% interval is \(\widehat P_{M,h}\pm1.96s_h/\sqrt{M}\). This is based on the classical iid central limit theorem, a standard result whose proof is not given here. The theorem requires independent, identically distributed payoffs with finite variance; it gives asymptotic coverage for \(P_h\). It does not automatically give coverage for \(P\) when \(P_h\ne P\). In this example the payoff has finite moments of every order, since the terminal state is Gaussian.

A Practical Diagnosis

The exact decomposition \(\widehat P_{M,h}-P=(\widehat P_{M,h}-P_h)+(P_h-P)\) separates the observed error into a centered sampling fluctuation and a systematic time-step difference. The first term is assessed by the standard error; the second can be calculated here from the analytic formulas, or studied by refining \(h\). The sampling standard deviation scales as \(M^{-1/2}\), while the discretization bias is controlled by the time step and the scheme, not by the path count.

A common mistake is to increase the path count whenever the estimate misses the benchmark. For example, in the two-step calculation above, the Euler expectation is \(0.3125\) while the benchmark is about \(0.2162\). More Euler paths concentrate the estimate more tightly around \(0.3125\); they do not move that center to the continuous-time price. Conversely, refining the time step does not eliminate the random variation from a finite path count.

Finally, distinguish a diagnostic from a guarantee. A single confidence interval may miss its target even when the simulation is correctly implemented, and a single interval that contains the benchmark does not establish that the discretization bias is negligible. A persuasive assessment reports the analytic or high-accuracy benchmark, the time step, the number of paths, the estimated standard error, and how the estimate changes when the time step is refined.